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

Rasputin
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    It seems to be explained pretty clearly in the answer. You show that $B_t^2 - t$ is a martingale and then you apply the optional stopping theorem. Can you explain which part of that you are asking about? – Nate Eldredge Dec 14 '18 at 16:57

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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.)

Myunghyun Song
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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 $Var(B_t)=t=E(B_t^2)-E(B_t)^2=E(B_t^2)$.

Rasputin
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    It's not true that $B_t^2 = t$. It's true that $E[B_t^2] = t$. But this property does not automatically carry over when you replace a constant time $t$ by a stopping time. One has to prove it. – Nate Eldredge Dec 14 '18 at 16:54