Calculate $E(S_T^2)$.

Question

Let $Y_1,Y_2,\dots \,$ i.i.d and centered random variables in $L^2$ in the probability space $(\Omega,F,P)$. Let $F_n:=\sigma(Y_1,\dots,Y_n), F_0:=\{\Omega,\emptyset\}$ and $S_0:=0$, $S_n=Y_1+\dots Y_n \,(n=1,2,\dots)$.

$\dots$if needed, I have shown that $S_n^2$ is a $F_n$-Submartingale and I have found a previsible process $A_n$ such that $(S_n^2-A_n)$ is a $F_n$-martingale.

Let T a $F_n$-stopping time in $L^1$. Calculate $E(S_T^2)$.

I started like this, using the submartingale property $$E(S_T^2)=E(E(S_T^2)\mid F_o)\ge E(S_0^2)=0.$$ I am not sure how to continue here. Maybe one can use this $\textbf{theorem}$: 'Let $X$ a $F$-submartingale with $sup_{n\in N_0}E(X_n^+) < \infty$. Then there exists a $F_{\infty}$ measerable integrable random variable $X_{\infty}(\omega)=lim_{n \to \infty}X_n(\omega) \,$a.s.. Furtheremore $E \mid X_{\infty}\mid \le \text{liminf}_{n\to \infty}E\mid X_n\mid$.'

I tried to use this theorem and concluded $E(S_T^2)=E(S_{T\land n}^2)$. So actually I do not get anywhere $\dots$

So how should one solve this problem? Any help is much appreciated!

P.S. I believe that $E(S_T^2)=0$, but this is just a guess.

Answer 1

Hints:

1. Recall the optional stopping theorem.
2. Set $M_n := S_n^2-A_n$. Show that $(M_{n \wedge T})_{n \geq 0}$ is a martingale and conclude that $$\mathbb{E}(M_{n \wedge T})=0,$$ i.e. $$\mathbb{E}(S_{n \wedge T}^2) = \mathbb{E}(A_{n \wedge T}) . \tag{1}$$
3. You already computed $A_n$ - plug it into $(1)$: $$\mathbb{E}(S_{n \wedge T}^2) = \mathbb{E}(Y_1^2) \mathbb{E}(T \wedge n). \tag{2}$$
4. Prove that $$\mathbb{E}((S_{n \wedge T}-S_{m \wedge T})^2) \xrightarrow[]{m,n \to \infty} 0.$$ Hint: Use (2) and $$\mathbb{E}((S_{n \wedge T}-S_{m \wedge T})^2) = \mathbb{E}(S_{n \wedge T}^2)-\mathbb{E}(S_{m \wedge T}^2)$$
5. Conclude that $S_{n \wedge T} \to S_T$ in $L^2$. In particular, $$\mathbb{E}(S_{n \wedge T}^2) \xrightarrow[]{n \to \infty} \mathbb{E}(S_T^2). \tag{3}$$
6. Combine $(2)$, $(3)$ and the monotone convergence theorem to conclude that $$\mathbb{E}(S_T^2) = \mathbb{E}(Y_1^2) \mathbb{E}(T).$$