Mixing
Expectation of $A^TXA$ for random $A$ and $X$
$begingroup$
Suppose there are two random matrices (distribution unknown), denoted as $A$ and $X$, both in the $mathbb{R}^{n times n}$ space. It is known that $|A| leq 1$ (for any $p$-norm) and $E[X]geq0$. I hope to evaluate the following expectation
$$E[A^TXA]$$
to be in terms of $E[X]$. My first instinct is to break it down into element-wise expression but it is quite messy and difficult to reconstruct the matrix form. Is there any smarter way to do this?
Thanks a lot!
expected-value random-matrices
asked Jan 11 at 2:10
PenthritePenthrite
377
$endgroup$
|
show 4 more comments
$begingroup$
Suppose there are two random matrices (distribution unknown), denoted as $A$ and $X$, both in the $mathbb{R}^{n times n}$ space. It is known that $|A| leq 1$ (for any $p$-norm) and $E[X]geq0$. I hope to evaluate the following expectation
$$E[A^TXA]$$
to be in terms of $E[X]$. My first instinct is to break it down into element-wise expression but it is quite messy and difficult to reconstruct the matrix form. Is there any smarter way to do this?
Thanks a lot!
expected-value random-matrices
asked Jan 11 at 2:10
PenthritePenthrite
377
$endgroup$
$begingroup$
Curious, what book do you use for random matrices?
$endgroup$
– Hermit with Adjoint
Jan 11 at 2:22
$begingroup$
$A$ and $X$ are independent of each other?
$endgroup$
– Lee
Jan 11 at 2:44
$begingroup$
@Lee May I ask how did you get this conclusion?
$endgroup$
– Penthrite
Jan 11 at 2:50
$begingroup$
@Penthrite I think it was wrong
$endgroup$
– Lee
Jan 11 at 2:53
$begingroup$
oh yes, maybe this one: $E[A^TXA]=trE[XAA^T]=tr{E[X]E[AA^T]}$
$endgroup$
– Lee
Jan 11 at 2:56
|
show 4 more comments
$begingroup$
Suppose there are two random matrices (distribution unknown), denoted as $A$ and $X$, both in the $mathbb{R}^{n times n}$ space. It is known that $|A| leq 1$ (for any $p$-norm) and $E[X]geq0$. I hope to evaluate the following expectation
$$E[A^TXA]$$
to be in terms of $E[X]$. My first instinct is to break it down into element-wise expression but it is quite messy and difficult to reconstruct the matrix form. Is there any smarter way to do this?
Thanks a lot!
expected-value random-matrices
asked Jan 11 at 2:10
PenthritePenthrite
377
$endgroup$
Suppose there are two random matrices (distribution unknown), denoted as $A$ and $X$, both in the $mathbb{R}^{n times n}$ space. It is known that $|A| leq 1$ (for any $p$-norm) and $E[X]geq0$. I hope to evaluate the following expectation
$$E[A^TXA]$$
to be in terms of $E[X]$. My first instinct is to break it down into element-wise expression but it is quite messy and difficult to reconstruct the matrix form. Is there any smarter way to do this?
Thanks a lot!
expected-value random-matrices
expected-value random-matrices
asked Jan 11 at 2:10
PenthritePenthrite
377
asked Jan 11 at 2:10
PenthritePenthrite
377
edited Jan 11 at 4:13
Penthrite
asked Jan 11 at 2:10
PenthritePenthrite
377
asked Jan 11 at 2:10
PenthritePenthrite
377
asked Jan 11 at 2:10
PenthritePenthrite
377
377
$begingroup$
Curious, what book do you use for random matrices?
$endgroup$
– Hermit with Adjoint
Jan 11 at 2:22
$begingroup$
$A$ and $X$ are independent of each other?
$endgroup$
– Lee
Jan 11 at 2:44
$begingroup$
@Lee May I ask how did you get this conclusion?
$endgroup$
– Penthrite
Jan 11 at 2:50
$begingroup$
@Penthrite I think it was wrong
$endgroup$
– Lee
Jan 11 at 2:53
$begingroup$
oh yes, maybe this one: $E[A^TXA]=trE[XAA^T]=tr{E[X]E[AA^T]}$
$endgroup$
– Lee
Jan 11 at 2:56
|
show 4 more comments
$begingroup$
Curious, what book do you use for random matrices?
$endgroup$
– Hermit with Adjoint
Jan 11 at 2:22
$begingroup$
$A$ and $X$ are independent of each other?
$endgroup$
– Lee
Jan 11 at 2:44
$begingroup$
@Lee May I ask how did you get this conclusion?
