# Central Limit Theorem for geometric mean

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!

## 2 Answers

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.

• Many thanks! This is very clear! – dogthepeter Oct 6 '18 at 5:12

Figured out the problem by the hint here https://stats.stackexchange.com/questions/7471/can-the-standard-deviation-be-calculated-for-harmonic-mean. Having used CLT to find $$\sqrt{n}(log\tilde{X_n}+1)$$ 's convergence to distribution, directly applying delta method can solve the problem.