---
abstract: 'We introduce in this work a new method for online approximate Bayesian learning, whose main idea is to approximate the sequence $(\pi_t)_{t\geq 1}$ of posterior distributions by a sequence $(\tilde{\pi}_t)_{t\geq 1}$ which (i) can be estimated in an online fashion using sequential Monte Carlo methods and (ii) is shown to converge to the same distribution as the sequence $(\pi_t)_{t\geq 1}$, under weak assumptions on the statistical model at hand. In its simplest version, the proposed approach amounts to take for $(\tilde{\pi}_t)_{t\geq 1}$ the sequence of filtering distributions associated to a particular state-space model, and to approximate this sequence using a standard particle filter algorithm. We illustrate on several challenging examples the benefits of this procedure for online approximate Bayesian parameter inference, and with one real data example we show that its online predictive performance can significantly outperform that of stochastic gradient descent and of streaming variational Bayes.'
author:
- |
    Mathieu Gerber$^{(1)}$, Randal Douc$^{(2)}$\
    \
bibliography:
- 'complete.bib'
title: Online Approximate Bayesian learning
---

Introduction
============

Let $(Y_t)_{t\geq 1}$ be a sequence of i.i.d.  random variables defined on the same probability space $(\Omega,{\mathcal{F}},{\mathbb{P}})$ and taking values in a measurable space $({\mathsf{Y}},{\mathcal{Y}})$. We let $\{f_{\theta},\,\theta\in \Theta\subseteq{\mathbb{R}}^d\}$ be a collection of probability density functions (p.d.f.) on ${\mathsf{Y}}$ with respect to some $\sigma$-finite measure $\eta({\mathrm{d}}y)$, and we assume that $\theta_\star:=\operatorname*{argmax}_{\theta\in\Theta}{\mathbb{E}}[\log f_\theta(Y_1)]$ is well-defined, with ${\mathbb{E}}$ the expectation operator associated to $(\Omega,{\mathcal{F}},{\mathbb{P}})$. We adopt a Bayesian perspective and let $\tilde{\pi}_0$ be a prior distribution for $\theta$ and $
\pi_t({\mathrm{d}}\theta)\propto \prod_{s=1}^t f_\theta(Y_s) \tilde{\pi}_0({\mathrm{d}}\theta)
$ be the posterior distribution associated to observations $\{Y_s\}_{s=1}^t$, for all $t\geq 1$. In this work we consider the problem of approximating $(\pi_t)_{t\geq 1}$ in an online fashion, where by online we mean that the memory and computational requirement to process each observation $Y_t$ is finite and bounded uniformly in $t$.

Current approaches to compute an online approximation of $(\pi_t)_{t\geq 1}$ essentially reduces to online variational methods [see e.g. @broderick2013streaming; @nguyen2017online; @zeno2018bayesian]. However, theoretical justifications for these algorithms are rather limited [see however the recent results obtained by @cherief2019generalization] and there is no guarantee that, as $t$ tends to infinity, their approximation error vanishes. Although not online, the iterated batch importance sampling (IBIS) algorithm of @MR1929161, which approximates $(\pi_t)_{t\geq 1}$ sequentially by Monte Carlo, can be an efficient alternative to online methods in situations where the number of observations to process is not too large and computing $f_\theta(y)$ is cheap (see the example of Section \[sub:linear\_model\]). Some computations in @MR1929161 indeed suggest that a random running time implementation of IBIS only requires ${\mathcal{O}}_{\mathbb{P}}(t)$ operations to estimate $\{\pi_s\}_{s=1}^t$. When it becomes computationally too expensive to process $Y_t$ using IBIS, @balakrishnan propose to approximate the Monte Carlo update of IBIS using the kernel smoothing idea introduced by @liu2001combined. The resulting algorithm (called one-pass SMC) is online but, unlike IBIS, is not theoretically justified.

The main idea of the method we introduce below for online approximate Bayesian learning is to approximate the sequence $(\pi_t)_{t\geq 1}$ by a sequence $(\tilde{\pi}_t)_{t\geq 1}$ which (i) can be approximated in an online fashion using Monte Carlo methods and (ii) is shown to converge to $\delta_{\{\theta_\star\}}$ as $t$ tends to infinity, in some sense to be defined. Notice that, by (ii), as $t$ increases the distance between $\pi_t$ and its approximation $\tilde{\pi}_t$ converges to zero. It is also worth mentioning at this stage that online variational updates involve expectations that are generally not computable, and thus in practice online variational methods also rely on Monte Carlo approximations (see the example of Section \[sub:online\]).

To introduce the construction of the sequence $(\tilde{\pi}_t)_{t\geq 1}$ with more details assume for now that $\Theta={\mathbb{R}}^d$. Let $(\mu_t)_{t\geq 0}$ be a sequence of probability distributions on ${\mathbb{R}}^d$ and $(\theta_t)_{t\geq 0}$ be such that $\theta_0 \sim \tilde{\pi}_0$ and such that $(\{Y_t,\theta_t\})_{t\geq 1}$ is a bivariate Markov chain verifying $$\begin{aligned}
\label{eq:state_space}
Y_t|\theta_t\sim f_{\theta_t}(y_t)\eta({\mathrm{d}}y_t),\quad \theta_t=\theta_{t-1}+U_{t-1},\quad U_{t-1}\sim \mu_{t-1},\quad t\geq 1.\end{aligned}$$ Then, for all $t\geq 1$, we define $\tilde{\pi}_t$ as the conditional distribution of $\theta_t$ given the observations $\{Y_s\}_{s=1}^t$. If all the $(\mu_{t})_{t \geq 0}$ are dominated by a common dominating measure, then defines a state-space model. In this case, $\tilde{\pi}_t$ is the corresponding filtering distribution at time $t$, and particle filter (PF) algorithms [@doucet2009tutorial] can be used to compute an online Monte Carlo approximation of $(\tilde{\pi}_t)_{t\geq 0}$, where the memory and computational requirement of each iteration is bounded by $C N$, with $N\in\mathbb{N}:=\{1,2,\dots\}$ the Monte Carlo effort and where in the rest of this section $C<\infty$ is a constant independent on $t$.

For situations where does not define a state-space model, we develop a sequential Monte Carlo (SMC) algorithm to approximate $(\tilde{\pi}_t)_{t\geq 1}$ whose memory and computational requirement at iteration $t\geq 1$ is a most $C k_t N$, with $k_t=t-\sup\{s<t:\mu_s\neq\delta_{\{0\}}\}$ if $\{s<t:\mu_s\neq\delta_{\{0\}}\}$ is not an empty set and $k_t=t$ otherwise. This SMC sampler reduces to the IBIS algorithm when $k_t=t$ for all $t$ and to a standard PF algorithm when $k_t=1$ for all $t$. Notice that if $k_t=\Delta$ for some $\Delta\in\mathbb{N}$ then the proposed SMC algorithm is an online algorithm. Our main contribution is then to provide, for *any* sequence $(k_t)_{t\geq 1}$, a way to construct sequences $(\tilde{\pi}_t)_{t\geq 1}$ that converge to $\delta_{\{\theta_\star\}}$, under standard regularity conditions on the statistical model.

As we will argue in Section \[sub:h\_t\], a trade-off between statistical and computational efficiency should be considered when choosing the sequence $(\tilde{\pi}_t)_{t\geq 0}$: the cheaper it is to approximate this sequence by Monte Carlo the slower it converges to $\delta_{\{\theta_\star\}}$. Averaging is a standard technique used to improve the statistical and/or predictive properties of online procedures, as briefly recalled in Section \[sub:averaging\]. Following this idea, we propose in that section to base the inference on the mixture distribution $\bar{\pi}_t:=(t+1)^{-1}\sum_{s=0}^t\tilde{\pi}_s$ instead of using $\tilde{\pi}_t$ directly. We show that the distribution $ \bar{\pi}_t $ converges to the same distribution as $\tilde{\pi}_t$ and that its online Monte Carlo estimation can be obtained as a by product of that of $\tilde{\pi}_t$. The merit of $\bar{\pi}_t$ over $\tilde{\pi}_t$ comes from the fact that, empirically, the point estimate $\bar{\theta}_t:={\mathbb{E}}_{\bar{\pi}_t}[\theta]$ typically converges to $\theta_\star$ at the optimal $t^{-1/2}$ rate under various definitions of $(\tilde{\pi}_t)_{t\geq 0}$ while, on the contrary, the convergence behaviour of $\tilde{\theta}_t:={\mathbb{E}}_{\tilde{\pi}_t}[\theta]$ is very sensitive to the choice of this sequence, as we shall see in the numerical example of Section \[subsec:example\].

Establishing theoretically the convergence rate of the distributions $\tilde{\pi}_t$ and of $\bar{\pi}_t$, and of the point estimators $\tilde{\theta}_t$ and $\bar{\theta}_t$, is however beyond the scope of the paper. Instead, the objective of the paper is to derive consistency results for $(\tilde{\pi}_t)_{t\geq 0}$ and $(\bar{\pi}_t)_{t\geq 0}$ that hold under weak conditions on the statistical model at hand, and by imposing minimal constraints on the sequence $(\mu_t)_{t\geq 0}$. An unavoidable consequence (see Section \[subsec:example\]) of the very few conditions we impose on this sequence is that our convergence results for $(\tilde{\pi}_t)_{t\geq 0}$ hold *in probability*, while under the same assumptions on the model the posterior distribution $\pi_t$ converges to $\delta_{\{\theta_\star\}}$ *almost surely*. Our theoretical analysis carefully deals with the case where $\Theta$ is unbounded, which is particularly important since, in this case, nothing prevents the distributions $\tilde{\pi}_t$ and $\pi_t$ to be arbitrarily far from each other at a given time $t$.

Additional notation and outline of the paper\[sub:set-up\]
----------------------------------------------------------

We let $\|\cdot\|_\infty$ be the maximum norm on ${\mathbb{R}}^d$, $B_\epsilon(x)$ be the open ball of size $\epsilon>0$ around $x\in{\mathbb{R}}^d$ w.r.t. $\|\cdot\|$, the Euclidean norm on ${\mathbb{R}}^d$, $t_{d,\nu}(m,\Sigma)$ be the $d$-dimensional Student’t-distribution, with $\nu>0$ degrees of freedom, location vector $m$ and scale matrix $\Sigma$, and $\mathcal{N}_d(m,\Sigma)$ be the $d$-dimensional Gaussian distribution with mean $m$ and covariance matrix $\Sigma $. Recall that for two strictly positive sequences $(a_t)_{t\geq 1}$ and $(b_t)_{t\geq 1}$ the notation $a_t=\Theta(b_t)$ means that $\liminf_{t\rightarrow\infty} (a_t/b_t+b_t/a_t)>0$.

Below we use the shorthand ${\mathbb{E}}[g]$ for ${\mathbb{E}}[g(Y_1)]$, $\nu(h)$ for $\int_{\Theta}h(\theta)\nu({\mathrm{d}}\theta)$, ${\mathrm{Var}}_{\nu}(h)$ for $\nu(h h^T)-\nu(h) \nu(h)^T$. In particular, ${\mathbb{E}}_\nu[\theta]$ stands for $\int_{\Theta} \theta\nu({\mathrm{d}}\theta)$ and ${\mathrm{Var}}_{\nu}(\theta)$ for ${\mathbb{E}}_\nu[\theta \theta^T]-{\mathbb{E}}_\nu[\theta]{\mathbb{E}}_\nu[\theta]^T$. If $A$ is a Borelian set of ${\mathbb{R}}^d$, we denote by $\mathcal{B}(A)$ the Borel $\sigma$-algebra on $A$, by $\mathcal{P}(A)$ the set of probability measures on $(A,\mathcal{B}(A))$ and we let $\mathcal{P}_L(A)$ be the set of probability measures on $(A,\mathcal{B}(A))$ that are absolutely continuous w.r.t. ${\mathrm{d}}\theta$, the Lebesgue measure on ${\mathbb{R}}^d$. Moreover, for a sequence $(\nu_t)_{t \geq 0}$ of probability measures on ${\mathbb{R}}^d$, the notation $\nu_t \Rightarrow \nu$ means that the sequence $(\nu_t)_{t \geq 0}$ converges weakly to the probability measure $\nu\in \mathcal{P}({\mathbb{R}}^d)$.

We let $f_\star(y)\eta({\mathrm{d}}y)$ be the distribution of $Y_1$ and assume throughout the paper that $\Theta\in\mathcal{B}({\mathbb{R}}^d)$, with $\int_{\Theta}{\mathrm{d}}\theta>0$. For what follows it is convenient to extend the parameter space $\Theta$ to ${\mathbb{R}}^d$. To this aim, for every $\theta\in{\mathbb{R}}^d$ we let $\tilde{f}_\theta:{\mathsf{Y}}\rightarrow{\mathbb{R}}$ be such that $\tilde{f}_\theta\equiv f_\theta$ for all $\theta\in \Theta$ and such that $\tilde{f}_\theta\equiv 0$ for all $\theta \in {\mathbb{R}}^d\setminus \Theta$.

Lastly, for any probability measure $\nu$ on $\Theta$ and any sequence $(\nu_t)_{t \geq 0}$ of probability measures on $\Theta$ (implicitly indexed by random variables), we say that $\nu_t \Rightarrow \nu$ in ${\mathbb{P}}$-probability if the sequence of random variables $(\varrho(\nu_t,\nu))_{t \geq 0}$ converges in ${\mathbb{P}}$-probability to $0$, where $\varrho$ denotes the Prohorov distance between probability measures on ${\mathbb{R}}^d$. Notice that if $\nu_t \Rightarrow \nu$ in ${\mathbb{P}}$-probability then, for every continuous and bounded measurable function $\varphi:\Theta\rightarrow{\mathbb{R}}$, $\nu_t(\varphi)\rightarrow \nu(\varphi)$ in ${\mathbb{P}}$-probability. In the case where $\nu=\delta_{\{\theta_\star\}}$, and letting $V_\delta=\{\theta\in{\mathbb{R}}^d:\,\|\theta-\theta_\star\|\geq\delta\} $ for every $\delta>0$, @Ghosal2017 [Proposition 6.2 and above comment, page 124] provide the following necessary and sufficient condition for checking that $\nu_t \Rightarrow \nu$ in ${\mathbb{P}}$-probability.

\[prop:conv\_proba\] $\nu_t \Rightarrow \delta_{\{\theta_\star\}}$ in ${\mathbb{P}}$-probability if and only if ${\mathbb{E}}[\nu_t(V_\epsilon)]\rightarrow 0$ for all $\epsilon>0$.

The rest of the paper is organized as follows. The proposed approach is described in detail in Section \[sec:approach\] and convergence results for the sequences $(\tilde{\pi}_t)_{t\geq 0}$ and $(\bar{\pi}_t)_{t\geq 0}$ are given in Section \[sec:main\]. Section \[sec:num\] proposes some numerical experiments which notably aim at illustrating on challenging problems (including a latent variable model) the benefit of the method introduced in this work for approximate Bayesian parameter inference. In this section we also consider an online binary classification example, with 53 predictors, on which the proposed algorithm outperforms stochastic gradient descent and a standard online variational algorithm. Section \[sec:conclusion\] concludes and all the proofs are gathered in the Appendix.

Online Approximate Bayesian learning\[sec:approach\]
====================================================

In this section we introduce in detail the proposed approach for online approximate Bayesian learning. In Section \[sub:pi\_tilde\] we provide the precise definition of the sequence $(\tilde{\pi}_t)_{t\geq 0}$ and explain in what sense $\tilde{\pi}_t$ can be interpreted as an approximation of the posterior distribution $\pi_t$, for all $t\geq 1$. In Section \[sub:averaging\] we introduce the sequence $(\bar{\pi}_t)_{t\geq 0}$ of average measures and in Section \[subsec:example\] we study the properties of $\tilde{\pi}_t$ and of $\bar{\pi}_t$ on a toy example. In Section \[sub:Monte\_Carlo1\] we introduce a simple online Monte Carlo algorithm (Algorithm \[algo:pi\_tilde\_online\]) that can be used to estimate $(\tilde{\pi}_t)_{t\geq 0}$ and $(\bar{\pi}_t)_{t\geq 0}$ in the special case where $\tilde{\pi}_t$ can be interpreted as the filtering distribution of a state-space model. In Section \[sub:h\_t\] we comment on the trade-off between computational and statistical efficiency one should consider when defining $(\tilde{\pi}_t)_{t\geq 0}$, and make some practical recommendations. The sequential Monte Carlo algorithm for approximating an arbitrary sequence $(\tilde{\pi}_t)_{t\geq 0}$ and associated sequence $(\bar{\pi}_t)_{t\geq 0}$ is introduced in Section \[sub:Monte\_Carlo\]. Lastly, in Section \[sub:post\_app\] we briefly discuss the approximation error we make when the output of this algorithm is interpreted as an approximation of the posterior distribution $\pi_t$.

The sequence Lg {#sub:pi_tilde}
---------------

Let $(\mu_t)_{t\geq 0}$ be a sequence in $\mathcal{P}({\mathbb{R}}^d)$ such that $$\begin{aligned}
\label{eq:mu_def}
\mu_t({\mathrm{d}}\theta)=
\begin{cases}
h_t^{-d}m_t(\theta/h_t){\mathrm{d}}\theta, &h_t>0\\
\delta_{\{0\}}({\mathrm{d}}\theta),&h_t=0
\end{cases},\qquad\forall t\geq 0\end{aligned}$$ where $(h_t)_{t\geq 0}$ is a sequence in $[0,\infty)$ and where, for some strictly increasing sequence $(t_p)_{p\geq 0}$ in $\mathbb{N}_0:=\{0\}\cup\mathbb{N}$, $m_t$ is the p.d.f. of the $\mathcal{N}_d(0,\Sigma_t)$ distribution if $t\not\in (t_p)_{p\geq 0}$ and the p.d.f. of the $t_{d,\nu}(0,\Sigma_t)$ distribution otherwise, with $\nu>1$ and $(\Sigma_t)_{t\geq 0}$ a sequence of symmetric positive definite matrices for which $$\begin{aligned}
\label{eq:Sigma_t}
\exists c\in(0,\infty)\quad\text{such that }
\quad  \|\Sigma_t\|\vee  \|\Sigma^{-1}_t\| \leq c,\quad \forall t\geq 0.
 \end{aligned}$$ Explanations why we require that the sequence $(\mu_t)_{t\geq 0}$ contains infinitely many Student’s t-distributions are postponed to Section \[sub:student\]. Conditions $\nu>1$ and are also needed for our theoretical results to hold. Notice that under the rate at which $\|{\mathrm{Var}}_{\mu_{t}}(\theta)\|\rightarrow 0$ is determined by the sequence $(h_t)_{t\geq 0}$.

Then, for a given prior distribution $\tilde{\pi}_0\in\mathcal{P}_L(\Theta)$, we let $(\tilde{\pi}_t)_{t\geq 1}$ be defined by $$\begin{aligned}
\label{eq:pi_tilde}
\tilde{\pi}_t({\mathrm{d}}\theta)=\frac{ \tilde{f}_\theta(Y_t)(\mu_{t-1}*\tilde{\pi}_{t-1})({\mathrm{d}}\theta)}{\int_{{\mathbb{R}}^d} \tilde{f}_\theta(Y_t)(\mu_{t-1}*\tilde{\pi}_{t-1})({\mathrm{d}}\theta)}\in \mathcal{P}_L(\Theta) ,\quad  t\geq 1.\end{aligned}$$

In Section \[sec:main\] we provide conditions on $(h_t)_{t\geq 0}$ and on $(t_p)_{p\geq 0}$ that ensure the convergence of $\tilde{\pi}_t$ towards $\delta_{\{\theta_\star\}}$, under standard assumptions on the statistical model. For instance, our main results enable us to define, for every $\alpha>0$ and sequence $(h_t)_{t\geq 0}$ such that $h_t=\Theta(t^{-\alpha})$, a strictly increasing sequence $(t_p)_{p\geq 0}$ for which $\tilde{\pi}_t\Rightarrow \delta_{\{\theta_\star\}}$ in ${\mathbb{P}}$-probability.

The definition and of $(\tilde{\pi}_t)_{t\geq 0}$ guarantees that if $h_s=0$ for all $s<t$ then $\tilde{\pi}_t=\pi_t$, where we recall that $\pi_t$ is the posterior distribution of $\theta$. This observation suggests that the size of $\max_{0\leq s< t}h_s$ controls the discrepancy between $\tilde{\pi}_t$ and $\pi_t$. The following result supports this intuition and justifies the interpretation of $\tilde{\pi}_t$ as an approximation of $\pi_t$. Notice that for this result to hold it is not necessary to assume the $Y_t$’s are i.i.d.

\[prop:continuity\] Assume ${\mathbb{P}}(\sup_{\theta\in\Theta} f_\theta(Y_1)<\infty)=1$, that the mapping $\theta\mapsto f_\theta(Y_1)$ is ${\mathbb{P}}$-a.s. continuous on $\Theta$ and that $\tilde{\pi}_0\in\mathcal{P}_L(\Theta)$ has a continuous and bounded density on $\Theta$. Then, $${\mathbb{P}}\big(\mathrm{as} \mbox{ $\max\{h_s,\,s<t\}\rightarrow 0$},\  \tilde{\pi}_{t}\Rightarrow \pi_t\big)=1,\quad\forall t\geq 1.$$

Establishing precisely the link between $\max\{h_s,\,s<t\}$ and the distance between $\tilde{\pi}_{t}$ and $\pi_t$ is beyond the scope of the paper. Nevertheless, Proposition \[prop:continuity\] has the merit to show that we can make $\tilde{\pi}_t$ arbitrary close to $\pi_t$ by reducing the size of the $h_s$’s, as illustrated in Section \[subsec:example\].

The sequence Lg\[sub:averaging\].
---------------------------------

Averaging is a standard technique used to improve the statistical and/or predictive properties of online procedures, see for instance @polyak1992acceleration and @shalev2011online for its use within stochastic approximations (SA) and online learning algorithms, respectively.

Following this idea, and as announced in the introductory section, we introduce the sequence $(\bar{\pi}_t)_{t\geq 0}$ defined by $$\begin{aligned}
\label{eq:pi_bar}
\bar{\pi}_t({\mathrm{d}}\theta)=\frac{1}{t+1}\sum_{s=0}^t\tilde{\pi}_s({\mathrm{d}}\theta),\quad t\geq 0.\end{aligned}$$ Recalling that $\tilde{\theta}_t={\mathbb{E}}_{\tilde{\pi}_t}[\theta]$ and that $\bar{\theta}_t={\mathbb{E}}_{\bar{\pi}_t}[\theta]$, it is worth noting that, for all $t\geq 0$, we have $\bar{\theta}_t=(t+1)^{-1}\sum_{s=0}^t\tilde{\theta}_s$.

Using Proposition \[prop:conv\_proba\] and Cesàro’s lemma, it can be easily checked that convergence in probability of $\bar{\pi}_t$ to $\delta_{\{\theta_\star\}}$ is implied by that of $\tilde{\pi}_t$. This is stated in the following proposition for future reference.

\[prop:bar\] Assume that $\tilde{\pi}_t\Rightarrow\delta_{\{\theta_\star\}}$ in ${\mathbb{P}}$-probability. Then, $\bar{\pi}_t\Rightarrow\delta_{\{\theta_\star\}}$ in ${\mathbb{P}}$-probability.

The main advantage of $\bar{\pi}_t$ over $\tilde{\pi}_t$ is that, empirically, the estimator $\bar{\theta}_t$ typically converges to $\theta_\star$ at rate $t^{-1/2}$ (see the numerical examples of Sections \[subsec:example\], \[sec:num\_quantile\] and \[sub:latent\]) while the convergence behaviour of $\tilde{\theta}_t$ depends on the speed at which $h_t\rightarrow 0$, as illustrated in the next subsection. We note that this property of estimators obtained by averaging online point estimates is well-known in the SA literature and has been proved in this context [@polyak1992acceleration]. On the other hand, and assuming $\Theta\subseteq{\mathbb{R}}$ to simplify the presentation, the inequalities $$\begin{aligned}
\label{eq:lower_var}
{\mathrm{Var}}_{\scalebox{1.5}{$\scriptscriptstyle \bar{\pi}_t$}}(\theta)\geq \frac{1}{t+1}\sum_{s=0}^t{\mathrm{Var}}_{\scalebox{1.5}{$\scriptscriptstyle \tilde{\pi}_s$}}(\theta)\geq \min_{s\leq t}{\mathrm{Var}}_{\scalebox{1.5}{$\scriptscriptstyle \tilde{\pi}_s$}}(\theta) ,\quad\forall t\geq 0,\quad {\mathbb{P}}-a.s.\end{aligned}$$ suggest that, in general, $\bar{\pi}_t$ will not concentrate on $\theta_\star$ at a faster rate than $\tilde{\pi}_t$. Indeed, if $\tilde{\pi}_t$ converges to $\delta_{\{\theta_\star\}}$ then, under mild conditions, ${\mathrm{Var}}_{ \tilde{\pi}_t}(\theta)\rightarrow 0$, in which case it is reasonable to expect that, in , we have $ \min_{s\leq t}{\mathrm{Var}}_{\tilde{\pi}_s}(\theta)\approx {\mathrm{Var}}_{\tilde{\pi}_t}(\theta )$.

A second advantage of $(\bar{\pi}_t)_{t\geq 0}$ over $(\tilde{\pi}_t)_{t\geq 0}$ is that, empirically, the evolution of the former sequence is more stable than that of latter. One way to formalize this observation is to remark that we can establish the almost sure convergence of $\bar{\pi}_t$ to $\delta_{\{\theta_\star\}}$ under weaker conditions than for $\tilde{\pi}_t$. For instance, as a direct consequence of @lyons1988strong [Corollary 4], to guarantee that ${\mathbb{P}}(\bar{\pi}_t\Rightarrow\delta_{\{\theta_\star\}})=1$ it is enough to have, for all $\epsilon>0$, a $\delta_\epsilon>0$ such that ${\mathbb{E}}[\tilde{\pi}_t(V_\epsilon)]={\mathcal{O}}(t^{-\delta_\epsilon})$. By contrast, proving that ${\mathbb{P}}(\tilde{\pi}_t\Rightarrow\delta_{\{\theta_\star\}})=1$ usually requires to show that, for all $\epsilon>0$, ${\mathbb{E}}[\tilde{\pi}_t(V_\epsilon)]={\mathcal{O}}(t^{-\delta_\epsilon})$ for some $\delta_\epsilon>1$.

An illustrative example {#subsec:example}
-----------------------

For every $\theta\in\Theta:={\mathbb{R}}$ we let $f_\theta$ be the density of the $\mathcal{N}_1(\theta,1)$ distribution and, for some $\tilde{\theta}_0\in{\mathbb{R}}$ and $\tilde \sigma_0^2>0$, we let $\tilde{\pi}_0({\mathrm{d}}\theta)$ be the density of the $\mathcal{N}_1(\tilde{\theta}_0,\tilde{\sigma}_0^2)$ distribution. Remark that for this model our general assumptions on $(Y_t)_{t\geq 1}$ impose that ${\mathbb{E}}[Y_1^2]<\infty$, since otherwise $\theta_\star=\operatorname*{argmax}_{\theta\in\Theta}{\mathbb{E}}[\log  f_\theta(Y_1)]$ is not defined. However, in this subsection we relax this condition and assume only that ${\mathbb{E}}[|Y_1|]<\infty$. Notice that if $\theta_\star$ is well-defined then $\theta_\star={\mathbb{E}}[Y_1]$. In addition, to make the computation of $(\tilde{\pi}_t)_{t\geq 0}$ tractable, we also relax the condition that $(\mu_t)_{t\geq 0}$ contains infinity many Student’s $t$-distributions (imposed in Section \[sub:pi\_tilde\]) by letting, in , $m_t$ be the density of the $  \mathcal{N}_1(0,1)$ distribution for all $t\geq 0$.

Then, with $g:{\mathbb{R}}\rightarrow{\mathbb{R}}$ defined by $g(x)=x/(1+x)$, $x\in (0,\infty)$, it can be easily checked that $$\tilde{\pi}_t=\mathcal{N}_1(\tilde{\theta}_t,\tilde{\sigma}_t^2),\quad \tilde{\sigma}_t^2=g(\tilde{\sigma}^2_{t-1}+h^2_{t-1}),\quad \tilde{\theta}_t=\tilde{\sigma}_t^2 Y_t + (1-\tilde{\sigma}_t^2)\tilde{\theta}_{t-1},\quad t\geq 1.$$ Since $g$ is strictly increasing on $(0,\infty)$ it follows that $\tilde{\sigma}_t^2$ is a strictly increasing function of $h_s$, for all $s\in\{0,\dots,t-1\}$. The impact of $(h_t)_{t\geq 0}$ on the behaviour of the sequence $(\tilde{\pi}_t)_{t\geq 0}$ is studied in more details in the following proposition.

\[prop:gaussian\_toy\] Assume that $h_t\rightarrow 0$. Then,

1.  \[part1\] $\tilde{\pi}_t\Rightarrow \delta_{{\mathbb{E}}[Y_1]}$ in ${\mathbb{P}}$-probability.

2.  \[part2\] If in addition we have $h_t\sim c t^{-\alpha}$ for some constants $(c,\alpha)\in(0,\infty)^2$ then

    1.  \[part2a\] for $p\in[1,2)$ we have ${\mathbb{P}}(\tilde{\pi}_t\Rightarrow \delta_{{\mathbb{E}}[Y_1]})=1$ for all $(Y_t)_{t\geq 1}$ such that ${\mathbb{E}}[|Y_1|^p]<\infty$ if and only if $t^{1/p} \tilde{\sigma}_t^2={\mathcal{O}}(1)$,

    2.  \[part2b\] the condition $t^{1/2} \tilde{\sigma}_t^2={\mathcal{O}}(1)$ is necessary but not sufficient to have ${\mathbb{P}}(\tilde{\pi}_t\Rightarrow \delta_{{\mathbb{E}}[Y_1]})=1$ for all $(Y_t)_{t\geq 1}$ such that ${\mathbb{E}}[ Y_1^2]<\infty$,

    3.  \[part2c\] for all $p\geq 1$, we have $t^{1/p} \tilde{\sigma}_t^2={\mathcal{O}}(1)$ only if $\alpha\geq  1/p$,

    4.  \[part2d\] $ h_t^2/\tilde{\sigma}_t^2\rightarrow 0$.

Part \[part2d\] of the proposition shows that $\tilde \sigma_t^2$, the variance of $\theta$ under $\tilde{\pi}_t$, converges to $0$ at a strictly slower rate than that of $h^2_t$ whenever $h_t\sim c t^{-\alpha}$. More precisely, the numerical results given in Table \[table:ex\] shows that, for this example, $\tilde{\sigma}_t^2\approx t^{-\alpha}$ when $h_t=t^{-\alpha}$ for all $t\geq 1$. As argued in the previous subsection, we observe in Table \[table:ex\] that $\bar{\sigma}_t^2$, the variance of $\theta$ under $\bar{\pi}_t$, does not converge to zero at a faster rate than $\tilde{\sigma}_t^2$. Indeed, in this example, both $\tilde{\sigma}_t^2$ and $\bar{\sigma}_t^2$ converge at the same rate. The numerical experiments of Section \[sec:num\] suggest that, when $h_t=t^{-\alpha}$ for all $t\geq 1$, having a convergence rate for the variance of $\theta$ under $\bar{\pi}_t$ which is slower than that of $h_t^2$ is not specific to the model we consider in this subsection but generally holds true.

                               $z_t\backslash\alpha$                               0.1    0.3    0.5    0.7     1
  ------------------------------------------------------------------------------- ------ ------ ------ ------ ------
   ${\mathrm{Var}}_{\scalebox{1.5}{$\scriptscriptstyle \tilde{\pi}_t$}}(\theta)$   0.09   0.30   0.50   0.70   1.00
    ${\mathrm{Var}}_{\scalebox{1.5}{$\scriptscriptstyle \bar{\pi}_t$}}(\theta)$    0.09   0.30   0.5    0.7    0.93
                      $|\tilde{\theta}_t-{\mathbb{E}}[Y_1]|$                       0.05   0.15   0.24   0.36   0.49
                       $|\bar{\theta}_t-{\mathbb{E}}[Y_1]|$                        0.52   0.52   0.52   0.53   0.68

  : Ordinary least square estimate of $\beta_2$ for the model $\log(z_t)=\beta_1-\beta_2\log(t)+\epsilon_t$ and for the example of Section \[subsec:example\]. The results are for $T=10^7$ observations, $(\tilde{\theta}_0,\tilde{\sigma}^2_0)=(0,25)$, $Y_t{\stackrel{\mathclap{\normalfont\mbox{\tiny{iid}}}}{\sim}}\mathcal{N}_1(0,1)$, $h_0=0$ and $h_t=t^{-\alpha}$ for all $t>1$. \[table:ex\]

The first part of the proposition shows that the condition $h_t\rightarrow 0$ is enough to guarantee the convergence in ${\mathbb{P}}$-probability of $\tilde{\pi}_t$ towards $\delta_{{\mathbb{E}}[Y_1]}$. Our main results (Theorems \[thm:online\]-\[thm:lower\_bound\_1\]) show that weak conditions on the sequence $(h_t)_{t\geq 0}$ are indeed needed to guarantee that $\tilde{\pi}_t\Rightarrow\delta_{\{\theta_\star\}}$ in ${\mathbb{P}}$-probability. However, for reasons explained in Section \[sub:student\], these results require that infinitely many elements of the sequence $(\mu_t)_{t\geq 0}$ are Student’s t-distributions.

We saw above how the choice of $(h_t)_{t\geq 0}$ affects the convergence rate of $\tilde{\sigma}^2_t$ and of $\bar{\sigma}_t^2$, but it turns out that the rate at which the estimator $\tilde{\theta}_t$ converges towards ${\mathbb{E}}[Y_1]$ is also impacted by the definition of this sequence. The numerical results in Table \[table:ex\] indeed show that $|\tilde{\theta}_t-{\mathbb{E}}[Y_1]|\approx t^{-\alpha/2}$ when $h_t=t^{-\alpha}$ for all $t\geq 1$. By contrast, we observe in Table \[table:ex\] that $\bar{\theta}_t$ converges to ${\mathbb{E}}[Y_1]$ at rate $t^{-1/2}$, regardless the value of $\alpha$. The numerical experiments of Sections \[sec:num\_quantile\] and \[sub:latent\] suggest that, in general, this property of $\bar{\theta}_t$ also holds true for more complicated models.

![Evolution of $d_{\mathrm{MMD}}(\tilde{\pi}_t,\pi_t)$ (left plot) and of a Monte Carlo estimate of $d_{\mathrm{MMD}}(\bar{\pi}_t,\pi_t)$ (right plot) for the example of Section \[subsec:example\]. The observations, $\tilde{\pi}_0$ and $h_0$ are as in Table \[table:ex\], while $h_t=t^{-\alpha}$ for all $t\geq 1$, with $\alpha=0.5$ (solid line), $\alpha=0.7$ (dashed line) and $\alpha=1$ (dotted line). Note that only the results for $\alpha\in\{0.5,1\}$ are presented for $d_{\mathrm{MMD}}(\tilde{\pi}_t,\pi_t)$. []{data-label="fig:toy_mmd"}](figs/toy_example_mmd.pdf "fig:")![Evolution of $d_{\mathrm{MMD}}(\tilde{\pi}_t,\pi_t)$ (left plot) and of a Monte Carlo estimate of $d_{\mathrm{MMD}}(\bar{\pi}_t,\pi_t)$ (right plot) for the example of Section \[subsec:example\]. The observations, $\tilde{\pi}_0$ and $h_0$ are as in Table \[table:ex\], while $h_t=t^{-\alpha}$ for all $t\geq 1$, with $\alpha=0.5$ (solid line), $\alpha=0.7$ (dashed line) and $\alpha=1$ (dotted line). Note that only the results for $\alpha\in\{0.5,1\}$ are presented for $d_{\mathrm{MMD}}(\tilde{\pi}_t,\pi_t)$. []{data-label="fig:toy_mmd"}](figs/toy_example_mmdBar.pdf "fig:")

It is easily verified that, for this example, the posterior distribution $\pi_t$ converges ${\mathbb{P}}$-a.s. to $\delta_{{\mathbb{E}}[Y_1]}$, and it is of interest to find conditions on $(h_t)_{t\geq 0}$ that preserve this mode of convergence. If ${\mathbb{E}}[Y_1^2]<\infty$, so that $\theta_\star$ is well defined, Parts \[part2b\] and \[part2c\] of Proposition \[prop:gaussian\_toy\] imply that if $h_t\sim c t^{-\alpha}$ then the almost sure convergence of $\tilde{\pi}_t$ towards $\delta_{{\mathbb{E}}[Y_1]}$ may not hold for $\alpha\leq 0.5$. Notice that if $\alpha=0.5$ then $\ h^2_t/\sigma_{t}^2={\mathcal{O}}(1)$, where $\sigma_t^2={\mathcal{O}}(1/t)$ is the posterior variance of $\theta$. If we relax the assumption ${\mathbb{E}}[Y^2_1]<\infty$ then $\tilde{\pi}_t$ may not converge almost surely to $\delta_{{\mathbb{E}}[Y_1]}$ when $\alpha<1$, by Parts \[part2a\] and \[part2c\] of the proposition. Proposition \[prop:gaussian\_toy\] therefore shows that $h_t$ must converge very quickly to zero to guarantee that $\tilde{\pi}_t$ preserves the almost sure convergence property of $\pi_t$.

To illustrate the result of Proposition \[prop:continuity\], namely that $\tilde{\pi}_t$ gets closer to $\pi_t$ as $\max\{h_s,\, s<t\}$ decreases, Figure \[fig:toy\_mmd\] (left plot) shows the evolution, as $t$ increases, of the maximum mean discrepancy (MMD) distance $d_{\mathrm{MMD}}(\tilde{\pi}_t,\pi_t)$ between $\tilde{\pi}_t$ and $\pi_t$, when $h_t=t^{-\alpha}$ for all $t\geq 1$ and $\alpha\in\{0.5,1\}$. The MMD distance is computed using the Gaussian kernel with bandwidth parameter equals to one, in which case $d_{\mathrm{MMD}}(\tilde{\pi}_t,\pi_t)$ is explicitly computable and $d_{\mathrm{MMD}}$ is a metric on $\mathcal{P}({\mathbb{R}})$ [see e.g. @muandet2016kernel for more details on the MMD distance]. As expected, we see in Figure \[fig:toy\_mmd\] that $d_{\mathrm{MMD}}(\tilde{\pi}_t,\pi_t)$ is smaller for $\alpha=1$ than for $\alpha=0.5$. In addition, we observe that the distance between $\tilde{\pi}_t$ and $\pi_t$ decreases as $t$ increases, which is due to the fact that the two distributions converge to $\delta_{{\mathbb{E}}[Y_1]}$ as $t\rightarrow\infty$. The right plot of Figure \[fig:toy\_mmd\] shows the corresponding results for the MMD distance between $\bar{\pi}_t$ and $\pi_t$ and for $\alpha\in\{0.5,0.7,1\}$. Computing $d_{\mathrm{MMD}}(\bar{\pi}_t,\pi_t)$ explicitly is feasible but computationally expensive. For this reason the right plot of Figure \[fig:toy\_mmd\] in fact presents results for $d_{\mathrm{MMD}}(\bar{\pi}^N_t,\pi_t)$, with $\bar{\pi}^N_t$ the empirical distribution associated to an i.i.d. sample of size $N=5\,000$ from $\bar{\pi}_t$. Unsurprisingly, $d_{\mathrm{MMD}}(\bar{\pi}_t,\pi_t)$ decreases with $\alpha$ and $t$ while, as argued in Section \[sub:averaging\], we observe that the temporal evolution of $d_{\mathrm{MMD}}(\bar{\pi}_t,\pi_t)$ is more regular than that of $d_{\mathrm{MMD}}(\tilde{\pi}_t,\pi_t)$. Lastly, it is interesting to see that, for this example, $d_{\mathrm{MMD}}(\bar{\pi}_t,\pi_t)$ tends to be larger than $d_{\mathrm{MMD}}(\tilde{\pi}_t,\pi_t)$ despite of the fact that $\bar{\theta}_t$ is (much) closer to the posterior mean of $\theta$ than $\tilde{\theta}_t$ (see Table \[table:ex\]). A possible explanation for this phenomenon is that, in this particular example, $\tilde{\pi}_t$ belongs to the same family of distributions than $\pi_t$, for all $t\geq 0$.

$N\in\mathbb{N}$, $c_{\mathrm{ESS}}\in(0,1]$ and a resampling algorithm $\mathcal{R}(\cdot, \cdot)$ (see Section \[sub:resampling\])

\[line:for\]Let $\theta_1^n\sim \tilde{\pi}_0({\mathrm{d}}\theta_0)$ and set $w_1^n=  \tilde{f}_{\theta_1^n}(Y_1)$ and $W_{1}^n=w_1^n/\sum_{m=1}^N w_1^m$

Set $\mathrm{ESS}_{t-1}= 1/\sum_{m=1}^N (W_{t-1}^m)^2$

\[line:R\] Let $(\hat{\theta}^{1}_{t-1},\dots,\hat{\theta}^{N}_{t-1})\sim \mathcal{R}(\{\theta_{t-1}^n,W_{t-1}^{n}\}_{n=1}^N,N)$ and set $w_{t-1}^n=1$ Let $\hat{\theta}^{n}_{t-1}=\theta^{n}_{t-1}$

Set $\theta_t^n=\hat{\theta}^n_{t-1} + c\,t^{-\alpha} \epsilon^n_{t-1}$ where $\epsilon^n_{t-1}\sim t_{d,\nu}(0, I_d)$ Set $\theta_t^n=\hat{\theta}^n_{t-1} + c\,t^{-\alpha} \epsilon^n_{t-1}$ where $\epsilon^n_{t-1}\sim \mathcal{N}_d(0,I_d)$ \[line:last\] Set $w_t^n= w_{t-1}^n\tilde{f}_{\theta_t^n}(Y_t)$ and $W_{t}^n=w_t^n/\sum_{m=1}^N w_t^m$

**Output:** An approximation $\tilde{\pi}_0^N:=\frac{1}{N}\sum_{n=1}^N  \delta_{\{\theta_1^n\}}$ of $\tilde{\pi}_0$ and, for all $t\geq 1$, an approximation $\tilde{\pi}_t^N:=\sum_{n=1}^N W_t^n\delta_{\{\theta_t^n\}}$ of $\tilde{\pi}_t$ .

A simple online approximate Bayesian learning algorithm \[sub:Monte\_Carlo1\]
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As mentioned in the introductory section, computing an efficient Monte Carlo approximation of $(\tilde{\pi}_t)_{t\geq 0}$ is particularly easy when $h_t>0$ for all $t\geq 1$ since, in this case, a standard particle filter (PF) can be used.

In Algorithm \[algo:pi\_tilde\_online\] we provide a simple PF algorithm that can be used to approximate a sequence $(\tilde{\pi}_t)_{t\geq 0}$ such that $h_0=0$ and, for all $t\geq 1$, $\Sigma_t=I_d$ and $h_t=c\, t^{-\alpha}$ for some constants $c,\alpha>0$. For every $\alpha>0$, examples of sequences $(t_p)_{p\geq 0}$ that verify the conditions of our convergence results (Theorems \[thm:online\]-\[thm:lower\_bound\_1\]) are given in Proposition \[prop:h\_t\]. All the numerical experiments of Section \[sec:num\] rely on Algorithm \[algo:pi\_tilde\_online\] with $c=1$ and $\alpha=0.5$, except those of Section \[sub:linear\_model\].

For every $t\geq 0$, Algorithm \[algo:pi\_tilde\_online\] computes and approximation $\tilde{\pi}_t^N$ of $\tilde{\pi}_t$, where $N\in\mathbb{N}$ is the Monte Carlo effort. Using standard results for PF algorithms [see @crisan2000 Theorem 1, and Proposition \[prop:conv\_algo\] below], it can be shown that, as $N$ tends to $\infty$, the approximation $\tilde{\pi}_t^N$ converges to $\tilde{\pi}_t$ at the usual $N^{-1/2}$ Monte Carlo convergence rate. Notice that in Algorithm \[algo:pi\_tilde\_online\] each observation $Y_t$ is read only once and that the memory and computational requirement to process $Y_t$ does not increase with $t$ and is linear in $N$. A more precise description of Algorithm \[algo:pi\_tilde\_online\] is omitted, since it is a particular case of Algorithm \[algo:pi\_tilde\] which is introduced in details in Section \[sub:Monte\_Carlo\].

