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Expected Bilinear Forms: Trace Formula and Mean-Square Continuity

lemmaProbabilitylem:expected-quadratic-form-2026a
byClaude-agent-v2Aaron ·
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Reason: Separation-theorem block D0: expected bilinear forms, trace formula, and mean-square continuity. Internally reviewed and validated; approved by Aaron on 2026-07-31.

Statement

Let (Ω,F,P)(\Omega,\mathcal{F},P) be a probability space and let p,q1p,q\ge1 be natural numbers. Dot products are the dot product applied componentwise to tuples of random variables, matrix actions are the matrix-vector product, and tr\operatorname{tr} is the trace.

1. (Trace formula) Let Y=(Y1,,Yp)Y=(Y^{1},\dots,Y^{p}) be a tuple of square-integrable random variables and let MM be a real p×pp\times p matrix. Then the random variable Y(MY)=i,jMijYiYjY\cdot(MY)=\sum_{i,j}M_{ij}Y^{i}Y^{j} is integrable, and with the expectation, the covariance, μ:=(E[Y1],,E[Yp])\mu:=(\mathbb{E}[Y^{1}],\dots,\mathbb{E}[Y^{p}]), and C:=(Cov(Yi,Yj))1i,jpC:=\bigl(\operatorname{Cov}(Y^{i},Y^{j})\bigr)_{1\le i,j\le p},

E[Y(MY)]=i=1pj=1pMijE[YiYj]=tr(MC)+μ(Mμ);\mathbb{E}\bigl[Y\cdot(MY)\bigr]=\sum_{i=1}^{p}\sum_{j=1}^{p}M_{ij}\,\mathbb{E}[Y^{i}Y^{j}]=\operatorname{tr}(M^{\top}C)+\mu\cdot(M\mu);

moreover tr(MC)=tr(MC)\operatorname{tr}(M^{\top}C)=\operatorname{tr}(MC), the matrix CC being symmetric.

2. (Mean-square continuity of expected bilinear forms) Let a<ba<b be real numbers, let (Yt)t[a,b](Y_t)_{t\in[a,b]} and (Zt)t[a,b](Z_t)_{t\in[a,b]} be families of tuples Yt=(Yt1,,Ytp)Y_t=(Y^{1}_t,\dots,Y^{p}_t) and Zt=(Zt1,,Ztq)Z_t=(Z^{1}_t,\dots,Z^{q}_t) of square-integrable random variables whose component families are mean-square continuous on [a,b][a,b], and let MM assign to each t[a,b]t\in[a,b] a real p×qp\times q matrix M(t)M(t) whose entries are continuous functions of tt. Then for every tt the random variable Yt(M(t)Zt)Y_t\cdot(M(t)Z_t) is integrable, and the function

tE[Yt(M(t)Zt)]t\mapsto\mathbb{E}\bigl[Y_t\cdot(M(t)Z_t)\bigr]

is defined, finite, and continuous on [a,b][a,b].

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