Mixing
Given that $X,Y$ are independent $N(0,1)$ , show that $frac{XY}{sqrt{X^2+Y^2}},frac{X^2-Y^2}{2sqrt{X^2+Y^2}}$...
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It is given that $X,Y overset{text{i.i.d.}}{sim} N(0,1)$
Show that $frac{XY}{sqrt{X^2+Y^2}},frac{X^2-Y^2}{2sqrt{X^2+Y^2}} overset{text{i.i.d.}}{sim} N(0,frac{1}{4})$
I was thinking of making polar transformations $X=r cos theta, Y=r sin theta$
Then I am getting stuck at ranges of $theta$
probability probability-distributions
edited Aug 20 '18 at 20:40
Michael Hardy
1
asked Feb 12 '18 at 13:55
Legend KillerLegend Killer
1,6061523
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add a comment |
$begingroup$
It is given that $X,Y overset{text{i.i.d.}}{sim} N(0,1)$
Show that $frac{XY}{sqrt{X^2+Y^2}},frac{X^2-Y^2}{2sqrt{X^2+Y^2}} overset{text{i.i.d.}}{sim} N(0,frac{1}{4})$
I was thinking of making polar transformations $X=r cos theta, Y=r sin theta$
Then I am getting stuck at ranges of $theta$
probability probability-distributions
edited Aug 20 '18 at 20:40
Michael Hardy
1
asked Feb 12 '18 at 13:55
Legend KillerLegend Killer
1,6061523
$endgroup$
$begingroup$
You're on the right track: don't get discouraged.
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– kimchi lover
Feb 12 '18 at 13:58
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Actually I can show each of the distributions $frac{XY}{sqrt{X^2+Y^2}}$ and the other one to be N(0,1/4) but I cannot show their independence
$endgroup$
– Legend Killer
Feb 12 '18 at 14:00
add a comment |
$begingroup$
It is given that $X,Y overset{text{i.i.d.}}{sim} N(0,1)$
Show that $frac{XY}{sqrt{X^2+Y^2}},frac{X^2-Y^2}{2sqrt{X^2+Y^2}} overset{text{i.i.d.}}{sim} N(0,frac{1}{4})$
I was thinking of making polar transformations $X=r cos theta, Y=r sin theta$
Then I am getting stuck at ranges of $theta$
probability probability-distributions
edited Aug 20 '18 at 20:40
Michael Hardy
1
asked Feb 12 '18 at 13:55
Legend KillerLegend Killer
1,6061523
$endgroup$
It is given that $X,Y overset{text{i.i.d.}}{sim} N(0,1)$
Show that $frac{XY}{sqrt{X^2+Y^2}},frac{X^2-Y^2}{2sqrt{X^2+Y^2}} overset{text{i.i.d.}}{sim} N(0,frac{1}{4})$
I was thinking of making polar transformations $X=r cos theta, Y=r sin theta$
Then I am getting stuck at ranges of $theta$
probability probability-distributions
probability probability-distributions
edited Aug 20 '18 at 20:40
Michael Hardy
1
asked Feb 12 '18 at 13:55
Legend KillerLegend Killer
1,6061523
edited Aug 20 '18 at 20:40
Michael Hardy
1
asked Feb 12 '18 at 13:55
Legend KillerLegend Killer
1,6061523
edited Aug 20 '18 at 20:40
Michael Hardy
1
edited Aug 20 '18 at 20:40
Michael Hardy
1
edited Aug 20 '18 at 20:40
Michael Hardy
1
1
asked Feb 12 '18 at 13:55
Legend KillerLegend Killer
1,6061523
asked Feb 12 '18 at 13:55
Legend KillerLegend Killer
1,6061523
asked Feb 12 '18 at 13:55
Legend KillerLegend Killer
1,6061523
1,6061523
$begingroup$
You're on the right track: don't get discouraged.
$endgroup$
– kimchi lover
Feb 12 '18 at 13:58
$begingroup$
Actually I can show each of the distributions $frac{XY}{sqrt{X^2+Y^2}}$ and the other one to be N(0,1/4) but I cannot show their independence
$endgroup$
– Legend Killer
Feb 12 '18 at 14:00
add a comment |
$begingroup$
You're on the right track: don't get discouraged.
$endgroup$
– kimchi lover
Feb 12 '18 at 13:58
$begingroup$
Actually I can show each of the distributions $frac{XY}{sqrt{X^2+Y^2}}$ and the other one to be N(0,1/4) but I cannot show their independence
$endgroup$
– Legend Killer
Feb 12 '18 at 14:00
$begingroup$
You're on the right track: don't get discouraged.
$endgroup$
– kimchi lover
Feb 12 '18 at 13:58
$begingroup$
You're on the right track: don't get discouraged.
$endgroup$
– kimchi lover
Feb 12 '18 at 13:58
$begingroup$
Actually I can show each of the distributions $frac{XY}{sqrt{X^2+Y^2}}$ and the other one to be N(0,1/4) but I cannot show their independence
$endgroup$
– Legend Killer
Feb 12 '18 at 14:00
$begingroup$
Actually I can show each of the distributions $frac{XY}{sqrt{X^2+Y^2}}$ and the other one to be N(0,1/4) but I cannot show their independence
$endgroup$
– Legend Killer
Feb 12 '18 at 14:00
add a comment |
3 Answers
3
active
oldest
votes
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If you transform $(X,Y)mapsto(R,Theta)$ where $X=RcosTheta,Y=RsinTheta$,
you should end up with the joint density of $(R,Theta)$ as $f_{R,Theta}(r,theta)=dfrac{r}{2pi}e^{-r^2/2}mathbf1_{{r>0,,0<theta<2pi}}$.
This implies $R$ and $Theta$ are independent, where $R$ has the Rayleigh distribution and $Thetasimmathcal{U}(0,2pi)$.
Now changing variables $(R,Theta)mapsto(U,V)$ such that $U=Rsin(2Theta),V=Rcos(2Theta)$,
you should be able to show that $U$ and $V$ are independent $mathcal{N}(0,1)$ variables.
Note that $U=dfrac{2XY}{sqrt{X^2+Y^2}}$ and $V=dfrac{X^2-Y^2}{sqrt{X^2+Y^2}}$ are independent, which in turn means that
$dfrac{U}{2}=dfrac{XY}{sqrt{X^2+Y^2}}$ and $dfrac{V}{2}=dfrac{X^2-Y^2}{2sqrt{X^2+Y^2}}$ are independent $mathcal{N}(0,1/4)$ variables.
