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Suppose that $X_k(1 \leqslant k \leqslant n )$ are positive i.i.d random variable with finite mean $\mu$ and variance $\sigma^2$. Define $P_n= (\prod \limits _{k=1}^n X_k)^{\frac{1}{n}}$ $(n\geqslant1)$and $\Pi_n=\frac{P_n-\mu}{\sigma / {\sqrt{n}}}$ $(n\geqslant1)$ Find the limit of $\Pi_n$ in distribution(i.e.,$\Pi_n \rightarrow_d ?$).

There is a hint for the question: Use the Delta Method.

First I take log on both side for $P_n= (\prod \limits _{k=1}^n X_k)^{\frac{1}{n}}$ and get: $$\log(P_n)=\frac{1}{n}\sum \limits_{k=1}^n \log(X_k)$$

I honestly don't have the slightest inkling on what to do for the next steps.

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  • $\begingroup$ Are you sure it's not the geometric mean instead of $\mu$ in $\Pi_n$ expression? $\endgroup$ Commented Nov 12, 2021 at 14:04
  • $\begingroup$ That's the reason why I feel a little bit awkward @Harnak $\endgroup$ Commented Nov 12, 2021 at 14:38
  • $\begingroup$ well, yeah, it's pretty weird. I'm sure there's an error because with geometric mean it converges, but with mean it doesn't $\endgroup$ Commented Nov 13, 2021 at 9:54

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