Write $Z=X-\mu$, so the <covariance operator> is
$$
C_Xh=\mathbb E[\langle Z,h\rangle Z].
$$
For $g,h\in L^2[0,1]$,
$$
\langle C_Xh,g\rangle
=\mathbb E[\langle Z,h\rangle\langle Z,g\rangle]
=\langle h,C_Xg\rangle,
$$
so $C_X$ is <self-adjoint operator>[self-adjoint]. Moreover,
$$
\langle C_Xh,h\rangle
=\mathbb E[\langle Z,h\rangle^2]\geq0,
$$
so it is a <positive operator>.

Let $(e_j)$ be any <orthonormal basis>. <Tonelli theorem> and <Parseval identity> give
$$
\operatorname{tr}C_X
=\sum_j\langle C_Xe_j,e_j\rangle
=\mathbb E\sum_j|\langle Z,e_j\rangle|^2
=\mathbb E\lVert Z\rVert^2<\infty.
$$
A positive operator with finite trace is a <trace-class operator>, completing the proof.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 225 / 1 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-225/1/a/solution