How to minimize the minimum mean square error of this difference
$begingroup$
I am trying to minimize the mean square error. More precisely, I am trying to minimize the following optimization problem
$$arg min _{bf{w_1},bf{w_2}}mathbb{E} ,,[|{bf s} - {bf Wy}|^2 ]$$
$${bf W} = begin{bmatrix}
{bf w_1} &{bf 0 } \
{bf 0 } & {bf w_2 }
end{bmatrix}$$
where ${bf W}$ is a $2times N$ matrix and where ${bf w_i}$ is a $1times N/2$ for $iin[1:2]$ and ${bf 0}$ is $1times N/2$ vector while ${bf s}$ is 2 $times $ 1 and ${bf y}$ is $N times 1$ vectors.
Any hints or ideas on finding the minimizing vectors of this problem?
It is also given that
$${bf y = A F s + z}$$ where ${bf A}$ is $Ntimes N$ matrix while
$${bf F} = begin{bmatrix}
{bf f_1} &{bf 0} \
{bf 0} & {bf f_2}
end{bmatrix}$$
where ${bf f_i}$ are $N/2times 1$ vector and ${bf Z}$ is $Ntimes 1$ vector
I am thinking of starting to start the solution as following
$$argmin _{bf{w_1},bf{w_2}}mathbb{E} ,,[[{bf s} - {bf Wy}]^H[{bf s} - {bf Wy}]]$$
But I assume given the specific block diagonal structure of matrix ${bf W}$ and ${bf F}$ it should be easier to solve ...
Thanks
optimization mean-square-error
asked Oct 21 '15 at 23:46
TyroneTyrone
244418
$endgroup$
add a comment |
$begingroup$
I am trying to minimize the mean square error. More precisely, I am trying to minimize the following optimization problem
$$arg min _{bf{w_1},bf{w_2}}mathbb{E} ,,[|{bf s} - {bf Wy}|^2 ]$$
$${bf W} = begin{bmatrix}
{bf w_1} &{bf 0 } \
{bf 0 } & {bf w_2 }
end{bmatrix}$$
where ${bf W}$ is a $2times N$ matrix and where ${bf w_i}$ is a $1times N/2$ for $iin[1:2]$ and ${bf 0}$ is $1times N/2$ vector while ${bf s}$ is 2 $times $ 1 and ${bf y}$ is $N times 1$ vectors.
Any hints or ideas on finding the minimizing vectors of this problem?
It is also given that
$${bf y = A F s + z}$$ where ${bf A}$ is $Ntimes N$ matrix while
$${bf F} = begin{bmatrix}
{bf f_1} &{bf 0} \
{bf 0} & {bf f_2}
end{bmatrix}$$
where ${bf f_i}$ are $N/2times 1$ vector and ${bf Z}$ is $Ntimes 1$ vector
I am thinking of starting to start the solution as following
$$argmin _{bf{w_1},bf{w_2}}mathbb{E} ,,[[{bf s} - {bf Wy}]^H[{bf s} - {bf Wy}]]$$
But I assume given the specific block diagonal structure of matrix ${bf W}$ and ${bf F}$ it should be easier to solve ...
Thanks
optimization mean-square-error
asked Oct 21 '15 at 23:46
TyroneTyrone
244418
$endgroup$
$begingroup$
There must be something wrong with the dimensions. $mathbf W$ is a $2times N$-matrix and $mathbf y$ is a $2times 1$-vector, Then $mathbf{ Wy}$ is not possible.
$endgroup$
– callculus
Oct 22 '15 at 0:17
$begingroup$
sorry had typo thanks
$endgroup$
– Tyrone
Oct 22 '15 at 0:19
$begingroup$
I hope I´m not wrong, but it seems very similar to linear regression. Here it is $(s-Wy)'(s-Wy)=(s'-y'W')(s-Wy)=s's-s'Wy-y'W's-y'W'Wy$ But at linear regression it is optimized w.r.t $y$, not $W$. But I think it wouldn´t be bad to have a look on this subject.
