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Stopped density-scaled jump versus chemical Langevin weak model error

Statement

Assume Countable Choice and Dependent Choice. Use the density-scaled generators in Density-scaled reaction jump and chemical Langevin generators. Let K⊂U be compact and the closure of a bounded open stopping region with nonempty interior, let x0∈int⁡K, and let LV=x0+1VZd be a containing lattice. For fixed V, the set D=int⁡K∩LV is finite: boundedness of K bounds each integer coordinate index in the lattice. Let XV be the jump process stopped at τX=inf⁡{t≥0:XtV∉int⁡K} and let ZV be a chemical Langevin diffusion stopped at τZ=inf⁡{t≥0:ZtV∉int⁡K}. Assume both stopped processes exist with X0V=Z0V=x0, and write Xˉt=Xt∧τXV and Zˉt=Zt∧τZV.

Assume aj∈C3(U), 0≤aj≤Aj on K, and every segment [x,x+νj/V] for x∈D with aj(x)>0 lies in U. For a bounded payoff g with ∣g∣≤G, suppose the stopped CLE backward solution u(s,x)=Ex[g(Z(T−s)∧τZV)] has an extension to an open neighborhood of K and all these jump segments that is C3 in space and C1 in time, with u(T,y)=g(y) on all states reached by the stopped jump chain and u(s,y)=g(y) on every exit state (including diffusion boundary points), and ∂su+GVu=0on [0,T]×int⁡K,∥D3u(s,y)∥≤M, and ∂su is continuous in time on the stated neighborhood. The third-derivative bound holds along every jump segment. The derivatives are spatial Fréchet derivatives with operator norm. Then ∣Eg(XˉT)−Eg(ZˉT)∣≤TM6V2∑j=1RAj∥νj∥3. If the original local jump and diffusion models are defined through T and are coupled to agree with their stopped versions up to their respective exits, then ∣Eg(XTV)−Eg(ZTV)∣≤TM6V2∑j=1RAj∥νj∥3+2G(P(τX≤T)+P(τZ≤T)). The V−2 term is density-model error from the generator expansion; the exit term is separate. Neither term is time-discretization error. The existence and regularity of the CLE backward solution are assumptions, not consequences of this estimate.

Facts & Assumptions

Given: Countable Choice, Dependent Choice, the density-scaled generators, the compact stopping region, bounded propensities on K, and the stopped CLE backward solution with the stated boundary and derivative conditions, with X0V=Z0V=x0.

[F1]

Density-scaled reaction jump and chemical Langevin generators: for a state x before exit, LVϕ(x)=V∑j:aj(x)>0aj(x)(ϕ(x+νj/V)−ϕ(x)), GVϕ(x)=∑jaj(x)νj⋅∇ϕ(x)+12V∑jaj(x)νjTD2ϕ(x)νj.

[F2]

Dynkin identity for a bounded-rate finite stopped reaction chain: a finite-state reaction chain stopped on its finite exit set obeys the time-dependent Dynkin identity. For this fixed V, the lattice states in int⁡K form a finite set, and each has at most R one-jump destinations, so their exit set is finite. The total reaction rate before exit is bounded by V∑jAj. Relabeling a density state by y=V(x−x0)∈Zd transports the channel increments to νj and the rates to Vaj(x0+y/V), so the supplier applies to this chain.

[A1]

Countable Choice is assumed for the exponential event law in [F2]. The Axiom of Countable Choice (ACω)

[A3]

The stopped CLE and its stated backward solution exist; the backward equation, boundary values, and third-derivative bound hold. They are not inferred from Taylor expansion of the jump generator.

[A4]

For the unstopped comparison, each original process agrees with its stopped version up to its first exit, and ∣g∣≤G globally.

Proof

Given: Countable Choice, Dependent Choice, the stopped processes starting at x0, the derivative and propensity bounds, the backward equation, and the finite stopped-chain Dynkin identity.

1.1F1givenalgebra

Fix an interior lattice state x and a time s. For channel j, apply one-variable Taylor's theorem to r↦u(s,x+rνj/V) on [0,1]. Its third-order remainder has absolute value at most M∥νj∥3/(6V3). Multiplying by Vaj(x) and summing over channels gives LVu(s,x)=∑jaj(x)νj⋅∇u(s,x)+12V∑jaj(x)νjTD2u(s,x)νj+RV(s,x), where ∣RV(s,x)∣≤M6V2∑jAj∥νj∥3. The first two terms are exactly GVu(s,x) by [F1].

1.2F2A1A2A3given

Set ΨV(x)=V(x−x0), DV=ΨV(D), and let FV be the images under ΨV of the one-jump exit destinations from D. These are disjoint finite subsets of Zd. The transported stopped chain X~=ΨV(Xˉ) moves from y∈DV by channel increment νj at rate Vaj(x0+y/V) and is fixed on FV. For u~(s,y)=u(s,x0+y/V), its time derivative is unchanged and its reaction generator on DV equals LVu(s,x0+y/V). Apply [F2] to this integer-lattice chain and u~. Before τX, the integrand is ∂su+LVu=(LV−GV)u by the backward equation [A3]. After exit, Xˉ is fixed at an exit destination where u(s,Xˉs)=g(Xˉs) is independent of s, so both its time derivative and stopped generator vanish. At time T, u(T,XˉT)=g(XˉT), and the common initial condition gives Xˉ0=Zˉ0=x0 and u(0,x0)=Eg(ZˉT). Therefore Eg(XˉT)−Eg(ZˉT)=E∫0T∧τX(LV−GV)u(s,XsV) ds.

2.1step 1.1step 1.2A4algebra∎

Bound the integrand in step 1.2 by step 1.1 and use T∧τX≤T. This proves the stopped-process estimate. Under [A4], the stopped and original terminal payoffs agree unless their respective process exits by time T; on that event their difference is at most 2G. The triangle inequality for the two expectations adds at most 2GP(τX≤T)+2GP(τZ≤T), which proves the unstopped bound.

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