$endgroup$
– Penthrite
Jan 11 at 2:50
$begingroup$
@Penthrite I think it was wrong
$endgroup$
– Lee
Jan 11 at 2:53
$begingroup$
oh yes, maybe this one: $E[A^TXA]=trE[XAA^T]=tr{E[X]E[AA^T]}$
$endgroup$
– Lee
Jan 11 at 2:56
$begingroup$
Curious, what book do you use for random matrices?
$endgroup$
– Hermit with Adjoint
Jan 11 at 2:22
$begingroup$
Curious, what book do you use for random matrices?
$endgroup$
– Hermit with Adjoint
Jan 11 at 2:22
$begingroup$
$A$ and $X$ are independent of each other?
$endgroup$
– Lee
Jan 11 at 2:44
$begingroup$
$A$ and $X$ are independent of each other?
$endgroup$
– Lee
Jan 11 at 2:44
$begingroup$
@Lee May I ask how did you get this conclusion?
$endgroup$
– Penthrite
Jan 11 at 2:50
$begingroup$
@Lee May I ask how did you get this conclusion?
$endgroup$
– Penthrite
Jan 11 at 2:50
$begingroup$
@Penthrite I think it was wrong
$endgroup$
– Lee
Jan 11 at 2:53
$begingroup$
@Penthrite I think it was wrong
$endgroup$
– Lee
Jan 11 at 2:53
$begingroup$
oh yes, maybe this one: $E[A^TXA]=trE[XAA^T]=tr{E[X]E[AA^T]}$
$endgroup$
– Lee
Jan 11 at 2:56
$begingroup$
oh yes, maybe this one: $E[A^TXA]=trE[XAA^T]=tr{E[X]E[AA^T]}$
$endgroup$
– Lee
Jan 11 at 2:56
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Let $X = [x_1,x_2,cdots, x_n].$ Then $AX = [sum_{k=1}^{n}a_{1k}x_{k},sum_{k=1}^{n}a_{2k}x_{k},cdots, sum_{k=1}^{n}a_{nk}x_{k}]$
and so $$X^{t}AX= sum_{i=1}^{n}sum_{k=1}^{n}a_{ik}x_{i}x_{k}$$
and so $$E[X^tAX] = Eleft[sum_{i=1}^{n}sum_{k=1}^{n}a_{ik}x_{i}x_{k}right] = sum_{i=1}^{n}sum_{k=1}^{n}Eleft[a_{ik}x_{i}x_{k}right].$$
Since
$$E(XYZ) = E(XY)bar{Z} + Cov(XY, Z)$$
for any three $X,Y,Z$ random variables we have that,
$$E[x_{i}x_{k}a_{ik}] = E[x_{i}x_{k}]E[a_{ik}]+Cov(x_ix_k,a_{ik})$$
$$=E[a_{ik}]c_{ki}+Cov(a_{ik},c_{ki}),$$
where $C$ is the covariance matrix of $X.$
So we have that
$$E[X^tAX] =sum_{i=1}^{n}sum_{k=1}^{n}E[a_{ik}]c_{ki}+Cov(a_{ik},c_{ki}) $$
$$ = text{Trace}(E[A]C)+text{Trace}(K)$$
where $K$ is the covariance matrix of with each entry as $text{Cov}(a_{ij},c_{mn}).$
answered Jan 11 at 4:32
Hello_WorldHello_World
4,12621731
$endgroup$
$begingroup$
From the question, $X$ is an $ntimes n$ matrix, not a vector
$endgroup$
– nemo
Jan 11 at 10:03
$begingroup$
I think I got the notation wrong. $X$ stands for $A$
$endgroup$
– Hello_World
Jan 11 at 10:06
$begingroup$
Both $A$ and $X$ are matrices according to the OP
$endgroup$
– nemo
Jan 11 at 14:55
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
Let $X = [x_1,x_2,cdots, x_n].$ Then $AX = [sum_{k=1}^{n}a_{1k}x_{k},sum_{k=1}^{n}a_{2k}x_{k},cdots, sum_{k=1}^{n}a_{nk}x_{k}]$
and so $$X^{t}AX= sum_{i=1}^{n}sum_{k=1}^{n}a_{ik}x_{i}x_{k}$$
and so $$E[X^tAX] = Eleft[sum_{i=1}^{n}sum_{k=1}^{n}a_{ik}x_{i}x_{k}right] = sum_{i=1}^{n}sum_{k=1}^{n}Eleft[a_{ik}x_{i}x_{k}right].$$
Since
$$E(XYZ) = E(XY)bar{Z} + Cov(XY, Z)$$
for any three $X,Y,Z$ random variables we have that,
