Consider one dimensional brownian motion. Let's suppose it starts at point $a \in [0,L]$. If particles reach the bounds of segment, then they are falling down.

Let's suppose that it goes a long time and all particles are falling down, then what part of particles fall in left bound ?

I've thought about stochastic solution, but it more looks like something else. Any hints?

  • $\begingroup$ Is this brownian motion supposed to be a Weiner process, or a 1-D discrete random walk? Or something else? $\endgroup$ – Mark Fischler Mar 14 at 15:24
  • $\begingroup$ Let $T$ be the time the particle falls down. On the one hand, $E[B_T]=a$. On the other hand, $B_T$ equals either $0$ or $L$, so $E[B_T]$ can also be written as$\dots$ $\endgroup$ – Mike Earnest Mar 14 at 15:25
  • $\begingroup$ @MarkFischler I guess it's supposed to be a Wiener process $\endgroup$ – openspace Mar 14 at 15:35
  • $\begingroup$ @MikeEarnest $B_T $ is position of particles in moment $T$ ? $\endgroup$ – openspace Mar 14 at 15:36
  • 1
    $\begingroup$ @MarkFischler $B_t$ is the position of a single particle in the ensemble. It is assumed to follow a standard wiener process until it hits the boundary $\{0,L\}$ and hence, by definition of the Wiener process basically, $\mathbb{E}[B_t] = a$ for all $t$. You can extend this to the hitting time of the boundary $T$ using martingale techniques (by an optional stopping argument) since $B_{t \wedge T}$ is a bounded martingale. So we do have $\mathbb{E}[B_T] = a$ under these assumptions. $\endgroup$ – Rhys Steele Mar 14 at 15:51

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