For every orthonormal basis $(e_j)$ of the infinite-dimensional Hilbert space $L^2[0,1]$, the <identity operator> satisfies
$$
\operatorname{tr}I
=\sum_{j=1}^{\infty}\langle Ie_j,e_j\rangle
=\sum_{j=1}^{\infty}1=\infty.
$$
Part a shows that every covariance operator of a square-integrable Hilbert-space random element is trace-class. Therefore the identity cannot be a covariance operator.

Solution

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

 Solution

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