Proof of Observation-Measurability of the Cascade Good Sets and the Restricted Filtering Bound

Throughout we use linearity and monotonicity of the integral freely. Every random variable occurring below is bounded: $|\mathfrak{s}_{t}|\le2\sqrt{N}$ at every point by the solution definition ($\Sigma_{t}$ and $S_{t}$ lying in the probability simplex), $|\mathfrak{a}_{t}|$ is bounded because $\alpha_{t}$ and $A_{t}$ take values in the control set, and the entries of $R_{t},R_{t}^{-1},W_{t}$ are continuous in $t$, hence bounded on $[0,T]$, by conclusion (a) of the completion-of-squares theorem and the extreme value theorem. In particular all components of $\mathfrak{s}_{t}$ and $\mathfrak{a}_{t}$, and hence of $u_{t}$, are square-integrable, and every expectation written below is finite.

Step 1: proof of claim 1. By claim 1 of the anchored good-set clocks lemma, each $\sigma^{(k)}$ is a stopping time of $(\mathcal{G}_{t})_{t\in[0,T]}$ — as well as of the system filtration — and $\{t<\sigma^{(k)}\}\in\mathcal{G}_{t}$ for every $t\in[0,T]$.

Claim 0. By the solution definition, each $\mathcal{G}_{t}$ is the $\sigma$-algebra generated by the observation processes up to time $t$ together with every event of $\mathcal{F}$ of probability zero. The regular event $\Omega_{0}$ has probability $1$, so its complement $\Omega\setminus\Omega_{0}$ is an event of $\mathcal{F}$ of probability zero and therefore belongs to $\mathcal{G}_{0}$; a $\sigma$-algebra being closed under complementation, $\Omega_{0}\in\mathcal{G}_{0}$.

We show $G_{k}\in\mathcal{G}_{t_{k}}$ by induction on $k$. For $k=0$: $G_{0}=\Omega_{0}\in\mathcal{G}_{0}$ by claim 0. Assume $G_{k}\in\mathcal{G}_{t_{k}}$ for some $k\le K-1$. Then $G_{k}\in\mathcal{G}_{t_{k+1}}$, a filtration being nondecreasing and $t_{k}\le t_{k+1}$; and $\{\sigma^{(k)}\ge t_{k+1}\}\in\mathcal{G}_{t_{k+1}}$ by claim 1 of the stopping-time toolkit applied to the filtration $(\mathcal{G}_{t})_{t\in[0,T]}$, of which $\sigma^{(k)}$ is a stopping time. A $\sigma$-algebra being closed under intersection, $G_{k+1}=G_{k}\cap\{\sigma^{(k)}\ge t_{k+1}\}\in\mathcal{G}_{t_{k+1}}$, completing the induction.

Finally, let $k\le K-1$ and $t\ge t_{k}$. Then $G_{k}\in\mathcal{G}_{t_{k}}\subseteq\mathcal{G}_{t}$ and $\{t<\sigma^{(k)}\}\in\mathcal{G}_{t}$, so $\mathcal{T}_{k}(t)=G_{k}\cap\{t<\sigma^{(k)}\}\in\mathcal{G}_{t}$.

Step 2: proof of claim 2. Fix $t$. By conclusion (a) of the completion-of-squares theorem, $R_{t}$ is symmetric and, under (H1), symmetric positive definite, hence invertible with $R_{t}^{-1}$ symmetric: from $R_{t}R_{t}^{-1}=I$ and the transpose identity $(MM')^{T}=M'^{T}M^{T}$ of claim 3 of the componentwise toolkit we get $(R_{t}^{-1})^{T}R_{t}=(R_{t}^{-1})^{T}R_{t}^{T}=(R_{t}R_{t}^{-1})^{T}=I$, so, right-multiplying this identity by $R_{t}^{-1}$, $(R_{t}^{-1})^{T}=(R_{t}^{-1})^{T}R_{t}R_{t}^{-1}=R_{t}^{-1}$. Hence

$$\Xi_{t}^{T}=\bigl(W_{t}R_{t}^{-1}W_{t}^{T}\bigr)^{T}=W_{t}\,(R_{t}^{-1})^{T}\,W_{t}^{T}=\Xi_{t},$$

using the transpose identity twice and $(W^{T})^{T}=W$; so $\Xi_{t}$ is symmetric. For positive semidefiniteness, let $x\in\mathbb{R}^{l}$ and put $y=W_{t}^{T}x\in\mathbb{R}^{m}$. By the transpose identity of claim 3 of the componentwise toolkit,

$$x\cdot\Xi_{t}x=x\cdot W_{t}\bigl(R_{t}^{-1}y\bigr)=\bigl(W_{t}^{T}x\bigr)\cdot\bigl(R_{t}^{-1}y\bigr)=y\cdot R_{t}^{-1}y\ \ge\ 0,$$

the last inequality because $R_{t}^{-1}$ is symmetric positive definite, being the inverse of a symmetric positive definite matrix by the lemma on inverses of positive definite matrices. Continuity and boundedness of the entries of $\Xi_{t}$ follow from those of $W_{t}$ and $R_{t}^{-1}$, recorded in conclusion (a) of the completion-of-squares theorem, together with continuity of sums and products and the extreme value theorem.

