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 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 at 1:36

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.