Write , so the covariance operator is
For ,
so is self-adjoint. Moreover,
so it is a positive operator.
Let be any orthonormal basis. Tonelli theorem and Parseval identity give
A positive operator with finite trace is a trace-class operator, completing the proof.
For every orthonormal basis of the infinite-dimensional Hilbert space , the identity operator satisfies
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.
Choose a unit vector and define the rank-one operator
It is bounded, self-adjoint, and Hilbert-Schmidt, with . But
so it is not positive. Since every covariance operator is positive, is the required counterexample.
Linearity of the Bochner integral gives
Independence and centering of imply
Since and the eigenfunctions can be chosen orthonormally, independence of the sample gives
This remains valid for repeated eigenvalues after choosing an orthonormal eigenbasis within each eigenspace.

Articles by others on the same topic (0)

There are currently no matching articles.