Timeline for Given a correlation matrix $B$. What correlation matrix A (maximizes / minimizes) the following: det(A+B)
Current License: CC BY-SA 3.0
15 events
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| Oct 1, 2013 at 13:20 | comment | added | user40607 | However, before I go on there is something I do not understand about @suvrit reasoning to fulfill the equality. You can have a function f(x) that is always lesser or equal than g(x), $f(x)\leq g(x)$. It might so happen that there is an xo for which the equality is fulfilled $f(xo)=g(xo)$, and yet there might be an x1 such that $f(x1)>f(xo)$. As an example you can take $f(x)=x^4,g(x)=x^2$ in the interval [-.5,.5]. | |
| Sep 30, 2013 at 18:45 | comment | added | user40607 | A family of matrices A that fulfills the equality is $A=I-\alpha (B-I)$ with $\alpha \geq 0$. A has the same eigenvectors than B, but the order of the eigenvalues is inverted because for each eigenvector: $λ(A)=1−α(λ(B)−1)$. <br/> A has 1s in the diagonal and stays semi-definite positive as long as its minimum eigenvalue is: $λ_{min}(A)=1−α(λ_{max}(B)−1)≥0$.<br/> Based on the example I gave above I suspect that is better to use the greatest possible α. @suvrit | |
| Sep 29, 2013 at 15:57 | history | edited | Suvrit | CC BY-SA 3.0 |
corrected the answer in light of comments
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| Sep 29, 2013 at 15:54 | comment | added | Suvrit | well, I guess then your only hope is to solve a convex optimization problem to maximize the determinant; there might not be a closed form way to maximize the eigenvalues of $A$. | |
| Sep 29, 2013 at 15:51 | comment | added | user40607 | The A as constructed above is not semi-definite positive. So it can not be a correlation matrix. @suvrit | |
| Sep 29, 2013 at 5:20 | comment | added | user40607 | Yes it is symmetric. | |
| Sep 29, 2013 at 2:53 | comment | added | Suvrit | why is this $A$ not symmetric? $B$ was symmetric to begin with, so $A$ as constructed above will also be symmetric! | |
| Sep 29, 2013 at 2:46 | comment | added | user40607 | A(i,j)=-B(i,j) works in 2 dimensions. But for larger dimensions this could be impossible because A needs to be hermitian. For example: if B=[1 1 1;1 1 1;1 1 1] then A should be A=[1 -1 -1;-1 1 -1;-1 -1 1] which is not hermitian. | |
| Sep 29, 2013 at 2:15 | history | edited | Suvrit | CC BY-SA 3.0 |
modified answer to fix prior incorrect conclusion.
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| Sep 29, 2013 at 2:11 | comment | added | Suvrit | I thought a bit more about it; as the above inequality shows, $A$ will have the same eigenvectors as $B$ (just permuted in opposite order); but it is not immediate what the eigenvalues should be---without that, a closed form solution won't be possible, and this turns into a convex optimization problem, which seems overkill.... | |
| Sep 29, 2013 at 1:51 | comment | added | Suvrit | Actually that is true---the simple diagonal $A$ is not going to work because correlation matrices can easily have eigenvalues larger than 1. So $A=I_n$ is not the answer, but the answer will be obtained by finding an $A$ for which the 2nd inequality listed above becomes an equality. | |
| Sep 28, 2013 at 22:31 | comment | added | user40607 | Actually I think there might be something wrong in that proof. Consider B=[1 .5;.5 1], if A=[1 0;0 1] the det(A+B) is less than if A=[1 -.5;-.5 1]. | |
| Sep 28, 2013 at 18:50 | vote | accept | user40607 | ||
| Sep 28, 2013 at 22:37 | |||||
| Sep 28, 2013 at 18:37 | comment | added | user40607 | And what A would minimize det(A+B)? | |
| Sep 28, 2013 at 17:26 | history | answered | Suvrit | CC BY-SA 3.0 |