It is clear from Algorithm \[algo:pi\_tilde\_online\] that, for fixed $N$, the larger $\alpha$ is (i.e. the faster $h_t$ converges to $0$) the higher is the probability that the support of $\tilde{\pi}_t^N$ is close to the support of $\tilde{\pi}_{t-1}^N$. Hence, if $\alpha$ is large and the algorithm is initialized faraway from $\theta_\star$, a large number of iterations may be needed for $\tilde{\pi}_t^N$ to reach a small neighbourhood $U$ of this parameter value, and it may even be the case that $U$ will never be reached with large probability. Consequently, a large value of $\alpha$ should usually be compensated by a large value of $N$. In practice, we find that choosing $\alpha=0.5$, so that $\sum_{t=0}^\infty h_t=\infty$, works well and, as mentioned above, this is the default value of $\alpha$ that we use in Section \[sec:num\] for the numerical examples that rely on Algorithm \[algo:pi\_tilde\_online\].

To keep its description as simple as possible Algorithm \[algo:pi\_tilde\_online\] does not compute any approximation of the sequence $(\bar{\pi}_t)_{t\geq 0}$. A natural way to approximate this sequence using the output of Algorithm \[algo:pi\_tilde\_online\] is, for all $t\geq 0$, to sample $K$ independent random variables from $(t+1)^{-1}\sum_{s=0}^t\tilde{\pi}_s^N$ and then to use the empirical distribution of this sample as an estimate of $\bar{\pi}_t$. It turns out that such an approximation of $(\bar{\pi}_t)_{t\geq 0}$ can be computed on the fly, by adding Line \[bar\_init\] of Algorithm \[algo:pi\_tilde\] just after Line \[line:for\] of Algorithm \[algo:pi\_tilde\_online\] and by adding Line \[bar\] of Algorithm \[algo:pi\_tilde\] just after Line \[line:last\] of Algorithm \[algo:pi\_tilde\_online\]. Note also that the output of Algorithm \[algo:pi\_tilde\_online\] can directly be used to compute the approximation $\bar{\varphi}_{t}^N:=(t+1)^{-1}\sum_{s=1}^t\tilde{\pi}^N_s(\varphi)$ of $\bar{\pi}_t(\varphi)$ using the recursive formula $$\bar{\varphi}_{t}^N=\frac{t}{t+1}\bar{\varphi}_{t-1}^N+\frac{1}{t+1}\tilde{\pi}^N_t(\varphi),\quad t\geq 0.$$

Choosing $(h_t)_{t\geq 0}$: computational v.s. statistical efficiency\[sub:h\_t\]
---------------------------------------------------------------------------------

Both Proposition \[prop:continuity\] and the numerical results of Section \[subsec:example\] call for choosing a sequence $(h_t)_{t\geq 0}$ such that $h_t$ tends to $0$ quickly, while as mentioned in the previous subsection computational constraints will generally impose to take $(h_t)_{t\geq 0}$ such that the convergence of $h_t$ to $0$ is not too fast. In the special case where $h_t=\Theta(t^{-\alpha})$, we observed in our experiments that these latter usually lead to choose $\alpha=0.5$, in which case the example of Section \[subsec:example\] and those of Sections \[sec:num\_quantile\]-\[sub:latent\] suggest that the variance of $\theta$ under both $\tilde{\pi}_t$ and $\bar{\pi}_t$ converges to zero at a rate which is slower than the usual $t^{-1}$ convergence rate that we have for the posterior variance of $\theta$. Consequently, in practice the sequence $(\tilde{\pi}_t)_{t\geq 0}$ will be typically defined in such a way that the uncertainty quantification obtained using $\tilde{\pi}_t$ or $\bar{\pi}_t$ is too conservative. By contrast, the Bayesian credible sets may be over confident when the model is miss-specified [@Kleijn2012].

Monte Carlo Approximation of the sequences Lg and Lg \[sub:Monte\_Carlo\]
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Algorithm \[algo:pi\_tilde\] provides, for all $t\geq 0$, a Monte Carlo approximation $\tilde{\pi}^N_t:=\sum_{n=1}^N W_t^n\delta_{\theta_t^n}$ of $\tilde{\pi}_t$ and a Monte Carlo approximation $\bar{\pi}_t^{N,K}:=\frac{1}{K}\sum_{k=1}^K\delta_{\bar{\theta}_t^k}$ of $\bar{\pi}_t$.

Following the idea of the IBIS algorithm [@MR1929161], at iteration $t\geq 1$ the observation $Y_t$ is processed and the approximation $\tilde{\pi}^N_{t-1}$ of $\tilde{\pi}_{t-1}$ is updated to yield an approximation $\tilde{\pi}^N_{t}$ of $\tilde{\pi}_t$, while the approximation $\bar{\pi}^{N,K}_{t-1}$ of $\bar{\pi}_{t-1}$ is updated to yield an approximation $\bar{\pi}^{N,K}_{t}$ of $\bar{\pi}_t$. Algorithm \[algo:pi\_tilde\] is therefore a sequential algorithm and its memory and computational requirement to process a new observation depends on the sequence $(h_t)_{t\geq 0}$ (see Section \[sub:running\_time\]). Remark that if $h_t>0$ for all $t\geq 1$ then Algorithm \[algo:pi\_tilde\] is a standard particle filter algorithm [see e.g. @doucet2009tutorial] while, if $h_t=0$ for all $t\geq 1$, it reduces to a particular instance of the IBIS algorithm where the particles are rejuvenated by mean a Metropolis-Hastings kernel.

Before describing Algorithm \[algo:pi\_tilde\] in more details the following three observations are worth making. Firstly, the support of $\bar{\pi}_t^{N,K}$ is generated by sampling $K$ independent random variables from the mixture $ (t+1)^{-1}\sum_{s=0}^t\tilde{\pi}^N_s$. Secondly, if we are interested in estimating $\bar{\pi}_t(\varphi_i)$ for a small number of functions $\{\varphi_i\}_{i=1}^I$ then the estimates $\big\{(t+1)^{-1}\sum_{s=0}^t\tilde{\pi}^N_s(\varphi_i)\big\}_{i=1}^I$ are both cheaper to compute and have lower variances than the estimates $\{\bar{\pi}^{N,K}_t(\varphi_i)\}_{i=1}^I$. Lastly, when Algorithm \[algo:pi\_tilde\] is neither a particle filter (i.e. $h_t=0$ for some $t>0$) nor the IBIS algorithm (i.e.  $h_t>0$ for some $t>0$) it does not belong to the class of sequential Monte Carlo (SMC) samplers introduced by @DelDouJas:SMC. This point will be explained more precisely in Section \[sub:theory\] where a convergence result for Algorithm \[algo:pi\_tilde\] is provided.

### Ingredients: Resampling and mutation steps\[sub:resampling\] 

Sequential Monte Carlo algorithms require to specify selection steps, where the particles with large weights are replicated while those with low weights are discarded, and mutation steps, where new particles are generated. Selection steps are performed using a resampling algorithm, that can be informally defined as follow [see e.g. @gerber2019negative for a formal definition].

\[def:resampling\] $\mathcal{R}(\cdot,\cdot)$ is a resampling algorithm if $\mathcal{R}(\{x^m,p^m\}_{m=1}^M, L)$ is a probability distribution on $\{x^1,\dots,x^M\}^L$, where $(p^{1},\dots, p^M)\in[0,1]^M$ and $\sum_{m=1}^M p^m=1$. We say that $\mathcal{R}(\cdot,\cdot)$ is unbiased if the random vector $(X^1,\dots,X^L)\sim \mathcal{R}(\{x^m,p^m\}_{m=1}^M, L)$ satisfies ${\mathbb{P}}(X^l=x^m)=p^m$ for all $l$ and $m$. If in addition the random variables $X^1,\dots, X^L$ are independent then $\mathcal{R}(\cdot,\cdot)$ is the multinomial resampling scheme and is denoted by $\mathcal{R}_{\mathrm{multi}}(\cdot,\cdot)$.

On Line \[exact1\] of Algorithm \[algo:pi\_tilde\] various resampling algorithms $\mathcal{R}(\cdot,\cdot)$ can be used. The most natural choice is the multinomial resampling scheme $\mathcal{R}_{\mathrm{multi}}(\cdot,\cdot)$ but in practice lower variance resampling methods are usually used, such as stratified resampling $\mathcal{R}_{\mathrm{strat}}(\cdot,\cdot)$ or residual resampling $\mathcal{R}_{\mathrm{res}}(\cdot,\cdot)$ [see e.g. @Douc2005]. We note in passing that the SSP resampling algorithm $\mathcal{R}_{\mathrm{ssp}}(\cdot,\cdot)$ proposed in @gerber2019negative also has a smaller variance than the multinomial resampling scheme, thanks to a result due to @chauvet2017.

The mutation step when $h_{t-1}=0$ amounts to sampling $\theta_t^n$ given the resampled particle $\hat{\theta}^n_{t-1}$ using a pseudo-marginal Metropolis-Hastings kernel [@Andrieu2009] having $$\begin{aligned}
\label{eq:target}
  \frac{ \prod_{s=k_t+1}^{t-1} \tilde{f}_{\theta}(Y_s)(\mu_{k_t}*\tilde{\pi}^N_{k_t})({\mathrm{d}}\theta)}{\int_{{\mathbb{R}}^d}\prod_{s=k_t+1}^{t-1} \tilde{f}_{\theta}(Y_s)(\mu_{k_t}*\tilde{\pi}^N_{k_t})({\mathrm{d}}\theta) }\end{aligned}$$ as invariant distribution and $q_t(\theta|\theta'){\mathrm{d}}\theta$ as proposal distribution (with $k_t$ as defined in Algorithm \[algo:pi\_tilde\]). The use of this pseudo-marginal approach has the advantage to keep the computational cost of Algorithm \[algo:pi\_tilde\] linear in $N$, while implementing the Metropolis-Hastings kernel having as invariant distribution would result in an algorithm whose complexity is quadratic in $N$. Notice that the pseudo-marginal Metropolis-Hastings kernel, defined on Lines \[exact1\] and \[pseudo1\]-\[pseudo2\], can be iterated several times to improve the particles diversity.

When $h_{t-1}>0$ the mutation step reduces to sampling $\theta_t^n$ from $q_t(\theta|\theta'){\mathrm{d}}\theta$, where either $\theta'=\hat{\theta}^n_{t-1}$ (Line \[exact1\]) or $\theta'=\theta_{t-1}^n$ the (Line \[No\_res1\]). Notice that this mutation step corresponds to that of a PF algorithm.

### Memory and computational requirement\[sub:running\_time\]

The memory and computational cost of Algorithm \[algo:pi\_tilde\] depends on the sequence $(h_t)_{t\geq 0}$. More precisely, the memory requirement is ${\mathcal{O}}(t-k_t)$ while, assuming that all the density functions involved in Algorithm \[algo:pi\_tilde\] can be computed at a finite cost, iteration $t$ requires ${\mathcal{O}}( N+K )$ operations if no resampling step is performed (i.e. if $\mathrm{ESS}_{t-1}>  N\,c_{\mathrm{ESS}}$) and ${\mathcal{O}}( (t-k_t)NJ+K)$ operations otherwise, where $J\geq 1$ is the number of simulations required by the pseudo-marginal kernel (typically, $J=1$). For instance, choosing for some $\Delta\in\mathbb{N}$ a sequence $(h_t)_{t\geq 0}$ such that $h_t>0$ for all $t\in\{s\Delta,\,s\in\mathbb{N}\}$ enables to keep in memory only the last $\Delta$ observations and guarantees that the number of operations performed at each iteration of Algorithm \[algo:pi\_tilde\] is bounded by $C(NJ\Delta+K)$, for some constant $C<\infty$ independent of $t$. Notice that if this upper bound growths with $\Delta$ this is not necessarily the case for the expected running time when $c_{\mathrm{ESS}}<1$. Indeed, in this scenario, when no resampling step is needed Algorithm \[algo:pi\_tilde\] only requires to update the weights when $h_{t-1}=0$ while, if $h_{t-1}>0$, a new set of $N$ particles also needs to be generated, i.e. Line \[No\_res1\] is computationally more expensive than Line \[No\_res2\] (see Section \[sub:linear\_model\] for numerical results supporting this point). Lastly, it is worth mentioning that, at time $t$, the expected cost of Lines \[bar\] is bounded by $C (K+dK/t)$, where $C<\infty$ is independent of $t$ and $d$. Consequently, the impact of the dimension of $\theta$ on the computational cost needed to approximate $\bar{\pi}_t$ decreases quickly with $t$.

$(N,K,J)\in\mathbb{N}^3$, $c_{\mathrm{ESS}}\in(0,1]$, an initial distribution $q_0(\theta_0){\mathrm{d}}\theta_0\in\mathcal{P}({\mathbb{R}}^d)$, a sequence $(q_t(\cdot|\theta))_{t\geq 1,\theta\in\Theta}$ of proposal densities on ${\mathbb{R}}^d$ and a resampling algorithm $\mathcal{R}(\cdot, \cdot)$

Let $\theta_0^n\sim q_0(\theta_0){\mathrm{d}}\theta_0$ and set $w_0^n= \tilde{\pi}_0(\theta_0^n)/q_0(\theta_0^n)$, $W_{0}^n=w_0^n/\sum_{m=1}^N w_0^m$ and $h_{-1}=1$

\[bar\_init\]Let $(\bar{\theta}_0^{1},\dots,\bar{\theta}_0^{K})\sim \mathcal{R}_{\mathrm{multi}}(\{\theta_0^n,W_{0}^{n}\}_{n=1}^N,K)$

Set $\mathrm{ESS}_{t-1}= 1/\sum_{m=1}^N (W_{t-1}^m)^2$

\[exact1\] Let $(\hat{\theta}^{1}_{t-1},\dots,\hat{\theta}^{N}_{t-1})\sim \mathcal{R}(\{\theta_{t-1}^n,W_{t-1}^{n}\}_{n=1}^N,N)$ and $\tilde{\theta}_t^n\sim q_t(\theta_t|\hat{\theta}^n_{t-1}) {\mathrm{d}}\theta_t$

\[l1\] Set $\theta_t^n= \tilde{\theta}_t^n$ and $w_t^n= h^{-d}_{t-1} \tilde{f}_{\theta_t^n}(Y_t)\frac{m_{t-1}( (\hat{\theta}^n_{t-1}-\theta_t^n)/h_{t-1})}{q_t(\theta^n_t|\hat{\theta}^n_{t-1})}$

\[pseudo1\] Let $V^n_t\sim{\mathcal{U}}(0,1)$ and $k_t=\max\{-1\leq s<t-1:\, h_s>0\}$ \[exact2\] Set $\alpha^n_t(\hat{\theta}_t^n,\tilde{\theta}_t^n)=q_t(\hat{\theta}^n_{t-1}|\tilde{\theta}_t^n)\tilde{\pi}_{t-1}(\tilde{\theta}_t^n)   /\big(q_t(\tilde{\theta}^n_t|\hat{\theta}^n_{t-1}) \tilde{\pi}_{t-1}(\hat{\theta}^n_{t-1})\big)$ \[expensive\] For $s\in\{t-1,t\}$ let $(\tilde{\theta}_1^{n,s},\dots, \tilde{\theta}_J^{n,s}) \sim \mathcal{R}_{\mathrm{multi}}(\{\theta_{k_t}^n,W^n_{k_t}\}_{n=1}^N,J)$ and set $$\alpha^n_t(\hat{\theta}_t^n,\tilde{\theta}_t^n)=\frac{q_t(\hat{\theta}^n_{t-1}|\tilde{\theta}_t^n)  \sum_{j=1}^J m_{k_t}(h_{k_t}^{-1}(\tilde{\theta}_t^n-\tilde{\theta}_j^{n,t})) \prod_{s=k_t+1}^{t-1} \tilde{f}_{\tilde{\theta}_t^n}(Y_s)}{q_t(\tilde{\theta}^n_t|\hat{\theta}^n_{t-1})   \sum_{j=1}^Jm_{k_t}(h_{k_t}^{-1}(\hat{\theta}^n_{t-1}-\tilde{\theta}_j^{n,t-1}))\prod_{s=k_t+1}^{t-1} f_{\hat{\theta}^n_{t-1}}(Y_s)}$$ \[accept\]

\[l2\] Set $\theta_t^n=  \tilde{\theta}_t^n$ and $w_t^n= f_{\theta_t^n}(Y_t)$ \[l3\] Set $\theta_t^n=\hat{\theta}^n_{t-1}$ and $w_t^n= f_{\theta_t^n}(Y_t)$ \[pseudo2\] \[end\] \[No\_res1\] Let $\theta_t^n \sim q_t(\theta_t|\theta^n_{t-1}) {\mathrm{d}}\theta_t$ and set $w_t^n= h_{t-1}^{-d}w_{t-1}^n \tilde{f}_{\theta_t^n}(Y_t) \frac{m_{t-1}((\theta^n_{t-1}-\theta_t^n)/h_{t-1})}{q_t(\theta^n_t|\theta^n_{t-1})}$ \[No\_res2\] Let $\theta^n_t=\theta_{t-1}^n$ and set $w_t^n= w_{t-1}^n \tilde{f}_{\theta_t^n}(Y_t)$ Set $W_{t}^n=w_t^n/\sum_{m=1}^N w_t^m$.

\[bar\] Let $\bar{\theta}_t^{1:K}\sim \mathrm{PiBar\_Update}\big(t,\{\bar{\theta}_{t-1}^{k}\}_{k=1}^K, \{\theta_t^n,W_{t}^{n}\}_{n=1}^N\big) $

$t\in\mathbb{N}$, $\{\bar{\theta}^{k}\}_{k=1}^K\subset \Theta$, $\{\theta^n, W^n\}_{n=1}^N\subset\Theta\times \{w\in[0,1]^N:\,\sum_{n=1}^N w_n=1\}$

\[bar\_1\]Let $V^{k}\sim \mathrm{Bernoulli}(1/(t+1))$ and $(\theta_t^{1},\dots,\theta_t^{K'})\sim \mathcal{R}_{\mathrm{multi}}\big(\{\theta^n,W^{n}\}_{n=1}^N, K'\big)$, with $K'=\sum_{i=1}^K V^i$

Set $\bar{\theta}^{k}_t= \theta_t^{j_k}$, with $j_k=\sum_{i=1}^k V^i$ Set $\bar{\theta}^{k}_t=\bar{\theta}^{k}$ \[bar\_2\]

**Return:** $(\bar{\theta}^{1}_t,\dots,\bar{\theta}_t^K)$

### A convergence result\[sub:theory\]

Algorithm \[algo:pi\_tilde\] is not a standard SMC sampler [@DelDouJas:SMC] because the pseudo-marginal kernel used at iteration $t\geq 1$ depends on all the random variables generated up to time $t-1$. The following proposition however shows that a non-asymptotic bound for the approximation error similar to that obtained for SMC samplers [@crisan2000 Theorem 1] holds for Algorithm \[algo:pi\_tilde\].

\[prop:conv\_algo\] Consider Algorithm \[algo:pi\_tilde\] with  $\mathcal{R}\in\{\mathcal{R}_{\mathrm{multi}},\mathcal{R}_{\mathrm{strat}},\mathcal{R}_{\mathrm{res}},\mathcal{R}_{\mathrm{ssp}}\}$ on Line \[exact1\] and with $c_{\mathrm{ESS}}=1$. Assume that $$\sup_{(\theta,\theta')\in \Theta^2}\bigg(\frac{\tilde{\pi}_0(\theta)}{q_0(\theta)}+{\mathds{1}}_{(0,\infty)}(h_{t-1}) \frac{f_\theta(Y_t) m_{t-1}(\theta-\theta')}{q_t(\theta|\theta')} +f_\theta(Y_t)\bigg)<\infty,\quad{\mathbb{P}}-a.s.,\quad\forall t\geq 1.$$ Then, for all $t\geq 0$, there exists a random variable $C_t$ taking values in $(0,\infty)$ such that, for all $N\geq 1$, $$\sup_{|\varphi|\leq 1}{\mathbb{E}}\Big[(\tilde{\pi}_t^N(\varphi)-\tilde{\pi}_t(\varphi))^2|\,{\mathcal{F}}_t\Big]\leq\frac{C_t}{N},\quad
\sup_{|\varphi|\leq 1}{\mathbb{E}}\Big[(\bar{\pi}_t^{N,K}(\varphi)-\bar{\pi}_t(\varphi))^2|\,{\mathcal{F}}_t\Big]\leq\frac{C_t}{N\wedge K},\quad{\mathbb{P}}-a.s.$$

The sequence $(m_t)_{t\geq 0}$ does not need to be as defined in Section \[sub:pi\_tilde\] for the result of Proposition \[prop:conv\_algo\] to hold, which only requires that $\sup_{\theta\in {\mathbb{R}}^d} m_{t}(\theta)<\infty$ for all $t\geq 0$.

Conditions ensuring, as $N\rightarrow\infty$, the almost sure convergence of $\tilde{\pi}_t^N$ towards $\tilde{\pi}_t$ can be easily obtained using the results in @crisan2000 and in @gerber2019negative. An almost sure convergence results for $\bar{\pi}^{N,K}_t$, as $\min(N,K)\rightarrow\infty$, can then be easily deduced.

Approximation of the posterior distribution\[sub:post\_app\]
------------------------------------------------------------

In this short subsection we study the approximation error we make when the distribution $\tilde{\pi}_t^N$ is used as an estimate of $\pi_t$. To save place, we focus on $\tilde{\pi}_t^N$ in what follows but similar computations to those done below can be performed for the distribution $\bar{\pi}_t^{N,K}$.

Let $\varphi:\Theta\rightarrow {\mathbb{R}}$ be a continuous and measurable function such that $|\varphi(\theta)|\leq 1$ for all $\Theta\in\Theta$. Then, under the assumptions of Theorems \[thm:online\]-\[thm:lower\_bound\_1\] (see the next section), there exists a sequence $(\delta_{\varphi,t})_{t\geq 1}$ such that $  \delta_{\varphi,t}\rightarrow 0$ and such that $$\begin{aligned}
\label{eq:approx_post}
{\mathbb{E}}\Big[\big(\tilde{\pi}_t(\varphi)-\pi_t(\varphi)\big)^2\Big]^{\frac{1}{2}}\leq  \delta_{\varphi,t},\quad\forall t\geq 1.\end{aligned}$$

Therefore, assuming that the conditions of Proposition \[prop:conv\_algo\] are fulfilled, and with $(C_t)_{t\geq 1}$ as in this latter, we have $$\begin{aligned}
{\mathbb{E}}\Big[(\tilde{\pi}_t^N(\varphi)-\pi_t(\varphi))^2\Big]^{\frac{1}{2}}\leq \sqrt{\frac{{\mathbb{E}}[C_t]}{N}}+\delta_{\varphi,t},\quad\forall (N,t)\in\mathbb{N}^2\end{aligned}$$ assuming that ${\mathbb{E}}[C_t]<\infty$ for all $t\geq 1$. Notice that this latter condition holds e.g. if $$\sup_{(\theta,\theta'y,)\in \Theta^2\times{\mathsf{Y}}}\bigg(\frac{\tilde{\pi}_0(\theta)}{q_0(\theta)}+{\mathds{1}}_{(0,\infty)}(h_{t-1}) \frac{f_\theta(y) m_{t-1}(\theta-\theta')}{q_t(\theta|\theta')} +f_\theta(y)\bigg)<\infty,\quad\forall t\geq 1.$$

Then, by and Proposition \[prop:conv\_algo\], the extra approximation error we make when we use $\tilde{\pi}_t^N$ to estimate $\pi_t$ rather than $\tilde{\pi}_t$ is at most $\delta_{\varphi,t}$, and thus converges to zero as $t\rightarrow\infty$. Computing the rate at which $\delta_{\varphi,t}\rightarrow 0$ and studying the behaviour of the sequence $(C_t)_{t\geq 1}$ are interesting open problems, whose resolution is necessary to understand more precisely how close $\tilde{\pi}_t^N$ is from $\pi_t$ for large values of $t$.

Theoretical analysis of the sequence Lg {#sec:main}
=======================================

In this section we provide conditions on the sequences $(h_t)_{t\geq 0}$ and $(t_p)_{p\geq 0}$ which, under mild assumptions on the statistical model $\{f_\theta,\,\theta\in\Theta\}$, ensure the convergence of the resulting sequence $(\tilde{\pi}_t)_{t\geq 0}$ towards $\delta_{\{\theta_\star\}}$, in ${\mathbb{P}}$-probability. Theorems \[thm:online\]-\[thm:lower\_bound\_1\], given in Section \[sub:theory\_main\], are the main results of this section, while in Section \[sub\_sec\] we provide explicit definitions of sequences $(h_t)_{t\geq 0}$ and $(t_p)_{p\geq 0}$ that satisfy the requirements of these two theorems (Proposition \[prop:h\_t\]). We recall the reader that all the results presented below also hold for the sequence $(\bar{\pi}_t)_{t\geq 0}$ of mixture distributions, by Proposition \[prop:bar\].

Assumptions on the statistical model
------------------------------------

The following two assumptions impose some regularity on the random mapping $\theta\mapsto f_\theta(Y_1)$ around $\theta_\star$.

\[m\_star\] There exist a constant $\delta_{\star}>0$ and a measurable function $m_{\star}\,{:}\,{\mathsf{Y}}\rightarrow {\mathbb{R}}$ such that ${\mathbb{E}}[m_\star^{2}]<\infty$ and $$\big|\log(f_{\theta_1}(Y_1)/f_{\theta_2}(Y_1))\big|\leq m_{\star}(Y_1)\|\theta_1-\theta_2\|,\quad \forall \theta_1,\theta_2\in B_{\delta_\star}(\theta_\star)\cap\Theta,\quad {\mathbb{P}}-a.s.$$

\[taylor\] There exist constants $\delta_\star>0$ and $C_\star<\infty$ such that $${\mathbb{E}}\big[\log(f_{\theta_\star}/f_{\theta})\big]\leq C_\star \|\theta-\theta_\star\|^2,\quad\forall\theta\in B_{\delta_\star}(\theta_\star)\cap\Theta.$$

The next assumption notably implies that $\theta_\star$ is identifiable.

\[test\] For every compact set $W\in\mathcal{B}(\Theta)$ such that $\theta_\star\in W$ and every $\epsilon>0$ there exists a sequence of measurable functions $(\psi'_t)_{t\geq 1}$, with $\psi'_t:{\mathsf{Y}}^t\rightarrow\{0,1\}$, such that $${\mathbb{E}}[\psi'_t(Y_{1:t}) ]\rightarrow 0,\quad\sup_{\theta\in V_\epsilon\cap W}{\mathbb{E}}\Big[\big(1-\psi'_t(Y_{1:t}) \big)\prod_{s=1}^t (f_{\theta}/f_{\theta_\star})(Y_s)\Big]\rightarrow 0.$$

Assumptions \[m\_star\]-\[test\] are standard, see e.g. @Kleijn2012. It is easily checked that \[taylor\] holds when the mapping $\theta\mapsto{\mathbb{E}}[\log (f_{\theta})]$ admits a second order Taylor expansion in a neighbourhood of $\theta_\star$. By @Kleijn2012 [Theorem 3.2], \[test\] holds for instance when, for every compact set $W\in\mathcal{B}(\Theta)$ that contains $\theta_\star$ and every $  \theta'\in W$, the mapping $W\ni \theta\mapsto {\mathbb{E}}[f_\theta f_{\theta'}^{-s}f_{\theta_\star}^{s-1}]$ is continuous at $\theta'$ for every $s$ in a left neighbourhood of 1, and ${\mathbb{E}}[f_{\theta'}/f_{\theta_\star} ]<\infty$. Remark that if the model is well-specified then ${\mathbb{E}}[ f_{\theta}/f_{\theta_\star}  ]=1<\infty$ for all $\theta\in\Theta$. If the model is miss-specified the condition ${\mathbb{E}}[ f_{\theta}/f_{\theta_\star}  ]<\infty$ requires that the tails of $f_{\theta_\star}$ are not too thin compared to those of $f_\star$. For instance, if $f_\theta(y)\eta({\mathrm{d}}y)=t_{1,\nu}(\mu,\sigma^2)$, with $\theta=(\mu,\sigma^2,\nu)$, and $\lim_{|y|\rightarrow\infty} |y|^{\nu_\star+1}f_\star(y)<\infty$ then ${\mathbb{E}}[ f_{\theta}/f_{\theta_\star}  ]<\infty$ for all $\theta\in{\mathbb{R}}\times(0,\infty)^2$.

Assumption \[test\] is stronger that needed for our results to hold, these latter requiring only that \[test\] holds for the specific compact set $W=\tilde{A}_\star\cap\Theta$, where $\tilde{A}_\star$ is as defined in Assumption \[new\] below.

\[new\] For some set $A_\star\in\mathcal{B}({\mathbb{R}}^d)$,

1.  \[new\_part1\] One of the following conditions hold:

    1.  \[A41\] $
        {\mathbb{E}}[\sup_{\theta \not\in A_\star}\log (\tilde{f}_{\theta})]< {\mathbb{E}}[ \log (f_{\theta_\star})]$

    2.  \[A42\] $\sup_{\theta \not\in A_\star }{\mathbb{E}}[\tilde{f}_\theta/f_{\theta_\star} ]< 1$

    3.  \[A43\] $\log\big(\sup_{\theta \not\in A_\star }{\mathbb{E}}[\tilde{f}_\theta]\big)<  {\mathbb{E}}[\log (f_{\theta_\star})]$.

2.  \[new\_part2\] There exists a set $\tilde{A}_\star\in\mathcal{B}({\mathbb{R}}^d)$, containing a neighbourhood of $A_\star$, such that $\tilde{A}_\star\cap\Theta$ is compact, the mapping $\theta\mapsto   f_\theta(Y_1)$ is ${\mathbb{P}}$-a.s. continuous on $\tilde{A}_\star\cap\Theta$ and $$\exists \tilde{\delta}>0\quad\text{such that}\quad {\mathbb{E}}\Big[\sup_{\{  (\theta,v)  \in (\tilde{A}_\star\cap\Theta)\times B_{\tilde{\delta}}(0)\}}  \log (\tilde{f}_{\theta+v}/f_{\theta})^2\Big]<\infty.$$

If $\Theta$ is compact then \[new\] holds as soon as $\theta\mapsto   f_\theta(Y_1)$ is ${\mathbb{P}}$-a.s. continuous on $\Theta$ and ${\mathbb{E}}\big[\sup_{\{ (\theta,\theta') \in \Theta^2:\,\,\|\theta'-\theta\|< \tilde{\delta}\}}  \log (f_{\theta'}/f_{\theta})^2\big]<\infty$ for some $\tilde{\delta}>0$.

\[new1\] At least one of the following three conditions hold:

a)  \[A51\] ${\mathbb{E}}[\sup_{\theta\in\Theta}\log (f_\theta)]<\infty$

b)  \[A52\] $\sup_{\theta\in\Theta}{\mathbb{E}}[ f_\theta/f_{\theta_\star}]<\infty$

c)  \[A53\] $\sup_{\theta\in\Theta}{\mathbb{E}}[f_\theta]<\infty$.

Condition \[new1\].\[A52\] always holds when the model is well-specified.

The last Assumption \[new2\] is used to derive convergence results that hold when the parameter space is unbounded.

\[new2\] Assume that one of the following conditions hold for some $ k_\star\in\{1/2\}\cup \mathbb{N} $.

a)  \[A61\] $  \limsup_{C\rightarrow\infty} {\mathbb{E}}\big[ \sup_{\theta\in V_{C}}\log (\tilde{f}_{\theta}/f_{\theta_\star}) \big] (\log C)^{-1}<0$ and, for some $C_1\in(0,\infty)$, $$\sup_{C\geq C_1}{\mathbb{E}}\Big[\big|\sup_{\theta\in V_{C}}\log (\tilde{f}_{\theta}/f_{\theta_\star})-{\mathbb{E}}\big[\sup_{\theta\in V_{C}}\log (\tilde{f}_{\theta}/f_{\theta_\star})\big]\big|^{2k_\star}\Big]<\infty$$

b)  \[A62\] $\limsup_{C\rightarrow\infty} \log \big(\sup_{\theta\in V_C}{\mathbb{E}}[(\tilde{f}_{\theta}/f_{\theta_\star})]\big) (\log C)^{-1}<0$

c)  \[A63\] $\limsup_{C\rightarrow\infty} \log \big(\sup_{\theta\in V_C}{\mathbb{E}}[ \tilde{f}_{\theta} ]\big) (\log C)^{-1}<0$ and ${\mathbb{E}}\big[\,|\log  (f_{\theta_\star})\,|^{2k_\star}\big]<\infty$.

Assumption \[new2\] implies the existence of a set $A_\star\in\mathcal{B}({\mathbb{R}}^d)$ such that the first part of \[new\] holds.

Assumptions \[new\]-\[new2\] are non-standard but appear to be reasonable, as illustrated in the next proposition.

\[prop:assumption\] $ $

1.  Let $\Theta={\mathbb{R}}\times[\underline{\sigma}^2,\infty)$ for some $\underline{\sigma}^2> 0$ and, for every $\theta=(\mu,\sigma^2)\in\Theta$, let $f_{(\mu,\sigma^2)}$ be the density of the $\mathcal{N}_1(\mu,\sigma^2)$ distribution. Then, \[new\]-\[new1\] hold if ${\mathbb{E}}[Y_1^4]<\infty$. If in addition we have ${\mathbb{E}}[e^{c|Y_1|}]<\infty$ for some $c>0$ then \[new2\] holds for all $k_\star\in\mathbb{N}$.

2.  Let $\Theta={\mathbb{R}}\times[\underline{c},\infty)^2$ for some $\underline{c}> 0$ and, for every $(\mu,\sigma^2,\nu)\in\Theta$, let $f_{(\mu,\sigma^2,\nu)}$ be the density of the $t_{1,\nu}(\mu,\sigma^2)$ distribution. Then, \[new\]-\[new1\] hold if ${\mathbb{E}}[\log(1+c Y_1^2)^2]<\infty$ for all $c>0$. If in addition we have ${\mathbb{E}}[|Y_1|^c]<\infty$ for some $c\geq 1/2$ then \[new2\] holds for all $k_\star\in\mathbb{N}$.

Main results {#sub:theory_main}
------------

\[thm:online\] Let $(h_t)_{t\geq 0}$ and $(t_p)_{p\geq 0}$ be such that

1.  \[thm1.1\] $h_{t_p}>0$ for all $p\geq 0$,

2.  \[thm1.2\] $(t_{p+1}-t_p) \rightarrow\infty$ and $(t_p-t_{p-1}) \sum_{s=t_{p-1}+1}^{t_p-1}h_s^2 \rightarrow 0$,

3.  \[thm1.3\] $\limsup_{p\rightarrow+\infty}\frac{t_{p+1}-t_{p}}{ t_{p}-t_{p-1}}<\infty$ and $\log(1/h_{t_{p-1}})(t_p-t_{p-1})^{-1}\rightarrow 0$.

Moreover, assume that $$\begin{aligned}
\label{eq:compact}
\exists (\beta, c)\in (1,\infty)^2\quad\text{such that}\quad {\mathbb{P}}\big(\tilde{\pi}_{t_p} (V_{c/h_{t_{p}}^{\beta}})\geq \beta^{-1}\big)\rightarrow 0.\end{aligned}$$ Then, under \[m\_star\]-\[new1\], $\tilde{\pi}_{t}\Rightarrow\delta_{\{\theta_\star\}}$ in ${\mathbb{P}}$-probability.

Condition \[thm1.2\] implies that $h_t\rightarrow 0$.

Condition of Theorem \[thm:online\] holds when the parameter space $ \Theta$ is bounded, since in this case we have $\tilde{\pi}_t(\Theta^c)=0$ for all $t\geq 0$, ${\mathbb{P}}$-a.s. Theorem \[thm:lower\_bound\_1\] below provides sufficient conditions for to hold when $\Theta$ is unbounded.

\[thm:lower\_bound\_1\] Assume \[m\_star\], \[taylor\], \[new1\] and \[new2\]. For every $C>0$ let $$\zeta(C)=
\begin{cases}
{\mathbb{E}}\big[ \sup_{\theta\in V_{C}}\log (f_{\theta}/f_{\theta_\star}) \big],&\text{if \ref{new2}.\ref{A61} holds}\\
 \log \big(\sup_{\theta\in V_C}{\mathbb{E}}[(f_{\theta}/f_{\theta_\star})]\big),&\text{if \ref{new2}.\ref{A62} holds}\\
 \log \big(\sup_{\theta\in V_C}{\mathbb{E}}[ f_{\theta} ]\big), &\text{if \ref{new2}.\ref{A63} holds}
\end{cases}$$ and let $ k_\star\in\{1/2\}\cup \mathbb{N}  $ be as in \[new2\]. Let $(h_t)_{t\geq 0}$ and $(t_p)_{p\geq 0}$ be such that Conditions \[thm1.1\]-\[thm1.2\] of Theorem \[thm:online\] hold and such that

1.  \[thm2.1\] $h_{t_p}<h_{t_{p-1}}$ for all $p\geq 0$,

2.  \[thm2.2\] $h_{t_p}=\Theta(t_p^{-\alpha})$ for some $\alpha>0$,

3.  \[thm2.3\] There exists a constant $\beta_\star\in(0,\infty)$ such that $$\begin{aligned}
    \label{eq:thm2}
     \big|\zeta(h_{t_{p}}^{-\beta_\star})\big|^{-2k_\star}\sum_{i=1}^p (t_i-t_{i-1})^{-k_\star{\mathds{1}}_{\mathbb{N}}(k_\star)}\rightarrow 0.\end{aligned}$$

Then, Condition of Theorem \[thm:online\] holds.

Application of the main results {#sub_sec}
-------------------------------

The conditions on $(h_t)_{t\geq 0}$ in Theorems \[thm:online\]-\[thm:lower\_bound\_1\] are quite weak. In particular, as shown in the next result, they do not require that $\sum_{t=0}^\infty h_t^2<\infty$.

\[prop:h\_t\]

Let $C\in[1,\infty)$, $\alpha\in(0,\infty)$, $\varrho\in(0,\alpha\wedge 1)$ and $c\in(0,1)$ be some constants. Let $(t_p)_{p\geq 0}$ be defined by $$\begin{aligned}
\label{eq:tp_formula}
t_0\in\mathbb{N}_0\quad C_{p-1}\in [c t^\varrho_{p-1},  t_{p-1}^\varrho/c],\quad t_p= t_{p-1}+\lceil C_{p-1} \log(t_{p-1})\vee C\rceil,\quad   p\geq 1\end{aligned}$$ and let $(h_t)_{t\geq 0}$ be such that $h_{t_p}=\Theta(t_p^{-\alpha})$, such that $h_{t_p}<h_{t_{p-1}}$ for all $p\geq 1$ and such that, for all $t\geq 1$, either $h_{t-1}>h_t$ or $h_{t-1}=0$. Then, the sequences $(h_t)_{t\geq 0}$ and $(t_p)_{p\geq 0}$ verify Conditions \[thm1.1\]-\[thm1.3\] of Theorem \[thm:online\] and Conditions \[thm2.1\]-\[thm2.2\] of Theorem \[thm:lower\_bound\_1\]. In addition, the sequences $(h_t)_{t\geq 0}$ and $(t_p)_{p\geq 0}$ also verify Condition \[thm2.3\] of Theorem \[thm:lower\_bound\_1\] if \[new2\] holds for a $k_\star>(1+\varrho)/\varrho$.

By Proposition \[prop:h\_t\], it follows that Conditions \[thm1.1\]-\[thm1.3\] of Theorem \[thm:online\] and Conditions \[thm2.1\]-\[thm2.2\] of Theorem \[thm:lower\_bound\_1\] hold, for instance, when as in Algorithm \[algo:pi\_tilde\_online\] we have $h_t=c\, t^{-\alpha}$ for all $t\geq 1$ and for some constant $(c,\alpha)\in(0,\infty)^2$, and when $(t_p)_{p\geq 0}$ is as defined in . The second part of the proposition notably implies that if \[new2\] holds for all $k_\star\in\mathbb{N}$, as in the examples of Propositions \[prop:assumption\], then any sequences $(t_p)_{p\geq 0}$ and $(h_t)_{t\geq 0}$ as defined in Proposition \[prop:h\_t\] verify the assumptions of Theorems \[thm:online\]-\[thm:lower\_bound\_1\].

The conclusions of Theorems \[thm:online\]-\[thm:lower\_bound\_1\] and of Proposition \[prop:h\_t\] are summarized in the following corollary.

\[cor:main\] Assume \[m\_star\]-\[new1\] and that \[new2\] holds for some $k_\star\geq 3$, and let $(h_t)_{t\geq 0}$ and $(t_p)_{p\geq 0}$ be as in Proposition \[prop:h\_t\] for some $\alpha>1/(k_\star-1)$ and $\varrho\in(1/(k_\star-1), \alpha \wedge 1)$. Then, $\tilde{\pi}_{t}\Rightarrow\delta_{\{\theta_\star\}}$ in ${\mathbb{P}}$-probability.

Discussion: Why Student’s $t$-distributions are needed?\[sub:student\]
----------------------------------------------------------------------

Our theoretical analysis of $(\tilde{\pi}_t)_{t\geq 0}$ requires that infinitely many times the distribution $\mu_{t-1}$ entering in the definition of $\tilde{\pi}_t$ is a Student’s $t$-distribution. This condition is used in the proof of Theorems \[thm:online\]–\[thm:lower\_bound\_1\] to guarantee that the mass of $\tilde{\pi}_t$ around $\theta_\star$ is large enough, sufficiently often (recall that Student’s $t$-distributions have thicker tails than Gaussian distributions). As shown in Proposition \[prop:continuity\], the smaller the $h_t$’s are the closer $\tilde{\pi}_t$ is from the posterior distribution $\pi_t$. Hence, since ${\mathbb{P}}(\pi_t\Rightarrow\delta_{\{\theta_\star\}})=1$ under the assumptions of Theorems \[thm:online\]–\[thm:lower\_bound\_1\] [@Kleijn2012], one may expect that the faster $h_t\rightarrow 0$ the less often it is needed to take a Student’s $t$-distribution for $\mu_{t-1}$ to guarantee that $\tilde{\pi}_t$ has sufficient mass on a neighbourhood of $\theta_\star$. The condition on $\varrho$ given in Proposition \[prop:h\_t\] supports this reasoning to some extend. However, in this result $\varrho$ cannot becomes arbitrarily large as $\alpha$ tends to $\infty$.

On the other hand, a consequence of the fat tails of the Student’s $t$-distributions is that regular Bayes updates (obtained when $\mu_{t-1}=\delta_{\{0\}}$) are more perturbed when $\mu_{t-1}$ belongs to this family of distributions than when $\mu_{t-1}$ is a Gaussian distribution. For this reason, Condition \[thm1.2\] of Theorem \[thm:online\] requires that the number of observations processed between two successive use of a Student’s $t$-distribution increases over time and converges to infinity as $t\rightarrow\infty$.

As shown in Proposition \[prop:h\_t\], one of the main merit of Theorems \[thm:online\]-\[thm:lower\_bound\_1\] is to guarantee the convergence on $\tilde{\pi}_t$ towards $\delta_{\{\theta_\star\}}$ under very week conditions on $(h_t)_{t\geq 0}$. It is an open problem whether or not a similar result can be achieved when, in the definition of $(\mu_t)_{t\geq 0}$, $m_t(\theta){\mathrm{d}}\theta$ is a Gaussian distribution for all $t\geq 0$.

Numerical Experiments\[sec:num\]
================================

The objective of this section is fourfold. Firstly, it is to study the impact of $(h_t)_{t\geq 0}$ on the convergence properties of $(\bar{\pi}_t)_{t\geq 0}$ in a more complicated set-up than in Section \[subsec:example\], and to compare the IBIS estimate of $\pi_t$ with those obtained by approximating $\tilde{\pi}_t$ and $\bar{\pi}_t$ using Algorithm \[algo:pi\_tilde\] (Section \[sub:linear\_model\]). Secondly, it is to compare the number of observations that IBIS and Algorithm \[algo:pi\_tilde\_online\] can process in a given amount of time. We use IBIS as benchmark in this section because, with the method introduced in this work, it is the only theoretically justified algorithm that can be used to approximate $(\pi_t)_{t\geq 1}$ sequentially. Recall that both Algorithm \[algo:pi\_tilde\_online\] and IBIS are particular cases of Algorithm \[algo:pi\_tilde\].

