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Let $\bar{X_n}$ denote the mean of a random sample of size n from a distribution that has pdf $f(x) = e^{-x}$, $0<x<\infty$, zero elsewhere.

a) Show that the mgf of $Y_n=\sqrt{n}(\bar{X_n}-1)$ is $M_{Y_n}(t) = [e^{t/\sqrt{n}} - (t/\sqrt{n})e^{t/\sqrt{n}}]^{-n}$, $t < \sqrt{n}$

b) Find the limiting distribution $Y_n$ as $n\rightarrow \infty$

a) We know that the pdf of $\bar{X_n}$ is $\Gamma(n,1)$

$M_{Y_n}(t) = (\frac{1}{1-t})^n$ $M_{Y_n}(t) = E(e^{tY_n}) = E(e^{t\sqrt{n}(\bar{X_n}-1)} = e^{-t\sqrt{n}}E(e^{t\sqrt{n}\bar{X_n}}$

MGF of $\bar{X_n}$ evaluated at $t\sqrt{n}$

$=e^{-t\sqrt{n}}(\frac{1}{1-t\sqrt{n}})^n = ((e^{t/\sqrt{n}})(1-t\sqrt{n}))^{-n}$

This does not match the answer they give because t is multiplied by $\sqrt{n}$ instead of being divided by it. But I just can't see where I went wrong...can anybody help me?

Thanks in advance

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I think $\sum X_i$ should be $\Gamma(n,1)$ right? – blitzer Feb 2 '13 at 19:37
up vote 2 down vote accepted

You forgot to divide the sample mean by $n$. This replaces $t$ by $t/n$ in the mgf to give $$ M_{\bar X_n}(t) = (1-\frac tn)^{-n}. $$ Then $$ M_{Y_n}(t) = M_{\bar X_n}(t{\sqrt{n}}) e^{-t\sqrt{n}} = \left(e^{t/\sqrt{n}} (1-\frac{t}{\sqrt{n}})\right)^{-n}. $$

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Thanks. For the second part of the question, this is what I did: $((1-\frac{t}{\sqrt{n}})^{\sqrt{n}})^{-\sqrt{n}}$. The part inside the brackets converges to $e^{-t}$. But $(e^{-t})^{-\sqrt{n}}$ converges to I'm a bit unsure about my answer. Do you think I did something wrong? – user58289 Feb 2 '13 at 22:05
Well, the population distribution has mean $1$ and finite variance, so by the Central Limit Theorem, $Y_n$ has to converge to a normal distribution. So, you should expect the limit of the mgfs to be the mgf of a normal distribution. – David Moews Feb 2 '13 at 22:24
One last question: I'm not sure how you got the mgf of $\bar{X_n}$ after dividing it by n. In other words, how can you "divide" $\Gamma(n,1)$ by n? Do you divide each of the alpa and the beta by n...or how does it work? – user58289 Feb 3 '13 at 19:17
Dividing a random variable $X$ by something corresponds to dividing the $t$ in the mgf $M_X(t)$ by the same thing. So, I took the original mgf and replaced $t$ by $t/n$. – David Moews Feb 3 '13 at 19:23
Oh so you just divide the beta...ok thanks. – user58289 Feb 3 '13 at 19:45

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