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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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