Let $X_1, X_2, \ldots$ be a sequence of i.i.d. random variables with distribution concentrated on $[1,\infty)$ and finite second moment. We assume that $a=E\ln X_1$, $\sigma^2=\operatorname{Var}\ln X_1$.
How to evaluate a limit of sequence of probabilities
$$\Pr\left(\prod_{i=1}^{n}X_i\leq \left(\prod_{i=1}^{n}X_i^2\right)^{\frac{1}{\sqrt n}}e^{na}\right) ? $$
I have no idea how to begin. I guess it can be associated with Central Limit Theorem, but I am not sure.
Best Answer
Taking logarithms and letting $Y_{i} = \ln(X_{i})$:
$$\prod_{i=1}^{n}X_i\leq \left(\prod_{i=1}^{n}X_i^2\right)^{\frac{1}{\sqrt n}}e^{na}$$
$$\begin{align} &\Longleftrightarrow \sum_{i=1}^{n} Y_{i} \leq \frac{2}{\sqrt{n}}\sum_{i=1}^{n} Y_{i} + na\\ &\Longleftrightarrow \sum_{i=1}^{n} (Y_{i}-a) - \frac{2}{\sqrt{n}}\sum_{i=1}^{n} Y_{i} \leq 0 \\ &\Longleftrightarrow A_{n} \equiv \frac{1}{\sqrt{n}}\sum_{i=1}^{n} (Y_{i}-a) - \frac{2}{n}\sum_{i=1}^{n} Y_{i} \leq 0 \end{align}$$
The first term converges in distribution to $N(0, \sigma^{2})$ by the central limit theorem, and the second term converges in probability to $-2a$ by the weak law of large numbers, therefore $A_n$ converges in distribution to $N(-2a, \sigma^{2})$.
$$\mathbb{P}(A_{n} \leq 0 ) \rightarrow \Phi\left(\frac{2a}{\sigma}\right)$$
Where $\Phi$ is the standard normal cdf.