Consider a sequence $(X_n)_{n\ge1}$ of i.i.d random variables with distribution $\mathcal{B}_{1/2}$, and set

$$S_n := X_1 + \cdots + X_n$$

Fix $p \in (0,1),\,q=1-p,$ and define

$$M_n := (2p)^{S_n}(2q)^{n-S_n}$$

$(M_n)_{n\ge1}$ is a martingale w.r.t the filtration generated by $(X_n)_{n\ge1}$. Using the appropriate martingale convergence theorem, show that $(M_n)_{n\ge1}$ admits a.s. limit $M$.

I've already shown that admits a.s limit $M$, but I can't determine it.

Thanks in advance for any help!


The limit is $1$ if $p=q$. This is trivial.

If $p \ne q$, then using $S_n/n \to 1/2$ almost surely (Law of Large number), we have $M_n \to 0$ almost surely.

To see, consider any fixed sequence $s_n$ such that $s_n/n \to 1/2$. It's not difficult to see that $$ (2p)^{s_n} (2(1-p))^{n-s_n} \to 0 $$ unless $p=1/2$.

  • $\begingroup$ I do not follow this argument. Can you elaborate? $\endgroup$ – Math1000 Dec 11 '19 at 20:34
  • $\begingroup$ @Math1000 See update $\endgroup$ – ablmf Dec 11 '19 at 20:37
  • $\begingroup$ Here we have $S_n$ and not $S_n/n$ though. $\endgroup$ – Math1000 Dec 11 '19 at 20:40
  • $\begingroup$ @Math1000 We don't need $S_n$ to converge. Just replace $S_n$ by $(1+o(1))n/2$. This is enough to make the sequence converge to $0$. $\endgroup$ – ablmf Dec 11 '19 at 20:42

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.