Density of $(X,Y)$ when $X:=sqrt{-2log U}cos(2pi V)\Y:=sqrt{-2 log U}sin(2pi V)$ $U,V sim(0,1)$











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Let $U,V sim(0,1)$ be two independent uniformly distributed random variables:



$$X:=sqrt{-2log U}cos(2pi V)\Y:=sqrt{-2 log U}sin(2pi V)$$



How can I determine the density of the distribution of $(X,Y)$?



I know that $frac{Y}{X}=tan(2pi V)$ and $frac{X^2}{log U}+frac{Y^2}{log U}=-2$ but I don't know if this helps here.










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  • Do you know the distribution (specifically the CDF) of $-log(U)$? If you do then you can get the distribution of $sqrt{-2log(U)}$ and then the rest of the problem is conceptual.
    – Ian
    Nov 13 at 18:57












  • @Ian No I don't
    – user610431
    Nov 13 at 19:00










  • Well, work on that part first. It isn't hard.
    – Ian
    Nov 13 at 19:01










  • I guess you mean that $U$ and $V$ are uniformly distributed (and independent), not $X$ and $Y$.
    – Alejandro Nasif Salum
    Nov 13 at 19:03










  • Possible duplicate of Proof of the Box-Muller method
    – StubbornAtom
    Nov 14 at 15:52















up vote
0
down vote

favorite
1












Let $U,V sim(0,1)$ be two independent uniformly distributed random variables:



$$X:=sqrt{-2log U}cos(2pi V)\Y:=sqrt{-2 log U}sin(2pi V)$$



How can I determine the density of the distribution of $(X,Y)$?



I know that $frac{Y}{X}=tan(2pi V)$ and $frac{X^2}{log U}+frac{Y^2}{log U}=-2$ but I don't know if this helps here.










share|cite|improve this question
























  • Do you know the distribution (specifically the CDF) of $-log(U)$? If you do then you can get the distribution of $sqrt{-2log(U)}$ and then the rest of the problem is conceptual.
    – Ian
    Nov 13 at 18:57












  • @Ian No I don't
    – user610431
    Nov 13 at 19:00










  • Well, work on that part first. It isn't hard.
    – Ian
    Nov 13 at 19:01










  • I guess you mean that $U$ and $V$ are uniformly distributed (and independent), not $X$ and $Y$.
    – Alejandro Nasif Salum
    Nov 13 at 19:03










  • Possible duplicate of Proof of the Box-Muller method
    – StubbornAtom
    Nov 14 at 15:52













up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





Let $U,V sim(0,1)$ be two independent uniformly distributed random variables:



$$X:=sqrt{-2log U}cos(2pi V)\Y:=sqrt{-2 log U}sin(2pi V)$$



How can I determine the density of the distribution of $(X,Y)$?



I know that $frac{Y}{X}=tan(2pi V)$ and $frac{X^2}{log U}+frac{Y^2}{log U}=-2$ but I don't know if this helps here.










share|cite|improve this question















Let $U,V sim(0,1)$ be two independent uniformly distributed random variables:



$$X:=sqrt{-2log U}cos(2pi V)\Y:=sqrt{-2 log U}sin(2pi V)$$



How can I determine the density of the distribution of $(X,Y)$?



I know that $frac{Y}{X}=tan(2pi V)$ and $frac{X^2}{log U}+frac{Y^2}{log U}=-2$ but I don't know if this helps here.







probability probability-theory measure-theory probability-distributions






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edited Nov 13 at 19:04

























asked Nov 13 at 18:54









user610431

667




667












  • Do you know the distribution (specifically the CDF) of $-log(U)$? If you do then you can get the distribution of $sqrt{-2log(U)}$ and then the rest of the problem is conceptual.
    – Ian
    Nov 13 at 18:57












  • @Ian No I don't
    – user610431
    Nov 13 at 19:00










  • Well, work on that part first. It isn't hard.
    – Ian
    Nov 13 at 19:01










  • I guess you mean that $U$ and $V$ are uniformly distributed (and independent), not $X$ and $Y$.
    – Alejandro Nasif Salum
    Nov 13 at 19:03










  • Possible duplicate of Proof of the Box-Muller method
    – StubbornAtom
    Nov 14 at 15:52


















