# Showing that $ES_N=0$ for a random walk where $N$ is a stopping time and $EN^{1/2}<\infty$

Question Let $$\xi_{1}, \xi_{2}, \dotsc$$ be i.i.d random variables with $$E \xi_i=0$$ and $$E\xi_i^2<\infty$$. Let $$S_n=\sum_{i=1}^n \xi_i$$ and $$N$$ be a stopping time. If $$EN^{1/2}<\infty$$, then $$ES_N=0$$.

Context The question is exercise $$5.4.10$$ from Durrett and claims that the following theorem is useful to answer the question.

Theorem Let $$(X_n)$$ be a martingale with $$X_0=0$$ and $$EX_n^2<\infty$$ for all $$n$$. Let $$A_n=\sum_{m=1}^n E((X_m-X_{m-1})^2\mid \mathcal{F}_{m-1})=\sum_{m=1}^n E(X_m^2\mid \mathcal{F}_{m-1})-X_{m-1}^2$$ be the increasing process associated with $$X_n$$ and let $$A_{\infty}=\lim A_n$$. Then $$E(\sup_n |X_n|)\leq 3EA_{\infty}^{1/2}\tag{0}.$$

My Attempt Because $$S_{N\wedge n}$$ is a martingale, $$ES_{N\wedge n}=E{S_0}=0.\tag{1}$$ Further, because $$EN^{1/2}<\infty$$, we have that $$P(N<\infty)=1$$. In particular, $$S_{N\wedge n}\to S_N$$ a.s. Hence, we would be done if we can invoke some sort of convergence theorem.

Based on (0), my idea is to show that $$E(\sup_n |S_{N\wedge n}|))<\infty$$ whence it would follow that $$S_{N\wedge n}$$ is a uniformly integrable martingale and the result would follow.

Let $$A_{\infty}$$ be the limit of the increasing process associated with $$S_{N\wedge n}$$.

My problem Assuming this is the right approach, I am having trouble showing that $$E{A_{\infty}^{1/2}}<\infty$$. I haven't been able to bound this expectation and haven't made it past writing the expectation down.

Any help is appreciated. Other methods/proofs not invoking the theorem above are welcome too.

• Opening question defines N as stopping time, without saying what stopping means and what is stopping. – herb steinberg Mar 30 at 0:28

It is not specified what the filtration is. Let us set $$\mathcal F_n:=\sigma(\xi_1,...,\xi_n)$$ (we could also take the natural filtration for $$S_n$$). Then $$S_n$$ is a $$\mathcal F_n$$-martingale.
Now we have $$A_n=\sum_{k=1}^n\mathbb E((S_k-S_{k-1})^2\mid\mathcal F_{k-1})=\sum_{k=1}^n\mathbb E(\xi_k^2\mid\mathcal F_{k-1})=\sum_{k=1}^n\mathbb E(\xi_k^2)$$where we have used the fact that $$\xi_k$$ is independent of $$\mathcal F_{k-1}$$ (would this argument change if we had the natural filtration?). This leads to $$A_n=n\mathbb E(\xi_1^2)$$ which is nice because now you can bound $$\mathbb E(\sup_n|S_{n\wedge N}|)$$ as follows: $$\mathbb E\left(\sup_n|S_{n\wedge N}|\right)\leq 3\mathbb E\left(\lim_{n\to\infty}\sqrt{( n\wedge N)\mathbb E(\xi_1^2)} \right)\leq 3\sqrt{\mathbb E(\xi_1^2)} \mathbb E(\sqrt N)$$