If $R$ is uniform $(0,1)$ and $Psi$ is uniform $(0,2pi)$, what is the density law of $(X,Y)=(RcosPsi...
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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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add a comment |
$begingroup$
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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@LeeDavidChungLin : You should really read the question to the end, and you'll see that it's absolutely not duplicate ;)
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– NewMath
Nov 30 '18 at 13:33
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Presumably, $R$ and $Psi$ are independent.
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– StubbornAtom
Nov 30 '18 at 14:03
add a comment |
$begingroup$
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
$endgroup$
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
asked Nov 30 '18 at 12:35
NewMathNewMath
4059
4059
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@LeeDavidChungLin : You should really read the question to the end, and you'll see that it's absolutely not duplicate ;)
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– NewMath
Nov 30 '18 at 13:33
$begingroup$
Presumably, $R$ and $Psi$ are independent.
$endgroup$
– StubbornAtom
Nov 30 '18 at 14:03
add a comment |
$begingroup$
@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
add a comment |
1 Answer
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$begingroup$
Looking at the area element $dA$ of the unit disk, we see that
$$dA = rdrdpsi = dxdy $$
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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1 Answer
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$begingroup$
Looking at the area element $dA$ of the unit disk, we see that
$$dA = rdrdpsi = dxdy $$
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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add a comment |
$begingroup$
Looking at the area element $dA$ of the unit disk, we see that
$$dA = rdrdpsi = dxdy $$
Hence
$$f_{X, Y}left(x, yright)dxdy = f_{R, Psi}left(r, psiright)drdpsi $$
$$f_{X, Y}left(x, yright) = dfrac{f_{R, Psi}left(r, psiright)}{r} $$
$$f_{X, Y}left(x, yright) = dfrac{1}{2pisqrt{x^{2} + y^{2}}}, 0 < x^{2} + y^{2} < 1 $$
$endgroup$
add a comment |
$begingroup$
Looking at the area element $dA$ of the unit disk, we see that
$$dA = rdrdpsi = dxdy $$
Hence
$$f_{X, Y}left(x, yright)dxdy = f_{R, Psi}left(r, psiright)drdpsi $$
$$f_{X, Y}left(x, yright) = dfrac{f_{R, Psi}left(r, psiright)}{r} $$
$$f_{X, Y}left(x, yright) = dfrac{1}{2pisqrt{x^{2} + y^{2}}}, 0 < x^{2} + y^{2} < 1 $$
$endgroup$
Looking at the area element $dA$ of the unit disk, we see that
$$dA = rdrdpsi = dxdy $$
Hence
$$f_{X, Y}left(x, yright)dxdy = f_{R, Psi}left(r, psiright)drdpsi $$
$$f_{X, Y}left(x, yright) = dfrac{f_{R, Psi}left(r, psiright)}{r} $$
$$f_{X, Y}left(x, yright) = dfrac{1}{2pisqrt{x^{2} + y^{2}}}, 0 < x^{2} + y^{2} < 1 $$
answered Nov 30 '18 at 13:41
rzchrzch
1363
1363
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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