This is independent of the post above:
Joint density of $(X,Y)$ is $displaystyle f_{X,Y}(x,y)=frac{1}{2pi}e^{-frac{1}{2}(x^2+y^2)},quad,(x,y)inmathbb{R^2}$
We transform $(X,Y)mapsto(R,Theta)mapsto(U,V)$ where
$x=rcostheta,,y=rsintheta$ and $u=frac{r}{2}sin(2theta),,v=frac{r}{2}cos(2theta)$
$(x,y)inmathbb{R^2}implies r>0,, 0<theta<2piimplies (u,v)inmathbb{R^2}$.
Note that this transformation is not one to one.
Jacobian of the transformation is $Jleft(frac{x,y}{u,v}right) = Jleft(frac{x,y}{r,theta}right)Jleft(frac{r,theta}{u,v}right)=J_1J_2$, say.
Also, $x^2+y^2=r^2=4(u^2+v^2)$ and $|J_1||J_2|=rtimesfrac{2}{r}=2$
Now $left(U=frac{XY}{sqrt{X^2+Y^2}},V=frac{X^2-Y^2}{2sqrt{X^2+Y^2}}right)$ has the preimages $(X,Y)$ and $(-X,-Y)$.
Moreover, $X,Ystackrel{text{i.i.d.}}{sim}mathcal{N}(0,1)iff -X,-Ystackrel{text{i.i.d.}}{sim}mathcal{N}(0,1)$.
Hence the joint density of $(U,V)$ is given by $$f_{U,V}(u,v)=f_{X,Y}(g_1(u,v),h_1(u,v))|J_1||J_2| + f_{X,Y}(g_2(u,v),h_2(u,v))|J_1||J_2|$$
$$=frac{1}{2pi}e^{-frac{1}{2} 4(u^2+v^2)}|J_1||J_2|times 2$$
$$=frac{1}{sqrt{frac{1}{4}}sqrt{2pi}}expleft(-frac{u^2}{2cdotfrac{1}{4}}right)cdotfrac{1}{sqrt{frac{1}{4}}sqrt{2pi}}expleft(-frac{v^2}{2cdotfrac{1}{4}}right)quad ,(u,v)inmathbb{R^2}$$
(We have the multiplier $2$ in the second step due to the preimages of $(x,y)$ namely $(g_i(u,v),h_i(u,v))$ for $i=1,2$, 'contributing' equally to the joint density).
This implies $U$ and $V$ are independent $mathcal{N}(0,1/4)$ variables.
edited Aug 20 '18 at 20:49
Michael Hardy
1
answered Feb 12 '18 at 14:18
StubbornAtomStubbornAtom
5,67411138
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Yes,I am facing difficulty in showing $U,V $ are N(0,1)
$endgroup$
– Legend Killer
Feb 12 '18 at 14:28
$begingroup$
Ok done! Good solution!
$endgroup$
– Legend Killer
Feb 12 '18 at 14:39
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What about the preimages (u, -v) and (-u,v)?
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– Legend Killer
Feb 24 '18 at 2:13
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@AyanBiswas Let me modify my answer to make it clear.
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– StubbornAtom
Feb 24 '18 at 6:36
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Hi @StubbornAtom. Thank you for this link. Can you please explain as to why the transformation (x,y) -> (u,v) is not one-to-one ? Also, we are mapping from X and Y. Why have you considered -X and -Y here ?
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– Dwaipayan Gupta
Mar 23 '18 at 19:24
add a comment |
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Setting $U=dfrac{XY}{sqrt{X^2+Y^2}}, V=dfrac{X^2-Y^2}{2sqrt{X^2+Y^2}}$ and using moment generating functions:
begin{align}
M_{(U,V)}(u,v) & =mathbb{E}left[mathrm{e}^{langle,(u,v),;,(U,V),rangle}right] \[10pt]
& =iint_{mathbb{R}^2}expleft(ufrac{xy}{sqrt{x^2+y^2}}+vfrac{x^2-y^2}{2sqrt{x^2+y^2}}right)cdot f_{(X,Y)}(x,y),mathrm{d}x,mathrm{d}y.
end{align}
You can see here that $$M_{(U,V)}(u,v)=expleft(frac18(u^2+v^2)right)=expleft(frac12(u,v)
begin{pmatrix}1/4 & 0\ 0 & 1/4end{pmatrix}
(u,v)^Tright)$$
so that $(U,V)$ is normal multivariate with $mathbf{0}$ mean and diagonal covariance matrix, meaning $U, V$ are independant $mathcal{N}(0,1/4)$.
edited Aug 20 '18 at 20:43
Michael Hardy
1
answered Feb 24 '18 at 21:37
dequedeque
413111
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If you have already proved $frac{X Y}{sqrt{X^2 + Y^2}} $ and $frac{X^2 - Y^2}{sqrt{X^2 + Y^2}}$ are gaussian, as long as that pair of them is jointly gaussian, then you may use their property: $u,, v text{ independent} iff operatorname{cov}(u,v) = 0$.
$$ operatorname{cov} left( frac{X Y}{sqrt{X^2 + Y^2}},frac{X^2 - Y^2}{sqrt{X^2 + Y^2}} right)$$
$$ = operatorname{cov} left( frac{X Y}{sqrt{X^2 + Y^2}},frac{X^2}{sqrt{X^2 + Y^2}} right) - operatorname{cov} left( frac{X Y}{sqrt{X^2 + Y^2}},frac{Y^2}{sqrt{X^2 + Y^2}} right)$$
Now make use of the symmetry of the expressions we have:
$$ operatorname{cov}(cdots) = 0 $$
To make this more rigorous one may rewrite the covariance as follows:
$$ operatorname{cov} left( frac{X Y}{sqrt{X^2 + Y^2}},frac{X^2}{sqrt{X^2 + Y^2}} right) - operatorname{cov} left( frac{X Y}{sqrt{X^2 + Y^2}},frac{Y^2}{sqrt{X^2 + Y^2}} right) $$
$$ = E left( frac{X^3 Y}{X^2 + Y^2} right) - E left( frac{X Y^3}{X^2 + Y^2} right) $$
Now renaming $X to Y, , Y to X$ under the first expectation sign (note that this is the same as renaming variables under integral) we get the result.
