Proof of Multiplier Identities and Interval Estimates for the Compensated Counters of the Controlled N-Agent Dynamics

Fix a solution and adopt the notation of the statement; in particular $n_c$ is the number of clock labels, $\bar{\beta}=\max(B,\tilde{B})$, and we set $\theta=1+\bar{\beta}T\ge1$. Throughout, $\Omega_0$ is the regular event of the solution; it has probability $1$, so expectations are unchanged when integrands are modified off $\Omega_0$, and we use this silently. Several pathwise comparisons and the frozen-consumption device below are adapted from Steps 1 and 3 of the published proof of the compensated-counters lemma (version with that label), extended here from indicator multipliers $\mathbf{1}_D$ to general multipliers $Z$.

Step 0: Poisson moment inequalities. Let $W$ be a random variable with the Poisson distribution with parameter $\mu\in[0,\bar{\beta}T]$, and let $j\ge1$ be a natural number with $j\le4$. By part (a) of the Poisson factorial-moments lemma, $\mathbb{E}[\prod_{q=0}^{j-1}(W-q)]=\mu^{j}$; in particular $\mathbb{E}[W]=\mu$ and $\mathbb{E}[W(W-1)]=\mu^2$. A Poisson variable takes values in the nonnegative integers almost surely, so the pointwise inequalities for nonnegative integers below may be taken in expectation. We also record, from the value $P(W=0)=e^{-\mu}$ of the Poisson distribution: $P(W\ge1)=1-e^{-\mu}\le\mu$, where the inequality holds for $\mu\le1$ by the alternating series of the exponential and for $\mu>1$ trivially, since $1-e^{-\mu}<1<\mu$. We use four consequences.

(P1) For every nonnegative integer $x$,

Indeed, the product vanishes for integer $0\le x\le j-1$ (the factor $x-x$ occurs) and is positive for $x\ge j$, hence is nonnegative for every nonnegative integer $x$; for $x=0$ both sides are $0$; for $1\le x\le 2j-1$ the first term alone suffices; and for $x\ge 2j$ each factor satisfies $x-q\ge x-j+1>x/2$, so $2^{j}\prod_{q=0}^{j-1}(x-q)\ge2^{j}(x/2)^{j}=x^{j}$.

(P2) $\mathbb{E}[W^{j}]\le m_j\,\mu$ with $m_j=(2j)^{j}+2^{j}\theta^{\,j-1}$: take expectations in (P1), use $P(W\ge1)\le\mu$, the factorial moment $\mu^{j}\le\theta^{\,j-1}\mu$, and monotonicity of the expectation. Explicitly $m_1=4$, $m_2=16+4\theta$, $m_3=216+8\theta^{2}$, $m_4=4096+16\theta^{3}$, and also $\mathbb{E}[W]=\mu$ exactly.

(P3) $P(W\ge2)\le\mu^{2}/2$: for nonnegative integers, $\mathbf{1}_{\{x\ge2\}}\le x(x-1)/2$, and $\mathbb{E}[W(W-1)]=\mu^{2}$.

(P4) For $2\le j\le4$, $\mathbb{E}[W^{j}\mathbf{1}_{\{W\ge2\}}]\le m'_j\,\mu^{2}$ with $m'_j=(2j)^{j}/2+2^{j}\theta^{\,j-2}$: as in (P1), for nonnegative integers $x$ and $j\ge2$, $x^{j}\mathbf{1}_{\{x\ge2\}}\le(2j)^{j}\mathbf{1}_{\{x\ge2\}}+2^{j}\prod_{q=0}^{j-1}(x-q)$ (the product also vanishes at $x=1$ since $j\ge2$); take expectations and use (P3) and $\mu^{j}\le\theta^{\,j-2}\mu^{2}$.

Step 1: part (c). By the event count bound (part (iii) of the existence theorem), almost surely $\Xi_t\le\Xi^{\sharp}_t$ for all $t\in[0,T]$, where $\Xi^{\sharp}_t$ is the sum, over all $n_c$ clock labels, of the variables $Y^{i,\sigma\gamma}_{Bt}$ and $\tilde{Y}^{i,\upsilon}_{\tilde{B}t}$, each of which has the Poisson distribution with parameter at most $\bar{\beta}t$, as recorded there. For nonnegative real numbers $x_1,\dots,x_n$ one has $(x_1+\cdots+x_n)^{p}\le n^{p}\max_q x_q^{p}\le n^{p}\sum_{q}x_q^{p}$, since the sum is at most $n$ times the maximum and the $p$-th power is nondecreasing on $[0,\infty)$. Hence, by monotonicity of the expectation and part (b) of the Poisson factorial-moments lemma,

