# Expected value of non-standard quadratic form

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.

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Matrix cookbook is a very good reference for such types of questions. –  mpiktas Jun 2 '11 at 14:02
@mpiktas: it is a good reference, I have it on my desktop. Unfortunately, it only has expressions for standard quadratic forms. Not to offend you, but I find the type of comment you made, i.e., referring somebody to a widely known and very general reference without knowing/checking that it actually contains information relevant to the problem inappropriate. All that does is clutter up the page... –  fabians Jun 2 '11 at 14:12
sorry, I just figured that maybe you did not know that reference, since you did not mention that you already checked it. In this case I felt that giving this reference in comment would serve well for somebody who comes across this question without knowing it before hand. If I remember correctly this is how I found out about it. –  mpiktas Jun 2 '11 at 14:34
In fact, it does follow from (296) in the Matrix cookbook by taking $x = (f(t), f(s))$ and $A$ the block matrix with three zero blocks and $\Sigma$ in the top right corner. The variance-covariance matrix of $x$ is given in terms of $\Sigma$ and $\Omega$ using the notation from @mpiktas's answer below. –  NRH Jun 2 '11 at 20:09
Right, thanks for pointing that out. –  fabians Jun 3 '11 at 10:17

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.

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Thank you! You made it seem easy... –  fabians Jun 2 '11 at 14:46