Covariance method can be carried even further, following the last line and the trick $ X = -X $ ($X$ is symmetric distribution) which we apply to the second $E$:
$$ = E left( frac{X^3 Y}{X^2 + Y^2} right) - E left( frac{X Y^3}{X^2 + Y^2} right) = E left( frac{X^3 Y}{X^2 + Y^2} right) + E left( frac{X Y^3}{X^2 + Y^2} right) $$
$$ = E left( frac{X Y(X^2 + Y^2)}{X^2 + Y^2} right) = E(XY) = 0$$
The last equation is due to independence of $X, Y$ $implies 0=operatorname{cov}(X,Y)=E(XY) - E(X)E(Y)=E(XY).$
answered Feb 24 '18 at 8:47
Al PrihodkoAl Prihodko
436
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3
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There is an error in the first sentence here. Being Gaussian and having zero covariance does NOT imply independnce. It does imply independence if they are JOINTLY Gaussian. But you haven't mentioned that.
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– Michael Hardy
Aug 20 '18 at 20:43
add a comment |
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3 Answers
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3 Answers
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$begingroup$
If you transform $(X,Y)mapsto(R,Theta)$ where $X=RcosTheta,Y=RsinTheta$,
you should end up with the joint density of $(R,Theta)$ as $f_{R,Theta}(r,theta)=dfrac{r}{2pi}e^{-r^2/2}mathbf1_{{r>0,,0<theta<2pi}}$.
This implies $R$ and $Theta$ are independent, where $R$ has the Rayleigh distribution and $Thetasimmathcal{U}(0,2pi)$.
Now changing variables $(R,Theta)mapsto(U,V)$ such that $U=Rsin(2Theta),V=Rcos(2Theta)$,
you should be able to show that $U$ and $V$ are independent $mathcal{N}(0,1)$ variables.
Note that $U=dfrac{2XY}{sqrt{X^2+Y^2}}$ and $V=dfrac{X^2-Y^2}{sqrt{X^2+Y^2}}$ are independent, which in turn means that
$dfrac{U}{2}=dfrac{XY}{sqrt{X^2+Y^2}}$ and $dfrac{V}{2}=dfrac{X^2-Y^2}{2sqrt{X^2+Y^2}}$ are independent $mathcal{N}(0,1/4)$ variables.
This is independent of the post above:
Joint density of $(X,Y)$ is $displaystyle f_{X,Y}(x,y)=frac{1}{2pi}e^{-frac{1}{2}(x^2+y^2)},quad,(x,y)inmathbb{R^2}$
We transform $(X,Y)mapsto(R,Theta)mapsto(U,V)$ where
$x=rcostheta,,y=rsintheta$ and $u=frac{r}{2}sin(2theta),,v=frac{r}{2}cos(2theta)$
$(x,y)inmathbb{R^2}implies r>0,, 0<theta<2piimplies (u,v)inmathbb{R^2}$.
Note that this transformation is not one to one.
Jacobian of the transformation is $Jleft(frac{x,y}{u,v}right) = Jleft(frac{x,y}{r,theta}right)Jleft(frac{r,theta}{u,v}right)=J_1J_2$, say.
Also, $x^2+y^2=r^2=4(u^2+v^2)$ and $|J_1||J_2|=rtimesfrac{2}{r}=2$
Now $left(U=frac{XY}{sqrt{X^2+Y^2}},V=frac{X^2-Y^2}{2sqrt{X^2+Y^2}}right)$ has the preimages $(X,Y)$ and $(-X,-Y)$.
Moreover, $X,Ystackrel{text{i.i.d.}}{sim}mathcal{N}(0,1)iff -X,-Ystackrel{text{i.i.d.}}{sim}mathcal{N}(0,1)$.
Hence the joint density of $(U,V)$ is given by $$f_{U,V}(u,v)=f_{X,Y}(g_1(u,v),h_1(u,v))|J_1||J_2| + f_{X,Y}(g_2(u,v),h_2(u,v))|J_1||J_2|$$
$$=frac{1}{2pi}e^{-frac{1}{2} 4(u^2+v^2)}|J_1||J_2|times 2$$
$$=frac{1}{sqrt{frac{1}{4}}sqrt{2pi}}expleft(-frac{u^2}{2cdotfrac{1}{4}}right)cdotfrac{1}{sqrt{frac{1}{4}}sqrt{2pi}}expleft(-frac{v^2}{2cdotfrac{1}{4}}right)quad ,(u,v)inmathbb{R^2}$$
(We have the multiplier $2$ in the second step due to the preimages of $(x,y)$ namely $(g_i(u,v),h_i(u,v))$ for $i=1,2$, 'contributing' equally to the joint density).
This implies $U$ and $V$ are independent $mathcal{N}(0,1/4)$ variables.
edited Aug 20 '18 at 20:49
Michael Hardy
1
answered Feb 12 '18 at 14:18
StubbornAtomStubbornAtom
5,67411138
$endgroup$
$begingroup$
Yes,I am facing difficulty in showing $U,V $ are N(0,1)
$endgroup$
– Legend Killer
Feb 12 '18 at 14:28
$begingroup$
Ok done! Good solution!
$endgroup$
– Legend Killer
Feb 12 '18 at 14:39
$begingroup$
What about the preimages (u, -v) and (-u,v)?
$endgroup$
– Legend Killer
Feb 24 '18 at 2:13
$begingroup$
@AyanBiswas Let me modify my answer to make it clear.
$endgroup$
– StubbornAtom
Feb 24 '18 at 6:36
$begingroup$
Hi @StubbornAtom. Thank you for this link. Can you please explain as to why the transformation (x,y) -> (u,v) is not one-to-one ? Also, we are mapping from X and Y. Why have you considered -X and -Y here ?
$endgroup$
– Dwaipayan Gupta
Mar 23 '18 at 19:24
add a comment |
$begingroup$
If you transform $(X,Y)mapsto(R,Theta)$ where $X=RcosTheta,Y=RsinTheta$,
you should end up with the joint density of $(R,Theta)$ as $f_{R,Theta}(r,theta)=dfrac{r}{2pi}e^{-r^2/2}mathbf1_{{r>0,,0<theta<2pi}}$.
This implies $R$ and $Theta$ are independent, where $R$ has the Rayleigh distribution and $Thetasimmathcal{U}(0,2pi)$.
Now changing variables $(R,Theta)mapsto(U,V)$ such that $U=Rsin(2Theta),V=Rcos(2Theta)$,
you should be able to show that $U$ and $V$ are independent $mathcal{N}(0,1)$ variables.
Note that $U=dfrac{2XY}{sqrt{X^2+Y^2}}$ and $V=dfrac{X^2-Y^2}{sqrt{X^2+Y^2}}$ are independent, which in turn means that
$dfrac{U}{2}=dfrac{XY}{sqrt{X^2+Y^2}}$ and $dfrac{V}{2}=dfrac{X^2-Y^2}{2sqrt{X^2+Y^2}}$ are independent $mathcal{N}(0,1/4)$ variables.
