Mean stopping time of a Brownian motion

Question

I came across the following proof of the fact that the mean stopping time of a Brownian motion to hit $-1$ or $1$ is $1$:

Let $B$ be a Brownian motion. We already know $B_t^2-t$ is a martingale. Let $T=\min(t:B_t\in\{-1,1\})$. A martingale stopped at a stopping time is still a martingale, so $B_T^2-T$ is a martingale. It follows that $E(B_T^2-T)=0$ so $E(T)=E(B_T^2)=1$.

I understand everything except for the notation $B_T^2-T$. What exactly is meant by this? Is it the martingale defined by $W_t=B_t^2-t$ if $t<T$ and $W_t=B_T^2-T$ otherwise? If so, what does $E(B_T^2-T)$ mean? Is it $E(W_T)$ or $\lim_{t\to\infty}E(W_t)$?

Answer 1

If $(X_t)_{t \geq0}$ is any stochastic process and $\tau$ is any non-negative random variable then $X_{\tau}$ is defined by $X_{\tau}(\omega)=X_{\tau(\omega)} (\omega)$. This is well defined but it is not a random variable in general. If the process $(X_t)_{t \geq0}$ has continuous paths and $\tau$ is a stopping time it can be shown that this is a random variable. Of course, if $\tau$ happens to be a constant $t$ the $X_{\tau}$ is same as $X_t$.

Answer 2

The proof skipped a few steps. Applying the optional stopping theorem to $W$ with condition c makes it more complete.