Proof of Stopped Completion of Squares on a Cascade Block

Throughout we use linearity and monotonicity of the integral freely. All random variables below are bounded — the entries of $Z_{s},\dot{z}(s),E_{s},\mathsf{B}_{s},Q_{s},V_{s},R_{s},R_{s}^{-1},W_{s}$ are continuous in $s$, hence bounded on $[0,T]$ by conclusion (a) of the completion-of-squares theorem, while $|\mathfrak{s}_{s}|\le2\sqrt{N}$ everywhere and $|\mathfrak{a}_{s}|$ is bounded by the adopted setting, and $|e_{s}|\le c_{e}N^{-1/2}(|\mathfrak{s}_{s}|^{2}+|\mathfrak{a}_{s}|^{2})$ is bounded in turn by conclusion (b) of that theorem, whence so is $g_{s}=e_{s}+E_{s}\mathfrak{s}_{s}+\mathsf{B}_{s}\mathfrak{a}_{s}$ — so every expectation appearing exists and is finite, and the finiteness assertions of claim 3 hold.

Step 1: proof of claim 1. That $\tau\ge t_{\flat}$ everywhere is immediate, since $\sigma\ge t_{\flat}$. Let $s\in[0,T]$. If $s<t_{\flat}$ then $\tau>s$ everywhere, so $\{\tau\le s\}=\emptyset\in\mathcal{F}^{\mathrm{sys}}_{s}$. If $s\ge t_{\flat}$ then

$$\{\tau\le s\}=\bigl(\mathcal{G}\cap\{\sigma\le s\}\bigr)\cup\bigl(\Omega\setminus\mathcal{G}\bigr),$$

because off $\mathcal{G}$ one has $\tau=t_{\flat}\le s$. Here $\mathcal{G}\in\mathcal{F}^{\mathrm{sys}}_{t_{\flat}}\subseteq\mathcal{F}^{\mathrm{sys}}_{s}$, a filtration being nondecreasing, and $\{\sigma\le s\}\in\mathcal{F}^{\mathrm{sys}}_{s}$ by the definition of a stopping time; a $\sigma$-algebra being closed under complements, intersections and unions, $\{\tau\le s\}\in\mathcal{F}^{\mathrm{sys}}_{s}$. Thus $\tau$ is a stopping time.

For $s\in[t_{\flat},T]$ and $\omega\in\Omega$: if $\omega\notin\mathcal{G}$ then $\tau(\omega)=t_{\flat}\le s$, so $\mathbf{1}_{\{s<\tau\}}(\omega)=0=\mathbf{1}_{\mathcal{G}}(\omega)\mathbf{1}_{\{s<\sigma\}}(\omega)$; if $\omega\in\mathcal{G}$ then $\tau(\omega)=\sigma(\omega)$ and both sides equal $\mathbf{1}_{\{s<\sigma\}}(\omega)$. Since $\tau\ge t_{\flat}$, $\min(t_{\flat},\tau)=t_{\flat}$ everywhere; and for $t\in[t_{\flat},T]$ and $\omega\notin\mathcal{G}$, $\min(t,\tau(\omega))=\min(t,t_{\flat})=t_{\flat}$.

Step 2: proof of claim 2. Fix $s$ and $\omega$ and drop them from the notation. Since $Z$ is symmetric, $\mathfrak{s}\cdot(E^{T}Z)\mathfrak{s}=\mathfrak{s}\cdot(ZE)\mathfrak{s}=\mathfrak{s}\cdot Z(E\mathfrak{s})$, by the transpose identity $y\cdot(Mz)=(M^{T}y)\cdot z$ of claim 3 of the componentwise toolkit together with the entry definitions; hence $\mathfrak{s}\cdot(E^{T}Z+ZE)\mathfrak{s}=2\,\mathfrak{s}\cdot Z(E\mathfrak{s})$. Substituting the density $\dot{z}=-(E^{T}Z+ZE-WR^{-1}W^{T}+Q)$ of (H2) and the definition $e=g-E\mathfrak{s}-\mathsf{B}\mathfrak{a}$, so that $g=E\mathfrak{s}+\mathsf{B}\mathfrak{a}+e$,