This is independent of the post above:
Joint density of $(X,Y)$ is $displaystyle f_{X,Y}(x,y)=frac{1}{2pi}e^{-frac{1}{2}(x^2+y^2)},quad,(x,y)inmathbb{R^2}$
We transform $(X,Y)mapsto(R,Theta)mapsto(U,V)$ where
$x=rcostheta,,y=rsintheta$ and $u=frac{r}{2}sin(2theta),,v=frac{r}{2}cos(2theta)$
$(x,y)inmathbb{R^2}implies r>0,, 0<theta<2piimplies (u,v)inmathbb{R^2}$.
Note that this transformation is not one to one.
Jacobian of the transformation is $Jleft(frac{x,y}{u,v}right) = Jleft(frac{x,y}{r,theta}right)Jleft(frac{r,theta}{u,v}right)=J_1J_2$, say.
Also, $x^2+y^2=r^2=4(u^2+v^2)$ and $|J_1||J_2|=rtimesfrac{2}{r}=2$
Now $left(U=frac{XY}{sqrt{X^2+Y^2}},V=frac{X^2-Y^2}{2sqrt{X^2+Y^2}}right)$ has the preimages $(X,Y)$ and $(-X,-Y)$.
Moreover, $X,Ystackrel{text{i.i.d.}}{sim}mathcal{N}(0,1)iff -X,-Ystackrel{text{i.i.d.}}{sim}mathcal{N}(0,1)$.
Hence the joint density of $(U,V)$ is given by $$f_{U,V}(u,v)=f_{X,Y}(g_1(u,v),h_1(u,v))|J_1||J_2| + f_{X,Y}(g_2(u,v),h_2(u,v))|J_1||J_2|$$
$$=frac{1}{2pi}e^{-frac{1}{2} 4(u^2+v^2)}|J_1||J_2|times 2$$
$$=frac{1}{sqrt{frac{1}{4}}sqrt{2pi}}expleft(-frac{u^2}{2cdotfrac{1}{4}}right)cdotfrac{1}{sqrt{frac{1}{4}}sqrt{2pi}}expleft(-frac{v^2}{2cdotfrac{1}{4}}right)quad ,(u,v)inmathbb{R^2}$$
(We have the multiplier $2$ in the second step due to the preimages of $(x,y)$ namely $(g_i(u,v),h_i(u,v))$ for $i=1,2$, 'contributing' equally to the joint density).
This implies $U$ and $V$ are independent $mathcal{N}(0,1/4)$ variables.
edited Aug 20 '18 at 20:49
Michael Hardy
1
answered Feb 12 '18 at 14:18
StubbornAtomStubbornAtom
5,67411138
$endgroup$
$begingroup$
Yes,I am facing difficulty in showing $U,V $ are N(0,1)
$endgroup$
– Legend Killer
Feb 12 '18 at 14:28
$begingroup$
Ok done! Good solution!
$endgroup$
– Legend Killer
Feb 12 '18 at 14:39
$begingroup$
What about the preimages (u, -v) and (-u,v)?
$endgroup$
– Legend Killer
Feb 24 '18 at 2:13
$begingroup$
@AyanBiswas Let me modify my answer to make it clear.
$endgroup$
– StubbornAtom
Feb 24 '18 at 6:36
$begingroup$
Hi @StubbornAtom. Thank you for this link. Can you please explain as to why the transformation (x,y) -> (u,v) is not one-to-one ? Also, we are mapping from X and Y. Why have you considered -X and -Y here ?
$endgroup$
– Dwaipayan Gupta
Mar 23 '18 at 19:24
add a comment |
$begingroup$
If you transform $(X,Y)mapsto(R,Theta)$ where $X=RcosTheta,Y=RsinTheta$,
you should end up with the joint density of $(R,Theta)$ as $f_{R,Theta}(r,theta)=dfrac{r}{2pi}e^{-r^2/2}mathbf1_{{r>0,,0<theta<2pi}}$.
This implies $R$ and $Theta$ are independent, where $R$ has the Rayleigh distribution and $Thetasimmathcal{U}(0,2pi)$.
Now changing variables $(R,Theta)mapsto(U,V)$ such that $U=Rsin(2Theta),V=Rcos(2Theta)$,
you should be able to show that $U$ and $V$ are independent $mathcal{N}(0,1)$ variables.
Note that $U=dfrac{2XY}{sqrt{X^2+Y^2}}$ and $V=dfrac{X^2-Y^2}{sqrt{X^2+Y^2}}$ are independent, which in turn means that
$dfrac{U}{2}=dfrac{XY}{sqrt{X^2+Y^2}}$ and $dfrac{V}{2}=dfrac{X^2-Y^2}{2sqrt{X^2+Y^2}}$ are independent $mathcal{N}(0,1/4)$ variables.
This is independent of the post above:
Joint density of $(X,Y)$ is $displaystyle f_{X,Y}(x,y)=frac{1}{2pi}e^{-frac{1}{2}(x^2+y^2)},quad,(x,y)inmathbb{R^2}$
We transform $(X,Y)mapsto(R,Theta)mapsto(U,V)$ where
$x=rcostheta,,y=rsintheta$ and $u=frac{r}{2}sin(2theta),,v=frac{r}{2}cos(2theta)$
$(x,y)inmathbb{R^2}implies r>0,, 0<theta<2piimplies (u,v)inmathbb{R^2}$.
Note that this transformation is not one to one.
Jacobian of the transformation is $Jleft(frac{x,y}{u,v}right) = Jleft(frac{x,y}{r,theta}right)Jleft(frac{r,theta}{u,v}right)=J_1J_2$, say.
Also, $x^2+y^2=r^2=4(u^2+v^2)$ and $|J_1||J_2|=rtimesfrac{2}{r}=2$
Now $left(U=frac{XY}{sqrt{X^2+Y^2}},V=frac{X^2-Y^2}{2sqrt{X^2+Y^2}}right)$ has the preimages $(X,Y)$ and $(-X,-Y)$.
Moreover, $X,Ystackrel{text{i.i.d.}}{sim}mathcal{N}(0,1)iff -X,-Ystackrel{text{i.i.d.}}{sim}mathcal{N}(0,1)$.
