# Proof that a Gamma distribution is finite almost surely

The problem I am working on is the following. Let $X_1, X_2; ...$ be i.i.d. random variables with uniform distribution on $[0; 1]$. We want to show that $\sum_{n=1}^{\infty}\prod_{i=1}^{n} X_i$ is finite almost surely, i.e. that $$P\left( \sum_{n=1}^{\infty}\prod_{i=1}^{n} X_i < \infty\right) = 1$$.

What I have been doing is defining $Y_n = \prod_{i=1}^{n} X_i$ and now I am trying to prove that $Y_n$ is distributed as a Gamma (a result I found in some textbook). At this point, I aim at showing the the sum of Gamma random variables is a Gamma random variable as well. Finally, I should show that a Gamma variable is finite almost surely, but I am not sure on how to procede on this.

Plus, is there, in your opinion, a more efficient way of proceeding?

• math.stackexchange.com/questions/659254/… Note that $-\ln X_1$ is exponentially distributed, therefore $\displaystyle Z_n = -\ln\prod_{i=1}^n X_i = -\sum_{i=1}^n \ln X_i$ will follows a gamma distribution. Thus the product of uniform $Y_n$, has the same distribution as $e^{-Z_n}$. (Actually easy to check as the support of $Y_n$ is also $(0, 1)$). – BGM Oct 24 '17 at 9:30

$$E(\sum_{n=1}^{\infty}\prod_{i=1}^{n} X_i) = \sum_{n=1}^{\infty}E(\prod_{i=1}^{n} X_i) = \sum_{n=1}^{\infty}\prod_{i=1}^{n} E(X_i)= \sum_{n=1}^{\infty}\frac 1{2^n} <\infty$$
Since $\sum_{n=1}^{\infty}\prod_{i=1}^{n} X_i$ is non-negative almost surely and has finite expectation, it is finite almost surely.
All this can be rewritten in the framework of measure theory using integrals instead of $E$. You might be more familiar with the results I used in a measure-theoretic outlook.
• @Elle It's a very standard result of measure theory. Here's a possible proof: $$P(X=\infty) = P(\bigcap_n X\geq n) = \lim_n P(X\geq n)$$ and $$P(X\geq n) \leq \frac{E(X)}{n}$$ – Gabriel Romon Oct 25 '17 at 10:43