# An equivalent of a central limit theorem for a geometric mean

I'm taking a probability theory actuary exam next week and came accross an interesting problem - consider a sequence $(X_n)_{n=1}^{\infty}$ of i.i.d. random variables with density $$f(x) = \frac{3}{x^4}\mathbb{1}_{(1, \infty)}(x)$$ and let $$U_n = (X_1 \cdot \ldots \cdot X_n)^{\frac{1}{n}}$$ I have to prove that $$P(3(U_n - e^{\frac{1}{3}})\sqrt{n} > e^{\frac{1}{3}}) \approx 0.1587$$ but I don't know how to approach it. My first thought was to attain some kind of an equivalent of a central limit theorem for the geometric mean. Taking the random variable $\ln U_n$ and proving $\mathbb{E} \ln U_n = \frac{1}{3}$, therefore $\ln U_n \rightarrow \frac{1}{3}$ almost surely and $U_n \rightarrow e^{\frac{1}{3}}$ a.s., alas I don't know what to do next.

• By the CLT, $\ln U_n=\frac13+\sigma Z_n\frac1{\sqrt{n}}$ where $\sigma^2$ is the variance of $\ln X_1$ and $Z_n$ converges in distribution to a standard normal random variable $Z$, hence $U_n=e^{1/3}e^{\sigma Z_n/\sqrt{n}}$, which yields $(U_n-e^{1/3})\sqrt{n}\sim e^{1/3}\sigma Z_n$, and your event has probability roughly $P(3e^{1/3}\sigma Z_n>e^{1/3})\to P(3\sigma Z>1)=\Phi(-1/(3\sigma))$. Since $\sigma^2=\frac19$, the limit is $\Phi(-1)\approx.1587$. – Did Sep 23 '15 at 21:15
• Oh my God, it was straightforward :P thanks a lot! – tosi3k Sep 23 '15 at 21:17