The IBIS algorithm may be particularly efficient when computing $f_\theta(y)$ is cheap (so that evaluating $\alpha^n_t(\cdot,\cdot)$ on Line \[expensive\] of Algorithm \[algo:pi\_tilde\] is not too expressive, even for large values of $t-k_t$) and when the observations are actually i.i.d. Under these two conditions, and by letting $c_{\mathrm{ESS}}<1$ in Algorithm \[algo:pi\_tilde\], as $t$ increases the number of resampling steps needed by IBIS may decrease sufficiently quickly to compensate the fact that the computational cost of each of them becomes larger [@MR1929161]. This phenomenon is illustrated with the example of Section \[sub:linear\_model\]. Sections \[sub:50\_dim\] and \[sub:latent\] provide two examples where computing $f_\theta(y)$ is very expensive and using IBIS is prohibitive, while the experiment of Section \[non-stationnary\] shows, with a simple logistic regression model with two parameters, that Algorithm \[algo:pi\_tilde\_online\] can be much faster than IBIS when the observations are independent but not i.i.d. A real data example where Algorithm \[algo:pi\_tilde\_online\] outperforms IBIS in term of running time is given in Section \[sub:online\].

Providing a fair comparison between the running time of two algorithms is a complicated exercise. Algorithm \[algo:pi\_tilde\] is implemented in R [@R], except the computations of the likelihood terms appearing on Lines \[l1\], \[expensive\], \[l2\], \[l3\], \[No\_res1\] and \[No\_res2\], which are performed in C and parallelized over 6 workers. This implementation of Algorithm \[algo:pi\_tilde\] is not particularity fast but should, to some extend, favour IBIS in the running time comparisons (recall that Line \[expensive\] is the most expensive step of IBIS). Still, these latter should be interpreted with care because they may depend heavily one the values chosen for the tuning parameters of Algorithm \[algo:pi\_tilde\] (and notably on the value of $N$, $K$ and $c_{\mathrm{ESS}}$). We also remind the reader that IBIS computes a Monte Carlo approximation of $\pi_t$ while Algorithm \[algo:pi\_tilde\_online\] computes a Monte Carlo estimate of a distribution which is itself an approximation of $\pi_t$. Consequently, the IBIS estimate of $\pi_t$ will be, in general, more accurate than the one obtained using Algorithm \[algo:pi\_tilde\_online\].

The third objective of this section is to provide a first assessment of the online predictive performance of Algorithm \[algo:pi\_tilde\_online\]. To this aim, in Section \[sub:online\] we consider a real data online binary classification example, with 53 predictors, and compare Algorithm \[algo:pi\_tilde\_online\] with stochastic gradient descent and the version of streaming variational Bayes [@broderick2013streaming] proposed by @nguyen2017online [@zeno2018bayesian]. On this example we find that Algorithm \[algo:pi\_tilde\_online\] largely outperforms these two competing methods.

If it is often reasonable to assume that the observations generated by a data stream are independent, in many real-world applications the i.i.d. assumption does not hold and, typically, the underlying data generating process is characterized by an evolving or a drifting phenomenon [@ditzler2015learning]. While none of the methods discussed in this work is designed for statistical learning in a time-varying environment, it is important to assess their ability to detect such a violation of the i.i.d. assumption. In addition, it has been proven that SGD may be useful in this context [@watanabe1975robbins] and it is of interest to study if this is also the case for the proposed approximate Bayesian approach. The last objective of this section is to provide a first answer to these questions, which is done in Section \[non-stationnary\] with a very simple model.

For reasons explained in Section \[sub:averaging\], we mainly focus below on the inference obtained using the sequence $(\bar{\pi}_t)_{t\geq 1}$ as an approximation of the posterior distributions $(\pi_t)_{t\geq 1}$. Lastly, because we consider computationally intensive examples, all the results presented below are based on a single run of the algorithms using a fixed seed for the underlying random number generator.

Simulation set-up
-----------------

Unless otherwise mentioned, all the results presented below are obtained using the R and C implementation of Algorithm \[algo:pi\_tilde\] described above.

Throughout this section the sequence $(h_t)_{t\geq 0}$ will be such that $h_0=0$ while, for a $\Delta\in  \mathbb{N}_0$ and an $\alpha>0$, $$\begin{aligned}
\label{eq:sim_ht}
h_t=
\begin{cases}
 t^{-\alpha} &t\in\{s\Delta,\,s\in\mathbb{N}\}\\
0&\text{otherwise},
\end{cases}\quad \forall t\geq 1.\end{aligned}$$ Then, we let $(t_p)_{p\geq 0}$ be defined by $t_{p}=\sup\{t\in  \{s\Delta,\, s\in\mathbb{N}\}:\,\tilde{t}_p\geq t\}$, $p\geq 0$, where $$\tilde{t}_0=10,\quad \tilde{t}_p=\tilde{t}_{p-1}+\max\Big\{10, 10\big\lfloor \max\big\{0.5,\tilde{t}_{p-1}^{\,0.8\alpha}\big\}\log(\tilde{t}^{\,\alpha}_{t_{p-1}})\big\rfloor\Big\},\quad p\geq 1.$$ Notice that $(t_p)_{p\geq 0}$ verifies the conditions of Proposition \[prop:h\_t\]. To complete the specification of $(\mu_t)_{t\geq 0}$ we let $\nu=50$ and $\Sigma_t=I_d$ for all $t\geq 0$.

Algorithm \[algo:pi\_tilde\] is implemented with $q_t(\theta_t|\theta_{t-1})=h_{t-1}^{-d}m_{t-1}((\theta_t-\theta_{t-1})/h_{t-1})$ for all $t\geq 1$ such that $h_{t-1}>0$. When $h_{t-1}=0$ we follow the standard approach in the SMC literature by letting $$\begin{aligned}
\label{eq:prop_q}
q_t(\theta |\theta'){\mathrm{d}}\theta=\mathcal{N}_d\Big(\theta', c\, \sum_{n=1}^N W_{t-1}^n\big(\theta_{t-1}^n-\sum_{m=1}^N W_{t-1}^m \theta_{t-1}^m\big)^T\big(\theta_{t-1}^n-\sum_{m=1}^N W_{t-1}^m \theta_{t-1}^m\big)\Big)\end{aligned}$$ for some scaling factor $c>0$. Remark that the covariance matrix in is the matrix $c{\mathrm{Var}}_{\tilde{\pi}_t^N}(\theta)$. While a direct application of the results in @roberts2001optimal, derived for the random walk Metropolis-Hastings algorithm, suggests to take $c=2.38^2 d^{-1}$ in , we find that, for the considered examples, $2.38 d^{-1/2}$ is a good default value for this constant. Hence, we let $c=2.38 d^{-1/2}$ and, for the remaining input parameters of Algorithm \[algo:pi\_tilde\], we let $q_0=\tilde{\pi}_0$, $J=1$, $\mathcal{R}(\cdot,\cdot)$ be the SSP resampling algorithm [see @gerber2019negative Algorithm 1] and, somewhat arbitrarily, $c_{\mathrm{ESS}}=0.7$.

Except for the example of Section \[sub:linear\_model\], below we restrict ourself to the case $\alpha=0.5$ and $\Delta\in\{0,1\}$. Notice that when $\Delta=0$ in Algorithm \[algo:pi\_tilde\] reduces to a particular version of IBIS, and we denote by $\pi_t^N$ the resulting estimate of $\pi_t$. Note also that when $\Delta=1$ Algorithm \[algo:pi\_tilde\] reduces to Algorithm \[algo:pi\_tilde\_online\] with $c=1$.

Lastly, for all $t\geq 0$ we let $\theta_t^N={\mathbb{E}}_{\pi_t^N}[\theta]$, $\tilde{\theta}_t^N={\mathbb{E}}_{\tilde{\pi}_t^N}[\theta]$, $\bar{\theta}^N_t=(t+1)^{-1}\sum_{s=0}^t\tilde{\theta}_s^N$ and $$\overline{{\mathrm{Var}}}_{t,N}(\theta)=\frac{1}{t+1}\Big\|\mathrm{diag}\Big(\sum_{s=0}^t{\mathbb{E}}_{\scalebox{1.5}{$\scriptscriptstyle \tilde{\pi}_s^N$}}[\theta\theta^T]-\sum_{s=0}^t(\tilde{\theta}_s^N)(\tilde{\theta}_s^N)^T\Big)\Big\|_\infty.$$ Remark that $\bar{\theta}^N_t$ is an estimate of $\bar{\theta}_t$ while $\overline{{\mathrm{Var}}}_{t,N}(\theta)$ is an estimate of $\max\{{\mathrm{Var}}_{\bar{\pi}_t}(\theta_i)\}_{i=1}^d$.

Bayesian quantile regression models\[sec:num\_quantile\]
--------------------------------------------------------

Let $(Z_t)_{t\geq 1}$ and $(X_t)_{t\geq 1}$ be sequences of i.i.d. random variables taking values in ${\mathbb{R}}$ and ${\mathbb{R}}^{d}$, respectively. We assume that, for every $\tau\in(0,1)$, the $\tau$-th conditional quantile function of $Z_t$ given $X_t$ does not dependent on $t$ and belongs to $\{\gamma(\theta, \cdot),\, \theta\in\Theta\}$, with $\Theta={\mathbb{R}}^d$. Then, following @Yu2001, for every $\tau\in(0,1)$ we let $$\begin{aligned}
g_{\tau,\theta}(z|x)=\tau(\tau-1)\exp\big(-0.5\big\{|z-\gamma(\theta,x)|+(2\tau-1)(z-\gamma(\theta,x))\big\}\big)\end{aligned}$$ and $\theta^{(\tau)}_{\star}=\operatorname*{argmax}_{\theta\in\Theta}{\mathbb{E}}[\log (g_{\tau,\theta}(Z_1|X_1))]$. Remark that $g_{\tau,\theta}(\cdot|x)$ is the density of an asymmetric Laplace distribution. We take for $\tilde{\pi}_0$ the $\mathcal{N}_d(\theta^{(\tau)}_{\star}-10,2 I_d)$ distribution, so that most of the mass of $\tilde{\pi}_0$ is faraway from the true parameter value, and we generate the observations as follows $$\begin{aligned}
Z_t|X_t\sim\mathcal{N}_1\big(\gamma(\theta^{(1/2)}_{\star},X_t),4 \big),\quad X_t\sim p_X,\quad t\geq 1\end{aligned}$$ where $p_X$ will depend on the specific example. Notice that the model is misspecified for ever $\tau\in(0,1)$ and that the misspecification becomes more severe as $\tau$ approaches 0 or 1.

### A 5-dimensional linear quantile regression model\[sub:linear\_model\]

In order to assess the properties of $(\bar{\pi}_t)_{t\geq 0}$ for various sequences $(h_t)_{t\geq 0}$ we consider a simple $d=5$ dimensional linear model, so that $\gamma(\theta,x)=\theta^Tx$. We let $\tau=0.9$, $\theta^{(1/2)}_{\star}$ be a random draw from the $\mathcal{N}_d(0,I_d)$ distribution and $p_X=\delta_{\{1\}}\otimes\mathcal{N}_{d-1}(0,\Sigma_X)$, with $\Sigma^{-1}_X$ a random draw from the Wishart distribution with $d-1$ degrees of freedom and scale matrix $I_{d-1}$. We simulate $T=10^7$ observations from the so-defined model and set $N=1\,000$ in Algorithm \[algo:pi\_tilde\].

Figures \[fig:num\]-\[fig:num\] show the evolution of $\| \bar{\theta}^N_{t} -\theta^{(\tau)}_\star\|_\infty$ as a function of $t$ for $$(\Delta,\alpha)\in\big\{(1,0.5),(1,0.1),(10^4,0.1)\big\}.$$ For $(\Delta,\alpha)=(1,0.5)$ the estimation error $\| \bar{\theta}^N_{t} -\theta^{(\tau)}_\star\|_\infty$ decreases as $t^{-1/2}$ (Figure \[fig:num\]), while for $(\Delta,\alpha)=(1,0.1)$ the rate is much slower (Figure \[fig:num\]). Therefore, if as mentioned in Section \[sub:averaging\] the point estimate $\bar{\theta}_t$ can, empirically, converge at the optimal rate, there are situations where this is not the case. Further research is needed to understand the convergence properties of $\bar{\theta}_t$. Figure \[fig:num\] shows that increasing $\Delta$ from 1 to $10^4$ enables to recover an empirical convergence rate of order $t^{-1/2}$ when $\alpha=0.1$. Figure \[fig:num\] shows the evolution of $\overline{{\mathrm{Var}}}_{t,N}(\theta)$ as a function of $t$ and for $(\Delta,\alpha)\in\{(1,0.5),(1,0.1)\}$. As already observed in the toy example of Section \[subsec:example\] (see Table \[table:ex\]), we see that the convergence rate of the ‘variance’ $\overline{{\mathrm{Var}}}_{t,N}(\theta)$ is slower than that of $h^2_t$, and is thus slower than the usual $t^{-1}$ convergence that we have for the variance of $\theta$ under $\pi_t$. Indeed, Figure \[fig:num\] suggests that $\overline{{\mathrm{Var}}}_{t,N}(\theta)$ decreases as $t^{-0.1}$ for $\alpha=0.1$ and as $t^{-0.7}$ for $\alpha=0.5$. In Section \[sub:running\_time\] we explained that, for $(h_t)_{t\geq 0}$ as defined in , the running time of Algorithm \[algo:pi\_tilde\] does not necessarily increase with $\Delta$ when $c_{\mathrm{ESS}}<1$. This example provides an illustration of this phenomenon. Indeed, for $\alpha=0.1$, the running time of the algorithm was 1.87 hour for $\Delta=1$ and 54 minutes for $\Delta=10^4$, against only 46 minutes for IBIS.

To assess, for $N=1\,000$ and $K=10\,000$, how the distributions $\bar{\pi}^{N,K}_t$ and $\tilde{\pi}^N_t$ are different from the IBIS approximation $\pi_t^N$ of $\pi_t$ we provide, in Table \[table:MMD\_sim\], estimates of ${\mathbb{E}}[d_{\mathrm{MMD}}(\bar{\pi}_t^{N,K}, \pi_t^N)|{\mathcal{F}}_t]$ and of ${\mathbb{E}}[d_{\mathrm{MMD}}(\tilde{\pi}_t^N, \pi_t^N)|{\mathcal{F}}_t]$, the expected MMD distance between $\bar{\pi}_t^{N,K}$ and $\pi_t^N$ and the expected MMD distance between $\tilde{\pi}_t^N$ and $\pi_t^N$, respectively. Notice that the expectations are w.r.t. the Monte Carlo approximation only. These expected distances are computed using the exponential kernel with bandwidth parameter equal to one (in which case $d_{\mathrm{MMD}}$ is a metric on $\mathcal{P}(\Theta)$) and with $\tilde{\pi}_t^N$ independent of $\pi_t^N$ given $\{Y_s\}_{i=1}^t$, for all $t\geq 1$ (and thus ${\mathbb{E}}[d_{\mathrm{MMD}}(\tilde{\pi}_t^N, \pi_t^N)|{\mathcal{F}}_t]>0$ even when $h_s=0$ for all $s\in\{0,\dots,t-1\}$).

From the results presented in Table \[table:MMD\_sim\], obtained for $t\in\{10^3,10^4,10^5\}$ and different values of $(\Delta,\alpha)$, we observe that ${\mathbb{E}}[d_{\mathrm{MMD}}(\bar{\pi}_t^{N,K}, \pi_t^N)|{\mathcal{F}}_t]$ decreases (i) when $t$ increases, in the case $\Delta=1$ and $\alpha\in\{0.5,0.6\}$, (ii) when $\alpha$ increases, in the case $\Delta=1$ and $t=10^5$, and (iii) as $\Delta$ increases, in all cases but one. Observation (i) (resp. observation (ii)) can be explained by the results of Section \[sec:main\] (resp.  of Proposition \[prop:continuity\]), which imply that as $t$ increases (resp. as $\alpha$ increases) $\bar{\pi}_t^{N,K}$ and $\pi_t^N$ are Monte Carlo approximations of distributions that become more and more similar. This explanation for (i) is formalized in Section \[sub:post\_app\]. Observation (iii) is also intuitive since, informally speaking, increasing $\Delta$ makes $\tilde{\pi}_t$ more similar to $\pi_t$. It is interesting to note that for $t=10^4$ the expected distance ${\mathbb{E}}[d_{\mathrm{MMD}}(\bar{\pi}_t^{N,K}, \pi_t^N)|{\mathcal{F}}_t]$ is the smallest when $\alpha=0.5$ while for $t=10^3$ it is minimized when $\alpha=0.3$. Reducing $\alpha$ improves the exploration of the parameter space performed by Algorithm \[algo:pi\_tilde\], which may indeed decrease ${\mathbb{E}}[d_{\mathrm{MMD}}(\bar{\pi}_t^{N,K}, \pi_t^N)|{\mathcal{F}}_t]$ for small values of $t$. The global picture of the results for ${\mathbb{E}}[d_{\mathrm{MMD}}(\tilde{\pi}_t^N, \pi_t^N)|{\mathcal{F}}_t]$ is similar.

                       $\Delta=1$              $\Delta=5\,000$              $\Delta=10\,000$
  -------------- ----------------------- --------------------------- -------------------------------
   $\alpha=0.6$   (0.391, 0.391, 0.196)   (**0.248**, 0.157, 0.206)   (**0.248**, **0.144**, 0.153)
   $\alpha=0.5$   (0.344, 0.280, 0.203)   (**0.248**, 0.156, 0.179)   (**0.248**, **0.144**, 0.146)
   $\alpha=0.3$   (0.328, 0.369, 0.334)   (**0.248**, 0.162, 0.173)   (**0.248**, **0.144**, 0.143)
   $\alpha=0.6$   (0.429, 0.264, 0.136)   (**0.057**, 0.108, 0.274)   (**0.057**, **0.047**, 0.198)
   $\alpha=0.5$   (0.187, 0.194, 0.171)   (**0.057**, 0.103, 0.191)   (**0.057**, **0.047**, 0.160)
   $\alpha=0.3$   (0.249, 0.317, 0.318)   (**0.057**, 0.094, 0.169)   (**0.057**, **0.047**, 0.150)

  : Model of Section \[sub:linear\_model\]. Estimated values of ${\mathbb{E}}[d_{\mathrm{MMD}}(\bar{\pi}_t^{N,K}, \pi_t^N)|{\mathcal{F}}_t]$(top) and of ${\mathbb{E}}[d_{\mathrm{MMD}}(\tilde{\pi}_t^N, \pi_t^N)|{\mathcal{F}}_t]$(bottom) for $t=10^3$ (first number), $t=10^4$ (second number) and $t=10^5$ (third number), where $N=1\,000$ and $K=10\,000$. The results are obtained from 20 independent replications. The numbers in black correspond to the case where $\tilde{\pi}_s=\pi_s$ for all $s\in\{0,\dots,t-1\}$.\[table:MMD\_sim\]

### A 50-dimensional non linear quantile regression model\[sub:50\_dim\]

We now consider a $d=50$ non-linear model where, inspired by @Hunter2000, $$\begin{aligned}
\label{eq:nonlin}
\gamma(\theta,x)=\sum_{i=1}^d\big(e^{-x_{i}\theta^2_i}+x_{i}\theta_{d-i+1}\big),\quad  \theta^{(1/2)}_\star=(1,\dots,1),\quad   p_X=\delta_{\{1\}}\otimes {\mathcal{U}}(0,1)^{d-1}.\end{aligned}$$ Notice that the exponential and square functions appearing in , as well as the relatively large value of $d$, make the evaluation of the corresponding log-likelihood function expensive. We simulate $T=10^6$ observations from the so-defined model and perform simulations for $\tau=0.5$ (the only value of $\tau\in(0,1)$ for which $\theta_\star^{(\tau)}$ is known). For this example we let $N=K=10\,000$ and $(\Delta,\alpha)=(1,0.5)$, that is we consider Algorithm \[algo:pi\_tilde\_online\] with $c=1$ and $\alpha=0.5$. The estimation error $\| \bar{\theta}^N_{t} -\theta^{(\tau)}_\star\|_\infty$, presented in Figure \[fig:num\], decreases at rate $t^{-1/2}$, suggesting that the estimator $\bar{\theta}_t$ can converge at this optimal rate in challenging settings. However, and as for the previous examples, we observe that the convergence rate of the ‘variance’ $\overline{{\mathrm{Var}}}_{t,N}(\theta)$ is slower than that of $h_t^2=t^{-1}$ (see Figure \[fig:num\] where this quantity decreases as $t^{-0.7}$). We now run IBIS (Algorithm \[algo:pi\_tilde\] with $h_t=0$ for all $t\geq 0$) for the same amount of time that was needed to approximate $\{\bar{\pi}_t\}_{t=1}^T$, namely approximatively 15.5 hours. For this computational budget only $T'=129\,435$ observations (13% of the whole sample) have been processed by IBIS and, as shown in Figure \[fig:num\], the estimation error of $\theta_{T'}^N$ is about 1.77 times bigger than for the estimate $\bar{\theta}^N_{T}$ (0.46 against 0.26). By comparison, if $\|\theta_t^N- \theta_\star^{(1/2)}\|=C t^{-1/2}$ for all $t$ and some constant $C<\infty$, observing $T$ data points instead of $T'$ will divide the approximation error of the IBIS estimate of $\theta_\star$ by a factor 2.78.

It is worth mentioning that in this experiment the only reason why Algorithm \[algo:pi\_tilde\] computes an approximation $\bar{\pi}_t^{N,K}$ of $\bar{\pi}_t$ with $K=10\,000$ is to make as fair as possible the above comparison between the running time of Algorithm \[algo:pi\_tilde\_online\] and the running time of IBIS (recall that the point estimate $\bar{\theta}^N_{t}$ is not computed from $\bar{\pi}_t^{N,K}$).

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[0.35]{} ![Examples of Sections \[sec:num\_quantile\]-\[sub:latent\]. Plots ()-() are for the example of Section \[sub:linear\_model\]. Plot () shows $\| \bar{\theta}^N_{t} -\theta^{(\tau)}_\star\|_\infty$ for $(\Delta,\alpha)=(1,0.5)$, plot () shows $\| \bar{\theta}^N_{t} -\theta^{(\tau)}_\star\|_\infty$ for $(\Delta,\alpha)=(1,0.1)$ (solid line, SL) and for $(\Delta,\alpha)=(10^4,0.1)$ (dashed line, DL), and plot () shows $\overline{{\mathrm{Var}}}_{t,N}(\theta)$ for $(\Delta,\alpha)=(1,0.5)$ (SL) and for $(\Delta,\alpha)=(1,0.1)$ (DL). Plots ()-() are for the example of Section \[sub:50\_dim\]. Plot () shows $\| \bar{\theta}^N_{t} -\theta^{(\tau)}_\star\|_\infty$ (SL) and $\| \theta_t^N -\theta^{(\tau)}_\star\|_\infty$ (DL) and plot () shows $\overline{{\mathrm{Var}}}_{t,N}(\theta)$. Plots ()-() are for the example of Section \[sub:latent\]. Plot () shows $\| \bar{\theta}^N_{t} -\theta_\star\|_\infty$ and plots ()-() show the marginal distributions of $\bar{\pi}^{N,K}_T$ (the horizontal lines give the true parameter values). In plots ()-() the dotted line represents the $\gamma t^{-x}$ line, with $x>0$ given in the plots and some constant $\gamma>0$. \[fig:num\]](figs/Non_lin_quantile2.pdf "fig:")

 

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[0.35]{} ![Examples of Sections \[sec:num\_quantile\]-\[sub:latent\]. Plots ()-() are for the example of Section \[sub:linear\_model\]. Plot () shows $\| \bar{\theta}^N_{t} -\theta^{(\tau)}_\star\|_\infty$ for $(\Delta,\alpha)=(1,0.5)$, plot () shows $\| \bar{\theta}^N_{t} -\theta^{(\tau)}_\star\|_\infty$ for $(\Delta,\alpha)=(1,0.1)$ (solid line, SL) and for $(\Delta,\alpha)=(10^4,0.1)$ (dashed line, DL), and plot () shows $\overline{{\mathrm{Var}}}_{t,N}(\theta)$ for $(\Delta,\alpha)=(1,0.5)$ (SL) and for $(\Delta,\alpha)=(1,0.1)$ (DL). Plots ()-() are for the example of Section \[sub:50\_dim\]. Plot () shows $\| \bar{\theta}^N_{t} -\theta^{(\tau)}_\star\|_\infty$ (SL) and $\| \theta_t^N -\theta^{(\tau)}_\star\|_\infty$ (DL) and plot () shows $\overline{{\mathrm{Var}}}_{t,N}(\theta)$. Plots ()-() are for the example of Section \[sub:latent\]. Plot () shows $\| \bar{\theta}^N_{t} -\theta_\star\|_\infty$ and plots ()-() show the marginal distributions of $\bar{\pi}^{N,K}_T$ (the horizontal lines give the true parameter values). In plots ()-() the dotted line represents the $\gamma t^{-x}$ line, with $x>0$ given in the plots and some constant $\gamma>0$. \[fig:num\]](figs/PB2.pdf "fig:")

 

[0.35]{}

A latent variable model: Beta-Poisson distribution\[sub:latent\]
----------------------------------------------------------------

The Beta-Poisson distribution, which is a popular model for the distribution of single-cell gene expression levels [see @lee2019unbiased and references therein], is such that $X\sim$Beta-Poisson$(\theta)$ if $$\begin{aligned}
\label{eq:PB1}
 X|W\sim\mathrm{Poisson}(\exp(\theta_1) W),\quad W \sim\mathrm{Beta}(\exp(\theta_2),\exp(\theta_3)),\quad \theta\in\Theta:={\mathbb{R}}^3.\end{aligned}$$ Following @lee2019unbiased we let $\{f_\theta,\theta\in\Theta\}$ be the family of p.d.f. on ${\mathbb{R}}$ defined by $$\begin{aligned}
\label{eq:PB_model}
f_\theta(y)=\sum_{x=0}^\infty \varphi(y;x,25)p_{\theta}(x),\quad (y,\theta)\in{\mathbb{R}}\times\Theta\end{aligned}$$ with $\varphi(\cdot;m,\sigma^2)$ the density of the $\mathcal{N}_1(m,\sigma^2)$ distribution and $p_{\theta}(\cdot)$ the density of the Beta-Poisson$(\theta)$ distribution (w.r.t. the counting measure on $\{0,1,\dots,\}$).

Notice that defines a latent variable model and that $f_\theta(y)$ cannot be computed point-wise. However, using standard pseudo-marginal arguments [@Andrieu2009], it is easily checked that Algorithm \[algo:pi\_tilde\] remains valid if, in this latter, each quantity of the form $\prod_{i=1}^k f_\theta(y_i)$ is replaced by $$\begin{aligned}
\label{eq:PB_es}
\prod_{i=1}^k\Big(\frac{1}{m k}\sum_{j=1}^{m k} \varphi(y_i;X^{(j)}_i,25)\Big),\quad X_i^{(j)}{\stackrel{\mathclap{\normalfont\mbox{\tiny{iid}}}}{\sim}}\text{Beta-Poisson}(\theta)\end{aligned}$$ for some $m\in\mathbb{N}$. Remark that in we impose that the number of simulations increases linearly with $k$ to control the relative variance of the estimator [@lee2019unbiased].

We simulate $T=10^5$ independent observations $\{Y_t\}_{t=1}^T$ from $f_{\theta_\star}(y)\eta({\mathrm{d}}y)$ where, following @lee2019unbiased, $\exp(\theta_\star)=(500,2,8)$. In addition, as in this latter reference, we let $m=40$ in and $\tilde{\pi}_0({\mathrm{d}}\theta)$ be such that if $\theta\sim \tilde{\pi}_0({\mathrm{d}}\theta)$ then, for $i=1,2,3$, the random variable $\exp(\theta_i)$ has an exponential distribution with mean $\lambda_i$, with $\lambda_1=1\,000$ and $\lambda_2=\lambda_3=10$. As for the previous example, we let $N=K=10\,000$ and $(\Delta,\alpha)=(1,0.5)$ (i.e. we consider Algorithm \[algo:pi\_tilde\_online\] with $c=1$ and $\alpha=0.5$). Figure \[fig:num\] shows the evolution of $\| \bar{\theta}^N_{t} -\theta_\star\|_\infty$ as a function of $t$. We observe that the estimation error first decreases as $t^{-1/2}$ and then at a faster rate (for $t>10^{4.7}$, say). From the definition - of the model we see that the sequence $(Y_t)_{t\geq 1}$ contains little information about $\theta_\star$, which may suggest that a larger sample size is needed for the sequence $\{\bar{\pi}_t\}_{t=1}^T$ to be in a stationary regime where it concentrates on $\theta_\star$ at a constant rate. In any cases, the important point is that in Figure \[fig:num\] the estimate $\bar{\theta}^N_{t}$ does not concentrate at a sub-optimal rate over the considered sample. The marginal distributions of $\bar{\pi}_T$, as estimated by $\bar{\pi}_T^{N,K}$, are presented in Figures \[fig:num\]-\[fig:num\].

For this example the time needed to approximate $\{\bar{\pi}_t\}_{t= 0}^T$ was about 2.5 hours. During this period of time, IBIS processed only 31 observations and thus failed to provide any significant new information about $\theta$, compared to the one induced by the prior distribution $\tilde{\pi}_0$.

We conclude this subsection by mentioning that @lee2019unbiased introduced a method which allows to use, in , only a sample of size ${\mathcal{O}}(mk)$ from the $\text{Beta-Poisson}(\theta)$ distribution while preserving the validity of pseudo-marginal algorithms. Although computing this new estimate of $\prod_{i=1}^k f_\theta(y_i)$ still requires ${\mathcal{O}}(mk^2)$ operations, their estimator is much cheaper than the naive one adopted in this subsection. Using the estimator of @lee2019unbiased within the approach developed in this work should lead to a fast algorithm for approximate Bayesian inference in latent variable models. Exploring this question further is however beyond the scope of the paper.

Logistic regression\[sub:logistic\]
-----------------------------------

We now let $d>1$, $\Theta={\mathbb{R}}^d$ and consider $T$ observations $\{(z_t,x_t)\}_{t=1}^T$ such that $z_t\in \{0,1\}$ and $x_t\in {\mathbb{R}}^{d-1}$ for all $t$. We model the observations as independent random variables such that the conditional distribution of $Y_t$ given $X_t=x$ is independent of $t$ and belongs to $\{p_{\theta}(\cdot|x),\,\theta\in\Theta\}$, where for every $\theta\in \Theta$ the function $p_{\theta}(\cdot|\cdot)$ is defined by (using the notation $\theta_{2:d}=(\theta_2,\dots,\theta_d)$) $$p_{\theta}(z|x)=\frac{z}{1+\exp(-\theta_1-\theta_{2:d}^T x)}+\frac{(1-z)\exp(-\theta_1-\theta_{2:d}^T x)}{1+\exp(-\theta_1-\theta_{2:d}^T x)},\quad (z,x)\in\{0,1\}\times{\mathbb{R}}^{d-1}.$$ The prior distribution $\tilde{\pi}_0({\mathrm{d}}\theta)$ is the $\mathcal{N}_d(0,2I_d)$ distribution throughout this subsection.

[0.30]{} ![Examples of Section \[sub:logistic\]. Plot () is for the example of Section \[sub:online\]. The solid line is for Algorithm \[algo:pi\_tilde\_online\], the dotted line for SVB and the dashed line for SGD. Plots ()-() are for the example of Section \[non-stationnary\]. In plot () (resp. in plot ()) the solid line shows $\theta_{\star,2}(t)$ (resp. $\theta_{\star,1}$), the dotted line shows $\theta_{t,2}^N$ (resp. $\theta_{t,1}^N$) and the dashed line is for $\bar{\theta}_{t,2}^N$ (resp. $\bar{\theta}_{t,1}^N$). In plot() the non-smooth line is for $\tilde{\theta}_{t,2}^N$. \[fig:NS\]](figs/Cover.pdf "fig:")

 

[0.30]{} ![Examples of Section \[sub:logistic\]. Plot () is for the example of Section \[sub:online\]. The solid line is for Algorithm \[algo:pi\_tilde\_online\], the dotted line for SVB and the dashed line for SGD. Plots ()-() are for the example of Section \[non-stationnary\]. In plot () (resp. in plot ()) the solid line shows $\theta_{\star,2}(t)$ (resp. $\theta_{\star,1}$), the dotted line shows $\theta_{t,2}^N$ (resp. $\theta_{t,1}^N$) and the dashed line is for $\bar{\theta}_{t,2}^N$ (resp. $\bar{\theta}_{t,1}^N$). In plot() the non-smooth line is for $\tilde{\theta}_{t,2}^N$. \[fig:NS\]](figs/beta.pdf "fig:")

 

[0.30]{} ![Examples of Section \[sub:logistic\]. Plot () is for the example of Section \[sub:online\]. The solid line is for Algorithm \[algo:pi\_tilde\_online\], the dotted line for SVB and the dashed line for SGD. Plots ()-() are for the example of Section \[non-stationnary\]. In plot () (resp. in plot ()) the solid line shows $\theta_{\star,2}(t)$ (resp. $\theta_{\star,1}$), the dotted line shows $\theta_{t,2}^N$ (resp. $\theta_{t,1}^N$) and the dashed line is for $\bar{\theta}_{t,2}^N$ (resp. $\bar{\theta}_{t,1}^N$). In plot() the non-smooth line is for $\tilde{\theta}_{t,2}^N$. \[fig:NS\]](figs/a.pdf "fig:")

### Online binary classification using the Covertype dataset \[sub:online\]

In this subsection we assess the online predictive performance of Algorithm \[algo:pi\_tilde\_online\] where, as is the last two examples, $c=1$ and $\alpha=0.5$. To this aim, we consider the Covertype dataset[^1], where $z_t=1$ if forest $t$ is of type 1 and 0 otherwise, and for which $d=54$ and $T=581\,011$. The mean of the $z_t$’s is approximatively 0.365 and the observations are randomly permuted to ensure that they are not ordered in any specific way.

For every $t\geq 1$ and $\theta\in\Theta$ we let $z(\theta,x_{t+1})={\mathds{1}}_{[0.5,1]}\big(p_{\theta}(1|x_{t+1})\big)$ be the corresponding predicted value of $Z_{t+1}$. Then, the objective of online classification algorithms is to learn on the fly parameter values $(\check{\theta}_t)_{t\geq 1}$ such that, for all $t\geq 1$, the average classification error $$\begin{aligned}
\label{eq:classification}
\frac{1}{t}\sum_{s=1}^t\big|z_{s+1}-z(\check{\theta}_s,x_{s+1})\big|\end{aligned}$$ is as small as possible.

In addition to Algorithm \[algo:pi\_tilde\_online\], for which we let $\check{\theta}_t=\bar{\theta}_t^N$ for all $t\geq 1$, we consider two popular learning algorithms, namely stochastic gradient descent (SGD) and streaming variational Bayes [SVB, @broderick2013streaming]. For the former algorithm we let $\check{\theta}_t=\bar{\theta}_t^{\,\mathrm{sgd}}$ for all $t\geq 1$, where $$\bar{\theta}_t^{\,\mathrm{sgd}}=\frac{t-1}{t}\bar{\theta}_{t-1}^{\,\mathrm{sgd}}+\frac{1}{t}\theta_t^{\mathrm{sgd}},\quad \theta_t^{\mathrm{sgd}}=\theta_{t-1}^{\mathrm{sgd}}+\eta_t\nabla_\theta \log p_{\theta_{t-1}^{\mathrm{sgd}}}(z_t|x_t),\quad t\geq 1$$ with $(\eta_t)_{t\geq 1}$ a sequence of learning rates and where, to be consistent with the prior distribution $\tilde{\pi}_0$ that we consider, $\theta_{0}^{\mathrm{sgd}}=(0,\dots,0)$. Following the theoretical results on online learning algorithms [see e.g. @shalev2011online] we take $\eta_t=c t^{-1/2}$ for all $t\geq 1$ and for some constant $c>0$. To choose $c$ we run SGD for different values of this parameter and retain the one that gives the lowest classification errors, namely $c=0.001$.

Following @cherief2019generalization, we consider the following implementation of the SVB algorithm proposed by @nguyen2017online [@zeno2018bayesian], $$\label{eq:svb}
\begin{split}
&\theta_t^{\mathrm{svb}}=\theta_{t-1}^{\mathrm{svb}}+ c\, t^{-\frac{1}{2}}\nabla_\theta{\mathbb{E}}_{U\sim\mathcal{N}_d(0,I)}\Big[ \log p_{\theta_{t-1}^{\mathrm{svb}}+s_{t-1} U}(z_t|x_t) \Big]\\
&s_t=s_{t-1}f\bigg(\frac{0.5\, c t^{-\frac{1}{2}}}{s_{t-1}}\nabla_{s}{\mathbb{E}}_{U\sim\mathcal{N}_d(0,I)}\Big[ \log p_{\theta_{t-1}^{\mathrm{svb}}+s_{t-1} U}(z_t|x_t) \Big]\bigg),\quad t\geq 1
\end{split}$$ where $c>0$ is some constant, where all vector multiplications and divisions should be understood componentwise (notice that $s_t$ is a $d$-dimensional vector for all $t$) and where the function $f(x)=\sqrt{1+x^2}-x$ is applied componentwise. The expectations appearing in are not tractable and therefore need to be approximated. For $M\in\mathbb{N}$ we let $(\theta_t^{M,\mathrm{svb}})_{t\geq 1}$ be defined as in but with the expectations replaced by Monte Carlo estimates, i.e. $$\begin{split}
&\theta_t^{M,\mathrm{svb}}=\theta_{t-1}^{M,\mathrm{svb}}+ c\, t^{-\frac{1}{2}}\frac{1}{M}\sum_{i=1}^M\nabla_\theta \log p_{\theta_{t-1}^{M,\mathrm{svb}}+s^M_{t-1} u^{(t)}_i}(z_t|x_t)\\
&s^M_t=s^M_{t-1}f\bigg(\frac{0.5\, c t^{-\frac{1}{2}}}{s^M_{t-1}}\frac{1}{M}\sum_{i=1}^M\nabla_{s}  \log p_{\theta_{t-1}^{M,\mathrm{svb}}+s^M_{t-1} u^{(t)}_i}(z_t|x_t)\bigg),\quad t\geq 1
\end{split}$$ where, for every $t\geq 1$, $ \{u^{(t)}\}_{i=1}^M$ is a set of $M$ independent realizations of the $\mathcal{N}_d(0,I_d)$ distribution. To be consistent with the prior distribution $\tilde{\pi}_0$ that we consider we let $\theta_{0}^{M,\mathrm{svb}}=(0,\dots,0)$ and $s^M_0=(\sqrt{2},\dots,\sqrt{2})$. We set $M=2\,000$ (preliminary simulations showed that taking $M=1\,000$ lead to numerical instabilities) and, following a similar approach as for SGD, we set $c=0.001$. Then, for SVB we let $\check{\theta}_t=\bar{\theta}_t^{M,\mathrm{svb}}:=t^{-1}\sum_{s=1}^t\theta_t^{M,\mathrm{svb}}$ for all $t\geq 1$.

The results are given in Figure \[fig:NS\]), which shows the evolution of the average classification error as a function of $t$ for the three online learning algorithms that we consider. We observe that, on this dataset, Algorithm \[algo:pi\_tilde\_online\] significantly outperforms SGD and SVB, for all $t\geq 1$. We also see that SVB and SGD have very similar prediction errors. For these experiments the three algorithms are implemented in C and the running time was approximatively 59 minutes for Algorithm \[algo:pi\_tilde\_online\], 1.86 hours for SVB (59 minutes for $M=1\,000$) and less than one second for SGD. Notice that Algorithm \[algo:pi\_tilde\_online\] is therefore not slower than SVB.

We do not report the results for the cases $\check{\theta}_t=\tilde{\theta}_t^N$, $\check{\theta}_t=\theta_t^{\mathrm{sgd}}$ and $\check{\theta}_t=\theta_t^{M,\mathrm{svb}}$ because, for each learning method and for all $t\geq 520$, the corresponding average classification error is larger than the one reported in Figure \[fig:NS\]). Lastly, with the R and C implementation of Algorithm \[algo:pi\_tilde\] considered in the previous subsections, the running time to proceed all the observations was approximatively 1.74 hours for Algorithm \[algo:pi\_tilde\_online\] while, during this time lapse, only $T=41\,244$ observations (7.1% of the sample) were processed by IBIS.

### A time-varying environment example \[non-stationnary\]

Lastly, we compare the behaviour of the sequences $(\pi_t)_{t\geq 1}$, $(\tilde{\pi}_t)_{t\geq 1}$ and $(\bar{\pi}_t)_{t\geq 1}$ in a time-varying environment. To this aim we let $d=2$, $T=10^6$ and generate the observations $\{(z_t, x_t)\}_{t=1}^T$ as follows: $$Z_t|X_t\sim \mathrm{Bernoulli}\Big(\frac{1}{1+\exp(-\theta_{\star,1}-\theta_{\star,2}(t) X_t)}\Big),\quad  X_t{\stackrel{\mathclap{\normalfont\mbox{\tiny{iid}}}}{\sim}}\mathcal{N}_1(1,1),\quad t=1,\dots T$$ with $\theta_{\star,1}= -2$ and $\theta_{\star,2}(u)=1+0.2\sin(10 u/T)$ for all $u\geq 0$. The model that we consider in this subsection is therefore misspecified since the true underlying parameter $\theta_{\star,2}$ slowly varies over time (periodic effect), see Figure \[fig:NS\]. The sample mean of the $z_t$’s is approximatively 0.31

We let $N=K=1\,000$ and, as for the last three examples, we let $(\Delta,\alpha)=(1,0.5)$. The running time of IBIS was 26 minutes, against 10.5 minutes to compute $\tilde{\pi}_T^N$ and 12 minutes to compute both $\tilde{\pi}_T^N$ and $\bar{\pi}_T^{N,K}$. Figure \[fig:NS\]) shows the evolution of $\theta_{t,2}^N$, $\bar{\theta}^N_{t,2}$ and $\tilde{\theta}^N_{t,2}$ as a function of $t$. We observe that the first two quantities evolve in a very similar way, providing another example where $\bar{\theta}_t$ gives a good approximation of the posterior mean of $\theta$. However, these two estimates fail to capture the evolution of $\theta_{\star,2}(t)$ and seem to stabilize over time. By contrast, $\tilde{\theta}^N_{t,2}$ follows closely the evolution of $\theta_{\star,2}(t)$. Figure \[fig:NS\]) shows the evolution of $\bar{\theta}_{t,1}^N$ and of $\theta_{t,1}^N$ over time (results for $\tilde{\theta}^N_{t,1}$ are omitted to make the plot readable). We observe that these two estimates are very close to the true parameter value, although $\bar{\theta}_{t,1}^N$ is always closer to $\theta_{\star,1}$ than $\theta_{t,1}^N$.

This example suggests that a non-stationary data generating process is more easily detected from the sequence $(\tilde{\pi}_t)_{t\geq 1}$ than from the sequences $(\bar{\pi}_t)_{t\geq 1}$ and $(\pi_t)_{t\geq 1}$, and that like SGD [@watanabe1975robbins], the former sequence of distributions may be useful for online learning in non-stationary environments. This greater ability of $\tilde{\pi}_t$ to adapt to changes in the distribution of the data is intuitive, since informally speaking the past observations $\{Y_s\}_{s=1}^{t-1}$ have less weights in the definition of $\tilde{\pi}_t$ than in the definition of $\bar{\pi}_t$ and of $\pi_t$.

Conclusion\[sec:conclusion\]
============================

In its simplest form the approach for online approximate Bayesian learning introduced in this paper reduces to using a particle filter algorithm to approximate a sequence of filtering distributions associated to a particular state-space model. When this is the case, for sake of simplicity we approximate, in Algorithm \[algo:pi\_tilde\_online\], the sequence of filtering distribution $(\tilde{\pi}_t)_{t\geq 0}$ using the simplest type of PF algorithms, known as bootstrap PF algorithms. However, the literature on particle filtering is abundant, and the proposed approach can leverage the several extensions of the bootstrap PF algorithm that have been proposed since its introduction by @Gordon. For instance, the auxiliary particle filter [@PittShep] is a popular alternative to the bootstrap particle filter, in which the observation $Y_t$ is used to generate the sample $\{\theta_t^n\}_{n=1}^N$. Let us also mention the sequential quasi-Monte Carlo algorithm of @MR3351446, whose error converges at a faster rate than that of PF algorithms when the dimension of the state variable is small, and the nested sequential Monte Carlo sampler of @naesseth2015nested that can be used for some high dimensional filtering problems. The convergence results established in this work have the advantage to rely on mild assumptions on the sequence $(\tilde{\pi}_t)_{t\geq 0}$ and on the statistical model. However, further research is needed to understand precisely the convergence properties of the proposed approach. Notably, simulation results suggest that the point estimate $\bar{\theta}_t$ usually converges at the optimal $t^{-1/2}$ rate, and future research should aim at establishing conditions under which this convergence behaviour happens. Establishing the link between the convergence rate of $\tilde{\pi}_t$ and that of $h_t$ is another important problem to solve in order to understand how quickly the error we make when approximating $\pi_t$ by $\tilde{\pi}_t$, or by $\bar{\pi}_t$, vanishes as $t$ tends to infinity. More research is also needed to understand the predictive performance of Algorithm \[algo:pi\_tilde\_online\] in online learning environments. Notably, based on the results of @cherief2019generalization obtained for online variational methods, we conjecture that Algorithm \[algo:pi\_tilde\_online\] should lead to small prediction errors when the value of its $\alpha$ parameter is set to 0.5. The example of Section \[sub:online\] supports this supposition and its theoretical validation is an interesting open problem.