Hence the joint density of $(U,V)$ is given by $$f_{U,V}(u,v)=f_{X,Y}(g_1(u,v),h_1(u,v))|J_1||J_2| + f_{X,Y}(g_2(u,v),h_2(u,v))|J_1||J_2|$$
$$=frac{1}{2pi}e^{-frac{1}{2} 4(u^2+v^2)}|J_1||J_2|times 2$$
$$=frac{1}{sqrt{frac{1}{4}}sqrt{2pi}}expleft(-frac{u^2}{2cdotfrac{1}{4}}right)cdotfrac{1}{sqrt{frac{1}{4}}sqrt{2pi}}expleft(-frac{v^2}{2cdotfrac{1}{4}}right)quad ,(u,v)inmathbb{R^2}$$
(We have the multiplier $2$ in the second step due to the preimages of $(x,y)$ namely $(g_i(u,v),h_i(u,v))$ for $i=1,2$, 'contributing' equally to the joint density).
This implies $U$ and $V$ are independent $mathcal{N}(0,1/4)$ variables.
edited Aug 20 '18 at 20:49
Michael Hardy
1
answered Feb 12 '18 at 14:18
StubbornAtomStubbornAtom
5,67411138
$endgroup$
If you transform $(X,Y)mapsto(R,Theta)$ where $X=RcosTheta,Y=RsinTheta$,
you should end up with the joint density of $(R,Theta)$ as $f_{R,Theta}(r,theta)=dfrac{r}{2pi}e^{-r^2/2}mathbf1_{{r>0,,0<theta<2pi}}$.
This implies $R$ and $Theta$ are independent, where $R$ has the Rayleigh distribution and $Thetasimmathcal{U}(0,2pi)$.
Now changing variables $(R,Theta)mapsto(U,V)$ such that $U=Rsin(2Theta),V=Rcos(2Theta)$,
you should be able to show that $U$ and $V$ are independent $mathcal{N}(0,1)$ variables.
Note that $U=dfrac{2XY}{sqrt{X^2+Y^2}}$ and $V=dfrac{X^2-Y^2}{sqrt{X^2+Y^2}}$ are independent, which in turn means that
$dfrac{U}{2}=dfrac{XY}{sqrt{X^2+Y^2}}$ and $dfrac{V}{2}=dfrac{X^2-Y^2}{2sqrt{X^2+Y^2}}$ are independent $mathcal{N}(0,1/4)$ variables.
This is independent of the post above:
Joint density of $(X,Y)$ is $displaystyle f_{X,Y}(x,y)=frac{1}{2pi}e^{-frac{1}{2}(x^2+y^2)},quad,(x,y)inmathbb{R^2}$
We transform $(X,Y)mapsto(R,Theta)mapsto(U,V)$ where
$x=rcostheta,,y=rsintheta$ and $u=frac{r}{2}sin(2theta),,v=frac{r}{2}cos(2theta)$
$(x,y)inmathbb{R^2}implies r>0,, 0<theta<2piimplies (u,v)inmathbb{R^2}$.
Note that this transformation is not one to one.
Jacobian of the transformation is $Jleft(frac{x,y}{u,v}right) = Jleft(frac{x,y}{r,theta}right)Jleft(frac{r,theta}{u,v}right)=J_1J_2$, say.
Also, $x^2+y^2=r^2=4(u^2+v^2)$ and $|J_1||J_2|=rtimesfrac{2}{r}=2$
Now $left(U=frac{XY}{sqrt{X^2+Y^2}},V=frac{X^2-Y^2}{2sqrt{X^2+Y^2}}right)$ has the preimages $(X,Y)$ and $(-X,-Y)$.
Moreover, $X,Ystackrel{text{i.i.d.}}{sim}mathcal{N}(0,1)iff -X,-Ystackrel{text{i.i.d.}}{sim}mathcal{N}(0,1)$.
Hence the joint density of $(U,V)$ is given by $$f_{U,V}(u,v)=f_{X,Y}(g_1(u,v),h_1(u,v))|J_1||J_2| + f_{X,Y}(g_2(u,v),h_2(u,v))|J_1||J_2|$$
$$=frac{1}{2pi}e^{-frac{1}{2} 4(u^2+v^2)}|J_1||J_2|times 2$$
$$=frac{1}{sqrt{frac{1}{4}}sqrt{2pi}}expleft(-frac{u^2}{2cdotfrac{1}{4}}right)cdotfrac{1}{sqrt{frac{1}{4}}sqrt{2pi}}expleft(-frac{v^2}{2cdotfrac{1}{4}}right)quad ,(u,v)inmathbb{R^2}$$
(We have the multiplier $2$ in the second step due to the preimages of $(x,y)$ namely $(g_i(u,v),h_i(u,v))$ for $i=1,2$, 'contributing' equally to the joint density).
This implies $U$ and $V$ are independent $mathcal{N}(0,1/4)$ variables.
edited Aug 20 '18 at 20:49
Michael Hardy
1
answered Feb 12 '18 at 14:18
StubbornAtomStubbornAtom
5,67411138
edited Aug 20 '18 at 20:49
Michael Hardy
1
edited Aug 20 '18 at 20:49
Michael Hardy
1
edited Aug 20 '18 at 20:49
Michael Hardy
1
1
answered Feb 12 '18 at 14:18
StubbornAtomStubbornAtom
5,67411138
answered Feb 12 '18 at 14:18
StubbornAtomStubbornAtom
5,67411138
answered Feb 12 '18 at 14:18
StubbornAtomStubbornAtom
5,67411138
5,67411138
$begingroup$
Yes,I am facing difficulty in showing $U,V $ are N(0,1)
$endgroup$
– Legend Killer
Feb 12 '18 at 14:28
$begingroup$
Ok done! Good solution!
$endgroup$
– Legend Killer
Feb 12 '18 at 14:39
$begingroup$
What about the preimages (u, -v) and (-u,v)?
$endgroup$
– Legend Killer
Feb 24 '18 at 2:13
$begingroup$
@AyanBiswas Let me modify my answer to make it clear.
$endgroup$
– StubbornAtom
Feb 24 '18 at 6:36
$begingroup$
Hi @StubbornAtom. Thank you for this link. Can you please explain as to why the transformation (x,y) -> (u,v) is not one-to-one ? Also, we are mapping from X and Y. Why have you considered -X and -Y here ?
$endgroup$
– Dwaipayan Gupta
Mar 23 '18 at 19:24
add a comment |
$begingroup$
Yes,I am facing difficulty in showing $U,V $ are N(0,1)
$endgroup$
– Legend Killer
Feb 12 '18 at 14:28
$begingroup$
Ok done! Good solution!
$endgroup$
– Legend Killer
Feb 12 '18 at 14:39
$begingroup$
What about the preimages (u, -v) and (-u,v)?
