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?