$$E[x_{i}x_{k}a_{ik}] = E[x_{i}x_{k}]E[a_{ik}]+Cov(x_ix_k,a_{ik})$$
$$=E[a_{ik}]c_{ki}+Cov(a_{ik},c_{ki}),$$
where $C$ is the covariance matrix of $X.$
So we have that
$$E[X^tAX] =sum_{i=1}^{n}sum_{k=1}^{n}E[a_{ik}]c_{ki}+Cov(a_{ik},c_{ki}) $$
$$ = text{Trace}(E[A]C)+text{Trace}(K)$$
where $K$ is the covariance matrix of with each entry as $text{Cov}(a_{ij},c_{mn}).$
answered Jan 11 at 4:32
Hello_WorldHello_World
4,12621731
$endgroup$
$begingroup$
From the question, $X$ is an $ntimes n$ matrix, not a vector
$endgroup$
– nemo
Jan 11 at 10:03
$begingroup$
I think I got the notation wrong. $X$ stands for $A$
$endgroup$
– Hello_World
Jan 11 at 10:06
$begingroup$
Both $A$ and $X$ are matrices according to the OP
$endgroup$
– nemo
Jan 11 at 14:55
add a comment |
$begingroup$
Let $X = [x_1,x_2,cdots, x_n].$ Then $AX = [sum_{k=1}^{n}a_{1k}x_{k},sum_{k=1}^{n}a_{2k}x_{k},cdots, sum_{k=1}^{n}a_{nk}x_{k}]$
and so $$X^{t}AX= sum_{i=1}^{n}sum_{k=1}^{n}a_{ik}x_{i}x_{k}$$
and so $$E[X^tAX] = Eleft[sum_{i=1}^{n}sum_{k=1}^{n}a_{ik}x_{i}x_{k}right] = sum_{i=1}^{n}sum_{k=1}^{n}Eleft[a_{ik}x_{i}x_{k}right].$$
Since
$$E(XYZ) = E(XY)bar{Z} + Cov(XY, Z)$$
for any three $X,Y,Z$ random variables we have that,
$$E[x_{i}x_{k}a_{ik}] = E[x_{i}x_{k}]E[a_{ik}]+Cov(x_ix_k,a_{ik})$$
$$=E[a_{ik}]c_{ki}+Cov(a_{ik},c_{ki}),$$
where $C$ is the covariance matrix of $X.$
So we have that
$$E[X^tAX] =sum_{i=1}^{n}sum_{k=1}^{n}E[a_{ik}]c_{ki}+Cov(a_{ik},c_{ki}) $$
$$ = text{Trace}(E[A]C)+text{Trace}(K)$$
where $K$ is the covariance matrix of with each entry as $text{Cov}(a_{ij},c_{mn}).$
answered Jan 11 at 4:32
Hello_WorldHello_World
4,12621731
$endgroup$
$begingroup$
From the question, $X$ is an $ntimes n$ matrix, not a vector
$endgroup$
– nemo
Jan 11 at 10:03
$begingroup$
I think I got the notation wrong. $X$ stands for $A$
$endgroup$
– Hello_World
Jan 11 at 10:06
$begingroup$
Both $A$ and $X$ are matrices according to the OP
$endgroup$
– nemo
Jan 11 at 14:55
add a comment |
$begingroup$
Let $X = [x_1,x_2,cdots, x_n].$ Then $AX = [sum_{k=1}^{n}a_{1k}x_{k},sum_{k=1}^{n}a_{2k}x_{k},cdots, sum_{k=1}^{n}a_{nk}x_{k}]$
and so $$X^{t}AX= sum_{i=1}^{n}sum_{k=1}^{n}a_{ik}x_{i}x_{k}$$
and so $$E[X^tAX] = Eleft[sum_{i=1}^{n}sum_{k=1}^{n}a_{ik}x_{i}x_{k}right] = sum_{i=1}^{n}sum_{k=1}^{n}Eleft[a_{ik}x_{i}x_{k}right].$$
Since
$$E(XYZ) = E(XY)bar{Z} + Cov(XY, Z)$$
for any three $X,Y,Z$ random variables we have that,
$$E[x_{i}x_{k}a_{ik}] = E[x_{i}x_{k}]E[a_{ik}]+Cov(x_ix_k,a_{ik})$$
$$=E[a_{ik}]c_{ki}+Cov(a_{ik},c_{ki}),$$
where $C$ is the covariance matrix of $X.$
So we have that
$$E[X^tAX] =sum_{i=1}^{n}sum_{k=1}^{n}E[a_{ik}]c_{ki}+Cov(a_{ik},c_{ki}) $$
$$ = text{Trace}(E[A]C)+text{Trace}(K)$$
where $K$ is the covariance matrix of with each entry as $text{Cov}(a_{ij},c_{mn}).$
answered Jan 11 at 4:32
Hello_WorldHello_World
4,12621731
$endgroup$
