A Schur-complement check

Topic: Covariance and the multivariate normal · more questions on this topic
‹ The bivariate conditional Conditional normal prediction ›

For the matrix in the trivariate conditional example, verify the determinant identity |\Sigma|=|\Sigma_{BB}|\,|\Sigma_{A\mid B}| with A=\{1,2\} and B=\{3\}, by computing |\Sigma| directly and comparing it with |\Sigma_{BB}|\,|\Sigma_{A\mid B}|.

Worked answer (click to show)

The determinant factorisation is easiest to check by calculating its two sides independently. The same arithmetic checks the conditional covariance from the trivariate example.

Expanding along the first row,

\begin{aligned} |\Sigma| &=4(3\cdot 2-0.25)-2(2\cdot 2-0.5)+1(1-3)\\ &=4(5.75)-2(3.5)-2=23-7-2=14. \end{aligned}

Also |\Sigma_{BB}|=2 and

|\Sigma_{A\mid B}|=3.5\times 2.875-1.75^2=10.0625-3.0625=7,

so |\Sigma_{BB}|\,|\Sigma_{A\mid B}|=14, as required.