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What is the expected value, $ \mathbb{E}\left[ I \otimes \left( \operatorname{diag}(ZZ^T\mathbf{1}) - ZZ^T\right)\right]$ where $Z \sim N(0, \sigma^2I) $? The kronecker product is where the confusion is setting in.

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$I \otimes B$ is just a block diagonal matrix with diagonal blocks equal to $B$. So $\mathbb{E}[I \otimes B] = I \otimes \mathbb{E}[B]$. –  Steve Huntsman Apr 9 '13 at 17:48
    
Can you go further and continue with the part involving $Z's$, which you refer to indirectly as $B$? I ask because I believe $ZZ^T$ may be Wishart distributed given the Gaussian assumptions. I'd like to see the extent to which one might compute the expectation. –  Pron Apr 9 '13 at 19:56

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This problem appears straightforward....? From Steve's comment,
$$\mathbb{E}[\mathrm{diag}(\bf{ZZ}^T)\bf{1}-ZZ^T]=\mathbb{E}[\mathrm{diag}(\bf{ZZ}^T)\bf{1}]-\mathbb{E}[ZZ^T]=0.$$

There must be some typos in the statement of this problem.

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