Find limit of $(X_1\cdots X_n)^\frac{1}{n}$ as $n$ goes to infinity, where $X_1,X_2,...$ are iid random variables uniformly distributed on interval $(1,2]$.
So first I need to show $ \lim_{n\to\infty} (X_1...X_n)^\frac{1}{n} $ exists almost surely, and then find the value of this almost sure limit.
So first of all, since $X_1,X_2,...$ are iid, this means that $Y_i=log(X_i)$ are iid random variable. Since $X_i$ are uniformly distribued on the interval $(1,2]$, this means $|X_i|\leq C$ and it also means $|Y_i|\leq C$. Now let $A_n=(X_1...X_n)^\frac{1}{n}$ and then $log(A_n)=\frac{1}{n}\sum_{i=1}^{n} Y_i$. Now by using strong law of large number, when $n$ goes to infinity, $\frac{1}{n}\sum_{i=1}^{n} Y_i$ almost surely converge to $E(Y_1)$.
$E(Y_1)=E(logX_1)=\int_1^2logX_1dx=log(4)-1$. Thus, $\frac{1}{n}\sum_{i=1}^{n} Y_i$ almost surely converge to $log(4)-1$.
Am I correct?