$endgroup$
– callculus
Oct 22 '15 at 0:43
$begingroup$
thnk you for the reference
$endgroup$
– Tyrone
Oct 22 '15 at 1:41
add a comment |
$begingroup$
I am trying to minimize the mean square error. More precisely, I am trying to minimize the following optimization problem
$$arg min _{bf{w_1},bf{w_2}}mathbb{E} ,,[|{bf s} - {bf Wy}|^2 ]$$
$${bf W} = begin{bmatrix}
{bf w_1} &{bf 0 } \
{bf 0 } & {bf w_2 }
end{bmatrix}$$
where ${bf W}$ is a $2times N$ matrix and where ${bf w_i}$ is a $1times N/2$ for $iin[1:2]$ and ${bf 0}$ is $1times N/2$ vector while ${bf s}$ is 2 $times $ 1 and ${bf y}$ is $N times 1$ vectors.
Any hints or ideas on finding the minimizing vectors of this problem?
It is also given that
$${bf y = A F s + z}$$ where ${bf A}$ is $Ntimes N$ matrix while
$${bf F} = begin{bmatrix}
{bf f_1} &{bf 0} \
{bf 0} & {bf f_2}
end{bmatrix}$$
where ${bf f_i}$ are $N/2times 1$ vector and ${bf Z}$ is $Ntimes 1$ vector
I am thinking of starting to start the solution as following
$$argmin _{bf{w_1},bf{w_2}}mathbb{E} ,,[[{bf s} - {bf Wy}]^H[{bf s} - {bf Wy}]]$$
But I assume given the specific block diagonal structure of matrix ${bf W}$ and ${bf F}$ it should be easier to solve ...
Thanks
optimization mean-square-error
asked Oct 21 '15 at 23:46
TyroneTyrone
244418
$endgroup$
I am trying to minimize the mean square error. More precisely, I am trying to minimize the following optimization problem
$$arg min _{bf{w_1},bf{w_2}}mathbb{E} ,,[|{bf s} - {bf Wy}|^2 ]$$
$${bf W} = begin{bmatrix}
{bf w_1} &{bf 0 } \
{bf 0 } & {bf w_2 }
end{bmatrix}$$
where ${bf W}$ is a $2times N$ matrix and where ${bf w_i}$ is a $1times N/2$ for $iin[1:2]$ and ${bf 0}$ is $1times N/2$ vector while ${bf s}$ is 2 $times $ 1 and ${bf y}$ is $N times 1$ vectors.
Any hints or ideas on finding the minimizing vectors of this problem?
It is also given that
$${bf y = A F s + z}$$ where ${bf A}$ is $Ntimes N$ matrix while
$${bf F} = begin{bmatrix}
{bf f_1} &{bf 0} \
{bf 0} & {bf f_2}
end{bmatrix}$$
where ${bf f_i}$ are $N/2times 1$ vector and ${bf Z}$ is $Ntimes 1$ vector
I am thinking of starting to start the solution as following
$$argmin _{bf{w_1},bf{w_2}}mathbb{E} ,,[[{bf s} - {bf Wy}]^H[{bf s} - {bf Wy}]]$$
But I assume given the specific block diagonal structure of matrix ${bf W}$ and ${bf F}$ it should be easier to solve ...
Thanks
optimization mean-square-error
optimization mean-square-error
asked Oct 21 '15 at 23:46
TyroneTyrone
244418
asked Oct 21 '15 at 23:46
TyroneTyrone
244418
edited Oct 22 '15 at 3:00
Tyrone
asked Oct 21 '15 at 23:46
TyroneTyrone
244418
asked Oct 21 '15 at 23:46
TyroneTyrone
244418
asked Oct 21 '15 at 23:46
TyroneTyrone
244418
244418
$begingroup$
There must be something wrong with the dimensions. $mathbf W$ is a $2times N$-matrix and $mathbf y$ is a $2times 1$-vector, Then $mathbf{ Wy}$ is not possible.
