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

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 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.

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

Looking at the area element $$dA$$ of the unit disk, we see that $$dA = rdrd\psi = dxdy$$ Hence $$f_{X, Y}\left(x, y\right)dxdy = f_{R, \Psi}\left(r, \psi\right)drd\psi$$ $$f_{X, Y}\left(x, y\right) = \dfrac{f_{R, \Psi}\left(r, \psi\right)}{r}$$ $$f_{X, Y}\left(x, y\right) = \dfrac{1}{2\pi\sqrt{x^{2} + y^{2}}}, 0 < x^{2} + y^{2} < 1$$