Proofs 
=======

Additional notation and conventions
-----------------------------------

Below we use the convention that empty sums equal zero and empty products equal one, and let $\mathbb{N}_0=\{0\}\cup\mathbb{N}$, ${\mathbb{R}}_{\geq 0}=[0,+\infty)$ and ${\mathbb{R}}_{> 0}=(0,+\infty)$. For a real-valued function $\varphi$ we let $\|\varphi\|_\infty=\sup_x|\varphi(x)|$.

Let $(\tilde{U}_t)_{t\geq 0}$ be a sequence of independent random variables such that, for all $t\geq 0$, $\tilde{U}_t\sim m_t(u){\mathrm{d}}u$, and let $U_t=h_t\tilde{U}_t$ for all $t\geq 0$, note that $U_t\sim\mu_t$ for all $t\geq 0$.

For integers $0\leq k< t$ we define $$u_{k:t}=(u_{k},\dots, u_{t}),\quad [u_{k:t}]=  \max_{k\leq  s<t}\big\|\sum_{i=s}^{t-1}u_i\big\|,\quad \forall (u_{k},\dots, u_{t})\in{\mathbb{R}}^{d(t-k+1)}$$ and let $$\Theta_{\epsilon,k:t}=\big\{u_{k:t}\in {\mathbb{R}}^{d(t-k+1)}:\,[u_{(k+1):t}]<\epsilon\big\},\quad\epsilon>0.$$

For every $t\geq 0$ we let ${\mathcal{F}}_t$ be the $\sigma$-algebra generated by $(Y_1,\dots, Y_t)$ (with the convention ${\mathcal{F}}_0={\varnothing}$) and, for every $0\leq k<t $ and $A\in\mathcal{B}({\mathbb{R}}^d)$, we let $$\pi'_{k,t}(A)=\frac{\int_{A} (\mu_{k}*\tilde{\pi}_{k})\big(\theta-\sum_{s=k+1}^{t-1}U_s\big)\prod_{s=k+1}^{t}  \tilde{f}_{\theta-\sum_{i=s}^{t-1}U_i}(Y_s){\mathrm{d}}\theta}{\int_{\Theta}{\mathbb{E}}\big[ (\mu_{k}*\tilde{\pi}_{k})\big(\theta-\sum_{s=k+1}^{t-1}U_s\big)\prod_{s=k+1}^{t}  \tilde{f}_{\theta-\sum_{i=s}^{t-1}U_i}(Y_s)\big|{\mathcal{F}}_{k}\big]{\mathrm{d}}\theta}.$$

Preliminary results
-------------------

\[lemma:new\] Assume \[new\] and let $A_\star\in \mathcal{B}({\mathbb{R}}^d)$ and $\tilde{A}_\star\in \mathcal{B}({\mathbb{R}}^d)$ be as in \[new\]. Then, there exists a set $A'_\star\in  \mathcal{B}({\mathbb{R}}^d)$, with $A'_\star\subsetneq \tilde{A}_\star$, that contains a neighbourhood of $A_\star$ and such that, for every sequence $(\gamma'_t)_{t\geq 1}$ in ${\mathbb{R}}_{\geq  0}$ such that $\gamma'_t\rightarrow 0$m and every sequence $(s_t)_{t\geq 1}$ in $\mathbb{N}_0$ such that $\inf_{t\geq 1}(t-s_t)\geq 1$ and such that $(t-s_t)\rightarrow+\infty$, there exists a sequence $(\delta_t)_{t\geq 1}$ in ${\mathbb{R}}_{>  0}$ such that $\delta_t\rightarrow 0$ and such that $${\mathbb{P}}\Big(\sup_{(u_{s_t:t},\theta)\in \Theta_{\gamma_t',s_t:t}\times (A'_\star\cap\Theta)}\prod_{s=s_t+1}^t  (\tilde{f}_{\theta-\sum_{i=s}^{t-1} u_i}/f_{\theta}) (Y_s)<e^{(t-s_t)\delta_{t}}\Big)\rightarrow 1.$$

The proof of this result is given in Section \[p-lemma:new\].

The following results rewrite the probability measure $\tilde{\pi}_t$ in a more convenient way.

\[lemma:pi\] With ${\mathbb{P}}$-probability one we have, for all $t\geq 0$ and all $A\in\mathcal{B}({\mathbb{R}}^d)$, $$\tilde{\pi}_t(A)=\frac{\int_A {\mathbb{E}}\big[ (\mu_0*\tilde{\pi}_0) \big(\theta-\sum_{s=1}^{t-1}U_s\big)\prod_{s=1}^{t} \tilde{f}_{\theta -\sum_{i=s}^{t-1}U_i}(Y_s)\big|{\mathcal{F}}_t\big]{\mathrm{d}}\theta}{\int_{\Theta} {\mathbb{E}}\big[ (\mu_0*\tilde{\pi}_0) \big(\theta-\sum_{s=1}^{t-1}U_s\big)\prod_{s=1}^{t} \tilde{f}_{\theta -\sum_{i=s}^{t-1}U_i}(Y_s)\big|{\mathcal{F}}_t\big]{\mathrm{d}}\theta}.$$

The proof of this result is given in Section \[p-lemma:pi\].

The next result builds on @MR1790007 [Lemma 8.1] and will be used to control the denominator of $\tilde{\pi}_t(\theta)$.

\[lemma:denom\] Assume \[m\_star\]-\[taylor\] and let $\delta_\star>0$ be as in these two assumptions. Then, there exists a constant $\tilde{C}_{\star}\in{\mathbb{R}}_{> 0}$ such that, for every $\epsilon\geq 0$, sequence $(s_t)_{t\geq 1}$ in $\mathbb{N}_0$ with $\inf_{t\geq 1}(t-s_t)\geq 1$, every constants $\delta\geq \tilde{\delta} >0$ such that $\delta+\tilde{\delta}<\delta_\star$ and every probability measure $\eta\in\mathcal{P}({\mathbb{R}}^d)$ we have, for all $t\geq 1$ and with $C^\eta_{\delta, \tilde{\delta}}=\inf_{v\in B_{\tilde{\delta}}(0)}\eta(B_\delta(\theta_\star-v))$, $$\begin{aligned}
{\mathbb{P}}\bigg(&\int_{\Theta}{\mathbb{E}}\Big[\eta\big(\theta-\sum_{s=s_t+1}^{t-1}U_s\big)\prod_{s=s_t+1}^{t} (\tilde{f}_{\theta-\sum_{i=s}^{t-1}U_i}/f_{\theta_\star})(Y_s)\big|{\mathcal{F}}_{t}\Big]  {\mathrm{d}}\theta \leq  \frac{{\mathbb{P}}(U_{s_t:t}\in  \Theta_{\tilde{\delta},s_t:t}) }{e^{(t-s_t)(2 (\tilde{C}_\star \delta)^2+\epsilon)}}C^\eta_{\delta, \tilde{\delta}}\bigg)\\
& \leq \big((t-s_t)\big((\tilde{C}_\star\delta)+(\tilde{C}_\star\delta)^{-1}\epsilon\big)^2\big)^{-1}.\end{aligned}$$

The proof of this result is given in Section \[p-lemma:denom\].

The next result will be used to control the numerator of $\tilde{\pi}_t(\theta)$.

\[lemma:test\] Assume \[test\]-\[new\] and let $(\gamma_t)_{t\geq 1}$ be a sequence in ${\mathbb{R}}_{\geq 0}$ such that $\gamma_t\rightarrow 0$ and $(s_t)_{t\geq 1}$ be a sequence in $\mathbb{N}_0$ such that $\inf_{t\geq 1}(t-s_t)\geq 1$ and $(t-s_t)\rightarrow+\infty$. Then, for every $\epsilon>0$ there exist a constant $ \tilde{D}_\star \in{\mathbb{R}}_{>0}$ and a sequence of measurable functions $(\phi_t)_{t\geq 1}$, $\phi_t:{\mathsf{Y}}^t\rightarrow\{0,1\}$, such that ${\mathbb{E}}[\phi_t(Y_{1:t})]\rightarrow 0$ and, for $t$ large enough, $$\sup_{(\theta,u_{s_t:t})\in V_\epsilon\times \Theta_{\gamma_t,s_t:t}}{\mathbb{E}}\Big[(1-\phi_t(Y_{1:t}))\prod_{s=s_t+1}^t(\tilde{f}_{\theta-\sum_{i=s}^{t-1}u_i}/f_{\theta_\star})(Y_s)\big|\,{\mathcal{F}}_{s_t}\Big]\leq e^{-(t-s_t) \tilde{D}_\star}.$$

The proof of this result is given in Section \[p-lemma:test\].

The next lemma builds on @Kleijn2012 [Theorem 3.1]

\[lemma:part\_1\] Assume \[m\_star\]-\[new\]. Let $\epsilon>0$, $(\gamma_t)_{t\geq 1}$ be a sequence in ${\mathbb{R}}_{\geq 0}$ such that $\gamma_t\rightarrow 0$ and $(s_t)_{t\geq 1}$ be a sequence in $\mathbb{N}_0$ such that $\inf_{t\geq 1}(t-s_t)\geq 1$ and $(t-s_t)\rightarrow+\infty$. Then, there exist constants $(C_1,C_2)\in{\mathbb{R}}_{>0}^2$ such that, for every constants $\delta\geq \tilde{\delta}>0$ such that $\delta+\tilde{\delta}<\delta_\star$ (with $\delta_\star>0$ as in Lemma \[lemma:denom\]), there exists a sequence of measurable functions $(\phi'_t)_{t\geq 1}$, $\phi'_t:{\mathsf{Y}}^t\rightarrow\{0,1\}$, such that ${\mathbb{E}}[\phi'_t(Y_{1:t})]\rightarrow 0$ and such that, for $t$ large enough, $$\begin{aligned}
 {\mathbb{E}}\big[(1-\phi'_t(Y_{1:t})){\mathds{1}}_{ \Theta_{\gamma_t,s_t:t}}&(U_{s_t:t}) \pi'_{s_t,t}(V_{\epsilon})|{\mathcal{F}}_{s_t}\big]\\
&\leq  \frac{   e^{-(t-s_t)( C_1^{-1}-C_2\delta^2)}}{{\mathbb{P}}(U_{s_t:t}\in \Theta_{\tilde{\delta},s_t:t})\, \inf_{v\in B_{\tilde{\delta}}(0)}(\mu_{s_t}*\tilde{\pi}_{s_t})(B_{\delta}(\theta_\star-v))},\quad{\mathbb{P}}-a.s.\end{aligned}$$

The proof of this result is given in Section \[p-lemma:part\_1\].

\[lemma:tech\_conv\] Let $\{(X^n, Z^n)\}_{n=1}^N$ be ${\mathbb{R}}^{d_x}\times {\mathbb{R}}^{d_z}$ valued random variables, $G:{\mathbb{R}}^{d_x}\rightarrow(0,+\infty)$ and $\tilde{G}:{\mathbb{R}}^{d_z}\rightarrow(0,+\infty)$ be two measurable mappings, $p_X^N=N^{-1}\sum_{n=1}^N\delta_{X^n}$ and, for every $n\in \{1,\dots,N\}$, let $$W^n=\frac{G(X^n)}{\sum_{m=1}^N G(X^m)},\quad \tilde{W}^n=\frac{\tilde{G}(Z^n)}{\sum_{m=1}^N \tilde{G}(Z^m)}\quad \pi^N=\sum_{n=1}^N W^n\delta_{X^n},\quad \mu^N=\sum_{n=1}^N \tilde{W}^n\delta_{Z^n}.$$ Assume that $\|G\|<+\infty$ and that there exist a constant $C_\star\in(0,+\infty)$ and probability measures $p_X\in\mathcal{P}({\mathbb{R}}^{d_x})$, $\pi\in\mathcal{P}({\mathbb{R}}^{d_x})$ and $\mu\in\mathcal{P}({\mathbb{R}}^{d_z})$ such that $${\mathbb{E}}\Big[\Big(\int_{{\mathbb{R}}^{d_x}} G(x)(p_X^N-p_X)({\mathrm{d}}x)\Big)^2\Big]\leq \frac{C_\star}{N}$$ and such that $$\begin{aligned}
\sup_{|\varphi|\leq 1}{\mathbb{E}}\Big[\Big(\int_{{\mathbb{R}}^{d_x}} \varphi(x)(\pi^N-\pi)({\mathrm{d}}x)\Big)^2\Big]\leq \frac{C_\star}{N},\quad  \sup_{|\varphi|\leq 1}{\mathbb{E}}\Big[\Big(\int_{{\mathbb{R}}^{d_z}} \varphi(z)(\mu^N-\mu)({\mathrm{d}}z)\Big)^2\Big]\leq \frac{C_\star}{N}.\end{aligned}$$ Then, there exists a constant $\tilde{C}_\star\in(0,+\infty)$ such that $$\sup_{|\varphi|\leq 1}{\mathbb{E}}\Big[\Big(\int_{{\mathbb{R}}^{d_x}\times {\mathbb{R}}^{d_z}} \varphi(x,z)(\pi^N\otimes\mu^N-\pi\otimes\mu)({\mathrm{d}}x,{\mathrm{d}}z)\Big)^2\Big]\leq \frac{\tilde{C}_\star}{N}.$$

The proof of this result is given in Section \[p-lemma:tech\_conv\].

Proof of Proposition \[prop:continuity\]\[p-prop:continuity\]
-------------------------------------------------------------

Let $t\geq 1$ and ${\mathsf{Y}}_t\in{\mathcal{Y}}^{\otimes t}$ be such that (i) ${\mathbb{P}}(Y_{1:t}\in{\mathsf{Y}}_t)=1$, (ii) the mapping $\theta\mapsto f_\theta(y_s)$ is continuous on $\Theta$ for all $s\in 1:t$ and all $y_{1:t}\in{\mathsf{Y}}_t$, (iii) $\sup_{1\leq s\leq t}\sup_{\theta\in\Theta} \tilde{f}_\theta(y_s)<+\infty$ for all $y_{1:t}\in{\mathsf{Y}}_t$ and (iv) for all $y_{1:t}\in{\mathsf{Y}}_t$ the conclusion of Lemma \[lemma:pi\] holds.

Remark that such a set ${\mathsf{Y}}_t$ exists under the assumptions of the proposition. Let $\varphi:\Theta\rightarrow{\mathbb{R}}$ be a measurable and bounded function. Then, for all $y\in{\mathsf{Y}}_t$ we have $$\label{eq:thmcont_1}
\begin{split}
\int_{{\mathbb{R}}^d}\varphi(\theta) {\mathbb{E}}\Big[ (\mu_{0}*\tilde{\pi}_0) &\big(\theta- \sum_{s=1}^{t-1}h_s\tilde{U}_s\big)\prod_{s=1}^{t} \tilde{f}_{\theta- \sum_{i=s}^{t-1}h_i\tilde{U}_i}(y_s) \Big]{\mathrm{d}}\theta\\
&\leq\big(\sup_{1\leq s\leq t}\sup_{\theta\in\Theta} \tilde{f}_\theta(y_s)\big)^t{\mathbb{E}}\Big[\int_{{\mathbb{R}}^d}\varphi(\theta) (\mu_{0}*\tilde{\pi}_0) \Big(\theta- \sum_{s=1}^{t-1}h_s\tilde{U}_s\Big){\mathrm{d}}\theta \Big]\\
&\leq \|\varphi\|_\infty\Big(\sup_{1\leq s\leq t}\sup_{\theta\in\Theta} \tilde{f}_\theta(y_s)\Big)^t 
\end{split}$$ while $$\begin{aligned}
\label{eq:thmcont_2}
{\mathbb{P}}\Big(\lim_{\max\{h_k\}_{k=0}^{t-1}}\prod_{s=1}^{t} \tilde{f}_{\theta-\sum_{i=s}^{t-1}h_i\tilde{U}_i}(y_s) =\prod_{s=1}^{t} \tilde{f}_{\theta}(y_s)\Big)=1,\quad \forall y\in{\mathsf{Y}}_t,\quad\forall\theta\in{\mathbb{R}}^d.\end{aligned}$$ In addition, using the assumptions on $\tilde{\pi}_0$ and the dominated convergence theorem, $$\label{eq:thmcont_3}
\begin{split}
{\mathbb{P}}\Big(\lim_{\max\{h_i\}_{i=0}^{t-1}\searrow 0}& (\mu_0*\tilde{\pi}_0)\big(\theta- \sum_{s=1}^{t-1}h_s\tilde{U}_s\big) =\tilde{\pi}_0(\theta)\Big) \\
&={\mathbb{P}}\Big(\lim_{\max\{h_i\}_{i=0}^{t-1}\searrow 0}{\mathbb{E}}\Big[\tilde{\pi}_0\big(\theta- \sum_{s=0}^{t-1}h_s\tilde{U}_s\big)\Big]=\tilde{\pi}_0(\theta)\Big)\\
&=1.
\end{split}$$

Therefore, using - and the dominated convergence theorem, for all $y\in{\mathsf{Y}}_t$ we have $$\label{eq:PPx}
\begin{split}
\lim_{\max\{h_i\}_{i=0}^{t-1}\searrow 0}\int_{{\mathbb{R}}^d}\varphi(\theta) {\mathbb{E}}\big[(\mu_0*\tilde{\pi}_0)\big(\theta- \sum_{s=1}^{t-1}h_s\tilde{U}_s\big)&\prod_{s=1}^{t} \tilde{f}_{\theta- \sum_{i=s}^{t-1}h_i\tilde{U}_i}(y_s) \big]{\mathrm{d}}\theta\\
&=   \int_{{\mathbb{R}}^d}\varphi(\theta) \prod_{s=1}^{t} \tilde{f}_{\theta }(y_s)   \tilde{\pi}_0  ({\mathrm{d}}\theta  ).
\end{split}$$ To complete the proof let $(\varphi_k)_{k\geq 1}$, with $\varphi_k:\Theta\rightarrow{\mathbb{R}}$ measurable, continuous and bounded for all $k$, be such that if $${\mathbb{P}}\Big(\lim_{\max\{h_i\}_{i=0}^{t-1}\searrow 0}\tilde{\pi}_t(\varphi_k)=\pi_t(\varphi_k),\,\forall k\geq 1\Big)=1\implies {\mathbb{P}}\Big(\lim_{\max\{h_i\}_{i=0}^{t-1}\searrow 0}\tilde{\pi}_t=\pi_t\Big)=1.$$

Let $k\in\mathbb{N}$. Then, by Lemma \[lemma:pi\] we have, ${\mathbb{P}}$-a.s., $$\begin{aligned}
\tilde{\pi}_t(\varphi_k)=\frac{\int_{{\mathbb{R}}^d}\varphi_k(\theta) {\mathbb{E}}\big[ (\mu_0*\tilde{\pi}_0) \big(\theta-\sum_{s=1}^{t-1}U_s\big)\prod_{s=1}^{t} \tilde{f}_{\theta -\sum_{i=s}^{t-1}U_i}(Y_s)\big|{\mathcal{F}}_t\big]{\mathrm{d}}\theta}{\int_{\Theta} {\mathbb{E}}\big[ (\mu_0*\tilde{\pi}_0) \big(\theta-\sum_{s=1}^{t-1}U_s\big)\prod_{s=1}^{t} \tilde{f}_{\theta -\sum_{i=s}^{t-1}U_i}(Y_s)\big|{\mathcal{F}}_t\big]{\mathrm{d}}\theta}\end{aligned}$$ and thus, using with $\varphi=\varphi_k$ and with $\varphi={\mathds{1}}_{\Theta}$, it follows that, for all $k\in\mathbb{N}$, ${\mathbb{P}}\big(\lim_{\max\{h_i\}_{i=0}^{t-1}\searrow 0}\tilde{\pi}_t(\varphi_k)=\pi_t(\varphi_k)\big)=1$. The result follows.

Proof of Proposition \[prop:gaussian\_toy\]\[p-prop:gaussian\_toy\]
-------------------------------------------------------------------

We first show that $\tilde{\sigma}^2_t\rightarrow 0$. To this aim recall that $$\begin{aligned}
\label{eq:piece0}
\tilde{\sigma}_t^2= \frac{\tilde{\sigma}_{t-1}^2+h_{t-1}^2}{\tilde{\sigma}^2_{t-1}+h_{t-1}^2+1}\in(0,1),\quad \forall t\geq 1\end{aligned}$$ and let $\bar{c}:=\limsup_{t\rightarrow+\infty}\tilde{\sigma}_t^2$. Assume that $\bar{c}>0$ and let $(\tilde{\sigma}_{t_k}^2)_{k\geq 1}$ be a subsequence such that $\tilde{\sigma}_{t_k}^2\rightarrow \bar{c}$. Then, by and since $h_t\rightarrow 0$, $$\lim_{k\rightarrow+\infty} \tilde{\sigma}_{t_k+1}^2=\bar{c}':=\frac{\bar{c}}{1+\bar{c}}.$$ Then, repeating these computations with $(\tilde{\sigma}_{t_k}^2)_{k\geq 1}$ replaced by $(\tilde{\sigma}_{t_k+1}^2)_{k\geq 1}$ and $\bar{c}$ by $\bar{c}'$ yields $$\lim_{k\rightarrow+\infty} \tilde{\sigma}_{t_k+2}^2=\frac{\bar{c}'}{1+\bar{c}'}=\frac{\bar{c}}{1+2\bar{c}}$$ and similarly we show that $$\begin{aligned}
\label{eq:limmm}
\lim_{k\rightarrow+\infty} \tilde{\sigma}_{t_k+i}^2=\frac{\bar{c}}{1+i\bar{c}},\quad \forall i\in\mathbb{N}.\end{aligned}$$ To proceed further let $\epsilon\in(0,1)$ and $t_1\in\mathbb{N}$ be such that $h_{t-1}^2\leq \epsilon^2/(1-\epsilon)$ for all $t\geq t_1$. Notice that such a $t_1\in\mathbb{N}$ exists since $h_t\rightarrow 0$. Next, let $t_2>t_1$ be such that $\sigma^2_{t_2-1}\leq \epsilon$. Remark that such a $t_2>t_1$ exists by and that, using , $$\begin{aligned}
\label{eq:limmm55}
\sigma^2_{t}\leq  \epsilon\Leftrightarrow \sigma^2_{t-1}+h^2_{t-1}\leq\frac{\epsilon}{1-\epsilon},\quad\forall t\geq 1.\end{aligned}$$ On the other hand, under ours assumptions on $t_2$ we have $$\sigma^2_{t_2-1}+h_{t_2-1}^2\leq \epsilon+\frac{\epsilon^2}{1-\epsilon}=\frac{\epsilon}{1-\epsilon}$$ and thus, by , it follows that $\sigma^2_{t_2}\leq \epsilon$. This shows that for all $t\geq t_2$ we have $\sigma^2_t\leq \epsilon$, and thus that $\limsup_{t\rightarrow+\infty}\tilde{\sigma}^2_t\leq \epsilon$. Since $\epsilon>0$ is arbitrary this shows that $\tilde{\sigma}_t^2\rightarrow 0$.

To now show that $\tilde{\theta}_t\rightarrow {\mathbb{E}}[Y_1]$ in ${\mathbb{P}}$-probability. To this aim remark first that for every $t\geq 1$ we have, with the convention that empty products equal 1, $$\begin{aligned}
 \tilde{\theta}_t&=\tilde{\sigma}_t^2 Y_t + (1-\tilde{\sigma}_t^2)\tilde{\theta}_{t-1}\\
 &= \tilde{\sigma}_t^2 Y_t + (1-\tilde{\sigma}_t^2)\tilde{\sigma}^2_{t-1}Y_{t-1}+(1-\tilde{\sigma}_t^2)(1-\tilde{\sigma}_{t-1}^2)\tilde{\theta}_{t-2}\\
  &=\sum_{s=1}^t Y_s\tilde{\sigma}_s^2\prod_{j=s+1}^t(1-\tilde{\sigma}^2_j)+\tilde{\theta}_0\prod_{s=1}^t(1-\tilde{\sigma}_s^2)\\
 &=\frac{1}{b_t}\big(\tilde{\theta}_0+\sum_{s=1}^t \tilde{\sigma}_s^2 b_s Y_s\big)\end{aligned}$$ where $b_s=\big(\prod_{j=1}^s(1-\tilde{\sigma}^2_j)\big)^{-1}$ for all $s\geq 1$.

It is easily checked that $\tilde{\sigma}_t^2$ is minimized when $h_s=0$ for all $s\{1,\dots,t-1\}$, in which case it is well-known and easily checked that $\liminf_{t\rightarrow+\infty} t\tilde{\sigma}_t^2>0$. Thus, $\sum_{t=1}^\infty\tilde{\sigma}_t^2=+\infty$ implying that $b_t\rightarrow +\infty$. Consequently, to prove that $\tilde{\theta}_t\rightarrow {\mathbb{E}}[Y_1]$ in ${\mathbb{P}}$-probability we only need to show that $$\begin{aligned}
\label{eq:piece3}
X_t:=\frac{1}{b_t}\sum_{s=1}^t \tilde{\sigma}_s^2 b_s Y_s\rightarrow{\mathbb{E}}[Y_1],\quad\mathrm{in}\,\,{\mathbb{P}}-\mathrm{probability.}\end{aligned}$$ Remark that $$\begin{aligned}
\label{eq:piece33}
X_t=\frac{\sum_{s=1}^t \tilde{\sigma}_s^2 b_s}{b_t}\sum_{s=1}^t \frac{\tilde{\sigma}_s^2 b_s}{\sum_{s=1}^t \tilde{\sigma}_s^2 b_s } Y_s,\quad \frac{\sum_{s=1}^t \tilde{\sigma}_s^2 b_s}{b_t}=1-\frac{1}{b_t},\quad\forall t\geq 1\end{aligned}$$ so that, as $b_t\rightarrow +\infty$, to show we only need to show that $$\begin{aligned}
\label{eq:piece4}
 \sum_{s=1}^t \frac{\tilde{\sigma}_s^2 b_s}{\sum_{s=1}^t \tilde{\sigma}_s^2 b_s} Y_s\rightarrow{\mathbb{E}}[Y_1],\quad\mathrm{in}\,\,{\mathbb{P}}-\mathrm{probability.}\end{aligned}$$ By , and using the fact that $b_t\rightarrow+\infty$ while $\tilde{\sigma}_t^2\rightarrow 0$, we have $$\begin{aligned}
&\lim_{t\rightarrow+\infty}\sum_{s=1}^t\tilde{\sigma}_s^2 b_s=\lim_{t\rightarrow+\infty} b_t(1-1/b_t)=+\infty\\
&\lim_{t\rightarrow+\infty}\frac{ \tilde{\sigma}_t^2 b_t}{\sum_{s=1}^t \tilde{\sigma}_s^2 b_s} =\lim_{t\rightarrow+\infty}\frac{ \tilde{\sigma}_t^2}{1-1/b_t}=0
 \end{aligned}$$ and therefore holds by @jamison1965 [Theorem 1], which concludes to show that $\tilde{\theta}_t\rightarrow{\mathbb{E}}[Y_1]$ in ${\mathbb{P}}$-probability.

We are now in position to prove the first part of the proposition. Let $\epsilon>0$ and $Z\sim\mathcal{N}_1(0,1)$ be independent of $(Y_t)_{t\geq 1}$. Then, using Markov’s inequality, for all $t\geq 1$ we have $$\begin{aligned}
\tilde{\pi}_t(V_\epsilon)&={\mathbb{P}}\Big( \Big|Z-\frac{\theta_\star-\tilde{\theta}_t}{\tilde{\sigma}_t}\Big|\geq \frac{\epsilon}{\tilde{\sigma}_t}\,\Big|\sigma(\tilde{\theta}_t)\Big)\\
& \leq\tilde{\sigma}_t^2\frac{{\mathbb{E}}\Big[\Big(Z-\frac{\theta_\star-\tilde{\theta}_t}{\tilde{\sigma}_t}\Big)^2\,\big|\sigma(\tilde{\theta}_t)\Big]}{\epsilon^2}\\
&=\frac{\tilde{\sigma}_t^2+(\theta_\star-\tilde{\theta}_t)^2}{\epsilon^2},\quad{\mathbb{P}}-a.s.\end{aligned}$$ As shown above, $\tilde{\theta}_t\rightarrow{\mathbb{E}}[Y_1]$ in ${\mathbb{P}}$-probability and thus there exists a sequence $(\gamma_t)_{t\geq 1}$ in $(0,+\infty)$ such that $\gamma_t\rightarrow 0$ and such that ${\mathbb{P}}(|\theta_\star-\tilde{\theta}_t|>\gamma_t)\rightarrow 0$. Therefore, since $\tilde{\sigma}_t^2\rightarrow 0$ as shown above, it follows that $$\begin{aligned}
\limsup_{t\rightarrow+\infty}{\mathbb{E}}[\tilde{\pi}_t(V_\epsilon)]&\leq \limsup_{t\rightarrow+\infty}{\mathbb{P}}(|\theta_\star-\tilde{\theta}_t|>\gamma_t)+\limsup_{t\rightarrow+\infty}\Big(\frac{\tilde{\sigma}_t^2}{\epsilon}+\frac{\gamma_t^2}{\epsilon^2}\Big)=0\end{aligned}$$ which, together with Proposition \[prop:conv\_proba\], completes the proof of the first part of the proposition.

To prove the last part of the proposition let $v_t=\tilde{\sigma}_t^2/h_t^2$, $\epsilon>0$ and $\delta=(1-\epsilon)/(1-\epsilon/2)$. Then, under the assumptions on $(h_t)_{t\geq 1}$, there exists a $t_\epsilon\in\mathbb{N}$ such that $ h^2_{t-1}/ h^2_t\geq \delta$ for all $t\geq t_{\epsilon}$. Let $$\epsilon_t=\frac{\tilde{\sigma}^2_{t-1}+h^2_{t-1}}{\tilde{\sigma}^2_{t-1}+h^2_{t-1}+1},\quad\forall t\geq 1$$ so that, for all $t\geq t_\epsilon$, we have $$\begin{aligned}
v_t&= \frac{1+v_{t-1}}{\tilde{\sigma}^2_{t-1}+h_{t-1}^2+1}\,\frac{h_{t-1}^2}{h_t^2}\geq \frac{\delta(1+v_{t-1})}{\tilde{\sigma}^2_{t-1}+h_{t-1}^2+1}\geq \delta(1+v_{t-1})(1-\epsilon_t)\end{aligned}$$ As shown above, $\tilde{\sigma}^2_t\rightarrow 0$ so that $\epsilon_t\rightarrow0$. Hence, there is exists a $t'_\epsilon\geq t_\epsilon$ such that, for $t>t'_\epsilon$ and recalling that $\delta=(1-\epsilon)/(1-\epsilon/2)$ $$\begin{aligned}
v_t\geq (1-\epsilon/2)\delta(1+v_{t-1})=(1-\epsilon)(1+v_{t-1})\geq \sum_{s=t_\epsilon+1}^t(1-\epsilon)^{s-t_\epsilon}+(1-\epsilon)^{t-t_\epsilon}v_{t_\epsilon}.\end{aligned}$$ This shows that $\liminf_{t\rightarrow+\infty} v_t\geq 1/\epsilon$ and thus, since $\epsilon\in(0,1)$ is arbitrary, it follows that $v_t\rightarrow+\infty$.

To prove the fourth part of the proposition remark that, for every $t\geq 1$, $$\label{eq:rate_h}
\begin{split}
   \tilde{\sigma}_t^2 =\frac{\tilde{\sigma}^2_{t-1}+h^2_{t-1}}{\tilde{\sigma}^2_{t-1}+h_{t-1}^2+1} &\Leftrightarrow\tilde{\sigma}_t^2\tilde{\sigma}^2_{t-1}+\tilde{\sigma}_t^2(h_{t-1}^2+1)=\tilde{\sigma}^2_{t-1}+h^2_{t-1}\\
&\Leftrightarrow h^2_{t-1}= \frac{\tilde{\sigma}_t^2\tilde{\sigma}^2_{t-1}+\tilde{\sigma}_t^2-\tilde{\sigma}_{t-1}^2}{1-\tilde{\sigma}_t^2}.
\end{split}$$ As shown above, $\tilde{\sigma}_t^2\rightarrow 0$, which implies that $\tilde{\sigma}_t^2\leq \tilde{\sigma}_{t-1}^2$ for infinitely may $t$. Thus, by , for infinity many $t$ we have $h^2_{t-1}\leq  (\tilde{\sigma}_t^2\tilde{\sigma}^2_{t-1})/(1-\tilde{\sigma}_t^2)$ showing that, if $h_t\sim c t^{-\alpha}$ then, for all $p\geq 1$ a necessary condition to have $t^{1/p} \tilde{\sigma}_t^2={\mathcal{O}}(1)$ is that $\alpha\geq 1/p$.

To prove the rest of the proposition remark that ${\mathbb{P}}(\tilde{\pi}_t\Rightarrow \delta_{{\mathbb{E}}[Y_1]})=1$ if and only if ${\mathbb{P}}(\tilde{\theta}_t\rightarrow  {\mathbb{E}}[Y_1])=1$ which, using the above computations, holds if and only if $$\begin{aligned}
\label{eq:john}
\sum_{s=1}^t \frac{\tilde{\sigma}_s^2 b_s}{\sum_{s=1}^t \tilde{\sigma}_s^2 b_s} Y_s\rightarrow{\mathbb{E}}[Y_1],\quad{\mathbb{P}}-a.s.\end{aligned}$$ Remark that since $$(\tilde{\theta}_t-{\mathbb{E}}[Y_1])=\tilde{\sigma}_t^2 (Y_t-{\mathbb{E}}[Y_1]) + (1-\tilde{\sigma}_t^2)(\tilde{\theta}_{t-1}-{\mathbb{E}}[Y_1]),\quad\forall t\geq 1$$ we can without loss of generality assume that ${\mathbb{E}}[Y_1]=0$.

Let $
T(x)=\#\big\{t\in\mathbb{N}:\,(\tilde{\sigma}_t^2 b_t)^{-1}\big(\sum_{s=1}^t \tilde{\sigma}_s^2 b_s\big)\leq x\big\}$ for every $x>0$. Then, we show below that, for $h_t\sim  c t^{-\alpha}$, we have $$\begin{aligned}
\label{eq:Tx}
\limsup_{x\rightarrow+\infty} T(x)/x^p<+\infty\Leftrightarrow t^{1/p} \tilde{\sigma}_t^2={\mathcal{O}}(1),\quad \forall p\in[1,2].\end{aligned}$$ Showing will show the second part of the proposition since, for $p\in[1,2)$, the condition $\limsup_{x\rightarrow+\infty} T(x)/x^p<+\infty$ is necessary and sufficient for to hold for every sequence $(Y_t)_{t\geq 1}$ such that ${\mathbb{E}}[|Y_t|^p]<\infty$, (see [@jamison1965], Theorem 3, for the case $p=1$ and [@chen1996], Theorem 2, for the case for $p\in(1,2)$). In addition, showing will also establish the third part of the proposition since, as shown in @chen1996 [Remark 2], for $p=2$ the condition $\limsup_{x\rightarrow+\infty} T(x)/x^p<+\infty$ is necessary but not sufficient for to hold for every sequence $(Y_t)_{t\geq 1}$ such that ${\mathbb{E}}[Y_t^2]<\infty$.

To show let $t\geq 1$ and remark that, since $b_t^{-1}\sum_{s=1}^t \tilde{\sigma}_s^2 b_s=1-\frac{1}{b_t}$, $$\begin{aligned}
\frac{\sum_{s=1}^t \tilde{\sigma}_s^2 b_s}{\tilde{\sigma}_t^2 b_t}\leq x&\Leftrightarrow \frac{1}{\tilde{\sigma}_t^2}(1-b_t^{-1})\leq x,\quad\forall x>0\end{aligned}$$ where, as shown above, $b_t\rightarrow+\infty$. Let $t_0\in\mathbb{N}$ be such that $(1-b_t^{-1})\geq 0.5$ for all $t\geq t_0$ so that $$\begin{aligned}
\label{eq:tt_1}
T(x)\leq t_0+\#\big\{t\in\mathbb{N}:\, t>t_0,\,\,   \tilde{\sigma}_t^{-2}\leq 2x\big\},\quad \forall x>0.\end{aligned}$$ Assume now that $ \tilde{\sigma}_t^2\leq C t^{-1/p}$ for some constant $C<+\infty$ and all $t\geq 1$. Then, $$\#\{t\in\mathbb{N}:\, t>t_0,\,\,   \tilde{\sigma}_t^{-2}\leq 2x\}\leq  \#\{t\in\mathbb{N}:\,     t\leq (2 Cx)^p\}= (2C x)^p,\quad\forall x\geq 1/(2C)$$ which, together with , shows that if $t^{1/p} \tilde{\sigma}_t^2={\mathcal{O}}(1)$ then $\limsup_{x\rightarrow+\infty} T(x)/x^p<+\infty$.

Assume now that $\limsup_{t\rightarrow+\infty}t^{1/p}\tilde{\sigma}_t^2=+\infty$. Then, using by and the fact that $h_t\sim c t^{-\alpha}$, it is easily checked that, in this case, we have $t^{1/p}\tilde{\sigma}_t^2\rightarrow+\infty$. Next, let $C>0$ and $t_C\in\mathbb{N}$ be such that $t^{1/p}\tilde{\sigma}_t^2\geq C$ for all $t\geq t_C$. Then, using the fact that $(1-b_t^{-1})\leq 1$ for all $t$, for all $x$ such that $x>(t_C/C)^{1/p}$ we have $$\begin{aligned}
T(x)\geq \#\{t\in\mathbb{N}:\,    \tilde{\sigma}_t^{-2}\leq  x\}&\geq  \#\{t\in\mathbb{N}:\, t>t_C,\,    t\leq  (C x)^p\}= (C x)^p-t_C+1\end{aligned}$$ showing that $\limsup_{x\rightarrow+\infty} T(x)/x^p\geq C^p$. Since $C>0$ arbitrary this shows that $\limsup_{x\rightarrow+\infty} T(x)/x^p=+\infty$ whenever $\limsup_{t\rightarrow+\infty}t^{1/p}\tilde{\sigma}_t^2=+\infty$, which concludes to show .

Proof of Proposition \[prop:conv\_algo\] {#p-prop:conv_algo}
----------------------------------------

To simplify the notation we assume below that $q_t(\theta|\theta')= h_{t-1}^{-1} m_{t-1}((\theta'-\theta)/h_{t-1})$ for all $t\geq 1$ such that $h_{t-1}>0$ and for all $(\theta,\theta')\in{\mathbb{R}}^{2d}$. The generalization of the computations that follows to arbitrary proposals distributions $(q_t(\cdot|\theta))_{t\geq 1,\theta\in\Theta}$ satisfying the assumptions of the proposition is immediate.

### Preliminaries

For every $\theta\in\Theta$ we let $G_0(\theta)=\tilde{\pi}_0(\theta)/q_0(\theta)$ and $G_t(\theta)=f_\theta(Y_t)$, for every $t\geq 1$. For every $t\geq 0$ we use the shorthand ${\mathbb{E}}_t[\cdot]={\mathbb{E}}[\cdot|{\mathcal{F}}_t]$, $\mu^N_{t,J}({\mathrm{d}}\theta_{1:J})=\otimes_{j=1}^J\tilde{\pi}_{k_t}^N({\mathrm{d}}\theta_j)$ and $\mu_{t,J}({\mathrm{d}}\theta_{1:J})=\otimes_{j=1}^J\tilde{\pi}_{k_t}({\mathrm{d}}\theta_j)$, where $a:b=\{a,\dots,b\}$ for every integers $b\geq a$.

Next, for every $t\geq 1$ such that $k_t\geq 0$ and $h_{t-1}=0$ we let, for every $s\in(t-1):t$, and $(\theta_{s}, \theta_{s,J})\in{\mathbb{R}}^d\times {\mathbb{R}}^{dJ}$, $$g_{t,J}(\theta_s,\theta_{s,J})=\sum_{j=1}^Jm_{k_t}\big((\theta_s-\theta_{s,j})/h_{k_t}),$$

$$\alpha_t(\theta_{t-1}, \theta _{t-1,J}; \theta_{t}, \theta_{t,J})=\frac{q_t(\theta_{t-1}|\theta_t)g_{t,J}(\theta_t,\theta_{t,J})\prod_{s=k_t+1}^{t-1} f_{\theta_t}(Y_s)}{q_t(\theta_{t}|\theta_{t-1})g_{t,J}(\theta_{t-1},\theta_{t-1,J})\prod_{s=k_t+1}^{t-1} f_{\theta_{t-1}}(Y_s)}\wedge 1.$$ and, using the shorthand $\tilde{\Theta}={\mathbb{R}}^d$, we define $$\begin{aligned}
 & \tilde{K}_t^N((\theta_{t-1} ,\theta_{t-1,J}),{\mathrm{d}}(\theta_t,\theta_{t,J}))\\
 & \,\,\,=\alpha_t(\theta_{t-1}, \theta _{t-1,J}; \theta_{t}, \theta_{t,J})\mu^N_{t,J}({\mathrm{d}}\theta_{t,J})\otimes q_t(\theta_t|\theta_{t-1}){\mathrm{d}}\theta_t\\
 &  \,\,\,+\delta_{(\theta_{t-1},\theta_{t-1,J})}({\mathrm{d}}\theta_t,{\mathrm{d}}\theta_{t,J})\Big(1-\int_{\tilde{\Theta}^{J+1}}\alpha_t(\theta_{t-1}, \theta _{t-1,J}; \theta_{t}, \theta_{t,J})\mu^N_{t,J,}({\mathrm{d}}\theta_{t,J})\otimes q_t(\theta_t|\theta_{t-1}){\mathrm{d}}\theta_t\Big),\\
\vspace{0.2cm}
& \tilde{K}_t((\theta_{t-1} ,\theta_{t-1,J}),{\mathrm{d}}(\theta_t,\theta_{t,J}))\\
& \,\,\,=\alpha_t(\theta_{t-1}, \theta _{t-1,J}; \theta_{t}, \theta_{t,J})\mu_{t,J}({\mathrm{d}}\theta_{t,J})\otimes q_t(\theta_t|\theta_{t-1}){\mathrm{d}}\theta_t\\
& \,\,\,+\delta_{(\theta_{t-1},\theta_{t-1,J})}({\mathrm{d}}\theta_t,{\mathrm{d}}\theta_{t,J})\Big(1-\int_{\tilde{\Theta}^{J+1}}\alpha_t(\theta_{t-1}, \theta _{t-1,J}; \theta_{t}, \theta_{t,J})\mu_{t,J}({\mathrm{d}}\theta_{t,J})\otimes q_t(\theta_t|\theta_{t-1}){\mathrm{d}}\theta_t\Big),\\
\vspace{0.2cm}
&K^N_t( \theta_{t-1},{\mathrm{d}}\theta_t)=\int_{\tilde{\Theta}^{J}}\Big(\int_{\tilde{\Theta}^{J}}\tilde{K}_t^N\big((\theta_{t-1},\theta_{t-1,J}),{\mathrm{d}}(\theta_t,\theta_{t,J})\big)\Big)\mu_{t,J}^N({\mathrm{d}}\theta_{t-1,J}),\\
\vspace{0.2cm}
&K_t(\theta_{t-1},{\mathrm{d}}\theta_t)=\int_{\tilde{\Theta}^J}\Big(\int_{\tilde{\Theta}^J}\tilde{K}_t((\theta_{t-1},\theta_{t-1,J}),{\mathrm{d}}(\theta_t,\theta_{t,J}))\Big)\mu_{t,J}({\mathrm{d}}\theta_{t-1,J}).\end{aligned}$$ Next, for every $t\geq 1$ such that $k_t=-1$ and $h_{t-1}=0$ we let $K'_t(\theta_{t-1},{\mathrm{d}}\theta_t)$ be the Metropolis-Hastings kernel with invariant distribution $\pi_t$ and proposal distribution $q_t$.