$endgroup$
– callculus
Oct 22 '15 at 0:17
$begingroup$
sorry had typo thanks
$endgroup$
– Tyrone
Oct 22 '15 at 0:19
$begingroup$
I hope I´m not wrong, but it seems very similar to linear regression. Here it is $(s-Wy)'(s-Wy)=(s'-y'W')(s-Wy)=s's-s'Wy-y'W's-y'W'Wy$ But at linear regression it is optimized w.r.t $y$, not $W$. But I think it wouldn´t be bad to have a look on this subject.
$endgroup$
– callculus
Oct 22 '15 at 0:43
$begingroup$
thnk you for the reference
$endgroup$
– Tyrone
Oct 22 '15 at 1:41
add a comment |
$begingroup$
There must be something wrong with the dimensions. $mathbf W$ is a $2times N$-matrix and $mathbf y$ is a $2times 1$-vector, Then $mathbf{ Wy}$ is not possible.
$endgroup$
– callculus
Oct 22 '15 at 0:17
$begingroup$
sorry had typo thanks
$endgroup$
– Tyrone
Oct 22 '15 at 0:19
$begingroup$
I hope I´m not wrong, but it seems very similar to linear regression. Here it is $(s-Wy)'(s-Wy)=(s'-y'W')(s-Wy)=s's-s'Wy-y'W's-y'W'Wy$ But at linear regression it is optimized w.r.t $y$, not $W$. But I think it wouldn´t be bad to have a look on this subject.
$endgroup$
– callculus
Oct 22 '15 at 0:43
$begingroup$
thnk you for the reference
$endgroup$
– Tyrone
Oct 22 '15 at 1:41
$begingroup$
There must be something wrong with the dimensions. $mathbf W$ is a $2times N$-matrix and $mathbf y$ is a $2times 1$-vector, Then $mathbf{ Wy}$ is not possible.
$endgroup$
– callculus
Oct 22 '15 at 0:17
$begingroup$
There must be something wrong with the dimensions. $mathbf W$ is a $2times N$-matrix and $mathbf y$ is a $2times 1$-vector, Then $mathbf{ Wy}$ is not possible.
$endgroup$
– callculus
Oct 22 '15 at 0:17
$begingroup$
sorry had typo thanks
$endgroup$
– Tyrone
Oct 22 '15 at 0:19
$begingroup$
sorry had typo thanks
$endgroup$
– Tyrone
Oct 22 '15 at 0:19
$begingroup$
I hope I´m not wrong, but it seems very similar to linear regression. Here it is $(s-Wy)'(s-Wy)=(s'-y'W')(s-Wy)=s's-s'Wy-y'W's-y'W'Wy$ But at linear regression it is optimized w.r.t $y$, not $W$. But I think it wouldn´t be bad to have a look on this subject.
$endgroup$
– callculus
Oct 22 '15 at 0:43
$begingroup$
I hope I´m not wrong, but it seems very similar to linear regression. Here it is $(s-Wy)'(s-Wy)=(s'-y'W')(s-Wy)=s's-s'Wy-y'W's-y'W'Wy$ But at linear regression it is optimized w.r.t $y$, not $W$. But I think it wouldn´t be bad to have a look on this subject.
$endgroup$
– callculus
Oct 22 '15 at 0:43
$begingroup$
thnk you for the reference
$endgroup$
– Tyrone
Oct 22 '15 at 1:41
$begingroup$
thnk you for the reference
$endgroup$
– Tyrone
Oct 22 '15 at 1:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I assume you wish to find the non random matrix $W$ and hence its elements are not boldface in my solution. $mathbf s$ and $mathbf y$ are random vectors, hence they are boldface. First let me write the matrix ${W}
= begin{bmatrix}
{w_{1}^*} &{ 0 } \
{ 0 } & { w_{2}^* }
end{bmatrix}
$
where $*$ denotes the hermitian operation. Now all lower case vectors are column vectors.
