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Uniform finite-horizon moments for tamed Euler approximations

Statement

Assume AC. Let d,m≥1 be finite, T>0, and let b,σ,ξ,W satisfy the coefficient, Brownian, and initial-moment hypotheses of Global existence and finite-horizon moments for one-sided Lipschitz SDEs. For each integer N≥max⁡(1,⌈T⌉), put h=T/N and define Y0,…,YN by Tamed Euler scheme for a superlinear drift. Then for every finite p≥1, sup⁡N≥max⁡(1,⌈T⌉) max⁡0≤n≤NE∣Yn∣p<∞. The bound may depend on p,T, the coefficient constants, and the initial moments, but not on N or n.

Facts & Assumptions

Given: AC; the autonomous coefficient and initial-moment hypotheses in the Statement; and the deterministic grid tn=nh.

[F1]

The tamed update is Yn+1=Yn+h b(Yn)/(1+h∣b(Yn)∣)+σ(Yn)ΔWn; in particular ∣h b(Yn)/(1+h∣b(Yn)∣)∣≤1. Tamed Euler scheme for a superlinear drift

[F2]

The sampled filtration (Ftn)n=0N is increasing. Filtration and filtered probability space

[F3]

Each increment ΔWn is independent of Ftn and has law Nm(0,hIm). d-dimensional Brownian motion

[F4]

For a nonnegative submartingale Z and r>1, ∥max⁡k≤NZk∥r≤r/(r−1) ∥ZN∥r. Doob Lp maximal inequality

[F5]

The coordinates of ΔWn are independent N(0,h) variables by Standard normal and normal laws. Their density gives, for 2ah<1, Eea∣ΔWn∣2=(1−2ah)−m/2: in one coordinate, integrate (2πh)−1/2e−(1/(2h)−a)x2 and rescale to the normalized Gaussian density. The same density and x=h u show E∣ΔWn∣r=hr/2E∣G∣r<∞ for every finite r, where G∼Nm(0,Im).

[F6]

For predictable G and p≥2, ∥sup⁡t≤T∣∫0tGs dWs∣∥p≤Cp,d∥(∫0T∥Gs∥F2ds)1/2∥p. Lp maximal moment bound for Brownian Itô integrals

[F7]

Hölder's inequality yields the Lr triangle inequality for r≥1, the three-factor bound E∣UVW∣≤(E∣U∣3E∣V∣3E∣W∣3)1/3, and E[1A∣Z∣p]≤P(A)1/2(E∣Z∣2p)1/2. Holder's inequality for integrals, including the endpoint cases Cauchy-Schwarz for random variables

[F8]

For a nonnegative random variable Z and a>0, P(Z≥a)≤EZ/a. Markov's inequality for random variables

[F9]

For a finite family of events, the probability of their union is at most the sum of their probabilities. The finite union bound

[A1]

AC is inherited through the Brownian, normal, filtration, and Itô interfaces; the estimates below use only countable grid constructions. The Axiom of Choice

Proof

Given: The hypotheses and suppliers in the Statement and Facts.

1.1F1givenalgebra

The derivative bound, integrated on the segment from 0 to x, gives ∣b(x)∣≤B(1+∣x∣q+1) for a finite B. One-sided Lipschitzness with y=0 gives ⟨x,b(x)⟩≤B(1+∣x∣2), and global Lipschitzness of σ gives ∥σ(x)∥F≤B(1+∣x∣). Increase B so these bounds hold with one constant. For ∣x∣≥1 they imply ∥σ(x)∥F2≤B1∣x∣2, ⟨x,b(x)⟩≤B1∣x∣2, and ∣b(x)∣2≤B1∣x∣2q+2. Thus, whenever 1≤∣x∣≤N1/(2q) and h=T/N, h2∣b(x)∣2≤B1Th∣x∣2. Fix once and for all Λ≥max⁡{1,2B1(1+T),1+sup⁡∣x∣≤1(∣b(x)∣+∥σ(x)∥F)}; continuity and compactness make this choice finite.

