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I have narrow escape problem here. But for the sake of completeness, it measures the time it takes for a Brownian particle, $\alpha$, escape through a narrow hole, $A$, on an otherwise reflecting surface, enclosing a volume $V$, provided that $a\ll V^{1/3}$, where $a$ is the largest dimension of $A$.

Assume that know there is a probability distribution $p(t)$ that gives the probability that $\alpha \in V$ at $t$. We learn that $\alpha \in V$ at $T$, but we don't know its exact location in $V$.

Can we assume that $p(t+T|(\alpha \in V)_{t=T})=p(t)$, without an explicit knowledge of $p(t)$, relying on the fact that Brownian motion itself is time independent?

ck1987pd
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  • Is $T$ fixed or is it a random hitting time? – Ian Mar 03 '22 at 16:30
  • $T$ is fixed, i.e., it is just the time we observe the system. – ck1987pd Mar 03 '22 at 16:31
  • And the unconditional start point is at the "center" of the domain, in some appropriate sense? – Ian Mar 03 '22 at 16:33
  • @Ian Yes, the start point is at the very centre of the domain for my simulation. But is the start point important since we are time averaging? – ck1987pd Mar 03 '22 at 16:36
  • Are you time averaging? I don't see any indication of that here. I actually look at this with an eye for counterexamples and so I think of very small $t$, in which case the conditional one is likely to be near the boundary and the unconditional one is not. – Ian Mar 03 '22 at 16:37
  • I think time averaging was unnecessary but as long as the start point is the same, the problem should be either time independent or not for all start points in $V$. – ck1987pd Mar 03 '22 at 16:39
  • @Ian why would the conditional one is more likely to be near the boundary, unless start point is the furthest possible point in the domain? Even for small $t$, this doesn't make sense. – ck1987pd Mar 03 '22 at 16:41
  • Sorry, I thought you were conditioning on hitting the boundary at time $T$. What is the meaning of conditioning on being in the volume? You're in the volume all the time right? You just stick to the absorbing piece of the boundary. Or is the condition on the RHS that you haven't absorbed yet at time $T$? – Ian Mar 03 '22 at 16:43
  • Let me try to clarify. If $\alpha$ leaves $V$, it can't come back, it is absorbed. With physics terminology, we measure if $\alpha$ is still in $V$ at $t=T$ and we know that it really is, but we don't know where in $V$. I ask about the probability of still being in $V$ at $t+T$, if we are certain that it was in $V$ at $T$. – ck1987pd Mar 03 '22 at 16:48
  • OK, so ${ \alpha \in V } = { \alpha \not \in A }$. – Ian Mar 03 '22 at 17:11
  • And you actually did want $p(t \mid \dots)$ not $p(t+T \mid \dots)$? I don't really see how $p(t \mid \dots)$ is guaranteed to make sense, since $t$ could be less than $T$. – Ian Mar 03 '22 at 17:32
  • @Ian Yes, I have to fix it. – ck1987pd Mar 03 '22 at 17:45
  • OK. Then no, these are still not the same, because the conditional distribution at time $T$, even when conditioned on not escaping, is different from the starting distribution. – Ian Mar 03 '22 at 17:47
  • @Ian Why? We have no knowledge of its location. But I assume you now have enough to carry on with an answer :) – ck1987pd Mar 03 '22 at 17:48
  • You started with some initial distribution, it evolves, even if you discard the trajectories where it hits the boundary before time $T$. I can't explicitly write down how it evolves in this reflecting boundary setting, but it does. – Ian Mar 03 '22 at 17:50
  • Unfortunately my best proof of this would be a numerical calculation that I don't really have the spare time to implement. – Ian Mar 03 '22 at 18:15

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