Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 225 / 1 / a

This is the Brownian covariance kernel, so its integral operator is a covariance operator. To obtain its eigendecomposition, suppose $Uf=\lambda f$ with $\lambda \ne 0$. Splitting the integral at $s$ givesDifferentiation yields $\lambda f^{\prime }(s)={\int }_s^{1}f(t)dt$ and $\lambda f^{\prime \prime }(s)=-f(s)$, with boundary conditions $f(0)=0$ and $f^{\prime }(1)=0$. Hence the normalized eigenpairs areThe eigenvalues are positive and summable, consistently with positivity and the trace-class operator property of a covariance operator.

The identity operator is positive and self-adjoint, but on the infinite-dimensional Hilbert space $L^2[0,1]$ it has eigenvalue one with infinite multiplicity. ConsequentlyA square-integrable Hilbert-space random variable has a trace-class covariance operator with trace $\mathbb{E}\Vert X{\Vert }^2<\infty$. The identity therefore cannot be such a covariance operator.

This symmetric finite-rank integral operator is not positive. For $f(t)=1-\frac{3}{2}t$, set $A_j={\int }_0^1{t}^{j}f(t)dt$. Directly,The associated quadratic form isEvery covariance operator is a positive operator, so this operator is not a covariance operator.