Probability of sum of product of uniform R.V.

Question

Suppose uniform[0,1] random variables $X_i$, $i=1,...,\infty$ are also independent. Whats the following probability?

\begin{split} P\Big(\sum_{n=1}^\infty\prod_{i=1}^nX_i<\infty\Big) \end{split}

Answer 1

Hint: Show that $$\mathbb{E} \left( \sum_{n \geq 1} \prod_{i=1}^n X_i \right) = \sum_{n \geq 1} \mathbb{E} \left( \prod_{i=1}^n X_i \right) < \infty.$$ What does this tell you about $$\mathbb{P} \left( \sum_{n \geq 1} \prod_{i=1}^n X_i < \infty \right)$$ ....?

Answer 2

Following saz's hint, we have that since the $X_i$ are nonnegative, we can interchange the order of the expectation and summation, in which case, \begin{align*} E \sum_{n \geq 1} \prod_{i=1}^n X_i = \sum_{n \geq 1} E \prod_{i=1}^n X_i. \end{align*} Since the $X_i$ are independent, each with mean $1/2$, the above is equal to \begin{align*} = \sum_{n \geq 1} \frac{1}{2^n} = 1. \end{align*} Hence $\sum_{n\geq 1} \prod_{i=1}^n X_i$ has finite expectation. From here it is clear that $\sum_{n\geq 1} \prod_{i=1}^n X_i < \infty$ with probability $1$, since otherwise, the expectation would be infinite. Hence, \begin{align*} P(\sum_{n\geq 1} \prod_{i=1}^n X_i < \infty) = 1. \end{align*}