Let $X = [x_1,x_2,cdots, x_n].$ Then $AX = [sum_{k=1}^{n}a_{1k}x_{k},sum_{k=1}^{n}a_{2k}x_{k},cdots, sum_{k=1}^{n}a_{nk}x_{k}]$
and so $$X^{t}AX= sum_{i=1}^{n}sum_{k=1}^{n}a_{ik}x_{i}x_{k}$$
and so $$E[X^tAX] = Eleft[sum_{i=1}^{n}sum_{k=1}^{n}a_{ik}x_{i}x_{k}right] = sum_{i=1}^{n}sum_{k=1}^{n}Eleft[a_{ik}x_{i}x_{k}right].$$
Since
$$E(XYZ) = E(XY)bar{Z} + Cov(XY, Z)$$
for any three $X,Y,Z$ random variables we have that,
$$E[x_{i}x_{k}a_{ik}] = E[x_{i}x_{k}]E[a_{ik}]+Cov(x_ix_k,a_{ik})$$
$$=E[a_{ik}]c_{ki}+Cov(a_{ik},c_{ki}),$$
where $C$ is the covariance matrix of $X.$
So we have that
$$E[X^tAX] =sum_{i=1}^{n}sum_{k=1}^{n}E[a_{ik}]c_{ki}+Cov(a_{ik},c_{ki}) $$
$$ = text{Trace}(E[A]C)+text{Trace}(K)$$
where $K$ is the covariance matrix of with each entry as $text{Cov}(a_{ij},c_{mn}).$
answered Jan 11 at 4:32
Hello_WorldHello_World
4,12621731
edited Jan 11 at 4:50
answered Jan 11 at 4:32
Hello_WorldHello_World
4,12621731
answered Jan 11 at 4:32
Hello_WorldHello_World
4,12621731
answered Jan 11 at 4:32
Hello_WorldHello_World
4,12621731
4,12621731
$begingroup$
From the question, $X$ is an $ntimes n$ matrix, not a vector
$endgroup$
– nemo
Jan 11 at 10:03
$begingroup$
I think I got the notation wrong. $X$ stands for $A$
$endgroup$
– Hello_World
Jan 11 at 10:06
$begingroup$
Both $A$ and $X$ are matrices according to the OP
$endgroup$
– nemo
Jan 11 at 14:55
add a comment |
$begingroup$
From the question, $X$ is an $ntimes n$ matrix, not a vector
$endgroup$
– nemo
Jan 11 at 10:03
$begingroup$
I think I got the notation wrong. $X$ stands for $A$
$endgroup$
– Hello_World
Jan 11 at 10:06
$begingroup$
Both $A$ and $X$ are matrices according to the OP
$endgroup$
– nemo
Jan 11 at 14:55
$begingroup$
From the question, $X$ is an $ntimes n$ matrix, not a vector
$endgroup$
– nemo
Jan 11 at 10:03
$begingroup$
From the question, $X$ is an $ntimes n$ matrix, not a vector
$endgroup$
– nemo
Jan 11 at 10:03
$begingroup$
I think I got the notation wrong. $X$ stands for $A$
$endgroup$
– Hello_World
Jan 11 at 10:06
$begingroup$
I think I got the notation wrong. $X$ stands for $A$
$endgroup$
– Hello_World
Jan 11 at 10:06
$begingroup$
Both $A$ and $X$ are matrices according to the OP
$endgroup$
– nemo
Jan 11 at 14:55
$begingroup$
Both $A$ and $X$ are matrices according to the OP
$endgroup$
– nemo
Jan 11 at 14:55
add a comment |
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$begingroup$
Curious, what book do you use for random matrices?
$endgroup$
– Hermit with Adjoint
Jan 11 at 2:22
$begingroup$
$A$ and $X$ are independent of each other?
$endgroup$
– Lee
Jan 11 at 2:44
$begingroup$
@Lee May I ask how did you get this conclusion?
$endgroup$
– Penthrite
Jan 11 at 2:50
$begingroup$
@Penthrite I think it was wrong
$endgroup$
– Lee
Jan 11 at 2:53
$begingroup$
oh yes, maybe this one: $E[A^TXA]=trE[XAA^T]=tr{E[X]E[AA^T]}$
$endgroup$
– Lee
Jan 11 at 2:56