Find the maximum value without computing matrix
$begingroup$
I have a optimization problem, which is to find a certain $h^*$ that:$$h^* = argmax(h'alpha-frac{kappa}{2}h'Sigma h)$$
where $alpha$ is a $(n times 1)$ vector and $Sigma$ is a $(ntimes n)$ matrix.
The $Sigma$ matrix is of form $$Sigma = XFX' + D$$
where $X$ is a $(ntimes k)$ matrix and $F$ is a covariance matrix of factor loadings, and $D$ is a diagonal matrix. $D = diag(sigma_1^2,..,sigma_n^2)$.
How can I write a function to compute $h^*$ without computing the a $(ntimes n)$ matrix?
linear-algebra optimization numerical-optimization
asked Dec 10 '18 at 20:05
QingQing
508
$endgroup$
|
show 1 more comment
$begingroup$
I have a optimization problem, which is to find a certain $h^*$ that:$$h^* = argmax(h'alpha-frac{kappa}{2}h'Sigma h)$$
where $alpha$ is a $(n times 1)$ vector and $Sigma$ is a $(ntimes n)$ matrix.
The $Sigma$ matrix is of form $$Sigma = XFX' + D$$
where $X$ is a $(ntimes k)$ matrix and $F$ is a covariance matrix of factor loadings, and $D$ is a diagonal matrix. $D = diag(sigma_1^2,..,sigma_n^2)$.
How can I write a function to compute $h^*$ without computing the a $(ntimes n)$ matrix?
linear-algebra optimization numerical-optimization
asked Dec 10 '18 at 20:05
QingQing
508
$endgroup$
$begingroup$
What kind of matrix is $Sigma$? Is it perchance diagonal, symmetric, or hermitian?
$endgroup$
– I like Serena
Dec 10 '18 at 20:18
$begingroup$
@IlikeSerena I have modified the question
$endgroup$
– Qing
Dec 10 '18 at 20:48
$begingroup$
BFGS only requires function evaluations and gradients, which can be computed efficiently for your problem
$endgroup$
– LinAlg
Dec 10 '18 at 21:18
$begingroup$
@LinAlg Hi, Can you explain it more details? Thank you very much!
$endgroup$
– Qing
Dec 10 '18 at 21:23
$begingroup$
What's this $h^*$? If it's the $text{argmax}$ of the given expression, shouldn't it just be $h$?
$endgroup$
– I like Serena
Dec 10 '18 at 22:00
|
show 1 more comment
$begingroup$
I have a optimization problem, which is to find a certain $h^*$ that:$$h^* = argmax(h'alpha-frac{kappa}{2}h'Sigma h)$$
where $alpha$ is a $(n times 1)$ vector and $Sigma$ is a $(ntimes n)$ matrix.
The $Sigma$ matrix is of form $$Sigma = XFX' + D$$
where $X$ is a $(ntimes k)$ matrix and $F$ is a covariance matrix of factor loadings, and $D$ is a diagonal matrix. $D = diag(sigma_1^2,..,sigma_n^2)$.
How can I write a function to compute $h^*$ without computing the a $(ntimes n)$ matrix?
linear-algebra optimization numerical-optimization
asked Dec 10 '18 at 20:05
QingQing
508
$endgroup$
I have a optimization problem, which is to find a certain $h^*$ that:$$h^* = argmax(h'alpha-frac{kappa}{2}h'Sigma h)$$
where $alpha$ is a $(n times 1)$ vector and $Sigma$ is a $(ntimes n)$ matrix.
The $Sigma$ matrix is of form $$Sigma = XFX' + D$$
where $X$ is a $(ntimes k)$ matrix and $F$ is a covariance matrix of factor loadings, and $D$ is a diagonal matrix. $D = diag(sigma_1^2,..,sigma_n^2)$.
How can I write a function to compute $h^*$ without computing the a $(ntimes n)$ matrix?
linear-algebra optimization numerical-optimization
linear-algebra optimization numerical-optimization
asked Dec 10 '18 at 20:05
QingQing
508
asked Dec 10 '18 at 20:05
QingQing
508
edited Dec 10 '18 at 20:45
Qing
asked Dec 10 '18 at 20:05
QingQing
508
asked Dec 10 '18 at 20:05
QingQing
508
asked Dec 10 '18 at 20:05
QingQing
508
508
$begingroup$
What kind of matrix is $Sigma$? Is it perchance diagonal, symmetric, or hermitian?
