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.

  • $\begingroup$ Opening question defines N as stopping time, without saying what stopping means and what is stopping. $\endgroup$ – 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: $$\mathbb E\left(\sup_n|S_{n\wedge N}|\right)\leq 3\mathbb E\left(\lim_{n\to\infty}\sqrt{( n\wedge N)\mathbb E(\xi_1^2)} \right)\leq 3\sqrt{\mathbb E(\xi_1^2)} \mathbb E(\sqrt N) $$


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