If $R$ is uniform $(0,1)$ and $\Psi$ is uniform $(0,2\pi)$, what is the density law of $(X,Y)=(R\cos\Psi ,R\sin\Psi)$ ?

I tried as follow :

I separate the cases where $\Psi\in (0,\pi/2)$, $\Psi\in (\pi/2,\pi)$, $\Psi\in (\pi,3\pi/2)$ and $\Psi\in (3\pi/2,2\pi)$.

  • For the first one, $(x,y)=g(r,\psi )=(r\cos\psi ,r\sin\psi )$ implies that $(r,\psi )=h(x,y)=(\sqrt{x^2+y^2},\arctan(y/x))$ and thus $$f_{X,Y}(x,y)=f_{R, \Psi}(h(x,y))|J_{h}(x,y)|,\ \ if\ x^2+y^2\leq 1, x,y>0,$$

  • For the second one, $(x,y)=k(r,\psi)=(r\cos\psi,r\sin\psi)$ implies $(r,\psi)=\ell(x,y)=(\sqrt{x^2+y^2}, \pi-\arctan(y/x))$, and thus $$f_{X,Y}(x,y)=f_{R,\psi}(\ell(x,y))J_{\ell}(x,y),\ \ if\ x^2+y^2\leq 1, x<0, y<0.$$

But now, how can I know $f_{X,Y}(x,y)$ when $x=0$ ? i.e. what is $f_{X,Y}(0,y)$ for $y\in [0,1]$ ? Because I didn't considered in my manipulation since I avoid $\psi=\pi/2$. I can compute $F_{X,Y}(0,y)$, but to get $f_{X,Y}(0,y)$ I can't derivate, so I'm a bit in truble.

  • $\begingroup$ @LeeDavidChungLin : You should really read the question to the end, and you'll see that it's absolutely not duplicate ;) $\endgroup$ – NewMath Nov 30 '18 at 13:33
  • $\begingroup$ Presumably, $R$ and $\Psi$ are independent. $\endgroup$ – StubbornAtom Nov 30 '18 at 14:03

Looking at the area element $dA$ of the unit disk, we see that $$dA = rdrd\psi = dxdy $$

enter image description here

Hence $$f_{X, Y}\left(x, y\right)dxdy = f_{R, \Psi}\left(r, \psi\right)drd\psi $$ $$f_{X, Y}\left(x, y\right) = \dfrac{f_{R, \Psi}\left(r, \psi\right)}{r} $$ $$f_{X, Y}\left(x, y\right) = \dfrac{1}{2\pi\sqrt{x^{2} + y^{2}}}, 0 < x^{2} + y^{2} < 1 $$


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