1.2F2F3F4F5F7given

The recursion and Brownian adaptedness show by induction that Yk is Ftk-measurable for every k. For 0≤k<N, define αk=1{∣Yk∣≥1}⟨Yk,σ(Yk)ΔWk⟩∣Yk∣2,S0=0,Sj=∑k<jαk (1≤j≤N). Since αk is Ftk+1-measurable, (Sj) is adapted to the sampled filtration. For 0≤n≤N, also define Dn=(Λ+∣ξ∣)exp⁡ ⁣(ΛT+max⁡0≤u≤n∑k=un−1(Λh+Λ∣ΔWk∣2+αk)). On ∣Yk∣≥1, write αk=vk⋅ΔWk with vk=σ(Yk)TYk/∣Yk∣2; global Lipschitzness of σ gives ∣vk∣≤∥σ(Yk)∥F/∣Yk∣≤C. Conditionally on Ftk its variance is h∣vk∣2≤C2h; on ∣Yk∣<1, αk=0. Brownian independence and the Gaussian density therefore give E[αk∣Ftk]=0 and, by completing the square in each coordinate, E[erαk∣Ftk]≤eC2r2h/2(r∈R). Conditional Jensen shows that eQSj and e−QSj are nonnegative submartingales for each Q>0. Iterating the preceding bound with r=2Q gives terminal second moments at most e2C2Q2T for both processes. Doob's L2 inequality [F4] then gives sup⁡NEeQmax⁡j≤N∣Sj∣<∞: the exponential of the maximum is bounded by the sum of the maxima of eQSj and e−QSj, whose expectations are bounded by their L2 maxima. For any r>0, choose Q≥max⁡(1,r); since xr/Q≤1+x for x≥0, this also bounds sup⁡NEermax⁡j≤N∣Sj∣. Finally, for N≥4rΛT, [F5] and −log⁡(1−u)≤2u on 0≤u≤1/2 give Eexp⁡ ⁣(rΛ∑k<N∣ΔWk∣2)=∏k<N(1−2rΛh)−m/2≤e2rmΛT. Every suffix sum ∑k=uj−1αk=Sj−Su is bounded in absolute value by 2max⁡i≤N∣Si∣. Since max⁡n≤NDnp≤(Λ+∣ξ∣)pe2pΛTexp⁡ ⁣(pΛ∑k<N∣ΔWk∣2+2pmax⁡j≤N∣Sj∣), the three-factor Hölder bound [F7], the all-moment assumption on ξ, the square-exponential estimate with r=3p, and the exponential-martingale bound with r=6p give a uniform moment for max⁡n≤NDnp when N≥N0(p):=max⁡{1,⌈T⌉,⌈12pΛT⌉}. The finitely many smaller N will be handled separately.

1.3F1F3F6F7givenalgebra

Set an=(E∣Yn∣p)1/p for p≥2. Decompose the recursion as Yn=ξ+∑k<nhbh(Yk)+σ(0)Wtn+∑k<n(σ(Yk)−σ(0))ΔWk. The drift sum has norm at most N by [F1]. The last sum is a continuous Brownian Itô integral with a predictable step integrand. By [F6], global Lipschitzness of σ, and the Lp/2 triangle inequality from [F7], its Lp norm is at most Cp(∑k<nh ∥σ(Yk)−σ(0)∥Lp2)1/2≤Cp(h∑k<nak2)1/2. The pointwise inequality ∣u1+⋯+u4∣p≤4p−1∑i∣ui∣p and the bounded pth moments of Wt on [0,T] therefore give an2≤Cp(1+∥ξ∥Lp2+N2)+Cph∑k<nak2. Iterating this finite recurrence yields max⁡n≤Nan2≤Cp(1+∥ξ∥Lp2+N2)eCpT, hence max⁡n≤N∥Yn∥Lp≤Cp(1+N).

2.1F1step 1.1step 1.2givenalgebraconstruct

Write ΔWk=Wtk+1−Wtk and use αk,Dn from step 1.2. Define Ωn={max⁡k<nDk≤N1/(2q), max⁡k<n∣ΔWk∣≤1}. On ∣Yk∣≤1 and ∣ΔWk∣≤1, [F1] gives ∣Yk+1∣≤1+h∣b(Yk)∣+∥σ(Yk)∥F≤Λ. On 1≤∣Yk∣≤N1/(2q), expand the squared update. The drift cross term is at most 2B1h∣Yk∣2; the squared taming increment is at most B1Th∣Yk∣2 by step 1.1; the diffusion square is at most B1∣Yk∣2∣ΔWk∣2; and the drift-noise cross term is at most the sum of those two square terms. The remaining cross term is exactly 2∣Yk∣2αk. With the fixed Λ chosen in step 1.1 large enough for these coefficients, 1+u≤eu therefore gives ∣Yk+1∣≤∣Yk∣exp⁡(Λh+Λ∣ΔWk∣2+αk). Induction on n, starting either from Y0=ξ or from the last index with ∣Yk∣≤1, now gives 1Ωn∣Yn∣≤Dn: the first step after such an index is at most Λ, and the later factors are bounded by the suffix sum in Dn.

2.2F3F5F8F9step 1.2givenalgebra

Markov's inequality [F8] and step 1.2 imply, for any r>0, P ⁣(max⁡n≤NDn>N1/(2q))≤N−r/(2q)Emax⁡n≤NDnr. Also, by the finite union bound [F9] and the rth Gaussian moment, P(max⁡k<N∣ΔWk∣>1)≤CrNhr/2=Cr,TN1−r/2. Choosing r arbitrarily large in these bounds shows that for every s>0 there is Cs with P(ΩNc)≤CsN−s for all N.

3.1A1F1F7step 2.1step 1.2step 2.2step 1.3algebra∎

For p≥1, on ΩN step 2.1 bounds E[1ΩN∣Yn∣p]≤EDnp≤Cp; here ΩN⊆Ωn. On the complement, [F7] and step 1.3 give E[1ΩNc∣Yn∣p]≤P(ΩNc)1/2∥Yn∥L2pp≤CsN−s/2(1+N)p. Choose s>2p. This is uniformly bounded. For the finitely many small N excluded in step 1.2, the finite recursion and linear growth of σ give finite moments by induction; taking their finite maximum completes the uniform bound. For 1≤p<2, apply Hölder on the probability space to the p=2 result. AC is exactly [A1].

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