Past exam of the advanced topics in statistical machine learning course of the University of Oxford / 2025 / 3 / c / iv / Solution

For $x_n=x_{\ast }$, the notation of part (i) gives $b=a$, $d=k_{\ast \ast }$, and $c=k_{\ast \ast }+{\sigma }^2$. Write $v={\mathrm{Var}}_{n-1}(f_{\ast })=k_{\ast \ast }-a^T{A}^{-1}a$. The remaining covariance is then $v$ and the new observation's remaining variance is $v+{\sigma }^2$. ThusEquivalently,The expression is valid also when $v=0$ and ${\sigma }^2>0$. For $v>0$, a noiseless observation, interpreted as the limit ${\sigma }^2\downarrow 0$, removes all remaining variance at the observed target.

The source's assertion that the target input always gives the greatest reduction needs an additional hypothesis. The reduction formula above does not require that assertion. For a counterexample, use the linear kernel $k(x,x^{\prime })=x{x}^{\prime }$ with ${\sigma }^2=1$ and target $x_{\ast }=1$. With no informative previous observations, $v=1$. This can also be arranged with $n-1=1$ by observing first at $x_1=0$, which has zero covariance with all latent values. Observing at the target gives reduction $1/2$, whereas observing at $x_n=2$ givesIndeed this Gaussian process is $f(x)=xZ$ for a standard normal random variable $Z$, so the measurement at $2$ observes the same unknown coefficient with a larger signal relative to its noise.

One sufficient extra condition is that the existing latent posterior variance be maximal at $x_{\ast }$ among all candidate inputs. To prove it, let $w=v_{n-1}(x)$ and $r=\mathrm{Cov}(f(x),f_{\ast }\mid {\mathbf{y}}_{1:n-1},\theta )$. The Cauchy-Schwarz inequality gives $r^2\le vw$. If $w\le v$, thenbecause $w\mapsto w/(w+{\sigma }^2)$ is increasing. Thus target sampling is optimal under this extra condition, but it is not implied by the paper's arbitrary-covariance function assumptions.