which is the display of (c). For the products: every counter is nonnegative and nondecreasing in $t$ (off $\Omega_0$ the counters vanish, and on $\Omega_0$ each agrees with the restriction of a counting path by condition 3 of the solution definition), and every consumed clock time lies in $[0,\bar{\beta}T]$ by condition 2, so $|M^{a}_{s}|\le N^{a}_{s}+\bar{\beta}T\le\Xi_T+\bar{\beta}T$ for every label $a$ and $s\in[0,T]$, almost surely. Hence $|M^{a_1}_{t_1}\cdots M^{a_k}_{t_k}|\le(\Xi_T+\bar{\beta}T)^{k}\le2^{k}\big(\Xi_T^{\,k}+(\bar{\beta}T)^{k}\big)$ (the two-term case of the sum-power inequality above), which is integrable by the display with $p=k$. This proves (c).

Step 2: part (a). Fix a clock label $a$ and $t\in[r,T]$. The increment $M^a_t-M^a_r$ is square-integrable by part (a) of the compensated-counters lemma, so $Z(M^a_t-M^a_r)$ is integrable by the Cauchy--Schwarz inequality for the mean-square norm. By that lemma and the averaged form of the martingale property in the definition of a square-integrable martingale, $\mathbb{E}[(M^a_t-M^a_r)\mathbf{1}_D]=0$ for every event $D\in\mathcal{F}^{\mathrm{sys}}_r$, where $\mathbf{1}_D$ is the function equal to $1$ on $D$ and $0$ off $D$; by linearity the same holds with $\mathbf{1}_D$ replaced by any simple $\mathcal{F}^{\mathrm{sys}}_r$-measurable random variable (a finite linear combination of such indicators). Truncating $Z$ at level $n$ and discretizing the truncation's values on a dyadic grid produces simple $\mathcal{F}^{\mathrm{sys}}_r$-measurable variables $Z_n$ with $|Z_n|\le|Z|+1$ and $Z_n\to Z$ pointwise (the device of Step 3 of the published proof cited above). Then $\mathbb{E}[(M^a_t-M^a_r)Z]=\mathbb{E}[(M^a_t-M^a_r)(Z-Z_n)]$ for every $n$, and by the Cauchy--Schwarz inequality this is bounded in absolute value by the mean-square norm of $M^a_t-M^a_r$ times that of $Z-Z_n$, which tends to $0$ by dominated convergence applied to $|Z-Z_n|^{2}\le(2|Z|+1)^{2}$. Hence $\mathbb{E}[Z(M^a_t-M^a_r)]=0$.

Step 3: part (b). All products of at most four compensated counters at times in $[0,T]$ are integrable, and squares of products of two of them are integrable, by part (c); in particular $M^a_tM^b_t$ and $(M^a_t-M^a_r)(M^b_t-M^b_r)$ are square-integrable, so all products below involving the square-integrable multipliers $Z$, $ZM^a_r$, $ZM^b_r$ are integrable by the Cauchy--Schwarz inequality, and $Z(A^a_t-A^a_r)$ is integrable since $0\le A^a_t-A^a_r\le\bar{\beta}T$ everywhere by condition 2 of the solution definition together with the additivity and monotonicity of the Lebesgue integral on a compact interval. Part (b) of the compensated-counters lemma gives, for every $D\in\mathcal{F}^{\mathrm{sys}}_r$,

by linearity the identity holds with $\mathbf{1}_D$ replaced by any simple $\mathcal{F}^{\mathrm{sys}}_r$-measurable variable, and passing to the limit along the approximating sequence $Z_n$ of Step 2 extends it to the square-integrable $Z$: on the left, $|\mathbb{E}[(M^a_tM^b_t-M^a_rM^b_r)(Z-Z_n)]|$ tends to $0$ by the Cauchy--Schwarz inequality and dominated convergence as in Step 2, using square-integrability of $M^a_tM^b_t$ and $M^a_rM^b_r$; on the right, $\mathbb{E}[(Z-Z_n)(A^a_t-A^a_r)]\to0$ by dominated convergence with dominating function $(2|Z|+1)\,\bar{\beta}T$. Now the algebraic identity

holds pointwise (expand the right-hand side). Multiplying by $Z$ and taking expectations termwise (each term integrable as noted), the last two terms vanish by part (a) applied with the square-integrable multipliers $ZM^a_r$ and $ZM^b_r$, which are $\mathcal{F}^{\mathrm{sys}}_r$-measurable since $M^a_r$ and $M^b_r$ are (adaptedness, part (iv) of the existence theorem). This proves (b).

