# Let $Y_n$ be i.i.d with $EY_n = 1, P(Y_n = 1) < 1, X_n = \prod_{k=1}^n Y_k$. Show martingale convergence of $X_n \to 0$ a.s.

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 $$X_n = E(X \mid \mathcal{F}_n)$$.

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, $$E\left(\sqrt{X}\right)=0$$.
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.
• 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?
• @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 Feb 19, 2022 at 12:08