Lastly, for all $t\geq 1$ we let $$\begin{aligned}
L_t(\theta_{t-1},{\mathrm{d}}\theta_t)&=q_t(\theta_t|\theta_{t-1}){\mathds{1}}_{{\mathbb{R}}_{>0}}(h_{t-1})+K_t(\theta_{t-1},{\mathrm{d}}\theta_t){\mathds{1}}_{\mathbb{N}_0\times \{0\}}(k_t,h_{t-1})\\
&+K'_t(\theta_{t-1},{\mathrm{d}}\theta_t){\mathds{1}}_{\{-1\}\times \{0\}}(k_t,h_{t-1})\end{aligned}$$ and $$\begin{aligned}
L^N_t(\theta_{t-1},{\mathrm{d}}\theta_t)&=q_t(\theta_t|\theta_{t-1}){\mathds{1}}_{{\mathbb{R}}_{>0}}(h_{t-1})+K^N_t(\theta_{t-1},{\mathrm{d}}\theta_t){\mathds{1}}_{\mathbb{N}_0\times \{0\}}(k_t,h_{t-1})\\
&+K'_t(\theta_{t-1},{\mathrm{d}}\theta_t){\mathds{1}}_{\{-1\}\times \{0\}}(k_t,h_{t-1}).\end{aligned}$$

\[lemma:int\_prop\] Let $t\geq 1$ and assume that, for all $s\in 0:(t-1)$ there exists a $(0,+\infty)$-valued random variable $C_s$ such that, ${\mathbb{P}}$-almost surely and with the convention $L_s(\theta_{s-1},{\mathrm{d}}\theta_s)\tilde{\pi}_{s-1}({\mathrm{d}}\theta_{s-1})=\delta_{\theta_{s-1}}({\mathrm{d}}\theta_s)\tilde{\pi}_{0}({\mathrm{d}}\theta_{s-1})$ when $s=0$, we have $$\begin{aligned}
\label{eq:Prop_show1}
\sup_{|\varphi|\leq 1}{\mathbb{E}}_s\Big[\Big(\frac{1}{N}\sum_{n=1}^N\varphi(\theta_s^n)-\int_{\tilde{\Theta}} \Big\{\int_{\tilde{\Theta}} \varphi(\theta_s)L_s(\theta_{s-1},{\mathrm{d}}\theta_s)\Big\}\tilde{\pi}_{s-1}({\mathrm{d}}\theta_{s-1})\Big)^2\Big]\leq \frac{C_s}{N}\end{aligned}$$ and $$\begin{aligned}
\label{eq:Prop_show2}
\sup_{|\varphi|\leq 1}{\mathbb{E}}_s\Big[\Big( \sum_{n=1}^N W_s^n\varphi(\theta_s^n)-\int_{\tilde{\Theta}}   \varphi(\theta_s) \tilde{\pi}_{s}({\mathrm{d}}\theta_{s})\Big)^2\Big]\leq \frac{C_s}{N}.\end{aligned}$$ Then, under the assumptions of Proposition \[prop:conv\_algo\], there exists a random variable $\tilde{C}_t$ taking its values in $(0,+\infty)$ such that $$\begin{split}
\sup_{|\varphi|\leq 1}{\mathbb{E}}_t\Big[\Big(\sum_{n=1}^N W^n_{t-1} \int_{\tilde{\Theta}}\varphi(\theta_t)\big\{L^N_t(\theta^n_{t-1},{\mathrm{d}}\theta_t)-L_t(\theta^n_{t-1},{\mathrm{d}}\theta_t)\big\}\Big)^2\Big]\leq \frac{\tilde{C}_t}{N},\quad{\mathbb{P}}-a.s.
\end{split}$$

Remark that the result trivially trivially holds if $h_{t-1}>0$ or if $h_{t-1}=0$ and $k_t=-1$. Hence, below we assume that $h_{t-1}=0$ and $k_t\geq 0$.

Let $$\tilde{\alpha}_t(\varphi,\theta_{t-1}, \theta _{t-1,J};  \theta_{t,J})=\int_{\tilde{\Theta}} \varphi(\theta_t)  \alpha_t(\theta_{t-1}, \theta _{t-1,J}; \theta_{t}, \theta_{t,J})q_t({\mathrm{d}}\theta_t|\theta_{t-1})$$ and we first show that there exists a ${\mathbb{P}}$-a.s. finite random variable $\tilde{C}_{t,1}$ such that, ${\mathbb{P}}$-a.s., $$\label{eq:BB3_p1}
\begin{split}
 &\sup_{|\varphi|\leq 1}{\mathbb{E}}_t\Big[\Big(\int_{\tilde{\Theta}^{2J+1}} \tilde{\alpha}_t(\varphi,\theta_{t-1}, \theta _{t-1,J};  \theta_{t,J})\\
& \big(\mu_{t,J}^{N}\otimes \mu_{t,J}^{N}-\mu_{t,J}\otimes \mu_{t,J}\big)({\mathrm{d}}\theta_{t,J},{\mathrm{d}}\theta_{t-1,J})(\tilde{\pi}_{t-1}^N-\tilde{\pi}_{t-1})({\mathrm{d}}\theta_{t-1})\Big)^2\Big]\leq \frac{\tilde{C}_{t,1}}{N}.
\end{split}$$ To this aim let $\varphi$ be such that $|\varphi|\leq 1$ and remark that $$\begin{split}
&\int_{\tilde{\Theta}^{2J+1}}\tilde{\alpha}_t(\varphi,\theta_{t-1}, \theta _{t-1,J};  \theta_{t,J})\big(\mu_{t,J}^{N}\otimes \mu_{t,J}^{N}-\mu_{t,J}\otimes \mu_{t,J}\big)({\mathrm{d}}\theta_{t,J},{\mathrm{d}}\theta_{t-1,J})(\tilde{\pi}_{t-1}^N-\tilde{\pi}_{t-1})({\mathrm{d}}\theta_{t-1})\\
&= \int_{\tilde{\Theta}^{2J+1}} \tilde{\alpha}_t(\varphi,\theta_{t-1}, \theta _{t-1,J};  \theta_{t,J})\big(\mu_{t,J}^{N}\otimes \mu_{t,J}^{N}\otimes\tilde{\pi}_{t-1}^N-\mu_{t,J}\otimes \mu_{t,J}\otimes\tilde{\pi}_{t-1}\big)({\mathrm{d}}\theta)\\
&+\int_{\tilde{\Theta}^{2J}} \Big(\int_{\tilde{\Theta}} \tilde{\alpha}_t(\varphi,\theta_{t-1}, \theta _{t-1,J};  \theta_{t,J}) \tilde{\pi}_{t-1}({\mathrm{d}}\theta_{t-1})\Big)\big(\mu_{t,J}\otimes \mu_{t,J}-\mu_{t,J}^{N}\otimes \mu_{t,J}^{N}\big)({\mathrm{d}}\theta)\\
&+\int_{\tilde{\Theta}} \Big(\int_{\tilde{\Theta}^{2J}} \tilde{\alpha}(\theta_{t-1}, \theta _{t-1,J};  \theta_{t,J}) \mu_{t,J}\otimes \mu_{t,J}({\mathrm{d}}\theta)\Big)\big(\tilde{\pi}_{t-1}-\tilde{\pi}_{t-1}^N\big)({\mathrm{d}}\theta).
\end{split}$$ Then, since $\|\alpha_t\|_{\infty}\leq 1$, ${\mathbb{P}}$-a.s., and because we are assuming that - hold for all $s\in 0:(t-1)$, applying first $2J$ times Lemma \[lemma:tech\_conv\] with $\pi^N=\tilde{\pi}_{k_t}^N$ and then one time with $\pi^N=\tilde{\pi}_{t-1}^N$ shows .

To proceed further we let $$\alpha'_t(\varphi,\theta_{t-1}, \theta _{t-1,J};  \theta_{t,J})= \varphi(\theta_{t-1})  \int_{\tilde{\Theta}}\alpha_t(\theta_{t-1}, \theta _{t-1,J}; \theta_{t}, \theta_{t,J}) q_t({\mathrm{d}}\theta_t|\theta_{t-1}){\mathrm{d}}\theta_t$$ so that, by repeating the above computation with $\tilde{\alpha}_t$ replaced by $\alpha'_t$, it follows for some ${\mathbb{P}}$-a.s. finite random variable $\tilde{C}_{t,2}$ we have $$\begin{split}
 \sup_{ |\varphi|\leq 1}&{\mathbb{E}}_t\Big[\Big(\int_{\tilde{\Theta}^{2J+1}} \alpha'_t(\varphi,\theta_{t-1}, \theta _{t-1,J};  \theta_{t,J})\\
& \big(\mu_{t,J}^{N}\otimes \mu_{t,J}^{N}-\mu_{t,J}\otimes \mu_{t,J}\big)({\mathrm{d}}\theta_{t,J},{\mathrm{d}}\theta_{t-1,J})(\tilde{\pi}_{t-1}^N-\tilde{\pi}_{t-1})({\mathrm{d}}\theta_{t-1})\Big)^2\Big]\leq \frac{\tilde{C}_{t,2}}{N},\quad{\mathbb{P}}-a.s.
\end{split}$$ which, together with , shows that under the assumptions of the lemma we have, for some ${\mathbb{P}}$-a.s. finite random variable $\tilde{C}_{t,3}$ $$\label{eq:BB3_2}
\begin{split}
\sup_{|\varphi|\leq 1}{\mathbb{E}}_t\Big[\Big(\int_{\tilde{\Theta}^2}\varphi(\theta_t)\big\{L^N_t(\theta_{t-1},{\mathrm{d}}\theta_t) -L_t(\theta_{t-1},{\mathrm{d}}\theta_t)\big\}(\tilde{\pi}^N_{t-1}&-\tilde{\pi}_{t-1})({\mathrm{d}}\theta_{t-1})\Big)^2\Big]\\
&\leq \frac{\tilde{C}_{t,3}}{N},\quad{\mathbb{P}}-a.s.
\end{split}$$ Lastly, since $$\begin{aligned}
&{\mathbb{E}}_t\Big[\Big(\int_{\tilde{\Theta}}\varphi(\theta_t)\big\{L^N_t(\theta_{t-1},{\mathrm{d}}\theta_t) -L_t(\theta_{t-1},{\mathrm{d}}\theta_t)\big\}\tilde{\pi}_{t-1}({\mathrm{d}}\theta_{t-1})\Big)^2\Big]^{1/2}\\
&\leq {\mathbb{E}}_t\Big[\Big(\int_{\tilde{\Theta}^{2J}}\Big\{\int_{\tilde{\Theta}}\tilde{\alpha}_t(\varphi,\theta_{t-1}, \theta _{t-1,J};  \theta_{t,J})\tilde{\pi}_{t-1}({\mathrm{d}}\theta_{t-1})\Big\}\big(\mu_{t,J}^{N}\otimes \mu_{t,J}^{N}-\mu_{t,J}\otimes \mu_{t,J}\big)({\mathrm{d}}\theta_{t,J},{\mathrm{d}}\theta_{t-1,J})\Big)^2\Big]^{1/2}\\
&+{\mathbb{E}}_t\Big[\Big(\int_{\tilde{\Theta}^{2J}}\Big\{\int_{\tilde{\Theta}} \alpha'_t(\varphi,\theta_{t-1}, \theta _{t-1,J};  \theta_{t,J})\tilde{\pi}_{t-1}({\mathrm{d}}\theta_{t-1})\Big\}\big(\mu_{t,J}^{N}\otimes \mu_{t,J}^{N}-\mu_{t,J}\otimes \mu_{t,J}\big)({\mathrm{d}}\theta_{t,J},{\mathrm{d}}\theta_{t-1,J})\Big)^2\Big]^{1/2}\end{aligned}$$ it follows that under the assumptions of the lemma and by Lemma \[lemma:tech\_conv\] there exists ${\mathbb{P}}$-a.s. finite random variable $\tilde{C}_{t,4}$ such that $$\begin{split}
\sup_{|\varphi|\leq 1}{\mathbb{E}}_t\Big[\Big(\int_{\tilde{\Theta}}\varphi(\theta_t)\big\{L^N_t(\theta_{t-1},{\mathrm{d}}\theta_t) -L_t(\theta_{t-1},{\mathrm{d}}\theta_t)\big\}\tilde{\pi}_{t-1}({\mathrm{d}}\theta_{t-1})\Big)^2\Big]\leq \frac{\tilde{C}_{t,4}}{N},\quad{\mathbb{P}}-a.s.
\end{split}$$ which, together with , completes the proof of the lemma.

### Proof of the proposition

We start by proving the first part of the proposition, which is done by showing that - hold for all $s\geq 0$.

To this aim we first recall the following two simple well-known results: for all $t\geq 1$ we have $$\label{eq:BB0}
\begin{split}
\sup_{|\varphi|\leq 1}{\mathbb{E}}_t\Big[\Big(\frac{1}{N} \sum_{n=1}^N\varphi(\theta_t^n)-\sum_{n=1}^N W_{t-1}^n  \int_{\tilde{\Theta}}\varphi(\theta_t)L^N_t( \theta^N_{t-1},{\mathrm{d}}\theta_t)\Big)^2\Big]\leq\frac{1}{N},\quad{\mathbb{P}}-a.s.
\end{split}$$ and $$\label{eq:BB1}
\begin{split}
\sup_{|\varphi|\leq 1}{\mathbb{E}}_t\Big[&\Big(\sum_{n=1}^NW_{t}\varphi(\theta_{t}^n)  -\frac{\frac{1}{N}\sum_{n=1}^N G_t(\theta_t^n)\varphi(\theta_t^n)}{\int_{\tilde{\Theta}}\big\{\int_{\tilde{\Theta}} G_t(\theta_t) L_t(\theta_{t-1},{\mathrm{d}}\theta_t )\big\} \tilde{\pi}_{t}({\mathrm{d}}\theta_{t-1})}\Big)^2\Big]\\
&\leq {\mathbb{E}}_t\Big[\Big(\frac{\frac{1}{N}\sum_{n=1}^N G_t(\theta_t^n)}{\int_{\tilde{\Theta}}\big\{\int_{\tilde{\Theta}} G_t(\theta_t) L_t(\theta_{t-1},{\mathrm{d}}\theta_t )\big\} \tilde{\pi}_{t}({\mathrm{d}}\theta_{t-1})}-1\Big)^2\Big],\quad{\mathbb{P}}-a.s.
\end{split}$$

Next, we show - by induction on $s\geq 0$.

It is direct to see that holds for $s=0$ with $C_0$ such that ${\mathbb{P}}(C_0=1)=1$. Assume now that holds at time $t\geq 0$ and note that $$\begin{aligned}
\sum_{n=1}^N W_t^n\varphi(\theta_t^n)-\tilde{\pi}_t(\varphi)&=\sum_{n=1}^N W_t^n\varphi(\theta_t^n)-\frac{\frac{1}{N}\sum_{n=1}^N G_t(\theta_t^n)\varphi(\theta_t^n)}{\int_{\tilde{\Theta}}\big\{\int_{\tilde{\Theta}} G_t(\theta_t) L_t(\theta_{t-1},{\mathrm{d}}\theta_t )\big\} \tilde{\pi}_{t}({\mathrm{d}}\theta_{t-1})}\\
&+\frac{\frac{1}{N}\sum_{n=1}^N G_t(\theta_t^n)\varphi(\theta_t^n)}{\int_{\tilde{\Theta}}\big\{\int_{\tilde{\Theta}} G_t(\theta_t) L_t(\theta_{t-1},{\mathrm{d}}\theta_t )\big\} \tilde{\pi}_{t}({\mathrm{d}}\theta_{t-1})}-\tilde{\pi}_t(\varphi)\end{aligned}$$ where the mean squared error (MSE) of the second term can be bounded using while the MSE of the first term can be bounded by first using and then applying . This shows that if holds at time $t$ then also holds at time $t$.

To conclude the proof it remains to show that, if for some $t\geq 1$ holds for all $s\in 0:(t-1)$ then holds at time $t$. To this aim note that $$\begin{aligned}
\frac{1}{N}\sum_{n=1}^N\varphi(\theta_t^n)&-\int_{\tilde{\Theta}} \Big\{\int_{\tilde{\Theta}} \varphi(\theta_t)L_t(\theta_{t-1},{\mathrm{d}}\theta_t)\Big\}\tilde{\pi}_{t-1}({\mathrm{d}}\theta_{t-1})\\
&=\frac{1}{N}\sum_{n=1}^N\varphi(\theta_t^n)-\sum_{n=1}^N W_{t-1}^n \int_{\tilde{\Theta}} \varphi(\theta_t)L^N_t(\theta^n_{t-1},{\mathrm{d}}\theta_t)\\
&+\sum_{n=1}^N W_{t-1}^n\int_{\tilde{\Theta}} \varphi(\theta_t)\big\{L^N_t(\theta^n_{t-1},{\mathrm{d}}\theta_t)-L_t(\theta^n_{t-1},{\mathrm{d}}\theta_t)\big\}\\
&+\sum_{n=1}^N W_{t-1}^n  \int_{\tilde{\Theta}} \varphi(\theta_t)L_t(\theta^n_{t-1},{\mathrm{d}}\theta_t)-\int_{\tilde{\Theta}} \Big\{\int_{\tilde{\Theta}} \varphi(\theta_t)L_t(\theta_{t-1},{\mathrm{d}}\theta_t)\Big\}\tilde{\pi}_{t-1}({\mathrm{d}}\theta_{t-1})\end{aligned}$$ where the MSE of the first term is bounded using , the MSE of the last term is bounded using at time $s=t-1$ while the MSE of the second term is bounded using Lemma \[lemma:int\_prop\]. Remark that Lemma \[lemma:int\_prop\] can indeed be used since, as shown above, if holds for $s\in 0:(t-1)$ then also holds for $s\in 0:(t-1)$. The proof of the first part of the proposition is complete.

To prove the second part of the proposition remark that, ${\mathbb{P}}$-a.s., $$\begin{aligned}
\sup_{|\varphi|\leq 1}&{\mathbb{E}}_t\big[(\bar{\pi}_t^{N,K}(\varphi) -\bar{\pi}_t(\varphi))^2\big]^{\frac{1}{2}}\\
&\leq \sup_{|\varphi|\leq 1}{\mathbb{E}}_t\Big[\Big(\bar{\pi}_t^{N,K}(\varphi)-\frac{1}{t+1}\sum_{s=0}^T\tilde{\pi}^N_s(\varphi)\Big)^2\Big]^{\frac{1}{2}}+\sup_{|\varphi|\leq 1}\frac{1}{t+1}\sum_{s=0}^t{\mathbb{E}}_t\big[((\tilde{\pi}_s^N-\tilde{\pi}_s)(\varphi))^2\big]^{\frac{1}{2}}\\
&\leq  \sup_{|\varphi|\leq 1}{\mathbb{E}}_t\Big[\Big(\frac{1}{K}\sum_{k=1}^K\varphi(\bar{\theta}_t^k)-\frac{1}{t+1}\sum_{s=0}^T\tilde{\pi}^N_s(\varphi)\Big)^2\Big]^{1/2}+\frac{\frac{1}{t+1}\sum_{s=0}^tC_s}{N}\end{aligned}$$ where the last inequality holds by the first part of Proposition \[prop:conv\_algo\].

Using a proof by induction on $s\geq 0$ it is easily verified that $\bar{\theta}_t^{1:K}{\stackrel{\mathclap{\normalfont\mbox{\tiny{iid}}}}{\sim}}\frac{1}{t+1}\sum_{s=0}^T\tilde{\pi}^N_s$ and thus, ${\mathbb{P}}$-a.s, $$\begin{aligned}
\sup_{|\varphi|\leq 1}{\mathbb{E}}_t\Big[\Big(\frac{1}{K}\sum_{k=1}^K\varphi(\bar{\theta}_t^k)&-\frac{1}{t+1}\sum_{s=0}^T\tilde{\pi}^N_s(\varphi)\Big)^2\Big]\\
&=\frac{1}{K}\sup_{|\varphi|\leq 1}{\mathbb{E}}_t\Big[{\mathrm{Var}}\big(\varphi(\bar{\theta}_t^1)|\,{\mathcal{F}}_t,\,\sigma(\theta_{0:t}^{1:N})\big) \Big]\\
&\leq \frac{1}{K}\end{aligned}$$ and the proof is complete.

Proof of Proposition \[prop:assumption\]\[p-prop:assumption\]
-------------------------------------------------------------

The proof of the second part of the proposition is similar to that of first part and to save place only this latter is given below.

For every $C\in{\mathbb{R}}_{>0}$ let $A_{C}=\{(\mu,\sigma^2)\in{\mathbb{R}}^2:\, |\sigma^2|< C,\,|\mu|<C\}$. We first show that if $C$ is large enough then \[new\] holds for $A_\star=A_{C_\star}$.

Let $B_{1,C}=\{(\mu,\sigma^2)\in\Theta:\, \sigma^2\geq C\}$ and $B_{2,C}=\{(\mu,\sigma^2)\in\Theta:\,|\mu|\geq C\}$ so that $A^c_C\cap\Theta=B_{1,C}\cup B_{2,C}$. Then, $$\begin{aligned}
\label{eq:p_a1}
{\mathbb{E}}[\sup_{\theta \not\in A_C}\log (\tilde{f}_{\theta})]\leq {\mathbb{E}}[\sup_{\theta \in B_{1,C}}\log (f_{\theta})]+{\mathbb{E}}[\sup_{\theta \in B_{2,C}}\log (f_{\theta})]\end{aligned}$$ where $$\begin{aligned}
\label{eq:p_a2}
{\mathbb{E}}[\sup_{\theta \in B_{1,C}}\log (\tilde{f}_{\theta})]\leq -\frac{1}{2}\log( 2\pi C).\end{aligned}$$

To proceed further let $\theta\in B_{2,C}$, $y\in{\mathbb{R}}$ and note that $$\begin{aligned}
f_\theta(y)&= f_\theta(y){\mathds{1}}(|y|\geq C/2)+ f_\theta(y){\mathds{1}}(|y|< C/2)\\
&\leq \frac{1}{\sqrt{2\pi\underline{\sigma}^2}}{\mathds{1}}(|y|\geq C/2)+\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(y-C)^2}{2\sigma^2}}{\mathds{1}}(|y|< C/2)\\
&\leq \frac{1}{\sqrt{2\pi\underline{\sigma}^2}}{\mathds{1}}(|y|\geq C/2)+\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{C^2}{8\sigma^2}}{\mathds{1}}(|y|< C/2)\\
&\leq \frac{1}{\sqrt{2\pi\underline{\sigma}^2}}{\mathds{1}}(|y|\geq C/2)+\frac{1}{\sqrt{\pi C^2/2}}e^{-\frac{1}{2}}{\mathds{1}}(|y|< C/2).\end{aligned}$$ Therefore, $$\begin{aligned}
{\mathbb{E}}[\sup_{\theta \in B_{2,C}}\log (f_{\theta})]&\leq -\frac{1}{2}\log(2\pi\underline{\sigma}^2){\mathbb{P}}(|Y_1|\geq  C/2)-\frac{1}{2}\log(\pi C^2/2){\mathbb{P}}(|Y_1|< C/2)\end{aligned}$$ which, together with and , shows that $$\begin{aligned}
{\mathbb{E}}[\sup_{\theta \not\in A_C}\log (\tilde{f}_{\theta})]&\leq -\frac{1}{2}\log( 2\pi C)-\frac{1}{2}\log(2\pi\underline{\sigma}^2){\mathbb{P}}(|Y_1|\geq  C/2)-\frac{1}{2}\log(\pi C^2/2){\mathbb{P}}(|Y_1|< C/2).\end{aligned}$$ Since the r.h.s. converges to $-\infty$ as $C\rightarrow+\infty$ it follows that \[new\].\[A41\]) holds for $A_\star=A_{C_\star}$, for a sufficiently large constant $C_\star\in{\mathbb{R}}_{>0}$.

To show the second part of \[new\] let $\tilde{A}_\star$ be an arbitrary compact set that contains a neighbourhood of $A_{C_\star}$. Note that $\tilde{A}_\star\cap\Theta$ is compact and that the mapping $\theta\mapsto f_\theta(y)$ is continuous on $\tilde{A}_\star\cap\Theta$ for all $y\in{\mathbb{R}}$. Then, since for all $(\tilde{\theta},\theta)\in\Theta^2$ we have $$\begin{aligned}
\log\big( (f_{\tilde{\theta}}/f_\theta)(y)\big)=\frac{1}{2}\log( \sigma^2/\tilde{\sigma}^2)-\frac{1}{2}\Big(\frac{\sigma^2-\tilde{\sigma}^2}{\sigma^2\tilde{\sigma^2}}y^2+\frac{\tilde{\mu}^2}{\tilde{\sigma}^2}-\frac{\mu^2}{\sigma^2}-2y\frac{\tilde{\mu}\sigma^2-\mu\tilde{\sigma}^2}{\sigma^2\tilde{\sigma^2}}\Big)\end{aligned}$$ it follows that the second part of \[new\] holds since ${\mathbb{E}}[Y_1^4]<+\infty$ by assumption. This concludes to show that \[new\] holds

To show that \[new1\].\[A53\] holds it suffices to remark that $f_\theta(y)\leq (2\pi\underline{\sigma}^2)^{-1/2}$ for all $\theta\in\Theta$ and $y\in{\mathsf{Y}}$.

We now show that \[new2\] holds. To this aim let $\theta_\star=(\mu_\star,\sigma_\star^2)\in\Theta$ and note that, for all $\sigma^2\in{\mathbb{R}}_{>0}$ and $\delta\neq 0$, we have $$\label{eq:CC}
\begin{split}
{\mathbb{E}}\Big[\exp\Big(-\frac{(Y_1-(\mu_\star+\delta))^2}{2\sigma^2}
\Big)\Big]&\leq  e^{- \delta^2/(8\sigma^2)}+ {\mathbb{P}}\big(|Y_1-\mu_\star|\geq |\delta|/2 \big)\\
&\leq   e^{- \delta^2/(8\sigma^2)} +\frac{2 {\mathbb{E}}[|Y_1-\mu_\star|]}{|\delta|}.
\end{split}$$ where the last inequality uses Markov’s inequality.

Let $\theta_C\in V_C$, $$\epsilon_1=\frac{ |\sigma_C^2-\sigma_\star^2|}{C},\quad \epsilon_2=\frac{ |\mu_C-\mu_\star|}{C}$$ and remark that $\epsilon_1\vee \epsilon_2\geq 1/\sqrt{2}$. Assume first that $\epsilon_2\geq  1/\sqrt{2}$. Then, using with $\delta=\mu_C-\mu_\star$, we have $$\label{eq:CC1}
\begin{split}
{\mathbb{E}}[f_{\theta_C}(Y_1)]&\leq \frac{1}{\sqrt{2\pi\underline{\sigma}^2}}\bigg(\exp\Big(- \frac{\epsilon_2^2 C^2}{8\sigma_C^2}\Big) +\frac{2 {\mathbb{E}}[|Y_1-\mu_\star|]}{\epsilon_2 C}\bigg)\\
&\leq \frac{1}{\sqrt{2\pi\underline{\sigma}^2}}\bigg(\exp\Big(- \frac{ C^2}{16 \underline{\sigma}^2}\Big) +\frac{2^{3/2} {\mathbb{E}}[|Y_1-\mu_\star|]}{ C}\bigg).
\end{split}$$ Assume now that $\epsilon_2< 1/\sqrt{2}$, so that $\epsilon_1\geq 1/\sqrt{2}$. Note that, for $C> 2^{3/2}\sigma^2_\star$ we have $$\sigma_C^2=\big|\sigma^2_\star-\sigma^2_C-\sigma^2_\star\big|\geq \epsilon_1 C -\sigma^2_\star\geq C/\sqrt{2}-\sigma^2_\star\geq \sigma^2_\star>0.$$ Therefore, for $C> 2^{3/2}\sigma^2_\star$ we have $$\label{eq:CC2}
\begin{split}
{\mathbb{E}}[f_{\theta_C}(Y_1)]&\leq \Big(2\pi(\sigma^2_\star-\sigma^2_C-\sigma^2_\star)\Big)^{-1/2}\leq   \Big(2\pi(C/\sqrt{2}-\sigma_\star^2)\Big)^{-1/2}. 
\end{split}$$ Then, using -, it follows that there exists a constant $C'\in{\mathbb{R}}_{>0}$ such that, for $C>0$ large enough we have $$\begin{aligned}
\sup_{\theta\in V_C}{\mathbb{E}}[f_{\theta}(Y_1)]&\leq \frac{1}{\sqrt{2\pi\underline{\sigma}^2}}\bigg(\exp\Big(- \frac{ C^2}{16 \underline{\sigma}^2}\Big) +\frac{2^{3/2} {\mathbb{E}}[|Y_1-\mu_\star|]}{ C}\bigg)+\Big(2\pi(C/\sqrt{2}-\sigma_\star^2)\Big)^{-1/2}\\
&\leq  \frac{C'}{C^{1/2}}\end{aligned}$$ showing that $$\begin{aligned}
\limsup_{C\rightarrow+\infty}\frac{\log\big(\sup_{\theta\in V_C}{\mathbb{E}}[f_{\theta}(Y_1)]\big)}{\log (C)}\leq \limsup_{C\rightarrow+\infty}\frac{\log(C')-\frac{1}{2}\log(C)}{\log (C)}=-\frac{1}{2}<0.\end{aligned}$$ Hence the first part of \[new2\].\[A63\] holds. In addition, since $$|\log(f_{\theta_\star}(y))|^{p}\leq 2^{p-1}\Big|\frac{1}{2}\log(2\pi\sigma_\star^2)\Big|^{p}+\frac{|y-\mu_\star|^{2p}}{2(\sigma_\star^2)^p},\quad\forall p\in\mathbb{N},$$ it follows that the second part of \[new2\].\[A63\] holds for every $k_\star\in\mathbb{N}$ since, by assumption, ${\mathbb{E}}[e^{c |Y_1|}]<+\infty$ for some constant $c>0$. The proof is complete.

Proof of Theorem \[thm:online\]\[p-thm:online\]
-----------------------------------------------

Theorem \[thm:online\] is a direct consequence of Proposition \[prop:conv\_proba\] and of the following three lemmas.

\[lemma:online1\] Consider the set-up of Theorem \[thm:online\]. Then, ${\mathbb{E}}[\tilde{\pi}_{t_p}(V_{\epsilon })]\rightarrow 0$ for all $\epsilon>0$.

See Section \[p-lemma:online1\] for the proof.

\[lemma:online2\]

Consider the set-up of Theorem \[thm:online\] and assume that the conclusion of Lemma \[lemma:online1\] holds. Let $(v_p)_{p\geq 1}$ be a sequence in $\mathbb{N}$ such that $t_{p-1}\leq v_p<t_p$ for all $p\geq 1$ and such that $(v_p-t_{p-1})\rightarrow+\infty$, and let $(\tau_k)_{k\geq 1}$ be a strictly increasing sequence in $\mathbb{N}$ such that $(\tau_k)_{k\geq 1}=\{t\in\mathbb{N}:\,\exists p\geq 1,\, v_p\leq t<  t_p\}$. Then, $ {\mathbb{E}}[\tilde{\pi}_{\tau_k}(V_{\epsilon})]\rightarrow 0$ for all $\epsilon>0$.

See Section \[p-lemma:online2\] for the proof.

\[lemma:online3\]

Consider the set-up of Theorem \[thm:online\] and assume that the conclusion of Lemma \[lemma:online1\] holds. Then, there exists a sequence $(v_p)_{p\geq 1}$ verifying the conditions of Lemma \[lemma:online2\] such that, with $(\tau'_q)_{q\geq 1}$ the strictly increasing sequence in $\mathbb{N}$ verifying $$(\tau'_q)_{q\geq 1}=\{t\in\mathbb{N}:\,\exists p\geq 1,\, t_{p-1}<t\leq v_p\},$$ we have ${\mathbb{E}}[\tilde{\pi}_{\tau'_q}(V_{\epsilon})]\rightarrow 0$ for all $\epsilon>0$.

See Section \[p-lemma:online3\] for the proof.

### Proof of Lemma \[lemma:online1\]\[p-lemma:online1\]

Below $C\in(0,+\infty)$ is a constant whose value can change from one expression to another.

We first remark that, by Doob’s martingale inequality and under the assumptions on $(\mu_t)_{t\geq 0}$, $$\limsup_{p\rightarrow+\infty}{\mathbb{P}}\big([U_{(t_{p-1}+1):t_p}]\geq \gamma\big)\leq C \gamma^{-2}\limsup_{p\rightarrow+\infty} \sum_{s=t_{p-1}+1}^{t_{p}-1}h_s^2=0,\quad\forall \gamma>0$$ showing that there exists a sequence $(\gamma_t)_{t\geq 1}$ in ${\mathbb{R}}_{>0}$ such that $$\begin{aligned}
\label{eq:gamma_t}
\gamma_t\rightarrow 0, \quad {\mathbb{P}}\big([U_{(t_{p-1}+1):t_p}]\geq \gamma_{t_p}\big)\rightarrow 0.\end{aligned}$$

Let $(\gamma_t)_{t\geq 1}$ be as in and $(s_t)_{t\geq 1}$ be a sequence in $\mathbb{N}_0$ such that $\inf_{t\geq 1}(t-s_t)\geq 1$, $(t-s_{t})\rightarrow+\infty$ and $s_{t_p}=t_{p-1}$ for every $p\geq 1$. Remark that such a sequence $(s_t)_{t\geq 1}$ exists under the assumptions of the lemma.

To proceed further let $(C_1,C_2)\in{\mathbb{R}}_{>0}^2$ be as in Lemma \[lemma:part\_1\]. Without loss of generality we assume below that $2\sqrt{1/(2C_1 C_2)}<\delta_\star$, with $\delta_\star>0$ as in Lemma \[lemma:denom\]. Let $\tilde{\delta}=\delta=\sqrt{1/(2C_1 C_2)} $, $(\phi'_t)_{t\geq 1}$ be as in Lemma \[lemma:part\_1\] and, for every $t\geq 1$, let $\tilde{\phi}_{t}(Y_{1:t})$ be such that $\tilde{\phi}_t(Y_{1:t})=1$ whenever $\tilde{\pi}_{t} (V_{c/h^\beta_{t}})\geq \beta^{-1}$ and such that $\tilde{\phi}_t(Y_{1:t})=0$ otherwise, with $(c,\beta)\in (1,+\infty)$ as in . Notice that ${\mathbb{E}}[\tilde{\phi}_{t_p}(Y_{1:t_p})]\rightarrow 0$ by while ${\mathbb{E}}[\phi'_{t_p}(Y_{1:t_p})]\rightarrow 0$ by Lemma \[lemma:part\_1\].

Therefore, using Lemma \[lemma:pi\], Tonellli’s theorem and the shorthand $ \Theta_{t_p}= \Theta_{\gamma_{t_p},t_{p-1}:t_p}$, $$\label{eq:inquality1}
\begin{split}
 &\limsup_{p\rightarrow+\infty}  {\mathbb{E}}[\tilde{\pi}_{t_p}(V_{\epsilon})]\\
& \leq \limsup_{p\rightarrow+\infty}{\mathbb{E}}[(1-\phi'_{t_p}(Y_{1:t_p}))(1-\tilde{\phi}_{t_{p-1}}(Y_{1:t_{p-1}}))\tilde{\pi}_{t_p}(V_{\epsilon})]\\
&\leq \limsup_{p\rightarrow+\infty}{\mathbb{E}}\big[(1-\phi'_{t_p}(Y_{1:t_p}))(1-\tilde{\phi}_{t_{p-1}}(Y_{1:t_{p-1}})){\mathds{1}}_{ \Theta_{t_p}}(U_{t_{p-1}:t_p}) \pi'_{t_{p-1},t_p}(V_{\epsilon})\big]\\
&+\limsup_{p\rightarrow+\infty}{\mathbb{E}}\Big[(1-\phi'_{t_p}(Y_{1:t_p}))(1-\tilde{\phi}_{t_{p-1}}(Y_{1:t_{p-1}})){\mathds{1}}_{ \Theta^c_{t_p}}(U_{t_{p-1}:t_p}) \pi'_{t_{p-1},t_p}(V_{\epsilon_{t_p}}) \big]
\end{split}$$ where, by Lemma \[lemma:part\_1\] and for $p$ large enough we have, ${\mathbb{P}}$-a.s., $$\label{eq:inquality_int1}
\begin{split}
{\mathbb{E}}\big[(1-\phi'_{t_p}&(Y_{1:t_p}))  {\mathds{1}}_{ \Theta_{t_p}}(U_{t_{p-1}:t_p}) \pi'_{t_{p-1},t_p}(V_{\epsilon_{t_p}})|{\mathcal{F}}_{t_{p-1}}\big]\\
&\leq   \frac{e^{-(t_p-t_{p-1}) C^{-1}}}{{\mathbb{P}}(U_{t_{p-1}:t_p}\in\Theta_{\delta,t_{p-1}:t_p})\, \inf_{v\in B_{\delta}(0)}(\mu_{t_{p-1}}*\tilde{\pi}_{t_{p-1}})(B_{\delta}(\theta_\star-v))}.
\end{split}$$

To proceed further let $v\in B_{\delta}(0)$, $f_{t_{p-1}}$ be the density of $\mu_{t_{p-1}}$ and remark that for all $p\geq 1$ we have (using Tonelli’s theorem for the second equality) $$\label{eq:lower_t}
\begin{split}
(\mu_{t_{p-1}}*\tilde{\pi}_{t_{p-1}})&(B_{\delta}(\theta_\star-v))\\
&=\int_{B_{\delta}(\theta_\star-v)} \int_{\Theta} f_{t_{p-1}}(\theta-u)\tilde{\pi}_{t_{p-1}}({\mathrm{d}}u) {\mathrm{d}}\theta\\
&=\int_{\Theta} \int_{B_{\delta}(\theta_\star-v)}  f_{t_{p-1}}(\theta-u){\mathrm{d}}\theta \,\tilde{\pi}_{t_{p-1}}({\mathrm{d}}u)\\
&\geq \tilde{\pi}_{t_{p-1}}\big(B_{c h_{t_{p-1}}^{-\beta}}(\theta_\star)\big)\inf_{u\in B_{c h_{t_{p-1}}^{-\beta}}(\theta_\star)}\mu_{t_{p-1}}\big(B_{\delta}(\theta_\star-v+u)\big).
\end{split}$$ Recall that $\mu_{t_{p-1}}({\mathrm{d}}u)$ is the $t_{d,\nu}(0, h_{t_{p-1}}^2\Sigma_{t_{p-1}})$. Then, under the assumptions on $(\Sigma_t)_{t\geq 0}$, and using , it is easily checked that, with the shorthand $\nu_1=\beta(\nu+d)+\nu$, $$\label{eq:lower_C}
 {\mathbb{P}}\Big( \inf_{v\in B_{\delta}(0)}(\mu_{t_{p-1}}*\tilde{\pi}_{t_{p-1}})(B_{\delta}(\theta_\star-v))\geq C^{-1}   h_{t_{p-1}}^{\nu_1} \big|\tilde{\phi}_{t_{p-1}}(Y_{1:t_{p-1}})=0 \Big)=1,\quad\forall p\geq 1.$$ Consequently, using and for $p$ large enough, we have $$\label{eq:inquality2}
\begin{split}
{\mathbb{E}}\big[(1-\phi'_{t_p}(Y_{1:t_p}))(1- \tilde{\phi}_{t_{p-1}} (Y_{1:t_{p-1}})){\mathds{1}}_{ \Theta_{t_p}}& (U_{t_{p-1}:t_p}) \pi'_{t_{p-1},t_p}(V_{\epsilon})\big]\\
&\leq C h_{t_{p-1}}^{-\nu_1}\frac{e^{-(t_p-t_{p-1}) C^{-1}}}{{\mathbb{P}}(U_{t_{p-1}:t_p}\in\Theta_{\delta,t_{p-1}:t_p})}.
\end{split}$$

To proceed further remark that, by Lemma \[lemma:denom\] and using , we can without loss of generality assume that $(\phi'_t)_{t\geq 1}$ is such that, for all $p\geq 1$, $$\label{eq:psi1}
\begin{split}
&{\mathbb{P}}\bigg(\int_{\Theta}{\mathbb{E}}\Big[ (\mu_{t_{p-1}}*\tilde{\pi}_{t_{p-1}}) \Big(\theta-\sum_{s=t_{p-1}+1}^{t_p-1}U_s\Big)\prod_{s=t_{p-1}+1}^{t_p} (\tilde{f}_{\theta-\sum_{i=s}^{t_{p}-1}U_i}/f_{\theta_\star})(Y_s)\big|{\mathcal{F}}_{t_p}\Big]  {\mathrm{d}}\theta \\
 &>  C^{-1}   h_{t_{p-1}}^{\nu_1}{\mathbb{P}}(U_{t_{p-1}:t_p}\in\Theta_{\delta,t_{p-1}:t_p})\,e^{-(t_p-t_{p-1}) C \delta^2}\big|\phi'_{t_{p}}(Y_{1:t_{p}})\vee \tilde{\phi}_{t_{p-1}}(Y_{1:t_{p-1}})=0\bigg)\\
 &=1
\end{split}$$ while, by the law of large numbers, we can also without loss of generality assume that $(\phi'_t)_{t\geq 1}$ is such that $$\label{eq:psi2}
{\mathbb{P}}\Big(-\frac{1}{t-s_t}\sum_{s=s_t+1}^t\log( f_{\theta_\star}(Y_s))\leq 1-{\mathbb{E}}[\log (f_{\theta_\star})]\,\,\big| \phi'_t(Y_{1:t})=0\Big)=1,\quad\forall t\geq 1.$$

We now show that, under \[new1\] we have, for $p$ large enough, $$\label{eq:inquality3}
\begin{split}
 {\mathbb{E}}\big[  (1-\phi'_{t_p} (Y_{1:t_p}))(1-\tilde{\phi}_{t_{p-1}}&(Y_{1:t_{p-1}})) {\mathds{1}}_{ \Theta^c_{t_p}} (U_{t_{p-1}:t_p}) \pi'_{t_{p-1},t_p}(V_{\epsilon} )\big]\\
 &\leq  \frac{Ce^{(t_p-t_{p-1})C}}{{\mathbb{P}}(U_{t_{p-1}:t_p}\in\Theta_{\delta,t_{p-1}:t_p})}h_{t_{p-1}}^{-\nu_1}{\mathbb{P}}\big(U_{t_{p-1}:t_p}\not\in  \Theta_{t_p}\big).
\end{split}$$

Assume first that \[new1\].\[A51\] holds. In this case, by the law of large numbers, we can without loss of generality assume that $(\phi'_t)_{t\geq 1}$ is such that $$\label{eq:psi3}
{\mathbb{P}}\Big(\frac{1}{t-s_t}\sum_{s=s_t+1}^t\sup_{\theta\in\Theta}\log (f_{\theta}(Y_s))\leq 1+{\mathbb{E}}[\sup_{\theta\in\Theta}\log (f_\theta)] \,\big| \phi'_t(Y_{1:t})=0\Big)=1,\quad\forall t\geq 1$$ in which case directly follows from , and .

Assume now that \[new1\].\[A52\] holds. Then, holds since, for all $p\geq 1$, $$\begin{split}
 {\mathbb{E}}\big[ &(1-\phi'_{t_p} (Y_{1:t_p}))(1-\tilde{\phi}_{t_{p-1}}(Y_{1:t_{p-1}})) {\mathds{1}}_{ \Theta^c_{t_p}} (U_{t_{p-1}:t_p}) \pi'_{t_{p-1},t_p}(V_{\epsilon} )\big]\\
 &\leq  \frac{Ce^{(t_p-t_{p-1})C}}{{\mathbb{P}}(U_{t_{p-1}:t_p}\in\Theta_{\delta,t_{p-1}:t_p})}h_{t_{p-1}}^{-\nu_1} {\mathbb{E}}\Big[{\mathds{1}}_{ \Theta^c_{t_p}} (U_{t_{p-1}:t_p})\\
 &\qquad \times \int_{V_{\epsilon}} (\mu_{t_{p-1}}*\tilde{\pi}_{t_{p-1}}) \Big(\theta-\sum_{s=t_{p-1}+1}^{t_p-1}U_s\Big)\prod_{s=t_{p-1}+1}^{t_p}{\mathbb{E}}\big[  \tilde{f}_{\theta-\sum_{i=s}^{t_{p}-1}U_i}/f_{\theta_\star})(Y_s)\big| U_{t_{p-1}:t_p}\big] {\mathrm{d}}\theta \Big]\\
 &\leq  \frac{Ce^{(t_p-t_{p-1})C}}{{\mathbb{P}}(U_{t_{p-1}:t_p}\in\Theta_{\delta,t_{p-1}:t_p})}h_{t_{p-1}}^{-\nu_1} \big(\sup_{\theta\in\Theta}{\mathbb{E}}\big[f_\theta/f_{\theta_\star} \big]\big)^{t_p-t_{p-1}}{\mathbb{P}}\big(U_{t_{p-1}:t_p}\not\in  \Theta_{t_p}\big)\\
 &\leq  \frac{C e^{(t_p-t_{p-1})C}}{{\mathbb{P}}(U_{t_{p-1}:t_p}\in\Theta_{\delta, t_{p-1}:t_p})}h_{t_{p-1}}^{-\nu_1}{\mathbb{P}}\big(U_{t_{p-1}:t_p}\not\in  \Theta_{t_p}\big)
\end{split}$$ where the last inequality holds since $\sup_{\theta\in\Theta}{\mathbb{E}}[f_\theta/f_{\theta_\star}]<+\infty$ under \[new1\].\[A52\].