$$\mathfrak{s}\cdot\dot{z}\,\mathfrak{s}+2\,\mathfrak{s}\cdot Zg=-2\,\mathfrak{s}\cdot Z(E\mathfrak{s})+\mathfrak{s}\cdot(WR^{-1}W^{T})\mathfrak{s}-\mathfrak{s}\cdot Q\mathfrak{s}+2\,\mathfrak{s}\cdot Z(E\mathfrak{s})+2\,\mathfrak{s}\cdot Z(\mathsf{B}\mathfrak{a})+2\,\mathfrak{s}\cdot Ze,$$

that is, $\mathfrak{s}\cdot\dot{z}\mathfrak{s}+2\mathfrak{s}\cdot Zg=\mathfrak{s}\cdot(WR^{-1}W^{T})\mathfrak{s}+2\,\mathfrak{s}\cdot(Z\mathsf{B})\mathfrak{a}-\mathfrak{s}\cdot Q\mathfrak{s}+2\,\mathfrak{s}\cdot Ze$. By $W=Z\mathsf{B}+\tfrac{1}{2}V$ we have $2\,\mathfrak{s}\cdot(Z\mathsf{B})\mathfrak{a}=2\,\mathfrak{s}\cdot W\mathfrak{a}-\mathfrak{s}\cdot V\mathfrak{a}$, so

$$\mathfrak{s}\cdot\dot{z}\mathfrak{s}+2\mathfrak{s}\cdot Zg=\Bigl(\mathfrak{s}\cdot(WR^{-1}W^{T})\mathfrak{s}+2\,\mathfrak{s}\cdot W\mathfrak{a}\Bigr)-\Bigl(\mathfrak{s}\cdot Q\mathfrak{s}+\mathfrak{s}\cdot V\mathfrak{a}\Bigr)+2\,\mathfrak{s}\cdot Ze .$$

On the other hand $R$ is symmetric by conclusion (a) of the completion-of-squares theorem and invertible under (H1), so $R^{-1}$ is symmetric as well, and expanding $u=\mathfrak{a}+R^{-1}W^{T}\mathfrak{s}$,

$$u\cdot Ru=\mathfrak{a}\cdot R\mathfrak{a}+2\,\mathfrak{a}\cdot R\bigl(R^{-1}W^{T}\mathfrak{s}\bigr)+\bigl(R^{-1}W^{T}\mathfrak{s}\bigr)\cdot R\bigl(R^{-1}W^{T}\mathfrak{s}\bigr)=\mathfrak{a}\cdot R\mathfrak{a}+2\,\mathfrak{a}\cdot(W^{T}\mathfrak{s})+\mathfrak{s}\cdot(WR^{-1}W^{T})\mathfrak{s},$$

using $RR^{-1}=I$ and $R^{-1}RR^{-1}=R^{-1}$, and $\mathfrak{a}\cdot(W^{T}\mathfrak{s})=\mathfrak{s}\cdot(W\mathfrak{a})$ by the transpose identity. Hence the first bracket above equals $u\cdot Ru-\mathfrak{a}\cdot R\mathfrak{a}$. Finally, by the first display of conclusion (a) of the completion-of-squares theorem applied to $z=\mathfrak{z}=(\mathfrak{s},\mathfrak{a})$,

$$\tfrac{1}{2}\sum_{i,j=1}^{l+m}H_{ij}\,\mathfrak{z}^{i}\mathfrak{z}^{j}=\mathfrak{s}\cdot Q\mathfrak{s}+\mathfrak{s}\cdot V\mathfrak{a}+\mathfrak{a}\cdot R\mathfrak{a},$$

so the second bracket equals $\tfrac{1}{2}\sum H_{ij}\mathfrak{z}^{i}\mathfrak{z}^{j}-\mathfrak{a}\cdot R\mathfrak{a}$. Substituting both brackets, the terms $\mathfrak{a}\cdot R\mathfrak{a}$ cancel and claim 2 follows.

