Proof of Fresh-Start Property for Independent Poisson Clocks Read at Levels Satisfying a Clock-Reading Bound

Throughout, $a$ ranges over the clock labels $\mathsf{A}$, $\mathcal{T}^a$ is the consumed level of clock $a$, an $\mathfrak{F}$-measurable random variable with $0\le\mathcal{T}^a\le\bar{c}$, and $\hat{Y}^a$ is the residual clock of the statement. This argument is adapted from the proof of Fresh-Start Property of the Controlled N-Agent Dynamics, of which the present lemma is the abstract form: there $\mathfrak{F}$ is the system filtration at time $r$, $\mathcal{I}$ is generated by the initial states, and the clock-reading bound is clause (vi) of Existence, Uniqueness, and Regularity for the Controlled N-Agent Dynamics.

Step 1: pathwise part of (a). For every $\omega$ and every clock, $u\mapsto\hat{Y}^a_u=Y^a_{\mathcal{T}^a+u}-Y^a_{\mathcal{T}^a}$ starts at $0$, is nondecreasing, integer-valued, right-continuous, and has unit jumps, all inherited from the counting path of $Y^a$; so every residual path is a counting path. Each $\hat{Y}^a_u$ is a random variable: for a natural number $m$ and $s\ge0$, $\{Y^a_s\ge m\}=\{\tau_m(Y^a)\le s\}$ with $\tau_m(Y^a)$ the $m$-th jump time, by monotonicity and right-continuity of counting paths, so $\tau_m(Y^a)$ is a $[0,\infty]$-valued measurable map, and, since $\mathcal{T}^a+u\le\bar{c}+u$, $\{Y^a_{\mathcal{T}^a+u}\ge m\}=\{\tau_m(Y^a)\le\mathcal{T}^a+u\}=\{\min(\tau_m(Y^a),\bar{c}+u+1)\le\mathcal{T}^a+u\}$ is an event, the minimum being a real-valued measurable map compared with another one (Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions); differences of random variables are random variables.

Step 2: grid approximation. Fix $\delta>0$. For each clock let $J_a$ be the smallest nonnegative integer with $\mathcal{T}^a\le J_a\delta$; then $J_a\le\lceil\bar{c}/\delta\rceil$, and the events $C_{\mathbf{j}}=\{J_a=j_a\ \text{for all}\ a\}$, over the finitely many integer vectors $\mathbf{j}$, partition $\Omega$ and lie in $\mathfrak{F}$, since each $\mathcal{T}^a$ is $\mathfrak{F}$-measurable. For an integer vector $\mathbf{k}$ write $C'_{\mathbf{k}}=\{\mathcal{T}^a\le k_a\delta\ \text{for all}\ a\}$ and

as in the statement, so that $\mathcal{H}_{\mathbf{k}}\subseteq\mathcal{H}_{\mathbf{j}}$ whenever $\mathbf{k}\le\mathbf{j}$ componentwise. By the clock-reading bound with caps $k_a\delta$, each $C'_{\mathbf{k}}$ agrees up to a null event with an event of $\mathcal{H}_{\mathbf{k}}$. Since

(where $\mathbf{j}-e_a$ decrements the $a$-th cap; the set subtracted for $j_a=0$ is empty), the cell $C_{\mathbf{j}}$ agrees up to a null event with an event $G_{\mathbf{j}}\in\mathcal{H}_{\mathbf{j}}$. Moreover, for $F\in\mathfrak{F}$, the clock-reading bound with caps $j_a\delta$ gives $H'\in\mathcal{H}_{\mathbf{j}}$ with $F\cap C'_{\mathbf{j}}$ equal to $H'\cap C'_{\mathbf{j}}$ up to a null event; intersecting with $C_{\mathbf{j}}\subseteq C'_{\mathbf{j}}$ and replacing $C_{\mathbf{j}}$ by $G_{\mathbf{j}}$,

