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 = $-\sqrt{n}$
Then
1) $\Psi _{Y_{n}}(t)$= $e^{-\sqrt{n}t}$$\Psi _{\bar{X_{n}}}\left ( \sqrt{n}t \right)$
If I can find $\Psi _{\bar{X_{n}}}\left ( \sqrt{n}t \right)$ then I should obtain $\Psi _{{Y_{n}}}$
I found:
2) MGF of Exponential = $\frac{1}{1-t}$
3) MGF of $\Psi _{\bar{X_{n}}}\left ( \sqrt{n}t \right)$ is $\left ( \frac{n}{n-\sqrt{n}t}\right )^{n}$
However, when I applied (3) back to (1), I am unable to obtain the final answer. Where did I go wrong, and how can I fix this?
thanks.
$$ \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 $$