= Solution
The <Schur complement> can be written
$$
s=\underbrace{k(x_n,x_n)-b^TA^{-1}b}_{v_{n-1}(x_n)}+\sigma^2>0,
$$
because the first term is a <conditional variance> and $\sigma^2>0$. Therefore $c_1=s^{-1}>0$ and the squared term is nonnegative, so
$$
\boxed{\operatorname{Var}_{n-1}(f_*)\geq\operatorname{Var}_n(f_*)}.
$$
Equality holds exactly when the remaining <conditional covariance> $k(x_n,x_*)-b^TA^{-1}a$ is zero. In particular, the reduction need not be strict even at a previously unseen input.
Back to article page
