moment generating function of normal distribution I know this question relates to the chi-squared distribution, but I think what the question wants me to do is somehow derive this distribution from the information given.
I have a normally distributed function $X$ with $E(X)=0$ and $\mathrm{Var}(X)=\sigma^2$ where $\sigma>0$.
If $X_1,.....X_n$ are independent and are distributed the same as X, how would I find the moment generating function of $Z=X_1^2,.....X_n^2$?
I believe here, $Z$ is the chi-squared distribution. Any help would be great.

If $Y_{1} \ldots Y_{n}$ are standard normally distributed random variables, then $Y_{1}^{2}+Y_{2}^{2}+\ldots+Y_{n}^{2}\sim\chi_{n}^{2}$
Note that your $X_{i}$ are not standard normal variables, but can be transformed to that by writing $Y_{i}=\frac{X_{i}}{\sigma}$. Now, each individual $Y_{i}^{2}$ has a $\chi_{1}^{2}$ distribution, and the variable $Z$ formed by adding each of the $Y_{i}^{2}$ follows the $\chi^{2}_{n}$ distribution. However, the distribution of the variable formed by adding all the $X_{i}^{2}=\sigma^{2}Z$ follows a Gamma distribution.  
If we let $Y\sim \chi^{2}_{n}$, and $Z=\sum_{i=1}^{n}X_{i}^{2}$, then $M_{Y}(t)=(1-2t)^{-k/2}$ and $Z=\sigma^{2}Y$
Then $$M_{Z}(t)=M_{\sigma^{2}Y}(t)=E(e^{t\sigma^{2}Y})=M_{Y}(\sigma^{2}t)=(1-2\sigma^{2}t)^{-k/2}$$
Which is the mgf of a Gamma distribution with parameters $2\sigma^{2}$ and $k/2$.
