Mixing
$V(X|Y)=Sigma_{XX}-Sigma_{XY}Sigma_{YY}^{-1}Sigma_{YX}$
$begingroup$
We know that the conditional variance of a multivariate normal vector $(X,Y)$ is equal to the Schur complement:
$$V(X|Y)=Sigma_{XX}-Sigma_{XY}Sigma_{YY}^{-1}Sigma_{YX}$$
However, $Sigma_{XX}-Sigma_{XY}Sigma_{YY}^{-1}Sigma_{YX}$ is a matrix of scalars, whereas $V(X|Y)$ is a matrix of random variables function of Y. Therefore, I guess that the Schur complement is the variance of X conditional to a certain realization of Y, $V(X|Y=(y_1,...,y_n))$, is it correct? If so, who is this $(y_1,...,y_n)$?
Thank you very much.
covariance variance schur-complement
asked Jan 13 at 16:46
DadooDadoo
212
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add a comment |
$begingroup$
We know that the conditional variance of a multivariate normal vector $(X,Y)$ is equal to the Schur complement:
$$V(X|Y)=Sigma_{XX}-Sigma_{XY}Sigma_{YY}^{-1}Sigma_{YX}$$
However, $Sigma_{XX}-Sigma_{XY}Sigma_{YY}^{-1}Sigma_{YX}$ is a matrix of scalars, whereas $V(X|Y)$ is a matrix of random variables function of Y. Therefore, I guess that the Schur complement is the variance of X conditional to a certain realization of Y, $V(X|Y=(y_1,...,y_n))$, is it correct? If so, who is this $(y_1,...,y_n)$?
Thank you very much.
covariance variance schur-complement
asked Jan 13 at 16:46
DadooDadoo
212
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Please do not cross-post.
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– StubbornAtom
Jan 13 at 17:35
add a comment |
$begingroup$
We know that the conditional variance of a multivariate normal vector $(X,Y)$ is equal to the Schur complement:
$$V(X|Y)=Sigma_{XX}-Sigma_{XY}Sigma_{YY}^{-1}Sigma_{YX}$$
However, $Sigma_{XX}-Sigma_{XY}Sigma_{YY}^{-1}Sigma_{YX}$ is a matrix of scalars, whereas $V(X|Y)$ is a matrix of random variables function of Y. Therefore, I guess that the Schur complement is the variance of X conditional to a certain realization of Y, $V(X|Y=(y_1,...,y_n))$, is it correct? If so, who is this $(y_1,...,y_n)$?
Thank you very much.
covariance variance schur-complement
asked Jan 13 at 16:46
DadooDadoo
212
$endgroup$
We know that the conditional variance of a multivariate normal vector $(X,Y)$ is equal to the Schur complement:
$$V(X|Y)=Sigma_{XX}-Sigma_{XY}Sigma_{YY}^{-1}Sigma_{YX}$$
However, $Sigma_{XX}-Sigma_{XY}Sigma_{YY}^{-1}Sigma_{YX}$ is a matrix of scalars, whereas $V(X|Y)$ is a matrix of random variables function of Y. Therefore, I guess that the Schur complement is the variance of X conditional to a certain realization of Y, $V(X|Y=(y_1,...,y_n))$, is it correct? If so, who is this $(y_1,...,y_n)$?
Thank you very much.
covariance variance schur-complement
covariance variance schur-complement
asked Jan 13 at 16:46
DadooDadoo
212
asked Jan 13 at 16:46
DadooDadoo
212
asked Jan 13 at 16:46
DadooDadoo
212
asked Jan 13 at 16:46
DadooDadoo
212
asked Jan 13 at 16:46
DadooDadoo
212
212
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Please do not cross-post.
$endgroup$
– StubbornAtom
Jan 13 at 17:35
add a comment |
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Please do not cross-post.
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– StubbornAtom
Jan 13 at 17:35
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Please do not cross-post.
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– StubbornAtom
Jan 13 at 17:35
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Please do not cross-post.
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– StubbornAtom
Jan 13 at 17:35
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1 Answer
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You are right when you said:
Therefore, I guess that the Schur complement is the variance of X conditional to a certain realization of Y.
