Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 219 / 3

Put $\rho =e^{-\Delta t/\tau }=e^{-x}$ and retain $R=k(0,0)$, which equals one for the stated exponential covariance function. The covariance matrix of $(y_1,y_2,y_3)$ isConditioning a multivariate normal distribution and simplifying givesThe absence of $y_1$ from both conditional moments shows that $y_3$ and $y_1$ are conditionally independent given $y_2$. Equivalently, the exponential-kernel Gaussian process is the stationary Ornstein-Uhlenbeck process, which is Markov.

As $\Delta t/\tau \to \infty$, $\rho \to 0$. Therefore the posterior predictive mean tends to $\mu$ and its variance tends to $R$: a sufficiently distant observation has reverted to the stationary prior distribution.

The Markov factorization isDifferentiating its log-likelihood with respect to $\mu$ givesIts coefficients sum to one, so it is unbiased. Direct covariance calculation, equivalently inversion of its Fisher information, givesThus the variance tends to $R$ as $\Delta t/\tau \to 0$, because the observations become perfectly correlated, and to $R/3$ as $\Delta t/\tau \to \infty$, because they become three independent draws.