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Let $\mathbf{A}\in\mathbb{R}^{p\times n} (n\ge p)$ be a positive definite symmetric matrix having a Wishart distribution with mean $\mathbf{0}$ and covariance $\boldsymbol\Sigma\otimes \mathbf{I}$. Let the eigenvalue-eigenvector pair of $\boldsymbol\Sigma$ be written as \begin{equation} \boldsymbol\Sigma\Gamma = \Gamma\Delta \end{equation} where, $\Gamma$ is the eigenvector matrix and $\Delta$ is the eigenvalue matrix with diagonal entries $\left(\delta_1,\cdots,\delta_p\right)$. Now, let $\mathbf{U} = \sqrt{n}\left(\Gamma'\mathbf{A}\Gamma-n\Delta\right)$. One can see that the mean of $\mathbf{U}$ is $\mathbf{0}$.

How to prove the following \begin{equation} \text{E}\left[u_{ij}u_{gh}\right] = \delta_i\delta_j\left(\delta_{ih}\delta_{jg} + \delta_{ig}\delta_{jh}\right) \end{equation} where, $\text{E}$ indicates expected value.

The following equation will be handy in finding the above expectation: \begin{equation} \mathbf{S} = \mathbf{AA}^T\\ \text{cov}\left(\text{vec }\mathbf{S}\right) = n\left(\mathbf{I}+K_p\right)\left(\boldsymbol\Sigma\otimes\boldsymbol\Sigma\right) \end{equation} where, $\mathbf{K}_p$ is a commutation matrix of size $p^2\times p^2$ and $\mathbf{I}$ is an identity matrix of same size. where $\left(\text{vec }\mathbf{S}\right)$ indicates that the columns of $\mathbf{S}$ have been stacked on top of each other forming a vector.

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    $\begingroup$ What is delta with double subscripts? From your explanation of your notation you only refer to single subscripted delta. $\endgroup$ – hejseb Sep 2 '15 at 19:18
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    $\begingroup$ Forgot to mention. $\delta$ with single subscript is the eigenvalue while with double subscript is kronecker delta function $\endgroup$ – user146290 Sep 2 '15 at 19:33
  • $\begingroup$ Makes sense. You also mention that the covariance of A is something (in the beginning), and then later on that E(AA') (which is the covariance in this case) is something else. Are these expressions correct? I suspect that the final expression should be E(UU')? $\endgroup$ – hejseb Sep 2 '15 at 19:41
  • $\begingroup$ Corrected the mistake. Thanks for pointing out. "cov" indicates covariance. $\endgroup$ – user146290 Sep 2 '15 at 19:53

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