Regard the vector $$Z = begin{bmatrix} X \ Y end{bmatrix} sim N( underbrace{begin{bmatrix} mu_X \ mu_Y end{bmatrix}}_{mu}; underbrace{begin{bmatrix} Sigma_{XX} & Sigma_{XY} \ Sigma_{XY} & Sigma_{YY} end{bmatrix}}_{Sigma} )$$
All we did is partition a given Normal vector into $2$ subvectors. Now, if you want to find $V(X|Y)$, it turns out to be (using Schurs' complement), nothing other than the equation you have provided.
answered Jan 13 at 16:54
Ahmad BazziAhmad Bazzi
8,2212824
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1 Answer
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1 Answer
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$begingroup$
You are right when you said:
Therefore, I guess that the Schur complement is the variance of X conditional to a certain realization of Y.
Regard the vector $$Z = begin{bmatrix} X \ Y end{bmatrix} sim N( underbrace{begin{bmatrix} mu_X \ mu_Y end{bmatrix}}_{mu}; underbrace{begin{bmatrix} Sigma_{XX} & Sigma_{XY} \ Sigma_{XY} & Sigma_{YY} end{bmatrix}}_{Sigma} )$$
All we did is partition a given Normal vector into $2$ subvectors. Now, if you want to find $V(X|Y)$, it turns out to be (using Schurs' complement), nothing other than the equation you have provided.
answered Jan 13 at 16:54
Ahmad BazziAhmad Bazzi
8,2212824
$endgroup$
add a comment |
$begingroup$
You are right when you said:
Therefore, I guess that the Schur complement is the variance of X conditional to a certain realization of Y.
Regard the vector $$Z = begin{bmatrix} X \ Y end{bmatrix} sim N( underbrace{begin{bmatrix} mu_X \ mu_Y end{bmatrix}}_{mu}; underbrace{begin{bmatrix} Sigma_{XX} & Sigma_{XY} \ Sigma_{XY} & Sigma_{YY} end{bmatrix}}_{Sigma} )$$
All we did is partition a given Normal vector into $2$ subvectors. Now, if you want to find $V(X|Y)$, it turns out to be (using Schurs' complement), nothing other than the equation you have provided.
answered Jan 13 at 16:54
Ahmad BazziAhmad Bazzi
8,2212824
$endgroup$
add a comment |
$begingroup$
You are right when you said:
Therefore, I guess that the Schur complement is the variance of X conditional to a certain realization of Y.
Regard the vector $$Z = begin{bmatrix} X \ Y end{bmatrix} sim N( underbrace{begin{bmatrix} mu_X \ mu_Y end{bmatrix}}_{mu}; underbrace{begin{bmatrix} Sigma_{XX} & Sigma_{XY} \ Sigma_{XY} & Sigma_{YY} end{bmatrix}}_{Sigma} )$$
All we did is partition a given Normal vector into $2$ subvectors. Now, if you want to find $V(X|Y)$, it turns out to be (using Schurs' complement), nothing other than the equation you have provided.
answered Jan 13 at 16:54
Ahmad BazziAhmad Bazzi
8,2212824
$endgroup$
You are right when you said:
Therefore, I guess that the Schur complement is the variance of X conditional to a certain realization of Y.
Regard the vector $$Z = begin{bmatrix} X \ Y end{bmatrix} sim N( underbrace{begin{bmatrix} mu_X \ mu_Y end{bmatrix}}_{mu}; underbrace{begin{bmatrix} Sigma_{XX} & Sigma_{XY} \ Sigma_{XY} & Sigma_{YY} end{bmatrix}}_{Sigma} )$$
All we did is partition a given Normal vector into $2$ subvectors. Now, if you want to find $V(X|Y)$, it turns out to be (using Schurs' complement), nothing other than the equation you have provided.
answered Jan 13 at 16:54
Ahmad BazziAhmad Bazzi
8,2212824
answered Jan 13 at 16:54
Ahmad BazziAhmad Bazzi
8,2212824
answered Jan 13 at 16:54
Ahmad BazziAhmad Bazzi
8,2212824
answered Jan 13 at 16:54
Ahmad BazziAhmad Bazzi
8,2212824
8,2212824
add a comment |
add a comment |
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– StubbornAtom
Jan 13 at 17:35