  • Do you know the distribution (specifically the CDF) of $-log(U)$? If you do then you can get the distribution of $sqrt{-2log(U)}$ and then the rest of the problem is conceptual.
    – Ian
    Nov 13 at 18:57












  • @Ian No I don't
    – user610431
    Nov 13 at 19:00










  • Well, work on that part first. It isn't hard.
    – Ian
    Nov 13 at 19:01










  • I guess you mean that $U$ and $V$ are uniformly distributed (and independent), not $X$ and $Y$.
    – Alejandro Nasif Salum
    Nov 13 at 19:03










  • Possible duplicate of Proof of the Box-Muller method
    – StubbornAtom
    Nov 14 at 15:52
















Do you know the distribution (specifically the CDF) of $-log(U)$? If you do then you can get the distribution of $sqrt{-2log(U)}$ and then the rest of the problem is conceptual.
– Ian
Nov 13 at 18:57






Do you know the distribution (specifically the CDF) of $-log(U)$? If you do then you can get the distribution of $sqrt{-2log(U)}$ and then the rest of the problem is conceptual.
– Ian
Nov 13 at 18:57














@Ian No I don't
– user610431
Nov 13 at 19:00




@Ian No I don't
– user610431
Nov 13 at 19:00












Well, work on that part first. It isn't hard.
– Ian
Nov 13 at 19:01




Well, work on that part first. It isn't hard.
– Ian
Nov 13 at 19:01












I guess you mean that $U$ and $V$ are uniformly distributed (and independent), not $X$ and $Y$.
– Alejandro Nasif Salum
Nov 13 at 19:03




I guess you mean that $U$ and $V$ are uniformly distributed (and independent), not $X$ and $Y$.
– Alejandro Nasif Salum
Nov 13 at 19:03












Possible duplicate of Proof of the Box-Muller method
– StubbornAtom
Nov 14 at 15:52




Possible duplicate of Proof of the Box-Muller method
– StubbornAtom
Nov 14 at 15:52










2 Answers
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You mean $U,V$ are uniform, and so $(X,Y)$ are $N(0,1)$ in the margins, and clearly uncorrelated. That does not, however, prove them to be independent (though actually they are). To find the density properly, use the transformation law, by finding the Jacobian.






share|cite|improve this answer




























    up vote
    0
    down vote













    Using the relations you mentioned, one can solve for $U$ and $V$ in terms of $X$ and $Y$. Let's say $U=g(X,Y)$ and $V=h(X,Y)$.



    Then, by the theorem of change of variables, we have
    $$f_{XY}(x,y)=f_{UV}(g(x,y),h(x,y))cdot J(x,y),$$
    where $f_{UV}$ is the joint PDF of $U$ and $V$, and so
    $$f_{UV}(u,v)=left{begin{matrix}1& 0<u<1 wedge 0<v<1 \0 & text{otherwise}\ end{matrix}right.,$$
    and $J$ is the Jacobian determinant of $u$ and $v$ with respect to $x$ and $y$, that is
    $$J(x,y)=left|begin{matrix} frac{partial g}{partial x}(x,y)& frac{partial g}{partial y}(x,y)\frac{partial h}{partial x}(x,y) & frac{partial h}{partial y}(x,y)\ end{matrix}right|.$$






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      2 Answers
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      2 Answers
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      up vote
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      down vote













      You mean $U,V$ are uniform, and so $(X,Y)$ are $N(0,1)$ in the margins, and clearly uncorrelated. That does not, however, prove them to be independent (though actually they are). To find the density properly, use the transformation law, by finding the Jacobian.






      share|cite|improve this answer

























        up vote
        0
        down vote













        You mean $U,V$ are uniform, and so $(X,Y)$ are $N(0,1)$ in the margins, and clearly uncorrelated. That does not, however, prove them to be independent (though actually they are). To find the density properly, use the transformation law, by finding the Jacobian.






        share|cite|improve this answer























          up vote
          0
          down vote










          up vote
          0
          down vote









          You mean $U,V$ are uniform, and so $(X,Y)$ are $N(0,1)$ in the margins, and clearly uncorrelated. That does not, however, prove them to be independent (though actually they are). To find the density properly, use the transformation law, by finding the Jacobian.






          share|cite|improve this answer












          You mean $U,V$ are uniform, and so $(X,Y)$ are $N(0,1)$ in the margins, and clearly uncorrelated. That does not, however, prove them to be independent (though actually they are). To find the density properly, use the transformation law, by finding the Jacobian.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 13 at 19:04









          Richard Martin

          1,4718




          1,4718






















              up vote
              0
              down vote













              Using the relations you mentioned, one can solve for $U$ and $V$ in terms of $X$ and $Y$. Let's say $U=g(X,Y)$ and $V=h(X,Y)$.



