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My book states that the variance-covariance matrix has the following formula:

$$\Sigma = V^{1/2}\times P\times V^{1/2}$$

where $V^{1/2}$ is the square-root of the variance matrix and $P$ is the correlation matrix

I don't see how this equality is true though because when I multiply it out, I get a matrix which only has values $\sigma( i,j)$ across the main diagonal and $0$'s for all off diagonals, which is not the variance-covariance matrix. Is there something that I'm not seeing?

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  • $\begingroup$ $V$ isn't the matrix with the standard deviations on the diagonal, it's the matrix with the variances on the diagonal. $V^{1/2}$, therefore, has the standard deviations on the diagonal. $\endgroup$ – Sean Lake Oct 8 '16 at 23:26
  • $\begingroup$ Caught that mistake. How do I show the formula though? $\endgroup$ – nodel Oct 9 '16 at 0:35
  • $\begingroup$ As far as I know, that formula is the definition of the correlation matrix. $\endgroup$ – Sean Lake Oct 9 '16 at 0:36
  • $\begingroup$ The correlation matrix is V^(-1/2) * Σ * V*(-1/2) $\endgroup$ – nodel Oct 9 '16 at 0:43
  • $\begingroup$ Right, so right and left multiply by $V^{1/2}$ to get from $P=\ldots$ to $\Sigma=\ldots$. $\endgroup$ – Sean Lake Oct 9 '16 at 0:53

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