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What is the value of $E[W\otimes W]$ for Wishart R.V. $W$?

$\otimes$ refers to Kronecker product

I found related formula for $E[WAW]$ on page 467 of Seber's Matrix handbook, wondering if $E[W\otimes W]$ can be extracted from it.

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I believe that there is an error in Seber's formula and the correct form is $$ \mathbf{E}(WAW) = m[\Sigma A \Sigma + \mathrm{Tr}(A\Sigma) \Sigma] + m^2 \Sigma A \Sigma $$ (try this and the orgiginal formula with $m = d = 1$).

Then, we can use the vec-trick to write, for any $A$, \begin{align*} \mathbf{E}(W \otimes W) \mathrm{vec}(A) &= \mathrm{vec}\{\mathbf{E}(W A W)\} \\ &= m(m+1) \mathrm{vec}(\Sigma A \Sigma) + m\mathrm{Tr}(A\Sigma) \mathrm{vec},(\Sigma). \end{align*} where we repeatedly commute the expectation with the $\mathrm{vec}$, since it is a linear operator.

But recall that $\mathrm{Tr}(A^{\mathrm{T}}B) = \mathrm{vec}(A)^{\mathrm{T}}\mathrm{vec}(B)$. Using this and the vec-trick again we have that \begin{align*} \mathbf{E}(W \otimes W) \mathrm{vec}(A) &= m(m+1) (\Sigma \otimes \Sigma) \mathrm{vec}(A) + m (\mathrm{vec}(\Sigma)^{\mathrm{T}} \mathrm{vec}(A) )\mathrm{vec}(\Sigma) \\ &= \{m(m+1) (\Sigma \otimes \Sigma) + m \mathrm{vec}(\Sigma)\mathrm{vec}(\Sigma)^{\mathrm{T}}\} \mathrm{vec}(A). \end{align*}

SInce this holds for any $A$, we conclude that $$ \mathbf{E}(W \otimes W) = m(m+1) (\Sigma \otimes \Sigma) + m \mathrm{vec}(\Sigma)\mathrm{vec}(\Sigma)^{\mathrm{T}}. $$

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  • $\begingroup$ \mathrm{vec} does not yield context-dependent horizontal spacing the way \operatorname{vec} does. That is why you see $m\mathrm{vec}(\Sigma)$ rather than $m\operatorname{vec}(\Sigma).$ $\endgroup$ Commented Jan 10 at 22:46

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