A random vector in $\Re^2$ is chosen as folllows: Its length, $Z$, and its angle, $\Theta$, with the positive $x$-axis, are independent random variables. $Z$ has density $$f(z)=ze^{-z^2/2}, z>0$$ and $\Theta \in Un(0,2\pi)$. Let $Q$ denote the point of the vector. Determine the joint distribution of the cartesian coordinates of $Q$.
I have been trying to solve this problem for two days without any success. I have tried using the transformation theorem, ending up with complicated integrals. I have tried to derive $F_Y(y)=P(Zsin(\Theta)<y)$ and $F_X(x)=P(Zcos(\Theta)<x)$ and then differentiating inside the integral to get the densities, and again i end up with integrals i can't solve.
I would appreciate alot to see how you smarter people out there would solve this. Thanks !
You should try to work directly with joint densities and change of variables here instead of finding the marginal distribution functions. This all too routine and straightforward.
Suppose $Q\equiv(X,Y)$.
You are given that $Z$ and $\Theta$ are independently distributed. So joint density of $(Z,\Theta)$ is $$f_{Z,\Theta}(z,\theta)=\frac{ze^{-z^2/2}}{2\pi}\mathbf1_{z>0,\,0<\theta<2\pi}$$
Transform $(Z,\Theta)\to(X,Y)$ such that $X=Z\cos\Theta$ and $Y=Z\sin\Theta$.
Clearly, $z>0,0<\theta<2\pi\implies x,y\in\mathbb R$, and $|\det J|=\frac{1}{\sqrt{x^2+y^2}}$ where $J$ is the jacobian matrix.
Joint density of $(X,Y)$ is thus $$f_{X,Y}(x,y)=\frac{1}{2\pi}\exp\left(-\frac{x^2+y^2}{2}\right),\quad(x,y)\in\mathbb R^2$$
Indeed, $X$ and $Y$ are independent standard normal variables and the joint distribution of $(X,Y)$ is trivially bivariate normal.