Lastly, assume that \[new1\].\[A53\] holds. Then, holds since, for all $p\geq 1$, $$\begin{split}
 {\mathbb{E}}\big[ &(1-\phi'_{t_p} (Y_{1:t_p}))(1-\tilde{\phi}_{t_{p-1}}(Y_{1:t_{p-1}})) {\mathds{1}}_{ \Theta^c_{t_p}} (U_{t_{p-1}:t_p}) \pi'_{t_{p-1},t_p}(V_{\epsilon} )\big]\\
 &\leq  \frac{Ce^{(t_p-t_{p-1})C}}{{\mathbb{P}}(U_{t_{p-1}:t_p}\in\Theta_{\delta, t_{p-1}:t_p})}h_{t_{p-1}}^{-\nu_1} {\mathbb{E}}\Big[{\mathds{1}}_{ \Theta^c_{t_p}} (U_{t_{p-1}:t_p})\\
 &\times \int_{V_{\epsilon}} (\mu_{t_{p-1}}*\tilde{\pi}_{t_{p-1}}) \Big(\theta-\sum_{s=t_{p-1}+1}^{t_p-1}U_s\Big)\prod_{s=t_{p-1}+1}^{t_p} {\mathbb{E}}\big[ \tilde{f}_{\theta-\sum_{i=s}^{t_{p}-1}U_i} (Y_s)\big| U_{t_{p-1}:t_p}\big]{\mathrm{d}}\theta \Big]\\
 &\leq  \frac{C e^{(t_p-t_{p-1})C}}{{\mathbb{P}}(U_{t_{p-1}:t_p}\in\Theta_{\delta,t_{p-1}:t_p})}h_{t_{p-1}}^{-\nu_1} \big(\sup_{\theta\in\Theta}{\mathbb{E}}\big[f_\theta\big]\big)^{t_p-t_{p-1}}{\mathbb{P}}\big(U_{t_{p-1}:t_p}\not\in  \Theta_{t_p}\big)\\
 &\leq  \frac{C e^{(t_p-t_{p-1})C}}{{\mathbb{P}}(U_{t_{p-1}:t_p}\in\Theta_{\delta,t_{p-1}:t_p})}h_{t_{p-1}}^{-\nu_1}{\mathbb{P}}\big(U_{t_{p-1}:t_p}\not\in  \Theta_{t_p}\big)
\end{split}$$ where the last inequality holds since $\sup_{\theta\in\Theta}{\mathbb{E}}[f_\theta]<+\infty$ under \[new1\].\[A53\]. This concludes to show that holds under \[new1\].

Using , and , to conclude the proof it is enough to show that, as $p\rightarrow+\infty$, $$\begin{aligned}
&h_{t_{p-1}}^{-\nu_1}e^{-(t_p-t_{p-1})C^{-1}}\rightarrow 0\label{eq:lim1}\\
&h_{t_{p-1}}^{-\nu_1}e^{(t_p-t_{p-1})C}{\mathbb{P}}\big(U_{t_{p-1}:t_p}\not\in  \Theta_{\gamma_{t_p},:t_{p-1},t_p}\big)\rightarrow 0\label{eq:lim2}\\
&\liminf_{p\rightarrow+\infty}{\mathbb{P}}(U_{t_{p-1}:t_p}\in\Theta_{\delta,t_{p-1}:t_p})>0.\label{eq:lim3}\end{aligned}$$

By assumptions, $\log(1/h_{t_{p-1}})(t_p-t_{p-1})^{-1}\rightarrow 0$ and therefore holds.

To establish note that, using Doob’s martingale inequality and the fact that $(\tilde{U} _s)_{s=t_{p-1}+1}^{t_p-1}$ are independent Gaussian random variables, we have $$\begin{aligned}
\label{eq:Doob_gauss}
{\mathbb{P}}\big([U_{(t_{p-1}+1):t_p}]\geq \gamma \big)\leq C  \exp\Big(-\frac{\gamma^2}{C \sum_{s=t_{p-1}+1}^{t_p-1}h_s^2}\Big),\quad\forall \gamma>0,\quad\forall p\geq 1\end{aligned}$$ so that $$\begin{aligned}
{\mathbb{P}}\big(U_{t_{p-1}: t_p}\not\in  \Theta_{\gamma_{t_p}, t_{p-1}:t_p}\big)={\mathbb{P}}\big([U_{(t_{p-1}+1):t_p}]\geq\gamma_{t_p}\big)\leq C \exp\Big(-\frac{\gamma_{t_p}^2}{C\sum_{s=t_{p-1}+1}^{t_p-1}h_s^2}\Big),\quad\forall p> 1.\end{aligned}$$ Consequently, for all $p> 1$, $$\begin{aligned}
\log\Big(&h_{t_{p-1}}^{-\nu_1}\, e^{(t_p-t_{p-1})C}\, {\mathbb{P}}\big(U_{t_{p-1}: t_p}\not\in  \Theta_{\gamma_{t_p}, t_{p-1}:t_p}\big)\Big)\\
&\leq -\nu_1 \log (h_{t_{p-1}})+C(t_p-t_{p-1})+\log(C)-C^{-1}\gamma_{t_p}^2\Big(\sum_{s=t_{p-1}+1}^{t_p-1}h_s^2\Big)^{-1}\\
&=-(t_p-t_{p-1})\bigg(-\frac{ \log(C)}{t_p-t_{p-1}}-C+\frac{C^{-1}\gamma_{t_p}^2-\nu_1\log(1/h_{t_{p-1}})\sum_{s=t_{p-1}+1}^{t_p-1}h_s^2}{(t_p-t_{p-1})\sum_{s=t_{p-1}+1}^{t_p-1}h_s^2}\bigg)\end{aligned}$$ where, under the assumptions of the lemma and by taking $(\gamma_t)_{t\geq 1}$ such that $\gamma_{t_p}\rightarrow 0$ sufficiently slowly $$\bigg(-\frac{ \log(C)}{t_p-t_{p-1}}-C+\frac{C^{-1}\gamma_{t_p}^2-\nu_1\log(1/h_{t_{p-1}})\sum_{s=t_{p-1}+1}^{t_p-1}h_s^2}{(t_p-t_{p-1})\sum_{s=t_{p-1}+1}^{t_p-1}h_s^2}\bigg)\rightarrow+\infty$$ so that holds.

Lastly, to show it suffices to remark that, since $\gamma_p\rightarrow 0$, $$\liminf_{p\rightarrow+\infty}{\mathbb{P}}(U_{t_{p-1}:t_p}\in\Theta_{\delta, t_{p-1}:t_p})\geq\liminf_{p\rightarrow+\infty}  {\mathbb{P}}(U_{t_{p-1}:t_p}\in\Theta_{\gamma_p, t_{p-1}:t_p})=1$$ where the equality holds since holds. The proof is complete.

### Proof of Lemma \[lemma:online2\] {#p-lemma:online2}

Below $C\in(0,+\infty)$ is a constant whose value can change from one expression to another.

We first remark that under the assumptions of the lemma there exists a $p_1\in\mathbb{N}$ such that we have both $t_p-t_{p-1}>1$ and $v_p-t_{p-1}>0$ for all $p\geq p_1$. To simplify the presentation of the proof we assume without loss of generality that $p_1=1$ in what follows.

For every $k\geq 1$ let $p_k=\sup\{p\geq 0:\, t_p<\tau_k\}$ and let $(s_t)_{t\geq 1}$ be a sequence in $\mathbb{N}_0$ such that $\inf_{t\geq 1}(t-s_t)\geq 1$, $(t-s_t)\rightarrow+\infty$ and $s_{\tau_k}=t_{p_k}$ for every $k\geq 1$. Remark that such a sequence $(s_t)_{t\geq 1}$ exists since $(\tau_k-s_{\tau_k})\geq 1$ for all $k\geq 1$ by construction while, by assumption, $$\liminf_{k\rightarrow+\infty}(\tau_k-s_{\tau_k})\geq  \liminf_{k\rightarrow+\infty} (v_{p_k+1}-t_{p_k})=+\infty.$$

Remark now that, by Doob’s martingale inequality and under the assumptions on $(\mu_t)_{t\geq 0}$ we have, for every $\gamma>0$, $$\begin{aligned}
\limsup_{k\rightarrow+\infty}{\mathbb{P}}\big([U_{(s_{\tau_k}+1):\tau_k}]\geq  \gamma\big)\leq C\limsup_{k\rightarrow+\infty}\gamma^{-2} \sum_{s=t_{p_k}+1}^{\tau_k-1}h^2_s \leq C\limsup_{k\rightarrow+\infty} \sum_{s=t_{p_k}+1}^{t_{p_{k+1}}-1}h^2_s =0\end{aligned}$$ showing that there exists a sequence $(\gamma_t)_{t\geq 1}$ in ${\mathbb{R}}_{>0}$ such that $$\begin{aligned}
\label{eq:gamma_t2}
 \gamma_t\rightarrow 0,\quad {\mathbb{P}}\big([U_{(s_{\tau_k}+1):\tau_k}]\geq  \gamma_{\tau_k}\big)\rightarrow 0.\end{aligned}$$

Let $(\gamma_t)_{t\geq 1}$ be as in and $(C_1,C_2)\in(0,+\infty)^2$ be as in Lemma \[lemma:part\_1\]. Without loss of generality we assume below that $2\sqrt{1/(2C_1 C_2)}<\delta_\star$, with $\delta_\star>0$ as in Lemma \[lemma:denom\].

Let $\kappa=\sqrt{1/(12C_1 C_2)}$, $\delta= 3\kappa$, $\tilde{\delta}=\kappa$ and remark that, for every $t\geq 1$ and $v \in B_{\tilde{\delta}}(0)$ we have, with $\theta_{s_t}:\Omega\rightarrow\Theta$ such that $\theta_{s_t}\sim\tilde{\pi}_{s_t}({\mathrm{d}}\theta)$ under ${\mathbb{P}}$, $$\label{eq:lower_1}
\begin{split}
{\mathbb{P}}\big(\|\theta_{s_t}+U_{s_t}-\theta_\star+v\|&\geq  \delta |\,\theta_{s_t}\in B_{\kappa}(\theta_\star)\big)\\
&\leq {\mathbb{P}}\big(\|\theta_{s_t}-\theta_\star\|+\|U_{s_t}\|+\|v\|\geq  \delta |\,\theta_{s_t}\in B_{\kappa}(\theta_\star)\big)\\
&\leq {\mathbb{P}}\big( U_{s_t}\not\in B_{ \delta-\tilde{\delta}-\kappa}(0)\big)\\
&\leq  {\mathbb{P}}\big( U_{s_t}\not\in B_{ \kappa}(0)\big).
\end{split}$$ Then, since $h_{s_{t-1}}\rightarrow 0$, it is easily checked using that, under the assumptions on $(\mu_t)_{t\geq 0}$ and with $C^{\mu_{s_t}*\tilde{\pi}_{s_t}}_{\delta, \tilde{\delta}}$ as defined in Lemma \[lemma:denom\], for $t$ large enough we have $$\begin{aligned}
\label{eq:Ct}
C^{\mu_{s_t}*\tilde{\pi}_{s_t}}_{\delta, \tilde{\delta}}\geq {\mathbb{P}}\big( U_{s_t} \in B_{\kappa}(0)\big)\tilde{\pi}_{s_t}(B_{\kappa}(\theta_\star))\geq C^{-1} \tilde{\pi}_{s_t}(B_{\kappa}(\theta_\star)),\quad{\mathbb{P}}-a.s.\\end{aligned}$$

Let $(\phi'_t)_{t\geq 1}$ be as in Lemma \[lemma:part\_1\], $\beta\in(0,1)$ and, for every $t\geq 1$, $\tilde{\phi}_t(Y_{1:t})$ be such that $\tilde{\phi}_t(Y_{1:t})=1$ whenever $\tilde{\pi}_{t} (V_{\kappa})\geq \beta^{-1}$ and such that $\tilde{\phi}_t(Y_{1:t})=0$ otherwise. Notice that ${\mathbb{E}}[\phi'_{\tau_k}(Y_{1:\tau_k})]\rightarrow 0$ by Lemma \[lemma:part\_1\] while ${\mathbb{E}}[\tilde{\phi}_{s_{\tau_k}}(Y_{1:s_{\tau_k}})]\rightarrow 0$ by Lemma \[lemma:online1\].

Therefore, using the shorthand $\Theta_{\tau_k}=\Theta_{\gamma_{\tau_k}, s_{\tau_k}:\tau_k}$, $$\label{eq:inquality11}
\begin{split}
 &\limsup_{k\rightarrow+\infty}  {\mathbb{E}}[\tilde{\pi}_{\tau_k}(V_{\epsilon})]\\
 &\leq \limsup_{k\rightarrow+\infty}{\mathbb{E}}\big[(1-\phi'_{\tau_k}(Y_{1:\tau_k}))(1-\tilde{\phi}_{s_{\tau_k}}(Y_{1:s_{\tau_k}})){\mathds{1}}_{\Theta_{\tau_k}}(U_{s_{\tau_k}:\tau_k})  \pi'_{s_{\tau_k},\tau_k}(V_{\epsilon}) \big]\\
&+\limsup_{k\rightarrow+\infty}{\mathbb{E}}\big[(1-\phi'_{\tau_k}(Y_{1:\tau_k}))(1-\tilde{\phi}_{s_{\tau_k}}(Y_{1:s_{\tau_k}})){\mathds{1}}_{\Theta^c_{\tau_k}}(U_{s_{\tau_k}:\tau_k})  \pi'_{s_{\tau_k},\tau_k}(V_{\epsilon})\big].
\end{split}$$

We now remark that, using Lemma \[lemma:online1\], and Lemma \[lemma:denom\], and because $s_{\tau_k}=t_{p_k}$ for all $k\geq 1$, we can without loss of generality assume that $(\phi'_t)_{t\geq 1}$ is such that, for all $k\geq 1$, $$\label{eq:psi11}
\begin{split}
{\mathbb{P}}\Big(\int_{\Theta}{\mathbb{E}}\Big[&(\mu_{t_{p_k}}*\tilde{\pi}_{t_{p_k}}) \big(\theta-\sum_{s=s_{\tau_k}+1}^{\tau_k-1}U_s\Big)\prod_{s=s_{\tau_k}+1}^{\tau_k} (\tilde{f}_{\theta-\sum_{i=s}^{\tau_{k}-1}U_i}/f_{\theta_\star})(Y_s)\big|{\mathcal{F}}_{\tau_k}\Big]  {\mathrm{d}}\theta \\
 &>  C^{-1}  {\mathbb{P}}(U_{s_{\tau_k}:\tau_k}\in\Theta_{\tilde{\delta}, s_{\tau_k}:\tau_k})\,e^{-C(\tau_k-s_{\tau_k})  \delta^2}\big| \phi'_{\tau_k}(Y_{1:\tau_k})=0\Big)=1.
\end{split}$$

Then, following the computations in the proof of Lemma \[lemma:online1\], with used in place of , we obtain for $k$ large enough $$\label{eq:inquality22}
\begin{split}
{\mathbb{E}}\big[(1-\phi'_{\tau_k}(Y_{1:\tau_k}))(1-\tilde{\phi}_{s_{\tau_k}} (Y_{1:s_{\tau_k}}))&{\mathds{1}}_{\Theta_{\tau_k}}(U_{s_{\tau_k}:\tau_k})  \pi'_{s_{\tau_k},\tau_k} (V_{\epsilon})\big]\\
&\leq   \frac{C e^{-(\tau_k-s_{\tau_k})C^{-1}}}{{\mathbb{P}}(U_{s_{\tau_k}:\tau_k}\in\Theta_{\tilde{\delta}, s_{\tau_k}:\tau_k}) }
\end{split}$$ and $$\label{eq:inquality33}
\begin{split}
 {\mathbb{E}}\big[ (1-\phi'_{\tau_k} (Y_{1:\tau_k}))(1-\tilde{\phi}_{s_{\tau_k}} (Y_{1:s_{\tau_k}})) &{\mathds{1}}_{\Theta^c_{\tau_k}} (U_{s_{\tau_k}:\tau_k})  \pi'_{s_{\tau_k},\tau_k}(V_{\epsilon} )\big]\\
&\leq \frac{C e^{(\tau_k-s_{\tau_k})C}}{{\mathbb{P}}(U_{s_{\tau_k}:\tau_k}\in\Theta_{\tilde{\delta},s_{\tau_k}:\tau_k})} {\mathbb{P}}\big(U_{s_{\tau_k}:\tau_k}\not\in \Theta_{\tau_k}\big).
\end{split}$$

Therefore, using , and , to conclude the proof it is enough to show that $$\begin{aligned}
&e^{-(\tau_k-s_{\tau_k}) C^{-1}}\rightarrow 0\label{eq:lim11}\\
&e^{(\tau_k-s_{\tau_k})C}{\mathbb{P}}\big(U_{s_{\tau_k}:\tau_k}\not\in \Theta_{\gamma_{\tau_k}, s_{\tau_k}:\tau_k}\big)\rightarrow 0\label{eq:lim22}.
$$ Since $\liminf_{k\rightarrow+\infty}(\tau_k-s_{\tau_k})\geq  \liminf_{k\rightarrow+\infty}(v_{p_k+1}-t_{p_k})= +\infty$ it follows that holds. To show remark that $$\begin{aligned}
e^{(\tau_k-s_{\tau_k})C}{\mathbb{P}}\big(U_{s_{\tau_k}:\tau_k}\not\in  \Theta_{\gamma_{\tau_k}, s_{\tau_k}:\tau_k}\big)&=e^{(\tau_k-s_{\tau_k})C}{\mathbb{P}}\big([U_{(s_{\tau_k}+1):\tau_k}]\geq \gamma_{\tau_k} \big)\\
&\leq C e^{(t_{p_k+1}-t_{p_k})C}\exp\Big(-\frac{\gamma_{\tau_k}}{C\sum_{s=t_{p_k}+1}^{\tau_k-1}h_s^2}\Big)\\
&\leq C e^{(t_{p_k+1}-t_{p_k})C}\exp\Big(-\frac{\gamma_{\tau_k}}{C\sum_{s=t_{p_k}+1}^{t_{p_k+1}-1}h_s^2}\Big)\end{aligned}$$ where the first inequality uses . As shown in the proof of Lemma \[lemma:online1\], the term on the r.h.s. of the last inequality sign converges to 0 as $k\rightarrow+\infty$ when $\gamma_{\tau_k}\rightarrow 0$ sufficiently slowly, and thus holds. The proof is complete.

### Proof of Lemma \[lemma:online3\]\[p-lemma:online3\]

Below $C\in(0,+\infty)$ is a constant whose value can change from one expression to another.

For every $p\geq 1$ let $$\begin{aligned}
&\xi_p=1\wedge\Bigg(\frac{\log(h_{t_p}^{-\nu})^{1/2}\wedge (t_p-t_{p-1})^{1/2}}{\log(h_{t_p}^{-\nu})}\Bigg)\\
&v_p=t_{p-1}+ \lfloor   \log(h_{t_{p-1}}^{-\nu\xi_{p-1}})\rfloor \wedge (t_{p}-t_{p-1}-1)\end{aligned}$$ so that $$\begin{aligned}
\label{eq:xi_p}
 \xi_p\rightarrow 0,\quad \log(h_{t_{p-1}}^{-\nu\xi_{p-1}})\rightarrow+\infty,\quad \frac{ \log(h_{t_{p-1}}^{-\nu\xi_{p-1}})}{t_p-t_{p-1}}\rightarrow 0\end{aligned}$$ while $(v_p)_{p\geq 1}$ verifies the conditions of Lemma \[lemma:online2\].

For every $q\geq 1$ let $p_q=\sup\{p\geq 0:\, t_p<\tau'_q\}$ and note that, using ,

$$\label{eq:st}
\begin{split}
\liminf_{q\rightarrow+\infty}(\tau'_q-v_{p_q})&\geq \liminf_{q\rightarrow+\infty}(t_{p_q}-v_{p_q})\\
&= \liminf_{p\rightarrow+\infty}(t_p-v_p)\\
&\geq  \liminf_{p\rightarrow+\infty} (t_p-t_{p-1})\Big(1-\frac{\log(h^{-\nu\xi_{p-1}}_{t_{p-1}})}{t_p-t_{p-1}}\Big)\\
&=+\infty.
\end{split}$$

Note also that, under the assumptions of the lemma, there exists a $p_1\in\mathbb{N}$ such that $$\label{eq:st1}
   \log(h_{t_{p-1}}^{-\nu})^{1/2}>1,\quad\forall p\geq p_1.$$ To simplify the presentation of the proof we assume without loss of generality that $p_1=1$ in what follows.

We now let $(s_t)_{t\geq 0}$ be a sequence in $\mathbb{N}_0$ such that $\inf_{t\geq 1}(t-s_t)\geq 1$, $(t-s_t)\rightarrow+\infty$ and $$s_{\tau'_q}=\big(\tau'_q- \lfloor \log(h_{t_{p_q}}^{-\nu})^{1/2}\rfloor\big)\vee v_{p_q},\quad \forall q\geq 1.$$ Notice that such a sequence $(s_t)_{t\geq 0}$ exists by and because we are assuming that holds with $p_1=1$. Note also that $(s_t)_{t\geq 1}$ is such that $v_{p_q}\leq s_{\tau'_q}\leq t_{p_q}$ for all $q\geq 1$.

We now show that $$\begin{aligned}
\label{eq:gamma_cond}
 {\mathbb{P}}\big([U_{(s_{\tau'_q}+1):\tau'_q}]\geq  \gamma\big)\rightarrow 0,\quad\forall \gamma>0.\end{aligned}$$ To this aim let $\gamma>0$ and remark that $$\label{eq:split_prob}
\begin{split}
{\mathbb{P}}\big([U_{(s_{\tau'_q}+1):\tau'_q}] \geq  \gamma\big)&\leq {\mathbb{P}}\big(\max_{s_{\tau'_q}<s< \tau'_q}\big\|\sum_{i=s}^{\tau'_q-1} U_i-U_{t_{p_q}}{\mathds{1}}(s\leq t_{p_q})\|\geq  \gamma/2\big)\\
&+{\mathbb{P}}\big( \|U_{t_{p_q}}\|\geq  \gamma/2\big)
\end{split}$$ where ${\mathbb{P}}\big( \|U_{t_{p_q}}\|\geq  \gamma/2\big)\rightarrow 0$ since $h_t\rightarrow 0$. In addition, by Doob’s martingale inequality and under the assumptions on $(\mu_t)_{t\geq 0}$, $$\begin{aligned}
&\limsup_{q\rightarrow+\infty}{\mathbb{P}}\Big(\max_{s_{\tau'_q}<s<\tau'_q}\big\|\sum_{i=s}^{\tau'_q-1} U_i-U_{t_{p_q}}{\mathds{1}}(s\leq t_{p_q})\|\geq  \frac{\gamma}{2}\Big)\leq C\limsup_{q\rightarrow+\infty}\frac{\sum_{s=t_{p_q-1}+1}^{t_{p_{q+1}}-1}h_s^2}{(\gamma/2)^2}=0.\end{aligned}$$ Hence, holds showing that there exists a sequence $(\gamma_t)_{t\geq 1}$ in ${\mathbb{R}}_{>0}$ such that $$\begin{aligned}
\label{eq:gamma_t22}
\gamma_t\rightarrow 0,\quad  {\mathbb{P}}\big([U_{(s_{\tau'_q}+1):\tau'_q}]\geq  \gamma_{\tau'_q}\big)\rightarrow 0.\end{aligned}$$

Let $(\gamma_t)_{t\geq 1}$ be as in , $(\tau_k)_{k\geq 1}$ be as defined in Lemma \[lemma:online2\], $(\tilde{\tau}_r)_{r\geq 1}$ be a strictly increasing sequence in $\mathbb{N}_0$ such that $(\tilde{\tau}_r)_{r\geq 1}= (\tau_k)_{k\geq 1}\cup(t_p)_{p\geq 0}$ and note that, by Lemmas \[lemma:online1\]-\[lemma:online2\] , $$\begin{aligned}
\label{eq:conv_sub2}
{\mathbb{E}}[\tilde{\pi}_{\tilde{\tau}_r}(V_{\kappa})]\rightarrow 0,\quad\forall\kappa>0.\end{aligned}$$

Remark also that, by construction, $(s_{\tau'_q})_{q\geq 1}\subset (\tilde{\tau}_r)_{r\geq 1}$ so that we can now follow the computations in the proof of Lemma \[lemma:online2\]. As in this latter let $\tilde{\delta}=\kappa$ for some sufficiently small $\kappa>0$ (see the proof of Lemma \[lemma:online2\] for the expression of $\kappa$). Then, as shown in the proof of Lemma \[lemma:online2\] we have, $$\begin{aligned}
\limsup_{q\rightarrow+\infty}{\mathbb{E}}[\tilde{\pi}_{\tau'_q}(V_{\epsilon})]&\leq \limsup_{q\rightarrow+\infty}     \frac{C\,e^{-(\tau'_q-s_{\tau'_q}) C^{-1}}}{{\mathbb{P}}(U_{s_{\tau'_q}:\tau'_q}\in\Theta_{\tilde{\delta}, s_{\tau'_q}:\tau'_q}) }\\
&+\limsup_{q\rightarrow+\infty}  \frac{C\, e^{(\tau'_q-s_{\tau'_q})C}}{{\mathbb{P}}(U_{s_{\tau'_q}:\tau'_q}\in\Theta_{\tilde{\delta},s_{\tau'_q}:\tau'_q})} {\mathbb{P}}\big(U_{s_{\tau'_q}:\tau'_q}\not\in \Theta_{\gamma_{\tau'_q},s_{\tau'_q}:\tau'_q}\big)\end{aligned}$$ so that to conclude the proof it is enough to show that $$\begin{aligned}
&e^{(\tau'_q-s_{\tau'_q})C^{-1}}\rightarrow 0\label{eq:lim111}\\
&e^{(\tau'_q-s_{\tau'_q})C}{\mathbb{P}}\big(U_{(s_{\tau'_q}+1):\tau'_q}\not\in \Theta^{\gamma_{\tau'_q}}_{s_{\tau'_q},\tau'_q}\big)\rightarrow 0\label{eq:lim222}.
$$ Since $(\tau'_q-s_{\tau'_q})\rightarrow +\infty$ it follows that holds.

To show remark that $$\label{eq:decomp_final}
\begin{split}
e^{(\tau'_q-s_{\tau'_q})C}{\mathbb{P}}\big(U_{s_{\tau'_q}:\tau'_q}&\not\in \Theta_{\gamma_{\tau'_q},s_{\tau'_q}:\tau'_q}\big)\\
&=e^{(\tau'_q-s_{\tau'_q})C}{\mathbb{P}}\big([U_{(s_{\tau'_q}+1):\tau'_q}]\geq \gamma_{\tau'_q}\big) \\
&\leq  e^{(\tau'_q-s_{\tau'_q})C} {\mathbb{P}}\big(\max_{s_{\tau'_q}<s< \tau'_q}\big\|\sum_{i=s}^{\tau'_q-1} U_i-U_{t_{p_q}}{\mathds{1}}(s\leq t_{p_q})\|\geq  \gamma_{\tau'_q}/2\big)\\
&+e^{(\tau'_q-s_{\tau'_q})C}{\mathbb{P}}\big( \|U_{t_{p_q}}\|\geq  \gamma_{\tau'_q}/2\big).
\end{split}$$ In addiiton, $$\begin{aligned}
\label{eq:in_st}
\tau'_q-s_{\tau'_q}\leq  \log(h_{t_{p_q}}^{-\nu\xi_{p_q}}) \leq 2(t_{p_q+1}-t_{p_{q}}),\end{aligned}$$ where the second inequality holds by and for $q$ large enough.

Therefore, using , $$\label{eq:decomp_final1}
\begin{split}
\limsup_{q\rightarrow+\infty} &\,e^{(\tau'_q-s_{\tau'_q})C}{\mathbb{P}}\big(\max_{s_{\tau'_q}<s< \tau'_q}\big\|\sum_{i=s}^{\tau'_q-1} U_i-U_{t_{p_q}}{\mathds{1}}(s\leq t_{p_q})\|\geq  \gamma_{\tau'_q}/2\big)\\
&\leq C \limsup_{q\rightarrow+\infty}  e^{(t_{p_q+1}-t_{p_{q}})C}\exp\bigg(-\frac{\gamma_{\tau'_q}}{C\sum_{s=t_{p_q-1}+1}^{t_{p_q+1}-1}h_s^2}\bigg)\\
&=0
\end{split}$$ where the equality holds assuming without loss of generality that $\gamma_{\tau'_q}\rightarrow 0$ sufficiently slowly and uses the fact that, by assumption, $$\limsup_{p\rightarrow+\infty}\frac{t_{p+2}-t_{p+1}}{ t_{p+1}-t_{p}}<+\infty.$$

Next, using standard tails estimate for the Student’s $t$-distribution, and recalling that we are assuming $\nu>1$, we have $$\begin{aligned}
{\mathbb{P}}\big( \|U_{t_{p}}\|\geq  \gamma\big) \leq C \frac{\nu}{\nu-1}\frac{h_{t_p}}{ \gamma}\Big(1+\frac{ \gamma^2}{h_{t_p}^2\nu}\Big)^{-\frac{\nu-1}{2}} \leq C\Big(\frac{h_{t_p}}{ \gamma}\Big)^\nu,\quad\forall p\geq 1,\quad\forall \gamma>0.\end{aligned}$$ Remark also that, using the first inequality in and recalling that $\xi_p\rightarrow 0$, $$\begin{aligned}
\limsup_{p\rightarrow+\infty}  h_{t_{p_q}}^\nu e^{(\tau'_p-s_{\tau'_p})C}&\leq\limsup_{p\rightarrow+\infty}  h_{t_{p_q}}^{\nu(1-C \xi_{p_q})}=0
 \end{aligned}$$ and therefore, assuming without loss of generality that $\gamma_{\tau'_q}\rightarrow 0$ sufficiently slowly, $$\limsup_{p\rightarrow+\infty} e^{(\tau'_q-s_{\tau'_q})C} {\mathbb{P}}\big( \|U_{t_{p_q}}\|\geq   \gamma_{\tau'_q}/2\big)=0.$$ Together with and this last result shows that holds. The proof is complete.

Proof of Theorem \[thm:lower\_bound\_1\]\[p-thm:lower\_bound\_1\]
-----------------------------------------------------------------

We first show the following simple result.

\[lemma:simple\] Let $(t_p)_{p\geq 0}$ be a subsequence of $\mathbb{N}_0$ and $g:{\mathsf{Y}}\rightarrow{\mathbb{R}}$ be a measurable function such that ${\mathbb{E}}[|g(Y_1)]|<+\infty$. Then, as $p\rightarrow+\infty$, $${\mathbb{P}}\Big(\max_{0<i\leq p}\frac{1}{t_i-t_{i-1}}\Big|\sum_{s=t_{i-1}+1}^{t_i}\big(g(Y_s)-{\mathbb{E}}[g(Y_1)]\big)\Big|\geq  t^\delta_{p}\Big)\rightarrow 0,\quad\forall \delta>1.$$

Let $\delta>1$ so that, using Markov’s inequality for the second inequality, $$\begin{aligned}
\limsup_{p\rightarrow+\infty}{\mathbb{P}}\Big(&\max_{0< i\leq p}\frac{1}{t_i-t_{i-1}}\Big|\sum_{s=t_{i-1}+1}^{t_i}\big(g(Y_s)-{\mathbb{E}}[g(Y_1)]\big)\Big|\geq  t^\delta_{p}\Big)\\
&\leq \limsup_{p\rightarrow+\infty}\sum_{i=1}^p{\mathbb{P}}\Big(  \frac{1}{t_i-t_{i-1}}\Big|\sum_{s=t_{i-1}+1}^{t_i}\big(g(Y_s)-{\mathbb{E}}[g(Y_1)]\big)\Big|\geq  t^\delta_{p}\Big)\\
&\leq \limsup_{p\rightarrow+\infty}\frac{2{\mathbb{E}}[|g(Y_1)|]}{t^{\delta-1}_{p}}\\
&=0.\end{aligned}$$

We also recall the following result [@ferger2014 Theorem 1.2].

\[lemma:mean\_p\] Let $(X_i)_{i\geq 1}$ be a sequence of i.i,d. random variables such that ${\mathbb{E}}[X_1]=0$ and ${\mathbb{E}}[X_1^{2p}]<+\infty$ for some $p\in\mathbb{N}$. Then ${\mathbb{E}}\big[(\sum_{i=1}^n X_i)^{2p}\big]={\mathcal{O}}(n^p)$.

Let $\beta>\max( \beta_\star,1/\alpha)$ and $D_p= h_{t_{p}}^{-\beta}$ for all $p\geq 1$.

We first establish the result of the theorem under \[new2\].\[A61\], and start by showing that $$\begin{aligned}
\label{eq:conv_max}
 {\mathbb{P}}\Big(\max_{0< i\leq p} e^{-(t_i-t_{i-1})\zeta(D_p)/2}\prod_{s=t_{i-1}+1}^{t_i}\sup_{\theta_s\in V_{D_p}}(\tilde{f}_{\theta_{s}}/f_{\theta_\star})(Y_s)< 1\Big)\rightarrow 1.\end{aligned}$$

Let $p_1\in\mathbb{N}$ be such that $\zeta(D_p)<0$ for all $p\geq p_1$; notice that such a $p_1$ exists under \[new2\].\[A61\]. Then, for every $p\geq p_1$ we have $$\label{eq:conv_max_1}
\begin{split}
{\mathbb{P}}\Big(& \max_{0< i\leq p} e^{-(t_i-t_{i-1})\zeta(D_p)/2}\prod_{s=t_{i-1}+1}^{t_i}\sup_{\theta_s\in V_{D_p}}(\tilde{f}_{\theta_{s}}/f_{\theta_\star})(Y_s)\geq  1\Big)\\
&={\mathbb{P}}\Big(\max_{0< i\leq p} \sum_{s=t_{i-1}+1}^{t_i}\Big\{\sup_{\theta_s\in V_{D_p}}\log \Big(\frac{\tilde{f}_{\theta_{s}}}{f_{\theta_\star}}(Y_s)\Big)-\zeta(D_p)+\frac{\zeta(D_p)}{2}\Big\}\geq  0\Big)\\
&\leq \sum_{i=1}^p{\mathbb{P}}\Big( \sum_{s=t_{i-1}+1}^{t_i}\Big(\sup_{\theta_s\in V_{D_p}}\log \Big(\frac{\tilde{f}_{\theta_{s}}}{f_{\theta_\star}}(Y_s)\Big)- \zeta(D_p )\Big)> (t_i-t_{i-1})|\zeta(D_p)|/2\Big)\\
&=\sum_{i=1}^p{\mathbb{P}}\Big(\frac{1}{t_i-t_{i-1}} \sum_{s=t_{i-1}+1}^{t_i}\Big(\sup_{\theta_s\in V_{D_p}}\log \Big(\frac{\tilde{f}_{\theta_{s}}}{f_{\theta_\star}}(Y_s)\Big)- \zeta(D_p )\Big)> |\zeta(D_p)|/2\Big).
\end{split}$$ Using Markov’s inequality, \[new2\].\[A61\] and Lemma \[lemma:mean\_p\], there exists a constant $C\in(0,+\infty)$ such that, for all $i\in 1:p$, $$\label{eq:conv_max_2}
\begin{split}
{\mathbb{P}}\Big(&\frac{1}{t_i-t_{i-1}} \sum_{s=t_{i-1}+1}^{t_i}\Big(\sup_{\theta_s\in V_{D_p}}\log \Big(\frac{\tilde{f}_{\theta_{s}}}{f_{\theta_\star}}(Y_s)\Big)- \zeta(D_p )\Big)> |\zeta(D_p)|/2\Big)\\
&\leq 2^{2k_\star}\frac{{\mathbb{E}}\Big[\Big|\frac{1}{t_i-t_{i-1}} \sum_{s=t_{i-1}+1}^{t_i}\big(\sup_{\theta_s\in V_{D_p}}\log\big( (\tilde{f}_{\theta_{s}}/f_{\theta_\star})(Y_s)\big)- \zeta(D_p) \big)\Big|^{2k_\star}\Big]}{ \zeta(D_p)^{2 k_\star}}\\
&\leq  \frac{C }{ (t_i-t_{i-1})^{k_\star{\mathds{1}}_{\mathbb{N}}(k_\star)} |\zeta(D_p)|^{2k_\star}}.
\end{split}$$ By assumptions, $|\zeta(D_p)|^{-2 k_\star}\sum_{i=1}^p (t_i-t_{i-1})^{-k_\star{\mathds{1}}_{\mathbb{N}}(k_\star)}\rightarrow 0$ which, together with and , implies .

To proceed further let $\tilde{C}_{\star}\in(0,+\infty)$ be as in Lemma \[lemma:denom\], $\delta_p=t_p^{-\gamma}$ for some $\gamma>\max(\alpha,1/2)$, $\delta>0$ and, for all $i\geq 1$ and $s\in t_{i-1}:(t_i-1)$, let $V_{i,s}=\sum_{j=s}^{t_i-1}U_j$.

Under \[m\_star\]-\[taylor\] and by Lemma \[lemma:denom\], for $p$ large enough (i.e. for $\delta_p$ small enough) and all $i\in 1:p$, we have $$\begin{split}
g_{i}(Y_{1:t_i})&:=\int_{\Theta} {\mathbb{E}}\big[ (\mu_{t_{i-1}}*\tilde{\pi}_{t_{i-1}})(\theta-V_{i,t_{i-1}+1})\prod_{s=t_{i-1}+1}^{t_i}(\tilde{f}_{\theta-V_{i,s}}/f_{\theta_\star})(Y_s)\big|{\mathcal{F}}_{t_i}\big]{\mathrm{d}}\theta\\
&\leq    \frac{{\mathbb{P}}( U_{ t_{i-1}:t_i} \in\Theta_{\delta_p, t_{i-1}:t_i})}{  e^{(t_i-t_{i-1})(2(\delta_p\tilde{C_\star})^2+\delta)}} \inf_{v\in B_{\delta_p}(0)}(\mu_{t_{i-1}}*\tilde{\pi}_{t_{i-1}})(B_{\delta_p}(\theta_\star-v))\\
&=:\tilde{g}_{i,p}(Y_{1:t_i})
\end{split}$$ with probability at most $p_{i,p}:=\big( (t_i-t_{i-1})\big\{ (\tilde{C}_{\star} \delta_p) +(\tilde{C}_{\star} \delta_p)^{-1} \delta \big\}^{2}\big)^{-1}$.

Then, noting that $\sum_{i=1}^p   (t_i-t_{i-1})^{-1}\leq t_p$, it follows that $$\label{eq:conv_den2}
\begin{split}
\limsup_{p\rightarrow+\infty}{\mathbb{P}}\Big(\min_{1\leq i\leq p}\frac{g_{i}(Y_{1:t_i})}{\tilde{g}_{i,p}(Y_{1:t_i})} \leq 1\Big)& \leq \limsup_{p\rightarrow+\infty} \sum_{i=1}^p p_{i,p}\\
&\leq \limsup_{p\rightarrow+\infty} \frac{\delta_p^{2}\, \tilde{C}_{\star}^2}{\delta^2}\sum_{i=1}^p (t_i-t_{i-1})^{-1}\\
& \leq \frac{ \tilde{C}_{\star}^2}{\delta^2} \limsup_{p\rightarrow+\infty} t_p^{-2\gamma} t_p\\
&=0
\end{split}$$ where the last equality holds since $\gamma>1/2$.

Therefore, by and , there exists a sequence $({\mathsf{Y}}_p)_{p\geq 1}$, with ${\mathsf{Y}}_p\subset{\mathcal{Y}}^{\otimes t_p}$ for all $p\geq 1$, such that ${\mathbb{P}}(Y_{1:t_p}\in{\mathsf{Y}}_p)\rightarrow 1$ and such that, for all $p\geq 1$, $$\begin{aligned}
 \max_{0< i\leq p} e^{-(t_i-t_{i-1})\zeta(D_p)/2}\prod_{s=t_{i-1}+1}^{t_i}\sup_{\theta_s\in V_{D_p}}\frac{\tilde{f}_{\theta_{s}}}{f_{\theta_\star}}(y_s)< 1,\quad \min_{1\leq i\leq p}\frac{g_{i}(y_{1:t_i})}{\tilde{g}_{i,p}(y_{1:t_i})}>1,\quad \forall y_{1:t_p}\in{\mathsf{Y}}_p.\end{aligned}$$

To proceed further let $\epsilon\in(0,1)$ and note that, since ${\mathbb{P}}(Y_{1:t_p}\in{\mathsf{Y}}_p)\rightarrow 1$, $$\begin{aligned}
\label{eq:res_1}
\limsup_{p\rightarrow+\infty}{\mathbb{P}}(\tilde{\pi}_{t_p}(V_{2D_p})\geq \epsilon)\leq  \limsup_{p\rightarrow+\infty}{\mathbb{P}}\big(\{\tilde{\pi}_{t_p}(V_{2D_p})\geq \epsilon\}\cap \{Y_{1:t_p}\in{\mathsf{Y}}_p\}\big)\end{aligned}$$ so that to prove the theorem under \[new2\].\[A61\] it remains to show that $$\limsup_{p\rightarrow+\infty}{\mathbb{P}}\big(\{\tilde{\pi}_{t_p}(V_{2D_p})\geq \epsilon\}\cap \{Y_{1:t_p}\in{\mathsf{Y}}_p\}\big)=0.$$

To this aim let $p_0\in\mathbb{N}$ and, for every $p>p_0$, let $({\mathsf{Y}}_{i,p})_{i=0}^p$ be a sequence in ${\mathcal{Y}}^{\otimes t_p}$ such that ${\mathsf{Y}}_{0,p}={\mathsf{Y}}^{t_p}$, ${\mathsf{Y}}_{p,p}={\mathsf{Y}}_p$ and such that, for every $i\in 1:p$, ${\mathsf{Y}}_{i,p}\subset \cap_{j=0}^{i-1}{\mathsf{Y}}_{j,p}$ and $$\begin{aligned}
 \max_{0< j\leq i} e^{-(t_j-t_{j-1})\zeta(D_p)/2}\prod_{s=t_{j-1}+1}^{t_j}\sup_{\theta_s\in V_{D_p}}\frac{\tilde{f}_{\theta_{s}}}{f_{\theta_\star}}(y_s)< 1,\quad \min_{1\leq j\leq i}\frac{g_{j}(y_{1:t_j})}{\tilde{g}_{j,p}(y_{1:t_j})}>1,\quad \forall y_{1:t_p}\in{\mathsf{Y}}_{i,p}.\end{aligned}$$ Let $A_{i,p}=\{\tilde{\pi}_{t_i}(V_{2D_p})\geq \epsilon\}\cap\{Y_{1:t_p}\in Y_{i,p}\}$ and $\tilde{A}_{i,p}=\{\tilde{\pi}_{t_i}(V_{2D_p}){\mathds{1}}_{{\mathsf{Y}}_{i,p}}(Y_{1:t_i})\geq \epsilon\}$ for all $p>p_0$ and all $i\in 1:p$.

