0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Opening question defines N as stopping time, without saying what stopping means and what is stopping. $\endgroup$ – herb steinberg Mar 30 at 0:28
1
$\begingroup$

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: $$\mathbb E\left(\sup_n|S_{n\wedge N}|\right)\leq 3\mathbb E\left(\lim_{n\to\infty}\sqrt{( n\wedge N)\mathbb E(\xi_1^2)} \right)\leq 3\sqrt{\mathbb E(\xi_1^2)} \mathbb E(\sqrt N) $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.