# Distribution of Brownian motion before stoping time.

Let $$B_t$$ be a standard Brownian motion, and define the stopping time time $$\tau_a=\inf\{t\geq0:|B_t|=a\}$$. How do we find find $$\mathbb E\left(B_{\frac{\tau_a}2}\right)$$? Where can I read about it? Thanks in advance.

We have $E[B_{\tau_a/2}]=0$, essentially by symmetry.

One should first prove, or accept on faith, that the question is well-posed: that the answer is completely determined from the properties of a Brownian motion, so that if we start with another standard Brownian motion $\{W_t\}$ and define $\sigma_a = \inf\{t \ge 0 : |W_t| = a\}$, then $E[B_{\tau_a/2}] = E[W_{\sigma_a/2}]$. (One way to do this would be to use the fact that every Brownian motion induces the same Wiener measure $\mu$ on $C([0,\infty))$, and then express the problem as the integral of a certain measurable function on this space.)

Once convinced of that, set $W_t = -B_t$, so that $W_t$ is also a standard Brownian motion. Then letting $\sigma_a = \inf\{t \ge 0 : |W_t| = a\}$, as above we have $E[B_{\tau_a/2}] = E[W_{\sigma_a/2}]$. But $$\sigma_a = \inf\{t \ge 0 : |W_t| = a\} = \inf\{t \ge 0 : |-B_t| = a\} = \inf\{t \ge 0 : |B_t| = a\} = \tau_a.$$ Then \begin{align*} E[B_{\tau_a/2}] &= E[W_{\sigma_a/2}] && \text{as above} \\ &= E[W_{\tau_a/2}] && \text{since \tau_a = \sigma_a} \\ &= E[-B_{\tau_a/2}] && \text{since W_t = -B_t} \\ &= -E[B_{\tau_a/2}]\end{align*} and therefore $E[B_{\tau_a/2}]=0$.

• You mean that $E[B_t] = E[-B_t]$, and so, it is 0? But here we have stopping time $\tau$, which is dependent with $B_t$, not just random time t.
– Ilya
Commented Mar 24, 2015 at 16:06
• @Ilya: So let $\sigma$ be a stopping time defined from $-B_t$ in the same way that $\tau$ is defined from $B_t$. Then replacing $B_t$ with $-B_t$ forces you to replace $\tau$ by $\sigma$. But for your particular $\tau$, how is it related to $\sigma$? Commented Mar 24, 2015 at 16:09
• @ Nate Eldredge If $\sigma$ is defined in the same way as $\tau$, so $\sigma = inf\{t\ge 0: |-B_t| = a\}$, then $\sigma = \tau$, right?
– Ilya
Commented Mar 24, 2015 at 16:25
• @Ilya: Right! By the way, use \inf instead of just inf to get the prettier $\inf$ instead of $inf$. Commented Mar 24, 2015 at 16:26
• @ Nate Eldredge :Okay, thank you! But I still do not understand why $E[-B_{\frac{\sigma}{2}}]$ which is equal to $-E[B_{\frac{\tau}{2}}]$ will be equal to $E[B_{\frac{\tau}{2}}]$ ?
– Ilya
Commented Mar 24, 2015 at 16:43