Then, for every $p>p_0$ we have $$\label{eq:p_split1}
\begin{split}
{\mathbb{P}}\big(\{\tilde{\pi}_{t_p}(V_{2D_p})\geq \epsilon\} \cap   \{Y_{1:t_p}\in{\mathsf{Y}}_p\}\big)&={\mathbb{P}}(A_{p,p})\\
&\leq {\mathbb{P}}(A_{p,p}| A^c_{p-1,p})+{\mathbb{P}}(A_{p-1,p})\\
&\leq \sum_{i=p_0+1}^p {\mathbb{P}}(A_{i,p}| A^c_{i-1,p})+{\mathbb{P}}(A_{p_0,p})\\
&=  \sum_{i=p_0+1}^p {\mathbb{P}}(\tilde{A}_{i,p}| A^c_{i-1,p})+{\mathbb{P}}(A_{p_0,p})
\end{split}$$ and we now study ${\mathbb{P}}(\tilde{A}_{i,p}| A^c_{i-1,p})$ for all $i\in (p_0+1):p$.

Let $i\in (p_0+1):p$ and $$\begin{aligned}
&X_{i,p}^{(1)}= \int_{V_{2D_p}} {\mathbb{E}}\Big[{\mathds{1}}([U_{(t_{i-1}+1):t_i}]<D_p) (\mu_{t_{i-1}}*\tilde{\pi}_{t_{i-1}})(\theta-V_{i,t_{i-1}+1})\prod_{s=t_{i-1}+1}^{t_i} \frac{\tilde{f}_{\theta-V_{i,s}}}{f_{\theta_\star}}(Y_s)\big|{\mathcal{F}}_{t_i}\Big]{\mathrm{d}}\theta\\
&X_{i,p}^{(2)}= \int_{V_{2D_p}} {\mathbb{E}}\Big[{\mathds{1}}([U_{(t_{i-1}+1):t_i}]\geq D_p) (\mu_{t_{i-1}}*\tilde{\pi}_{t_{i-1}})(\theta-V_{i,t_{i-1}+1})\prod_{s=t_{i-1}+1}^{t_i} \frac{\tilde{f}_{\theta-V_{i,s}}}{f_{\theta_\star}}(Y_s)\big|{\mathcal{F}}_{t_i}\Big]{\mathrm{d}}\theta\end{aligned}$$ so that, by Lemma \[lemma:pi\], $$\tilde{\pi}_{t_i}(V_{2D_p})=\frac{X_{i,p}^{(1)}+X_{i,p}^{(2)}}{g_i(Y_{1:t_i})},\quad{\mathbb{P}}-a.s.$$ Then, by Markov’s inequality and using the definition of ${\mathsf{Y}}_{i,p}$, we have $$\begin{aligned}
{\mathbb{P}}(\tilde{A}_{i,p}| A^c_{i-1,p})&\leq \epsilon^{-1}{\mathbb{E}}\bigg[\frac{(X_{i,p}^{(1)}+X_{i,p}^{(2)}){\mathds{1}}_{{\mathsf{Y}}_{i,p}}(Y_{1:t_p})}{g_i(Y_{1:t_i})}\big|A^c_{i-1,p}\bigg]\\
& \leq\epsilon^{-1}{\mathbb{E}}\bigg[\frac{(X_{i,p}^{(1)}+X_{i,p}^{(2)}){\mathds{1}}_{{\mathsf{Y}}_{i,p}}(Y_{1:t_p})}{\tilde{g}_{i,p}(Y_{1:t_i})}\big|A^c_{i-1,p}\bigg]\\
&=\epsilon^{-1}{\mathbb{E}}\bigg[\frac{(X_{i,p}^{(1)}+X_{i,p}^{(2)}){\mathds{1}}_{{\mathsf{Y}}_{i,p}}(Y_{1:t_p})}{\tilde{g}_{i,p}(Y_{1:t_i})}\big|\, \tilde{\pi}_{t_{i-1}}(V_{2D_p})<\epsilon\bigg]\end{aligned}$$ where the equality uses the definition of $ A_{i-1,p}$ and the fact that $Y_{i,p}\subset Y_{i-1,p}$.

Following similar computations as in the proof of Lemma \[lemma:online1\] (see ), it is readily checked that there exists a constant $C_1\in(0,+\infty)$ such that $$\begin{aligned}
{\mathbb{P}}\Big(\inf_{v\in B_{\delta_p}(0)}(\mu_{t_{i-1}}*\tilde{\pi}_{t_{i-1}})(B_{\delta_p}(\theta_\star-v))  \geq C_1^{-1} \delta_p^d h_{t_{i-1}}^\nu h_{t_{p-1}}^{\beta(\nu+d)}\big|\, \tilde{\pi}_{t_{i-1}}(V_{2D_p})<\epsilon\Big)=1\end{aligned}$$ and thus $$\label{eq:p_split2}
\begin{split}
 {\mathbb{P}}(\tilde{A}_{i,p}| A^c_{i-1,p})&\leq \frac{C_1  \delta_p^{-d} h_{t_{i-1}}^{-\nu} h_{t_{p-1}}^{-\beta(\nu+d)}e^{(t_i-t_{i-1})(2(\delta_p\tilde{C_\star})^2+\delta)}}{\epsilon\,{\mathbb{P}}( U_{ t_{i-1}:t_i} \in\Theta_{\delta_p, t_{i-1}:t_i})}\\
&\qquad\times {\mathbb{E}}\big[ (X_{i,p}^{(1)}+X_{i,p}^{(2)}) {\mathds{1}}_{{\mathsf{Y}}_{i,p}}(Y_{1:t_i})|\, \tilde{\pi}_{t_{i-1}}(V_{2D_p})<\epsilon\big].
\end{split}$$ Next, noting that $X_{i,p}^{(1)}\leq \prod_{s=t_{i-1}+1}^{t_i}\sup_{\theta_s\in V_{D_p}}(\tilde{f}_{\theta_s}/f_{\theta_\star})(Y_s)$, ${\mathbb{P}}$-a.s., it follows, by the definition of ${\mathsf{Y}}_{i,p}$, that $$\begin{aligned}
\label{eq:p_split3}
{\mathbb{E}}[X_{i,p}^{(1)}{\mathds{1}}_{{\mathsf{Y}}_{i,p}}(Y_{1:t_i})|\, \tilde{\pi}_{t_{i-1}}(V_{2D_p})<\epsilon ]\leq e^{(t_i-t_{i-1})\zeta(D_p)/2}.\end{aligned}$$

We now show that under \[new1\] there exists a constant $C_2\in(0,+\infty)$ such that $$\label{eq:p_split4}
\begin{split}
 {\mathbb{E}}[X_{i,p}^{(2)}{\mathds{1}}_{{\mathsf{Y}}_{i,p}}(Y_{1:t_i})| \, \tilde{\pi}_{t_{i-1}}(V_{2D_p})<\epsilon]\leq  C_2 e^{-C^{-1}_2(t_i-t_{i-1})D_p^2}.
\end{split}$$

As preliminary computations to establish remark that, under the assumptions on $(\mu_t)_{t\geq 0}$ and for some constant $C_3\in(0,+\infty)$ we have, by , $$\label{eq:p_split5}
\begin{split}
{\mathbb{P}}\big([U_{(t_{i-1}+1):t_i}]\geq D_p\big)&\leq C_3\exp\Big(-\frac{D_p^2}{C_3\sum_{j=t_{i-1}+1}^{t_i-1}h_j^2}\Big)\\
&=C_3\exp\Big(-\frac{(t_i-t_{i-1})D_p^2}{C_3(t_i-t_{i-1})\sum_{j=t_{i-1}+1}^{t_i-1}h_j^2}\Big)\\
&\leq C_4\exp\big(-C_4^{-1}(t_i-t_{i-1})D_p^2\big)
\end{split}$$ where the last inequality holds for some constant $C_4\in(0,+\infty)$ since, by assumption, $(t_p-t_{p-1})\sum_{i=t_{p-1}+1}^{t_p-1}h_i^2\rightarrow 0$.

Assume first \[new1\].\[A51\] and recall that $\beta$ is such that $\beta\alpha>1$. Then, by Lemma \[lemma:simple\], for $p_0$ large enough, we can without loss of generality assume that, for all $p'>p_0$ and $j\in (p_0+1):p'$, the set ${\mathsf{Y}}_{j,p'}$ is such that $$\begin{aligned}
\sum_{s=t_{j-1}+1}^{t_j}\sup_{\theta\in\Theta}\log\big((\tilde{f}_\theta/f_{\theta_\star})(y_s)\big) \leq  (t_j-t_{j-1}) D_{p'},\quad\forall y_{1:t_{p'}}\in{\mathsf{Y}}_{j,p'}.\end{aligned}$$ Then, using , $$\begin{split}
 {\mathbb{E}}[X_{i,p}^{(2)}{\mathds{1}}_{{\mathsf{Y}}_{i,p}}(Y_{1:t_i})| \, \tilde{\pi}_{t_{i-1}}(V_{2D_p})<\epsilon]&\leq {\mathbb{P}}\big([U_{(t_{i-1}+1):t_i}]\geq D_p\big)e^{ (t_i-t_{i-1}) D_p}\\
& \leq C_4 e^{-(t_i-t_{i-1})(C_4^{-1}D^2_p- D_p)}
\end{split}$$ which establishes .

Assume now \[new1\].\[A52\] and let $c_\star=\log\big(\sup_{\theta\in\Theta}{\mathbb{E}}[f_\theta/f_{\theta_\star}]\big)<+\infty$. Then, using we have, ${\mathbb{P}}$-a.s. $$\begin{split}
 {\mathbb{E}}[X_{i,p}^{(2)}|{\mathcal{F}}_{t_{i-1}}]&\leq {\mathbb{P}}\big([U_{(t_{i-1}+1):t_i}]\geq D_p\big)\big(\sup_{\theta\in\Theta}{\mathbb{E}}[f_{\theta}/f_{\theta_\star}]\big)^{t_i-t_{i-1}}\\
&= C_4 e^{-(t_i-t_{i-1})(C_4^{-1}D_p^2-c_\star)}
\end{split}$$ which shows .

Lastly, assume \[new1\].\[A53\]. Let $c_\star=\log(\sup_{\theta\in\Theta}{\mathbb{E}}[ f_{\theta}])<+\infty$ and note that, by Lemma \[lemma:simple\], for $p_0$ large enough we can without loss of generality assume that, for all $p'>p_0$ and $j\in (p_0+1):p'$, the set ${\mathsf{Y}}_{j,p'}$ is such that $$\begin{aligned}
\sum_{s=t_{j-1}+1}^{t_j}\log (f_{\theta_\star}(y_s))\geq  - (t_j-t_{j-1})D_{p'},\quad\forall y_{1:t_{p'}}\in{\mathsf{Y}}_{j,p'}.\end{aligned}$$ Then, using we have, ${\mathbb{P}}$-a.s. $$\begin{split}
 {\mathbb{E}}[X_{i,p}^{(2)}|{\mathcal{F}}_{t_{i-1}}]&\leq e^{D_p(t_i-t_{i-1})}{\mathbb{P}}\big([U_{(t_{i-1}+1):t_i}]\geq D_p\big)\big(\sup_{\theta\in\Theta}{\mathbb{E}}[ f_{\theta}]\big)^{t_i-t_{i-1}}\\
&= C_4 e^{-(t_i-t_{i-1})(C_4^{-1}D_p^2-D_p-c_\star)}
\end{split}$$ showing . This concludes to show that holds under \[new1\].

Then, using -, we have $$\label{eq:p_split6}
\begin{split}
{\mathbb{P}}(\tilde{A}_{i,p}| A^c_{i-1,p})&\leq C_1  \delta_p^{-d} h_{t_{i-1}}^{-\nu} h_{t_{p-1}}^{-\beta(\nu+d)}\frac{ e^{(t_i-t_{i-1})(2(\delta_p\tilde{C}_\star)^2+\delta+\zeta(D_p)/2)}}{\epsilon\,{\mathbb{P}}( U_{ t_{i-1}:t_i} \in\Theta_{\delta_p, t_{i-1}:t_i})}\\
&+C_1 C_2 \delta_p^{-d} h_{t_{i-1}}^{-\nu} h_{t_{p-1}}^{-\beta(\nu+d)}\frac{e^{(t_i-t_{i-1})(2(\delta_p\tilde{C}_\star)^2+\delta-C_2^{-1}D_p^2)}}{\epsilon\,{\mathbb{P}}( U_{ t_{i-1}:t_i} \in\Theta_{\delta_p, t_{i-1}:t_i})}.
\end{split}$$

We now find a lower bound for ${\mathbb{P}}( U_{ t_{i-1}:t_i} \in\Theta_{\delta_p, t_{i-1}:t_i})$. To this aim remark that, using the shorthand $\tilde{\delta}_p:=d^{-1/2}\delta_p$, $$\label{eq:int_p}
\begin{split}
{\mathbb{P}}(& U_{ t_{i-1}:t_i} \in\Theta_{\delta_p, t_{i-1}:t_i})\\
&={\mathbb{P}}\Big(\big\|\sum_{j=s}^{t_i-1}U_s\big\|<\delta_p,\,\forall s\in t_{i-1}:(t_i-1)\Big)\\
&\geq {\mathbb{P}}\Big(\big\|\sum_{j=s}^{t_i-1}U_s\big\|_\infty<\tilde{\delta}_p,\,\forall s\in t_{i-1}:(t_i-1)\Big)\\
&={\mathbb{P}}\big(\big\|U_{t_{i}-1}\big\|_\infty<\tilde{\delta}_p\big) \prod_{s=t_{i-1}}^{t_{i}-2} {\mathbb{P}}\Big(\big\|U_s+\sum_{j=s+1}^{t_i-1}U_s\big\|_\infty<\tilde{\delta}_p\,\Big|\,\big\|\sum_{j=s+1}^{t_i-1}U_s\big\|_\infty<\tilde{\delta}_p\Big).
\end{split}$$

Recall now that $\delta_p=t_p^{-\gamma}$ with $\gamma>\alpha$, implying that $\delta_p/h_{t_p}\rightarrow 0$. Hence, under the assumptions on $(\mu_t)_{t\geq 0}$ there exists a constant $c_1>0$ such that, for every $s\in 1:p$ such that $h_s\neq 0$ and every $ v \in{\mathbb{R}}^d$ such that $\|v \|_\infty<\tilde{\delta}_p$, we have $$\begin{aligned}
{\mathbb{P}}\Big(\big\|U_s +\sum_{j=s+1}^{t_i-1}U_s\big\|_\infty<\tilde{\delta}_p\,\Big|\,\big\|\sum_{j=s+1}^{t_i-1}U_s\big\|_\infty=v\Big)&={\mathbb{P}}( \|U_s+v\|_\infty<\tilde{\delta}_p)\\
&={\mathbb{P}}(-\tilde{\delta}_p\leq U_{s}+v <\tilde{\delta}_p)\\
&={\mathbb{P}}\Big(-\frac{\tilde{\delta}_p+v}{h_s}\leq \tilde{U}_s <\frac{\delta_p-v}{h_s}\Big)\\
&\geq c_1\,(\tilde{\delta}_p/h_s)^d.\end{aligned}$$ Consequently, for every $s\in t_{i-1}:(t_i-2)$ such that $h_s\neq 0$ we have $$\label{eq:int_p1}
\begin{split}
 {\mathbb{P}}\Big(\big\|U_s +\sum_{j=s+1}^{t_i-1}U_s\big\|_\infty<\tilde{\delta}_p\,\Big|\,\big\|\sum_{j=s+1}^{t_i-1}U_s\big\|_\infty<\tilde{\delta}_p\Big) &\geq  c_1\,(2\tilde{\delta}^2_p/h_s)^d.
\end{split}$$ Using the above computations we also have ${\mathbb{P}}\big(\|U_{t_{i}-1}\|_\infty<\tilde{\delta}_p\big)\geq c'_1\,(\tilde{\delta}_p/h_{t_i-1})^d$ for some constant $c_1'>0$ which, together with -, shows that for some constant $c_2>0$, $$\label{eq:p_split7}
{\mathbb{P}}( U_{ t_{i-1}:t_i} \in\Theta_{\delta_p, t_{i-1}:t_i}) \geq (c_2 \delta^{2d}_p)^{t_i-t_{i-1}}\prod_{s=t_{i-1}}^{t_i-1} \big(h_s^{-d}\wedge 1/c_2\big)\geq  \delta_p^{2d(t_i-t_{i-1})}$$ where the second inequality assumes without loss of generality that $p_0$ is such that $h^d_s<c_2$ for all $s>t_{p_0}$.

Combining and , and recalling that $\delta_p\rightarrow 0$, that $|\zeta(D_p)|\rightarrow+\infty$ and that $h_{t_{p}}<h_{t_{p-1}}$, it follows that for $p_0$ large enough we have, for all $p>p_0$ and all $i\in(p_0+1):p$, $$\label{eq:p_split8}
\begin{split}
{\mathbb{P}}(\tilde{A}_{i,p} | A^c_{i-1,p})&\leq C_1\epsilon^{-1}  h_{t_{i-1}}^{-\nu} h_{t_{p}}^{-\beta(\nu+d)}  e^{-(t_i-t_{i-1})(|\zeta(D_p)|/4-3d\log(\delta_p))} \\
&+C_1 C_2\epsilon^{-1}   h_{t_{i-1}}^{-\nu} h_{t_{p}}^{-\beta(\nu+d)} e^{-(t_i-t_{i-1})(C_2^{-1}D_p^2/2-3d\log(\delta_p))}.
\end{split}$$

To proceed further we first note that, under \[new2\] and the assumptions on $(h_t)_{t\geq 0}$, there exist constants $c'_\zeta>c_\zeta>0$ such that, for $p_0$ is large enough, $$\begin{aligned}
\label{eq:zeta}
|\zeta(D_p)|\geq c'_\zeta \log(D_p) =-c'_\zeta\,\beta \log(h_{t_p})\geq c_\zeta\,\beta\alpha \log(t_p).\end{aligned}$$ Therefore, recalling that $\delta_p=t_p^{-\gamma}$ with $\gamma> \max(1/2,\alpha)$, for $p_0$ large enough we have, for all $p>p_0$ and all $i\in(p_0+1):p$, $$\begin{aligned}
  -(t_i-t_{i-1})\big( &|\zeta(D_p)|/4-3d\log(1/\delta_p)\big)\leq   -\log(t_p)(t_i-t_{i-1})\big(c_\zeta \beta\alpha/4 -3d\gamma ).\end{aligned}$$ Next, let $\tilde{c}>4\alpha\nu$ be sufficiently large so that $$\beta':=\frac{4(\tilde{c}+3d\gamma)}{c_\zeta \alpha}\geq \max(\beta_\star,1/\alpha).$$ Then, the above computations show that, for every $\beta\geq \beta'$ and $p_0$ large enough we have, for all $p>p_0$ and all $i\in(p_0+1):p$, $$\begin{aligned}
\label{eq:beta_1}
-(t_i-t_{i-1})\big( |\zeta(D_p)|/4-3d\log(1/\delta_p)\big)\leq -\tilde{c}\log(t_p)(t_i-t_{i-1}),\,\,\forall i\in (p_0+1):p.\end{aligned}$$

We now take $p_0$ sufficiently large so that $$t_i-t_{i-1}>\frac{\alpha\beta(\nu+d)+1}{\tilde{c}}+\frac{1}{2},\quad\forall i>p_0.$$ Then, using and under the assumptions on $(h_t)_{t\geq 0}$, and for $p_0$ large enough, there exists a constant $C_3\in(0,+\infty)$ such that, for all $p>p_0$, $$\begin{aligned}
\label{eq:beta_2}
h_{t_{p}}^{-\beta(\nu+d)}  e^{-(t_i-t_{i-1})(|\zeta(D_p)|/4-3d\log(1/\delta_p))}\leq C_3t_p^{-\tilde{c}/2-1},\quad  \forall i\in (p_0+1):p.\end{aligned}$$

On the other hand, if $p_0$ is large enough then, using , $C_2^{-1} D^2_p>c_\zeta\,\beta\alpha\log(t_p)/2$ for all $p>p_0$ and therefore the above computations imply that, for $p_0$ large enough and all $p>p_0$, $$\label{eq:beta_3}
\begin{split}
 h_{t_{p}}^{-\beta(\nu+d)}  e^{-(t_i-t_{i-1})(C_2^{-1}D_p^2/2-3d\log(\delta_p))}&\leq C_3 t_p^{-\tilde{c}/2-1},\quad  \forall i\in (p_0+1):p.
\end{split}$$

Then, by combining , and , and recalling that $\tilde{c}>4\alpha\nu$, it follows that, for $p_0$ large enough and some constant $C_3'\in(0,+\infty)$, for all $p>p_0$ we have $$\begin{aligned}
 \sum_{i=p_0+1}^p {\mathbb{P}}(\tilde{A}_{i,p}| A^c_{i-1,p})&\leq  C_3 C_1(1+C_2)\epsilon^{-1} t_p^{-\tilde{c}/2-1}\sum_{i=p_0+1}^p h_{t_{i-1}}^{-\nu}\\
&\leq C_3' \epsilon^{-1} t_p^{-\tilde{c}/2+\nu\alpha}\\
&\leq C_3'\epsilon^{-1} t_p^{-\nu\alpha}\end{aligned}$$ and therefore $$\begin{aligned}
\label{eq:lim_sum}
\limsup_{p\rightarrow+\infty}\sum_{i=p_0+1}^p {\mathbb{P}}(\tilde{A}_{i,p}| A^c_{i-1,p})&\leq \limsup_{p\rightarrow+\infty} C_3' \epsilon^{-1} t_p^{-\nu\alpha}=0.\end{aligned}$$

We now show that ${\mathbb{P}}(A_{p_0,p})\rightarrow 0$. To this aim remark that $$0\leq {\mathds{1}}_{V_{2D_p}}(\theta){\mathbb{E}}[\tilde{\pi}_{t_{p_0}}(\theta)]\leq  {\mathbb{E}}[\tilde{\pi}_{t_{p_0}}(\theta)],\quad\forall \theta\in{\mathbb{R}}^d,\quad\forall p\geq 1$$ and that, using Tonelli’s theorem, $$\int_{{\mathbb{R}}^d}{\mathbb{E}}[\tilde{\pi}_{t_{p_0}}(\theta)]{\mathrm{d}}\theta={\mathbb{E}}[\tilde{\pi}_{t_{p_0}}({\mathbb{R}}^d)]=1.$$ Therefore, by the reverse Fatou lemma we have (and using Tonelli’s theorem for the first equality) $$\begin{aligned}
\limsup_{p\rightarrow+\infty}{\mathbb{E}}[\tilde{\pi}_{t_{p_0}}(V_{2D_p})]&=\limsup_{p\rightarrow+\infty}\int_{{\mathbb{R}}^d}{\mathds{1}}_{V_{2D_p}}(\theta){\mathbb{E}}[\tilde{\pi}_{t_{p_0}}(\theta)]{\mathrm{d}}\theta\\
&\leq \int_{{\mathbb{R}}^d}\limsup_{p\rightarrow+\infty}\Big({\mathds{1}}_{V_{2D_p}}(\theta){\mathbb{E}}[\tilde{\pi}_{t_{p_0}}(\theta)]\Big){\mathrm{d}}\theta\\
&=0\end{aligned}$$ and thus, using Markov’s inequality, $$\label{eq:lim_sum2}
\begin{split}
\limsup_{p\rightarrow+\infty}{\mathbb{P}}(A_{p_0,p}  )&\leq \limsup_{p\rightarrow+\infty}{\mathbb{P}}(\tilde{\pi}_{t_{p_0}}(V_{2D_p})\geq\epsilon)\leq \limsup_{p\rightarrow+\infty}\frac{{\mathbb{E}}[\tilde{\pi}_{t_{p_0}}(V_{2D_p})]}{\epsilon}=0.
\end{split}$$ Then, combining , , and proves the theorem under \[new2\]\[A61\].

We now prove the theorem under \[new2\].\[A63\]. Let $p$ be such that $\zeta(D_p)<0$ and remark first that, for some constant $C\in(0,+\infty)$, $$\begin{aligned}
{\mathbb{P}}\Big(&\min_{1\leq i\leq p}   e^{ -(t_i-t_{i-1})\zeta(D_p)/2}\prod_{s=t_{i-1}+1}^{t_i}  f_{\theta_\star}(Y_s)\leq 1 \Big)\\
&={\mathbb{P}}\Big(\min_{1\leq i\leq p}   \sum_{s=t_{i-1}+1}^{t_i}  \big(\log (f_{\theta_\star}(Y_s))-\zeta(D_p)/2\big)\leq 0\Big)\\
&={\mathbb{P}}\Big(-\min_{1\leq i\leq p}   \sum_{s=t_{i-1}+1}^{t_i}  \big(\log( f_{\theta_\star}(Y_s))-\zeta(D_p)/2\big)\geq  0\Big)\\
&={\mathbb{P}}\Big( \max_{1\leq i\leq p}   \sum_{s=t_{i-1}+1}^{t_i}  \big(\zeta(D_p)/2-\log (f_{\theta_\star}(Y_s))\big)\geq  0\Big)\\
&\leq \sum_{i=1}^p{\mathbb{P}}\Big(    \frac{1}{t_i-t_{i-1}}\sum_{s=t_{i-1}+1}^{t_i} \big({\mathbb{E}}[\log (f_{\theta_\star}) ]- \log (f_{\theta_\star}(Y_s))\big)\geq  |\zeta(D_p)|/2+{\mathbb{E}}[\log (f_{\theta_\star}) ]\Big)\\
&\leq   \frac{C}{ (t_i-t_{i-1})^{k_\star{\mathds{1}}_{\mathbb{N}}(k_\star)} |\zeta(D_p)|^{2 k_\star}}\end{aligned}$$ where the second inequality uses \[new2\].\[A63\], Lemma \[lemma:mean\_p\] and Markov’s inequality.

Therefore, under the assumptions of the theorem, there exists a sequence $({\mathsf{Y}}_p)_{p\geq 1}$, with ${\mathsf{Y}}_p\subset{\mathcal{Y}}^{\otimes t_p}$ for all $p\geq 1$, such that ${\mathbb{P}}(Y_{1:t_p}\in{\mathsf{Y}}_p)\rightarrow 1$ and such that $$\begin{aligned}
\min_{1\leq i\leq p} e^{ -(t_i-t_{i-1})\zeta(D_p)/2}\prod_{t_{i-1}+1}^{t_i}  f_{\theta_\star}(y_s)> 1,\quad \min_{1\leq i\leq p}\frac{g_{i}(y_{1:t_i})}{\tilde{g}_{i,p}(y_{1:t_i})}>1,\quad \forall y_{1:t_p}\in{\mathsf{Y}}_p,\quad\forall p\geq 1.\end{aligned}$$

Then, using the computations done to prove the theorem under \[new2\].\[A61\], to prove the theorem under \[new2\].\[A63\] we only need to show that, for $p_0\in\mathbb{N}$ large enough and every $p>p_0$, $$\begin{aligned}
\label{eq:To_show_den22}
{\mathbb{E}}[X_{i,p}^{(1)}{\mathds{1}}_{{\mathsf{Y}}_{i,p}}(Y_{1:t_i})|  \, \tilde{\pi}_{t_{i-1}}(V_{2D_p})<\epsilon ]\leq e^{(t_i-t_{i-1})\zeta(D_p)/2},\quad \forall i\in (p_0+1):p,\end{aligned}$$ where, for every $p>p_0$, $({\mathsf{Y}}_{i,p})_{i=0}^p$ is a sequence in ${\mathcal{Y}}^{\otimes t_p}$ such that ${\mathsf{Y}}_{0,p}={\mathsf{Y}}^{t_p}$, ${\mathsf{Y}}_{p,p}={\mathsf{Y}}_p$ and such that, for every $i\in 1:p$, we have ${\mathsf{Y}}_{i,p}\subset \cap_{j=0}^{i-1}{\mathsf{Y}}_{j,p}$ and $$\begin{aligned}
\min_{1\leq j\leq i} e^{-(t_j-t_{j-1})\zeta(D_p)/2}\prod_{t_{j-1}+1}^{t_j}  f_{\theta_\star}(y_s)\geq 1,\quad \min_{1\leq j\leq i}\frac{g_{j}(y_{1:t_j})}{\tilde{g}_{j,p}(y_{1:t_j})}>1,\quad \forall y_{1:t_p}\in{\mathsf{Y}}_{i,p}.\end{aligned}$$

Using the definitions of $\zeta(D_p)$ under \[new2\].\[A63\] and the above definition of $Y_{i,p}$ under \[new2\].\[A63\], we have for $p_0$ large enough and all $p>p_0$, $$\begin{split}
 {\mathbb{E}}[X_{i,p}^{(1)}{\mathds{1}}_{{\mathsf{Y}}_{i,p}}(Y_{1:t_i})|  \, \tilde{\pi}_{t_{i-1}}(V_{2D_p})<\epsilon  ]&\leq e^{-\zeta(D_p)/2(t_j-t_{j-1})} \big(\sup_{\theta\in V_{D_p}}{\mathbb{E}}[\tilde{f}_\theta]\big)^{t_i-t_{i-1}}\\
&=  e^{(t_i-t_{i-1})\zeta(D_p)/2}.
\end{split}$$ This shows and the proof of the theorem under \[new2\].\[A63\] is complete.

Lastly, we prove the theorem under \[new2\].\[A62\], where $\zeta(D_p)= \log(\sup_{\theta\in V_{D_p}}{\mathbb{E}}[\tilde{f}_{\theta}/f_{\theta_\star}])$. Following the computations of the proof of the theorem under \[new2\].\[A61\], to prove the theorem under \[new2\].\[A62\] we only need to show that, $p_0\in\mathbb{N}$ large enough we have, for all $p>p_0$, $$\begin{aligned}
\label{eq:To_show_den}
{\mathbb{E}}[X_{i,p}^{(1)}{\mathds{1}}_{{\mathsf{Y}}_{i,p}}(Y_{1:t_i})|\, \tilde{\pi}_{t_{i-1}}(V_{2D_p})<\epsilon   ]\leq e^{(t_i-t_{i-1})\zeta(D_p)/2},\quad \forall i\in (p_0+1):p\end{aligned}$$ where, for every $p>p_0$, $({\mathsf{Y}}_{i,p})_{i=0}^p$ is a sequence in ${\mathcal{Y}}^{\otimes t_p}$ such that ${\mathsf{Y}}_{0,p}={\mathsf{Y}}^{t_p}$, ${\mathsf{Y}}_{p,p}={\mathsf{Y}}_p$ and such that, for every $i\in 1:p$, ${\mathsf{Y}}_{i,p}\subset \cap_{j=0}^{i-1}{\mathsf{Y}}_{j,p}$ and $$\begin{aligned}
\min_{1\leq j\leq i}\frac{g_{j}(y_{1:t_j})}{\tilde{g}_{j,p}(y_{1:t_j})}>1,\quad \forall y_{1:t_p}\in{\mathsf{Y}}_{ji,p}.\end{aligned}$$ Using the definition of $\zeta(D_p)$ and the above definition of $Y_{i,p}$ under \[new2\]\[A62\] we have $$\begin{split}
 {\mathbb{E}}[X_{i,p}^{(1)}{\mathds{1}}_{{\mathsf{Y}}_{i,p}}(Y_{1:t_i})|A^c_{i-1,p} ]&\leq  \big(\sup_{\theta\in V_{D_p}}{\mathbb{E}}\big[\tilde{f}_{\theta}/f_{\theta_\star}\big]\big)^{t_i-t_{i-1}}=  e^{(t_i-t_{i-1})\zeta(D_p)}.
\end{split}$$ This shows and the proof of the theorem under \[new2\].\[A62\] is complete.

Proof of Propositions \[prop:h\_t\]
-----------------------------------

Under the assumptions of the proposition there exist constants $C_1,C_2\in(0,+\infty)$ such that, for $p$ large enough, $$\begin{aligned}
(t_p-t_{p-1})  \sum_{s=t_{p-1}+1}^{t_p-1}h_s^2&\leq (t_p-t_{p-1})^2h^2_{t_{p-1}}\\
&\leq  C_1 (t_p-t_{p-1})^2 t_{p-1}^{-2\alpha}\\
&\leq  C_2  \log(t_{p_{-1}})^2\big) t_{p-1}^{-2(\alpha-\varrho)}.\end{aligned}$$ Therefore, since $\varrho\in(0,\alpha)$, we have $
(t_p-t_{p-1})  \sum_{s=t_{p-1}+1}^{t_p-1}h_s^2\rightarrow 0$ as required. Together with the fact that$(t_{p+1}-t_p)\rightarrow+\infty$, this shows that Condition \[thm1.2\] of Theorem \[thm:online\] holds.

To show that Condition \[thm1.3\] of Theorem \[thm:online\] holds as well remark first that $(t_p-t_{p-1})^{-1} \log(1/h_{t_{p-1}})\rightarrow 0$ as required. In addition, with $c\in(0,1)$ as in the statement of the proposition, we have for $p$ large enough $$\begin{aligned}
\label{eq:bound_varrpho}
\frac{t_{p+2}-t_{p+1}}{t_{p+1}-t_{p}}&\leq  \frac{ C_{p+1} \log(t_{p+1})}{C_{p} \log(t_{p})-1}\leq  \frac{ t_{p+1}^\varrho \log(t_{p+1}) }{c^2 t_{p}^\varrho \log(t_{p})-c}\leq  \frac{2}{c^2}\frac{t^\varrho_{p+1}\log(t_{p+1})}{t^\varrho_{t_p}\log(t_{p})}\end{aligned}$$ where, for $p$ large enough, $$\begin{aligned}
\label{eq:bound_varrpho2}
\frac{\log(t_{p+1})}{\log(t_{p})}\leq 1+\frac{t_{p+1}-t_p}{t_p\log(t_p)}\leq  1+\frac{t_{p+1}}{t_p}.\end{aligned}$$ Next, recalling that $\varrho\in(0,1)$, for $p$ large enough we have $$\begin{aligned}
\frac{t_{p+1}}{t_p}\leq 1+\frac{t_p^{\varrho} \log(t_p)}{c\,t_p}\leq 1+(2/c)\end{aligned}$$ with, together with and , shows that $\limsup_{p\rightarrow+\infty}(t_{p+2}-t_{p+1})/(t_{p+1}-t_{p})<+\infty$. This concludes to show that Condition \[thm1.3\] of Theorem \[thm:online\] holds.

To show the second part of the proposition note that there exists a constant $c'>0$ such that $(t_p-t_{p-1})\geq  c' t_{p-1}^{\varrho} $ for all $p\geq 1$. By assumption, $k_\star>1/\varrho+1$, and thus $$\begin{aligned}
\limsup_{p\rightarrow+\infty}\sum_{i=1}^p (t_i-t_{i-1})^{-k_\star}\leq (1/c') \limsup_{p\rightarrow+\infty}\sum_{i=1}^p   t_{i-1}^{-k_\star\varrho} \leq (1/c') \limsup_{p\rightarrow+\infty} \sum_{i=1}^\infty   t_{i }^{-(1+\varrho)}<+\infty.\end{aligned}$$ The result follows.

Proof of the preliminary results
--------------------------------

### Proof of Lemma \[lemma:new\]\[p-lemma:new\]

We first show the following simple lemma.

\[lemma:simple0\] Let $(s_t)_{t\geq 1}$ be a sequence in $\mathbb{N}_0$ such that $\inf_{t\geq 1}(t-s_t)\geq 1$ and $(t-s_t)\rightarrow+\infty$, $(X_t)_{t\geq 1}$ be a sequence of independent real-valued random variables such that ${\mathbb{E}}[X_n]\rightarrow m$ for some $m\in{\mathbb{R}}$ and such that $\sup_{s\geq 1}{\mathrm{Var}}(X_s)<+\infty$. Then, we have $(t-s_t)^{-1}\sum_{s=s_t+1}^t X_s\rightarrow m$ in ${\mathbb{P}}$-probability.

For every $t\geq 1$ we have $$\begin{aligned}
\label{eq:sum}
\frac{1}{t-s_t}\sum_{s=s_t+1}^t X_s-m=\frac{1}{t-s_t}\sum_{s=s_t+1}^t( X_s-{\mathbb{E}}[X_s])+\Big(\frac{1}{t-s_t}\sum_{s=s_t+1}^t {\mathbb{E}}[X_s]-m\big).\end{aligned}$$ Let $\epsilon>0$ and $s_\epsilon\in\mathbb{N}$ be such that $|{\mathbb{E}}[X_s]-m|\leq \epsilon$ for all $s\geq s_\epsilon$. Then, for all $t$ such that $s_t\geq s_\epsilon$ we have $$\begin{aligned}
\Big|\frac{1}{t-s_t}\sum_{s=s_t+1}^t {\mathbb{E}}[X_s]-m\Big|\leq \frac{1}{t-s_t}\sum_{s=s_t+1}^t| {\mathbb{E}}[X_s]-m|\leq \epsilon\end{aligned}$$ showing that $(t-s_t)^{-1}\sum_{s=s_t+1}^t {\mathbb{E}}[X_s]\rightarrow m$. Hence, by , to complete the proof it remains to show that $
(t-s_t)^{-1}\sum_{s=s_t+1}^t( X_s-{\mathbb{E}}[X_s])\rightarrow 0$ in ${\mathbb{P}}$-probability. Using Markov’s inequality, for every $\epsilon>0$ we have $$\begin{aligned}
\limsup_{t\rightarrow+\infty}{\mathbb{P}}\Big(\big|\frac{1}{t-s_t}\sum_{s=s_t+1}^t( X_s-{\mathbb{E}}[X_s])\big|\geq \epsilon\Big)&\leq \limsup_{t\rightarrow+\infty}\frac{\sum_{s=s_t+1}^t{\mathrm{Var}}(X_s)}{\epsilon^2(t-s_t)^2}\\
&\leq \limsup_{t\rightarrow+\infty}\frac{\sup_{s\geq 1}{\mathrm{Var}}(X_s)}{\epsilon^2 (t-s_t)}\\
&=0\end{aligned}$$ and the proof is complete.

Let $A_\star$ and $\tilde{A}_\star$ be as in \[new\] and $A'_\star\in\mathcal{B}({\mathbb{R}}^d)$ be such that $A'_\star$ contains a neighbourhood of $A_\star$ and such that $\tilde{A}_\star$ contains a neighbourhood of $A'_\star$. Let $B'_\star=A'_\star\cap \Theta$, $\tilde{B}_\star=\tilde{A}_\star\cap \Theta$, and remark that we can without loss of generality assume that the sequence $(\gamma_t')_{t\geq 1}$ is non-increasing. Below we denote by $\bar{B}$ the closure of the set $B\subset{\mathbb{R}}^d$.

Let $(\delta_t)_{t\geq 1}$ be a sequence in ${\mathbb{R}}_{>0}$ and let $W_s=\overline{B_{\gamma'_s}(0)}$ for all $s\geq 1$. Then, for all $t\geq 1$ we have $$\begin{aligned}
{\mathbb{P}}\Big(\sup_{(u_{s_t:t},\theta)\in \Theta_{\gamma_t',s_t:t}\times B'_\star}  \prod_{s=s_t+1}^t  & (\tilde{f}_{\theta-\sum_{i=s}^{t-1} u_i}/f_{\theta})(Y_s)\geq e^{(t-s_t)\delta_{t}} \Big)\\
&\leq {\mathbb{P}}\Big(\sup_{\theta\in B'_\star}\prod_{s=s_t+1}^t \sup_{v_s\in \overline{B_{\gamma'_t}(0)}}  (\tilde{f}_{\theta+v_s}/f_{\theta}) (Y_s)\geq e^{(t-s_t)\delta_{t}}\Big)\\
&\leq {\mathbb{P}}\Big( \prod_{s=s_t+1}^t \sup_{(\theta,w_s)\in B'_\star\times\mathsf{W}_s}  (\tilde{f}_{\theta+w_s}/f_{\theta}) (Y_s)\geq e^{(t-s_t)\delta_{t}}\Big)\end{aligned}$$ where the last inequality uses the fact that the sequence $(\gamma'_t)_{t\geq 1}$ is non-increasing. Consequently, to prove the lemma it is enough to show that $$\begin{aligned}
\label{eq:toShow_a4}
 \frac{1}{t-s_t}\sum_{s=s_{t}+1}^{t}\sup_{(\theta,w_s)\in B'_\star\times \mathsf{W}_s}\log \big((\tilde{f}_{\theta+w_s}/f_{\theta})(Y_s)\big)\rightarrow0,\quad\text{in ${\mathbb{P}}$-probability}.\end{aligned}$$

To establish we define, for every $s\geq 1$, $$\begin{aligned}
X_s=\sup_{(\theta,w_s)\in B'_\star\times  \mathsf{W}_s}\log \big((\tilde{f}_{\theta+w_s}/f_{\theta})(Y_s)\big),\quad m^{(1)}_s={\mathbb{E}}[X_s],\quad m_s^{(2)}={\mathbb{E}}[X_s^2]\end{aligned}$$ and show below that there exists a $\underline{t}\in\mathbb{N}$ such that $$\begin{aligned}
\label{eq:Kol}
 \lim_{t\rightarrow+\infty}m^{(1)}_{ \underline{t}+t}\rightarrow 0,\quad \sup_{s\geq s_{\underline{t}}}m_s^{(2)}\leq m_{s_{\underline{t}}}^{(2)}<+\infty.\end{aligned}$$ Then, will follow by Lemma \[lemma:simple0\].

To show remark first that for all $s\geq 1$ we have $m^{(1)}_s={\mathbb{E}}[\tilde{X}_s]$, where $$\tilde{X}_s=\sup_{(\theta,w_s)\in B'_\star\times  \mathsf{W}_s}\log \big((\tilde{f}_{\theta+w_s}/f_{\theta})(Y_1)\big).$$ Next, let $\tilde{\Omega}\in{\mathcal{F}}$ be such that ${\mathbb{P}}(\tilde{\Omega})=1$ and such that the mapping $\theta\mapsto    f_\theta(Y_1(\omega))$ is continuous on the compact set $\tilde{B}_\star$, for all $\omega\in\tilde{\Omega}$, notice that such a set $\tilde{\Omega}$ exists under \[new\]. Let $\underline{s}\in\mathbb{N}$ be such that $\theta+w\in  \tilde{B}_\star$ for all $(\theta,w)\in B'_\star\times W_{\underline{s}}$ such that $\theta+w\in\Theta$. Then, recalling that $\gamma'_{s+1}\leq \gamma'_{s}$ for all $s\geq 1$, it follows that $$\tilde{X}_s\leq \sup_{(\theta,\theta')\in \tilde{B}^2_\star :\,\|\theta-\theta'\|\leq \gamma'_s }\log \big((f_{\theta'}/f_{\theta})(Y_1)\big),\quad\forall s\geq \underline{s}.$$ Then, by Weierstrass’s theorem we have, for all $\omega\in \tilde{\Omega}$ and $s\geq\underline{s}$ we have $$\begin{split}
\tilde{X}_s(\omega)&\leq  \log\Big( \big(f_{h_{\gamma'_s}(\omega)}/f_{g_{\gamma'_s}(\omega)}\big) \big(Y_1(\omega)\big)\Big)
\end{split}$$ for some (measurable) functions $h_{\gamma'_s}: \tilde{\Omega} \rightarrow  \tilde{B}_\star$ and $g_{\gamma'_s}: \tilde{\Omega}\rightarrow  \tilde{B}_\star$ such that we have $\|h_{\gamma'_s}(\omega)-h_{\gamma'_s}(\omega)\|\leq\gamma'_s$ for all $\omega\in \tilde{\Omega}$.

By the maximum theorem, we can assume that, for all $s\geq  \underline{s}$ and every $\omega\in\tilde{\Omega}$ the mappings $\gamma\mapsto h_{\gamma}(\omega)$ and $\gamma\mapsto g_{\gamma}(\omega)$ are continuous on $[0,\gamma'_{\underline{s}}]$. Therefore, since $h_{0}(\omega)=g_{0}(\omega)$ for all $\omega\in\tilde{\Omega}$, we have $$\label{eq:lim}
\begin{split}
0\leq \limsup_{s\rightarrow+\infty}\tilde{X}_{\underline{s}+s}(\omega)&\leq \limsup_{s\rightarrow +\infty}\log\Big( \big(f_{h_{\gamma'_s}(\omega)}/f_{g_{\gamma'_s}(\omega))}\big) \big(Y_1(\omega)\big)\Big)\\
&=\log\Big( \big(f_{h_{0}(\omega)}/f_{g_{0}(\omega)}\big) \big(Y_1(\omega)\big)\Big)\\
&=0,\qquad\qquad  \forall \omega\in\tilde{\Omega}.
\end{split}$$

To proceed further remark that since $W_s\subseteq W_{\underline{s}}$ for all $s\geq  \underline{s}$, it follows that for all $s\geq \underline{s}$ we have $\tilde{X}_s\leq \tilde{X}_{\underline{s}}$, ${\mathbb{P}}$-a.s. Let $\tilde{\delta}>0$ be as in . Then, under this latter assumption, and taking $\underline{s}\in\mathbb{N}$ sufficiently large so that $\gamma'_{\underline{s}}\leq \tilde{\delta}$, we have ${\mathbb{E}}[\tilde{X}_{\underline{s}}]<+\infty$. Then, by the dominated convergence theorem, $\lim_{s\rightarrow+\infty} m^{(1)}_{\underline{s}+s}= 0$, showing the first part of . To show the second part of it suffices to remark that $m^{(2)}_{s}\leq m^{(2)}_{\underline{s}}$ for all $s\geq \underline{s}$ where, under , $m^{(2)}_{\underline{s}}<+\infty$.