$endgroup$
– I like Serena
Dec 10 '18 at 20:18
$begingroup$
@IlikeSerena I have modified the question
$endgroup$
– Qing
Dec 10 '18 at 20:48
$begingroup$
BFGS only requires function evaluations and gradients, which can be computed efficiently for your problem
$endgroup$
– LinAlg
Dec 10 '18 at 21:18
$begingroup$
@LinAlg Hi, Can you explain it more details? Thank you very much!
$endgroup$
– Qing
Dec 10 '18 at 21:23
$begingroup$
What's this $h^*$? If it's the $text{argmax}$ of the given expression, shouldn't it just be $h$?
$endgroup$
– I like Serena
Dec 10 '18 at 22:00
|
show 1 more comment
$begingroup$
What kind of matrix is $Sigma$? Is it perchance diagonal, symmetric, or hermitian?
$endgroup$
– I like Serena
Dec 10 '18 at 20:18
$begingroup$
@IlikeSerena I have modified the question
$endgroup$
– Qing
Dec 10 '18 at 20:48
$begingroup$
BFGS only requires function evaluations and gradients, which can be computed efficiently for your problem
$endgroup$
– LinAlg
Dec 10 '18 at 21:18
$begingroup$
@LinAlg Hi, Can you explain it more details? Thank you very much!
$endgroup$
– Qing
Dec 10 '18 at 21:23
$begingroup$
What's this $h^*$? If it's the $text{argmax}$ of the given expression, shouldn't it just be $h$?
$endgroup$
– I like Serena
Dec 10 '18 at 22:00
$begingroup$
What kind of matrix is $Sigma$? Is it perchance diagonal, symmetric, or hermitian?
$endgroup$
– I like Serena
Dec 10 '18 at 20:18
$begingroup$
What kind of matrix is $Sigma$? Is it perchance diagonal, symmetric, or hermitian?
$endgroup$
– I like Serena
Dec 10 '18 at 20:18
$begingroup$
@IlikeSerena I have modified the question
$endgroup$
– Qing
Dec 10 '18 at 20:48
$begingroup$
@IlikeSerena I have modified the question
$endgroup$
– Qing
Dec 10 '18 at 20:48
$begingroup$
BFGS only requires function evaluations and gradients, which can be computed efficiently for your problem
$endgroup$
– LinAlg
Dec 10 '18 at 21:18
$begingroup$
BFGS only requires function evaluations and gradients, which can be computed efficiently for your problem
$endgroup$
– LinAlg
Dec 10 '18 at 21:18
$begingroup$
@LinAlg Hi, Can you explain it more details? Thank you very much!
$endgroup$
– Qing
Dec 10 '18 at 21:23
$begingroup$
@LinAlg Hi, Can you explain it more details? Thank you very much!
$endgroup$
– Qing
Dec 10 '18 at 21:23
$begingroup$
What's this $h^*$? If it's the $text{argmax}$ of the given expression, shouldn't it just be $h$?
$endgroup$
– I like Serena
Dec 10 '18 at 22:00
$begingroup$
What's this $h^*$? If it's the $text{argmax}$ of the given expression, shouldn't it just be $h$?
$endgroup$
– I like Serena
Dec 10 '18 at 22:00
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
A covariance matrix is symmetric. After multiplying with the presumed orthogonal matrix $X$ and its transpose, that will still be the case. The diagonal matrix is also symmetric. So their sum will be as well. In particular it means that $Sigma'=Sigma$.
When $h$ is optimal, each partial derivative of the expression must be $0$. Starting with the first:
$$frac{partial}{partial h_1}(h'alpha -frackappa 2h'Sigma h)
= e_1'alpha - frackappa 2e_1'Sigma h - frackappa 2h'Sigma e_1
= e_1'alpha - kappa e_1'Sigma h
= alpha_1 - kappa sigma_1 h = 0
$$
where $sigma_1$ is the first row vector of $Sigma$.
Extending to a vector and to a matrix again:
$$nabla (h'alpha -frackappa 2h'Sigma h) = alpha - kappa Sigma h = 0$$
We can find $h^*$ now by solving:
$$Sigma h^* = frac 1kappa alpha implies h^*=frac 1kappaSigma^{-1}alpha$$
This can be done with for instance Gaussian elimination, or otherwise with a more advanced method that is more robust.
answered Dec 11 '18 at 7:52
I like SerenaI like Serena
4,2221722
$endgroup$
1
$begingroup$
Thank you so much!
$endgroup$
– Qing
Dec 11 '18 at 12:55
$begingroup$
@Qing how does this answer your question if you do not want to compute $Sigma$? Consider using the generalized Sherman-Morrison-Woodbury formula.
