Proof that if $Z$ is standard normal, then $Z^2$ is distributed Chi-Square (1).

Solution 1:

$$\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{z^2(t-\frac{1}{2})}dz=\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-z^2(\frac{1}{2}-t)}dz$$

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

Solution 2:

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}