Mixing
Covariance Identity: $DeclareMathOperator{Cov}{Cov}Cov(X,Y) = E(Cov(X,Ymid Z)) + Cov(E(Xmid Z),E(Ymid Z))$
$begingroup$
How can I show that:
$Cov(X,Y) = E(Cov(X,Ymid Z)) + Cov(E(Xmid Z),E(Ymid Z))$?
With $X, Y$ and $Z;$ r.v with finite variances.
probability probability-distributions
edited Jan 26 at 16:31
Bernard
123k741117
asked Jan 26 at 16:02
LauraLaura
3518
$endgroup$
add a comment |
$begingroup$
How can I show that:
$Cov(X,Y) = E(Cov(X,Ymid Z)) + Cov(E(Xmid Z),E(Ymid Z))$?
With $X, Y$ and $Z;$ r.v with finite variances.
probability probability-distributions
edited Jan 26 at 16:31
Bernard
123k741117
asked Jan 26 at 16:02
LauraLaura
3518
$endgroup$
$begingroup$
If you want to try this yourself with a hint, search for 'This follows directly from the tower property' on this page isical.ac.in/~arnabc/prob1/condl.html
$endgroup$
– Anvit
Jan 26 at 16:36
add a comment |
$begingroup$
How can I show that:
$Cov(X,Y) = E(Cov(X,Ymid Z)) + Cov(E(Xmid Z),E(Ymid Z))$?
With $X, Y$ and $Z;$ r.v with finite variances.
probability probability-distributions
edited Jan 26 at 16:31
Bernard
123k741117
asked Jan 26 at 16:02
LauraLaura
3518
$endgroup$
How can I show that:
$Cov(X,Y) = E(Cov(X,Ymid Z)) + Cov(E(Xmid Z),E(Ymid Z))$?
With $X, Y$ and $Z;$ r.v with finite variances.
probability probability-distributions
probability probability-distributions
edited Jan 26 at 16:31
Bernard
123k741117
asked Jan 26 at 16:02
LauraLaura
3518
edited Jan 26 at 16:31
Bernard
123k741117
asked Jan 26 at 16:02
LauraLaura
3518
edited Jan 26 at 16:31
Bernard
123k741117
edited Jan 26 at 16:31
Bernard
123k741117
edited Jan 26 at 16:31
Bernard
123k741117
123k741117
asked Jan 26 at 16:02
LauraLaura
3518
asked Jan 26 at 16:02
LauraLaura
3518
asked Jan 26 at 16:02
LauraLaura
3518
3518
$begingroup$
If you want to try this yourself with a hint, search for 'This follows directly from the tower property' on this page isical.ac.in/~arnabc/prob1/condl.html
$endgroup$
– Anvit
Jan 26 at 16:36
add a comment |
$begingroup$
If you want to try this yourself with a hint, search for 'This follows directly from the tower property' on this page isical.ac.in/~arnabc/prob1/condl.html
$endgroup$
– Anvit
Jan 26 at 16:36
$begingroup$
If you want to try this yourself with a hint, search for 'This follows directly from the tower property' on this page isical.ac.in/~arnabc/prob1/condl.html
$endgroup$
– Anvit
Jan 26 at 16:36
$begingroup$
If you want to try this yourself with a hint, search for 'This follows directly from the tower property' on this page isical.ac.in/~arnabc/prob1/condl.html
$endgroup$
– Anvit
Jan 26 at 16:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here's a ugly looking jumble of Expectations
$$begin{align}E[DeclareMathOperator{Cov}{Cov}Cov[X,Y|Z]]&=E[E[XY|Z]-E[X|Z]E[Y|Z]]\[2ex]
&=E[E[XY|Z]]-E[E[X|Z]E[Y|Z]]\[2ex]
&=E[XY]-E[E[X|Z]E[Y|Z]]
end{align}$$
Similarly,
$$begin{align}
Cov[E[X|Z],E[Y|Z]]&=E[E[X|Z]E[Y|Z]]-E[E[X|Z]]E[E[Y|Z]]\[2ex]
&=E[E[X|Z]E[Y|Z]]-E[X]E[Y]\[2ex]
end{align}$$
Add the two results to get the desired result. This proof utilizes the tower property. A simple version of which is $E[E[X|Y]]=E[X].$ Here is the wikipage Tower Property
edited Jan 26 at 17:16
Antoni Parellada
