I'm currently studying some probability and I'm stuck with this question.
Let $R_1, . . . , R_n$ be independent continuous uniform over [0, 1] random variables.
The geometric mean of $R_1, . . . , R_n$ is defined by $$G_n = \sqrt[n]{R_1 ×...×R_n }=(R_1 ×...×R_n)^\frac{1}{n}.$$ Show that $G_n$ converges in probability as well as with probability 1 (i.e. almost surely) to a constant, and identify the limit. Make sure you state any theorem you use.
From what I gather, I'm pretty sure I'm supposed to use the Strong Law of Large Numbers. $$\mathbb{P}[\lim_{n \to \infty} M_n = \mu] = 1 $$ Where $M_n$ is the sample mean.
But I don't really understand how to show this, I'm completely stuck.
Any tips/help would be greatly appreciated
Consider $F_n = \ln(G_n)$. Then by logarithm rules $F_n = \frac{1}{n} \sum_{i=1}^n \ln(R_i)$. This is the arithmetic mean of the iid random variables $\ln(R_i)$. Now prove that these have finite mean $\mu$. Then the strong law says that $F_n$ converges almost surely to $\mu$.
Now $G_n = \exp(F_n)$. Note that $\exp$ is continuous. So if $F_n(\omega)$ converges to $\mu$ then $G_n(\omega)$ converges to $\exp(\mu)$. Hence $G_n$ converges almost surely to $\exp(\mu)$.