Let $(X_{n})_{n}$ be independent random variables that are $\mathcal{U}{[1,2]}$
Prove $(\prod_{i=1}^{n}X_{n})^{\frac{1}{n}}$ exists for $n \to \infty$ and that $\exists c \in \mathbb R$ such that
$(\prod_{i=1}^{n}X_{n})^{\frac{1}{n}}\to c$, a.s. for $n \to \infty$
I honestly do not even know where to begin because I've never worked with the product of random variables.
Any ideas and tips?
It is unclear what $\mathcal{U}[1,2]$ means, so I'll presume $\mathcal{U}[1,2]$ refers to the uniform distribution on $[1,2]$.
The answer is straightforward once we note that $$\ln \left([\prod_{i=1}^n X_i]^{1/n}\right) = \frac{1}{n}\sum_{i=1}^n\ln X_i$$ is an averaged sum of i.i.d. random variables $(\ln X_n)_{n=1}^\infty$. By strong law of large numbers, it follows $$ \frac{1}{n}\sum_{i=1}^n\ln X_i \to E[\ln X_1]= \int_1^2 \ln x \;dx = (x\ln x-x)\big|^2_1=2\ln 2-1 $$almost surely as $n\to\infty$. Therefore, we have $$ \left[\prod_{i=1}^n X_i\right]^{1/n}=\exp\left(\frac{1}{n}\sum_{i=1}^n\ln X_i\right)\to \exp(2\ln 2-1) = \frac{4}{e}. $$