Convergence in Probability am I doing something wrong?

[Artusartos]

Homework Statement

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 $n \rightarrow \infty$

Homework Equations

The Attempt at a Solution

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...

Thanks in advance

 

[csopi]

The sum of n independent Exp(1) is Gamma(n,1), so if I'm not mistaken, the distribution of the mean is 1/n*Gamma(n,1) instead of Gamma(n,1).

 

[Artusartos]

The sum of n independent Exp(1) is Gamma(n,1), so if I'm not mistaken, the distribution of the mean is 1/n*Gamma(n,1) instead of Gamma(n,1).

Thanks a lot. Can you explain a little more about why it needs to be $\frac{1}{n}\Gamma(n,1)$?

Is it because the sample mean is $\frac{X_1 + X_2 + ... + X_n}{n}$? And since we know that the sum of n independent Exp(1) is Gamma(n,1), $X_1 + ... + X_n = \Gamma(n,1)$, right? Is that why we divide by n?