Proof of Orthogonal Decomposition of Expected Quadratic Forms under Independence

Expanding pointwise on $\Omega$ with the definitions of the dot product and the matrix-vector product,

each term being of the form $\sum_{i,j}M_{ij}\,\eta^{i}\vartheta^{j}$, a finite sum of scalar multiples of products of two square-integrable random variables. Every such product is integrable by Cauchy-Schwarz and Triangle Inequalities for the Mean-Square Norm, so all four terms are integrable and expectations add by Linearity and Monotonicity of the Lebesgue Integral.

The cross terms vanish. Fix $i,j$. Let $\tilde\zeta^{i}$ be an $\mathcal{H}$-measurable random variable almost surely equal to $\zeta^{i}$ (hypothesis (i)). The component $\rho^{j}$ is measurable with respect to $\sigma(\rho^{1},\dots,\rho^{p})$, since by Sigma-Algebra Generated by Random Variables and Independence of Sigma-Algebras that $\sigma$-algebra contains the preimages of Borel sets under each component. By hypothesis (iii), $\rho^{j}$ and $\tilde\zeta^{i}$ are therefore independent, and both are integrable, so Expectation of a Product of Independent Random Variables gives $\mathbb{E}[\tilde\zeta^{i}\rho^{j}]=\mathbb{E}[\tilde\zeta^{i}]\,\mathbb{E}[\rho^{j}]=0$ by hypothesis (ii). Since $\zeta^{i}\rho^{j}=\tilde\zeta^{i}\rho^{j}$ almost surely, $\mathbb{E}[\zeta^{i}\rho^{j}]=0$. Hence

Taking expectations in the first display now gives the asserted decomposition.

The trace formulas. By hypothesis (ii) the mean tuple of $\rho$ is zero, so claim 1 of Expected Bilinear Forms: Trace Formula and Mean-Square Continuity gives $\mathbb{E}[\rho\cdot(M\rho)]=\operatorname{tr}(M^{\top}C_{\rho})$, and also $\operatorname{tr}(M^{\top}C_{\rho})=\operatorname{tr}(MC_{\rho})$, as stated there. $\square$