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Let $B_t$ be a standard browian motion without drift. I define a stopping time $\tau$ as the first exit time from the interval $[-c,c]$ (assuming for simplicity, symmetrical barriers). $$ \tau = \inf \{t\geq0 \mid \left| B_t\right|\geq c\} $$

Additionally I introduce a fixed time barrier $T$. I consider the bounded stopping time $\tilde\tau$, defined as the minimum of $T$ and $\tau$:

$$ \tilde\tau = T\wedge\tau $$

This can be seen as a "triple barrier" where I stop my process at a time T if no barrier has been hit. I want to compute $E[\tilde\tau]$.

Intuitively, as $T\rightarrow \infty$, then $E[\tilde\tau] \rightarrow E[\tau]=c²$

Can't really do much, I've been stuck but so far, since $B_t^2 -t$ is a martingale, $$ E[B_{\tilde\tau}^2] = E[\tilde\tau] $$

J. W. Tanner
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pduce
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    The distribution of $\tau$ is known. See https://math.stackexchange.com/questions/3550228/distribution-of-first-exit-time-of-brownian-motion – Andrea Aveni Nov 20 '24 at 23:01
  • @AndreaAveni I think one can compute the expected value directly without knowing the exact distribution. For the example in your link it should be $|ab|$. – Pedro Nov 20 '24 at 23:21
  • I know, but you are interested in the expectation of $\tilde\tau$, not in that of $\tau$. From the distribution of $\tau$, you can get the distribution of $\tilde\tau$, but from $\mathbb E[\tau]$ you cannot get $\mathbb E[\tilde\tau]$. – Andrea Aveni Nov 21 '24 at 00:11

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I am not aware of closed forms for $E[\tau \wedge T]$. One can however start from following. We have, since $\tau \wedge T\leq T$ and for all $s<T$, $\{\tau \wedge T>s\}=\{T>s\}\cap\{\tau>s\}=\{\tau>s\}$, that $E[\tau \wedge T]=\int_0^TP(\tau>s)ds$. The density of $\tau$ is known (see also e.g. Borodin-Salminen) in the form of an alternating sum of Levy distributions. The resulting integral for $E[\tau \wedge T]$ does not look easily tractable, but it can be studied.

Snoop
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