Suppose that $X_1,X_2,…$ be i.i.d. variable uniformly distributed on (0,1), and let $\tilde{X_n}$ denote the geometric average of $n$ of these variables, i.e.: $\tilde{X_n}=(X_1X_2\cdots X_n)^{1/n}$. I am trying to show that $\sqrt{n}(\tilde{X_n}-1/e)$ converges in distribution and identify what the limit is.
I have tried taking $log$ for $\tilde{X_n}$, using WLLN and continuous mapping theorem I am able to prove that $\tilde{X_n}\overset{p}{\to}1/e$. Nevertheless, I'm not able to derive the result for the mentioned question. It seems to me like it's related to using CLT, but I'm unable to figure out the result when applying it on $log\tilde{X_n}$. Can anyone help me out? Thanks in advance!
Best Answer
Here is a direct solution: Write $Y_k := 1 + \log X_k$ and $\bar{Y}_n = \frac{1}{n}\sum_{k=1}^{n} Y_k$ and $f(x) = (e^x - 1)/x$. Then $f$ is continuous and
$$ \sqrt{n}(\bar{X}_n - e^{-1}) = e^{-1} f(\bar{Y}_n) \sqrt{n} Y_n. $$
Then
By SLLN, $f(\bar{Y}_n)$ converges a.s. to $f(\mathbb{E} Y_1) = f(0) = 1$.
By CLT, $\sqrt{n}\bar{Y}_n$ converges in distribution to $\mathcal{N}(0, \operatorname{Var}(Y_1)) = \mathcal{N}(0, 1) $
By converging together theorem (a.k.a. Slutsky's theorem), $\sqrt{n}(\bar{X}_n - e^{-1})$ converges in distribution to $\mathcal{N}(0, e^{-2})$.
Of course, this is nothing but an elementary rendition of a more general argument that leads to delta method.