Proof of Mean-Square Error Covariance of the Approximate Kalman Filter

Throughout, expectations are unchanged when integrands are modified off events of probability $1$, and all uses of the Tonelli and Fubini theorems are on the product of $[0,T]$ (trace Borel $\sigma$-algebra, restricted Lebesgue measure) with $(\Omega,\mathcal{F},\mathbb{P})$, both finite measures. By conclusions 1 and 2 of the policy lemma all entries of $\mathcal{E}$, $\mathcal{B}$, $\tilde{\mathcal{E}}$, $\mathcal{G}$, $\tilde{\mathcal{K}}$, $\Pi$, $(\tilde{\Theta}^\star)^{-1}$, and of the closed-loop matrix $\mathcal{M}_t=\mathcal{E}_t-\mathcal{B}_t\mathcal{G}_t-\tilde{\mathcal{K}}_t\tilde{\mathcal{E}}_t$ (written $M_t$ in the policy lemma; renamed $\mathcal{M}_t$ here to avoid collision with the state martingale vector $M_t$ of the statement) are continuous, hence bounded: fix entry bounds $c_{\mathcal{E}}$, $c_{\mathcal{B}}$, $c_{\tilde{\mathcal{E}}}$, $C_{\mathcal{G}}$, $C_{\tilde{K}}$, $c_\Pi$, $c_{\tilde\Theta}$, $c_{\mathcal{M}}$ respectively. For a matrix $H$ with $p$ rows, $q$ columns, and entries bounded by $c$, the row-wise Cauchy-Schwarz estimate gives $|Hx|\le\sqrt{pq}\,c\,|x|$ for $x\in\mathbb{R}^q$ (as in the proof of the completion-of-squares theorem); write $g^*=\sqrt{ml}\,C_{\mathcal{G}}$, $k^*=\sqrt{l\tilde{l}}\,C_{\tilde{K}}$, $c^*=\sqrt{l\,l}\,(c_{\mathcal{E}}+\tilde{l}C_{\tilde{K}}c_{\tilde{\mathcal{E}}})$, $c_{\mathcal{M}}^*=l\,c_{\mathcal{M}}$ for the resulting operator bounds of $\mathcal{G}_t$, $\tilde{\mathcal{K}}_t$, $\mathcal{E}_t-\tilde{\mathcal{K}}_t\tilde{\mathcal{E}}_t$, $\mathcal{M}_t$ (the entries of $\mathcal{E}_t-\tilde{\mathcal{K}}_t\tilde{\mathcal{E}}_t$ being bounded by $c_1=c_{\mathcal{E}}+\tilde{l}C_{\tilde{K}}c_{\tilde{\mathcal{E}}}$). Recall the residual bounds: with $z_s=(\mathfrak{s}_s,\mathfrak{a}_s)$, $\Lambda=l(B+K)\sqrt{l(l+m)}$, $\tilde{\Lambda}=\sqrt{l\tilde{l}}(\tilde{B}+\tilde{K})$, $c_e=\tfrac{3}{2}l\sqrt{l}K(l+m)$, and $\tilde{c}_e=\tfrac{3}{2}l\sqrt{\tilde{l}}\tilde{K}$, part (a) of the state residual lemma gives $|e_s|\le\min(c_e N^{-1/2}|z_s|^2,\,2\Lambda|z_s|)$ and part (a) of the observation residual lemma gives $|\tilde{e}_s|\le\min(\tilde{c}_e N^{-1/2}|\mathfrak{s}_s|^2,\,2\tilde{\Lambda}|\mathfrak{s}_s|)$, at every point of $[0,T]\times\Omega$. Finally let $\Xi_t$ be the sum of all counters, with $\tilde{c}_t\le\Xi_t$ and $\mathbb{E}[\Xi_T^p]<\infty$ for every natural $p$, by part (c) of the multiplier lemma.

