Dynkin identity for a bounded-rate finite stopped reaction chain

Uniformly over $x\in D\cup F$, the no-event probability in a time interval of length $\delta$ is $e^{-\lambda (x)\delta }$, where $\lambda (x)=0$ on $F$ and $\lambda (x)\le A$ on $D$. Its difference from $1-\lambda (x)\delta$ is at most $A^2{\delta }^2\mathrm{/}2$. For a channel $j$ from $x\in D$, the probability that it fires once at time $r\in (0,\delta )$ and no later event occurs is ${\int }_0^{\delta }{a}_j(x)e^{-\lambda (x)r}{e}^{-\lambda (x+{\nu }_j)(\delta -r)}dr\mathrm{.}$ Both rates in the exponent are at most $A$, so its difference from $a_j(x)\delta$ is at most $a_j(x)A{\delta }^2$. Summing over channels gives at most $\lambda (x)A{\delta }^2\le A^2{\delta }^2$. After the first event, the conditional probability of another before $\delta$ is at most $A(\delta -r)$; integrating against the first-event rate, at most $A$, shows that the probability of at least two events is at most $A^2{\delta }^2\mathrm{/}2$. These estimates are uniform on the finite state set.

Since $D\cup F$ is finite and ${\mathrm{\partial }}_{t}v$ is continuous, uniformly in $s$ and $x$, $v(s+\delta ,x)=v(s,x)+\delta {\mathrm{\partial }}_{s}v(s,x)+o(\delta )\mathrm{.}$ The zero-event probability is $1-\lambda (x)\delta +O({\delta }^2)$; the single-channel probabilities are $a_j(x)\delta +O({\delta }^2)$ by step 1.1; and the multiple-event probability is $O({\delta }^2)$. In the zero-event case, the contribution is $v(s,x)+\delta ({\mathrm{\partial }}_{s}v(s,x)-\lambda (x)v(s,x))+o(\delta )$. The one-event channel contributions are $\delta {\sum }_j{a}_j(x)v(s,x+{\nu }_j)+o(\delta )$, because the time change in $v(s+\delta ,x+{\nu }_j)$ is multiplied by an $O(\delta )$ event probability. The multiple-event contribution is $O({\delta }^2)$. Adding the cases, the order-$\delta$ terms are $\delta {\mathrm{\partial }}_{s}v(s,x)+\delta {\sum }_j{a}_j(x)(v(s,x+{\nu }_j)-v(s,x)),$ with a uniform $o(\delta )$ remainder. This is $\delta ({\mathrm{\partial }}_{s}v(s,x)+Qv(s,\cdot )(x))+o(\delta )$; for $x\in F$ the chain is fixed and $Q=0$. Multiple-event payoffs contribute $O({\delta }^2)$ because $v$ is bounded on the finite set.

Sum the conditional identity in step 2.1 over a deterministic partition of $[0,t]$ and take expectations. The uniform $o(\delta )$ remainders sum to $o(1)$. By [F3] the chain has finitely many jumps almost surely on $[0,t]$; therefore $s\mapsto {\mathrm{\partial }}_{s}v(s,X_s)+Qv(s,\cdot )(X_s)$ is bounded and piecewise continuous, so its Riemann sums converge pathwise. Boundedness and dominated convergence let expectations pass to the integral. The left side telescopes to $\mathbb{E}[v(t,X_t)]-\mathbb{E}[v(0,X_0)]$, proving the identity.