3,10421341
answered Jan 26 at 16:51
AnvitAnvit
1,733419
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Here's a ugly looking jumble of Expectations
$$begin{align}E[DeclareMathOperator{Cov}{Cov}Cov[X,Y|Z]]&=E[E[XY|Z]-E[X|Z]E[Y|Z]]\[2ex]
&=E[E[XY|Z]]-E[E[X|Z]E[Y|Z]]\[2ex]
&=E[XY]-E[E[X|Z]E[Y|Z]]
end{align}$$
Similarly,
$$begin{align}
Cov[E[X|Z],E[Y|Z]]&=E[E[X|Z]E[Y|Z]]-E[E[X|Z]]E[E[Y|Z]]\[2ex]
&=E[E[X|Z]E[Y|Z]]-E[X]E[Y]\[2ex]
end{align}$$
Add the two results to get the desired result. This proof utilizes the tower property. A simple version of which is $E[E[X|Y]]=E[X].$ Here is the wikipage Tower Property
edited Jan 26 at 17:16
Antoni Parellada
3,10421341
answered Jan 26 at 16:51
AnvitAnvit
1,733419
$endgroup$
add a comment |
$begingroup$
Here's a ugly looking jumble of Expectations
$$begin{align}E[DeclareMathOperator{Cov}{Cov}Cov[X,Y|Z]]&=E[E[XY|Z]-E[X|Z]E[Y|Z]]\[2ex]
&=E[E[XY|Z]]-E[E[X|Z]E[Y|Z]]\[2ex]
&=E[XY]-E[E[X|Z]E[Y|Z]]
end{align}$$
Similarly,
$$begin{align}
Cov[E[X|Z],E[Y|Z]]&=E[E[X|Z]E[Y|Z]]-E[E[X|Z]]E[E[Y|Z]]\[2ex]
&=E[E[X|Z]E[Y|Z]]-E[X]E[Y]\[2ex]
end{align}$$
Add the two results to get the desired result. This proof utilizes the tower property. A simple version of which is $E[E[X|Y]]=E[X].$ Here is the wikipage Tower Property
edited Jan 26 at 17:16
Antoni Parellada
3,10421341
answered Jan 26 at 16:51
AnvitAnvit
1,733419
$endgroup$
add a comment |
$begingroup$
Here's a ugly looking jumble of Expectations
$$begin{align}E[DeclareMathOperator{Cov}{Cov}Cov[X,Y|Z]]&=E[E[XY|Z]-E[X|Z]E[Y|Z]]\[2ex]
&=E[E[XY|Z]]-E[E[X|Z]E[Y|Z]]\[2ex]
&=E[XY]-E[E[X|Z]E[Y|Z]]
end{align}$$
Similarly,
$$begin{align}
Cov[E[X|Z],E[Y|Z]]&=E[E[X|Z]E[Y|Z]]-E[E[X|Z]]E[E[Y|Z]]\[2ex]
&=E[E[X|Z]E[Y|Z]]-E[X]E[Y]\[2ex]
end{align}$$
Add the two results to get the desired result. This proof utilizes the tower property. A simple version of which is $E[E[X|Y]]=E[X].$ Here is the wikipage Tower Property
edited Jan 26 at 17:16
Antoni Parellada
3,10421341
answered Jan 26 at 16:51
AnvitAnvit
1,733419
$endgroup$
Here's a ugly looking jumble of Expectations
$$begin{align}E[DeclareMathOperator{Cov}{Cov}Cov[X,Y|Z]]&=E[E[XY|Z]-E[X|Z]E[Y|Z]]\[2ex]
&=E[E[XY|Z]]-E[E[X|Z]E[Y|Z]]\[2ex]
&=E[XY]-E[E[X|Z]E[Y|Z]]
end{align}$$
Similarly,
$$begin{align}
Cov[E[X|Z],E[Y|Z]]&=E[E[X|Z]E[Y|Z]]-E[E[X|Z]]E[E[Y|Z]]\[2ex]
&=E[E[X|Z]E[Y|Z]]-E[X]E[Y]\[2ex]
end{align}$$
Add the two results to get the desired result. This proof utilizes the tower property. A simple version of which is $E[E[X|Y]]=E[X].$ Here is the wikipage Tower Property
edited Jan 26 at 17:16
Antoni Parellada
3,10421341
answered Jan 26 at 16:51
AnvitAnvit
1,733419
edited Jan 26 at 17:16
Antoni Parellada
3,10421341
edited Jan 26 at 17:16
Antoni Parellada
3,10421341
edited Jan 26 at 17:16
Antoni Parellada
3,10421341
3,10421341
answered Jan 26 at 16:51
AnvitAnvit
1,733419
answered Jan 26 at 16:51
AnvitAnvit
1,733419
answered Jan 26 at 16:51
AnvitAnvit
1,733419
1,733419
add a comment |
add a comment |
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$begingroup$
If you want to try this yourself with a hint, search for 'This follows directly from the tower property' on this page isical.ac.in/~arnabc/prob1/condl.html
$endgroup$
– Anvit
Jan 26 at 16:36