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Let $\bar{X_{n}}$ be the sample mean of a random sample of size n from a distribution which has a pdf of:

f(x) = ${e^{-x}}$, for x> 0; 0, other wise.

a) show that the mgf of $_{Y_{n}}=\sqrt{n}(\bar{X_{n}}-1)$ is $$\Psi _{Y_{n}}\left ( t \right )=\left ( e^{\frac{t}{\sqrt{n}}} - \frac{te^{\frac{t}{\sqrt{n}}}}{\sqrt{n}}\right)^{-n}$$ ,$t< \sqrt{n}$

So far I have been able to obtain:

$Y_{n}=\sqrt{n}\bar{X_{n}}-\sqrt{n}$, where a =$\sqrt{n}$ , b = $-\sqrt{n}$


1) $\Psi _{Y_{n}}(t)$= $e^{-\sqrt{n}t}$$\Psi _{\bar{X_{n}}}\left ( \sqrt{n}t \right)$

If I can find $\Psi _{\bar{X_{n}}}\left ( \sqrt{n}t \right)$ then I should obtain $\Psi _{{Y_{n}}}$

I found:

2) MGF of Exponential = $\frac{1}{1-t}$

3) MGF of $\Psi _{\bar{X_{n}}}\left ( \sqrt{n}t \right)$ is $\left ( \frac{n}{n-\sqrt{n}t}\right )^{n}$

However, when I applied (3) back to (1), I am unable to obtain the final answer. Where did I go wrong, and how can I fix this?


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Hint: $\sum_i X_i$ is a Gamma random variable with parameters $(n,1)$ and so $\bar{X}_n$ is a Gamma random variable with parameters $(n,n)$ and therefore mean $1$, variance $n^{-1}$. Thus, $\bar{X}_n-1$ is a zero-mean random variable, and your $Y_n$ is a zero-mean unit-variance random variable. The MGF of $\sum_i X_i$ is $(1-t)^{-n}$, and so figure out the MGF of $\bar{X}_n$ and thus of $Y_n$. – Dilip Sarwate Apr 28 '13 at 2:23
Thanks. I found the mgf of $\bar{X_{n}}$ to be $\left ( 1-\frac{t}{n} \right )^{-n}$ but when I applied this back to (1), Iam still having troubles. – JGH Apr 28 '13 at 6:15

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

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