$endgroup$
– Legend Killer
Feb 24 '18 at 2:13
$begingroup$
@AyanBiswas Let me modify my answer to make it clear.
$endgroup$
– StubbornAtom
Feb 24 '18 at 6:36
$begingroup$
Hi @StubbornAtom. Thank you for this link. Can you please explain as to why the transformation (x,y) -> (u,v) is not one-to-one ? Also, we are mapping from X and Y. Why have you considered -X and -Y here ?
$endgroup$
– Dwaipayan Gupta
Mar 23 '18 at 19:24
$begingroup$
Yes,I am facing difficulty in showing $U,V $ are N(0,1)
$endgroup$
– Legend Killer
Feb 12 '18 at 14:28
$begingroup$
Yes,I am facing difficulty in showing $U,V $ are N(0,1)
$endgroup$
– Legend Killer
Feb 12 '18 at 14:28
$begingroup$
Ok done! Good solution!
$endgroup$
– Legend Killer
Feb 12 '18 at 14:39
$begingroup$
Ok done! Good solution!
$endgroup$
– Legend Killer
Feb 12 '18 at 14:39
$begingroup$
What about the preimages (u, -v) and (-u,v)?
$endgroup$
– Legend Killer
Feb 24 '18 at 2:13
$begingroup$
What about the preimages (u, -v) and (-u,v)?
$endgroup$
– Legend Killer
Feb 24 '18 at 2:13
$begingroup$
@AyanBiswas Let me modify my answer to make it clear.
$endgroup$
– StubbornAtom
Feb 24 '18 at 6:36
$begingroup$
@AyanBiswas Let me modify my answer to make it clear.
$endgroup$
– StubbornAtom
Feb 24 '18 at 6:36
$begingroup$
Hi @StubbornAtom. Thank you for this link. Can you please explain as to why the transformation (x,y) -> (u,v) is not one-to-one ? Also, we are mapping from X and Y. Why have you considered -X and -Y here ?
$endgroup$
– Dwaipayan Gupta
Mar 23 '18 at 19:24
$begingroup$
Hi @StubbornAtom. Thank you for this link. Can you please explain as to why the transformation (x,y) -> (u,v) is not one-to-one ? Also, we are mapping from X and Y. Why have you considered -X and -Y here ?
$endgroup$
– Dwaipayan Gupta
Mar 23 '18 at 19:24
add a comment |
$begingroup$
Setting $U=dfrac{XY}{sqrt{X^2+Y^2}}, V=dfrac{X^2-Y^2}{2sqrt{X^2+Y^2}}$ and using moment generating functions:
begin{align}
M_{(U,V)}(u,v) & =mathbb{E}left[mathrm{e}^{langle,(u,v),;,(U,V),rangle}right] \[10pt]
& =iint_{mathbb{R}^2}expleft(ufrac{xy}{sqrt{x^2+y^2}}+vfrac{x^2-y^2}{2sqrt{x^2+y^2}}right)cdot f_{(X,Y)}(x,y),mathrm{d}x,mathrm{d}y.
end{align}
You can see here that $$M_{(U,V)}(u,v)=expleft(frac18(u^2+v^2)right)=expleft(frac12(u,v)
begin{pmatrix}1/4 & 0\ 0 & 1/4end{pmatrix}
(u,v)^Tright)$$
so that $(U,V)$ is normal multivariate with $mathbf{0}$ mean and diagonal covariance matrix, meaning $U, V$ are independant $mathcal{N}(0,1/4)$.
edited Aug 20 '18 at 20:43
Michael Hardy
1
answered Feb 24 '18 at 21:37
dequedeque
413111
$endgroup$
add a comment |
$begingroup$
Setting $U=dfrac{XY}{sqrt{X^2+Y^2}}, V=dfrac{X^2-Y^2}{2sqrt{X^2+Y^2}}$ and using moment generating functions:
begin{align}
M_{(U,V)}(u,v) & =mathbb{E}left[mathrm{e}^{langle,(u,v),;,(U,V),rangle}right] \[10pt]
& =iint_{mathbb{R}^2}expleft(ufrac{xy}{sqrt{x^2+y^2}}+vfrac{x^2-y^2}{2sqrt{x^2+y^2}}right)cdot f_{(X,Y)}(x,y),mathrm{d}x,mathrm{d}y.
end{align}
You can see here that $$M_{(U,V)}(u,v)=expleft(frac18(u^2+v^2)right)=expleft(frac12(u,v)
begin{pmatrix}1/4 & 0\ 0 & 1/4end{pmatrix}
(u,v)^Tright)$$
so that $(U,V)$ is normal multivariate with $mathbf{0}$ mean and diagonal covariance matrix, meaning $U, V$ are independant $mathcal{N}(0,1/4)$.
edited Aug 20 '18 at 20:43
Michael Hardy
1
answered Feb 24 '18 at 21:37
dequedeque
413111
$endgroup$
add a comment |
$begingroup$
Setting $U=dfrac{XY}{sqrt{X^2+Y^2}}, V=dfrac{X^2-Y^2}{2sqrt{X^2+Y^2}}$ and using moment generating functions:
begin{align}
M_{(U,V)}(u,v) & =mathbb{E}left[mathrm{e}^{langle,(u,v),;,(U,V),rangle}right] \[10pt]
& =iint_{mathbb{R}^2}expleft(ufrac{xy}{sqrt{x^2+y^2}}+vfrac{x^2-y^2}{2sqrt{x^2+y^2}}right)cdot f_{(X,Y)}(x,y),mathrm{d}x,mathrm{d}y.
end{align}
You can see here that $$M_{(U,V)}(u,v)=expleft(frac18(u^2+v^2)right)=expleft(frac12(u,v)
begin{pmatrix}1/4 & 0\ 0 & 1/4end{pmatrix}
(u,v)^Tright)$$
so that $(U,V)$ is normal multivariate with $mathbf{0}$ mean and diagonal covariance matrix, meaning $U, V$ are independant $mathcal{N}(0,1/4)$.
edited Aug 20 '18 at 20:43
Michael Hardy
1
answered Feb 24 '18 at 21:37
dequedeque
413111
$endgroup$
Setting $U=dfrac{XY}{sqrt{X^2+Y^2}}, V=dfrac{X^2-Y^2}{2sqrt{X^2+Y^2}}$ and using moment generating functions:
begin{align}
M_{(U,V)}(u,v) & =mathbb{E}left[mathrm{e}^{langle,(u,v),;,(U,V),rangle}right] \[10pt]
& =iint_{mathbb{R}^2}expleft(ufrac{xy}{sqrt{x^2+y^2}}+vfrac{x^2-y^2}{2sqrt{x^2+y^2}}right)cdot f_{(X,Y)}(x,y),mathrm{d}x,mathrm{d}y.
