I am working on an exercising about proving Wald's second equation, which is as follows:
Let $S_n = \xi_1 + \cdots + \xi _n$ where $\xi _i$ are independent, $\mathbb{E} \xi_1 = 0$ and $\text{Var}(\xi_1) = \sigma^2 < \infty$. Show that if $T$ is a stopping time with $\mathbb{E} T < \infty$ then $\mathbb{E} S ^2 _T = \sigma^2 \mathbb{E} T$.
I am aware and understand proof of the above statement using an $L^2$ convergence theorem, see this. However, I have tried proving this with the Optional Stopping theorem but to no avail. The exact statement of the theorem is as follows:
Optional Stopping Theorem
Suppose $X_n$ is a martingale and $\mathbb{E}(|X_{n+1} - X_n| | \mathcal{F}_n) \leq B$ a.s. for some $B < \infty$. If $N$ is a stopping time with $\mathbb{E} N < \infty$ then $X _{N \wedge n}$ is uniformly integrable and hence $\mathbb{E} X_N = \mathbb{E} X_0$.
Suppose $\mathbb{F}_n = \sigma(\xi_1 , \dots, \xi _n)$. To use this theorem, $\mathbb{E}T < \infty$ had been given. Let $$ X_n = S ^2 _n - n \sigma^2, $$ which is a martingale, we need to show $\mathbb{E}(|X_{n+1} - X_n| | \mathcal{F}_n) \leq B$. Then $$ \begin{align*} \mathbb{E}(|X_{n+1} - X_n| | \mathcal{F}_n) &= \mathbb{E}(|S^2 _{n+1} - S^2 _n - \sigma^2 | | \mathcal{F}_n) \\ &\leq \mathbb{E}(|S^2 _{n+1} - S^2 _n| | \mathcal{F}_n) + \sigma^2 \\ &\leq \mathbb{E}(|(S _{n+1} - S _n) (S _{n+1} + S _n)| | \mathcal{F}_n) + \sigma^2 \\ &\leq \mathbb{E}(|\xi _{n+1} (\xi _{n+1} + 2 S _n)| | \mathcal{F}_n) + \sigma^2 \\ &\leq \mathbb{E} (\xi _{n+1} ^2 | \mathcal{F}_n ) + 2 \mathbb{E} (|\xi _{n+1} S _n| | \mathcal{F}_n ) + \sigma^2 \\ &\leq \mathbb{E} (\xi _{n+1} ^2 ) + 2 |S _n| \cdot \mathbb{E} (|\xi _{n+1}| ) + \sigma^2 \tag{*} \\ &\leq \sigma^2 + 2 |\sigma| |S _n| + \sigma^2 \end{align*} $$ where (*) follows from $\xi _{n + 1}$ is independent of $\mathcal{F}_n$ and $S_n$ is $\mathcal{F}_n$-measurable. This is where I get stuck as I cannot bound $|S_n|$ by a finite number.
I tried replacing $n$ by $n \wedge T$, but it would not help as neither $\xi_i$'s nor $T$ are bounded. I also thought about using another stopping time, say $$ N_a = \inf \{n : S_n \not \in (-a , a) \}. \tag{1} $$ I think it can be shown that $\mathbb{E}N_a < \infty$. Then $$ \mathbb{E}(|X_{(n+1) \wedge T \wedge N_a} - X_{n \wedge T \wedge N_a}| | \mathcal{F}_n) \leq 2\sigma^2 + 2|\sigma|a, \tag{2} $$ and we can use the Optional Stopping Theorem on it to get $$ 0 = \mathbb{E} X_0 = \mathbb{E} X_{T \wedge N_a}. $$ But to make the right-hand side $X_T$ one need to take $a \to \infty$. I am not sure if this violates the stopping time (1) or the bound (2).
Please contribute if you have any thought or idea on this. Any help is appreciated.