prove that $S_{\tau \land n} \to S_{\tau}$ in $L^1$ for a random walk with $E\tau^{1/2} < \infty$

Question

Let $X_i$ be a sequence of iid $L^2$ RVs with $EX_i = 0$ and define the martingale $S_n = \sum_1^n X_i$. I want to show that if $\tau$ is a stopping time and $E\tau^{1/2} < \infty$, then $ES_{\tau} =0$.

I have been given the hint that if $Y_n$ is an $L^2$ martingale, then $E(\sup_n | Y_n|) \leq 3E(\sqrt{A_{\infty}})$, where $A = \langle Y \rangle$ is the quadratic variation process of $Y$, i.e. $A_0 =0$, $$A_n = \sum_{k=1}^nE(Y_k^2 \mid \mathcal{F}_{k-1}) - E(Y_{k} \mid \mathcal{F}_{k-1})^2$$ and $\lim_n A_n =A_{\infty}$.

My work:

Following the idea in Doob's OST, clearly have that $S_{\tau \land n}$ is a martingale and therefore $E[S_{\tau \land n}] = E[S_0] = 0$. Since $\tau< \infty$ a.s, I need to show $\lim_n E[S_{\tau \land n}] = E[S_{\tau}]$ (that is, $S_{\tau \land n} \to S_{\tau}$ in $L^1$).

I have tried to follow this hint, and noted that since $X_i$ are $L^2$, $S_n = \sum_{i=1}^n X_i \in L^2$. Also, $$\begin{aligned} A_n &= \langle S \rangle_n =\sum_{k=1}^nE(S_k^2 \mid \mathcal{F}_{k-1}) -0\\ & = E \left(\sum_{k=1}^nX_k^2 + 2\sum_{i<j \leq n} X_iX_j \mid \mathcal{F}_{k-1} \right)\\ & = E(X_n^2) + \sum_{k=1}^{n-1}X_k^2 + 2 \sum_{i<j<n}X_iX_j \\ &= E(X_n^2)+S_{n-1}^2 \end{aligned}$$ I don't know that this has a limit. If I use the hint on the $X_i$s, I get that $$\langle X\rangle_n = E(X_n^2) \implies E(\sup_n|X_n|) \leq 3(X_{\infty}^2)$$ which I don't think is helpful.

I'm not sure exactly which route I should be going down to show convergence, there are so many theorems on it with slightly different conditions.

I'm so lost, any ideas would be much appreciated, thanks!

Answer 1

First of all: Your calculation of $A_n$ is not correct; note that

$$A_n = \sum_{k=1}^n \mathbb{E}(S_k^2 \color{red}{-S_{k-1}^2} \mid \mathcal{F}_{k-1})$$

and then you will end up with $$A_n = \sum_{k=1}^n \mathbb{E}(X_k^2) = n \mathbb{E}(X_1^2). \tag{1}$$

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If we consider the stopped process $Y_n := S_{n \wedge T}$, then it follows from $(1)$ that its compensator is given by $A_n := \mathbb{E}(X_1^2) \min\{n,T\}$. Applying the inequality which you were given in the hint, we thus find

$$\mathbb{E} \left( \sup_{n \geq 1} |S_{n \wedge T}| \right) \leq 3 \mathbb{E}(\sqrt{T}) < \infty.$$

Now it follows easily from the dominated convergence theorem that $S_{T \wedge n} \to S_T$ in $L^1$.