Step 4: the fresh-start frame for (d), (e), (f). Fix $r\in[0,T)$ and $0<\delta\le T-r$, write $\mu=\bar{\beta}\delta\le\bar{\beta}T\le\theta$, and for clock labels $b$ write $\Delta N^b=N^b_{r+\delta}-N^b_r$, $\Delta A^b=A^b_{r+\delta}-A^b_r$, $\Delta M^b=\Delta N^b-\Delta A^b$. (For $r=T$ there is no admissible $\delta$ and parts (d), (e), (f) are vacuous.) Let $\hat{Y}^b$ be the residual clocks of the fresh-start property at time $r$: rate-$1$ Poisson processes with counting paths, mutually independent and jointly independent of $\mathcal{F}^{\mathrm{sys}}_r$. Since the paths of $\hat{Y}^b$ are counting paths, $\hat{Y}^b_0=0$, so $\hat{Y}^b_{u}=\hat{Y}^b_{u}-\hat{Y}^b_{0}$ is an increment and has the Poisson distribution with parameter $u$ by the increment law of the Poisson process definition; thus Step 0 applies to $\hat{Y}^b_{\mu}$. Three pathwise facts: (i) $0\le\Delta A^b\le\mu$ at every outcome, by condition 2 of the solution definition (integrand in $[0,\bar{\beta}]$) and the integral toolkit; (ii) on $\Omega_0$, $0\le\Delta N^b\le\hat{Y}^b_{\mu}$, since by condition 3 the increments of $N^b$ over $(r,r+\delta]$ are increments of the clock path $Y^b$ over $(A^b_r,A^b_{r+\delta}]\subseteq(A^b_r,A^b_r+\mu]$; (iii) hence $|\Delta M^b|\le\hat{Y}^b_{\mu}+\mu$ on $\Omega_0$.

Step 5: part (d). Fix the label $a$ and suppose $|Z|\le\zeta$. Write $\Delta N=\Delta N^a$, $\Delta A=\Delta A^a$, $\Delta M=\Delta M^a$, $\hat{Y}=\hat{Y}^a$, and let $k\in\{3,4\}$. Integrability of $Z((\Delta M)^k-\Delta A)$ holds by part (c), boundedness of $Z$ and $\Delta A$, and the Cauchy--Schwarz inequality.

(d1) From $(\Delta M)^{k}$ to $\mathbf{1}_{\{\Delta N\ge1\}}$. Expand $(\Delta N-\Delta A)^{k}$ by the binomial theorem. For $0\le j\le k-1$, the term with $j$ factors $\Delta N$ consists of $\binom{k}{j}$ equal monomials, each bounded in absolute value on $\Omega_0$ by $\mu^{k-j}\hat{Y}_{\mu}^{\,j}$ (with $\hat{Y}^{0}_{\mu}=1$), whose expectation is at most $m_j\theta^{\,k-j-1}\mu^{2}$ by (P2) and $\mu\le\theta$ (for $j=0$, with $m_0=1$, since $\mu^{k}\le\theta^{\,k-2}\mu^{2}\le\theta^{\,k-1}\mu^{2}$); so the whole term contributes at most $\binom{k}{j}m_j\theta^{\,k-j-1}\mu^{2}$. For the term $j=k$: for nonnegative integers $x$, $0\le x^{k}-\mathbf{1}_{\{x\ge1\}}\le x^{k}\mathbf{1}_{\{x\ge2\}}$ (equality holds at $x=0,1$), and on $\Omega_0$, $(\Delta N)^{k}\mathbf{1}_{\{\Delta N\ge2\}}\le\hat{Y}_{\mu}^{\,k}\mathbf{1}_{\{\hat{Y}_{\mu}\ge2\}}$ since $\Delta N\le\hat{Y}_{\mu}$ are integers; so by (P4) the replacement of $(\Delta N)^{k}$ by $\mathbf{1}_{\{\Delta N\ge1\}}$ costs at most $m'_k\mu^{2}$ in expectation. Altogether

Numerically, $c^{(1)}_3\le\theta^{2}+12\theta+3(16+4\theta)+108+8\theta\le2^{9}\theta^{2}$ and $c^{(1)}_4\le\theta^{3}+16\theta^{2}+6(16+4\theta)\theta+4(216+8\theta^{2})+2048+16\theta^{2}\le2^{12}\theta^{3}$.