Step 1: part (a). By conclusion 4(c) of the policy lemma, $|\hat{\mathfrak{s}}^N_t|\le C^\circ(N^{1/2}+N^{-1/2}\tilde{c}_T)$ at every point of $\Omega_0\times[0,T]$, so $|\mathfrak{a}_t|=|\mathcal{G}_t\hat{\mathfrak{s}}^N_t|\le g^*C^\circ(N^{1/2}+N^{-1/2}\Xi_T)$ there. Hence, for $p\in\{2,4\}$, $\mathbb{E}[|\mathfrak{a}_t|^p]\le(g^*C^\circ)^p\,2^p\,\mathbb{E}[N^{p/2}+N^{-p/2}\Xi_T^{\,p}]<\infty$ uniformly in $t$, and $\mathcal{A}$ and $\mathcal{A}_4$ are finite, their integrands being measurable by clause (a) of the a priori second-moment bound and of the fourth-moment bound. So the state residual lemma applies. Adaptedness. $\Sigma_t$, hence $\mathfrak{s}_t$, is $\mathcal{F}^{\mathrm{sys}}_t$-measurable by part (iv) of the existence theorem ($S_t$ being constant). For the filter: by conclusion 4(a) of the policy lemma, $\hat{\mathfrak{s}}^N_t=f^N_{\tilde{c}_t}(t,(\tau_1,\dots,\tau_{\tilde{c}_t}),(\upsilon_1,\dots,\upsilon_{\tilde{c}_t}))$ on $\Omega_0$; by part (iv) of the existence theorem the observation-event count, event times, and channels up to $t$ are measurable for the observation filtration, contained in $\mathcal{F}^{\mathrm{sys}}_t$; on each event $\{\tilde{c}_t=k\}\in\mathcal{F}^{\mathrm{sys}}_t$ the value $f^N_k(t,\tau,\upsilon)$ is, for each of the finitely many channel words $\upsilon\in\{1,\dots,\tilde{l}\}^k$, a sequentially continuous function of $(\tau_1,\dots,\tau_k)$ (its defining formula in conclusion 3 of the policy lemma involving the continuous $\Phi$, $\Psi$, $\tilde{\mathcal{K}}$ and, on $\{\tilde{c}_t=k\}$, indicators $\mathbf{1}_{\{\tau_j\le t\}}$ identically $1$), hence composes measurably by the composition lemma; the countable sum over $k$ of the indicator-multiplied values is then $\mathcal{F}^{\mathrm{sys}}_t$-measurable, and off $\Omega_0$ the modification is absorbed because $\mathcal{F}^{\mathrm{sys}}_t$ contains all null events (solution definition). So $\varepsilon_t=\mathfrak{s}_t-\hat{\mathfrak{s}}^N_t$ is $\mathcal{F}^{\mathrm{sys}}_t$-measurable. Product-measurability of $\mathbf{1}_{\Omega_0}\varepsilon^\gamma$ holds because $\mathbf{1}_{\Omega_0}\mathfrak{s}^\gamma$ is product-measurable (joint measurability lemma, as in Step 0 of the published proof of the weighted second-moment lemma) and $\mathbf{1}_{\Omega_0}\hat{\mathfrak{s}}^{N,\gamma}$ is product-measurable by conclusion 4(c) of the policy lemma. Moments: $|\varepsilon_t|\le2\sqrt{N}+C^\circ(N^{1/2}+N^{-1/2}\Xi_T)$ almost surely, so $\mathbb{E}[|\varepsilon_t|^p]<\infty$ for all $p$, uniformly in $t$; $\Pi^{N,\gamma\delta}_t$ is then finite, symmetric by commutativity, measurable in $t$ by the Fubini theorem (bounded-by-integrable product-measurable integrand $\mathbf{1}_{\Omega_0}\varepsilon^\gamma\varepsilon^\delta$), and bounded in $t$ by the uniform moment bound; $\hat{\mathfrak{s}}^N_0=0$ (conclusion 4(a)) gives $\varepsilon_0=\mathfrak{s}_0$ on $\Omega_0$, hence $\Pi^{N,\gamma\delta}_0=\mathbb{E}[\mathfrak{s}^\gamma_0\mathfrak{s}^\delta_0]$.