end{align}
You can see here that $$M_{(U,V)}(u,v)=expleft(frac18(u^2+v^2)right)=expleft(frac12(u,v)
begin{pmatrix}1/4 & 0\ 0 & 1/4end{pmatrix}
(u,v)^Tright)$$
so that $(U,V)$ is normal multivariate with $mathbf{0}$ mean and diagonal covariance matrix, meaning $U, V$ are independant $mathcal{N}(0,1/4)$.
edited Aug 20 '18 at 20:43
Michael Hardy
1
answered Feb 24 '18 at 21:37
dequedeque
413111
edited Aug 20 '18 at 20:43
Michael Hardy
1
edited Aug 20 '18 at 20:43
Michael Hardy
1
edited Aug 20 '18 at 20:43
Michael Hardy
1
1
answered Feb 24 '18 at 21:37
dequedeque
413111
answered Feb 24 '18 at 21:37
dequedeque
413111
answered Feb 24 '18 at 21:37
dequedeque
413111
413111
add a comment |
add a comment |
$begingroup$
If you have already proved $frac{X Y}{sqrt{X^2 + Y^2}} $ and $frac{X^2 - Y^2}{sqrt{X^2 + Y^2}}$ are gaussian, as long as that pair of them is jointly gaussian, then you may use their property: $u,, v text{ independent} iff operatorname{cov}(u,v) = 0$.
$$ operatorname{cov} left( frac{X Y}{sqrt{X^2 + Y^2}},frac{X^2 - Y^2}{sqrt{X^2 + Y^2}} right)$$
$$ = operatorname{cov} left( frac{X Y}{sqrt{X^2 + Y^2}},frac{X^2}{sqrt{X^2 + Y^2}} right) - operatorname{cov} left( frac{X Y}{sqrt{X^2 + Y^2}},frac{Y^2}{sqrt{X^2 + Y^2}} right)$$
Now make use of the symmetry of the expressions we have:
$$ operatorname{cov}(cdots) = 0 $$
To make this more rigorous one may rewrite the covariance as follows:
$$ operatorname{cov} left( frac{X Y}{sqrt{X^2 + Y^2}},frac{X^2}{sqrt{X^2 + Y^2}} right) - operatorname{cov} left( frac{X Y}{sqrt{X^2 + Y^2}},frac{Y^2}{sqrt{X^2 + Y^2}} right) $$
$$ = E left( frac{X^3 Y}{X^2 + Y^2} right) - E left( frac{X Y^3}{X^2 + Y^2} right) $$
Now renaming $X to Y, , Y to X$ under the first expectation sign (note that this is the same as renaming variables under integral) we get the result.
Covariance method can be carried even further, following the last line and the trick $ X = -X $ ($X$ is symmetric distribution) which we apply to the second $E$:
$$ = E left( frac{X^3 Y}{X^2 + Y^2} right) - E left( frac{X Y^3}{X^2 + Y^2} right) = E left( frac{X^3 Y}{X^2 + Y^2} right) + E left( frac{X Y^3}{X^2 + Y^2} right) $$
$$ = E left( frac{X Y(X^2 + Y^2)}{X^2 + Y^2} right) = E(XY) = 0$$
The last equation is due to independence of $X, Y$ $implies 0=operatorname{cov}(X,Y)=E(XY) - E(X)E(Y)=E(XY).$
answered Feb 24 '18 at 8:47
Al PrihodkoAl Prihodko
436
$endgroup$
3
$begingroup$
There is an error in the first sentence here. Being Gaussian and having zero covariance does NOT imply independnce. It does imply independence if they are JOINTLY Gaussian. But you haven't mentioned that.
$endgroup$
– Michael Hardy
Aug 20 '18 at 20:43
add a comment |
$begingroup$
If you have already proved $frac{X Y}{sqrt{X^2 + Y^2}} $ and $frac{X^2 - Y^2}{sqrt{X^2 + Y^2}}$ are gaussian, as long as that pair of them is jointly gaussian, then you may use their property: $u,, v text{ independent} iff operatorname{cov}(u,v) = 0$.
$$ operatorname{cov} left( frac{X Y}{sqrt{X^2 + Y^2}},frac{X^2 - Y^2}{sqrt{X^2 + Y^2}} right)$$
$$ = operatorname{cov} left( frac{X Y}{sqrt{X^2 + Y^2}},frac{X^2}{sqrt{X^2 + Y^2}} right) - operatorname{cov} left( frac{X Y}{sqrt{X^2 + Y^2}},frac{Y^2}{sqrt{X^2 + Y^2}} right)$$
Now make use of the symmetry of the expressions we have:
$$ operatorname{cov}(cdots) = 0 $$
To make this more rigorous one may rewrite the covariance as follows:
$$ operatorname{cov} left( frac{X Y}{sqrt{X^2 + Y^2}},frac{X^2}{sqrt{X^2 + Y^2}} right) - operatorname{cov} left( frac{X Y}{sqrt{X^2 + Y^2}},frac{Y^2}{sqrt{X^2 + Y^2}} right) $$
$$ = E left( frac{X^3 Y}{X^2 + Y^2} right) - E left( frac{X Y^3}{X^2 + Y^2} right) $$
Now renaming $X to Y, , Y to X$ under the first expectation sign (note that this is the same as renaming variables under integral) we get the result.
Covariance method can be carried even further, following the last line and the trick $ X = -X $ ($X$ is symmetric distribution) which we apply to the second $E$:
$$ = E left( frac{X^3 Y}{X^2 + Y^2} right) - E left( frac{X Y^3}{X^2 + Y^2} right) = E left( frac{X^3 Y}{X^2 + Y^2} right) + E left( frac{X Y^3}{X^2 + Y^2} right) $$
$$ = E left( frac{X Y(X^2 + Y^2)}{X^2 + Y^2} right) = E(XY) = 0$$
The last equation is due to independence of $X, Y$ $implies 0=operatorname{cov}(X,Y)=E(XY) - E(X)E(Y)=E(XY).$
answered Feb 24 '18 at 8:47
Al PrihodkoAl Prihodko
436
$endgroup$
3
$begingroup$
There is an error in the first sentence here. Being Gaussian and having zero covariance does NOT imply independnce. It does imply independence if they are JOINTLY Gaussian. But you haven't mentioned that.
