Proof of Mean Deviation Bound for the Aggregate Fluctuation Covariance along a Mean-Field Trajectory Pair

Throughout, adopt the notation of the statement, and write $x_s=(S_s,A_s)$ and $y_s=(\Sigma_s,\alpha_s)$ as points of $\mathbb{R}^{l+m}$ under the coordinate identification of the extension definition, so that $y_s-x_s=z_s/\sqrt{N}$ pointwise. Products with an infinite factor are read with the convention $0\cdot\infty=0$. If $c_\Theta=0$ then $B=K=0$, the rate bound forces $\beta\equiv0$, hence $\Theta\equiv0$ by its entry formulas, every left-hand side of the statement vanishes, and all three clauses hold; assume $c_\Theta>0$ below.

Step 1: pointwise bound (clause (a)). Fix $\omega\in\Omega$, $s\in[0,T]$, and $\gamma,\delta\in\{1,\dots,l\}$. By the definition of the controlled $N$-agent dynamics, the state processes take values in $\{1,\dots,l\}$ at every point, so the empirical state measure $\Sigma_s(\omega)$ lies in the probability simplex $\Delta^l$; and $S_s\in\Delta^l$, the mean-field trajectory pair having $S:[0,T]\to\Delta^l$. Moreover $\alpha_s(\omega)\in\mathcal{A}$ at every point of $\Omega$ — the control process of a solution takes values in $\mathcal{A}$ by that definition's solution data, the policy being $\mathcal{A}$-valued — and $A_s\in\mathcal{A}$ (the trajectory pair having $A:[0,T]\to\mathcal{A}$). Hence $x_s$ and $y_s$ lie in $\Delta^l\times\mathcal{A}\subseteq\Delta^l\times V$, and every point of the segment $\{x_s+\tau(y_s-x_s):\tau\in[0,1]\}$ lies in $\Delta^l\times V$: the convex combination $(1-\tau)S_s+\tau\Sigma_s(\omega)$ of two points of the simplex has nonnegative entries with sum $(1-\tau)+\tau=1$, and the control coordinates form a convex combination of the two points $A_s$ and $\alpha_s(\omega)$ of $\mathcal{A}\subseteq V$, the set $V$ being convex by the extension definition. Fix an ordered pair $(\sigma,\gamma')$ with $\sigma\neq\gamma'$. By clause 1 of the extension definition, $\bar{\beta}(\sigma,\gamma',\cdot,\cdot)$ agrees with $\beta(\sigma,\gamma',\cdot,\cdot)$ on $\Delta^l\times\mathcal{A}$, which contains $x_s$ and $y_s$; by clause 2 it is a $C^1$ map on the open set $U\times V\supseteq\Delta^l\times V$; and by clause 3, $|\partial_i\bar{\beta}(\sigma,\gamma',x)|\le K$ for all $i$ and all $x\in U\times V$, in particular on the segment. Part (i) of the Taylor lemma, with $n=l+m$ and $M_1=K$, therefore gives

Moreover $|\Sigma^\sigma_s-S^\sigma_s|\le|y_s-x_s|$, a single coordinate of the difference being at most its Euclidean norm, and, by the transition-rate family definition, $0\le\beta(\sigma,\gamma',\cdot,\cdot)\le B$, while $0\le S^\sigma_s\le1$ on the simplex. Hence, for each ordered pair $(\sigma,\gamma')$ with $\sigma\neq\gamma'$,

the first step by the triangle inequality (the Euclidean distance is a metric, applied in $\mathbb{R}$) after adding and subtracting $S^\sigma_s\,\beta(\sigma,\gamma',y_s)$. By the entry formulas of the aggregate fluctuation covariance, the entry $\Theta^{\gamma\gamma}$ is a sum of $2(l-1)$ terms of this form (the pairs $(\sigma,\gamma)$ and $(\gamma,\sigma)$ for $\sigma\neq\gamma$), and an off-diagonal entry $\Theta^{\gamma\delta}$ ($\gamma\neq\delta$) is, up to sign, a sum of $2$ such terms; since $l\ge2$, in either case the triangle inequality gives

which is clause (a).

Step 2: mean bound (clause (b)). Fix $s\in[0,T]$ and $\gamma,\delta$. The map $\Theta^{\gamma\delta}(\Sigma_s,\alpha_s)$ is a random variable: the entry $\Theta^{\gamma\delta}$ is a finite sum of products of coordinate maps with members of the rate family, hence sequentially continuous on $\Delta^l\times\mathcal{A}$ by the joint-continuity clause of the transition-rate family definition, and measurability of sequentially continuous functions of measurable Euclidean maps applies to the components of $(\Sigma_s,\alpha_s)$, which are random variables by the controlled-dynamics definition (as in the proof of the martingale decomposition). By clause (a) of the weighted second-moment evolution lemma, applied with the constant matrix family $Z_t=0$ (whose densities $\dot{z}^{\gamma\delta}\equiv0$ are continuous), $\mathbb{E}[\Theta^{\gamma\delta}(\Sigma_s,\alpha_s)]$ is finite, and it is bounded and measurable as a function of $s$. Subtracting the constant $\Theta^{\gamma\delta}(S_s,A_s)$, whose expectation is itself, and using linearity of the expectation and the bound $|\mathbb{E}[X]|\le\mathbb{E}[|X|]$ from the integrable-case clause there,

