Seriously I dont have any idea what is this thing called. I know how to find Joint PDf of two variables.. But i dont know how to transform it in other variables ?
Do they require Jacobians?
Here is my doubt regarding this question: how to solve it?
If you tell me what is this topic called it will be very helpful i would like to see it first... but if you solve it also I will be highly obliged:)
Using the change of basis formula of PDFs: $$g_{Z,\Theta} (z,\theta) = \vert J \vert p_{X,Y}(f(z,\theta),g(z,\theta))$$ where $X = f(z,\theta)$ and $Y = g(z,\theta)$. With a bit of math, you can show that $$X = f(z,\theta)=\sqrt{z} \cos \theta$$ and $$Y = g(z,\theta) = \sqrt{z} \sin \theta$$ and the determinant of the Jacobian is
$$\vert J \vert = \det \begin{bmatrix} \frac{\partial f(z,\theta)}{\partial z} &\frac{\partial f(z,\theta)}{\partial \theta} \\ \frac{\partial g(z,\theta)}{\partial z} &\frac{\partial g(z,\theta)}{\partial \theta} \end{bmatrix} = \det \begin{bmatrix} \frac{1}{2\sqrt{z}}\cos \theta& -\sqrt{z}\sin\theta\\ \frac{1}{2\sqrt{z}}\sin \theta& \sqrt{z}\cos \theta \end{bmatrix} = \frac{1}{2}$$ So $$g_{Z,\Theta} (z,\theta) = \frac{1}{2} p_{X,Y}(\sqrt{z} \cos \theta,\sqrt{z} \sin \theta) \tag{1}$$ The link that @StubbornAtom shared works out the PDF for the normal case but for $Z = \sqrt{X^2 + y^2}$. Here we do $Z = X^2 + Y^2$, the PDF of a standard normal is $$p_{X,Y}(x,y)=\frac{1}{2\pi}e^{-\frac{x^2+y^2}{2}}$$ Replacing the above in equation $(1)$, we get $$g_{Z,\Theta} (z,\theta) = \frac{1}{2}\frac{1}{2\pi}e^{-\frac{(\sqrt{z} \cos \theta)^2+(\sqrt{z} \sin \theta)^2}{2}} = \frac{1}{4\pi}e^{-\frac{z}{2}} $$ independent of $\Theta $!!!!!!!!!!!!