PDF of $Z=\frac{X^2+Y^2}{2}$ where $X\sim N(0,1)$ and $Y\sim N(0,1)$ Say $X \sim N(0,1)$ and $Y\sim N(0,1)$ are independent random variables. So: $f_X(x) = \frac{1}{\sqrt{2\pi}}e^{\frac{-1}{2}x^2}$ and $f_Y(y) = \frac{1}{\sqrt{2\pi}}e^{\frac{-1}{2}y^2}$. Now I am interested in the probability density function (PDF) $f_Z(z)$ when $Z=\frac{1}{2}(X^2+Y^2)$. 
I know some things about the sum of two normally distributed random variables, for example: $Z_1 = X +Y$ gives that $Z_1 \sim N(0,2)$ by the use of convolution.
How to obtain (efficiently) the PDF for $Z$, again by the use of convolution?
 A: $$\mathbb{P}[X^2+Y^2\leq R^2]=\frac{1}{2\pi}\iint_{x^2+y^2\leq r^2}e^{-\frac{x^2+y^2}{2}}\,dx\,dy=\frac{1}{2\pi}\int_{0}^{2\pi}\int_{0}^{R}\rho e^{-\rho^2/2}\,d\rho\,d\theta$$
so:
$$\mathbb{P}[X^2+Y^2\leq R^2]=1-e^{-R^2/2}$$
and $X^2+Y^2$ has an exponential distribution with parameter $\lambda=\frac{1}{2}$.
A: At one point you say $Z=X^2+Y^2$ and at another you say $Z=\dfrac{X^2+Y^2}2$.  I'll go with the latter version.
Use polar coordinates:
\begin{align}
\Pr(Z>z) & = \iint\limits_{(x,y)\,:\,\frac{x^2+y^2}2>z} \frac 1 {2\pi} e^{-(x^2+y^2)/2} \,d(x,y) \\[10pt]
& = \frac 1 {2\pi} \int_0^{2\pi} \int_{\sqrt{2z}}^\infty e^{-r^2/2} \, r\, dr\, d\theta. \tag 1
\end{align}
The inside integral does not depend on $\theta$; therefore the outside integral is the integral of a constant.  Its value is therefore that constant times the length of the interval.  The length of the interval is $2\pi-0$.  That cancels the preceding denominator, and $(1)$ is then seen to be equal to the inside integral:
$$
\int_{\sqrt{2z}}^\infty e^{-r^2/2} \Big(r\,dr\Big) = \int_z^\infty e^{-u}\,du = e^{-z}. 
$$
Thus for $z\ge 0$ we have
$$
\Pr(Z>z) = e^{-z}.
$$
It's easy to find $f_Z(z)$ from there.
A: Maybe I'm making stuff up but

$\bf IF$ ${\color{red}{X }},{\color{Blue}{Y }}\overset{\tt iid}{\sim}\mathcal{\color{Orange}{N }} {\color{Orange}{(}}{\color{cyan}{0 }},{\color{cyan}{1 }}{\color{Orange}{)}}$, $\bf THEN$ ${\color{Purple}{R }}=\sqrt{{\color{red}{X }}^2+{\color{Blue}{Y }}^2}\sim{Ray}$ where a $\tt Rayleigh\,\, distribution$ is the square root of an ${Exp}({\color{Blue}{\lambda }}=\frac{\color{cyan}{1 }}{\color{cyan}{2 }})$
So ${\color{Purple}{R }}^2={\color{red}{X }}^2+{\color{Blue}{Y }}^2\sim{Exp}({\color{Blue}{\lambda }}=\frac{\color{cyan}{1 }}{\color{cyan}{2 }})$

Now ${\color{orange}{Z }}=\frac{\color{cyan}{1 }}{\color{cyan}{2 }}({\color{red}{X }}^2+{\color{Blue}{Y }}^2)=\frac{\color{cyan}{1 }}{\color{cyan}{2 }}{\color{Purple}{R }}^2\sim \frac{\color{cyan}{1 }}{\color{cyan}{2 }}{Exp}\left(\frac{\color{cyan}{1 }}{\color{cyan}{2 }} \right)={Exp}\left(\frac{\frac{\color{cyan}{1 }}{\color{cyan}{2 }}}{\frac{\color{cyan}{1 }}{\color{cyan}{2 }}} \right)={Exp}({\color{Blue}{\lambda }}={\color{cyan}{1 }})\Longrightarrow {\color{orange}{Z }}\sim {Exp}({\color{cyan}{1 }})$

So ${\color{blue}{f }}_{\color{orange}{Z }}{\color{blue}{(}}{\color{orange}{z }}{\color{blue}{)}}={\color{Blue}{\lambda }}{\color{Turquoise}{e }}^{-{\color{Blue}{\lambda }}{\color{orange}{z }}}={\color{Turquoise}{e }}^{-{\color{orange}{z }}}$

I guess Michael Hardy was right.
$\bf IF$ there are any issues with my answer, $\bf THEN$ let me know in the comments below or by sending a message to TheGreatJRB@Berkeley.edu
Thanks 
Jason