(d2) Freezing. Let $c^a\in[0,\mu]$ be the frozen consumption: the integral over $(r,r+\delta]$ of the rate obtained by freezing the time-$r$ states and observation record. As in Step 1 of the published proof cited above, $c^a$ is $\mathcal{F}^{\mathrm{sys}}_r$-measurable: the time-$r$ states are $\mathcal{F}^{\mathrm{sys}}_r$-measurable and the observation-event count, event times, and channels up to $r$ are measurable for the observation filtration, which is contained in $\mathcal{F}^{\mathrm{sys}}_r$, by part (iv) of the existence theorem; the frozen rate path is a measurable function of these data by composition measurability and the policy definition, and its integral is measurable by Tonelli. Let $R_b=\{\hat{Y}^b_{\mu}\ge1\}$, so $P(R_b)=1-e^{-\mu}\le\mu$ (Step 0), and $R=\bigcup_{b\neq a}R_b$. The pathwise claim of that proof applies verbatim:

Indeed, off $R$ no clock $b\neq a$ rings in $(r,r+\delta]$; then up to the first ring of $a$ in the interval (or up to $r+\delta$ if there is none) no counter jumps, the states and record stay at their time-$r$ values, and the consumed time of $a$ follows the frozen schedule, so $a$ rings in the interval if and only if $\hat{Y}^a_{c^a}\ge1$; thus on either difference set the outcome lies in $R$, and in both cases $\hat{Y}^a_{\mu}\ge1$ (once because $a$ rings within budget $\mu$, once because $c^a\le\mu$). By the pairwise independence of the residual clocks, $P(R\cap R_a)\le\sum_{b\neq a}P(R_b)P(R_a)\le n_c\mu^{2}$. Hence $|\mathbb{E}[Z\mathbf{1}_{\{\Delta N\ge1\}}]-\mathbb{E}[Z\mathbf{1}_{\{\hat{Y}^a_{c^a}\ge1\}}]|\le\zeta n_c\mu^{2}$.

(d3) Exponential formula. Splitting $Z=Z^{+}-Z^{-}$ into positive and negative parts, the joint measurability of $(c,\hat{\omega})\mapsto\hat{Y}^a_{c}$ (right-continuity and grid limits), the independence of $\hat{Y}^a$ from $\mathcal{F}^{\mathrm{sys}}_r$, and the Tonelli theorem on the product of the two laws give, as in the published Step 1, $\mathbb{E}[Z\mathbf{1}_{\{\hat{Y}^a_{c^a}\ge1\}}]=\mathbb{E}[Z(1-e^{-c^a})]$. Moreover $|1-e^{-c}-c|=c-(1-e^{-c})\le c^{2}/2$ for every $c\ge0$: for $0\le c\le2$ by the alternating series of the exponential (from the quadratic term on, the term magnitudes $c^{n}/n!$ are nonincreasing when $c\le3$), and for $c>2$ trivially, since then $0\le c-(1-e^{-c})\le c\le c^{2}/2$; the nonnegativity of $c-(1-e^{-c})$ is the bound $1-e^{-c}\le c$ of Step 0. Hence $|\mathbb{E}[Z\mathbf{1}_{\{\hat{Y}^a_{c^a}\ge1\}}]-\mathbb{E}[Z\,c^a]|\le\zeta\mu^{2}/2$, using $c^a\in[0,\mu]$.

(d4) Unfreezing the compensator. On the complement of $R\cup R_a$ no clock rings in $(r,r+\delta]$, so the states and record stay at their time-$r$ values there and $\Delta A=c^a$; always $|c^a-\Delta A|\le2\mu$, both lying in $[0,\mu]$. Hence $|\mathbb{E}[Z(c^a-\Delta A)]|\le\zeta\cdot2\mu\,P(R\cup R_a)\le2\zeta n_c\mu^{2}$.

Combining (d1)--(d4),

using $c^{(1)}_k+3n_c+1\le2^{12}\theta^{3}(n_c+1)+4(n_c+1)\le2^{13}(n_c+1)\theta^{3}$, $\bar{\beta}^{2}\le(1+\bar{\beta})^{2}$, and $\theta^{3}\le(1+\bar{\beta}T)^{4}$.