Define the grid residuals $\hat{Y}^{a,\delta}_u=Y^a_{J_a\delta+u}-Y^a_{J_a\delta}$, which on $C_{\mathbf{j}}$ equal $Y^a_{j_a\delta+u}-Y^a_{j_a\delta}$. For fixed $\mathbf{j}$, the family of $\sigma$-algebras consisting of $\mathcal{H}_{\mathbf{j}}$ together with $\sigma(Y^a_{j_a\delta+u}-Y^a_{j_a\delta}:u\ge0)$, one per clock, is independent: the family consisting of $\mathcal{I}$ and the full clock $\sigma$-algebras is independent by hypothesis, within each clock the pre-$j_a\delta$ $\sigma$-algebra and the shifted-increment $\sigma$-algebra are independent, which we check as follows. The lemma Increments Are Independent of the Natural Filtration Past gives, for any $s<t$, that the single increment $Y^a_t-Y^a_s$ is independent of the natural filtration of $Y^a$ up to time $s$; it does not by itself cover the whole shifted $\sigma$-algebra, so we build the latter from finitely many increments. Fix $0=u_0<u_1<\dots<u_p$, nonnegative integers $m_1,\dots,m_p$, an event $E$ of the $\sigma$-algebra generated by the variables $Y^a_u$ with $u\le j_a\delta$, and set $A_q=\{Y^a_{j_a\delta+u_q}-Y^a_{j_a\delta+u_{q-1}}=m_q\}$. The event $E\cap A_1\cap\dots\cap A_{p-1}$ lies in the natural filtration of $Y^a$ up to time $j_a\delta+u_{p-1}$, so the cited lemma applied at that time gives $P(E\cap A_1\cap\dots\cap A_p)=P(E\cap A_1\cap\dots\cap A_{p-1})\,P(A_p)$; descending induction on $p$ yields $P\big(E\cap\bigcap_qA_q\big)=P(E)\prod_qP(A_q)$. The sets $\bigcap_qA_q$, together with $\Omega$ and $\emptyset$, form a $\pi$-system generating $\sigma(Y^a_{j_a\delta+u}-Y^a_{j_a\delta}:u\ge0)$, so Dynkin's Pi-Lambda Theorem upgrades the displayed factorization to independence of the two $\sigma$-algebras. The two levels then combine by factorizing $P\big(D_0\cap\bigcap_a(P_a\cap R_a)\big)=P(D_0)\prod_aP(P_a\cap R_a)=P(D_0)\prod_aP(P_a)P(R_a)$ for $D_0\in\mathcal{I}$, $P_a$ in the $\pi$-system of finite-dimensional cylinder events of $Y^a$ at times $\le j_a\delta$ (together with $\Omega$), and $R_a$ in the $\pi$-system of the sets $\bigcap_qA_q$ above (together with $\Omega$); the sets $D_0\cap\bigcap_aP_a$ form a $\pi$-system generating $\mathcal{H}_{\mathbf{j}}$, so, fixing all other entries and applying Dynkin's Pi-Lambda Theorem to one entry at a time (first the entry from $\mathcal{H}_{\mathbf{j}}$, then each shifted-increment $\sigma$-algebra in turn), the factorization extends to arbitrary events of the respective $\sigma$-algebras, which is the independence of the whole family; each shifted process $u\mapsto Y^a_{j_a\delta+u}-Y^a_{j_a\delta}$ has independent increments, and its increment over $(u_{q-1},u_q]$ has the Poisson distribution with parameter $u_q-u_{q-1}$ whenever $j_a\delta+u_q\le R$ (properties 2 and 3 of Poisson Clock with a Horizon, a deterministic shift preserving both; for $R=+\infty$ this is the defining property of a rate-$1$ Poisson process, Stochastic Process, Independent Increments, and Inhomogeneous Poisson Process). The lemma Increments Are Independent of the Natural Filtration Past used above applies to $Y^a$ because a Poisson clock with horizon $R$ has independent increments and starts at $0$.

Step 3: the limit identity. Let $0=u_0<u_1<\dots<u_p<R^-$, let $m^a_q$ be nonnegative integers ($q\in\{1,\dots,p\}$, finitely many clocks), write $\pi(\mu;m)=e^{-\mu}\mu^m/m!$ for the Poisson probabilities, and fix $F\in\mathfrak{F}$. Take $\delta=\delta_n=2^{-n}$ with $n$ so large that $\bar{c}+\delta_n+u_p<R$ (possible since $u_p<R-\bar{c}$; no restriction when $R=+\infty$), so that $J_a\delta_n+u_p\le\bar{c}+\delta_n+u_p<R$ on every cell and the shifted increments of Step 2 over $(u_{q-1},u_q]$ have the Poisson distributions with parameters $u_q-u_{q-1}$. The caps $J_a\delta_n$ decrease to $\mathcal{T}^a$ pointwise as $n\to\infty$ (they lie in $[\mathcal{T}^a,\mathcal{T}^a+\delta_n)$), so by right-continuity of the clock paths, for every $\omega$ and every $u$,

