3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ @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
1
$\begingroup$

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

enter image description here

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.