Let $X$ be strictly positive r.v. such that $X$ not equal to 1 almost surely and $\mathbb{E}(X)=1$. Let $X_1, X_2,...$ be a sequence of i.i.d. random variables each having the same distribution as $X$. Let $M_0=1$ and $$M_n=\exp\left(\sum^n_{i=0} \ln(X_i)\right).$$

How to show that $M_n\to0$ almost surely?

  • $\begingroup$ This is false. Just let $X_n=1$ for all $n$. You need the added assumption that $X$ is not a.s. 1, or similar. $\endgroup$ – Shalop Oct 26 '15 at 18:22
  • $\begingroup$ Just thinking out loud: if $X_i$ are iid, so are $\ln(X_i)$, and so $$\frac{1}{n} \sum_{i=0}^n \ln (X_i) \to \mathbb{E}[X_i] = 0$$ a.s. by KSSLN. $\endgroup$ – gt6989b Oct 26 '15 at 18:22
  • $\begingroup$ Isn't it $M_n = X_1 \cdots X_n$? Why do you write that way? $\endgroup$ – Ant Oct 26 '15 at 18:23
  • $\begingroup$ See my answer here: math.stackexchange.com/questions/876399/… $\endgroup$ – user940 Oct 26 '15 at 18:28

As Shalop mentioned, you want to assume $X_i$ is not a.s. $1$.

$M_n \to 0$ iff $\ln(M_n) \to -\infty$, so you want to look at $$S_n = \sum_{i=0}^n \ln(X_i)$$ Note that $\ln(X_i) \le X_i - 1$, with equality iff $X_i = 1$, so $\mathbb E[\ln(X_i)] < 0$. Now use the Strong Law of Large Numbers.

EDIT: It is possible that $\mathbb E[\ln(X_i)]$ doesn't exist (i.e. the integral diverges to $-\infty$. In that case, replace $\ln(X_i)$ by a truncated version with a finite (and still negative) expectation.

Now SLLN says almost surely $$ S_n/(n+1) \to \mathbb E[\ln(X_i)] < 0$$ In particular, for some $\epsilon > 0$ we have almost surely $S_n/(n+1) < -\epsilon$ for all sufficiently large $n$. And then almost surely $S_n < -(n+1) \epsilon \to -\infty$.

  • $\begingroup$ thanks. So the original problem boils down to showing that actually $S_n \to -\infty$. Could you please explain more on the last step using SLLN. $\endgroup$ – Ruth90 Oct 26 '15 at 18:35

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