# MGF of chi square

1. Sep 28, 2008

### dim&dimmer

1. The problem statement, all variables and given/known data
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.
2. Relevant equations
3. 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

2. Sep 28, 2008

I'll use $$T$$ for the statistic of interest.
$$T = 2n\alpha \overline X = 2 \alpha \sum_{i=1}^n X_i$$
You know the m.g.f. of each $$X_i$$ (they are iid). When you begin to calculate
$$\int_{-\infty}^\infty e^{st} \, dt = E[e^{st}]$$
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.)