$endgroup$
– LinAlg
Dec 11 '18 at 17:45
$begingroup$
@LinAlg, we don't actually need to compute $Sigma$. Since $F$ is a covariance matrix, we're talking about a probability cloud around a mean. Statistically the most likely solution is if we substitute $D$ for $Sigma$. I only realized this just now though. My answer was just to solve the optimization problem with a solution that is as simple as possible.
$endgroup$
– I like Serena
Dec 11 '18 at 18:45
$begingroup$
@IlikeSerena there are many ways to cope with uncertainty, just inserting the most likely parameter value is not always the best strategy
$endgroup$
– LinAlg
Dec 11 '18 at 18:56
$begingroup$
@LinAlg I think you are right, it's not solving the major issue of the $Sigma$ here though.
$endgroup$
– Qing
Dec 11 '18 at 21:36
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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oldest
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$begingroup$
A covariance matrix is symmetric. After multiplying with the presumed orthogonal matrix $X$ and its transpose, that will still be the case. The diagonal matrix is also symmetric. So their sum will be as well. In particular it means that $Sigma'=Sigma$.
When $h$ is optimal, each partial derivative of the expression must be $0$. Starting with the first:
$$frac{partial}{partial h_1}(h'alpha -frackappa 2h'Sigma h)
= e_1'alpha - frackappa 2e_1'Sigma h - frackappa 2h'Sigma e_1
= e_1'alpha - kappa e_1'Sigma h
= alpha_1 - kappa sigma_1 h = 0
$$
where $sigma_1$ is the first row vector of $Sigma$.
Extending to a vector and to a matrix again:
$$nabla (h'alpha -frackappa 2h'Sigma h) = alpha - kappa Sigma h = 0$$
We can find $h^*$ now by solving:
$$Sigma h^* = frac 1kappa alpha implies h^*=frac 1kappaSigma^{-1}alpha$$
This can be done with for instance Gaussian elimination, or otherwise with a more advanced method that is more robust.
answered Dec 11 '18 at 7:52
I like SerenaI like Serena
4,2221722
$endgroup$
1
$begingroup$
Thank you so much!
$endgroup$
– Qing
Dec 11 '18 at 12:55
$begingroup$
@Qing how does this answer your question if you do not want to compute $Sigma$? Consider using the generalized Sherman-Morrison-Woodbury formula.
$endgroup$
– LinAlg
Dec 11 '18 at 17:45
$begingroup$
@LinAlg, we don't actually need to compute $Sigma$. Since $F$ is a covariance matrix, we're talking about a probability cloud around a mean. Statistically the most likely solution is if we substitute $D$ for $Sigma$. I only realized this just now though. My answer was just to solve the optimization problem with a solution that is as simple as possible.
$endgroup$
– I like Serena
Dec 11 '18 at 18:45
$begingroup$
@IlikeSerena there are many ways to cope with uncertainty, just inserting the most likely parameter value is not always the best strategy
$endgroup$
– LinAlg
Dec 11 '18 at 18:56
$begingroup$
@LinAlg I think you are right, it's not solving the major issue of the $Sigma$ here though.
$endgroup$
– Qing
Dec 11 '18 at 21:36
add a comment |
$begingroup$
A covariance matrix is symmetric. After multiplying with the presumed orthogonal matrix $X$ and its transpose, that will still be the case. The diagonal matrix is also symmetric. So their sum will be as well. In particular it means that $Sigma'=Sigma$.
When $h$ is optimal, each partial derivative of the expression must be $0$. Starting with the first:
$$frac{partial}{partial h_1}(h'alpha -frackappa 2h'Sigma h)
= e_1'alpha - frackappa 2e_1'Sigma h - frackappa 2h'Sigma e_1
= e_1'alpha - kappa e_1'Sigma h
= alpha_1 - kappa sigma_1 h = 0
$$
where $sigma_1$ is the first row vector of $Sigma$.
Extending to a vector and to a matrix again:
$$nabla (h'alpha -frackappa 2h'Sigma h) = alpha - kappa Sigma h = 0$$
We can find $h^*$ now by solving:
$$Sigma h^* = frac 1kappa alpha implies h^*=frac 1kappaSigma^{-1}alpha$$
This can be done with for instance Gaussian elimination, or otherwise with a more advanced method that is more robust.
answered Dec 11 '18 at 7:52
I like SerenaI like Serena
4,2221722
$endgroup$
1
$begingroup$
Thank you so much!