Step 2: part (b). Let $\Omega_1$ be the intersection of $\Omega_0$, the event $\Omega_{\mathfrak{a}}$ of clause (b) of the state residual lemma, and the almost-sure event of clause (a) of the martingale decomposition; $\mathbb{P}(\Omega_1)=1$. Fix $\omega\in\Omega_1$ and $t\in[0,T]$. Clause (c) of the state residual lemma (with $E_s=\mathcal{E}_s$, $\mathsf{B}_s=\mathcal{B}_s$ as noted in the statement) gives

all integrals existing componentwise at $\omega$ (the $e$-integral by clause (b) of that lemma, the others having bounded integrands on $\Omega_0$). Conclusion 4(b) of the policy lemma, with $N^{1/2}(\alpha_r-A_r)=\mathfrak{a}_r$, gives

where the jump sum of that conclusion is exactly the weighted observation sum $J^{\tilde{\mathcal{K}}}_t$ with weight $\tilde{\mathcal{K}}$. By the definition of the weighted compensated sum, $J^{\tilde{\mathcal{K}}}_t=\tilde{J}_t+N^{1/2}\int_{[0,t]}\tilde{\mathcal{K}}_r\tilde{b}(\Sigma_r)\,dr$. Subtracting the two displays, the $\mathcal{B}_r\mathfrak{a}_r$ terms cancel, and, with $\tilde{g}_r=\sqrt{N}(\tilde{b}(\Sigma_r)-\tilde{b}(S_r))$,

using the linearity of the Lebesgue integral (each summand separately integrable at $\omega$). By the definition of the observation linearization residual, $\tilde{g}_r=\tilde{\mathcal{E}}_r\mathfrak{s}_r+\tilde{e}_r$ pointwise, so $\tilde{\mathcal{K}}_r\tilde{\mathcal{E}}_r\hat{\mathfrak{s}}^N_r-\tilde{\mathcal{K}}_r\tilde{g}_r=-\tilde{\mathcal{K}}_r\tilde{\mathcal{E}}_r\varepsilon_r-\tilde{\mathcal{K}}_r\tilde{e}_r$, which yields the display of (b) on $\Omega_1$.

Step 3: part (c). Write $v_r=(\mathcal{E}_r-\tilde{\mathcal{K}}_r\tilde{\mathcal{E}}_r)\varepsilon_r+e_r-\tilde{\mathcal{K}}_r\tilde{e}_r$ and $m_t=\sqrt{N}M_t$, and define $V_t=\int_{[0,t]}v_r\,dr$ componentwise at each $\omega\in\Omega_1$ and $V_t=0$ off $\Omega_1$; each $V^\gamma_t$ is a random variable by the Tonelli theorem applied to the positive and negative parts of $\mathbf{1}_{\Omega_1}v^\gamma$, which is product-measurable: $\mathbf{1}_{\Omega_0}e^\gamma$ and $\mathbf{1}_{\Omega_0}\tilde{e}^\upsilon$ are product-measurable by clause (b) of the state and observation residual lemmas, $\mathbf{1}_{\Omega_0}\varepsilon^\gamma$ by Step 1, the coefficients are continuous in $r$, and products with the measurable $\mathbf{1}_{\Omega_1}$ preserve product-measurability (composition lemma). By Step 2, almost surely $\varepsilon_t=\varepsilon_0+V_t+m_t-\tilde{J}_t$ for all $t$, with $\varepsilon_0=\mathfrak{s}_0$. Moreover $v_r$ is $\mathcal{F}^{\mathrm{sys}}_r$-measurable up to modification on a null event (Step 1 for $\varepsilon_r$; $e_r=g_r-\mathcal{E}_r\mathfrak{s}_r-\mathcal{B}_r\mathfrak{a}_r$ and $\tilde{e}_r=\tilde{g}_r-\tilde{\mathcal{E}}_r\mathfrak{s}_r$ are compositions of continuous maps with the adapted $\Sigma_r$, $\alpha_r$ from part (iv) of the existence theorem), and $\sup_r\mathbb{E}[(v^\gamma_r)^2]<\infty$ by Step 1 and the linear residual bounds ($|v_r|\le c^*|\varepsilon_r|+2\Lambda|z_r|+2k^*\tilde{\Lambda}|\mathfrak{s}_r|$, and $|z_r|\le|\mathfrak{s}_r|+|\mathfrak{a}_r|$ has uniformly bounded second moments for the fixed $N$).

