Let $Z$~$N(0,1)$.
If I want to write down the definition of $M_{z^2}(t)$ which is the MGF of $Z^2$, I would write the following correct?
$E(e^{tZ^2})$
Assuming that the answer above is correct, if I wanted to find the MGF I would do the following:
I am having trouble with this next part. We can take the integral from $-\infty$ to $\infty$ of $e^{tz^2}$ $*$ $pdf of Z^2$ correct? But what would be the $pdf$? Would it be the $pdf$ of the Standard Normal Distribution multiplied by itself? I don't think that is correct. Or would we use the Chi Squared $pdf$ since it is proven that a Standard Normal R.V. squared is a Chi-Squared?
I am having quite a lot of trouble with MGFs and these types of problems and help would be greatly appreciated as the textbooks I have are quite rigorous and do not go into much depth of the exact mechanism behind each step or the intuition. Thanks!
EDIT: Here is what I have so far...
$\int$ $e^{tz^2}$ $*$ $e^{-0.5y^2}$ $/sqrt(2pi)$ which can be simplified to...
$sqrt(2pi)$ $\int$ $e^{-0.5z^2(-2t+1)}$
So there is a way to mold this integral to make it look like the $pdf$ of a Standard Normal so that we don't need to integrate right? Thanks!
You need to multiply $\exp(tz^2)$ by the density function of the standard normal, and integrate. A change of variable will then reduce the problem to a definite integral that you know. Apart from a constant, you will be integrating $\exp((-z^2/2)(1-2t))$. Let $u=z\sqrt{1-2t}$.
Remark: If $W=g(X)$, then $E(W)=\int_{-\infty}^\infty g(x)f_X(x)\,dx$.