$endgroup$
– Michael Hardy
Aug 20 '18 at 20:43
add a comment |
$begingroup$
If you have already proved $frac{X Y}{sqrt{X^2 + Y^2}} $ and $frac{X^2 - Y^2}{sqrt{X^2 + Y^2}}$ are gaussian, as long as that pair of them is jointly gaussian, then you may use their property: $u,, v text{ independent} iff operatorname{cov}(u,v) = 0$.
$$ operatorname{cov} left( frac{X Y}{sqrt{X^2 + Y^2}},frac{X^2 - Y^2}{sqrt{X^2 + Y^2}} right)$$
$$ = operatorname{cov} left( frac{X Y}{sqrt{X^2 + Y^2}},frac{X^2}{sqrt{X^2 + Y^2}} right) - operatorname{cov} left( frac{X Y}{sqrt{X^2 + Y^2}},frac{Y^2}{sqrt{X^2 + Y^2}} right)$$
Now make use of the symmetry of the expressions we have:
$$ operatorname{cov}(cdots) = 0 $$
To make this more rigorous one may rewrite the covariance as follows:
$$ operatorname{cov} left( frac{X Y}{sqrt{X^2 + Y^2}},frac{X^2}{sqrt{X^2 + Y^2}} right) - operatorname{cov} left( frac{X Y}{sqrt{X^2 + Y^2}},frac{Y^2}{sqrt{X^2 + Y^2}} right) $$
$$ = E left( frac{X^3 Y}{X^2 + Y^2} right) - E left( frac{X Y^3}{X^2 + Y^2} right) $$
Now renaming $X to Y, , Y to X$ under the first expectation sign (note that this is the same as renaming variables under integral) we get the result.
Covariance method can be carried even further, following the last line and the trick $ X = -X $ ($X$ is symmetric distribution) which we apply to the second $E$:
$$ = E left( frac{X^3 Y}{X^2 + Y^2} right) - E left( frac{X Y^3}{X^2 + Y^2} right) = E left( frac{X^3 Y}{X^2 + Y^2} right) + E left( frac{X Y^3}{X^2 + Y^2} right) $$
$$ = E left( frac{X Y(X^2 + Y^2)}{X^2 + Y^2} right) = E(XY) = 0$$
The last equation is due to independence of $X, Y$ $implies 0=operatorname{cov}(X,Y)=E(XY) - E(X)E(Y)=E(XY).$
answered Feb 24 '18 at 8:47
Al PrihodkoAl Prihodko
436
$endgroup$
If you have already proved $frac{X Y}{sqrt{X^2 + Y^2}} $ and $frac{X^2 - Y^2}{sqrt{X^2 + Y^2}}$ are gaussian, as long as that pair of them is jointly gaussian, then you may use their property: $u,, v text{ independent} iff operatorname{cov}(u,v) = 0$.
$$ operatorname{cov} left( frac{X Y}{sqrt{X^2 + Y^2}},frac{X^2 - Y^2}{sqrt{X^2 + Y^2}} right)$$
$$ = operatorname{cov} left( frac{X Y}{sqrt{X^2 + Y^2}},frac{X^2}{sqrt{X^2 + Y^2}} right) - operatorname{cov} left( frac{X Y}{sqrt{X^2 + Y^2}},frac{Y^2}{sqrt{X^2 + Y^2}} right)$$
Now make use of the symmetry of the expressions we have:
$$ operatorname{cov}(cdots) = 0 $$
To make this more rigorous one may rewrite the covariance as follows:
$$ operatorname{cov} left( frac{X Y}{sqrt{X^2 + Y^2}},frac{X^2}{sqrt{X^2 + Y^2}} right) - operatorname{cov} left( frac{X Y}{sqrt{X^2 + Y^2}},frac{Y^2}{sqrt{X^2 + Y^2}} right) $$
$$ = E left( frac{X^3 Y}{X^2 + Y^2} right) - E left( frac{X Y^3}{X^2 + Y^2} right) $$
Now renaming $X to Y, , Y to X$ under the first expectation sign (note that this is the same as renaming variables under integral) we get the result.
Covariance method can be carried even further, following the last line and the trick $ X = -X $ ($X$ is symmetric distribution) which we apply to the second $E$:
$$ = E left( frac{X^3 Y}{X^2 + Y^2} right) - E left( frac{X Y^3}{X^2 + Y^2} right) = E left( frac{X^3 Y}{X^2 + Y^2} right) + E left( frac{X Y^3}{X^2 + Y^2} right) $$
$$ = E left( frac{X Y(X^2 + Y^2)}{X^2 + Y^2} right) = E(XY) = 0$$
The last equation is due to independence of $X, Y$ $implies 0=operatorname{cov}(X,Y)=E(XY) - E(X)E(Y)=E(XY).$
answered Feb 24 '18 at 8:47
Al PrihodkoAl Prihodko
436
edited Jan 6 at 15:22
answered Feb 24 '18 at 8:47
Al PrihodkoAl Prihodko
436
answered Feb 24 '18 at 8:47
Al PrihodkoAl Prihodko
436
answered Feb 24 '18 at 8:47
Al PrihodkoAl Prihodko
436
436
3
$begingroup$
There is an error in the first sentence here. Being Gaussian and having zero covariance does NOT imply independnce. It does imply independence if they are JOINTLY Gaussian. But you haven't mentioned that.
$endgroup$
– Michael Hardy
Aug 20 '18 at 20:43
add a comment |
3
$begingroup$
There is an error in the first sentence here. Being Gaussian and having zero covariance does NOT imply independnce. It does imply independence if they are JOINTLY Gaussian. But you haven't mentioned that.
$endgroup$
– Michael Hardy
Aug 20 '18 at 20:43
3
3
$begingroup$
There is an error in the first sentence here. Being Gaussian and having zero covariance does NOT imply independnce. It does imply independence if they are JOINTLY Gaussian. But you haven't mentioned that.
$endgroup$
– Michael Hardy
Aug 20 '18 at 20:43
$begingroup$
There is an error in the first sentence here. Being Gaussian and having zero covariance does NOT imply independnce. It does imply independence if they are JOINTLY Gaussian. But you haven't mentioned that.
$endgroup$
– Michael Hardy
Aug 20 '18 at 20:43
add a comment |
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$begingroup$
You're on the right track: don't get discouraged.
$endgroup$
– kimchi lover
Feb 12 '18 at 13:58
$begingroup$
Actually I can show each of the distributions $frac{XY}{sqrt{X^2+Y^2}}$ and the other one to be N(0,1/4) but I cannot show their independence
$endgroup$
– Legend Killer
Feb 12 '18 at 14:00