# Central Limit Theorem and Sum of Random Variables

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:

$$\sqrt{n}(\ln(S_n)-E(\ln(X_1)))=\frac{\sum_{i-1}^n{\ln(X_i)}-n\alpha^{-1}}{\sqrt{n}}.$$

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.
• 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. – ereHsaWyhsipS Jan 27 '18 at 20:04
• If $\log S_n \xrightarrow{p} \alpha^{-1}$ then $\log S_n \xrightarrow{D} \alpha^{-1}$. – V. Vancak Jan 27 '18 at 22:22