Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Generalized variance is the determinant of correlation matrix. Does increasing the off-diagonal entries (correlation coefficients) decreases the determinant? Is a proof available? All elements are positive. Can we deduce from Hadamard inequality of determinant?

share|improve this question
    
The determinant of the covariance matrix could be considered a generalization of variance, in that it's equal to the scalar variance in the case of dimension 1. But the determinant of the correlation matrix, as opposed to the covariance matrix, is not in that sense a generalization of the variance. –  Michael Hardy Aug 27 '11 at 11:46
    
Thanks for the proper definition. –  shakera Aug 27 '11 at 12:15
add comment

1 Answer

It can do either. $\;$ Suppose the correlation matrix is $\begin{bmatrix} 1 & x \\ x & 1 \end{bmatrix}$.

$\operatorname{det}\left(\begin{bmatrix} 1 & x \\ x & 1 \end{bmatrix}\right) = 1\cdot 1-x\cdot x = 1-x^2$

If $x<0$ then increasing the off-diagonal entries increases the determinant.
If $0<x$ then increasing the off-diagonal entires decreases the determinant.

share|improve this answer
    
Thanks... Does it generalizes for $N$. Let us supposes all elements of the correlation matrix are positive. –  shakera Aug 27 '11 at 9:25
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.