- #1

Artusartos

- 247

- 0

## Homework Statement

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

a) Show that the mgf of [itex]Y_n=\sqrt{n}(\bar{X_n}-1)[/itex] is [itex]M_{Y_n}(t) = [e^{t/\sqrt{n}} - (t/\sqrt{n})e^{t/\sqrt{n}}]^{-n}[/itex], [itex]t < \sqrt{n}[/itex] b) Find the limiting distribution [itex]Y_n[/itex] as [itex]n \rightarrow \infty[/itex]

## Homework Equations

## The Attempt at a Solution

a) We know that the pdf of [itex]\bar{X_n}[/itex] is [itex]\Gamma(n,1)[/itex]

[itex]M_{Y_n}(t) = (\frac{1}{1-t})^n[/itex]

[itex]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}})[/itex]

MGF of [itex]\bar{X_n}[/itex] evaluated at [itex]t\sqrt{n}[/itex]

[itex]=e^{-t\sqrt{n}}(\frac{1}{1-t\sqrt{n}})^n = ((e^{t/\sqrt{n}})(1-t\sqrt{n}))^{-n}[/itex]

This does not match the answer they give because t is multiplied by [itex]\sqrt{n}[/itex] instead of being divided by it. But I just can't see where I went wrong...

Thanks in advance