Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 216 / 6 / c

All moments below are under the proper posterior distribution of part (b). Write the known covariance matrix in block formThe score covariance $S$ is positive definite in this model. Indeed, a further integration by parts gives $\mathbb{E}[g(\beta ){\beta }^T]=-I_p$. If $a^{T}Sa=0$, then $a^{T}g=0$ almost surely, because $\mathbb{E}[g]=0$. Multiplying by $a^T\beta$ would give $0=-\Vert a{\Vert }^2$, so $a=0$.

For a fixed coefficient vector $a$, the control variate summand has varianceThus the unique minimum is attained at $a_{\ast }=S^{-1}c$, givingThe coefficient is fixed because the covariance is assumed known, and the posterior score control variate has mean zero, so this estimator remains unbiased.

We can identify exactly when the improvement is strict. Integration by parts with the bounded smooth function $h$ givesThe scalar expectation is strictly positive. Therefore $c\ne 0$ whenever $x_{\mathrm{test}}\ne 0$, and positive definiteness gives a strictly smaller variance in that case. If $x_{\mathrm{test}}=0$, $h\equiv 1/2$ and both estimators already have variance zero. The printed request for a smaller variance therefore needs this nondegeneracy qualification; a non-increasing variance always holds.