$endgroup$
– Qing
Dec 11 '18 at 12:55
$begingroup$
@Qing how does this answer your question if you do not want to compute $Sigma$? Consider using the generalized Sherman-Morrison-Woodbury formula.
$endgroup$
– LinAlg
Dec 11 '18 at 17:45
$begingroup$
@LinAlg, we don't actually need to compute $Sigma$. Since $F$ is a covariance matrix, we're talking about a probability cloud around a mean. Statistically the most likely solution is if we substitute $D$ for $Sigma$. I only realized this just now though. My answer was just to solve the optimization problem with a solution that is as simple as possible.
$endgroup$
– I like Serena
Dec 11 '18 at 18:45
$begingroup$
@IlikeSerena there are many ways to cope with uncertainty, just inserting the most likely parameter value is not always the best strategy
$endgroup$
– LinAlg
Dec 11 '18 at 18:56
$begingroup$
@LinAlg I think you are right, it's not solving the major issue of the $Sigma$ here though.
$endgroup$
– Qing
Dec 11 '18 at 21:36
add a comment |
$begingroup$
A covariance matrix is symmetric. After multiplying with the presumed orthogonal matrix $X$ and its transpose, that will still be the case. The diagonal matrix is also symmetric. So their sum will be as well. In particular it means that $Sigma'=Sigma$.
When $h$ is optimal, each partial derivative of the expression must be $0$. Starting with the first:
$$frac{partial}{partial h_1}(h'alpha -frackappa 2h'Sigma h)
= e_1'alpha - frackappa 2e_1'Sigma h - frackappa 2h'Sigma e_1
= e_1'alpha - kappa e_1'Sigma h
= alpha_1 - kappa sigma_1 h = 0
$$
where $sigma_1$ is the first row vector of $Sigma$.
Extending to a vector and to a matrix again:
$$nabla (h'alpha -frackappa 2h'Sigma h) = alpha - kappa Sigma h = 0$$
We can find $h^*$ now by solving:
$$Sigma h^* = frac 1kappa alpha implies h^*=frac 1kappaSigma^{-1}alpha$$
This can be done with for instance Gaussian elimination, or otherwise with a more advanced method that is more robust.
answered Dec 11 '18 at 7:52
I like SerenaI like Serena
4,2221722
$endgroup$
A covariance matrix is symmetric. After multiplying with the presumed orthogonal matrix $X$ and its transpose, that will still be the case. The diagonal matrix is also symmetric. So their sum will be as well. In particular it means that $Sigma'=Sigma$.
When $h$ is optimal, each partial derivative of the expression must be $0$. Starting with the first:
$$frac{partial}{partial h_1}(h'alpha -frackappa 2h'Sigma h)
= e_1'alpha - frackappa 2e_1'Sigma h - frackappa 2h'Sigma e_1
= e_1'alpha - kappa e_1'Sigma h
= alpha_1 - kappa sigma_1 h = 0
$$
where $sigma_1$ is the first row vector of $Sigma$.
Extending to a vector and to a matrix again:
$$nabla (h'alpha -frackappa 2h'Sigma h) = alpha - kappa Sigma h = 0$$
We can find $h^*$ now by solving:
$$Sigma h^* = frac 1kappa alpha implies h^*=frac 1kappaSigma^{-1}alpha$$
This can be done with for instance Gaussian elimination, or otherwise with a more advanced method that is more robust.
answered Dec 11 '18 at 7:52
I like SerenaI like Serena
4,2221722
answered Dec 11 '18 at 7:52
I like SerenaI like Serena
4,2221722
answered Dec 11 '18 at 7:52
I like SerenaI like Serena
4,2221722
answered Dec 11 '18 at 7:52
I like SerenaI like Serena
4,2221722
4,2221722
1
$begingroup$
Thank you so much!
$endgroup$
– Qing
Dec 11 '18 at 12:55
$begingroup$
@Qing how does this answer your question if you do not want to compute $Sigma$? Consider using the generalized Sherman-Morrison-Woodbury formula.
$endgroup$
– LinAlg
Dec 11 '18 at 17:45
$begingroup$
@LinAlg, we don't actually need to compute $Sigma$. Since $F$ is a covariance matrix, we're talking about a probability cloud around a mean. Statistically the most likely solution is if we substitute $D$ for $Sigma$. I only realized this just now though. My answer was just to solve the optimization problem with a solution that is as simple as possible.
