Xrfgtjtk

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
$$X = BW(delta) + (1-B)W(2delta),$$
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 $t=2delta$, equally likely.

So given the observation $X$, I can infer something about the time of its occurrence: By Bayesian rule,
$$mathbb{P}(B=1|X) = frac{mathbb{P}(B=1)mathbb{P}(X|B=1)}{mathbb{P}(B=1)mathbb{P}(X|B=1)+mathbb{P}(B=0)mathbb{P}(X|B=0)}
= frac{frac{1}{2}mathbb{P}(W(delta)=X)}{frac{1}{2}mathbb{P}(W(delta)=X)+frac{1}{2}mathbb{P}(W(2delta)=X)}.
$$
Given that $W(t)$ is a Brownian motion, $W(delta)$ and $W(2delta)$ are normally distributed with respective densities
$$mathbb{P}(W(delta)=X) = frac{1}{sqrt{2pidelta}}e^{-X^2/(2delta)}$$
and
$$mathbb{P}(W(2delta)=X) = frac{1}{sqrt{4pidelta}}e^{-X^2/(4delta)}.$$
So the above translates to a function of the density ratio
$$mathbb{P}(B=1|X) = frac{1}{1+frac{1}{sqrt{2}}e^{X^2/(4delta)}},
$$
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 $W(2delta)$.

Next I want to consider what happens when $deltadownarrow0$, i.e., when the gap between the timing of the observation asymptotically vanishes.
Intuitively, when $deltadownarrow0$, $W(delta)=W(2delta)=W(0)$ and $X=W(0)$, leaving no room for inference for $B$.

Analytically, however, from the above expression, in the limit of $deltadownarrow0$, $mathbb{P}(B=1|X)=0$, 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 $deltadownarrow0$?

The probability $$mathbb{P}(B = 1 mid X) = frac{1}{1 + frac{1}{sqrt{2}}e^{X^2/(4delta)}}$$ only depends on $X$ and $delta$ through the term $X^2/(4delta)$. 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 $X^2/(4delta) sim chi^2_1/4$, the distribution of a $chi^2_1$ random variable divided by $4$.

When $B = 0$ then $X sim N(0, 2delta)$, so $X/sqrt{2delta} sim N(0, 1)$. Therefore $X^2/(2delta) sim chi^2_1$, and $X^2/(4delta) sim chi^2_1/2$, the distribution of a $chi^2_1$ random variable divided by $2$.

In either case, the distribution of $X^2/(4delta)$ does not depend on $delta$. So as $delta rightarrow 0$, $mathbb{P}(B = 1mid X)$ does not tend to any particular limit.