We record four facts, for $0\le s\le t\le T$ and all indices. (3a) If $X$ is square-integrable and $\mathcal{F}^{\mathrm{sys}}_s$-measurable then $\mathbb{E}[X(m^\delta_t-m^\delta_s)]=0$ and $\mathbb{E}[X(\tilde{J}^\delta_t-\tilde{J}^\delta_s)]=0$: the second is part (b) of the weighted-sums lemma; for the first, part (c) of the counter moment lemma gives, almost surely, $N M^\delta_u=\sum_{\sigma\neq\delta}(\mathfrak{M}^{\sigma\delta}_u-\mathfrak{M}^{\delta\sigma}_u)$ for all $u$, with $\mathfrak{M}^{\sigma\delta}=\sum_iM^{i,\sigma\delta}$ the aggregate compensated counters, so the claim follows from part (a) of the multiplier lemma applied per clock label and linearity. (3b) $\mathbb{E}[m^\gamma_tm^\delta_t]=\int_{[0,t]}\mathbb{E}[\Theta^{\gamma\delta}(\Sigma_s,\alpha_s)]\,ds$: part (c) of the martingale decomposition with $r=0$, $D=\Omega$, multiplied by $N$, followed by the Fubini theorem for the bounded product-measurable integrand (as in fact (1d) of the published proof of the weighted second-moment lemma). (3c) $\mathbb{E}[\tilde{J}^\gamma_t\tilde{J}^\delta_t]=\int_{[0,t]}\sum_\upsilon\tilde{\mathcal{K}}^{\gamma\upsilon}_s\tilde{\mathcal{K}}^{\delta\upsilon}_s\,\mathbb{E}[\tilde{b}^\upsilon(\Sigma_s)]\,ds$: part (c) of the weighted-sums lemma with $Z=1$, $r=0$, $F=G=\tilde{\mathcal{K}}$, followed by Fubini (the integrand bounded and product-measurable by its part (a)); the expectations $\mathbb{E}[\tilde{b}^\upsilon(\Sigma_s)]$ are bounded and measurable in $s$ by part (a) of the decomposition theorem and Fubini. (3d) $\mathbb{E}[m^\gamma_t\tilde{J}^\delta_t]=0$: part (d) of the weighted-sums lemma with $Z=1$, $r=0$ ($m_0=\tilde{J}_0=0$ almost surely).

