MGF Techniques for Chi-Square Distribution on 2n Degrees of Freedom

[dim&dimmer]

Homework Statement

rvX has f(x) = \alpha \exp^{-\alpha x} , and \ W = 2n \alpha \overline {X} defines a random sample from the distribution.
Use moment generating function techniques to show that the distribution of W is chi-square on 2n degrees of freedom.

Homework Equations

The Attempt at a Solution

Well...
Ive let \alpha = \frac {1}{\beta}, then f(x) ~ exp(\beta)
M_x(t) = (1 - \beta t)^{-1}
mgf of W with w~chisquare(2n)
M_w(t) = (1 - 2t)^{-2v}

I don't really know what to do after this. Any help appreciated

 

[statdad]

I'll use T for the statistic of interest.

<br /> T = 2n\alpha \overline X = 2 \alpha \sum_{i=1}^n X_i<br />

You know the m.g.f. of each X_i (they are iid). When you begin to calculate

<br /> \int_{-\infty}^\infty e^{st} \, dt = E[e^{st}]<br />

remember that t is a sum, and use properties of exponents and expected values. You should wind up with a product that will lead you to the answer. (This all relies on the fact that the moment-generating function uniquely identifies the \chi^2 distribution.)