Orthogonal Decomposition of Expected Quadratic Forms under Independence

Let $(\Omega,\mathcal{F},P)$ be a probability space, let $p\ge1$ be a natural number, and let $\mathcal{H}$ be a sub-$\sigma$-algebra of $\mathcal{F}$. Let $\zeta=(\zeta^{1},\dots,\zeta^{p})$ and $\rho=(\rho^{1},\dots,\rho^{p})$ be tuples of square-integrable random variables such that:

(i) each $\zeta^{i}$ is almost surely equal to an $\mathcal{H}$-measurable random variable;

(ii) $\mathbb{E}[\rho^{i}]=0$ for every $i$, with the expectation;

(iii) the $\sigma$-algebras $\sigma(\rho^{1},\dots,\rho^{p})$ and $\mathcal{H}$ are independent.

Let $M$ be a real $p\times p$ matrix, and let dot products and matrix actions be the dot product and matrix-vector product applied componentwise to tuples. Then all three expectations below are defined and finite, and with $C_{\rho}:=\bigl(\operatorname{Cov}(\rho^{i},\rho^{j})\bigr)_{1\le i,j\le p}$ (covariance) and the trace:

$$\mathbb{E}\bigl[(\zeta+\rho)\cdot\bigl(M(\zeta+\rho)\bigr)\bigr]=\mathbb{E}\bigl[\zeta\cdot(M\zeta)\bigr]+\mathbb{E}\bigl[\rho\cdot(M\rho)\bigr],\qquad \mathbb{E}\bigl[\rho\cdot(M\rho)\bigr]=\operatorname{tr}(M^{\top}C_{\rho})=\operatorname{tr}(MC_{\rho}),$$

where $\zeta+\rho$ is the componentwise sum.