Let each $X_i$ be i.i.d and $Y_i=\ln(X_i)$ is exponentially distributed with mean $\alpha^{-1}$ and variance $\alpha^{-2}$. Now let $$S_n=\left[ \prod_{i=1}^{n}X_i \right ]^{n^{-1}}.$$

Use the central limit theorem to find as $n \rightarrow \infty$ the limiting distribution of:


The way that I understand the central limit theorem, if the standard deviation appeared in the denominator, then this distribution would converge to $N(0,1)$, but since it does not appear, can I assume that it instead converges to $N(0,\alpha ^{-2})$? (The second parameter is the variance). I don't really understand how this part works.

Finally, if $N(0,\alpha^{-2})$ is correct, does it give me any information about the underlying $S_n$? Can I use that normal distribution to approximate, for example, $P(S_n>1.12)$?

I'm just really confused on how to properly apply the central limit theorem to situations like these.


Observe that $$ \log S_n=\frac{\log X_{1}+\dotsb+\log X_n}{n}=\frac{Y_{1}+\dotsb+Y_{n}}{n} $$ where the $Y_i$ are i.i.d since the $X_i$ are. Observe that $$ E(\log S_n)=\frac{1}{n}\times nEY_{1}=\alpha^{-1}=EY_{1};\quad \text{Var}(\log S_n)=\frac{1}{n^2}\times n\text{Var}(Y_{1})=\alpha^{-2}/n. $$ The central limit theorem implies that $$ \alpha\sqrt n(\log S_n-\alpha^{-1})\stackrel{D}{\rightarrow} N(0,1). $$ so that $$ \sqrt n(\log S_n-\alpha^{-1})\stackrel{D}{\rightarrow} N(0,\alpha^{-2}) $$ where the second parameter is variance.

To see this suppose that $Z_{n}$ is a sequence of random variables such that $Z_{n}\stackrel{D}{\rightarrow}Z\sim N(0,1)$. We claim that $Z_{n}/\alpha\stackrel{D}{\rightarrow}Z/\alpha\sim N(0,\alpha^{-2})$ (Here $\alpha >0)$. Indeed, $$ P(Z_{n}/\alpha\leq x)=P(Z_{n}\leq\alpha x)\stackrel{n\to \infty}{\rightarrow} P(Z\leq \alpha x)=P(Z/\alpha\leq x). $$ Using similar reasoning we can conclude that $\log S_n\stackrel{D}{\rightarrow} N(\alpha^{-1}, \alpha^{-2}/n)$. Then $S_n\stackrel{D}{\rightarrow} \text{Lognormal}(\alpha^{-1}, \alpha^{-2}/n)$. For more on the lognormal distribution see here.

  • $\begingroup$ Thank you, this is very helpful. However, I'm still not sure about the last part; does this give me any information that can help me determine the limiting distribution of $S_n$ itself? For example, to compute $P(S_n>1.12)$, if I knew that $S_n$ converged to a normal distribution, it would be simple; but instead I only know that particular function of $S_n$ converges to $N(0,\alpha^{-2})$, which isn't the same. $\endgroup$ – ereHsaWyhsipS Jan 27 '18 at 20:04
  • $\begingroup$ @ereHsaWyhsipS I added more to my answer that addresses your question. $\endgroup$ – Foobaz John Jan 27 '18 at 22:14
  • $\begingroup$ If $\log S_n \xrightarrow{p} \alpha^{-1}$ then $\log S_n \xrightarrow{D} \alpha^{-1}$. $\endgroup$ – V. Vancak Jan 27 '18 at 22:22

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