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Suppose that $X_i$ are iid positive random variables, with $\mathbb{E}(X_i) \leq 1$, and $\operatorname{Var}(X_i) > 0$. Let $P_n = \prod^n_{i = 1} X_i,\:$ I want to show that almost surely $\lim_{n \rightarrow \infty} P_n = 0$.

If $\mathbb{E}(X_i) < 1$ then I can use the Markov inequality to say that $P(|P_n| \geq \varepsilon) \leq \frac{\mathbb{E}(X_1)^n}{\varepsilon},\:$ for any $\varepsilon > 0$. Since $\mathbb{E}(X_1) < 1$, $\sum^{\infty}_{n = 1} P(|P_n| \geq \varepsilon) \leq \frac{1}{1 - \mathbb{E}(X_1)}$ and so the Borel-Cantelli Lemma tells me that the limit of $P_n$ is $0$ almost surely.

I’m wondering if there is an extension of this idea to when $E(X_i) = 1$, and if not looking for a proof of the case when the expectation is 1.

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  • $\begingroup$ Are $X_i$'s non-negative? $\endgroup$ Commented 2 days ago
  • $\begingroup$ Yes, I forgot to add that assumption, I will edit the question. $\endgroup$ Commented 2 days ago

1 Answer 1

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Let $Y_n=X_n^{1/2}$. If $EX_n=1$, then $EY_n<1$ (by Holder's/Cauchy-Schwarz inequlity, combined with the fact that $X_n$ is not a constant random variable). $[Var (X_n)>0$].

By your argument, $\prod_{i=1}^{n} Y_i \to 0$ a.s. But this also implies $P_n \to 0$ a.s.

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  • $\begingroup$ Thank you so much, I'm happy to know there is a simple extension of my idea. $\endgroup$ Commented yesterday

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