Step 6: part (e). Let $Z$ be square-integrable and let $\kappa=1/\sqrt{\delta}$. The truncation $Z^{(\kappa)}=\max(-\kappa,\min(\kappa,Z))$ is $\mathcal{F}^{\mathrm{sys}}_r$-measurable with $|Z^{(\kappa)}|\le \kappa$, so part (d) gives $|\mathbb{E}[Z^{(\kappa)}((\Delta M)^{k}-\Delta A)]|\le C_\star \kappa\delta^{2}=C_\star\delta^{3/2}$. For the remainder, $|Z-Z^{(\kappa)}|\le Z^{2}/\kappa$ pointwise (the left side vanishes when $|Z|\le \kappa$ and is at most $|Z|\le Z^{2}/\kappa$ otherwise), and $|(\Delta M)^{k}-\Delta A|\le(\hat{Y}_{\mu}+\mu)^{k}+\mu$ on $\Omega_0$ by Step 4(iii). By the independence of $\hat{Y}=\hat{Y}^a$ from $\mathcal{F}^{\mathrm{sys}}_r$ (fresh-start property) and the Tonelli theorem on the product of the laws (both factors nonnegative, $Z^{2}$ being $\mathcal{F}^{\mathrm{sys}}_r$-measurable),

By the two-term sum-power inequality of Step 1 and (P2), $\mathbb{E}[(\hat{Y}_{\mu}+\mu)^{k}]\le2^{k}(m_k+\theta^{\,k-1})\mu$, so $\mathbb{E}[(\hat{Y}_{\mu}+\mu)^{k}+\mu]\le(2^{k}(m_k+\theta^{\,k-1})+1)\mu\le2^{17}\theta^{3}\mu$ (for $k=4$: $16(4096+17\theta^{3})+1\le2^{17}\theta^{3}$; the case $k=3$ is smaller). With $\kappa=1/\sqrt{\delta}$ and $\mu=\bar{\beta}\delta$ the remainder is at most $2^{17}\theta^{3}\bar{\beta}\,\mathbb{E}[Z^{2}]\,\delta^{3/2}\le C_\star\,\mathbb{E}[Z^{2}]\,\delta^{3/2}$. Adding the truncated part proves (e). (All splittings are legitimate: $Z(\Delta M)^{k}$ and $Z\Delta A$ are integrable by (c) and the Cauchy--Schwarz inequality.)

Step 7: part (f). Let $a_1,\dots,a_j$ be clock labels, $2\le j\le4$, not all equal, and let $b_1,\dots,b_d$ be the distinct labels among them, with multiplicities $j_1+\cdots+j_d=j$; then $d\ge2$, each $j_i\le3$, and at most two of the $j_i$ exceed $1$. On $\Omega_0$, by Step 4(iii),

The family consisting of $\mathcal{F}^{\mathrm{sys}}_r$ and the $\sigma$-algebras of the individual residual clocks is independent (fresh-start property, part (b)), so by the Tonelli theorem on the product of the laws the expectation of the right-hand side factorizes as $\mathbb{E}[|Z|]\cdot\prod_{i=1}^{d}\mathbb{E}[(\hat{Y}^{b_i}_{\mu}+\mu)^{j_i}]$; in particular the left-hand side is integrable. By (P2) and the sum-power inequality, the factors satisfy: $\mathbb{E}[\hat{Y}_{\mu}+\mu]=2\mu$; $\mathbb{E}[(\hat{Y}_{\mu}+\mu)^{2}]\le4(16+5\theta)\mu\le2^{7}\theta\mu$; $\mathbb{E}[(\hat{Y}_{\mu}+\mu)^{3}]\le8(216+9\theta^{2})\mu\le2^{11}\theta^{2}\mu$. Running over the possible multiplicity patterns $(j_1,\dots,j_d)$ --- namely $(1,1)$, $(2,1)$, $(1,1,1)$, $(3,1)$, $(2,2)$, $(2,1,1)$, $(1,1,1,1)$ --- and using $\mu^{d}\le\theta^{\,d-2}\mu^{2}$, the product of the factors is in every case at most $2^{14}\theta^{2}\mu^{2}$ (the largest case being $(2,2)$ with $(2^{7}\theta\mu)^{2}$). Hence