This problem can be restated as
$arg min _{{w_1},{w_2}}mathbb{E} ,,left[leftVertbegin{bmatrix}
bf s_1^* \
bf s_{2}^*
end{bmatrix} - begin{bmatrix}
{bf y_{1}^*} &{bf 0 } \
{bf 0 } & {bf y_{2}^* }
end{bmatrix}begin{bmatrix}
w_1 \
w_{2}
end{bmatrix}rightVert^2right] $.
After this, the problem decouples to solving for $w_1$ and $w_2$. For eg. for $w_1$, we need to minimize $J =mathbb E[ ||mathbf s_1^* - mathbf y_1^* w_1||^2]$.
Just expand the inside argument and differentiate w.r.t. $w_1^*$ and put the gradient to $0$. We get the solution for $w_1$ as
$mathbb E [mathbf y_1 mathbf y_1^*] w_1 = mathbb E [mathbf y_1^* mathbf s_1^*]$.
Now, assuming you can find the correlation matrix of $mathbf y_1$ and it is invertible, and the cross correlation between $mathbf y_1$ and $mathbf s_1$, you can find $w_1$. Similarly, you can solve for $w_2$.
answered Oct 22 '15 at 0:29
shaktimanshaktiman
628
$endgroup$
$begingroup$
i am not sure... wht how did you decompose vector y?
$endgroup$
– Tyrone
Oct 22 '15 at 1:49
$begingroup$
The Nx1 vector $mathbf y$ is split into two vectors $mathbf y_1$ and $mathbf y_2$ each N/2 x1, stacked on top of each other. Hope that clears the confusion.
$endgroup$
– shaktiman
Oct 22 '15 at 3:31
add a comment |
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I assume you wish to find the non random matrix $W$ and hence its elements are not boldface in my solution. $mathbf s$ and $mathbf y$ are random vectors, hence they are boldface. First let me write the matrix ${W}
= begin{bmatrix}
{w_{1}^*} &{ 0 } \
{ 0 } & { w_{2}^* }
end{bmatrix}
$
where $*$ denotes the hermitian operation. Now all lower case vectors are column vectors.
This problem can be restated as
$arg min _{{w_1},{w_2}}mathbb{E} ,,left[leftVertbegin{bmatrix}
bf s_1^* \
bf s_{2}^*
end{bmatrix} - begin{bmatrix}
{bf y_{1}^*} &{bf 0 } \
{bf 0 } & {bf y_{2}^* }
end{bmatrix}begin{bmatrix}
w_1 \
w_{2}
end{bmatrix}rightVert^2right] $.
After this, the problem decouples to solving for $w_1$ and $w_2$. For eg. for $w_1$, we need to minimize $J =mathbb E[ ||mathbf s_1^* - mathbf y_1^* w_1||^2]$.
Just expand the inside argument and differentiate w.r.t. $w_1^*$ and put the gradient to $0$. We get the solution for $w_1$ as
$mathbb E [mathbf y_1 mathbf y_1^*] w_1 = mathbb E [mathbf y_1^* mathbf s_1^*]$.
Now, assuming you can find the correlation matrix of $mathbf y_1$ and it is invertible, and the cross correlation between $mathbf y_1$ and $mathbf s_1$, you can find $w_1$. Similarly, you can solve for $w_2$.
answered Oct 22 '15 at 0:29
shaktimanshaktiman
628
$endgroup$
$begingroup$
i am not sure... wht how did you decompose vector y?
$endgroup$
– Tyrone
Oct 22 '15 at 1:49
$begingroup$
The Nx1 vector $mathbf y$ is split into two vectors $mathbf y_1$ and $mathbf y_2$ each N/2 x1, stacked on top of each other. Hope that clears the confusion.
$endgroup$
– shaktiman
Oct 22 '15 at 3:31
add a comment |
$begingroup$
I assume you wish to find the non random matrix $W$ and hence its elements are not boldface in my solution. $mathbf s$ and $mathbf y$ are random vectors, hence they are boldface. First let me write the matrix ${W}
= begin{bmatrix}
{w_{1}^*} &{ 0 } \
{ 0 } & { w_{2}^* }
end{bmatrix}
$
where $*$ denotes the hermitian operation. Now all lower case vectors are column vectors.