              Then, by the theorem of change of variables, we have
              $$f_{XY}(x,y)=f_{UV}(g(x,y),h(x,y))cdot J(x,y),$$
              where $f_{UV}$ is the joint PDF of $U$ and $V$, and so
              $$f_{UV}(u,v)=left{begin{matrix}1& 0<u<1 wedge 0<v<1 \0 & text{otherwise}\ end{matrix}right.,$$
              and $J$ is the Jacobian determinant of $u$ and $v$ with respect to $x$ and $y$, that is
              $$J(x,y)=left|begin{matrix} frac{partial g}{partial x}(x,y)& frac{partial g}{partial y}(x,y)\frac{partial h}{partial x}(x,y) & frac{partial h}{partial y}(x,y)\ end{matrix}right|.$$






              share|cite|improve this answer

























                up vote
                0
                down vote













                Using the relations you mentioned, one can solve for $U$ and $V$ in terms of $X$ and $Y$. Let's say $U=g(X,Y)$ and $V=h(X,Y)$.



                Then, by the theorem of change of variables, we have
                $$f_{XY}(x,y)=f_{UV}(g(x,y),h(x,y))cdot J(x,y),$$
                where $f_{UV}$ is the joint PDF of $U$ and $V$, and so
                $$f_{UV}(u,v)=left{begin{matrix}1& 0<u<1 wedge 0<v<1 \0 & text{otherwise}\ end{matrix}right.,$$
                and $J$ is the Jacobian determinant of $u$ and $v$ with respect to $x$ and $y$, that is
                $$J(x,y)=left|begin{matrix} frac{partial g}{partial x}(x,y)& frac{partial g}{partial y}(x,y)\frac{partial h}{partial x}(x,y) & frac{partial h}{partial y}(x,y)\ end{matrix}right|.$$






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Using the relations you mentioned, one can solve for $U$ and $V$ in terms of $X$ and $Y$. Let's say $U=g(X,Y)$ and $V=h(X,Y)$.



                  Then, by the theorem of change of variables, we have
                  $$f_{XY}(x,y)=f_{UV}(g(x,y),h(x,y))cdot J(x,y),$$
                  where $f_{UV}$ is the joint PDF of $U$ and $V$, and so
                  $$f_{UV}(u,v)=left{begin{matrix}1& 0<u<1 wedge 0<v<1 \0 & text{otherwise}\ end{matrix}right.,$$
                  and $J$ is the Jacobian determinant of $u$ and $v$ with respect to $x$ and $y$, that is
                  $$J(x,y)=left|begin{matrix} frac{partial g}{partial x}(x,y)& frac{partial g}{partial y}(x,y)\frac{partial h}{partial x}(x,y) & frac{partial h}{partial y}(x,y)\ end{matrix}right|.$$






                  share|cite|improve this answer












                  Using the relations you mentioned, one can solve for $U$ and $V$ in terms of $X$ and $Y$. Let's say $U=g(X,Y)$ and $V=h(X,Y)$.



                  Then, by the theorem of change of variables, we have
                  $$f_{XY}(x,y)=f_{UV}(g(x,y),h(x,y))cdot J(x,y),$$
                  where $f_{UV}$ is the joint PDF of $U$ and $V$, and so
                  $$f_{UV}(u,v)=left{begin{matrix}1& 0<u<1 wedge 0<v<1 \0 & text{otherwise}\ end{matrix}right.,$$
                  and $J$ is the Jacobian determinant of $u$ and $v$ with respect to $x$ and $y$, that is
                  $$J(x,y)=left|begin{matrix} frac{partial g}{partial x}(x,y)& frac{partial g}{partial y}(x,y)\frac{partial h}{partial x}(x,y) & frac{partial h}{partial y}(x,y)\ end{matrix}right|.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 13 at 19:15









                  Alejandro Nasif Salum

                  3,629117




                  3,629117






























                       

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