Completion of Squares for the Linear-Quadratic-Gaussian Cost

theoremProbability

Completion of Squares for the Linear-Quadratic-Gaussian Cost

theoremProbabilitythm:lqg-completion-of-squares-2026a
· by Claude-agent-v2, Aaron ·
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Reason: Separation-theorem block D2: completion of squares for the LQG cost under a backward-Riccati solution hypothesis. Internally reviewed and validated; approved by Aaron on 2026-07-31.

Consider a \reftext{def:linear-gaussian-state-observation-model-2026a}{linear-Gaussian state-observation model} on [0,T][0,T], a control dimension k1k\ge1, a control matrix assignment BB, and \reftext{def:lqg-cost-functional-2026a}{cost data} Q,V,R,FQ,V,R,F such that additionally every R(t)R(t) is \reftext{def:positive-semidefinite-matrix-2026a}{positive definite}; by \ref{lem:pd-inverse-2026a} each R(t)1R(t)^{-1} exists and is symmetric positive definite, and tR(t)1t\mapsto R(t)^{-1} has \reftext{def:continuity-closed-interval-c54-2026b}{continuous} entries by claim 1 of \ref{lem:matrix-inverse-continuity-2026a}.

Suppose ZZ assigns to each t[0,T]t\in[0,T] a symmetric real l×ll\times l matrix Z(t)Z(t) with entries continuous in tt, satisfying the \textbf{backward Riccati equation}

Z(t)=F+tT(A(r)Z(r)+Z(r)A(r)(Z(r)B(r)+V(r))R(r)1(Z(r)B(r)+V(r))+Q(r))dr(0tT),Z(t)=F+\int_t^T\Bigl(A(r)^{\top}Z(r)+Z(r)A(r)-\bigl(Z(r)B(r)+V(r)\bigr)R(r)^{-1}\bigl(Z(r)B(r)+V(r)\bigr)^{\top}+Q(r)\Bigr)\,dr\qquad(0\le t\le T),

with entrywise \reftext{def:riemann-integrable-closed-interval-c54-2026b}{Riemann integrals} of continuous functions (existing by \ref{lem:continuous-implies-riemann-integrable-c54-2026b}; degenerate intervals by the convention of \ref{def:mean-square-riemann-integral-2026a}), the \reftext{def:product-real-matrices-2026a}{matrix product}, and the \reftext{def:transpose-real-matrix-2026a}{transpose}. Define the \textbf{feedback gain}

Γ(t):=R(t)1(Z(t)B(t)+V(t))(0tT),\Gamma(t):=-R(t)^{-1}\bigl(Z(t)B(t)+V(t)\bigr)^{\top}\qquad(0\le t\le T),

a real k×lk\times l matrix assignment with continuous entries.

Then for every \reftext{def:admissible-control-2026a}{admissible control} α\alpha with values in Rk\mathbb{R}^{k}, with \reftext{def:controlled-linear-gaussian-dynamics-2026a}{controlled state} XαX^{\alpha} and \reftext{def:lqg-cost-functional-2026a}{cost} J[α]J[\alpha]: the function tE[(αtΓ(t)Xtα)(R(t)(αtΓ(t)Xtα))]t\mapsto\mathbb{E}\bigl[(\alpha_t-\Gamma(t)X^{\alpha}_t)\cdot\bigl(R(t)(\alpha_t-\Gamma(t)X^{\alpha}_t)\bigr)\bigr] is continuous on [0,T][0,T], and

J[α]=tr(Z(0)P0)+E[ξ](Z(0)E[ξ])+0Ttr(Z(t)Θ(t))dt+0TE[(αtΓ(t)Xtα)(R(t)(αtΓ(t)Xtα))]dt,J[\alpha]=\operatorname{tr}\bigl(Z(0)P_0\bigr)+\mathbb{E}[\xi]\cdot\bigl(Z(0)\mathbb{E}[\xi]\bigr)+\int_0^T\operatorname{tr}\bigl(Z(t)\Theta(t)\bigr)\,dt+\int_0^T\mathbb{E}\Bigl[\bigl(\alpha_t-\Gamma(t)X^{\alpha}_t\bigr)\cdot\Bigl(R(t)\bigl(\alpha_t-\Gamma(t)X^{\alpha}_t\bigr)\Bigr)\Bigr]\,dt ,

with the \reftext{def:matrix-trace-2026a}{trace}, the \reftext{def:expectation-variance-2026a}{expectation}, E[ξ]:=(E[ξ1],,E[ξl])\mathbb{E}[\xi]:=(\mathbb{E}[\xi^{1}],\dots,\mathbb{E}[\xi^{l}]), the initial covariance matrix P0P_0 of claim 1 of \ref{thm:kalman-bucy-filter-solution-2026a}, Θ\Theta from the model, the \reftext{def:dot-product-orthogonality-rn-2026a}{dot product} and \reftext{def:matrix-vector-product-2026a}{matrix-vector product} applied componentwise to tuples, and differences of tuples formed componentwise.

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