Distribution of Brownian motion before stoping time.

Question

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.

Answer 1

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 $E[B_{\tau_a/2}]=0$.