Step 3: proof of claim 3. Apply conclusion (c) of the stopped weighted second-moment lemma with the stopping time $\tau$ of claim 1, at the two times $t_{\sharp}$ and $t_{\flat}$, and subtract. On the left, at $t_{\flat}$ we have $\min(t_{\flat},\tau)=t_{\flat}$ by claim 1, so the term is $\mathbb{E}[\mathbf{1}_{\Omega_{0}}\mathfrak{s}_{t_{\flat}}\cdot Z_{t_{\flat}}\mathfrak{s}_{t_{\flat}}]$; at $t_{\sharp}$, splitting $\mathbf{1}_{\Omega_{0}}=\mathbf{1}_{\mathcal{G}}+\mathbf{1}_{\Omega_{0}\setminus\mathcal{G}}$ (legitimate as $\mathcal{G}\subseteq\Omega_{0}$) and using $\min(t_{\sharp},\tau)=t_{\flat}$ off $\mathcal{G}$ (claim 1) and $\min(t_{\sharp},\tau)=\min(t_{\sharp},\sigma)$ on $\mathcal{G}$, the term is

$$\mathbb{E}\bigl[\mathbf{1}_{\mathcal{G}}\,\mathfrak{s}^{\sharp}\cdot Z^{\sharp}\mathfrak{s}^{\sharp}\bigr]+\mathbb{E}\bigl[\mathbf{1}_{\Omega_{0}\setminus\mathcal{G}}\,\mathfrak{s}_{t_{\flat}}\cdot Z_{t_{\flat}}\mathfrak{s}_{t_{\flat}}\bigr].$$

Subtracting, the common term $\mathbb{E}[\mathbf{1}_{\Omega_{0}\setminus\mathcal{G}}\mathfrak{s}_{t_{\flat}}\cdot Z_{t_{\flat}}\mathfrak{s}_{t_{\flat}}]$ cancels and the left-hand difference equals $\mathbb{E}[\mathbf{1}_{\mathcal{G}}\mathfrak{s}^{\sharp}\cdot Z^{\sharp}\mathfrak{s}^{\sharp}]-\mathbb{E}[\mathbf{1}_{\mathcal{G}}\mathfrak{s}_{t_{\flat}}\cdot Z_{t_{\flat}}\mathfrak{s}_{t_{\flat}}]$. On the right, the two integrals over $[0,t_{\sharp}]$ and $[0,t_{\flat}]$ differ by the integral over $[t_{\flat},t_{\sharp}]$, by the splitting of a Lebesgue integral over adjacent compact subintervals (obtained, as in the proof of the time-shift lemma, from claim 2 of the toolkit and claim 2 of the null-set lemma), the integrand being measurable and bounded by conclusion (a) of the stopped lemma. In that integrand, for $s\in[t_{\flat},t_{\sharp}]$ we have $\mathbf{1}_{\{s<\tau\}}=\mathbf{1}_{\mathcal{G}}\mathbf{1}_{\{s<\sigma\}}$ by claim 1, and $\mathbf{1}_{\Omega_{0}}\mathbf{1}_{\mathcal{G}}=\mathbf{1}_{\mathcal{G}}$; substituting claim 2 for $\mathfrak{s}_{s}\cdot\dot{z}(s)\mathfrak{s}_{s}+2\mathfrak{s}_{s}\cdot Z_{s}g_{s}$ and rearranging the resulting identity — moving the term carrying $\tfrac{1}{2}\sum H_{ij}\mathfrak{z}^{i}\mathfrak{z}^{j}$ to the left and the boundary terms to the right — gives exactly the display of claim 3.

Step 4: proof of claim 4. By claim 1 of the localized coercivity lemma, at every point

$$N\mathcal{D}_{s}\ \ge\ \tfrac{1}{2}\sum_{i,j=1}^{l+m}H_{ij}(s)\,\mathfrak{z}^{i}_{s}\mathfrak{z}^{j}_{s}\ -\ \tfrac{1}{2}(l+m)\bigl(\omega_{L}(\rho_{s})+C_{P}\omega_{b}(\rho_{s})\bigr)|\mathfrak{z}_{s}|^{2}.$$