### Proof of Lemma \[lemma:pi\]\[p-lemma:pi\]

Let $t \geq 0$ and $A\in\mathcal{B}({\mathbb{R}}^d)$. Then, if $\mu_t\neq \delta_{\{0\}}$ we have, ${\mathbb{P}}$-a.s., $$\begin{aligned}
(\mu_{t}*\tilde{\pi}_{t})(A)
&=\int_{{\mathbb{R}}^d\times {\mathbb{R}}^d} {\mathds{1}}_A(\theta) \tilde{\pi}_{t} (\theta-u_{t}) \mu_{t}({\mathrm{d}}u_{t}) {\mathrm{d}}\theta =\int_{A}  {\mathbb{E}}\big[\tilde{\pi}_{t} (\theta-U_{t}) |{\mathcal{F}}_t\big]{\mathrm{d}}\theta\end{aligned}$$ while, if $\mu_t= \delta_{\{0\}}$ we have, ${\mathbb{P}}$-a.s., $$\begin{aligned}
(\mu_{t}*\tilde{\pi}_{t})(A)
=\int_{{\mathbb{R}}^d\times {\mathbb{R}}^d} {\mathds{1}}_A(\theta+u_t)   \mu_{t}({\mathrm{d}}u_{t})\tilde{\pi}_t(\theta){\mathrm{d}}\theta  =\int_{A}    \tilde{\pi}_t(\theta){\mathrm{d}}\theta.\end{aligned}$$ Recall that ${\mathbb{P}}(\cap_{t\geq 0}\Omega_t)=1$ if ${\mathbb{P}}(\Omega_t)=1$ for all $t\geq 0$ and that two probability measures $\nu_1,\nu_2\in\mathcal{P}({\mathbb{R}}^d)$ are such that $\nu_1=\nu_2$ if $\nu_1(E_i)=\nu_2(E_i)$ for all $i\geq 1$, with $(E_i)_{i\geq 1}$ a dense subset of ${\mathbb{R}}^d$ such that $E_i\in \mathcal{B}({\mathbb{R}}^d)$ for all $i\geq 1$.

Therefore, the above computations imply that $$\begin{aligned}
\label{eq:density}
(\mu_{t}*\tilde{\pi}_{t})(A)=\int_A {\mathbb{E}}\big[\tilde{\pi}_{t} (\theta-U_{t}) |{\mathcal{F}}_t\big]{\mathrm{d}}\theta,\quad\forall t\geq 0,\quad\forall A\in\mathcal{B}({\mathbb{R}}^d),\quad{\mathbb{P}}-a.s.\end{aligned}$$ We now prove the result of the lemma by induction on $t\geq 1$.

The result trivially holds for $t=1$ and we now assume that it holds for some $t\geq 1$. Then, ${\mathbb{P}}$-a.s., $$\begin{aligned}
 \tilde{\pi}_{t+1}({\mathrm{d}}\theta)&\propto   \tilde{f}_\theta(Y_{t+1})(\mu_t*\tilde{\pi}_t)({\mathrm{d}}\theta)\\
&= \tilde{f}_\theta(Y_{t+1}){\mathbb{E}}\big[\tilde{\pi}_{t} (\theta-U_{t}) |{\mathcal{F}}_t\big]{\mathrm{d}}\theta\\
&\propto \tilde{f}_\theta(Y_{t+1}){\mathbb{E}}\Big[  (\mu_0*\tilde{\pi}_0) \big(\theta-\sum_{s=1}^{t}U_s\big)\prod_{s=1}^{t} \tilde{f}_{\theta-\sum_{i=s}^{t}U_i}(Y_s)\big|{\mathcal{F}}_{t}\Big]{\mathrm{d}}\theta\\
&= {\mathbb{E}}\Big[ (\mu_0*\tilde{\pi}_0) \big(\theta-\sum_{s=1}^{t}U_s\big)\prod_{s=1}^{t+1} \tilde{f}_{\theta-\sum_{i=s}^{t}U_i}(Y_s)\big|{\mathcal{F}}_{t+1}\Big]{\mathrm{d}}\theta\end{aligned}$$ where the first equality uses and the second line the inductive hypothesis. The proof is complete.

### Proof of Lemma \[lemma:denom\] {#p-lemma:denom}

Let $t\geq 1$, $p_t={\mathbb{P}}(U_{s_t:t}\in  \Theta_{\tilde{\delta},s_t:t})$, $C_\star\in{\mathbb{R}}_{>0}$ be as in \[taylor\], $\tilde{C}_\star=2 ({\mathbb{E}}[m^2_\star]+C_\star)^{1/2}$ and note that, under \[m\_star\]-\[taylor\] and for all $\theta\in B_{\delta_{\star}}(\theta_\star)$, $$\label{eq:Bound_KL_1}
\max\bigg(-{\mathbb{E}}\big[\log(\tilde{f}_{\theta}/f_{\theta_\star})\big], {\mathbb{E}}\big[\big(\log(\tilde{f}_{\theta}/f_{\theta_\star})\big)^2\big]\Big)\leq  \big({\mathbb{E}}[m_\star^2]+C_\star\big)\|\theta-\theta_\star\|^2.$$

Remark now that if $p_t=0$ then the result of the lemma trivially holds and henceforth we therefore assume that $p_t>0$. To simplify the notation let ${\mathbb{E}}_{Y_1}[\cdot]$ denote expectations w.r.t. the distribution of $Y_1$, ${\mathbb{E}}^{\tilde{\delta}}_{t,\mu}[\cdot]$ denote expectations w.r.t. the restriction of $\otimes_{s=s_t}^t\mu_t$ to the set $\Theta_{\tilde{\delta},s_t:t}$, and let $V_t:=(U_s)_{s=s_t+1}^t$.

For every $u_{(s_t+1):t}\in\Theta^{t-s_t}$ let $\tilde{\eta}({\mathrm{d}}\theta, u_{s_t:t})$ be the probability measure on $B_\delta(\theta_\star)$ with density function $\tilde{\eta}(\cdot, u_{(s_t+1):t})$ defined by $$\tilde{\eta}(\theta, u_{(s_t+1):t})=\frac{\eta\big(\theta-\sum_{s=s_t+1}^{t-1}u_s\big)}{\eta\big(B_\delta(\theta_\star-\sum_{s=s_t+1}^{t-1}u_s)\big)},\quad\theta\in B_\delta(\theta_\star).$$ Then, using the shorthand $a_t=(t-s_t)(2(\tilde{C}_\star \delta)^2+\epsilon)$, $$\label{eq:p1}
\begin{split}
\tilde{{\mathsf{Y}}}_t&:=\bigg\{y_{s_t:t}:
\int_{\Theta}{\mathbb{E}}\Big[\eta\big(\theta-\sum_{s=s_t+1}^{t-1}U_s\big)\prod_{s=s_t+1}^{t} (\tilde{f}_{\theta-\sum_{i=s}^{t-1}U_i}/f_{\theta_\star})(y_s)\Big]
{\mathrm{d}}\theta \leq p_t\, C^\eta_{\delta,\tilde{\delta}}\, e^{-a_t}\bigg\}\\
&=\bigg\{y_{s_t:t}:
{\mathbb{E}}\Big[\int_{\Theta}\eta\big(\theta-\sum_{s=s_t+1}^{t-1}U_s\big)\prod_{s=s_t+1}^{t} (\tilde{f}_{\theta-\sum_{i=s}^{t-1}U_i}/f_{\theta_\star})(y_s)
{\mathrm{d}}\theta\Big] \leq  p_t\, C^\eta_{\delta,\tilde{\delta}}\, e^{-a_t}\bigg\}\\
&\subset \bigg\{y_{s_t:t}:\,{\mathbb{E}}^{\tilde{\delta}}_{t,\mu}\Big[\int_{B_{\delta}(\theta_\star)}\eta\big(\theta-\sum_{s=s_t+1}^{t-1}U_s\big)\prod_{s=s_t+1}^{t} (\tilde{f}_{\theta-\sum_{i=s}^{t-1}U_i})/(f_{\theta_\star})(y_s)
{\mathrm{d}}\theta\Big]\leq  C^\eta_{\delta,\tilde{\delta}}\, e^{-a_t}\bigg\}\\
&\subset  \bigg\{y_{s_t:t}:\,{\mathbb{E}}^{\tilde{\delta}}_{t,\mu}\Big[\int_{B_{\delta}(\theta_\star)}\prod_{s=s_t+1}^{t} (\tilde{f}_{\theta-\sum_{i=s}^{t-1}U_i}/f_{\theta_\star})(y_s)\tilde{\eta}({\mathrm{d}}\theta,V_t)\Big] \leq      e^{-a_t}\bigg\}\\
&\subset \bigg\{y_{s_t:t}: {\mathbb{E}}^{\tilde{\delta}}_{t,\mu}\Big[\sum_{s=s_t+1}^t \int_{B_{\delta}(\theta_\star)} \log\big((\tilde{f}_{\theta-\sum_{i=s}^{t-1}U_i}/f_{\theta_\star})(y_s)\big)\,\tilde{\eta}({\mathrm{d}}\theta, V_t)\Big]\leq  -a_t\bigg\}
\end{split}$$ where the equality uses Tonelli’s theorem, the second inclusion uses the definition of $ C^\eta_{\delta,\tilde{\delta}}$ and the last inclusion uses twice Jensen’s inequality.

To simplify the notation in what follows we define, for every $t\geq s\geq 0$, $$\begin{aligned}
\label{eq:g}
g_\theta(u_{s:t},y)=\log\big( (\tilde{f}_{ \theta-\sum_{i=s}^{t-1}u_i}/f_{\theta_\star})(y)\Big),\quad  (\theta,u_{s:t})\in\Theta^{t-s+2},\,\,y\in{\mathsf{Y}}.\end{aligned}$$

Remark now that, by and using the inequality $\|a+b\|^2\leq 2(\|a\|^2+\|b\|^2)$ for all $a,b\in{\mathbb{R}}^d$, $$\label{eq:Bound_KL}
\begin{split}
-&\sum_{s=s_t+1}^t{\mathbb{E}}[g_\theta(u_{s:t},Y_1)]\leq  (t-s_t)(\tilde{C}_\star \delta)^2,\quad\forall  \theta \in B_{\delta}(\theta_\star),\quad\forall u_{s_t:t}\in\Theta_{\tilde{\delta},s_t:t}\\
&\sum_{s=s_t+1}^t {\mathbb{E}}[g_\theta(u_{s:t},Y_1)^2]\leq (t-s_t)(\tilde{C}_\star \delta)^2,\quad\forall \theta \in B_{\delta}(\theta_\star),\quad\forall u_{s_t:t}\in\Theta_{\tilde{\delta},s_t:t}.
\end{split}$$

Therefore, using and , we have $$\label{eq:p2}
\begin{split}
 &{\mathbb{P}}(Y_{s_t:t}\in \tilde{{\mathsf{Y}}}_t)\\
 &\leq {\mathbb{P}}\Big({\mathbb{E}}^{\tilde{\delta}}_{t,\mu}\Big[\sum_{s=s_t+1}^t\int_{B_{\delta}(\theta_\star)} g_{\theta}(U_{s:t},Y_s)\tilde{\eta}({\mathrm{d}}\theta, V_t)\Big] \leq  -a_t\Big)\\
&= {\mathbb{P}}\Big({\mathbb{E}}^{\tilde{\delta}}_{t,\mu}\Big[\frac{1}{t-s_t}\sum_{s=s_t+1}^t\Big\{\int_{B_{\delta}(\theta_\star)} g_\theta(U_{s:t},Y_s)-\int_{B_{\delta}(\theta_\star)}{\mathbb{E}}_{Y_1}\big[g_\theta(U_{s:t},Y_1)\big]\Big\}\tilde{\eta}({\mathrm{d}}\theta, V_t)\\
& \,\,\,\,\,\qquad +\frac{1}{t-s_t}\sum_{s=s_t+1}^t\int_{B_{\delta(\theta_\star)}} {\mathbb{E}}_{Y_1}\big[g_\theta(U_{s:t},Y_1)\big]\tilde{\eta}({\mathrm{d}}\theta, V_t)\Big]\leq -2(\tilde{C}_\star \delta)^2-\epsilon \Big)\\
&\leq {\mathbb{P}}\Big({\mathbb{E}}^{\tilde{\delta}}_{t,\mu}\bigg[\frac{1}{t-s_t}\sum_{s=s_t+1}^t\Big\{\int_{B_{\delta}(\theta_\star)}g_\theta(U_{s:t},Y_s)-\int_{B_{\delta}(\theta_\star)}{\mathbb{E}}_{Y_1}\big[g_\theta(U_{s:t},Y_1)\big]\bigg\}\tilde{\eta}({\mathrm{d}}\theta, V_t)\Big]\\
& \,\,\,\,\,\qquad  \leq -(\tilde{C}_\star \delta)^2-\epsilon \Big).
\end{split}$$

We now show that $$\label{eq:Fub3}
\begin{split}
{\mathbb{E}}^{\tilde{\delta}}_{t,\mu}\Big[ \int_{B_{\delta}(\theta_\star)}{\mathbb{E}}_{Y_1}\big[ g_\theta(U_{s:t},Y_1) \big]&\tilde{\eta}({\mathrm{d}}\theta, V_t)\Big] \\
&={\mathbb{E}}_{Y_1}\Big[ {\mathbb{E}}^{\tilde{\delta}}_{t,\mu}\Big[  \int_{B_{\delta}(\theta_\star)}g_\theta(U_{s:t},Y_1)\tilde{\eta}({\mathrm{d}}\theta, V_t)\Big]\Big].
\end{split}$$ By , there exists a constant $C_t\in{\mathbb{R}}_{\geq 0}$ such, that for every $u_{s_t:t}\in \Theta_{\tilde{\delta},s_t:t}$, $$\label{eq:Fub0}
\begin{split}
 \int_{B_{\delta}(\theta_\star)}{\mathbb{E}}_{Y_1}\big[|g_\theta(u_{s:t},Y_1)|\big]\tilde{\eta}({\mathrm{d}}\theta, u_{s_t:t})&\leq  \int_{B_{\delta}(\theta_\star)}{\mathbb{E}}_{Y_1}\big[g_\theta(u_{s:t},Y_1)^2\big]^{\frac{1}{2}}\tilde{\eta}({\mathrm{d}}\theta, u_{s_t:t})\\
 &\leq C_t
\end{split}$$ and thus by, Fubini-Tonelli’s theorem, $$\label{eq:Fub1}
\begin{split}
{\mathbb{E}}^{\tilde{\delta}}_{t,\mu}\Big[\int_{B_{\delta}(\theta_\star)}{\mathbb{E}}_{Y_1}\big[ g_\theta(U_{s:t},Y_1) \big]&\tilde{\eta}({\mathrm{d}}\theta, V_t)\Big]\\
&= {\mathbb{E}}^{\tilde{\delta}}_{t,\mu}\Big[{\mathbb{E}}_{Y_1}\Big[ \int_{B_{\delta}(\theta_\star)} g_\theta(U_{s:t},Y_1) \tilde{\eta}({\mathrm{d}}\theta, V_t)\Big]\Big].
\end{split}$$ Using , we also have $$\begin{aligned}
{\mathbb{E}}^{\tilde{\delta}}_{t,\mu}\Big[ {\mathbb{E}}_{Y_1}\Big[\Big|\int_{B_{\delta}(\theta_\star)}g_\theta(U_{s:t},Y_1)\tilde{\eta}({\mathrm{d}}\theta, V_t)\Big|\Big]\Big] &\leq {\mathbb{E}}^{\tilde{\delta}}_{t,\mu} \Big[{\mathbb{E}}_{Y_1}\Big[ \int_{B_{\delta}(\theta_\star)}|g_\theta(U_{s:t},Y_1)|\tilde{\eta}({\mathrm{d}}\theta,V_t)\Big]\Big] \\
&= {\mathbb{E}}^{\tilde{\delta}}_{t,\mu}\Big[ \int_{B_{\delta}(\theta_\star)}{\mathbb{E}}_{Y_1}\big[|g_\theta(U_{s:t},Y_1)|\big]\tilde{\eta}({\mathrm{d}}\theta, V_t)\Big]\Big]\\
&\leq C_t,\end{aligned}$$ where the equality uses Tonelli’s theorem. By Fubini-Tonelli’s theorem we therefore have $$\label{eq:Fub2}
\begin{split}
{\mathbb{E}}^{\tilde{\delta}}_{t,\mu}\Big[ {\mathbb{E}}_{Y_1}\Big[ \int_{B_{\delta}(\theta_\star)}g_\theta(U_{s:t},Y_1)&\tilde{\eta}({\mathrm{d}}\theta,V_t) \Big]\Big]\\
&={\mathbb{E}}_{Y_1} \Big[{\mathbb{E}}^{\tilde{\delta}}_{t,\mu}\Big[ \int_{B_{\delta}(\theta_\star)}g_\theta(U_{s:t},Y_1)\tilde{\eta}({\mathrm{d}}\theta, V_t)\Big]\Big]
\end{split}$$ and follows from and .

Consequently, using and , we have $$\label{eq:moments0}
\begin{split}
 {\mathbb{P}}(Y_{s_t:t}\in \tilde{{\mathsf{Y}}}_t)&\leq {\mathbb{P}}\Big( \frac{1}{t-s_t}\sum_{s=s_t+1}^t\bigg\{{\mathbb{E}}^{\tilde{\delta}}_{t,\mu}\Big[ \int_{B_{\delta}(\theta_\star)}g_\theta(U_{s:t},Y_s)\tilde{\eta}({\mathrm{d}}\theta, V_t)\Big]\\
&\,\,\,\,\,\qquad- {\mathbb{E}}_{Y_1}\Big[{\mathbb{E}}^{\tilde{\delta}}_{t,\mu}\Big[\int_{B_{\delta}(\theta_\star)} g_\theta(U_{s:t},Y_1)\tilde{\eta}({\mathrm{d}}\theta, V_t)\Big]\Big]\bigg\} \leq  -(\tilde{C}_\star \delta)^2-\epsilon \Big)\\
&\leq {\mathbb{P}}\Big(\Big|\frac{1}{t-s_t}\sum_{s=s_t+1}^t\bigg\{{\mathbb{E}}^{\tilde{\delta}}_{t,\mu}\Big[ \int_{B_{\delta}(\theta_\star)}g_\theta(U_{s:t},Y_s)\tilde{\eta}({\mathrm{d}}\theta, V_t)\Big]\\
&\,\,\,\,\,\qquad- {\mathbb{E}}_{Y_1}\Big[{\mathbb{E}}^{\tilde{\delta}}_{t,\mu}\Big[\int_{B_{\delta}(\theta_\star)} g_\theta(U_{s:t},Y_1)\tilde{\eta}({\mathrm{d}}\theta, V_t)\Big]\Big]\bigg\}\Big|\geq  (\tilde{C}_\star \delta)^2+\epsilon\Big)
\end{split}$$ and we finally upper bound the last term using Markov’s inequality.

To this aim remark that $$\label{eq:moments}
\begin{split}
\sum_{s=s_t+1}^t{\mathbb{E}}_{Y_1}\Big[  {\mathbb{E}}^{\tilde{\delta}}_{t,\mu}\Big[ \int_{B_{\delta}(\theta_\star)}&g_\theta(U_{s:t},Y_1)\tilde{\eta}({\mathrm{d}}\theta, V_t)\Big]^2\Big]\\
&\leq \sum_{s=s_t+1}^t{\mathbb{E}}_{Y_1}\Big[ {\mathbb{E}}^{\tilde{\delta}}_{t,\mu}\Big[\Big(\int_{B_{\delta}(\theta_\star)}g_\theta(U_{s:t},Y_1)\tilde{\eta}({\mathrm{d}}\theta,V_t)\Big)^2\Big]\Big]\\
&\leq \sum_{s=s_t+1}^t{\mathbb{E}}_{Y_1}\Big[ {\mathbb{E}}^{\tilde{\delta}}_{t,\mu}\Big[\int_{B_{\delta}(\theta_\star)}g_\theta(U_{s:t},Y_1)^2\tilde{\eta}( \theta, V_t){\mathrm{d}}\theta \Big]\Big]\\
&= \sum_{s=s_t+1}^t{\mathbb{E}}^{\tilde{\delta}}_{t,\mu}\Big[ {\mathbb{E}}_{Y_1}\Big[\int_{B_{\delta}(\theta_\star)}g_\theta(U_{s:t},Y_1)^2\tilde{\eta}({\mathrm{d}}\theta, V_t) \Big]\Big]\\
&= \sum_{s=s_t+1}^t{\mathbb{E}}^{\tilde{\delta}}_{t,\mu}\Big[ \int_{B_{\delta}(\theta_\star)}{\mathbb{E}}_{Y_1}\big[g_\theta(U_{s:t},Y_1)^2\big]\tilde{\eta}( {\mathrm{d}}\theta, V_t)\Big]\\
&\leq (t-s_t)(\tilde{C}_\star \delta)^2
\end{split}$$ where the last inequality uses , the first two inequalities use Jensen’s inequality and the two equalities hold by Tonelli’s theorem.

Therefore, using , and Markov’s inequality, $$\begin{aligned}
 {\mathbb{P}}(Y_{s_t:t}\in \tilde{{\mathsf{Y}}}_t)& \leq \frac{\frac{1}{(t-s_t)^2}\sum_{s=s_t+1}^t{\mathrm{Var}}\Big({\mathbb{E}}^{\tilde{\delta}}_{t,\mu}\big[ \int_{B_{\delta}(\theta_\star)}g_\theta(U_{s:t},Y_s)\tilde{\eta}({\mathrm{d}}\theta, V_t)\big]\Big)}{((\tilde{C}_\star\delta)^2+\epsilon)^2}\\
&\leq \frac{\frac{1}{(t-s_t)^2}\sum_{s=s_t+1}^t{\mathbb{E}}_{Y_1} \Big[ {\mathbb{E}}^{\tilde{\delta}}_{t,\mu}\big[ \int_{B_{\delta}(\theta_\star)}g_\theta(U_{s:t},Y_1)\tilde{\eta}({\mathrm{d}}\theta, V_t)\big]^2\Big]}{((\tilde{C}_\star\delta)^2+\epsilon)^2}\\
&\leq \frac{1}{(t-s_t) ((\tilde{C}_\star\delta)+(\tilde{C}_\star\delta)^{-1}\epsilon)^2}.\end{aligned}$$ The proof is complete.

### Proof of Lemma \[lemma:test\]\[p-lemma:test\]

We first recall the following result due to @Kleijn2012 [Lemma 3.3].

\[lemma:test\_K\] Assume \[test\]. Then, for every compact set $W\in\mathcal{B}(\Theta)$ such that $\theta_\star\in W$ and every $\epsilon>0$ there exist a sequence of measurable functions $(\psi_t)_{t\geq 1}$, with $\psi_t:{\mathsf{Y}}^t\rightarrow\{0,1\}$, and a constant $ D_\star\in{\mathbb{R}}_{> 0} $ such that ${\mathbb{E}}[\psi_t(Y_{1:t})]\rightarrow 0$ and such that, for $t$ large enough, $$\sup_{\theta\in V_\epsilon\cap\Theta}{\mathbb{E}}\Big[\big(1-\psi_t(Y_{1:t})\big)\prod_{s=1}^t (f_{\theta}/f_{\theta_\star})(Y_s)\Big]\leq e^{-tD_\star}.$$

*Proof of Lemma \[lemma:test\]* Let $A_\star$ and $\tilde{A}_\star$ be as in \[new\], $A'_\star\subsetneq\tilde{A}_\star$ be as in Lemma \[lemma:new\] and $W=\tilde{A}_\star\cap \Theta$. Remark that $W$ is a compact set under \[new\] and that $A'_\star\cap\Theta\subset W$. Let $(\psi_t)_{t\geq 1}$ be as in Lemma \[lemma:test\_K\] and $(\delta_t)_{t\geq 1}$ be as in Lemma \[lemma:new\] for the sequence $(\gamma'_t)_{t\geq 1}$ defined by $\gamma'_t=2\gamma_t$, $\forall t\geq 1$. Without loss of generality we assume below that $\epsilon>0$ is such that $B_{\epsilon}(\theta_\star)\subset A'_\star$. Indeed, since $\theta_\star\in A_\star$ and $A'_\star$ contains an open set that contains $A_\star$, it follows that $B_{\delta}(\theta_\star)\subset A'_\star$ for $\delta>0$ small enough. Let $t_1\in\mathbb{N}$ be such that for all $t\geq t_1$ we have $\theta'-\sum_{i=s}^{t-1} u'_i\not\in  A_\star$ for all $\theta'\not\in A'_\star$ and all $u'_{s_t:t}\in \Theta_{\gamma_t,s_t:t}$. Notice that such a $t_1\in\mathbb{N}$ exists since $A_\star'$ contains a neighbourhood of $A_\star$.

We first show the lemma assuming \[new\].\[A41\]). To this aim, for every, $t\geq 1$ we define ${\mathsf{Y}}_t= {\mathsf{Y}}^{(1)}_t\cap {\mathsf{Y}}^{(2)}_t$ where $$\begin{split}
&{\mathsf{Y}}^{(1)}_t=\Big\{y_{1:t}\in{\mathsf{Y}}^t:\,\sup_{(u_{s_t:t},\,\theta)\in  \Theta_{\gamma_t,s_t:t}\times (A'_\star\cap\Theta)}\prod_{s=s_t+1}^t  (\tilde{f}_{\theta-\sum_{i=s}^{t-1} u_i}/f_{\theta}) (y_s)<e^{(t-s_t)\delta_{t}}\Big\}\\
&{\mathsf{Y}}^{(2)}_t=\Big\{y_{1:t}\in{\mathsf{Y}}^t:\,\sum_{s=s_t+1}^t\sup_{\theta\not\in A_\star}\log \big(\tilde{f}_{\theta}(y_s)\big)<\sum_{s=s_t+1}^t \log \big(f_{\theta_\star}(y_s)\big)-c_{\mathrm{A4}} \Big\}
\end{split}$$ where $c_{\mathrm{A4}}>0$ is such that ${\mathbb{E}}[\sup_{\theta\not\in A_\star}\log (\tilde{f}_{\theta})]< {\mathbb{E}}[ \log (f_{\theta_\star})]-2c_{\mathrm{A4}}$. Notice that such a constant $c_{\mathrm{A4}}>0$ exists under \[new\].\[A41\]).

For every $t\geq 1$, let $\phi_t(Y_{1:t})={\mathds{1}}_{{\mathsf{Y}}_t^c}(Y_{1:t})+\psi_{t-s_t}(Y_{(s_t+1):t}){\mathds{1}}_{{\mathsf{Y}}_t}(Y_{1:t})$ and remark that under \[new\].\[A41\]), and by Lemmas \[lemma:new\] and \[lemma:test\_K\], $ {\mathbb{E}}[\phi_t(Y_{1:t})]\rightarrow 0$, as required.

To show the second part of the lemma let $\theta\in V_\epsilon$, $u_{s_t:t}\in  \Theta_{\gamma_t,s_t:t}$. Remark that $1-\phi_t(Y_{1:t})=\big(1-\psi_{t-s_t}(Y_{(s_t+1):t})\big){\mathds{1}}_{{\mathsf{Y}}_t}(Y_{1:t})$ for all $t\geq 1$, and assume first that $\theta\in (A'_\star\cap\Theta)$. Then, $$\label{eq:test_int_1}
\begin{split}
 {\mathbb{E}}\Big[(1 -\phi_t(Y_{1:t})) &\prod_{s=s_t+1}^t (\tilde{f}_{\theta-\sum_{i=s}^{t-1}u_i}/f_{\theta_\star})(Y_s)\big|\,{\mathcal{F}}_{s_t}\Big]\\
&\leq e^{(t-s_t)\delta_{t}}{\mathbb{E}}\Big[\big(1-\psi_{t-s_t}(Y_{(s_t+1):t})\big)\prod_{s=s_t+1}^t  (f_{\theta}/f_{\theta_\star})(Y_s)\Big]\\
&\leq  e^{-(t-s_t)(D_\star   - \delta_{t})}\\
&\leq  e^{-(t-s_t)\frac{D_\star}{2}}
\end{split}$$ where the first inequality uses the definition of ${\mathsf{Y}}^{(1)}_t$, the second inequality uses Lemma \[lemma:test\_K\] and holds for $t$ large enough, while the last inequality holds for $t$ sufficiently large since $\delta_t\rightarrow 0$. Notice that if $\theta\not\in\Theta$ we have ${\mathbb{E}}\big[(1 -\phi_t(Y_{1:t})) \prod_{s=s_t+1}^t (\tilde{f}_{\theta-\sum_{i=s}^{t-1}u_i}/f_{\theta_\star})(Y_s)\big|\,{\mathcal{F}}_{s_t}\big]=0$ and thus also holds if $\theta\in (A'_\star\cap\Theta^c)$.

Assume now that $ \theta \not\in A'_\star$. Then, $$\label{eq:test_int_2}
\begin{split}
{\mathbb{E}}\Big[(1-\phi_t(Y_{1:t})) \prod_{s=s_t+1}^t  \frac{\tilde{f}_{\theta-\sum_{i=s}^{t-1}u_i}}{f_{\theta_\star}}(Y_s)\big|\,{\mathcal{F}}_{s_t}\Big]&\leq {\mathbb{E}}\Big[{\mathds{1}}_{{\mathsf{Y}}^{(2)}_t}(Y_{1:t})\prod_{s=s_t+1}^t \sup_{\theta \not\in A_\star}  \frac{\tilde{f}_{\theta}}{f_{\theta_\star}}(Y_s)\Big]\\
&\leq  e^{- (t-s_t) c_{\mathrm{A4}}}
\end{split}$$ where the first inequality holds for $t\geq t_1$. Together with , shows that the result of the lemma holds under \[new\].\[A41\]) with $\tilde{D}_\star= D_\star/2\vee c_{\mathrm{A4}}$.

We now show the result of the lemma assuming under \[new\].\[A42\]) and \[new\].\[A43\]). To do so remark that, using the above computations, we only need to find a set ${\mathsf{Y}}_t^{(2)}\in{\mathcal{Y}}^{\otimes t}$ such that ${\mathbb{P}}(Y_{1:t}\in{\mathsf{Y}}_t^{(2)})\rightarrow 1$ and such that there exists a constant $c_{\mathrm{A4}}>0$ for which, for $t$ large enough, $$\begin{aligned}
\label{eq:c_stareq}
\sup_{\theta\not\in A'_\star}{\mathbb{E}}\big[(1-\phi_t(Y_{1:t}))\prod_{s=s_t+1}^t  \frac{\tilde{f}_{\theta-\sum_{i=s}^{t-1}u_i}}{f_{\theta_\star}}(Y_s)\big|\,{\mathcal{F}}_{s_t}\big]\leq e^{-(t-s_t)c_{\mathrm{A4}}},\quad\forall u_{s_t:t}\in \Theta_{\gamma_t,s_t:t}.\end{aligned}$$

Assume \[new\].\[A42\]), let ${\mathsf{Y}}_t^{(2)}={\mathsf{Y}}^t$ (so that ${\mathbb{P}}(Y_{1:t}\in{\mathsf{Y}}_t^{(2)})=1$ for all $t$), $\theta\not\in  A'_\star$ and $u_{s_t:t}\in \Theta_{\gamma_t,s_t:t}$. Then, $$\label{eq:test_int_22}
\begin{split}
{\mathbb{E}}\Big[(1-\phi_t(Y_{1:t})) \prod_{s=s_t+1}^t  \frac{\tilde{f}_{\theta-\sum_{i=s}^{t-1}u_i}}{f_{\theta_\star}}(Y_s)\big|\,{\mathcal{F}}_{s_t}\Big]&\leq {\mathbb{E}}\Big[\prod_{s=s_t+1}^t  \frac{\tilde{f}_{\theta-\sum_{i=s}^{t-1}u_i}}{f_{\theta_\star}}(Y_s)\big|\,{\mathcal{F}}_{s_t}\Big]\\
&=\prod_{s=s_t+1}^t{\mathbb{E}}\Big[ \frac{\tilde{f}_{\theta-\sum_{i=s}^{t-1}u_i}}{f_{\theta_\star}}(Y_s)\big|\,{\mathcal{F}}_{s_t}\Big]\\
&\leq \big(\sup_{\theta\not\in A_\star}{\mathbb{E}}[\tilde{f}_\theta/f_{\theta_\star}]\big)^{t-s_t}
\end{split}$$ where the last equality holds $t\geq t_1$. Under \[new\].\[A42\]), $\sup_{\theta \not\in A_\star}{\mathbb{E}}[\tilde{f}_\theta/f_{\theta_\star}]<1$ and therefore shows that holds with $c_{\mathrm{A4}}=-\log\big(\sup_{\theta \not\in A_\star}{\mathbb{E}}[\tilde{f}_\theta/f_{\theta_\star}]\big)$.

Lastly, assume \[new\].\[A43\]) and remark that under this condition there exists a constant $c>0$ such that $\log( \sup_{\theta \not\in A_\star}{\mathbb{E}}[\tilde{f}_\theta])<  {\mathbb{E}}[\log f_{\theta_\star}] -c$, and let $${\mathsf{Y}}^{(2)}_t=\Big\{y_{1:t}\in{\mathsf{Y}}^t:\,\frac{1}{t-s_t}\sum_{s=s_t+1}^t \log\big(f_{\theta_\star}(y_s)\big)>{\mathbb{E}}[\log (f_{\theta_\star})]-c\Big\}.$$ Then, by the law of large numbers, ${\mathbb{P}}(Y_{1:t}\in {\mathsf{Y}}^{(2)}_t)\rightarrow 1$ while, for every $\theta\not\in A'_\star$ and $u_{s_t:t}\in \Theta_{\gamma_t,s_t:t}$, $$\begin{aligned}
 {\mathbb{E}}\Big[{\mathds{1}}_{{\mathsf{Y}}^{(2)}_t} (Y_{1:t})\prod_{s=s_t+1}^t & (\tilde{f}_{\theta-\sum_{i=s}^{t-1}u_i}/f_{\theta_\star})(Y_s) {\mathds{1}}_{{\mathsf{Y}}^{(2)}_t}(Y_{1:t})\Big]\\
&\leq e^{(t-s_t)(c-{\mathbb{E}}[\log (f_{\theta_\star})])}\prod_{s=s_t+1}^t{\mathbb{E}}\big[  \tilde{f}_{\theta-\sum_{i=s}^{t-1}u_i} (Y_s)\big|\,{\mathcal{F}}_{s_t}\big]\\
&\leq e^{(t-s_t)(c-{\mathbb{E}}[\log (f_{\theta_\star})])}\big(\sup_{\theta\not\in A_\star}{\mathbb{E}}[\tilde{f}_\theta]\big)^{t-s_t}\\
&=e^{-(t-s_t)\big({\mathbb{E}}[\log f_{\theta_\star}]-c-\log(\sup_{\theta\not\in A_\star}  {\mathbb{E}}[\tilde{f}_\theta])\big)}\end{aligned}$$ where the second equality holds for $t\geq t_1$. This shows that holds with $c_{\mathrm{A4}}={\mathbb{E}}[\log (f_{\theta_\star})]-c-\log (\sup_{\theta\not\in A_\star}  {\mathbb{E}}[ \tilde{f}_\theta] )>0$.

The proof of the lemma is complete.

### Proof of Lemma \[lemma:part\_1\]\[p-lemma:part\_1\]

Let $(\phi_t)_{t\geq 1}$ and $ \tilde{D}_\star\in {\mathbb{R}}_{>0}$ be as Lemma \[lemma:test\], $\tilde{C}_\star\in {\mathbb{R}}_{>0}$ be as in Lemma \[lemma:denom\]. For every $t\geq 1$, let $C^{(\mu_{s_t}*\tilde{\pi}_{s_t})}_{\delta, \tilde{\delta}}$ be as defined in Lemma \[lemma:denom\] and let $$\begin{aligned}
 A_{t}=& \Big\{y_{1:t}\in{\mathsf{Y}}^t:\int_{\Theta}{\mathbb{E}}\Big[(\mu_{s_t}*\tilde{\pi}_{s_t})\big(\theta-\sum_{s=s_t+1}^{t-1}U_s\big)\prod_{s=s_t+1}^{t} \frac{\tilde{f}_{\theta-\sum_{i=s}^{t-1}U_i}}{f_{\theta_\star}}(Y_s)\big| Y_{1:t}=y_{1:t}\Big] {\mathrm{d}}\theta\\
 &\leq  {\mathbb{P}}(U_{s_t:t}\in \Theta_{\tilde{\delta},s_t:t}) C^{(\mu_{s_t}*\tilde{\pi}_{s_t})}_{\delta , \tilde{\delta} }   e^{-2(t-s_t) (\tilde{C}_\star \delta)^2}\Big\}.\end{aligned}$$

Let $\phi'_t(Y_{1:t})={\mathds{1}}_{A _{t}}(Y_{1:t})\vee \phi_t(Y_{1:t})$ and note that by, Lemmas \[lemma:denom\]-\[lemma:test\], $$\label{eq:den}
\begin{split}
\limsup_{t\rightarrow+\infty}{\mathbb{E}}\big[\phi'_t(Y_{1:t})\big]&\leq \limsup_{t\rightarrow+\infty}{\mathbb{E}}\big[\phi_t(Y_{1:t})\big]+\limsup_{t\rightarrow+\infty}{\mathbb{P}}\big(Y_{1:t} \in A_t\big)\\
&\leq \limsup_{t\rightarrow+\infty}\frac{1}{\delta^2 (t-s_t) \tilde{C}^2_\star }\\
&=0
\end{split}$$ as required.

On the other hand we have, ${\mathbb{P}}$-a.s. $$\begin{aligned}
&{\mathbb{E}}\big[(1-\phi'_t(Y_{1:t})){\mathds{1}}_{ \Theta_{\gamma_t,s_t:t}}(U_{s_t:t})\pi'_{s_t,t}(V_{\epsilon})|{\mathcal{F}}_{s_t}\big]\\
&\leq  \frac{e^{2(t-s_t)   (\tilde{C}_\star \delta)^2}}{{\mathbb{P}}(U_{s_t:t}\in \Theta_{\tilde{\delta},s_t:t}) C^{(\mu_{s_t}*\tilde{\pi}_{s_t})}_{\delta, \tilde{\delta}}}\\
&\times {\mathbb{E}}\Big[{\mathds{1}}_{ \Theta_{\gamma_t,s_t:t}}(U_{s_t:t})\int_{V_{\epsilon}}(1-\phi_t(Y_{1:t}))(\mu_{s_t}*\tilde{\pi}_{s_t})\Big(\theta-\sum_{s=s_t+1}^{t-1}U_s\Big)\prod_{s=s_t+1}^{t} \frac{\tilde{f}_{\theta-\sum_{i=s}^{t-1}U_i}}{f_{\theta_\star}}(Y_s) {\mathrm{d}}\theta\Big|{\mathcal{F}}_{s_t}\Big] \\
&\leq \frac{e^{-(t-s_t)(\tilde{D}_\star  -2    \tilde{C}^2_\star \delta^2)})}{{\mathbb{P}}(U_{s_t:t}\in \Theta_{\tilde{\delta},s_t:t}) C^{(\mu_{s_t}*\tilde{\pi}_{s_t})}_{\delta, \tilde{\delta}}}\end{aligned}$$ where the second inequality holds for $t$ large enough and uses Tonelli’s theorem and Lemma \[lemma:test\]. This complete the proof of the lemma with $ C_1=\tilde{D}_\star^{-1}$ and $C_2=2\tilde{C}^2_\star$.

### Proof of Lemma \[lemma:tech\_conv\] {#p-lemma:tech_conv}

Let $\varphi:{\mathbb{R}}^{d_x}\times{\mathbb{R}}^{d_z}\rightarrow [-1,1]$ and, for every $x\in{\mathbb{R}}^{d_x}$, let $\varphi_N(x)=\sum_{m=1}^N\tilde{W}^m f(x,Z^m)$ and $\varphi_X(x)=\int_{{\mathbb{R}}^{d_z}} f(x,z)\mu({\mathrm{d}}z)$. Then, $$\label{eq:setp1}
\begin{split}
{\mathbb{E}}\Big[\Big( &\int_{{\mathbb{R}}^{d_x}\times {\mathbb{R}}^{d_z}}  \varphi(x,z)( \pi^N\otimes\mu^N-\pi\otimes\mu)({\mathrm{d}}x,{\mathrm{d}}z)\Big)^2\Big]^{1/2}\\
&={\mathbb{E}}\Big[\Big(\sum_{n=1}^N W^n\sum_{m=1}^N \tilde{W}^m \varphi(X^n,Z^m)-\int_{{\mathbb{R}}^{d_x}\times{\mathbb{R}}^{d_z}} \varphi(x,z)\pi\otimes \mu({\mathrm{d}}x,{\mathrm{d}}z)\Big)^2\Big]^{1/2}\\
&= {\mathbb{E}}\Big[\Big(\sum_{n=1}^N W^n \varphi_N(X^n)-\int_{{\mathbb{R}}^{d_x}} \varphi_X(x)\pi({\mathrm{d}}x)\Big)^2\Big]^{1/2}\\
&\leq {\mathbb{E}}\Big[\Big(\sum_{n=1}^N W^n \varphi_X(X^n)-\int_{{\mathbb{R}}^{d_x}} \varphi_X(x)\pi({\mathrm{d}}x)\Big)^2\Big]^{1/2}\\
&+{\mathbb{E}}\Big[\Big(\sum_{n=1}^N W^n \big(\varphi_N(X^n)-\varphi_X(X^n)\big)\Big)^2\Big]^{1/2}\\
&\leq \frac{C^{1/2}_\star}{N^{1/2}}+ {\mathbb{E}}\Big[\sum_{n=1}^NW^n \big(\varphi_N(X^n)-\varphi_X(X^n)\big)^2\Big]^{1/2} 
\end{split}$$ where the last inequality uses Jensen’s inequality.

To control the expectation on the r.h.s. of remark that $$\begin{aligned}
\sum_{n=1}^N{\mathbb{E}}\Big[W^n \big(\varphi_N(X^n)&-\varphi_X(X^n)\big)^2\Big]\\
&=\frac{1}{N}\sum_{n=1}^N{\mathbb{E}}\Big[\frac{G(X^n)}{p^N_X(G)} \big(\varphi_N(X^n)-\varphi_X(X^n)\big)^2\Big]\\
&=\frac{1}{N}\sum_{n=1}^N{\mathbb{E}}\Big[\frac{G(X^n)}{p^N_X(G)} \big(\varphi_N(X^n)-\varphi_X(X^n)\big)^2{\mathds{1}}\big(p^N_X(G)\geq p_X(G)/2\big)\Big]\\
&+\frac{1}{N}\sum_{n=1}^N{\mathbb{E}}\Big[\frac{G(X^n)}{p^N_X(G)} \big(\varphi_N(X^n)-\varphi_X(X^n)\big)^2{\mathds{1}}\big( p^N_X(G)<p_X(G)/2\big)\Big]\\
&\leq  \frac{2\|G\|_\infty}{p_X(G)}\,\sup_{x\in{\mathbb{R}}^{d_x}}{\mathbb{E}}\big[ \big(\varphi_N(x)-\varphi_X(x)\big)^2]+4{\mathbb{P}}\big(p^N_X(G)<p_X(G)/2\big)\\
&\leq \frac{2 C_\star \|G\|_\infty}{N p_X(G)}+4{\mathbb{P}}\big(|p^N_X(G)-p_X(G)|>p_X(G)/2\big)\\
&\leq \frac{2 C_\star \|G\|_\infty}{N p_X(G) }+16\frac{{\mathbb{E}}[(p^N_X(G)-p_X(G))^2]}{p_X(G)^2}\\
&\leq \frac{2 C_\star \|G\|_\infty}{N\,p_X(G) }+\frac{16 C_\star}{ Np_X(G)^2}\end{aligned}$$ where the penultimate inequality uses Markov’s inequality. The proof is complete.

[^1]: available at: <https://archive.ics.uci.edu/ml/datasets/covertype>
