let $X, Y \stackrel{}{\sim} \mathscr{N}(0,1)$ be two independant r.v's
I want to find the joint PDF of $(X,Z)$ where $Z = X^2+Y^2$
how do I proceed ?
I have tried the following : ($F$ denotes the CDF and $f$ the PDF)
$$ \begin{align} F_{X,Z}(x,z) & = \mathbb{P}(X \leq x, Z \leq z) = \mathbb{P}(X \leq x, Y^2 \leq z - x^2) \\ & = \mathbb{P}(X \leq x, -\sqrt{z - x^2} \leq Y \leq \sqrt{z - x^2}) \\ & = \mathbb{P}(X \leq x, Y \leq \sqrt{z - x^2}) + \mathbb{P}(X \leq x, -\sqrt{z - x^2} \leq Y ) \\ & = F_{X,Y}(x,\sqrt{z - x^2}) + {\color{red}{\mathbb{P}(X \leq x, -\sqrt{z - x^2} \leq Y )}} \end{align} $$
I also know how to find $F_{X,Y}(x,y)$ but I don't know how to deal with the red term.
any help or other methods will be greatly appreciated.
Polar coordinates help here. \begin{align} \Pr(X^2+Y^2 > w) & = \iint\limits_{ \left( \begin{array}{c} \text{complement of} \\ \text{disk of radius }\sqrt w \end{array}\right)} \frac 1 {2\pi} e^{-(x^2+y^2)/2} \, d(x,y) \\[10pt] & = \Pr(R^2>x) = \int_0^{2\pi} \underbrace{ \left( \int_{\sqrt w}^\infty \frac 1 {2\pi} e^{-r^2/2} r \, dr \right) }_{\text{No $\theta$ appears here.}} \,\,d\theta \\[10pt] & = \int_{\sqrt w}^\infty e^{-r^2/2} r\, dr = \int_{w/2}^\infty e^{-u} \, du = e^{-w/2} \text{ for } w\ge 0. \end{align} Thus $X^2+Y^2$ has an exponential distribution with expected value $2.$