Stack the observations as $y=(y_1^T,y_2^T)^T$ and set
$$
\mu_\theta=
\begin{pmatrix}c\mathbf1\\(c+\Delta m)\mathbf1\end{pmatrix}.
$$
Let $K_f(a,b)$ and $K_g(a,b)$ denote matrices obtained by evaluating the two <Gaussian process> <covariance kernels>. Independence of the <quasar light curve>, <gravitational microlensing>, and <Gaussian noise> processes gives
$$
C_\theta=
\begin{pmatrix}
K_f(t,t)+K_g(t,t)+E_1&K_f(t,t-\Delta t)\\
K_f(t-\Delta t,t)&K_f(t-\Delta t,t-\Delta t)+K_g(t,t)+E_2
\end{pmatrix},
$$
where $E_i=\operatorname{diag}(\sigma_{i,1}^2,\ldots,\sigma_{i,N}^2)$. Thus $y\mid\theta$ is a <multivariate normal distribution> and its <Gaussian-process marginal likelihood> is
$$
\boxed{p(y_1,y_2\mid\theta)
=(2\pi)^{-N}|C_\theta|^{-1/2}
\exp\!\left[-\frac12(y-\mu_\theta)^TC_\theta^{-1}(y-\mu_\theta)\right].}
$$
The off-diagonal blocks are essential: both images contain the same delayed <Ornstein-Uhlenbeck process>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 219 / 3 / a / Solution

{c}

For Gaussian-process latent values with mean vector $m$ and covariance matrix $K$, observed with independent Gaussian noise covariance $E$, integrating out the latent values gives the marginal distribution $N(m,K+E)$. Its density as a function of the mean and kernel parameters is the Gaussian-process marginal likelihood.


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 Incoming links: Gaussian-process marginal likelihood