Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 219 / 2

The vector $(y_1,y_2,y_3)$ is Jointly Gaussian. Assuming $R>0$, Gaussian conditional independence givesThus the required condition is $R{R}_{13}=R_{12}{R}_{23}$. Under it, conditioning on $y_1$ supplies no further information after $y_2$, and the Gaussian process regression posterior isBoth the conditional expectation and conditional variance depend only on $y_2,t_2,t_3$; neither contains $y_1$ or $t_1$.

For $t_1<t_2<t_3$, the condition from part a becomesIt holds for arbitrary positive time gaps exactly when $\eta =1$. The resulting exponential covariance function is the covariance of a stationary Ornstein-Uhlenbeck process, hence has the Markov property.

Writing $a_{32}=e^{-(t_3-t_2)/\tau }$, the predictive law isAs $t_3\to \infty$, $a_{32}\to 0$, so the predictive mean tends to the stationary mean $\mu$ and the predictive variance tends to the stationary variance $1$.

SetThe Markov factorization and the conditional normal laws from part b give the fully univariate productwhere $\varphi (\cdot ;m,v)$ denotes the $N(m,v)$ density. This is a weighted least squares problem in $\mu$. Differentiating its log-likelihood givesEvery innovation in the numerator has expectation equal to its coefficient in the denominator times $\mu$. Therefore $\mathbb{E}\hat{\mu }=\mu$, so this maximum likelihood estimator is unbiased.