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Weak order two for additive-noise stochastic Heun

Statement

Assume AC and the Brownian filtration conventions in Continuous Brownian Ito processes. Consider dXt=b(Xt)dt+σdWt on [0,T], where b∈Cb6(R;R), σ∈R is constant, and X0=x0∈R is deterministic. For t≤T and x∈R, let Xt,x denote the solution of the same SDE started from x at time t. Let ϕ:R→R have a bounded classical backward solution u(t,x)=E[ϕ(XTt,x)] with u(T,x)=ϕ(x), ut+Lu=0, L=b∂x+12σ2∂xx, and all mixed derivatives ∂tr∂xju for 0≤r≤3 and 0≤j≤6 continuous and bounded. On a uniform grid h=T/N≤1, initialize Y0=x0 and apply the additive-noise stochastic Heun update in Explicit stochastic Heun update for additive noise. There is a constant CT, independent of h, such that ∣Eϕ(YN)−Eϕ(XT)∣≤CTh2. This is weak order two, meaning the expectation error is bounded by a constant times h2 for the stated payoff class. The claim is specific to constant additive diffusion; no multiplicative-noise order-two result is asserted.

Facts & Assumptions

Given: AC; the scalar additive-noise SDE, deterministic initial value x0, drift and constant diffusion assumptions; the backward solution and derivative bounds; and the uniform grid of the Statement.

[L1]

Explicit stochastic Heun update for additive noise: Z=Y+b(Y)h+σΔW and Y+=Y+h2(b(Y)+b(Z))+σΔW.

[L2]

One-step moment matching for additive-noise stochastic Heun: for every starting time/state, the expectation of u(t+h,Y+)−u(t,Y) over one Brownian increment has absolute value at most Ch3, uniformly in the starting state.

[F1]

Continuous Brownian Ito processes: each next Brownian increment is independent of the current filtration and has its stated normal law.

[A1]

The Axiom of Choice: AC is inherited through the Brownian interface [F1]; the telescoping argument uses no further choice.

Proof

Given: The assumptions above.

1.1F1L1L2given

For each n, the numerical value Yn is Ftn-measurable by induction from the update [L1]. The next increment ΔWn is independent of Ftn by [F1]. Therefore the joint law of (Yn,ΔWn) is the product of the law of Yn and N(0,h); integrating the uniform one-step bound [L2] against the law of Yn gives ∣Eu(tn+1,Yn+1)−Eu(tn,Yn)∣≤Ch3.

2.1step 1.1algebra

Sum the inequalities in step 1.1 over n=0,…,N−1. The intermediate expectations telescope, so ∣Eu(T,YN)−Eu(0,Y0)∣≤NCh3=CTh2, because Nh=T.

3.1A1step 2.1given∎

The terminal condition gives u(T,YN)=ϕ(YN). Since Y0=x0 and u(0,x0) is the expectation of ϕ for the exact SDE started at (0,x0), Eu(0,Y0)=u(0,x0)=Eϕ(XT). Substituting these identities into step 2.1 yields the claimed weak error. The argument uses only the admitted additive-noise backward solution and makes no inference for state-dependent diffusion.

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