Proof of Conditional-Variance Lower Bound for the Completed-Square Control Energy

Fix $t$ as in the statement and abbreviate $R=R_t$, $W=W_t$, $\mathsf{G}=R^{-1}W^{\top}$ (a real $m\times l$ matrix), $\mathfrak{s}=\mathfrak{s}_t$, $\mathfrak{a}=\mathfrak{a}_t$, $\hat{\mathfrak{s}}=\hat{\mathfrak{s}}_t$, $\varepsilon=\varepsilon_t$, and $u=u_t=\mathfrak{a}+\mathsf{G}\mathfrak{s}$. The sans-serif $\mathsf{G}$ is distinct from the terminal cost function $G$ of the adopted setting, and the plain $R$ from the cost remainder $R_N$ there; the letter $M$ is kept for the generic matrices of the entry notation.

Step 1 (Square-integrability). Each $\mathfrak{s}^\gamma$ is a random variable: $\Sigma^\gamma_t$ is one by the solution definition, and $\mathfrak{s}^\gamma=\sqrt{N}(\Sigma^\gamma_t-S^\gamma_t)$ is a sequentially continuous (affine) function of it, measurable by measurability of sequentially continuous functions of measurable maps. Moreover $(\mathfrak{s}^\gamma)^2\le2N$ at every point of $\Omega$: first, $(\mathfrak{s}^\gamma)^2\le\sum_{\gamma'}(\mathfrak{s}^{\gamma'})^2=|\mathfrak{s}|^2$ by claim 1 of the norm properties lemma, and $|\mathfrak{s}|^2=N\,d(\Sigma_t,S_t)^2$ by claims 5 and 2 of the same lemma. Second, all coordinates of a point $x$ of the probability simplex lie in $[0,1]$: they are nonnegative, and each is at most the total sum $1$, the remaining summands being nonnegative. For $s,s'\in[0,1]$ we have $(s-s')^2=s\,s-2\,s\,s'+s'\,s'\le s\,s+s'\,s'\le s+s'$, since $0\le s\,s'$ (either a factor is $0$, or both are positive and claim 5 of the order arithmetic lemma applies) and since $r\,r\le r$ for $r\in[0,1]$ (for $r=0$ or $r=1$ this holds with equality; for $0<r<1$, claim 10 of the same lemma gives $r\,r<r\cdot1=r$). Summing over coordinates, any $x,y\in\Delta^l$ satisfy $d(x,y)^2=\sum_\gamma(x^\gamma-y^\gamma)^2\le\sum_\gamma(x^\gamma+y^\gamma)=2$. Hence $\mathbb{E}[(\mathfrak{s}^\gamma)^2]\le2N<\infty$ by monotonicity of the integral, and $\mathfrak{s}^\gamma$ is square-integrable. Each $\hat{\mathfrak{s}}^\gamma$ is $\mathcal{G}_t$-measurable and square-integrable by conditions (i)--(ii) of the conditional-expectation definition, and each $\varepsilon^\gamma=\mathfrak{s}^\gamma-\hat{\mathfrak{s}}^\gamma$ is square-integrable by the closure properties of the square-integrability definition. Define tuples $X=(X^1,\dots,X^m)$ and $\hat{X}=(\hat{X}^1,\dots,\hat{X}^m)$ by

so that $X=-\mathsf{G}\mathfrak{s}$ and $\hat{X}=-\mathsf{G}\hat{\mathfrak{s}}$ componentwise in the entry notation. Each $X^j$ and each $\hat{X}^j$ is a finite linear combination of square-integrable random variables, hence square-integrable by the closure properties.

Step 2 ($\hat{X}$ is a conditional expectation of $X$). By claim 1 (linearity) of the basic properties of conditional expectation, applied repeatedly --- induction on the number of summands, the claim being stated for two --- each $\hat{X}^j$, a finite linear combination of the conditional expectations $\hat{\mathfrak{s}}^\gamma$ with coefficients $-\mathsf{G}^{j\gamma}$, is a conditional expectation of the corresponding combination $X^j$ of the $\mathfrak{s}^\gamma$ given $\mathcal{G}_t$.

Step 3 (Admissibility of $Y=\mathfrak{a}$). Each $\mathfrak{a}^j$ is square-integrable by hypothesis, and by conclusion 3 of the adaptedness lemma it is almost surely equal to a $\mathcal{G}_t$-measurable random variable, which is square-integrable with the same mean-square norm by conclusions 2--3 of that lemma. Thus the tuple $Y=(\mathfrak{a}^1,\dots,\mathfrak{a}^m)$ is admissible in the sense of the conditional mean-square optimality lemma: each component is square-integrable and almost surely equal to a $\mathcal{G}_t$-measurable square-integrable random variable.