Now expand $\mathbb{E}[\varepsilon^\gamma_t\varepsilon^\delta_t]$ using the four-term representation; all sixteen products are integrable (every factor lies in every mean-square space by Step 1, part (a) of the weighted-sums lemma, the boundedness of $M^\gamma$ on $\Omega_0$, and $|V_t|\le\int_{[0,T]}|v_r|dr$ with $\mathbb{E}[(\int|v|)^2]\le T\int\mathbb{E}|v_r|^2dr<\infty$ by the Cauchy-Schwarz inequality and Tonelli). The terms $\mathbb{E}[\varepsilon^\gamma_0m^\delta_t]$, $\mathbb{E}[\varepsilon^\gamma_0\tilde{J}^\delta_t]$ and their mirrors vanish by (3a) with $s=0$ ($\varepsilon_0=\mathfrak{s}_0$ bounded and $\mathcal{F}^{\mathrm{sys}}_0$-measurable). Next: $\mathbb{E}[\varepsilon^\gamma_0V^\delta_t]=\int_{[0,t]}\mathbb{E}[\varepsilon^\gamma_0v^\delta_s]ds$ by Fubini (dominated by $2\sqrt{N}\,\mathbf{1}_{\Omega_1}|v^\delta_s|$). Pathwise on $\Omega_1$, the integration by parts lemma with $u_0=v_0=0$ gives $V^\gamma_tV^\delta_t=\int_{[0,t]}(v^\gamma_sV^\delta_s+V^\gamma_sv^\delta_s)ds$, so $\mathbb{E}[V^\gamma_tV^\delta_t]=\int_{[0,t]}\mathbb{E}[v^\gamma_sV^\delta_s+V^\gamma_sv^\delta_s]ds$ by Fubini (dominated by $|v_s|\cdot\int_{[0,T]}|v_r|dr$, integrable on the product as just noted). Pathwise $V^\gamma_tm^\delta_t=\int_{[0,t]}v^\gamma_s\,m^\delta_t\,ds$, so by Fubini (domination by $\mathbf{1}_{\Omega_1}|v^\gamma_s||m^\delta_t|$, integrable on the product by Cauchy-Schwarz) and (3a) with $X=v^\gamma_s$,

and identically $\mathbb{E}[V^\gamma_t\tilde{J}^\delta_t]=\int_{[0,t]}\mathbb{E}[v^\gamma_s\tilde{J}^\delta_s]ds$ (using the second identity of (3a)), with the mirrored versions for the $(\delta,\gamma)$ pairs. Combining all terms with (3b), (3c), (3d), and re-assembling $\varepsilon^\delta_s=\varepsilon^\delta_0+V^\delta_s+m^\delta_s-\tilde{J}^\delta_s$ (almost surely) inside the integrands:

Finally, $\mathbb{E}[v^\gamma_s\varepsilon^\delta_s]=\sum_\rho(\mathcal{E}_s-\tilde{\mathcal{K}}_s\tilde{\mathcal{E}}_s)^{\gamma\rho}\,\Pi^{N,\rho\delta}_s+\mathbb{E}[(e_s-\tilde{\mathcal{K}}_s\tilde{e}_s)^\gamma\varepsilon^\delta_s]$ by linearity, and the pair of such terms produces $((\mathcal{E}-\tilde{\mathcal{K}}\tilde{\mathcal{E}})\Pi^N+\Pi^N(\mathcal{E}-\tilde{\mathcal{K}}\tilde{\mathcal{E}})^{\top})^{\gamma\delta}+X^{\gamma\delta}_s$, using the symmetry $\Pi^{N,\rho\delta}=\Pi^{N,\delta\rho}$ and the transpose convention. All integrands are bounded in $s$ (Step 1 moment bounds, fixed $N$) and measurable (Fubini on the product-measurable, bounded-by-integrable modified integrands), so $t\mapsto\Pi^{N,\gamma\delta}_t$ is continuous by the absolute continuity of the Lebesgue integral. This proves (c).

Step 4: part (d). On $\Omega_0$, subtracting as in Step 2 but keeping the filter alone, conclusion 4(b) of the policy lemma with $\mathfrak{a}_r=-\mathcal{G}_r\hat{\mathfrak{s}}^N_r$ and $J^{\tilde{\mathcal{K}}}_t=\tilde{J}_t+N^{1/2}\int_{[0,t]}\tilde{\mathcal{K}}_r\tilde{b}(\Sigma_r)dr$ gives, pathwise on $\Omega_0$, for all $t$:

