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.

  • $\begingroup$ Can you say anything about $\ln(\prod_1^n X_i)$? In particular, if it converges/blows up then what does this say about $\prod_1^n X_i$? $\endgroup$ – P.Windridge May 12 '15 at 14:21
  • $\begingroup$ Also, how about calculating $\mathbb{E}[\prod_1^n X_i] = \mathbb{E}[X_1]^n$? What would you expect to happen if the product blows up a.s.? $\endgroup$ – P.Windridge May 12 '15 at 14:33
  • $\begingroup$ What theorems do you know about convergence of random series? $\endgroup$ – P.Windridge May 12 '15 at 14:37
  • 1
    $\begingroup$ Noting that $\mathbb E X_i=1$ and finding a martingale may also lead you somewhere. Personally, I would go for a direct proof but you already have hints for that. $\endgroup$ – ekvall May 12 '15 at 14:56

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.


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.

  • $\begingroup$ Is the first one Jensens? But the inequality is not strict in Jensens right? $\endgroup$ – ObsidianDestroyer May 12 '15 at 15:26
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    $\begingroup$ In general it is not strict, no. You would have to show/argue it's strict in this case. $\endgroup$ – ekvall May 12 '15 at 17:43

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