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.   
 A: 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.
A: I think the easiest way is to simply start with a single squared gaussian:
$$E[e^{tX^2}]
= \int_{-\infty}^\infty e^{tx^2}\tfrac1{\sqrt{2\pi}}e^{-x^2/2}dx
= \tfrac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-(1-2t)x^2/2}dx
= \tfrac1{\sqrt{1-2t}},$$ for $t<1/2$.
Since the Chi-squared is just a sum of independent squared gaussians, you get the factor $k$ in the exponent.
A: You can also do this calculation by brute-force straight from the general chi-squared distribution, without appeal to any intermediate appeal to sums of random variables.  For $X \sim \chi_n^2$ we have moment generating function:
$$\begin{equation} \begin{aligned}
M_X(t) \equiv \mathbb{E}(\exp (tX)) 
&= \int \limits_0^\infty \exp(tx) \cdot \text{Chi-Sq}(x | n) dx \\[8pt]
&= \frac{1}{2^{n/2} \Gamma(n/2)} \int \limits_0^\infty \exp(tx) \cdot x^{n/2-1} \exp(-x/2) dx \\[8pt]
&= \frac{1}{2^{n/2} \Gamma(n/2)} \int \limits_0^\infty x^{n/2-1} \exp((t -\tfrac{1}{2})x) dx. \\[8pt]
\end{aligned} \end{equation}$$
For the case where $t < \tfrac{1}{2}$, using the change-of-variable $r = (\tfrac{1}{2} - t)x$ we have:
$$\begin{equation} \begin{aligned}
M_X(t) 
&= \frac{1}{2^{n/2} \Gamma(n/2)} \int \limits_0^\infty x^{n/2-1} \exp((t -\tfrac{1}{2})x) dx. \\[8pt]
&= (\tfrac{1}{2} - t)^{-n/2} \cdot  \frac{1}{2^{n/2} \Gamma(n/2)} \int \limits_0^\infty r^{n/2-1} \exp(-r) dr. \\[8pt]
&= (1 - 2t)^{-n/2} \cdot \frac{1}{\Gamma(n/2)} \int \limits_0^\infty r^{n/2-1} \exp(-r) dr. \\[8pt]
&= (1 - 2t)^{-n/2}. \\[8pt]
\end{aligned} \end{equation}$$
