The kernel is the <Brownian bridge covariance kernel>. If $C_X\phi=\lambda\phi$, differentiating the integral equation twice gives
$$
\lambda\phi''(t)=-\phi(t),\qquad \phi(0)=\phi(1)=0.
$$
The normalized solutions and eigenvalues are therefore
$$
\boxed{\phi_k(t)=\sqrt2\sin(k\pi t),\qquad
\lambda_k=\frac1{k^2\pi^2}},\qquad k\geq1.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 225 / 1 / i / Solution

The stated kernel is the <Brownian bridge covariance kernel>. Its eigenvalue equation is
$$
\lambda\phi(s)=\int_0^1\{\min(s,t)-st\}\phi(t)\,dt.
$$
The right-hand side vanishes at $s=0$ and $s=1$, and differentiating it twice gives
$$
\lambda\phi''(s)=-\phi(s),
\qquad
\phi(0)=\phi(1)=0.
$$
Thus the normalized eigenfunctions and eigenvalues are
$$
\phi_k(t)=\sqrt2\sin(k\pi t),
\qquad
\lambda_k=\frac1{k^2\pi^2},
\qquad k\geq1.
$$
The <Karhunen–Loève expansion> is consequently
$$
X(t)=\mu(t)+\sum_{k=1}^{\infty}\xi_k\sqrt2\sin(k\pi t)
=\mu(t)+\sum_{k=1}^{\infty}\frac{Z_k}{k\pi}\sqrt2\sin(k\pi t),
$$
with convergence in $L^2(\Omega;L^2[0,1])$, where $\mathbb EZ_k=0$ and $\mathbb E[Z_jZ_k]=\mathbf1_{\{j=k\}}$. Covariance alone does not imply that the $Z_k$ are independent or normal; they are independent standard normal variables when $X$ is Gaussian.

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 225 / 1 / b / Solution

{c}

The covariance kernel of a standard <Brownian bridge> on $[0,1]$ is $c(s,t)=\min(s,t)-st$. Its integral operator has eigenfunctions $\sqrt2\sin(k\pi t)$ and eigenvalues $(k\pi)^{-2}$ for $k\geq1$.

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