Step 4 (Optimality). The matrix $R$ is symmetric positive definite by conclusion (a) of the completion-of-squares theorem under (H1); in particular it is positive semidefinite, the quadratic form being $0$ at the origin and positive elsewhere. By claim 2 of the optimality lemma, applied with $k=m$, the tuple $X$, the conditional expectations $\hat{X}$ of Step 2, and the admissible tuple $Y$ of Step 3 (the tuple written $\varepsilon$ in that lemma being our $X-\hat{X}$, not our $\varepsilon$),

both sides being defined and finite by the integrability of such forms established in that lemma's preamble (the components of $Y-X$ and $X-\hat{X}$ being square-integrable by the closure properties). Componentwise, $Y-X=\mathfrak{a}+\mathsf{G}\mathfrak{s}=u$ and $X-\hat{X}=-\mathsf{G}(\mathfrak{s}-\hat{\mathfrak{s}})=-\mathsf{G}\varepsilon$.

Step 5 (Matrix algebra). At every point of $\Omega$, expanding $(\mathsf{G}\varepsilon)^j=\sum_\gamma \mathsf{G}^{j\gamma}\varepsilon^\gamma$ and rearranging the finite sums (the two sign factors cancelling),

since $(\mathsf{G}^{\top}R\mathsf{G})^{\gamma\delta}=\sum_{j,k}(\mathsf{G}^{\top})^{\gamma j}R^{jk}\mathsf{G}^{k\delta}=\sum_{j,k}\mathsf{G}^{j\gamma}R^{jk}\mathsf{G}^{k\delta}$ in the entry notation. Moreover $\mathsf{G}^{\top}=WR^{-1}$: entrywise, $(\mathsf{G}^{\top})^{\gamma j}=\mathsf{G}^{j\gamma}=\sum_{k}(R^{-1})^{jk}(W^{\top})^{k\gamma}=\sum_{k}(R^{-1})^{kj}W^{\gamma k}=(WR^{-1})^{\gamma j}$, using the symmetry of $R^{-1}$ (inverse of a positive definite matrix). Hence, by associativity of the matrix product and the defining property $R^{-1}R\,R^{-1}=R^{-1}$ of the inverse,

Combining with Step 4 yields conclusion 1:

Step 6 (Trace form). Claim 1 of the expected bilinear forms lemma, applied to the tuple $\varepsilon$ (components square-integrable by Step 1) and the matrix $WR^{-1}W^{\top}$, gives

Step 7 (Independence of the choice). Let $\hat{\mathfrak{s}}'^{\gamma}$ be any other conditional expectations of the $\mathfrak{s}^\gamma$ given $\mathcal{G}_t$, and set $\varepsilon'^{\gamma}=\mathfrak{s}^\gamma-\hat{\mathfrak{s}}'^{\gamma}$, square-integrable as in Step 1. By the uniqueness part of the existence and uniqueness theorem, $\hat{\mathfrak{s}}'^{\gamma}$ is almost surely equal to $\hat{\mathfrak{s}}^{\gamma}$ for each $\gamma$. Hence $\varepsilon'^{\gamma}-\varepsilon^{\gamma}=\hat{\mathfrak{s}}^{\gamma}-\hat{\mathfrak{s}}'^{\gamma}$ is a square-integrable random variable almost surely equal to $0$, so its mean-square norm vanishes, $\lVert\varepsilon'^{\gamma}-\varepsilon^{\gamma}\rVert_2=0$, by the null-equivalence property recorded in the square-integrability definition. Fix $\gamma,\delta$. All products below are integrable, as products of square-integrable random variables (closure properties of the same definition), and by linearity of the integral,

By the mean-square Cauchy-Schwarz inequality, the first term satisfies $|\mathbb{E}[\varepsilon'^{\gamma}(\varepsilon'^{\delta}-\varepsilon^{\delta})]|\le\lVert\varepsilon'^{\gamma}\rVert_2\,\lVert\varepsilon'^{\delta}-\varepsilon^{\delta}\rVert_2=0$, and likewise $|\mathbb{E}[(\varepsilon'^{\gamma}-\varepsilon^{\gamma})\varepsilon^{\delta}]|\le\lVert\varepsilon'^{\gamma}-\varepsilon^{\gamma}\rVert_2\,\lVert\varepsilon^{\delta}\rVert_2=0$. Hence $\mathbb{E}[\varepsilon'^{\gamma}\varepsilon'^{\delta}]=\mathbb{E}[\varepsilon^{\gamma}\varepsilon^{\delta}]$, which proves conclusion 2. $\blacksquare$