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

Write the expected value as ${\mu }^T=({\mu }_1^T,{\mu }_2^T)$ and assume the relevant covariance matrix $V_{22}$ is invertible. Remove the linear predictor from the first block by definingA linear image of a multivariate normal vector is again multivariate normal. Direct covariance calculation givesThus $Y$ and $X_2$ are independent: uncorrelated jointly Gaussian variables are independent. Expanding the covariance matrix of $Y$, using $V_{21}=V_{12}^T$, givesConditioning on $X_2=x_2$ translates $Y$ by the fixed vector ${\mu }_1+V_{12}{V}_{22}^{-1}(x_2-{\mu }_2)$ and leaves its covariance matrix unchanged. Consequently the conditional multivariate normal distribution isThe conditional covariance is the Schur complement and does not depend on $x_2$.

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 $V_{11}$.

Write ${\sigma }_i^2=V_{ii}$ and $V_{ij}={\rho }_{ij}{\sigma }_i{\sigma }_j$. Apply the Schur complement covariance to the first two coordinates, conditioning on the third. The conditional covariance and conditional variances areDivide the conditional covariance by the product of the conditional standard deviations. The conditional correlation is thereforeThis is the partial correlation after removing the linear effect of $X_3$. For a nonsingular multivariate normal distribution, both residual variances are positive and the answer is independent of $x_3$. If a residual variance vanishes, the requested correlation coefficient is undefined, even though the residual covariance formula remains valid.