with $\mathcal{M}_r$ the closed-loop matrix (as renamed above). By parts (i) and (ii) of the observation drift regularity lemma (restriction to the simplex and the Lipschitz bound, with $d(\Sigma_r,S_r)=N^{-1/2}|\mathfrak{s}_r|$), $|\tilde{g}_r|\le\sqrt{\tilde{l}}\,\sqrt{l}\,(\tilde{B}+\tilde{K})\,|\mathfrak{s}_r|=\tilde{\Lambda}|\mathfrak{s}_r|$. For a componentwise integral: for $t=0$ the bound $(\int_{[0,0]}f)^4\le0=t^3\int_{[0,0]}f^4$ is trivial; for $t\in(0,T]$, claim 4 of the toolkit (with $a=0<b=t$ and $g=1$) gives $(\int_{[0,t]}f)^2\le t\int_{[0,t]}f^2$ for measurable $f\ge0$ with $\int_{[0,t]}f^2<\infty$; applying this once to $f$ and once to $f^2$ in place of $f$ (valid for the bounded integrands used below, which have finite fourth moments) gives $(\int_{[0,t]}f^2)^2\le t\int_{[0,t]}f^4$, so $(\int_{[0,t]}f)^4=\big((\int_{[0,t]}f)^2\big)^2\le\big(t\int_{[0,t]}f^2\big)^2=t^2\big(\int_{[0,t]}f^2\big)^2\le t^2\cdot t\int_{[0,t]}f^4=t^3\int_{[0,t]}f^4$; together with $(a+b+c)^4\le27(a^4+b^4+c^4)$; hence, on $\Omega_0$,

Taking expectations (Tonelli, the integrands product-measurable as in Steps 1 and 3) and using $\mathbb{E}[|\tilde{J}_t|^4]\le l\sum_\gamma\mathbb{E}[(\tilde{J}^\gamma_t)^4]\le11\,l^2(1+C_{\tilde{K}})^4(1+\tilde{l}\tilde{B}T)^2=:c_7$ by part (e) of the weighted-sums lemma,

Part (b) of the fourth-moment bound, together with $|\mathfrak{a}_r|\le g^*|\hat{\mathfrak{s}}^N_r|$ (so $\mathbb{E}[|\mathfrak{a}_r|^4]\le(g^*)^4\mathbb{E}[|\hat{\mathfrak{s}}^N_r|^4]$), gives, with $c_{M'}=6l^2(2(l-1))^4$,

Set $\psi(t)=\mathbb{E}[|\mathfrak{s}_t|^4]+\mathbb{E}[|\hat{\mathfrak{s}}^N_t|^4]$. Adding the two displays, there are reals $a\le c_8\,\kappa_0$ and $b\ge0$, both determined by the data listed in (d) only ($a=27\mathbb{E}[|\mathfrak{s}_0|^4]+27c_{M'}(BT+(BT)^2)+27c_7\le c_8\kappa_0$), with $\psi(t)\le a+b\int_{[0,t]}\psi(s)ds$ for all $t$. The function $\psi$ is measurable (clause (a) of the fourth-moment lemma; conclusion 4(c) of the policy lemma with Tonelli) and bounded on $[0,T]$ for the fixed $N$ (Step 1). Let $u(t)=\int_{[0,t]}\psi(s)ds$; then $u$ is continuous by the absolute continuity of the integral, and $u(t)\le aT+b\int_{[0,t]}u(s)ds$ by monotonicity of the integral; since the Lebesgue and Riemann integrals of the continuous $u$ agree, Gronwall's lemma gives $u(t)\le aT\exp(bT)$, whence $\psi(t)\le a(1+bT\exp(bT))$ for every $t$. Since $|\varepsilon_t|^4\le8(|\mathfrak{s}_t|^4+|\hat{\mathfrak{s}}^N_t|^4)$ and $|\mathfrak{a}_t|^4\le(g^*)^4|\hat{\mathfrak{s}}^N_t|^4$, part (d) follows with $C_1=(8+(g^*)^4+1)\,c_8\,(1+bT\exp(bT))$; the displayed consequences follow from $\mathbb{E}[|\cdot|^2]\le1+\mathbb{E}[|\cdot|^4]$, $\kappa_0\ge1$, and integration over $[0,T]$.

