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One-step moment matching for additive-noise stochastic Heun

Statement

Let $b \in C_{b}^{6} (R; R)$ , let $σ \in R$ be constant, and define the generator $L f = b f^{'} + \frac{1}{2} σ^{2} f^{''}$ . Let $u : [0, T] \times R \to R$ satisfy $u_{t} + L u = 0$ . Assume all mixed derivatives $\partial_{t}^{r} \partial_{x}^{j} u$ for $0 \leq r \leq 3$ and $0 \leq j \leq 6$ are continuous and bounded. For $0 < h \leq 1$ , $t + h \leq T$ , fixed $x \in R$ , and $w \sim N (0, h)$ , take one Heun step from Explicit stochastic Heun update for additive noise and let $D = Y_{n + 1} - x$ . Then a constant $C$ , uniform in $t, x, h$ , satisfies $∣ E [u (t + h, x + D)] - u (t, x) ∣ \leq C h^{3} .$

Facts & Assumptions

Given: The additive-noise Heun update, $b \in C_{b}^{6}$ , the constant $σ$ , the backward equation $u_{t} + L u = 0$ , its bounded derivative assumptions, $w \sim N (0, h)$ , and $0 < h \leq 1$ .

[L1]

Explicit stochastic Heun update for additive noise: for increment $w$ , $Z = x + b (x) h + σ w$ and $D = σ w + \frac{h}{2} (b (x) + b (Z))$ ; $C_{b}^{6}$ means bounded continuous derivatives through order six.

[F1]

Standard normal and normal laws: $w = \sqrt{h} Z_{0}$ with $Z_{0}$ standard normal of density $(2 π)^{- 1 / 2} e^{- z^{2} / 2}$ ; its odd moments vanish and its even moments are obtained by integration by parts.

[A1]

The Axiom of Choice: AC is inherited through the Brownian/normal-law interface in [L1] and [F1]; the one-step Taylor calculation itself uses no further choice.

Proof

Given: The assumptions above.

1.1L1F1A1

Put $B = b (x)$ , $B_{j} = b^{(j)} (x)$ , $δ = B h + σ w$ , and $D = σ w + B h + \frac{h}{2} (b (x + δ) - B)$ . Taylor's formula through degree three gives $b (x + δ) = B + B_{1} δ + \frac{1}{2} B_{2} δ^{2} + \frac{1}{6} B_{3} δ^{3} + ρ_{4}$ , with $∣ ρ_{4} ∣ \leq ∥ b^{(4)} ∥_{\infty} ∣ δ ∣^{4} / 24$ . From [F1], $E w = 0$ , $E w^{2} = h$ , $E w^{3} = 0$ , $E w^{4} = 3 h^{2}$ , $E w^{5} = 0$ , and $E ∣ w ∣^{6} = 15 h^{3}$ . Substituting $δ = B h + σ w$ and collecting terms of order at most $h^{2}$ gives $\begin{aligned} E D & = B h + (\frac{1}{2} B B_{1} + \frac{1}{4} σ^{2} B_{2}) h^{2} + O (h^{3}), \\ E D^{2} & = σ^{2} h + (B^{2} + σ^{2} B_{1}) h^{2} + O (h^{3}), \\ E D^{3} & = 3 B σ^{2} h^{2} + O (h^{3}), \\ E D^{4} & = 3 σ^{4} h^{2} + O (h^{3}), E D^{5} = O (h^{3}) . \end{aligned}$ For the second moment, the $h^{2}$ terms are $(B h)^{2}$ and $2 (σ w) (\frac{1}{2} B_{1} σ h w)$ ; for the third they are $3 (σ w)^{2} (B h)$ ; for the fourth it is $E (σ w)^{4}$ . All other terms either contain an odd Gaussian moment or have expected magnitude $O (h^{3})$ by $E ∣ δ ∣^{4} = O (h^{2})$ and the displayed remainder bound. Since $b$ is bounded, $∣ D ∣ \leq ∣ σ ∣ ∣ w ∣ + h ∥ b ∥_{\infty}$ , so $E ∣ D ∣^{6} \leq C h^{3}$ .

2.1L1F1step 1.1

Taylor-expand $u (t + h, x + D)$ in its spatial argument through degree five. The sixth-derivative bound and step 1.1 yield $E u (t + h, x + D) = u + u_{x} E D + \frac{1}{2} u_{x x} E D^{2} + \frac{1}{6} u_{x x x} E D^{3} + \frac{1}{24} u_{x x x x} E D^{4} + \frac{1}{120} u_{x x x x x} E D^{5} + O (h^{3}),$ where the displayed derivatives are evaluated at $(t + h, x)$ . Substituting the moment formulas, the coefficients through order $h^{2}$ are $h L u$ and $\frac{1}{2} h^{2} L^{2} u$ , because $L^{2} u = (B B_{1} + \frac{1}{2} σ^{2} B_{2}) u_{x} + (B^{2} + σ^{2} B_{1}) u_{x x} + B σ^{2} u_{x x x} + \frac{1}{4} σ^{4} u_{x x x x} .$ Thus $E u (t + h, x + D) = u (t + h, x) + h L u (t + h, x) + \frac{1}{2} h^{2} L^{2} u (t + h, x) + O (h^{3})$ .

3.1givenstep 2.1algebra

Since $u_{t} = - L u$ and $L$ is independent of $t$ , differentiating the backward equation gives $u_{t t} = L^{2} u$ . Taylor's formula in time at fixed $x$ , with bounded $u_{t t t}$ , therefore gives $u (t, x) = u (t + h, x) + h L u (t + h, x) + \frac{1}{2} h^{2} L^{2} u (t + h, x) + O (h^{3}) .$ The coefficient and derivative bounds in the Statement make the constants in the preceding expansions uniform in $t, x, h$ .

4.1step 2.1step 3.1∎

The expansions in steps 2.1 and 3.1 have identical terms through order $h^{2}$ ; subtracting them leaves a remainder bounded by $C h^{3}$ , which is the asserted one-step weak defect.

Depends on

Used by

Weak order two for additive-noise stochastic Heun Theorem

Dependency tree · two levels

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Sources

M. Tretyakov, Lecture Notes for a Mini-Course on Numerical Methods for SDEs, Theorem 7.1 and Section 7.3, PDF pp. 39-42 (standard reference, not scraped)
Kloeden and Platen, Numerical Solution of Stochastic Differential Equations, Theorem 14.5.2 and Section 15.1, printed pp. 474-480 and 485-487 (standard reference, not scraped)