Proof of Expected Bilinear Forms: Trace Formula and Mean-Square Continuity

Throughout, $\lVert\cdot\rVert_{2}$ and $\langle\cdot,\cdot\rangle_{2}$ are the mean-square norm and inner product of Square-Integrable Random Variables and the Mean-Square Inner Product.

Claim 1. By the definitions of the dot product and the matrix-vector product, applied pointwise on $\Omega$,

For each pair $(i,j)$ the product $Y^{i}Y^{j}$ is integrable with $\mathbb{E}\bigl[|Y^{i}Y^{j}|\bigr]\le\lVert Y^{i}\rVert_{2}\lVert Y^{j}\rVert_{2}$, by Cauchy-Schwarz and Triangle Inequalities for the Mean-Square Norm applied to $|Y^{i}|$ and $|Y^{j}|$, which are square-integrable by the closure properties of Square-Integrable Random Variables and the Mean-Square Inner Product (their squares agree with those of $Y^{i},Y^{j}$). Hence the finite sum is integrable and, by linearity of the integral (Linearity and Monotonicity of the Lebesgue Integral),

By the definition of the covariance, $\mathbb{E}[Y^{i}Y^{j}]=\operatorname{Cov}(Y^{i},Y^{j})+\mathbb{E}[Y^{i}]\mathbb{E}[Y^{j}]=C_{ij}+\mu^{i}\mu^{j}$. Therefore

using claim 4 of Basic Properties of the Trace for the first sum and the definitions of dot product and matrix-vector product for the second. Finally, $C$ is symmetric ($\operatorname{Cov}(Y^{i},Y^{j})=\operatorname{Cov}(Y^{j},Y^{i})$ by the symmetry of its defining formula), so claims 2-3 of Basic Properties of the Trace give $\operatorname{tr}(M^{\top}C)=\operatorname{tr}\bigl((M^{\top}C)^{\top}\bigr)=\operatorname{tr}(C^{\top}M)=\operatorname{tr}(CM)=\operatorname{tr}(MC)$, with $(M^{\top}C)^{\top}=C^{\top}M$ by claim 3 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals.

Claim 2. As in claim 1, for each $t$,

all terms defined and finite. It therefore suffices, by the continuity of sums and products of continuous real functions (Sums and Products of Continuous Real-Valued Functions), to show that each function $t\mapsto\mathbb{E}[Y^{i}_tZ^{j}_t]=\langle Y^{i}_t,Z^{j}_t\rangle_{2}$ is continuous on $[a,b]$. Fix $s,t\in[a,b]$. Adding and subtracting $\mathbb{E}[Y^{i}_sZ^{j}_t]$ and applying Cauchy-Schwarz and Triangle Inequalities for the Mean-Square Norm twice,

By claim 4 of Basic Properties of the Mean-Square Riemann Integral, the functions $t\mapsto\lVert Y^{i}_t\rVert_{2}$ and $t\mapsto\lVert Z^{j}_t\rVert_{2}$ are continuous on the compact interval $[a,b]$, hence bounded there by Extreme Value Theorem on a Compact Interval; and $\lVert Y^{i}_t-Y^{i}_s\rVert_{2}\to0$, $\lVert Z^{j}_t-Z^{j}_s\rVert_{2}\to0$ as $t\to s$ by mean-square continuity. Hence the right-hand side tends to $0$ as $t\to s$, which is the asserted continuity at $s$. $\square$