Step 3: proof of claim 3. Fix $t$ and $\mathcal{H}\in\mathcal{G}_{t}$. Square-integrability of the components of $\mathfrak{s}_{t}$ and $\mathfrak{a}_{t}$ was recorded in the preamble, and the almost-sure $\mathcal{G}_{t}$-measurability of the components of $\mathfrak{a}_{t}$ is the observation-adaptedness lemma. Write $\hat{\mathfrak{s}}_{t}$ for the tuple with components $\mathbb{E}[\mathfrak{s}^{\gamma}_{t}\mid\mathcal{G}_{t}]$, so that $\mathfrak{s}_{t}=\hat{\mathfrak{s}}_{t}+\varepsilon_{t}$, and set

$$c_{t}=\mathfrak{a}_{t}+R_{t}^{-1}W_{t}^{T}\hat{\mathfrak{s}}_{t},\qquad\text{so that}\qquad u_{t}=c_{t}+R_{t}^{-1}W_{t}^{T}\varepsilon_{t}$$

by linearity of the matrix-vector product. Each component of $c_{t}$ is square-integrable and almost surely equal to a $\mathcal{G}_{t}$-measurable square-integrable random variable: this holds for the components of $\mathfrak{a}_{t}$ as just recalled, for those of $\hat{\mathfrak{s}}_{t}$ by the existence and uniqueness theorem for conditional expectation, and is preserved by the fixed linear combinations with the constant coefficients $(R_{t}^{-1}W_{t}^{T})^{j\gamma}$.

Expanding the symmetric bilinear form $x\cdot R_{t}y$ at $u_{t}=c_{t}+R_{t}^{-1}W_{t}^{T}\varepsilon_{t}$,

$$u_{t}\cdot R_{t}u_{t}=c_{t}\cdot R_{t}c_{t}+2\,c_{t}\cdot R_{t}\bigl(R_{t}^{-1}W_{t}^{T}\varepsilon_{t}\bigr)+\bigl(R_{t}^{-1}W_{t}^{T}\varepsilon_{t}\bigr)\cdot R_{t}\bigl(R_{t}^{-1}W_{t}^{T}\varepsilon_{t}\bigr).$$

For the third term, $R_{t}R_{t}^{-1}=I$ and the transpose identity give $\bigl(R_{t}^{-1}W_{t}^{T}\varepsilon_{t}\bigr)\cdot R_{t}\bigl(R_{t}^{-1}W_{t}^{T}\varepsilon_{t}\bigr)=\bigl(R_{t}^{-1}W_{t}^{T}\varepsilon_{t}\bigr)\cdot\bigl(W_{t}^{T}\varepsilon_{t}\bigr)=\varepsilon_{t}\cdot W_{t}R_{t}^{-1}W_{t}^{T}\varepsilon_{t}=\varepsilon_{t}\cdot\Xi_{t}\varepsilon_{t}$, the middle equality moving $W_{t}^{T}$ across the pairing by the transpose identity and using the symmetry of the dot product. For the second term, $R_{t}R_{t}^{-1}=I$ gives $2\,c_{t}\cdot\bigl(W_{t}^{T}\varepsilon_{t}\bigr)=2\,\bigl(W_{t}c_{t}\bigr)\cdot\varepsilon_{t}$ by the transpose identity. Hence, multiplying by $\mathbf{1}_{\mathcal{H}}$ and taking expectations,

$$\mathbb{E}\bigl[\mathbf{1}_{\mathcal{H}}u_{t}\cdot R_{t}u_{t}\bigr]=\mathbb{E}\bigl[\mathbf{1}_{\mathcal{H}}\,c_{t}\cdot R_{t}c_{t}\bigr]+2\sum_{\gamma=1}^{l}\mathbb{E}\bigl[\mathbf{1}_{\mathcal{H}}\,(W_{t}c_{t})^{\gamma}\,\varepsilon^{\gamma}_{t}\bigr]+\mathbb{E}\bigl[\mathbf{1}_{\mathcal{H}}\,\varepsilon_{t}\cdot\Xi_{t}\varepsilon_{t}\bigr].$$

