# Limit of a product of series of independent random variables uniformly distributed

A series of independent random variables is given, $X_n$, which is uniformly distributed in an interval $[0,1]$. We are supposed to determine a limit:

$$\lim_{n\to \infty} (\prod_{i=1}^n X_i )^{1/n}$$

Since we're doing measure theory, and we can use the law of large numbers (Kolmogorov), I was trying to transform the product into a sum and then determine the limit. I am not sure how to do that, this product looks like a geometric mean and I wanted to transform it to an average mean with this law with logarithms. It looks way too complicated, so I wanted to ask if I'm missing out on a certain property of uniformly distributed random variables so it can be done in an easier way.

• Why complicated? Do you know the properties of logarithms? – BGM Sep 18 '17 at 15:32

Let's take the logarithm of the limit and then undo that by taking the exponential, and we can nice $\ln$ inside $\lim$ so the desired result is $\exp\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}\ln X_i$. In other words, it's $e^\mu$ with $\mu:=\mathbb{E}(\ln X_1) =\int_0^1 \ln x dx$. I'll leave pricing $\mu=-1$ as an exercise.
• Why reproduce the typo $X_n$ standing for $X_i$ already in the question? – Did Sep 18 '17 at 15:41