the last step by clause (a) and monotonicity, $|z_s|$ being a random variable — a continuous function of the components of $\mathfrak{s}_s$ and $\mathfrak{a}_s$, which are random variables by the controlled-dynamics definition and the trajectory pair. If $\mathbb{E}[|\mathfrak{s}_s|^2]+\mathbb{E}[|\mathfrak{a}_s|^2]=\infty$, the right-hand side of clause (b) is $\infty$ ($c_\Theta>0$) and the clause is trivial. Otherwise $|z_s|$ is square-integrable, since $\mathbb{E}[|z_s|^2]=\mathbb{E}[|\mathfrak{s}_s|^2]+\mathbb{E}[|\mathfrak{a}_s|^2]$ by the pointwise identity $|z_s|^2=|\mathfrak{s}_s|^2+|\mathfrak{a}_s|^2$ and additivity; the constant $1$ is square-integrable with $\mathbb{E}[1^2]=1$, so the mean-square Cauchy--Schwarz inequality applied to the pair $(|z_s|,1)$ gives $\mathbb{E}[|z_s|]=\mathbb{E}[|z_s|\cdot1]\le\big(\mathbb{E}[|z_s|^2]\big)^{1/2}\cdot1$, the left side being nonnegative so the absolute value there is immaterial. This completes clause (b).

Step 3: integrated bound (clause (c)). The integrand $s\mapsto\big|\mathbb{E}[\Theta^{\gamma\delta}(\Sigma_s,\alpha_s)]-\Theta^{\gamma\delta}(S_s,A_s)\big|$ is measurable, by measurability of sequentially continuous functions of measurable maps applied to the continuous function $(u,v)\mapsto|u-v|$ and the two measurable functions of $s$: the bounded measurable function of Step 2, and $s\mapsto\Theta^{\gamma\delta}(S_s,A_s)$, which is continuous and bounded, as both are recorded in clause (c) of the statement (boundedness via the extreme value theorem), and hence measurable (claim 3). Being measurable and bounded on $[0,T]$, the integrand has a defined, finite Lebesgue integral, by monotonicity against a constant and claim 1 of the toolkit ($\lambda([0,T])=T$). If $\mathcal{A}_2=\infty$ the right-hand side of clause (c) is $\infty$ ($c_\Theta>0$) and the claim is trivial; assume $\mathcal{A}_2<\infty$, so that $\mathcal{S}+\mathcal{A}_2<\infty$ by $\mathcal{S}\le4NT$. Set $\phi(s)=\big(\mathbb{E}[|\mathfrak{s}_s|^2]+\mathbb{E}[|\mathfrak{a}_s|^2]\big)^{1/2}\in[0,\infty]$; the map $s\mapsto\mathbb{E}[|\mathfrak{s}_s|^2]+\mathbb{E}[|\mathfrak{a}_s|^2]$ is a measurable $[0,\infty]$-valued function by clause (a) of the a priori second-moment bound and additivity, and $\phi$ is measurable: for real $a\ge0$, $\{\phi>a\}=\{\mathbb{E}[|\mathfrak{s}_\cdot|^2]+\mathbb{E}[|\mathfrak{a}_\cdot|^2]>a^2\}$ is measurable, for $a<0$ the set $\{\phi>a\}$ is all of $[0,T]$, and the sets $(a,\infty]$ over real $a$ generate the $\sigma$-algebra of $[0,\infty]$ used for $[0,\infty]$-valued measurability. Moreover $\int_{[0,T]}\phi^2\,ds=\mathcal{S}+\mathcal{A}_2<\infty$ by additivity for nonnegative measurable integrands. Let $D=\{s:\phi(s)<\infty\}$ and $\phi_0=\phi\,\mathbf{1}_{D}$, where $\mathbf{1}_D$ is the function equal to $1$ on $D$ and $0$ off $D$; then $\phi_0$ is measurable and real-valued, and $D$ is co-null: for every natural number $n$, $\mathcal{S}+\mathcal{A}_2\ge n^2\,\lambda([0,T]\setminus D)$ by monotonicity and the integral of a simple function ($\lambda$ the trace Lebesgue measure), forcing $\lambda([0,T]\setminus D)=0$. By claim 6 of the interval toolkit applied to the nonnegative measurable functions $\phi$ and $\phi^2$ with the co-null set $D$,

since $\phi_0=\phi\mathbf{1}_D$ and $\phi_0^2=\phi^2\mathbf{1}_D$ pointwise under the convention $0\cdot\infty=0$. By clause (b), the integrand of clause (c) is at most $\tfrac{c_\Theta}{\sqrt{N}}\,\phi(s)$ at every $s$, so by monotonicity and linearity (the constant $\tfrac{c_\Theta}{\sqrt{N}}$ passing out of the integral), and then claim 4 (Cauchy--Schwarz) of the interval toolkit applied to the pair $(1,\phi_0)$ on $[0,T]$ (both measurable with finite integrals of their squares, $T>0$), whose squared form $\big(\int_{[0,T]}\phi_0\,ds\big)^2\le T\int_{[0,T]}\phi_0^2\,ds$ passes to square roots because the nonnegative square root is nondecreasing,

where the first inequality also uses that $\phi$ and $\phi_0$ have equal integrals. This is clause (c).\ $\square$