Stochastic Reaction Jump Simulation and Coupling — Examples

At state $n$, the only active channel has finite rate $n^2$ and moves to $n+1\in E$, so [F1] defines a valid countable-state network. With independent unit exponentials $E_n$, its $n$th holding time is $W_n=E_n\mathrm{/}{n}^2$; [F2] gives the stated law and the recursive path visits $1,2,3,\dots$ until its limiting time.

The density of $W_n$ is $n^2{e}^{-n^{2}t}$ for $t>0$. For $b>0$, [F5] and differentiation of the displayed antiderivative give $\mathbb{E}[W_n{\mathbf{1}}_{\{W_n\le b\}}]={\int }_0^{b}t{n}^2{e}^{-n^{2}t}dt=\frac{1-(n^{2}b+1)e^{-n^{2}b}}{n^2}\mathrm{.}$ As $b\to \mathrm{\infty }$, $(n^{2}b+1)e^{-n^{2}b}\to 0$: setting $u=n^{2}b$, the exponential series gives $e^u\ge u^3\mathrm{/}6$, so $(u+1)e^{-u}\le 6(u+1)\mathrm{/}{u}^3\to 0$. Applying [F4] to the increasing truncations yields $\mathbb{E}{W}_n=1\mathrm{/}{n}^2$.

For $n\ge 2$, $1\mathrm{/}{n}^2\le 1\mathrm{/}(n(n-1))=1\mathrm{/}(n-1)-1\mathrm{/}n$. Therefore $\mathbb{E}{\sum }_{n=1}^m{W}_n={\sum }_{n=1}^{m}1\mathrm{/}{n}^2\le 2$. Since the partial sums increase, [F4] gives $\mathbb{E}{\tau }_{\mathrm{\infty }}=\mathbb{E}{\lim }_m{\sum }_{n=1}^m{W}_n\le 2$. If $p=\mathbb{P}({\tau }_{\mathrm{\infty }}=\mathrm{\infty })>0$, then $\mathbb{E}{\tau }_{\mathrm{\infty }}\ge kp$ for every integer $k$, contradicting this finite bound. Thus ${\tau }_{\mathrm{\infty }}<\mathrm{\infty }$ almost surely.

For $V(n)=n$, the finite local generator sum is $QV(n)=n^2((n+1)-n)=n^2$. If $QV\le cV$ for all $n$, then $n\le c$ for all positive integers, impossible; thus [F7] does not apply although step 3.1 proves explosion.

The target system has constant total rate $2$; [F4] therefore gives its first marked-event survival $e^{-2t}$. This is an event even though $\nu =0$, by [F1].

A rate-one candidate process has no candidate by time $t$ with probability $e^{-t}$. The clipped rule accepts every candidate, hence its first accepted-event survival is $e^{-t}$; for $t>0$, $e^{-t}\mathrm{\ne }{e}^{-2t}$. Thus clipping changes the event law.

Here $a=2>\mathrm{\Lambda }=1$, so the required certificate is false. The specified procedure therefore lies outside the theorem's hypothesis and, whenever the violation is evaluated at a candidate, returns invalid-envelope; [F2] expressly forbids continuing by clipping.

The only positive rate is $a(1)=1$, and its poststate $1+(-1)=0$ lies in $E$, so the system is admissible. At $1$, [F2] gives $W\sim \mathrm{Exp}(1)$; at $0$, the total rate is $0$.

Once the first event occurs, the state is $0$ and every propensity is zero. By [F2] the process remains at $0$ at all later times. In an unlimited-budget call with finite horizon $T$, if $W<T$, the event is recorded and the next zero-propensity test returns status absorbed; if $W\ge T$, [F3] returns status horizon without recording the event. The underlying process therefore satisfies $\mathbb{P}(X_t=1)=e^{-t}$ and $\mathbb{P}(X_t=0)=1-e^{-t}$ for $t\ge 0$, independently of the call's horizon censoring.

The generator is $(Qf)(1)=f(0)-f(1)$ and $(Qf)(0)=0$ for every bounded $f:E\to \mathbb{R}$, so the zero-hazard state has no outgoing transition and the example satisfies the operator convention.

For $x<2$, the birth rate $2-x$ is positive and $x+1\in E$; at $x=2$ its rate is zero. For $x>0$, the death rate $x$ is positive and $x-1\in E$; at $x=0$ its rate is zero. Thus both channels are admissible, and $a_{+}(x)+a_{-}(x)=2$ for all $x\in E$.

At state $1$ the two propensities are each $1$. The direct method's waiting time is $E_1\mathrm{/}2$, and its cumulative selection rule makes the first event a birth exactly when $0\le U_1<1\mathrm{/}2$; after each event the same calculation applies at the new state.

The generator is $Qf(0)=2(f(1)-f(0))$, $Qf(1)=f(2)-2f(1)+f(0)$, and $Qf(2)=2(f(1)-f(2))$, obtained from the active channels without evaluating $f$ outside $E$. The rates are bounded by $\mathrm{\Lambda }=2$, so the finite-state theorem identifies the direct method with the exact path law and gives $\mathbb{E}N(T)\le 2T$.

For both systems, a positive birth rate occurs only at $x<2$ and moves to $x+1\in E$; a positive death rate occurs only at $x>0$ and moves to $x-1\in E$. All rates are nonnegative and finite, so [F1] gives valid finite-state systems.

At $(1,1)$ the birth rates are $1,1$, hence [F2] gives shared birth rate $1$. The death rates are $1,2$, hence shared death rate $1$ and $Y$-only death rate $1$; the $X$-only death rate is $0$. There are no other event types, so the total paired-event rate is $3$.

The algorithm's birth envelope is $M_{+}=\max \{{\max }_x{a}_{+}^X(x),{\max }_y{a}_{+}^Y(y)\}=2$, and its death envelope is $M_{-}=\max \{{\max }_x{a}_{-}^X(x),{\max }_y{a}_{-}^Y(y)\}=4$, so the total candidate rate is $M=M_{+}+M_{-}=6$. At the fixed pair $(1,1)$ the three disjoint event-mark portions have total length $3\mathrm{/}6$; [F4] gives first-event survival $e^{-6(3\mathrm{/}6)t}=e^{-3t}$ and assigns each equal-length portion probability $1\mathrm{/}3$ conditional on an event. Their state updates are respectively $(2,2)$, $(0,0)$, and $(1,0)$.

The marginal theorem [F3] applies because the budget is unlimited and the finite rates supply the bounds $M_{+}=2$ and $M_{-}=4$. It identifies each coordinate path with its own Gillespie reaction law, completing the example.