The problem is as follows:
If $Z_{1},Z_{2}\sim N\left(0,1\right)$ , show that the sum of their squares is ~exp(2)
And then the problem gives two hints:
First, find the distribution of $Z^{2}_{1} , Z^{2}_{2}$
$\sqrt{\pi}=\Gamma\left(\dfrac{1}{2}\right)$
I know that the square of a std normal variable is chi squared one, and that I could just write the PDF of the chi-squared distribution with k=2. But is there a way to show that squaring each z, and summing them leads to the Chi-squared PDF for k=2. I have tried to do that, because I feel I get a better understanding of the connection between the two distributions, but I can't seem to get it right.
It might be a weird question, but I always try to get an understanding these things instead of just "accepting it" and move on. Hope someone can show me how this is done.
Let $Z = N(0,1)^2$, and met $M$ by the CDF of this normal distribution. Then
$$ P(Z < x) = P(N^2 < x) = P(- \sqrt x < N < \sqrt x).$$
Writing in terms of the CDF function $M$ tells us:
$$ = M(\sqrt x) - M(- \sqrt x)$$
so that the PDF of $Z$ is given by:
$$ \frac{d}{dx} [ M(\sqrt x) - M(- \sqrt x)] = \frac{e^{-\frac{x}{2}}}{\sqrt{2\pi}} \frac{d}{dx} \sqrt x - \frac{e^{-\frac{x}{2}}}{\sqrt{2\pi}} \frac{d}{dx} (-\sqrt x) = \frac{x^{-\frac{1}{2}}e^{-\frac{x}{2}}}{\sqrt{2\pi}}$$
This should now look familiar to you if you note the value of $\Gamma$ at $\frac{1}{2}$.