Compensators, brackets, and jump-integral isometry for reaction counts

Let ${\mathcal{G}}_t$ be generated by candidate times and marks up to $t$; these sigma-algebras increase with $t$, so [F5] makes them a continuous-time filtration. If $\mathrm{\Lambda }=0$, all rates and all counts vanish. Otherwise the thinning realization has a rate-$\mathrm{\Lambda }$ candidate process, and [F2] gives the same reaction path law as Gillespie. By [F3], candidates have no ties and their number on $[0,T]$ is finite almost surely. Hence each $N_j$ is adapted, nondecreasing and integer-valued with unit jumps and finite terminal value; also $N_j(T)\le N^{\ast }(T)$ and $0\le A_j(T)\le \mathrm{\Lambda }T$. The state path is piecewise constant with finitely many jumps. Its left-limit process $X_{s-}$ is a pointwise limit of left-grid processes taking the value $X_{k{2}^{-n}}$ on $(k{2}^{-n},(k+1)2^{-n}]$; each is measurable for the generators in [F4]. Thus $a_j(X_{s-})$ is predictable. Because the density is bounded and piecewise constant along the path, for fixed $t$ its time integral is the limit of left-grid Riemann sums of values known by time $t$, so $A_j(t)$ is ${\mathcal{G}}_t$-measurable. The integral is continuous and bounded by $\mathrm{\Lambda }T$; continuous adapted processes are pointwise limits of their left-grid predictable approximations, so $A_j$ is predictable with finite mean.

Fix deterministic $s$ and $0<h\le T-s$, and condition on ${\mathcal{G}}_s$ with $X_s=x$. The first accepted channel-$j$ event has density $a_j(x)e^{-\lambda (x)u}$ at elapsed time $u$, by [F2]. Its probability over $0<u\le h$ differs from $a_j(x)h$ by at most $a_j(x)\lambda (x)h^2\mathrm{/}2\le {\mathrm{\Lambda }}^2{h}^2\mathrm{/}2$. Further channel-$j$ events require at least two candidates; if $N^{\ast }$ is their count and $\mu =\mathrm{\Lambda }h$, then [F3] gives $\mathbb{E}(N^{\ast }{1}_{\{N^{\ast }\ge 2\}})=\mu (1-e^{-\mu })\le {\mu }^2$. Therefore $\mid \mathbb{E}(N_j(s+h)-N_j(s)\mid {\mathcal{G}}_s)-a_j(x)h\mid \le \frac{3}{2}{\mathrm{\Lambda }}^2{h}^2\mathrm{.}$ If no candidate occurs in $(s,s+h]$, the state stays at $x$ and the compensator increment equals $a_j(x)h$. Otherwise both that increment and $a_j(x)h$ lie between $0$ and $\mathrm{\Lambda }h$, while the conditional probability of a candidate is at most $\mathrm{\Lambda }h$. Therefore $\mid \mathbb{E}(A_j(s+h)-A_j(s)\mid {\mathcal{G}}_s)-a_j(x)h\mid \le {\mathrm{\Lambda }}^2{h}^2,$ so $\mathrm{\mid }\mathbb{E}(M_j(s+h)-M_j(s)\mid {\mathcal{G}}_s)\mathrm{\mid }\le \frac{5}{2}{\mathrm{\Lambda }}^2{h}^2$. For $B\in {\mathcal{G}}_s$, partition $[s,t]$ into intervals of mesh at most $\delta$; conditional expectation on each left endpoint bounds $\mathrm{\mid }\mathbb{E}(1_B(M_j(t)-M_j(s)))\mathrm{\mid }$ by $\frac{5}{2}{\mathrm{\Lambda }}^2\delta (t-s)$. Letting $\delta \downarrow 0$ gives zero for every $B\in {\mathcal{G}}_s$, which is the all-pairs martingale identity. Since $N_j(T)\le N^{\ast }(T)$ and $A_j(T)\le \mathrm{\Lambda }T$, the bound $M_j(T{)}^2\le 2N_j(T{)}^2+2A_j(T{)}^2$ and [F3] give $\mathbb{E}{M}_j(T{)}^2<\mathrm{\infty }$; the same bound applies at every $t\le T$. Thus $M_j$ is square-integrable.

Candidates are simple, and each candidate mark selects at most one reaction channel. Therefore no two distinct channel counts jump together. Steps 1.1–2.1 verify the simple-count, finite-mean continuous predictable compensator, and square-integrable martingale hypotheses of [F6]. Applying [F7] gives $[M_j]=N_j$ and $⟨M_j⟩=A_j$; applying [F8] to each distinct pair gives $[M_i,M_j]=⟨M_i,M_j⟩=0$.

For bounded predictable step integrands $H_i,H_j$ on a common deterministic partition, expand $\mathbb{E}(I_i{I}_j)$, where $I_i=\int H_{i}d{M}_i$. Terms from distinct time intervals vanish by conditioning on the left endpoint of the later interval and using the martingale property from step 2.1. On an interval of length $h$, the same-interval product has conditional absolute expectation at most $5{\mathrm{\Lambda }}^2{h}^2$: the product of distinct channel counts is zero unless at least two candidates occur, whose $N^{\ast 2}$ expectation is ${\mu }^2+\mu (1-e^{-\mu })\le 2{\mu }^2$; the two count-compensator terms are each at most $\mathrm{\Lambda }h\mathbb{E}(N^{\ast }\mid {\mathcal{G}}_s)\le {\mathrm{\Lambda }}^2{h}^2$, and the compensator product is at most ${\mathrm{\Lambda }}^2{h}^2$. If the step coefficients are bounded in absolute value by $K$, multiplying by them gives a bound $5K^2{\mathrm{\Lambda }}^2{h}^2$. Refining the common partition leaves each step integral unchanged and makes the sum of these bounds tend to zero, so $\mathbb{E}(I_i{I}_j)=0$.

For general $H_j$ in the stated finite-energy class, [F10] supplies bounded predictable step approximations in each compensator $L^2$ norm. By [F9], their individual integrals converge in $L^2$ to $I_j={\int }_0^T{H}_{j}d{M}_j$ and satisfy $\mathbb{E}{I}_j^2=\mathbb{E}{\int }_0^T{H}_j^{2}d{A}_j$. For each $i\mathrm{\ne }j$, use the linearity and modulus inequality in [F12], then apply [F11] to the expanded product difference, to obtain $\mathrm{\mid }\mathbb{E}(I_i^{(n)}{I}_j^{(n)})-\mathbb{E}(I_i{I}_j)\mathrm{\mid }\le \mathrm{\parallel }{I}_i^{(n)}-I_i{\mathrm{\parallel }}_2\mathrm{\parallel }{I}_j^{(n)}{\mathrm{\parallel }}_2+\mathrm{\parallel }{I}_i{\mathrm{\parallel }}_2\mathrm{\parallel }{I}_j^{(n)}-I_j{\mathrm{\parallel }}_2\to 0$, since the isometry bounds the approximating norms. Step 3.2 therefore gives $\mathbb{E}(I_i{I}_j)=0$. Expanding the square of the finite sum now yields the displayed vector isometry because $d{A}_j=a_j(X_{s-})ds$. The proof uses AC only through [A1], the input-stream construction only through [A2]–[A3], and counts zero-increment reactions because they remain marked candidate acceptances.