Let there be $X, Y$, RV such that the have the joint PDF:
$f(x,y) = 4/(\pi R^2)$, for all $x^2 + y^2 \leq R^2, x<0, y<0$.
Let the constraints above be defined as a set P.
then let $\theta(x,y) = \arctan(Y/X)$, for all $x,y$ in $P$. Find the DF with respect to $\theta$.
I have an idea of have to do this, I had to find the DF with respect to $R$, where $R = \sqrt{X^2 + Y^2}$.
My idea right now is to find the area of which $\theta$ covers, and multiply it by the joint PDF to get the DF, obviously with the respective bounds for $\theta$. The question simply asks:
What is $P(\theta \leq a)$, for all $(x,y)$ in $P$? This is where I am stuck. If anyone could help me out it would be appreciated. Also some formatting errors in the question, feel free to tell me my errors
It is also important to mention that I am only looking for a way to find the DF of $\theta$ by the area of the region the probability covers, and without the joint PDF/DF of $R$, and $\theta$.
The long way:
Well the change of variables transformation would be:
$$f_{X,\Theta}(x, \theta) = {f_{X,Y}(x,x\tan\theta)\begin{Vmatrix}\partial x /\partial x&\partial x/\partial \theta \\ \partial (x\tan\theta)/\partial x&\partial (x\tan\theta)/\partial \theta\end{Vmatrix}}\;\mathbf 1_{x\in[-R;0]\wedge\theta\in[-\pi;\arccos(x/\sqrt R))}$$
Then find the marginal from the joint.
The short way:
Alternatively: You can simply observe that the joint of vector $(X,Y)$ is uniform over the quadrant, so the distribution of the angle will be...