If $R$ is uniform $(0,1)$ and $Psi$ is uniform $(0,2pi)$, what is the density law of $(X,Y)=(RcosPsi...












3












$begingroup$


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



I tried as follow :



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




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


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



But now, how can I know $f_{X,Y}(x,y)$ when $x=0$ ? i.e. what is $f_{X,Y}(0,y)$ for $yin [0,1]$ ? 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.






probability







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  • $begingroup$
    @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
















3












$begingroup$


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



I tried as follow :



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




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


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



But now, how can I know $f_{X,Y}(x,y)$ when $x=0$ ? i.e. what is $f_{X,Y}(0,y)$ for $yin [0,1]$ ? 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.






probability







share|cite|improve this question









$endgroup$












  • $begingroup$
    @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














3












3








3





$begingroup$


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



I tried as follow :



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




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


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



But now, how can I know $f_{X,Y}(x,y)$ when $x=0$ ? i.e. what is $f_{X,Y}(0,y)$ for $yin [0,1]$ ? 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.






probability







share|cite|improve this question









$endgroup$




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



I tried as follow :



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




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


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



But now, how can I know $f_{X,Y}(x,y)$ when $x=0$ ? i.e. what is $f_{X,Y}(0,y)$ for $yin [0,1]$ ? 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.






probability




probability






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asked Nov 30 '18 at 12:35









NewMathNewMath

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  • $begingroup$
    @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


















  • $begingroup$
    @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
















$begingroup$
@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




$begingroup$
@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




$begingroup$
Presumably, $R$ and $Psi$ are independent.
$endgroup$
– StubbornAtom
Nov 30 '18 at 14:03










1 Answer
1






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$begingroup$

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



enter image description here



Hence
$$f_{X, Y}left(x, yright)dxdy = f_{R, Psi}left(r, psiright)drdpsi $$
$$f_{X, Y}left(x, yright) = dfrac{f_{R, Psi}left(r, psiright)}{r} $$
$$f_{X, Y}left(x, yright) = dfrac{1}{2pisqrt{x^{2} + y^{2}}}, 0 < x^{2} + y^{2} < 1 $$






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    $begingroup$

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



    enter image description here



    Hence
    $$f_{X, Y}left(x, yright)dxdy = f_{R, Psi}left(r, psiright)drdpsi $$
    $$f_{X, Y}left(x, yright) = dfrac{f_{R, Psi}left(r, psiright)}{r} $$
    $$f_{X, Y}left(x, yright) = dfrac{1}{2pisqrt{x^{2} + y^{2}}}, 0 < x^{2} + y^{2} < 1 $$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

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



      enter image description here



      Hence
      $$f_{X, Y}left(x, yright)dxdy = f_{R, Psi}left(r, psiright)drdpsi $$
      $$f_{X, Y}left(x, yright) = dfrac{f_{R, Psi}left(r, psiright)}{r} $$
      $$f_{X, Y}left(x, yright) = dfrac{1}{2pisqrt{x^{2} + y^{2}}}, 0 < x^{2} + y^{2} < 1 $$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

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



        enter image description here



        Hence
        $$f_{X, Y}left(x, yright)dxdy = f_{R, Psi}left(r, psiright)drdpsi $$
        $$f_{X, Y}left(x, yright) = dfrac{f_{R, Psi}left(r, psiright)}{r} $$
        $$f_{X, Y}left(x, yright) = dfrac{1}{2pisqrt{x^{2} + y^{2}}}, 0 < x^{2} + y^{2} < 1 $$






        share|cite|improve this answer









        $endgroup$



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



        enter image description here



        Hence
        $$f_{X, Y}left(x, yright)dxdy = f_{R, Psi}left(r, psiright)drdpsi $$
        $$f_{X, Y}left(x, yright) = dfrac{f_{R, Psi}left(r, psiright)}{r} $$
        $$f_{X, Y}left(x, yright) = dfrac{1}{2pisqrt{x^{2} + y^{2}}}, 0 < x^{2} + y^{2} < 1 $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 30 '18 at 13:41









        rzchrzch

        1363




        1363






























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