If $\tau=\inf \{t \ge 0: B(t)= -a \text{ or } b\}$ is a stopping time of Brownian motion $B_t$, why does $E(\tau)=E(B_{\tau}^2)$?

Question

The solution to this question states that $E(\tau)=E(B_{\tau}^2)$. But how do we know this?

Answer 1

In fact, the hint given above only applies to deterministic $t$, not to random time $\tau$. The problem here is that $$ E[B^2_\tau |\tau = t] \neq t.$$ You should use the fact that $$ B^2_t - t$$ is a continuous $\{\mathcal{F}_t\}$-martingale. By optional sampling theorem, we have $$ E[B^2_{\tau\wedge t} - (\tau\wedge t)] = 0. $$ Now, let $t\to \infty$ and conclude that $$ E[B^2_{\tau}]=E[\tau]. $$ (by Lebesgue's dominated convergence and monotone convergence theorem.)

Answer 2

Ohh wait I see, it's because it's a property of standard Brownian motion that $B_t^2=t$ for all $t \ge 0$!!

Since $B(t) \sim N(0, t)$, so $Var(B_t)=t=E(B_t^2)-E(B_t)^2=E(B_t^2)$.