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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!

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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 $p=1/2$.

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  • $\begingroup$ 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

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