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Looking for the variance of $S=\sigma _{1,3}-\sigma _{1,4}-\sigma _{2,3}+\sigma _{2,4}$, where $\sigma_{i,j}$ are Wishart-distributed elements of the random matrix

$$\Sigma =\left( \begin{array}{cccc} \sigma _1^2 & \sigma _{1,2} & \sigma _{1,3} & \sigma _{1,4} \\ \sigma _{1,2} & \sigma _2^2 & \sigma _{2,3} & \sigma _{2,4} \\ \sigma _{1,3} & \sigma _{2,3} & \sigma _3^2 & \sigma _{3,4} \\ \sigma _{1,4} & \sigma _{2,4} & \sigma _{3,4} & \sigma _4^2 \\ \end{array} \right)$$ the $m$-sample estimation of the covariance matrix of 4 multivariate Gaussian distributed random variables with $n$ observations each.

(I tried Math Stack Exchange with no result).

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Let $a = (\begin{matrix} 1 & -1 & -1 &1\end{matrix})$ and $\sigma = (\begin{matrix} \sigma_{1,3} & \sigma_{1,4} &\sigma_{2,3} & \sigma_{2,4}\end{matrix})'$.

Then $S=a\sigma$. Let $V$ be the variance-covariance matrix of $\sigma$. We have $$\mathrm{Var}(S) = aVa'$$

About $V$, the variance-covariance matrix of 4 multivariate Gaussian distributed random variables is needed. Then referring to eqn (1) on page 4 of http://www.math.unm.edu/~fletcher/Wishart.pdf, you can get the answer.

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  • $\begingroup$ When you say "About V, the variance-covariance matrix of 4 multivariate Gaussian distributed random variables is needed " do you mean $\Sigma$ in the OP? Or V shd be the Kronecker product of $\Sigma$ by $\Sigma$ ? $\endgroup$ – Nero Nov 6 '18 at 21:09
  • $\begingroup$ I mean the Kronecker product of Σ by Σ in eqn (1) on page 4 linked paper. $\endgroup$ – user158565 Nov 6 '18 at 21:16
  • $\begingroup$ Thanks! Then a should have 16 elemts, s.a. (0,0,0,1,-1....), no? $\endgroup$ – Nero Nov 6 '18 at 21:32
  • $\begingroup$ depend on you. If you want to use that 16 $\sigma$s, yes, need 16 dimension vector $a$. If you want to get a specific 4x4 matrix from that 16x16 matrix, you can keep a=(1,-1,-1,1). Both ways will generate the exact same results. $\endgroup$ – user158565 Nov 6 '18 at 21:39
  • $\begingroup$ What I am getting from $a V a'$ is $S^2$. I am looking for $E(S^2)$. Any glitch? $\endgroup$ – Nero Nov 8 '18 at 5:45

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