Proof that if $Z$ is standard normal, then $Z^2$ is distributed Chi-Square (1). Suppose that $Z\sim N(0,1)$ and let $V=Z^2$.  Prove that $V\sim \chi^2(1)$.
I want to use the method of moment generating functions, because I already understand the proof using the method of distribution functions.  I will show my work, and then where I got stuck.
Since $Z\sim N(0,1)$, then $\mu=0$ and $\sigma^2=1$, and we have
$$M_V(t) =E[e^{tV}]=E[e^{tZ^2}]=\int_{-\infty}^\infty e^{tz^2}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}z^2}dz=\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}e^{z^2(t-\frac{1}{2})}dz.$$
At this point, I'm out of ideas.  I want to eventually get something that looks like $\frac{1}{(1-2t)^{\frac{1}{2}}}$.  Could I get a hint please?
 A: $$\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{z^2(t-\frac{1}{2})}dz=\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-z^2(\frac{1}{2}-t)}dz$$
The general PDF for a normal distribution is given by:
$$
f(x)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}
$$
You should attempt to solve the integral by fitting a normal distribution and cancelling it out by realising that it integrates to 1. Currently:
$$
\mu=0
$$
$$
\frac{1}{2\sigma^2}=\frac{1}{2}-t
$$
So, solve for $\sigma$ and multiply accordingly to make the integral the pdf of a normal distribution (integrates to 1) whatever is left over should give you the result you're looking for.
Hope this helps
A: If $a^2=1-2t$, then $t-\frac12= -\frac12\left(1-2t\right)=-\frac{a^2}{2}$, so
\begin{align}
\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}e^{z^2(t-\frac{1}{2})} \, dz
& = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} e^{-z^2 a^2/2} \, dz \\[10pt]
& = \frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-w^2/2} \left(\frac{dw}a\right) \\[10pt]
& = \frac1a\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-w^2/2} \, dw \\[10pt]
& = \frac1a = \frac{1}{\sqrt{1-2t}}. 
\end{align}
