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Coarse and fine Euler paths coupled by a bridge split

Example

Assume AC as in the Gaussian and Brownian interfaces.

For the scalar SDE $d X_{t} = X_{t} d t + d W_{t}$ on one interval of length $h > 0$ , let $Δ W \sim N (0, h)$ and independently $Z \sim N (0, h / 4)$ . Couple the two fine increments by the Brownian bridge split $Δ W_{1} = Δ W / 2 + Z$ and $Δ W_{2} = Δ W / 2 - Z$ . From initial value $y_{0}$ , one coarse Euler step and two fine Euler steps give $Y_{c} = (1 + h) y_{0} + Δ W,$ $Y_{f} = (1 + h / 2)^{2} y_{0} + (1 + h / 2) Δ W_{1} + Δ W_{2},$ and hence $Y_{f} - Y_{c} = \frac{h^{2}}{4} y_{0} + \frac{h}{4} Δ W + \frac{h}{2} Z .$ For $y_{0} = 0$ , the coupled difference is centered with variance $h^{3} / 8$ .

Facts & Assumptions

Given: AC, the Euler update for $b (x) = x$ , $σ (x) = 1$ , the parent $Δ W$ , and the independent midpoint innovation $Z$ .

[F1]

A scalar Euler–Maruyama step is $Y_{k + 1} = Y_{k} + b (Y_{k}) h + σ (Y_{k}) Δ W_{k}$ on an interval of length $h$ . Euler–Maruyama recursion

[F2]

The bridge children are $Δ W / 2 + Z$ and $Δ W / 2 - Z$ with a fresh independent $Z \sim N (0, h / 4)$ . Dyadic Brownian bridge increment tree

[F3]

The children have the exact joint independent $N (0, h / 2)$ law and sum to the parent. Exact increment laws in the dyadic Brownian bridge tree

[A1]

AC is inherited through the Euler, Gaussian, and bridge-sampling interfaces; the finite update calculation makes no further selection. The Axiom of Choice

Proof

1.1F1givenalgebra

Applying [F1] once with $b (y) = y$ and $σ (y) = 1$ gives $Y_{c} = y_{0} + h y_{0} + Δ W = (1 + h) y_{0} + Δ W$ .

2.1F1givenstep 1.1algebra

Applying [F1] on the first fine half-step gives $Y_{1 / 2} = (1 + h / 2) y_{0} + Δ W_{1}$ . The second half-step gives $Y_{f} = (1 + h / 2) Y_{1 / 2} + Δ W_{2}$ , so substitution yields the displayed fine formula.

3.1F2F3step 1.1step 2.1algebra

Substitute [F2] into $Y_{f}$ and subtract $Y_{c}$ from step 1.1: $Y_{f} - Y_{c} = ((1 + h / 2)^{2} - (1 + h)) y_{0} + ((1 + h / 2) (Δ W / 2 + Z) + (Δ W / 2 - Z) - Δ W) = \frac{h^{2}}{4} y_{0} + \frac{h}{4} Δ W + \frac{h}{2} Z .$ For $y_{0} = 0$ , independence and the variances in [F2] give $Var (Y_{f} - Y_{c}) = (h / 4)^{2} h + (h / 2)^{2} (h / 4) = h^{3} / 8$ . The identity $Δ W_{1} + Δ W_{2} = Δ W$ makes this a path-consistent coupling, not two independent simulations.

4.1A1F2F3step 3.1given∎

The case $h = 0$ is excluded; for every $h > 0$ the fine difference has strictly positive variance at $y_{0} = 0$ . A nonzero initial value changes the mean by $h^{2} y_{0} / 4$ but not that variance. This example checks an exact finite-dimensional coupling only; it is not a general MLMC or convergence rate theorem.

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Sources

Simo Särkkä and Arno Solin, Applied Stochastic Differential Equations, Algorithm 8.1 (standard reference, not scraped)
Mike Giles, Monte Carlo Methods, Module 4: Monte Carlo path simulation, Brownian bridge, pp. 27–30 (standard reference, not scraped)