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Let $Y_n$ be a sequence of non-negative i.i.d random variables with $EY_n = 1$ and $P(Y_n = 1) < 1$. Consider the martingale process formed by $X_n = \prod_{k=1}^n Y_k$. Use the martingale convergence theorem to show that $X_n \to 0$ almost surely.

I see that the Martingale convergence theorem says that $X_n \to X$ almost surely with $E \lvert X \rvert < \infty$.

I don't see how to reach the conclusion that $X = 0$ or $X_n \to 0$.

I see we can prove that $E \lvert X_n \rvert < \infty$ and that $X_n$ is uniformly integrable and $X_n \to X$ in $L^1$. And that $X_n = E(X \mid \mathcal{F}_n)$.

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2 Answers 2

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By Jensen's inequality, $b:=E\left(\sqrt{Y_1}\right)<\sqrt{E(Y_1)}=\sqrt{1}=1$. Therefore $E\left(\sqrt{X_n}\right)=b^n\to 0$ as $n\to\infty$. By Fatou's lemma, $E\left(\sqrt{X}\right)=0$.

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Since $X_n$ is a positive martingale, it is also a supermartingale bounded below by $0$, therefore $X_n\to X_\infty$ a.s. by supermartingale convergence. Now consider that $P(|Y_n-1|>\varepsilon \textrm{ i.o.})=1$ by Borel-Cantelli II. This implies that $X_\infty=0$ is the only admissible limit rv so that $X_n \to 0$ a.s.

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  • $\begingroup$ I follow up to $P(\lvert Y_n - 1 \rvert > \epsilon \text{i.o.}) = 1$ which means $Y_n$ isn't within $\epsilon$ of $1$ infinitely often. How do you get from there to the implication that $X_\infty = 0$ is the only admissible limit value? $\endgroup$
    – clay
    Feb 19, 2022 at 3:58
  • $\begingroup$ @clay by BC II we can claim that $Y_n$ is i.o. different from $1$. This makes the process $X_n$ unstable (like a usual random walk), unless it goes to $0$, which then is the only admissible a.s. limit $\endgroup$
    – Snoop
    Feb 19, 2022 at 12:08

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