Step 5: part (e). By conclusion 2 of the policy lemma, $\Pi_t=\Pi_0+\int_0^t(\mathcal{E}_r\Pi_r+\Pi_r\mathcal{E}_r^{\top}-\Pi_r\tilde{D}_r\Pi_r+\Theta^\star_r)dr$ with every $\Pi_r$ symmetric. We first record the algebraic identity, for each $r$:

Indeed, with $\tilde{\mathcal{K}}_r=\Pi_r\tilde{\mathcal{E}}_r^{\top}(\tilde{\Theta}^\star_r)^{-1}$, the reversal rule for the transpose, the symmetry of $\Pi_r$ and of $(\tilde{\Theta}^\star_r)^{-1}$ (a diagonal matrix, by conclusion 1 of the policy lemma): $\tilde{\mathcal{K}}_r\tilde{\mathcal{E}}_r\Pi_r=\Pi_r\tilde{D}_r\Pi_r$; $\Pi_r(\tilde{\mathcal{K}}_r\tilde{\mathcal{E}}_r)^{\top}=\Pi_r\tilde{\mathcal{E}}_r^{\top}(\tilde{\Theta}^\star_r)^{-1}\tilde{\mathcal{E}}_r\Pi_r=\Pi_r\tilde{D}_r\Pi_r$; and $\tilde{\mathcal{K}}_r\tilde{\Theta}^\star_r\tilde{\mathcal{K}}_r^{\top}=\Pi_r\tilde{\mathcal{E}}_r^{\top}(\tilde{\Theta}^\star_r)^{-1}\tilde{\Theta}^\star_r(\tilde{\Theta}^\star_r)^{-1}\tilde{\mathcal{E}}_r\Pi_r=\Pi_r\tilde{D}_r\Pi_r$, so the two sides agree, the three correction terms combining to $-\Pi_r\tilde{D}_r\Pi_r$. By clauses 6 and 7 of the fluctuation LQG data, $(\Theta^\star_r)^{\gamma\delta}=\Theta^{\gamma\delta}(S_r,A_r)$ and $(\tilde{\mathcal{K}}_r\tilde{\Theta}^\star_r\tilde{\mathcal{K}}_r^{\top})^{\gamma\delta}=\sum_\upsilon\tilde{\mathcal{K}}^{\gamma\upsilon}_r\tilde{\mathcal{K}}^{\delta\upsilon}_r\,\tilde{b}^\upsilon(S_r)$. Subtracting the resulting integral equation for $\Pi$ from the evolution identity of part (c) (the Riemann and Lebesgue integrals agreeing for the continuous integrand of the Riccati equation), the difference $D_t=\Pi^N_t-\Pi_t$ satisfies, entrywise,

where $\Delta^{\gamma\delta}_r=\mathbb{E}[\Theta^{\gamma\delta}(\Sigma_r,\alpha_r)]-\Theta^{\gamma\delta}(S_r,A_r)$ and $\tilde{\Delta}^{\gamma\delta}_r=\sum_\upsilon\tilde{\mathcal{K}}^{\gamma\upsilon}_r\tilde{\mathcal{K}}^{\delta\upsilon}_r(\mathbb{E}[\tilde{b}^\upsilon(\Sigma_r)]-\tilde{b}^\upsilon(S_r))$. We bound the three inhomogeneous terms using part (d); write $\bar{C}=(4+C_1)$, so all the second moments named in (d) are at most $\bar{C}\kappa_0$ and all the fourth moments at most $C_1\kappa_0$, and recall $\kappa_0\ge1$, so $\kappa_0^{1/2}\le\kappa_0$.

