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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.

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

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  • $\begingroup$ Why reproduce the typo $X_n$ standing for $X_i$ already in the question? $\endgroup$ – Did Sep 18 '17 at 15:41
  • $\begingroup$ @Did Coincidence. $\endgroup$ – J.G. Sep 18 '17 at 15:43

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