Convergence of a product of random variables Let $X_1, X_2, \dots $ be a sequence of I.I.D. random variables with pdf $f(x) = \frac{8x}{9}, 0 < x < 1.5$. What does the product $\prod_1^nX_i$ converge to in the almost sure sense?
Shouldn't this blow up to infinity, since larger values are more likely thus eventually it will keep growing? Thanks.
 A: The PDF of $Y_i = \log X_i$ (for which $X_i = \exp(Y_i)$) is given by
$$g(y)dy = f(\exp(y)) d(\exp(y)) = \frac{8}{9} \exp(2 y) dy$$
for $-\infty \lt y \le \log(3/2)$.  This distribution has mean
$$\mu = \int_{-\infty}^{\log(3/2)} y g(y) dy = \log(3/2) - 1$$
and variance
$$\sigma^2 = \int_{-\infty}^{\log(3/2)} (y-\mu)^2 g(y) dy = 1/4.$$
Consequently $Z_n = \sum_{i=1}^n Y_i = \log \prod_{i=1}^n X_i$ has a mean $n\mu$ and variance $n/4$.  In particular, given $k\gt 0$, Chebyshev's Inequality asserts
$$\eqalign{
\Pr\left(Z_n  \ge n\left(\mu + \frac{k}{2\sqrt{n}}\right)\right) &= \Pr\left(Z_n  - n\mu \ge k\frac{\sqrt{n}}{2}\right) \\&= \Pr\left(Z_n - \mathbb{E}(Z_n) \ge k\sqrt{\text{Var}({Z_n})}\right) \\&\le \frac{1}{k^2}.
}$$
Since $\mu = \log(3/(2e)) \lt \log(1) = 0$, $n\left(\mu + \frac{k}{4\sqrt{n}}\right)$ can be made arbitrarily negative for sufficiently large $n$, regardless of the value of $k$.  Consequently, almost all the probability of $Z_n$ can be pushed arbitrarily far to the left for large enough $n$. Equivalently, almost all the probability of $Y_n$ will then be arbitrarily close to $0$.
The conclusion should now be obvious.  Making it rigorous (if that's needed) is simply a matter of restating these last two sentences in terms of epsilons and deltas.
A: Some hints for a proof:
Prove, or invoke known theorems to show:
$$
-\infty<\mathbb E \log X_i< \log\mathbb E X_i = 0
$$
Then prove, or invoke a known theorem to argue that:
$$
\sum_{i=1}^n \log X_i \to -\infty \text{ (a.s.)}.
$$ 
Lastly, take some $\omega\in \Omega$ where this holds and say something about $\exp\{\sum_{i=1}^n \log X_i\}$ on that $\omega$.

Some hints for another proof:
First show that $\xi_n = \prod_{i=1}^nX_i$ is a non-negative martingale. 
If this is true, then a theorem about convergence of martingales gives $\xi_n \to \xi$ (a.s.), where $\xi$ is  some r.v. with finite expectation. Two steps remain:
Show that on sets where $\xi>0$, $\xi_n/\xi_{n-1}=X_n\to1$. 
Show that $| X_n - 1 | \geq \epsilon$, for some $\epsilon > 0$ infinitely often (a.s.).
The conclusion follows.  
