Variance of a combination of random vectors
$begingroup$
I was looking at the proof for optimal weighting matrix when using GMM (slide 35 here)
At one point, they take the variance of both sides of following expression (where Z is a random vector)
$A_1Z = A_0Z + (A_1-A_0)Z$
And get the following:
$var(A_1 Z) =var( A_0Z) + var((A_1-A_0)Z) + A_0 var(Z)(A_1'-A_0') + (A_1-A_0)Var(Z)A_0'$
which evalates to $var( A_0Z) + var((A_1-A_0)Z)$
After a lot of thinking, I don't seem to get how the term $var(Z)(A_1'-A_0') + (A_1-A_0)Var(Z)A_0'$ shows up. Can anyone help please, what part of the linear algebra class I missed?
linear-algebra matrices statistics linear-transformations matrix-calculus
asked Dec 14 '18 at 15:43
DjpengoDjpengo
204
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add a comment |
$begingroup$
I was looking at the proof for optimal weighting matrix when using GMM (slide 35 here)
At one point, they take the variance of both sides of following expression (where Z is a random vector)
$A_1Z = A_0Z + (A_1-A_0)Z$
And get the following:
$var(A_1 Z) =var( A_0Z) + var((A_1-A_0)Z) + A_0 var(Z)(A_1'-A_0') + (A_1-A_0)Var(Z)A_0'$
which evalates to $var( A_0Z) + var((A_1-A_0)Z)$
After a lot of thinking, I don't seem to get how the term $var(Z)(A_1'-A_0') + (A_1-A_0)Var(Z)A_0'$ shows up. Can anyone help please, what part of the linear algebra class I missed?
linear-algebra matrices statistics linear-transformations matrix-calculus
asked Dec 14 '18 at 15:43
DjpengoDjpengo
204
$endgroup$
add a comment |
$begingroup$
I was looking at the proof for optimal weighting matrix when using GMM (slide 35 here)
At one point, they take the variance of both sides of following expression (where Z is a random vector)
$A_1Z = A_0Z + (A_1-A_0)Z$
And get the following:
$var(A_1 Z) =var( A_0Z) + var((A_1-A_0)Z) + A_0 var(Z)(A_1'-A_0') + (A_1-A_0)Var(Z)A_0'$
which evalates to $var( A_0Z) + var((A_1-A_0)Z)$
After a lot of thinking, I don't seem to get how the term $var(Z)(A_1'-A_0') + (A_1-A_0)Var(Z)A_0'$ shows up. Can anyone help please, what part of the linear algebra class I missed?
linear-algebra matrices statistics linear-transformations matrix-calculus
asked Dec 14 '18 at 15:43
DjpengoDjpengo
204
$endgroup$
I was looking at the proof for optimal weighting matrix when using GMM (slide 35 here)
At one point, they take the variance of both sides of following expression (where Z is a random vector)
$A_1Z = A_0Z + (A_1-A_0)Z$
And get the following:
$var(A_1 Z) =var( A_0Z) + var((A_1-A_0)Z) + A_0 var(Z)(A_1'-A_0') + (A_1-A_0)Var(Z)A_0'$
which evalates to $var( A_0Z) + var((A_1-A_0)Z)$
After a lot of thinking, I don't seem to get how the term $var(Z)(A_1'-A_0') + (A_1-A_0)Var(Z)A_0'$ shows up. Can anyone help please, what part of the linear algebra class I missed?
linear-algebra matrices statistics linear-transformations matrix-calculus
linear-algebra matrices statistics linear-transformations matrix-calculus
asked Dec 14 '18 at 15:43
DjpengoDjpengo
204
asked Dec 14 '18 at 15:43
DjpengoDjpengo
204
asked Dec 14 '18 at 15:43
DjpengoDjpengo
204
asked Dec 14 '18 at 15:43
DjpengoDjpengo
204
asked Dec 14 '18 at 15:43
DjpengoDjpengo
204
204
add a comment |
add a comment |
1 Answer
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$begingroup$
In your case the random vector $Z$ has zero mean. So your calculation simplifies to $operatorname{Var}(X+Y)$ where $X:=A_0 Z$ and $Y:=(A_1-A_0)Z$ are mean-zero random vectors. Since $X+Y$ will also have mean zero, its variance is $operatorname{Var}(X+Y)=E(X+Y)(X+Y)'$, which equals
$$
E(XX' + YY' +XY' + YX')=operatorname{Var}(X)+operatorname{Var}(Y)+E(XY')+E(YX'),
$$
where the final two terms are
$$E(XY')=Eleft(A_0Z[(A_1-A_0)Z]'right)=A_0E(ZZ')(A_1'-A_0')$$
and
$$E(YX')=Eleft([(A_1-A_0)Z](A_0Z)'right)=(A_1-A_0)E(ZZ')A_0'.$$
To finish off, observe that $E(ZZ')=operatorname{Var}(Z)$ since $Z$ has zero mean.
answered Dec 14 '18 at 20:13
grand_chatgrand_chat
20.4k11327
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$begingroup$
All is clear now, thanks a lot!
