Finding the Moment Generating Function of chi-squared distribution

I’m tasked with deriving 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.

I don’t need it solved really just need to get down the track a little further.

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.

Attribution
Source : Link , Question Author : tshauck , Answer Author : Aniko

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