This problem can be restated as
$arg min _{{w_1},{w_2}}mathbb{E} ,,left[leftVertbegin{bmatrix}
bf s_1^* \
bf s_{2}^*
end{bmatrix} - begin{bmatrix}
{bf y_{1}^*} &{bf 0 } \
{bf 0 } & {bf y_{2}^* }
end{bmatrix}begin{bmatrix}
w_1 \
w_{2}
end{bmatrix}rightVert^2right] $.
After this, the problem decouples to solving for $w_1$ and $w_2$. For eg. for $w_1$, we need to minimize $J =mathbb E[ ||mathbf s_1^* - mathbf y_1^* w_1||^2]$.
Just expand the inside argument and differentiate w.r.t. $w_1^*$ and put the gradient to $0$. We get the solution for $w_1$ as
$mathbb E [mathbf y_1 mathbf y_1^*] w_1 = mathbb E [mathbf y_1^* mathbf s_1^*]$.
Now, assuming you can find the correlation matrix of $mathbf y_1$ and it is invertible, and the cross correlation between $mathbf y_1$ and $mathbf s_1$, you can find $w_1$. Similarly, you can solve for $w_2$.
answered Oct 22 '15 at 0:29
shaktimanshaktiman
628
$endgroup$
$begingroup$
i am not sure... wht how did you decompose vector y?
$endgroup$
– Tyrone
Oct 22 '15 at 1:49
$begingroup$
The Nx1 vector $mathbf y$ is split into two vectors $mathbf y_1$ and $mathbf y_2$ each N/2 x1, stacked on top of each other. Hope that clears the confusion.
$endgroup$
– shaktiman
Oct 22 '15 at 3:31
add a comment |
$begingroup$
I assume you wish to find the non random matrix $W$ and hence its elements are not boldface in my solution. $mathbf s$ and $mathbf y$ are random vectors, hence they are boldface. First let me write the matrix ${W}
= begin{bmatrix}
{w_{1}^*} &{ 0 } \
{ 0 } & { w_{2}^* }
end{bmatrix}
$
where $*$ denotes the hermitian operation. Now all lower case vectors are column vectors.
This problem can be restated as
$arg min _{{w_1},{w_2}}mathbb{E} ,,left[leftVertbegin{bmatrix}
bf s_1^* \
bf s_{2}^*
end{bmatrix} - begin{bmatrix}
{bf y_{1}^*} &{bf 0 } \
{bf 0 } & {bf y_{2}^* }
end{bmatrix}begin{bmatrix}
w_1 \
w_{2}
end{bmatrix}rightVert^2right] $.
After this, the problem decouples to solving for $w_1$ and $w_2$. For eg. for $w_1$, we need to minimize $J =mathbb E[ ||mathbf s_1^* - mathbf y_1^* w_1||^2]$.
Just expand the inside argument and differentiate w.r.t. $w_1^*$ and put the gradient to $0$. We get the solution for $w_1$ as
$mathbb E [mathbf y_1 mathbf y_1^*] w_1 = mathbb E [mathbf y_1^* mathbf s_1^*]$.
Now, assuming you can find the correlation matrix of $mathbf y_1$ and it is invertible, and the cross correlation between $mathbf y_1$ and $mathbf s_1$, you can find $w_1$. Similarly, you can solve for $w_2$.
answered Oct 22 '15 at 0:29
shaktimanshaktiman
628
$endgroup$
I assume you wish to find the non random matrix $W$ and hence its elements are not boldface in my solution. $mathbf s$ and $mathbf y$ are random vectors, hence they are boldface. First let me write the matrix ${W}
= begin{bmatrix}
{w_{1}^*} &{ 0 } \
{ 0 } & { w_{2}^* }
end{bmatrix}
$
where $*$ denotes the hermitian operation. Now all lower case vectors are column vectors.
