Dominating thinning has the prescribed reaction law

First assume $\mathrm{\Lambda }>0$. If the candidate interarrival increments are $E_n\mathrm{/}\mathrm{\Lambda }$, then for exactly $m$ candidates at ordered times $0<s_1<\cdots <s_m\le T$, the product of the $m$ exponential gap densities and the final no-candidate survival factor is ${\prod }_{i=1}^m\mathrm{\Lambda }{e}^{-\mathrm{\Lambda }(s_i-s_{i-1})}\cdot e^{-\mathrm{\Lambda }(T-s_m)}=e^{-\mathrm{\Lambda }T}{\mathrm{\Lambda }}^m,$ where $s_0=0$. This is the ordered-event density in [F4]. The independent uniform marks give the conditional product-uniform mark law, so [F5] identifies the finite marked candidate pattern with a Poisson point process of intensity $\mathrm{\Lambda }dsdu$ on $(0,T]\times [0,1]$. By the defining independent-count property in [F5], restrictions to disjoint time intervals are independent; after translating time, the restriction to any interval of length $v$ has ordered-time density $e^{-\mathrm{\Lambda }v}{\mathrm{\Lambda }}^m$. In particular $N^{\ast }(T)\sim \mathrm{Pois}(\mathrm{\Lambda }T)$.

Fix deterministic $t$, condition on ${\mathcal{F}}_t$ with $X_t=x$, and let $0\le u\le T-t$. By step 1.1 and [F5], future candidates in $(t,t+u]\times [0,1]$ are independent of ${\mathcal{F}}_t$ and their ordered-time density is $e^{-\mathrm{\Lambda }u}{\mathrm{\Lambda }}^m$. For $s\in (t,t+u]$ put $b(s)=\mathrm{\Lambda }-\lambda (s,x)\ge 0$. Conditional on $m$ ordered candidate times, the probability all marks are rejected is the product of their rejection fractions. Using that density and Tonelli [F6], the conditional probability of no accepted candidate is $e^{-\mathrm{\Lambda }u}{\sum }_{m=0}^{\mathrm{\infty }}{\int }_{t<s_1<\cdots <s_m\le t+u}{\prod }_{i=1}^{m}b(s_i)d{s}_1\cdots d{s}_m=e^{-\mathrm{\Lambda }u}{\sum }_{m=0}^{\mathrm{\infty }}\frac{({\int }_t^{t+u}b(s)ds{)}^m}{m!}=e^{-{\int }_t^{t+u}\lambda (s,x)ds}\mathrm{.}$ The middle identity holds because the symmetric product integrand has equal integrals on the $m!$ orderings of the cube; ties form a Lebesgue-null set.

Fix a channel $j$ and a Borel set $B\subseteq (t,T]$. By step 1.1 and [F5], future candidate points on $(t,T]$ are independent of ${\mathcal{F}}_t$ and have ordered-time density $e^{-\mathrm{\Lambda }(T-t)}{\mathrm{\Lambda }}^m$. For the first accepted candidate to occur at time $s\in B$ with mark in $D_j(s,x)$, all earlier candidates must be rejected, its mark must lie in a set of measure $a_j(s,x)\mathrm{/}\mathrm{\Lambda }$, and later candidates are unrestricted. For $k$ earlier rejected candidates, integration over their ordered times contributes $({\int }_t^{s}b(v)dv{)}^k\mathrm{/}k!$; for $\mathrm{\ell }$ later unrestricted candidates, integration contributes $(\mathrm{\Lambda }(T-s){)}^{\mathrm{\ell }}\mathrm{/}\mathrm{\ell }!$. Summing the nonnegative series and using [F6], the first-event measure on $B$ is $e^{-\mathrm{\Lambda }(T-t)}{\int }_B{a}_j(s,x)\left({\sum }_{k=0}^{\mathrm{\infty }}\frac{({\int }_t^{s}b(v)dv{)}^k}{k!}\right)\left({\sum }_{\mathrm{\ell }=0}^{\mathrm{\infty }}\frac{(\mathrm{\Lambda }(T-s){)}^{\mathrm{\ell }}}{\mathrm{\ell }!}\right)ds={\int }_B{a}_j(s,x)e^{-{\int }_t^s\lambda (v,x)dv}ds\mathrm{.}$ This proves the channel-specific first-event law for arbitrary Borel rates and for disjoint mark sets of the specified measures whose time-mark regions are product-measurable; their locations within $[0,1]$ do not enter the calculation. The endpoint mark conventions differ only on a null set.

At an accepted candidate time the new state depends only on the past and that candidate's time and mark. If its candidate index is $K$, the event $\{K=n\}$ is determined by the first $n$ gap and mark coordinates; partitioning over finite $n$ shows that the coordinates after $K$ retain their prescribed product law. Thus the same two calculations restart at the new state and time. Since $N(T)\le N^{\ast }(T)$ pathwise, [F7] and step 1.1 give $\mathbb{E}N(T)\le \mathrm{\Lambda }T$ and $N(T)<\mathrm{\infty }$ almost surely. Repeating the event-kernel construction proves the full path law through $T$.

If $\mathrm{\Lambda }=0$, the envelope forces every propensity to vanish everywhere, so the constant path is exact. If the global envelope hypothesis is false, this theorem asserts no path law; the algorithm returns invalid-envelope only if its pointwise guard encounters a violating candidate as specified in [F2]. Steps 1.1, 2.1, 3.1 and 4.1 establish the candidate law, both conditional event laws, the restart rule and the finite-count bound under the global envelope.