# Finding the Moment Generating Function of Chi-Squared Dist

I'm tasked with finding the MGF of a $\chi^2$ random variable.

I think the way to do is is by using the fact that $\Sigma_{j=1}^{m} Z^2_j$ is a $\chi^2$ R.V. and that MGF of a sum is the product of the MGFs of the individual terms. Although that may not be right and it may be $E(e^{tX})$ way.

Any help would be appreciated. I don't need it solved really just need to get down the track a little further.

Thanks

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What's wrong with the formula given by en.wikipedia.org/wiki/Chi_squared ? i.e. $(1-2t)^{-m/2}$ –  shabbychef Feb 16 '11 at 4:15
There's nothing wrong with it, but I'd like to derive it. –  tshauck Feb 16 '11 at 4:30
ah, gotcha. the question does not say 'derive', however, but 'find'. –  shabbychef Feb 16 '11 at 5:37
sure, but contextually it's obvious I'm not looking for a link –  tshauck Feb 16 '11 at 14:37
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## 1 Answer

Yes, since $\chi^2$ is a sum of $Z_i^2$ the MGF is a product of individual summands. But then you need the MGF of $Z_i^2$ which is $\chi^2$ with 1 degree of freedom. The obvious way of calculating the MGF of $\chi^2$ is by integrating. It is not that hard:

$$Ee^{tX}=\frac{1}{2^{k/2}\Gamma(k/2)}\int_0^\infty x^{k/2-1}e^{-x(1/2-t)}dx$$

Now do the change of variables $y=x(1/2-t)$, then note that you get Gamma function and the result is yours. If you want deeper insights (if there are any) try asking at http://math.stackexchange.com.

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Thanks for the help. –  tshauck Feb 16 '11 at 18:28
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