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?

Thank you in advance

  • 1
    $\begingroup$ 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 $(0, 1)$). $\endgroup$ – BGM Oct 24 '17 at 9:30

Since the terms in the sum are non-negative almost surely,

$$E(\sum_{n=1}^{\infty}\prod_{i=1}^{n} X_i) = \sum_{n=1}^{\infty}E(\prod_{i=1}^{n} X_i) = \sum_{n=1}^{\infty}\prod_{i=1}^{n} E(X_i)= \sum_{n=1}^{\infty}\frac 1{2^n} <\infty $$

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.

  • $\begingroup$ Thanks a lot for this answer. However, I still missing something: how do you prove that a random variable non negative a.s. with finite expectation is finite a.s.? $\endgroup$ – XYZ Oct 25 '17 at 7:52
  • 1
    $\begingroup$ @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 $$P(X\geq n) \leq \frac{E(X)}{n}$$ $\endgroup$ – Gabriel Romon Oct 25 '17 at 10:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.