Expected value of Dirac measure: $E[\delta_{\tau}(t)]$ Thx for the efforts! I tried to clean up the question and remove the ambiguous notations. For the dirac measure, I use the one found in the Wikipedia article.

I try to formally understand a statement found across literature (as @Did pointed out, assuming independence):

$E[\mathbb{I}_{[0,T]}(\tau)D(\tau)] = \int_{0}^{T}E[D(t)]dP_{\tau}(t)$.

To derive it, the authors do the following:

$E[\mathbb{I}_{[0,T]}(\tau)D(\tau)] = E[\int_{-\infty}^{\infty} \mathbb{I}_{[0,T]}(t) D( t) d\delta_{\tau}(t)] = E[\int_{0}^{T} D( t) d\delta_{\tau}(t)] = E[\int_{0}^{T} D( t) \delta_{\tau}(t)dt]$.

And then, what I don't get, they do the following (which might be a notational mess):

$\int_{0}^{T} E[D(t)] E[d\delta_{\tau}(t)] = \int_{0}^{T} E[D(t)] P(\tau \in [t,dt]) $

There are two questions


How does the distribution of $\tau$ enter the picture, i.e. what is the relation of $\delta_{\tau}$ to $P_{\tau}$, to get the proposition (1st equation) from the second equation? (MickG's explanations).
What is going on with the third equation? Which is hopefully resolved after answering the first question:-) Maybe, the authors mean something like $E[\delta_{\tau}(t)] = E[\mathbb{I}_{[t,dt]}(\tau)]=P(\tau \in [t,dt])$...


Thx!
On the RHS I get $\int_0^\infty E[D(t)]\mathbb1_{0&lt;t&lt;T}d\mu(t)=\int_0^TE[D(t)]d\mu(t)$. On the LHS, I get $E[E[D(t)]\mathbb1_{0&lt;t&lt;T}]$. So:

$$\int_0^TE[D(t)]d\mu(t)=E[E[D(t)]\mathbb1_{0&lt;t&lt;T}].$$

Seems we have one $E$ too many on the right… If I consider $D(t)\mathbb1_{[0,T]}(t)$ instead, I get:

$$E[D(t)\mathbb1_{[0,T]}(t)]=\int_0^\infty D(t)\mathbb1_{[0,T]}(t)d\mu(t)=\int_0^TD(t)d\mu(t).$$

Now there is a missing $E$. So the question I ask is: is the identity stated in your reference actually correct? Also, that integral of $E[D(t)]$ leaves me wondering whether $E[D(t)]$ actually depends on $t$. After all, it is an expected value, so why should it depend on anything? With that, I would have that the first equality has LHS equal to $P([0,T])E[D(t)]$.

Let us now look at how your authors derive their identity:

\begin{align*}
E[\mathbb{I}_{[0,T]}(\tau)D(\tau)] ={}&amp; E\left[\int_{-\infty}^{\infty} \mathbb{I}_{[0,T]}(t) D( t) \delta(t-\tau)dt\right] = E\left[\int_{0}^{T} D( t) \delta(t-\tau)dt\right] ={} \\
{}={}&amp; \int_{0}^{T} E[D(t)] E[\delta(t-\tau)]dt = \int_{0}^{T} E[D(t)] P(\tau \in [t,dt]).
\end{align*}

I restored the "$\delta$ function" since this is just a formal trick to convince readers of the equality. Let us try to see this formal derivation step by step:


Introduce the $\delta$ function to express evaluation as an integral;
Collapse the indicator into the integration bounds;
Bring the $E$ in (it is, at the end of the day, an integral, so you can surely swap it with $\int_0^T$, right?), then assume some sort of independence between $D$ and $\delta$ to turn $E$ of a product to a product of $E$; now this is very sloppy;
A very big problem; It seems they assume $E[\delta(t-\tau)]=P(\tau\in[t,dt])$; why is that? And what does that probability even mean? After all, infinitesimal are out of standard analysis, so what is $dt$? They are probably thinking of having $\tau$ very close to $t$; or, they may be thinking of approximating the $\delta$ function with some more treatable function; if $\delta$ is 0 everywhere except $\delta(0)=\infty$, we can see $\delta(x)=\lim_{n\to\infty}n\mathbb1_{[-\frac1n,\frac1n]}(x)$; so $\delta(t-\tau)=\lim_nn\mathbb1_{[-\frac1n,\frac1n]}(t-\tau)$; assuming limit and integral swap, $E[\delta(t-\tau)]=\int\delta(t-\tau)d\mu(\tau)=\lim_n\int n\mathbb1_{[-\frac1n,\frac1n]}(t-\tau)d\mu(\tau)=\lim_nnP(\tau\in[t-\frac1n,t+\frac1n]$; now $\tau$ has a density, and we know the density is the derivative of the cumulative distribution function, so since $P(\tau\in[t-\frac1n,t+\frac1n])=F_P(t+\frac1n)=F_P(t-\frac1n)$, where $F_P$ is the CDF, we get precisely $f(t)$ as that expected value of the $\delta$.


