Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 43 / 1 / ii / Solution

Put $K=V^{-1}$, the precision matrix, with $X_1$ scalar and $X_2$ containing all remaining coordinates. Holding $X_2=x_2$ in the multivariate normal density, the terms involving $x_1$ in its exponent areCompleting the square gives a normal distribution with mean ${\mu }_1-K_{11}^{-1}{K}_{12}(x_2-{\mu }_2)$ andEquivalently, the block matrix inverse identity gives $K_{11}=(V_{11}-V_{12}{V}_{22}^{-1}{V}_{21}{)}^{-1}$. The Gaussian conditional variance from precision is a reciprocal diagonal entry of the precision matrix, rather than the marginal variance $V_{11}$.