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 $(y_1,y_2,y_3)$ is
$$
R\begin{pmatrix}1&\rho&\rho^2\\\rho&1&\rho\\\rho^2&\rho&1\end{pmatrix}.
$$
Conditioning a <multivariate normal distribution> and simplifying gives
$$
y_3\mid y_1,y_2
\sim N\!\left(\mu+\rho(y_2-mu),R(1-\rho^2)\right).
$$
The 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.

Solution

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

The stationary exponential covariance function is
$$
k(s,t)=R\exp(-|t-s|/\tau).
$$
The corresponding <Gaussian process> is Markov and is a stationary <Ornstein-Uhlenbeck process>.

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 Incoming links: Exponential covariance function