To better (but always formally) treat the third step, we may assume Fubini-Tonelli works fine and try this:

$$E\left[\int_0^TD(t)\delta(t-\tau)dt\right]=\int_0^\infty\int_0^TD(t)\delta(t-\tau)dtd\mu(\tau)=\int_0^T\int_0^\infty\delta(t-\tau)d\mu(\tau)D(t)dt,$$

then we have that inner integral, and, if $d\mu(\tau)=f(\tau)d\tau$, we know that is $f(t)\mathbb1_{[0,\infty)}(t)$, and we are there.

I hope this answers your question. Though that identity in your reference seems incorrect, as I pointed out at the start considering two different functions.
On the RHS I get $\int_0^\infty E[D(t)]\mathbb1_{0&lt;t&lt;T}d\mu(t)=\int_0^TE[D(t)]d\mu(t)$. On the LHS, I get $E[E[D(t)]\mathbb1_{0&lt;t&lt;T}]$. So:

$$\int_0^TE[D(t)]d\mu(t)=E[E[D(t)]\mathbb1_{0&lt;t&lt;T}].$$

Seems we have one $E$ too many on the right… If I consider $D(t)\mathbb1_{[0,T]}(t)$ instead, I get:

$$E[D(t)\mathbb1_{[0,T]}(t)]=\int_0^\infty D(t)\mathbb1_{[0,T]}(t)d\mu(t)=\int_0^TD(t)d\mu(t).$$

Now there is a missing $E$. So the question I ask is: is the identity stated in your reference actually correct? Also, that integral of $E[D(t)]$ leaves me wondering whether $E[D(t)]$ actually depends on $t$. After all, it is an expected value, so why should it depend on anything? With that, I would have that the first equality has LHS equal to $P([0,T])E[D(t)]$.

Let us now look at how your authors derive their identity:

\begin{align*}
E[\mathbb{I}_{[0,T]}(\tau)D(\tau)] ={}&amp; E\left[\int_{-\infty}^{\infty} \mathbb{I}_{[0,T]}(t) D( t) \delta(t-\tau)dt\right] = E\left[\int_{0}^{T} D( t) \delta(t-\tau)dt\right] ={} \\
{}={}&amp; \int_{0}^{T} E[D(t)] E[\delta(t-\tau)]dt = \int_{0}^{T} E[D(t)] P(\tau \in [t,dt]).
\end{align*}

I restored the "$\delta$ function" since this is just a formal trick to convince readers of the equality. Let us try to see this formal derivation step by step:


Introduce the $\delta$ function to express evaluation as an integral;
Collapse the indicator into the integration bounds;
Bring the $E$ in (it is, at the end of the day, an integral, so you can surely swap it with $\int_0^T$, right?), then assume some sort of independence between $D$ and $\delta$ to turn $E$ of a product to a product of $E$; now this is very sloppy;
A very big problem; It seems they assume $E[\delta(t-\tau)]=P(\tau\in[t,dt])$; why is that? And what does that probability even mean? After all, infinitesimal are out of standard analysis, so what is $dt$? They are probably thinking of having $\tau$ very close to $t$; or, they may be thinking of approximating the $\delta$ function with some more treatable function; if $\delta$ is 0 everywhere except $\delta(0)=\infty$, we can see $\delta(x)=\lim_{n\to\infty}n\mathbb1_{[-\frac1n,\frac1n]}(x)$; so $\delta(t-\tau)=\lim_nn\mathbb1_{[-\frac1n,\frac1n]}(t-\tau)$; assuming limit and integral swap, $E[\delta(t-\tau)]=\int\delta(t-\tau)d\mu(\tau)=\lim_n\int n\mathbb1_{[-\frac1n,\frac1n]}(t-\tau)d\mu(\tau)=\lim_nnP(\tau\in[t-\frac1n,t+\frac1n]$; now $\tau$ has a density, and we know the density is the derivative of the cumulative distribution function, so since $P(\tau\in[t-\frac1n,t+\frac1n])=F_P(t+\frac1n)=F_P(t-\frac1n)$, where $F_P$ is the CDF, we get precisely $f(t)$ as that expected value of the $\delta$.


To better (but always formally) treat the third step, we may assume Fubini-Tonelli works fine and try this:

$$E\left[\int_0^TD(t)\delta(t-\tau)dt\right]=\int_0^\infty\int_0^TD(t)\delta(t-\tau)dtd\mu(\tau)=\int_0^T\int_0^\infty\delta(t-\tau)d\mu(\tau)D(t)dt,$$

then we have that inner integral, and, if $d\mu(\tau)=f(\tau)d\tau$, we know that is $f(t)\mathbb1_{[0,\infty)}(t)$, and we are there.

I hope this answers your question. Though that identity in your reference seems incorrect, as I pointed out at the start considering two different functions.