and since all values are integers the corresponding indicator variables converge pointwise. Fix $n$ and a cell $\mathbf{j}$. On $C_{\mathbf{j}}$ the grid residuals are the shifted increments of the cell, so $\mathbf{1}_{C_{\mathbf{j}}}\prod_{a,q}\mathbf{1}_{\{\hat{Y}^{a,\delta_n}_{u_q}-\hat{Y}^{a,\delta_n}_{u_{q-1}}=m^a_q\}}=\mathbf{1}_{C_{\mathbf{j}}}\Pi_{\mathbf{j}}$ with $\Pi_{\mathbf{j}}=\prod_{a,q}\mathbf{1}_{\{Y^a_{j_a\delta+u_q}-Y^a_{j_a\delta+u_{q-1}}=m^a_q\}}$, a product of indicators of events of the shifted-increment $\sigma$-algebras of the cell. Since $F\cap C_{\mathbf{j}}$ agrees with $H^F_{\mathbf{j}}\in\mathcal{H}_{\mathbf{j}}$ up to a null event, $\mathbb{E}[\mathbf{1}_{F\cap C_{\mathbf{j}}}\Pi_{\mathbf{j}}]=\mathbb{E}[\mathbf{1}_{H^F_{\mathbf{j}}}\Pi_{\mathbf{j}}]$, and the independence of Step 2 gives

Summing over $\mathbf{j}$ and using $\sum_{\mathbf{j}}P(H^F_{\mathbf{j}})=\sum_{\mathbf{j}}P(F\cap C_{\mathbf{j}})=P(F)$, then letting $n\to\infty$ with dominated convergence,

Step 4: conclusion. Taking $F=\Omega$ and one clock at a time (summing the product formula over all values of the increments of the remaining clocks, whose Poisson probabilities sum to $1$), the increments of $\hat{Y}^a$ over any finite partition $0=u_0<u_1<\dots<u_p<R^-$ anchored at $0$ are independent with the Poisson distributions of parameters $u_q-u_{q-1}$ (the joint probability mass function factorizes into the required product for every such partition, and events involving the increments of integer-valued variables are unions of such atoms). This is the increment assertion of (a); with Step 1 it proves (a). When $R=+\infty$, for times $0<v_1<\dots<v_p$ not anchored at $0$, apply this to the partition $0<v_1<\dots<v_p$ and sum the resulting product formula over all values of the leading increment $\hat{Y}^a_{v_1}-\hat{Y}^a_0$, whose Poisson probabilities sum to $1$; this marginalization gives the same factorization and the same Poisson laws for the increments over $v_1<\dots<v_p$, and together with $\hat{Y}^a_0=0$ it shows that each residual clock is a homogeneous Poisson process with rate $1$ all of whose paths are counting paths.

For (b): for each clock, the collection of events $\{\hat{Y}^a_{u_1}=w_1,\dots,\hat{Y}^a_{u_p}=w_p\}$ over finite time sets $0<u_1<\dots<u_p<R^-$ and integer values, together with $\Omega$ and $\emptyset$, is a $\pi$-system generating $\sigma(\hat{Y}^a_u:0\le u<R^-)$ (the variables are integer-valued, so these cylinder atoms generate). Step 3 shows that for every $F\in\mathfrak{F}$ and every choice of one such event $E_a$ per clock (finitely many clocks, the remaining ones taken to be $\Omega$; refine all the time sets involved to their common union, which still lies in $[0,R^-)$, write each $E_a$ as the disjoint union of the atoms of the refined grid it contains, and sum the product formula of Step 3 over these atoms),

since $P(E_a)$ is exactly the corresponding product of Poisson probabilities by (a). Fixing all but one entry and letting the remaining entry range over a $\pi$-system, Dynkin's $\pi$-$\lambda$ theorem upgrades each $\pi$-system in turn to the generated $\sigma$-algebra, one at a time; after finitely many applications this yields the factorization for arbitrary events from $\mathfrak{F}$ and from the residual clock $\sigma$-algebras, which is the asserted independence of the family (b). $\blacksquare$