Multiplying by $\mathbf{1}_{\mathcal{G}}\mathbf{1}_{\{s<\sigma\}}\ge0$, taking expectations, integrating over $[t_{\flat},t_{\sharp}]$ and using monotonicity,

$$\int_{[t_{\flat},t_{\sharp}]}\mathbb{E}\bigl[\mathbf{1}_{\mathcal{G}}\mathbf{1}_{\{s<\sigma\}}N\mathcal{D}_{s}\bigr]ds\ \ge\ \int_{[t_{\flat},t_{\sharp}]}\mathbb{E}\Bigl[\mathbf{1}_{\mathcal{G}}\mathbf{1}_{\{s<\sigma\}}\tfrac{1}{2}\sum_{i,j}H_{ij}(s)\mathfrak{z}^{i}_{s}\mathfrak{z}^{j}_{s}\Bigr]ds\ -\ \mathcal{X}^{\mathrm{quad}} .$$

Insert claim 3 for the first term on the right. In the resulting expression the term $\int\mathbb{E}[\mathbf{1}_{\mathcal{G}}\mathbf{1}_{\{s<\sigma\}}u_{s}\cdot R_{s}u_{s}]ds$ is at least $r\int\mathbb{E}[\mathbf{1}_{\mathcal{G}}\mathbf{1}_{\{s<\sigma\}}|u_{s}|^{2}]ds$ by hypothesis (H1) and monotonicity, and the term $2\int\mathbb{E}[\mathbf{1}_{\mathcal{G}}\mathbf{1}_{\{s<\sigma\}}\mathfrak{s}_{s}\cdot Z_{s}e_{s}]ds$ is at least $-\mathcal{X}^{\mathrm{lin}}$: indeed, by claim 1 of the componentwise toolkit and the bound $|Z^{\gamma\delta}_{s}|\le C_{Z}$,

$$\bigl|2\,\mathfrak{s}_{s}\cdot Z_{s}e_{s}\bigr|\le2\sum_{\gamma,\delta}|Z^{\gamma\delta}_{s}||\mathfrak{s}^{\gamma}_{s}||e^{\delta}_{s}|\le2\,C_{Z}\Bigl(\sum_{\gamma}|\mathfrak{s}^{\gamma}_{s}|\Bigr)\Bigl(\sum_{\delta}|e^{\delta}_{s}|\Bigr)\le2\,l^{2}\,C_{Z}\,|\mathfrak{s}_{s}|\,|e_{s}|\le2\,l^{2}\,C_{Z}\,c_{e}\,N^{-1/2}\,|\mathfrak{s}_{s}|\bigl(|\mathfrak{s}_{s}|^{2}+|\mathfrak{a}_{s}|^{2}\bigr),$$

using claim 1 of the componentwise toolkit twice, in the form $\sum_{\gamma}|x^{\gamma}|\le p\,|x|$ for $x\in\mathbb{R}^{p}$ with $p=l$, and then conclusion (b) of the completion-of-squares theorem; multiplying by the indicator, taking expectations and integrating gives the claimed bound. This proves the displayed inequality of claim 4.

For the final assertion, let $(s,\omega)$ satisfy $s\in[t_{\flat},t_{\sharp}]$ and $\mathbf{1}_{\mathcal{G}}(\omega)\mathbf{1}_{\{s<\sigma\}}(\omega)=1$, that is, $\omega\in\mathcal{G}$ and $s<\sigma(\omega)$. The confinement condition gives $\rho_{s}(\omega)\le\rho^{*}$, and claim 2 of the localized coercivity lemma gives $\tfrac{1}{2}(l+m)(\omega_{L}(\rho_{s})+C_{P}\omega_{b}(\rho_{s}))\le\tfrac{c_{J}}{2}$. Hence the integrand of $\mathcal{X}^{\mathrm{quad}}$ is bounded above pointwise by $\tfrac{c_{J}}{2}\mathbf{1}_{\mathcal{G}}\mathbf{1}_{\{s<\sigma\}}|\mathfrak{z}_{s}|^{2}$, and integrating gives $\mathcal{X}^{\mathrm{quad}}\le\tfrac{c_{J}}{2}\int_{[t_{\flat},t_{\sharp}]}\mathbb{E}[\mathbf{1}_{\mathcal{G}}\mathbf{1}_{\{s<\sigma\}}|\mathfrak{z}_{s}|^{2}]ds$. $\blacksquare$