Inferring observation time from a Brownian motion This might be a bit lengthy question. So let me proceed in steps.
General description: I have some observations, based on which I want to infer their occurring time.
Specific setting: Let $W(t)$ be a Brownian motion.
I have an observation $X$ defined as
$$X = BW(\delta) + (1-B)W(2\delta),$$
where $B$ is a Bernoulli draw with success rate $\frac{1}{2}$ and is independent of $W(t)$; and $\delta>0$ is some constant observation latency.
In words, the above is meant to capture the setting that I do not know when the observation $X$ occurred:
It might have occurred at time $t=\delta$ or at time $t=2\delta$, equally likely.
So given the observation $X$, I can infer something about the time of its occurrence: By Bayesian rule,
$$\mathbb{P}(B=1|X) = \frac{\mathbb{P}(B=1)\mathbb{P}(X|B=1)}{\mathbb{P}(B=1)\mathbb{P}(X|B=1)+\mathbb{P}(B=0)\mathbb{P}(X|B=0)}
= \frac{\frac{1}{2}\mathbb{P}(W(\delta)=X)}{\frac{1}{2}\mathbb{P}(W(\delta)=X)+\frac{1}{2}\mathbb{P}(W(2\delta)=X)}.
$$
Given that $W(t)$ is a Brownian motion, $W(\delta)$ and $W(2\delta)$ are normally distributed with respective densities
$$\mathbb{P}(W(\delta)=X) = \frac{1}{\sqrt{2\pi\delta}}e^{-X^2/(2\delta)}$$
and
$$\mathbb{P}(W(2\delta)=X) = \frac{1}{\sqrt{4\pi\delta}}e^{-X^2/(4\delta)}.$$
So the above translates to a function of the density ratio
$$\mathbb{P}(B=1|X) = \frac{1}{1+\frac{1}{\sqrt{2}}e^{X^2/(4\delta)}},
$$
which is decreasing in the magnitude of $|X|$.
That is, the more extreme $X$ realizes to be, the less likely it comes from the early $W(\delta)$.
This makes sense as $W(\delta)$ is less volatile than $W(2\delta)$.
Next I want to consider what happens when $\delta\downarrow0$, i.e., when the gap between the timing of the observation asymptotically vanishes.
Intuitively, when $\delta\downarrow0$, $W(\delta)=W(2\delta)=W(0)$ and $X=W(0)$, leaving no room for inference for $B$.
Analytically, however, from the above expression, in the limit of $\delta\downarrow0$,  $\mathbb{P}(B=1|X)=0$, suggesting that I should always infer that $B=0$.
So what is the correct result? Can I infer anything about $B$ in the limit of $\delta\downarrow0$?

The probability $$\mathbb{P}(B = 1 \mid X) = \frac{1}{1 + \frac{1}{\sqrt{2}}e^{X^2/(4\delta)}}$$ only depends on $X$ and $\delta$ through the term $X^2/(4\delta)$. Now let's think about the cases when $B = 1$ or $B = 0$.
When $B = 1$ then $X \sim N(0, \delta)$, so $X/\sqrt{\delta} \sim N(0, 1)$. Therefore $X^2/\delta \sim \chi^2_1$, a chi-squared distribution. And $X^2/(4\delta) \sim \chi^2_1/4$, the distribution of a $\chi^2_1$ random variable divided by $4$.
When $B = 0$ then $X \sim N(0, 2\delta)$, so $X/\sqrt{2\delta} \sim N(0, 1)$. Therefore $X^2/(2\delta) \sim \chi^2_1$, and $X^2/(4\delta) \sim \chi^2_1/2$, the distribution of a $\chi^2_1$ random variable divided by $2$.
In either case, the distribution of $X^2/(4\delta)$ does not depend on $\delta$. So as $\delta \rightarrow 0$, $\mathbb{P}(B = 1\mid X)$ does not tend to any particular limit.
