Central Limit Theorem and Strong Law of Large Numbers. Proof that converges in distribution to $N(0, e^2)$

Question

I have this exercise of my homework.

Let $\{ X_n: n \geq 1\}$ be independent random variables and identically distribuited with uniform distribution $U(0,1)$ and $Y_n=(\prod_{i=1}^n X_i)^{-1/n} $. Show that $\sqrt{n} (Y_n-e)$ converges in distribution to $N(0, e^2)$

I only know that $Y_n$ converges almost surely to $e$.

I do not know to do the proof using central limit theorem. I know the law of large numbers. I think that you can use delta method.

Answer 1

Note that if $X\sim U(0,1)$ then $-\ln(X)\sim Exp(1)$ and so $$E(-\ln(X))=V(-\ln(X))=1$$ This implies $\ln(Y_n)=\frac{1}{n}\sum_{i=1}^n\big[-\ln(X_i)\big]$ converges in distribution to a $N(1,1/n)$ random variable by the central limit theorem. Equivalently, $$\sqrt{n}\Big(\ln(Y_n)-1\Big)\rightarrow N(0,1)$$ Apply the delta$-$ method with $g(x)=e^x$ to get $$\sqrt{n}\Big(g\big[\ln(Y_n)\big]-g(1)\Big)\rightarrow N\Big(0,(1\cdot g'(1))^2\Big)$$ which is what you're looking for.

Answer 2

One direction that sounds promising is to note that $$ \ln Y_n = \ln \left( \left(\prod_{k=1}^n X_k\right)^{-1/n}\right) = \frac{-1}{n} \sum_{k=1}^n \ln X_k = \frac{1}{n} \sum_{k=1}^n L_k, $$ where $L_k = -\ln X_k$ and now you can apply the CLT directly.