This problem can be restated as
$arg min _{{w_1},{w_2}}mathbb{E} ,,left[leftVertbegin{bmatrix}
bf s_1^* \
bf s_{2}^*
end{bmatrix} - begin{bmatrix}
{bf y_{1}^*} &{bf 0 } \
{bf 0 } & {bf y_{2}^* }
end{bmatrix}begin{bmatrix}
w_1 \
w_{2}
end{bmatrix}rightVert^2right] $.
After this, the problem decouples to solving for $w_1$ and $w_2$. For eg. for $w_1$, we need to minimize $J =mathbb E[ ||mathbf s_1^* - mathbf y_1^* w_1||^2]$.
Just expand the inside argument and differentiate w.r.t. $w_1^*$ and put the gradient to $0$. We get the solution for $w_1$ as
$mathbb E [mathbf y_1 mathbf y_1^*] w_1 = mathbb E [mathbf y_1^* mathbf s_1^*]$.
Now, assuming you can find the correlation matrix of $mathbf y_1$ and it is invertible, and the cross correlation between $mathbf y_1$ and $mathbf s_1$, you can find $w_1$. Similarly, you can solve for $w_2$.
answered Oct 22 '15 at 0:29
shaktimanshaktiman
628
answered Oct 22 '15 at 0:29
shaktimanshaktiman
628
answered Oct 22 '15 at 0:29
shaktimanshaktiman
628
answered Oct 22 '15 at 0:29
shaktimanshaktiman
628
628
$begingroup$
i am not sure... wht how did you decompose vector y?
$endgroup$
– Tyrone
Oct 22 '15 at 1:49
$begingroup$
The Nx1 vector $mathbf y$ is split into two vectors $mathbf y_1$ and $mathbf y_2$ each N/2 x1, stacked on top of each other. Hope that clears the confusion.
$endgroup$
– shaktiman
Oct 22 '15 at 3:31
add a comment |
$begingroup$
i am not sure... wht how did you decompose vector y?
$endgroup$
– Tyrone
Oct 22 '15 at 1:49
$begingroup$
The Nx1 vector $mathbf y$ is split into two vectors $mathbf y_1$ and $mathbf y_2$ each N/2 x1, stacked on top of each other. Hope that clears the confusion.
$endgroup$
– shaktiman
Oct 22 '15 at 3:31
$begingroup$
i am not sure... wht how did you decompose vector y?
$endgroup$
– Tyrone
Oct 22 '15 at 1:49
$begingroup$
i am not sure... wht how did you decompose vector y?
$endgroup$
– Tyrone
Oct 22 '15 at 1:49
$begingroup$
The Nx1 vector $mathbf y$ is split into two vectors $mathbf y_1$ and $mathbf y_2$ each N/2 x1, stacked on top of each other. Hope that clears the confusion.
$endgroup$
– shaktiman
Oct 22 '15 at 3:31
$begingroup$
The Nx1 vector $mathbf y$ is split into two vectors $mathbf y_1$ and $mathbf y_2$ each N/2 x1, stacked on top of each other. Hope that clears the confusion.
$endgroup$
– shaktiman
Oct 22 '15 at 3:31
add a comment |
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$begingroup$
There must be something wrong with the dimensions. $mathbf W$ is a $2times N$-matrix and $mathbf y$ is a $2times 1$-vector, Then $mathbf{ Wy}$ is not possible.
$endgroup$
– callculus
Oct 22 '15 at 0:17
$begingroup$
sorry had typo thanks
$endgroup$
– Tyrone
Oct 22 '15 at 0:19
$begingroup$
I hope I´m not wrong, but it seems very similar to linear regression. Here it is $(s-Wy)'(s-Wy)=(s'-y'W')(s-Wy)=s's-s'Wy-y'W's-y'W'Wy$ But at linear regression it is optimized w.r.t $y$, not $W$. But I think it wouldn´t be bad to have a look on this subject.
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– callculus
Oct 22 '15 at 0:43
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thnk you for the reference
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– Tyrone
Oct 22 '15 at 1:41