Residual cross terms. By the componentwise Cauchy-Schwarz inequality, $|X^{\gamma\delta}_r|\le2\,\Vert\varepsilon_r\Vert_2\big(\Vert e_r\Vert_2+k^*\Vert\tilde{e}_r\Vert_2\big)$, where $\Vert\cdot\Vert_2$ denotes the mean-square norm of the Euclidean norm of the indicated vector. By the quadratic residual bounds and (d): since $|z_r|^4\le8(|\mathfrak{s}_r|^4+|\mathfrak{a}_r|^4)$ pointwise, $\mathbb{E}[|z_r|^4]\le16C_1\kappa_0$, so $\Vert e_r\Vert_2\le c_eN^{-1/2}\,\mathbb{E}[|z_r|^4]^{1/2}\le4c_e\sqrt{C_1}\,N^{-1/2}\kappa_0^{1/2}$; likewise $\Vert\tilde{e}_r\Vert_2\le\tilde{c}_e N^{-1/2}\mathbb{E}[|\mathfrak{s}_r|^4]^{1/2}\le\tilde{c}_e\sqrt{C_1}N^{-1/2}\kappa_0^{1/2}$; and $\Vert\varepsilon_r\Vert_2\le(\bar{C}\kappa_0)^{1/2}$. Hence $|X^{\gamma\delta}_r|\le c_9\,N^{-1/2}\kappa_0$ with $c_9=2\sqrt{\bar{C}}(4c_e+k^*\tilde{c}_e)\sqrt{C_1}$.

Covariance deviation. By part (b) of the covariance deviation lemma and (d), $|\Delta^{\gamma\delta}_r|\le c_\Theta N^{-1/2}(\mathbb{E}[|\mathfrak{s}_r|^2]+\mathbb{E}[|\mathfrak{a}_r|^2])^{1/2}\le c_\Theta\sqrt{2\bar{C}}\,N^{-1/2}\kappa_0$, with its constant $c_\Theta=2(l-1)(B+K\sqrt{l+m})$.

Observation drift deviation. By parts (i) and (ii) of the observation drift regularity lemma, pointwise $|\tilde{b}^\upsilon(\Sigma_r)-\tilde{b}^\upsilon(S_r)|\le\sqrt{l}(\tilde{B}+\tilde{K})N^{-1/2}|\mathfrak{s}_r|$, so $|\mathbb{E}[\tilde{b}^\upsilon(\Sigma_r)]-\tilde{b}^\upsilon(S_r)|\le\sqrt{l}(\tilde{B}+\tilde{K})N^{-1/2}\Vert\mathfrak{s}_r\Vert_2\le\sqrt{l}(\tilde{B}+\tilde{K})\sqrt{\bar{C}}\,N^{-1/2}\kappa_0$ and $|\tilde{\Delta}^{\gamma\delta}_r|\le\tilde{l}C_{\tilde{K}}^2\sqrt{l}(\tilde{B}+\tilde{K})\sqrt{\bar{C}}\,N^{-1/2}\kappa_0=:c_{10}N^{-1/2}\kappa_0$.

Gronwall. Let $u(t)=\max_{\gamma,\delta}|D^{\gamma\delta}_t|$, continuous on $[0,T]$ (each $\Pi^{N,\gamma\delta}$ continuous by (c), each $\Pi^{\gamma\delta}$ continuous by conclusion 2 of the policy lemma; maxima of finitely many continuous functions are continuous). Entrywise, $|((\mathcal{E}-\tilde{\mathcal{K}}\tilde{\mathcal{E}})D)^{\gamma\delta}_r|\le l\,c_1\,u(r)$ and likewise for the transposed product, so by monotonicity of the integral,

As in Part (d) of the published proof of the completion-of-squares theorem, the Riemann-Lebesgue agreement for the continuous $u$ and Gronwall's lemma give

and since $u(0)=\max_{\gamma,\delta}|\mathbb{E}[\mathfrak{s}^\gamma_0\mathfrak{s}^\delta_0]-\Pi^{\gamma\delta}_0|$ by part (a), the conclusion (e) holds with $C=\exp(2lc_1T)\max\big(1,\,T(c_9+c_\Theta\sqrt{2\bar{C}}+c_{10})\big)$, which depends only on the data named in the statement. $\blacksquare$