1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.