Expected value of non-standard quadratic form

Question

Say I have two $n$-vectors $f(t)=(f_1(t), \dots, f_n(t))$ and $f(s)=(f_1(s), \dots, f_n(s))$, both with expected value 0.
Let $\operatorname{Cov}(f(t)) = a \Sigma$ and $\operatorname{Cov}(f(s)) = b \Sigma$ with scalar $a,b$. Further assume that the two vectors are not independent and that $Cov(f_i(t), f_j(s)) = c \Sigma_{ij} \quad \forall i,j$.
I know that $\operatorname{E}(f(t)' \Sigma f(t)) = a \operatorname{trace}(\Sigma^2)$, but can any of you help me derive an expression for

$\operatorname{E}(f(t)' \Sigma f(s))$?

Background: This is not homework (i'm too old for that...), I'm working on an estimator for the temporal covariance structure of spatially correlated functional data.

Answer 1

The answer is not that hard to get directly (without resorting to references). Denote $\Sigma=(\sigma_{ij})$, $\Omega=(\Sigma_{ij})$. We have

$$Ef(t)'\Sigma f(s)=\sum_{i=1}^n\sum_{j=1}^n\sigma_{ij}Ef_i(t)f_j(s)=\sum_{i=1}^n\sum_{j=1}^n\sigma_{ij}c\Sigma_{ij}$$

Then it is a matter of figuring out what that means:

$$Ef(t)'\Sigma f(s)=c\operatorname{trace}(\Sigma\Omega)$$

where we exploit the fact that $\Sigma$ is symmetric.