$endgroup$
– Djpengo
Dec 15 '18 at 7:57
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
In your case the random vector $Z$ has zero mean. So your calculation simplifies to $operatorname{Var}(X+Y)$ where $X:=A_0 Z$ and $Y:=(A_1-A_0)Z$ are mean-zero random vectors. Since $X+Y$ will also have mean zero, its variance is $operatorname{Var}(X+Y)=E(X+Y)(X+Y)'$, which equals
$$
E(XX' + YY' +XY' + YX')=operatorname{Var}(X)+operatorname{Var}(Y)+E(XY')+E(YX'),
$$
where the final two terms are
$$E(XY')=Eleft(A_0Z[(A_1-A_0)Z]'right)=A_0E(ZZ')(A_1'-A_0')$$
and
$$E(YX')=Eleft([(A_1-A_0)Z](A_0Z)'right)=(A_1-A_0)E(ZZ')A_0'.$$
To finish off, observe that $E(ZZ')=operatorname{Var}(Z)$ since $Z$ has zero mean.
answered Dec 14 '18 at 20:13
grand_chatgrand_chat
20.4k11327
$endgroup$
$begingroup$
All is clear now, thanks a lot!
$endgroup$
– Djpengo
Dec 15 '18 at 7:57
add a comment |
$begingroup$
In your case the random vector $Z$ has zero mean. So your calculation simplifies to $operatorname{Var}(X+Y)$ where $X:=A_0 Z$ and $Y:=(A_1-A_0)Z$ are mean-zero random vectors. Since $X+Y$ will also have mean zero, its variance is $operatorname{Var}(X+Y)=E(X+Y)(X+Y)'$, which equals
$$
E(XX' + YY' +XY' + YX')=operatorname{Var}(X)+operatorname{Var}(Y)+E(XY')+E(YX'),
$$
where the final two terms are
$$E(XY')=Eleft(A_0Z[(A_1-A_0)Z]'right)=A_0E(ZZ')(A_1'-A_0')$$
and
$$E(YX')=Eleft([(A_1-A_0)Z](A_0Z)'right)=(A_1-A_0)E(ZZ')A_0'.$$
To finish off, observe that $E(ZZ')=operatorname{Var}(Z)$ since $Z$ has zero mean.
answered Dec 14 '18 at 20:13
grand_chatgrand_chat
20.4k11327
$endgroup$
$begingroup$
All is clear now, thanks a lot!
$endgroup$
– Djpengo
Dec 15 '18 at 7:57
add a comment |
$begingroup$
In your case the random vector $Z$ has zero mean. So your calculation simplifies to $operatorname{Var}(X+Y)$ where $X:=A_0 Z$ and $Y:=(A_1-A_0)Z$ are mean-zero random vectors. Since $X+Y$ will also have mean zero, its variance is $operatorname{Var}(X+Y)=E(X+Y)(X+Y)'$, which equals
$$
E(XX' + YY' +XY' + YX')=operatorname{Var}(X)+operatorname{Var}(Y)+E(XY')+E(YX'),
$$
where the final two terms are
$$E(XY')=Eleft(A_0Z[(A_1-A_0)Z]'right)=A_0E(ZZ')(A_1'-A_0')$$
and
$$E(YX')=Eleft([(A_1-A_0)Z](A_0Z)'right)=(A_1-A_0)E(ZZ')A_0'.$$
To finish off, observe that $E(ZZ')=operatorname{Var}(Z)$ since $Z$ has zero mean.
answered Dec 14 '18 at 20:13
grand_chatgrand_chat
20.4k11327
$endgroup$
In your case the random vector $Z$ has zero mean. So your calculation simplifies to $operatorname{Var}(X+Y)$ where $X:=A_0 Z$ and $Y:=(A_1-A_0)Z$ are mean-zero random vectors. Since $X+Y$ will also have mean zero, its variance is $operatorname{Var}(X+Y)=E(X+Y)(X+Y)'$, which equals
$$
E(XX' + YY' +XY' + YX')=operatorname{Var}(X)+operatorname{Var}(Y)+E(XY')+E(YX'),
$$
where the final two terms are
$$E(XY')=Eleft(A_0Z[(A_1-A_0)Z]'right)=A_0E(ZZ')(A_1'-A_0')$$
and
$$E(YX')=Eleft([(A_1-A_0)Z](A_0Z)'right)=(A_1-A_0)E(ZZ')A_0'.$$
To finish off, observe that $E(ZZ')=operatorname{Var}(Z)$ since $Z$ has zero mean.
answered Dec 14 '18 at 20:13
grand_chatgrand_chat
20.4k11327
answered Dec 14 '18 at 20:13
grand_chatgrand_chat
20.4k11327
answered Dec 14 '18 at 20:13
grand_chatgrand_chat
20.4k11327
answered Dec 14 '18 at 20:13
grand_chatgrand_chat
20.4k11327
20.4k11327
$begingroup$
All is clear now, thanks a lot!
$endgroup$
– Djpengo
Dec 15 '18 at 7:57
add a comment |
$begingroup$
All is clear now, thanks a lot!
$endgroup$
– Djpengo
Dec 15 '18 at 7:57
$begingroup$
All is clear now, thanks a lot!
$endgroup$
– Djpengo
Dec 15 '18 at 7:57
$begingroup$
All is clear now, thanks a lot!
$endgroup$
– Djpengo
Dec 15 '18 at 7:57
add a comment |
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