The cross term vanishes. Fix $\gamma$ and put $\Psi=\mathbf{1}_{\mathcal{H}}(W_{t}c_{t})^{\gamma}$, a square-integrable random variable. Each component of $c_{t}$ is almost surely equal to a $\mathcal{G}_{t}$-measurable square-integrable random variable, so, choosing such representatives and forming the same fixed linear combination, there is a $\mathcal{G}_{t}$-measurable square-integrable $\Psi'$ with $\Psi=\Psi'$ almost surely — here $\mathbf{1}_{\mathcal{H}}$ is itself $\mathcal{G}_{t}$-measurable and bounded, $\mathcal{H}$ belonging to $\mathcal{G}_{t}$, and square-integrability of $\Psi'$ follows from that of $\Psi$ by the lemma on almost sure equality and square-integrability. By the defining property of the conditional expectation $\mathbb{E}[\mathfrak{s}^{\gamma}_{t}\mid\mathcal{G}_{t}]$ — that $\mathbb{E}[(\mathfrak{s}^{\gamma}_{t}-\mathbb{E}[\mathfrak{s}^{\gamma}_{t}\mid\mathcal{G}_{t}])\,\Psi']=0$ for every square-integrable $\mathcal{G}_{t}$-measurable $\Psi'$, as in the existence and uniqueness theorem — we get $\mathbb{E}[\Psi'\varepsilon^{\gamma}_{t}]=0$. Since $\Psi=\Psi'$ almost surely, the products $\Psi\varepsilon^{\gamma}_{t}$ and $\Psi'\varepsilon^{\gamma}_{t}$ agree almost surely and are both integrable, so their expectations coincide and $\mathbb{E}[\mathbf{1}_{\mathcal{H}}(W_{t}c_{t})^{\gamma}\varepsilon^{\gamma}_{t}]=0$. Summing over $\gamma$, the cross term is $0$.

Conclusion. The first term is nonnegative: $R_{t}$ is positive semidefinite by conclusion (a) of the completion-of-squares theorem (indeed positive definite under (H1)), so $c_{t}\cdot R_{t}c_{t}\ge0$ pointwise, and $\mathbf{1}_{\mathcal{H}}\ge0$. Therefore

$$\mathbb{E}\bigl[\mathbf{1}_{\mathcal{H}}u_{t}\cdot R_{t}u_{t}\bigr]\ \ge\ \mathbb{E}\bigl[\mathbf{1}_{\mathcal{H}}\,\varepsilon_{t}\cdot\Xi_{t}\varepsilon_{t}\bigr]=\sum_{\gamma,\delta=1}^{l}\Xi^{\gamma\delta}_{t}\,\mathbb{E}\bigl[\mathbf{1}_{\mathcal{H}}\varepsilon^{\gamma}_{t}\varepsilon^{\delta}_{t}\bigr],$$

the last equality by linearity and the definition of the entry pairing. The right-hand side is nonnegative because $\Xi_{t}$ is positive semidefinite by claim 2, so $\varepsilon_{t}\cdot\Xi_{t}\varepsilon_{t}\ge0$ pointwise. (The same conclusion, in the abstract setting of a single quadratic form, is claim 2 of the restricted conditional mean-square optimality lemma; the direct computation above is given because the present situation involves the two distinct matrices $R_{t}$ and $\Xi_{t}$ and the intermediate linear map $R_{t}^{-1}W_{t}^{T}$.)

Independence of the choice. Let $M^{\gamma}$ and $M'^{\gamma}$ both be conditional expectations of $\mathfrak{s}^{\gamma}_{t}$ given $\mathcal{G}_{t}$, with errors $\varepsilon^{\gamma}_{t}=\mathfrak{s}^{\gamma}_{t}-M^{\gamma}$ and $\varepsilon'^{\gamma}_{t}=\mathfrak{s}^{\gamma}_{t}-M'^{\gamma}$. By the uniqueness clause of the existence and uniqueness theorem, $M^{\gamma}=M'^{\gamma}$ almost surely for each $\gamma$, hence $\varepsilon^{\gamma}_{t}=\varepsilon'^{\gamma}_{t}$ almost surely, hence $\mathbf{1}_{\mathcal{H}}\varepsilon^{\gamma}_{t}\varepsilon^{\delta}_{t}=\mathbf{1}_{\mathcal{H}}\varepsilon'^{\gamma}_{t}\varepsilon'^{\delta}_{t}$ almost surely for all $\gamma,\delta$ (a finite intersection of events of probability $1$ having probability $1$). Both products are integrable, being bounded, so their expectations agree, and the middle quantity of claim 3 is unchanged.

Step 4: proof of claim 4. Let $k\le K-1$ and $t\in[t_{k},T]$. By claim 1 the event $\mathcal{T}_{k}(t)=G_{k}\cap\{t<\sigma^{(k)}\}$ belongs to $\mathcal{G}_{t}$, and $\mathbf{1}_{\mathcal{T}_{k}(t)}=\mathbf{1}_{G_{k}}\mathbf{1}_{\{t<\sigma^{(k)}\}}$; claim 3 applied with $\mathcal{H}=\mathcal{T}_{k}(t)$ gives the assertion. $\blacksquare$