$endgroup$
– I like Serena
Dec 11 '18 at 18:45
$begingroup$
@IlikeSerena there are many ways to cope with uncertainty, just inserting the most likely parameter value is not always the best strategy
$endgroup$
– LinAlg
Dec 11 '18 at 18:56
$begingroup$
@LinAlg I think you are right, it's not solving the major issue of the $Sigma$ here though.
$endgroup$
– Qing
Dec 11 '18 at 21:36
add a comment |
1
$begingroup$
Thank you so much!
$endgroup$
– Qing
Dec 11 '18 at 12:55
$begingroup$
@Qing how does this answer your question if you do not want to compute $Sigma$? Consider using the generalized Sherman-Morrison-Woodbury formula.
$endgroup$
– LinAlg
Dec 11 '18 at 17:45
$begingroup$
@LinAlg, we don't actually need to compute $Sigma$. Since $F$ is a covariance matrix, we're talking about a probability cloud around a mean. Statistically the most likely solution is if we substitute $D$ for $Sigma$. I only realized this just now though. My answer was just to solve the optimization problem with a solution that is as simple as possible.
$endgroup$
– I like Serena
Dec 11 '18 at 18:45
$begingroup$
@IlikeSerena there are many ways to cope with uncertainty, just inserting the most likely parameter value is not always the best strategy
$endgroup$
– LinAlg
Dec 11 '18 at 18:56
$begingroup$
@LinAlg I think you are right, it's not solving the major issue of the $Sigma$ here though.
$endgroup$
– Qing
Dec 11 '18 at 21:36
1
1
$begingroup$
Thank you so much!
$endgroup$
– Qing
Dec 11 '18 at 12:55
$begingroup$
Thank you so much!
$endgroup$
– Qing
Dec 11 '18 at 12:55
$begingroup$
@Qing how does this answer your question if you do not want to compute $Sigma$? Consider using the generalized Sherman-Morrison-Woodbury formula.
$endgroup$
– LinAlg
Dec 11 '18 at 17:45
$begingroup$
@Qing how does this answer your question if you do not want to compute $Sigma$? Consider using the generalized Sherman-Morrison-Woodbury formula.
$endgroup$
– LinAlg
Dec 11 '18 at 17:45
$begingroup$
@LinAlg, we don't actually need to compute $Sigma$. Since $F$ is a covariance matrix, we're talking about a probability cloud around a mean. Statistically the most likely solution is if we substitute $D$ for $Sigma$. I only realized this just now though. My answer was just to solve the optimization problem with a solution that is as simple as possible.
$endgroup$
– I like Serena
Dec 11 '18 at 18:45
$begingroup$
@LinAlg, we don't actually need to compute $Sigma$. Since $F$ is a covariance matrix, we're talking about a probability cloud around a mean. Statistically the most likely solution is if we substitute $D$ for $Sigma$. I only realized this just now though. My answer was just to solve the optimization problem with a solution that is as simple as possible.
$endgroup$
– I like Serena
Dec 11 '18 at 18:45
$begingroup$
@IlikeSerena there are many ways to cope with uncertainty, just inserting the most likely parameter value is not always the best strategy
$endgroup$
– LinAlg
Dec 11 '18 at 18:56
$begingroup$
@IlikeSerena there are many ways to cope with uncertainty, just inserting the most likely parameter value is not always the best strategy
$endgroup$
– LinAlg
Dec 11 '18 at 18:56
$begingroup$
@LinAlg I think you are right, it's not solving the major issue of the $Sigma$ here though.
$endgroup$
– Qing
Dec 11 '18 at 21:36
$begingroup$
@LinAlg I think you are right, it's not solving the major issue of the $Sigma$ here though.
$endgroup$
– Qing
Dec 11 '18 at 21:36
add a comment |
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$begingroup$
What kind of matrix is $Sigma$? Is it perchance diagonal, symmetric, or hermitian?
$endgroup$
– I like Serena
Dec 10 '18 at 20:18
$begingroup$
@IlikeSerena I have modified the question
$endgroup$
– Qing
Dec 10 '18 at 20:48
$begingroup$
BFGS only requires function evaluations and gradients, which can be computed efficiently for your problem
$endgroup$
– LinAlg
Dec 10 '18 at 21:18
$begingroup$
@LinAlg Hi, Can you explain it more details? Thank you very much!
$endgroup$
– Qing
Dec 10 '18 at 21:23
$begingroup$
What's this $h^*$? If it's the $text{argmax}$ of the given expression, shouldn't it just be $h$?
$endgroup$
– I like Serena
Dec 10 '18 at 22:00