The question is as follows:
Let $W$ be Brownian motion and $T$ a stopping time with $\mathbb{E} T < ∞$. Show
(use stochastic integrals) that $\mathbb{E}W_T = 0$ and $\mathbb{E} W^2_T = \mathbb{E} T$.
Now, I have showed this result before (it was somewhere along these lines: Dominated convergence problems with Wald's identity for the Brownian Motion), but that proof did not include stochastic integrals and I do not see where it would come in handy.
At first, I figured that we have $W_t = \int_0^t \text{ d}W_s$ for any $t\geq0$, so the equality will also hold when we plug in a stopping time $T$. However, then we have
$$
\mathbb{E}W_T = \mathbb{E}\int_0^T 1 \text{ d}W_s,
$$
but we cannot swap integral and expectation here. Does anyone have an idea where we could use stochastic integrals in a meaningful way in this proof?
Best Answer
I guess that the problem is asking you to use stochastic integral in place of the optional stopping theorem. Notice that the Itô isometry applied to
$$ W_{T\wedge b} - W_{T\wedge a} = \int_{a}^{b} \mathbf{1}_{\{s < T\}} \, \mathrm{d}W_s, \quad (a < b) $$
yields
$$ \Bbb{E}( |W_{T\wedge b} - W_{T\wedge a}|^2 ) = \int_{a}^{b} \Bbb{P}(s < T) \, \mathrm{d}s.$$
From this and $\int_{0}^{\infty} \Bbb{P}(s < T) \, \mathrm{d}s = \Bbb{E}T < \infty$, we find that $\{W_{T\wedge t} : t \geq 0 \}$ is Cauchy in $L^2$ and hence convergent in $L^2$. But since $T$ is finite a.s., we know that $W_{T\wedge t} \to W_T$ a.s. This identifies the $L^2$-limit of $W_{T\wedge t}$ as $W_T$. Putting these altogether, we have
$$\Bbb{E} W_T^2 = \lim_{t\to\infty} \Bbb{E} W_{T\wedge t}^2 = \lim_{t\to\infty} \int_{0}^{t} \Bbb{P}(s < T) \, \mathrm{d}s = \int_{0}^{\infty} \Bbb{P}(s < T) \, \mathrm{d}s = \Bbb{E}T. $$
Finally, since $W_{T \wedge t} = \int_{0}^{t} \mathbf{1}_{ \{ s < T \}} \, \mathrm{d}W_s$ is an $L^2$-martingale which converges to $W_T$ in $L^2$, it also converges in $L^1$ and hence
$$ \Bbb{E}W_T = \lim_{t\to\infty} \Bbb{E}W_{T \wedge t} = 0. $$