Proof of The Separation Theorem for Partial-Information Linear-Quadratic-Gaussian Control

Fix an admissible control $\alpha$; write $Y:=X^{\alpha}$, $\widehat X:=\widehat X(\alpha)$, and let $e_t=X_t-m^{\mathrm f}_t$ be the estimation error, so that by claim 3 of Conditional Expectation and Estimation Error of the Controlled State: $Y_t-\widehat X_t=e_t$ almost surely componentwise, $\mathbb{E}[e^{i}_t]=0$, the covariance matrix of $e_t$ is $\Pi(t)$, and $\sigma(e^{1}_t,\dots,e^{l}_t)$ is independent of $\mathcal{G}_t$. Throughout, a real-valued function on a subinterval $I$ of the real numbers $\mathbb{R}$ is called continuous on $I$ when it is continuous relative to $I$, both $I$ and the codomain $\mathbb{R}$ carrying the metric of the real line. The integrands below are continuous in $t$; by claim 3 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval each is Riemann integrable on $[0,T]$ with Riemann integral equal to its Lebesgue integral against $\lambda_{[0,T]}$, and linearity and monotonicity are then claim 1 of Linearity and Monotonicity of the Lebesgue Integral.

Claim 1. By Completion of Squares for the Linear-Quadratic-Gaussian Cost,

Fix $t$ and decompose, componentwise and almost surely,

We apply Orthogonal Decomposition of Expected Quadratic Forms under Independence with $\mathcal{H}:=\mathcal{G}_t$ and $M:=R(t)$. Hypothesis (i): each component of $\zeta$ is almost surely equal to a $\mathcal{G}_t$-measurable square-integrable random variable, by condition (ii) of Admissible Control for the Linear-Gaussian State-Observation Model, claim 1 of Conditional Expectation and Estimation Error of the Controlled State, and closure under finite linear combinations (claim 2 of The Closed Mean-Square Span of a Family of Random Variables). Hypothesis (ii): each $\rho^{\kappa}=-\sum_i\Gamma_{\kappa i}(t)e^{i}_t$ has expectation $0$. Hypothesis (iii): each $\rho^{\kappa}$ is a Borel (linear) function of the tuple $(e^{1}_t,\dots,e^{l}_t)$, so every preimage of a Borel set under a component of $\rho$ lies in $\sigma(e^{1}_t,\dots,e^{l}_t)$; hence $\sigma(\rho^{1},\dots,\rho^{k})\subseteq\sigma(e^{1}_t,\dots,e^{l}_t)$, which is independent of $\mathcal{G}_t$. (The expectation of the quadratic form depends on $\alpha_t-\Gamma(t)Y_t$ only through its almost sure class, so the decomposition may be used after modification on a null set.) The lemma gives

where $C_{\rho}(t)$ is the covariance matrix of $\rho=-\Gamma(t)e_t$. By bilinearity of the covariance (linearity of the expectation applied to its defining formula),

by the index formulas of the matrix product and transpose. Now apply the cyclic property (claim 3 of Basic Properties of the Trace) to the pair of matrices $R(t)\Gamma(t)\Pi(t)$ (size $k\times l$) and $\Gamma(t)^{\top}$ (size $l\times k$), and compute, from the definition of the feedback gain in Completion of Squares for the Linear-Quadratic-Gaussian Cost: $R(t)\Gamma(t)=-(Z(t)B(t)+V(t))^{\top}$ (since $R(t)R(t)^{-1}$ is the identity, by associativity), and $\Gamma(t)^{\top}=-(Z(t)B(t)+V(t))R(t)^{-1}$ (transpose rules of claim 3 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals and symmetry of $R(t)^{-1}$, Invertibility of Symmetric Positive Definite Matrices), so that

Hence

Both $t\mapsto\mathbb{E}[(\alpha_t-\Gamma(t)\widehat X_t)\cdot(R(t)(\alpha_t-\Gamma(t)\widehat X_t))]$ (continuous by claim 2 of Expected Bilinear Forms: Trace Formula and Mean-Square Continuity, the components of $\alpha-\Gamma\widehat X$ being mean-square continuous by claim 1 of Conditional Expectation and Estimation Error of the Controlled State and claims 1-2 of Basic Properties of the Mean-Square Riemann Integral) and the trace term (entries continuous) are continuous, and the first is nonnegative at each $t$: pointwise $q\cdot(R(t)q)\ge0$ since $R(t)$ is positive definite, and expectations of nonnegative random variables are nonnegative (Linearity and Monotonicity of the Lebesgue Integral). Integrating the pointwise identity over $[0,T]$ and splitting the integral by linearity yields claim 1 with the stated $V^{*}$.

Claim 2. The inequality $J[\alpha]\ge V^{*}$ is immediate from claim 1, the integrand being continuous and nonnegative and the Riemann integral monotone. If $\alpha_t=\Gamma(t)\widehat X_t$ almost surely for every $t$, the integrand vanishes identically and $J[\alpha]=V^{*}$. Conversely suppose $J[\alpha]=V^{*}$, i.e. $\int_0^T g(t)\,dt=0$ for the continuous nonnegative integrand $g$. If $g(t_0)>0$ for some $t_0$, then by continuity $g>g(t_0)/2$ on a nondegenerate subinterval, and splitting the integral by additivity (claim 4 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals) and monotonicity gives $\int_0^Tg>0$, a contradiction; hence $g\equiv0$. Fix $t$ and put $q:=\alpha_t-\Gamma(t)\widehat X_t$; then $q\cdot(R(t)q)\ge0$ pointwise with $\mathbb{E}[q\cdot(R(t)q)]=0$. For every $\varepsilon>0$, Markov's inequality (Markov's and Chebyshev's Inequalities) gives $P\bigl(q\cdot(R(t)q)\ge\varepsilon\bigr)=0$, and taking the union over $\varepsilon=1/n$ ($n\in\mathbb{N}$) with countable subadditivity of the measure yields $q\cdot(R(t)q)=0$ almost surely. On the event where in addition $q$ is defined and $q\ne0$, positive definiteness of $R(t)$ would give $q\cdot(R(t)q)>0$; hence $q=0$ almost surely, i.e. $\alpha_t=\Gamma(t)\widehat X_t$ componentwise almost surely.

Claim 3. By claim 2 of Existence and Self-Consistency of the Closed-Loop Feedback Control, $\alpha^{*}$ is admissible and $\alpha^{*}_t=\Gamma(t)\widehat X_t(\alpha^{*})$ almost surely for every $t$; by claim 3 there it is determined by its own observations in the stated sense. By the equality case of claim 2, $J[\alpha^{*}]=V^{*}$. $\square$