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?
Generalized variance
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matrices
correlation
determinant
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0The 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. – 2011-08-27
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0Thanks for the proper definition. – 2011-08-27
1 Answers
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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
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1Thanks... Does it generalizes for $N$. Let us supposes all elements of the correlation matrix are positive. – 2011-08-27