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Explicit Euler moments diverge for a cubic dissipative SDE

Statement

Fix T>0 and consider dXt=−Xt3dt+dWt,X0=0. For N≥1, let h=T/N and let Y0=0, Yn+1=Yn−hYn3+(Wtn+1−Wtn),tn=nh. The exact solution has finite moments of every order at time T, but for every p≥1, lim⁡N→∞E∣YN∣p=+∞.

Facts & Assumptions

Given: AC, a one-dimensional standard Brownian motion, fixed T>0, and the SDE and Euler–Maruyama scheme in the Statement.

[F1]

Euler–Maruyama has grid update Yn+1=Yn+hb(Yn)+σ(Yn)ΔWn. Autonomous Euler–Maruyama scheme and interpolation

[F2]

Brownian increments on disjoint deterministic intervals are independent, and ΔWn has law N(0,h) with density (2πh)−1/2e−x2/(2h) and second moment h. d-dimensional Brownian motion Standard normal and normal laws

[F3]

If b is one-sided Lipschitz with polynomially bounded derivative, σ is globally Lipschitz, and the initial state has all finite moments, the local solution is global and has finite moments of every order on finite horizons. Global existence and finite-horizon moments for one-sided Lipschitz SDEs Locally Lipschitz Itô SDE on an open state domain

[A1]

AC is inherited through the Brownian and SDE interfaces. The Axiom of Choice

Proof

Given: The assumptions and supplier statements above.

1.1F3givenalgebra

The drift b(x)=−x3 satisfies (x−y)(b(x)−b(y))=−(x−y)2(x2+xy+y2)≤0, since x2+xy+y2=(x+y/2)2+3y2/4≥0. Also ∣b′(x)∣=3x2≤3(1+∣x∣2), and σ(x)=1 is globally Lipschitz. Thus [F3] applies with deterministic X0=0, so E∣XT∣p<∞ for every finite p≥1.

1.2F1F2givenalgebra

The algorithm [F1] gives the displayed recursion. For h≤1/4, set EN={h−1≤ΔW0≤2h−1}∩⋂n=1N−1{∣ΔWn∣≤1}. The increments are independent by [F2]. The normal density, bounded below on [h−1,2h−1] by its value at 2h−1, gives P(h−1≤ΔW0≤2h−1)≥(2π)−1/2h−3/2e−2h−3. Also P(∣ΔWn∣>1)≤E∣ΔWn∣2=h, since 1{∣z∣>1}≤∣z∣2. For h≤1/4, (1−h)N−1≥e−2h(N−1)≥e−2T. Hence P(EN)≥(2π)−1/2e−2Th−3/2e−2h−3.

2.1F1F2step 1.2givenalgebra

On EN, Y1=ΔW0≥h−1. If ∣y∣≥h−1 and ∣δ∣≤1, then ∣y−hy3+δ∣≥h∣y∣3−∣y∣−1=h∣y∣3(1−1h∣y∣2−1h∣y∣3)≥h2∣y∣3, because the two fractions are at most h and h2, whose sum is at most 5/16<1/2. The lower bound is at least h−1 when h≤1/2 and ∣y∣≥h−1. Induction therefore gives ∣Yn∣≥h−1 for all 1≤n≤N on EN. With zn=h/2∣Yn∣, the same inequality gives zn+1≥zn3, while z1≥(2h)−1/2>1. Thus ∣YN∣≥2/h (2h)−3N−1/2on EN.

3.1A1F3step 1.2step 2.1givenalgebra∎

Combining steps 1.2 and 2.1 yields E∣YN∣p≥(2π)−1/2e−2Th−3/2e−2h−3(2/h (2h)−3N−1/2)p. Here h=T/N. The positive logarithmic term p23N−1log⁡((2h)−1) grows faster than the negative term 2h−3=2N3/T3, so this lower bound tends to +∞. AC is exactly [A1].

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