# Variance of the sum of elements of a Wishart distributed matrix

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.

Answer: $$\left(\left(\sigma _{1,3}-\sigma _{1,4}-\sigma _{2,3}+\sigma _{2,4}\right)^2+\left(-2 \sigma _{1,2}+\sigma _1^2+\sigma _2^2\right) \left(-2 \sigma _{3,4}+\sigma _3^2+\sigma _4^2\right)\right)$$

• Probability is not my turf but this looks like you aren't using much of the matrix. Instead, if I let $X_1, X_2, X_3, X_4$ be your four random variables and set $\overline{X}_i := X_i - \mathbb{E}\left(X_i\right)$ to be their mean-to-zero translates, then $\sigma_{i, j} = \mathbb{E}\left(\overline{X}_i \overline{X}_j\right)$, so that a short computation shows that $S = \operatorname{Var}\left(\left(\overline{X}_1 - \overline{X}_2\right) \left(\overline{X}_3 - \overline{X}_4\right)\right)$. Do you expect it to simplify any further? – darij grinberg Nov 12 '18 at 1:22
• I've solved it in another way, but indeed it looks like I wasn't adding I+T the transfer function in the paper. After I did, I got the same as Monte Carlo. – Nero Nov 12 '18 at 1:36