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Discrete stochastic local-to-global RMS stability bound

Statement

Let $N \geq 1$ , $0 < h \leq 1$ , and $N h \leq T$ . Let $(F_{n})_{n = 0}^{N}$ be a filtration as in Filtration and filtered probability space. For every $n$ , let $e_{n}, u_{n}, r_{n}$ be square-integrable $R^{d}$ -valued, $F_{n}$ -measurable random vectors; let $M_{n + 1}, Q_{n + 1}$ be square-integrable and $F_{n + 1}$ -measurable. Say $Z$ is orthogonal to $F_{n}$ if $E ⟨ H, Z ⟩ = 0$ for every square-integrable $F_{n}$ -measurable vector $H$ . Suppose $e_{0} = 0, e_{n + 1} = e_{n} + h u_{n} + r_{n} + M_{n + 1} + Q_{n + 1},$ $∣ u_{n} ∣ \leq L ∣ e_{n} ∣, ∣ r_{n} ∣ \leq C_{0} h^{2},$ and $M_{n + 1}, Q_{n + 1}$ are orthogonal to $F_{n}$ , with $E ∣ M_{n + 1} ∣^{2} \leq C_{1} h E ∣ e_{n} ∣^{2}, E ∣ Q_{n + 1} ∣^{2} \leq C_{1} h^{3} .$ Then $E ∣ e_{N} ∣^{2} \leq C_{T} h^{2}$ , where $C_{T}$ depends only on $T, L, C_{0}, C_{1}$ and not on $h$ or $N$ . This is a local-to-global estimate for an order-one grid RMS error when the drift error is Lipschitz and the centered one-step defect has second moment $O (h^{3})$ .

Facts & Assumptions

Given: The filtered probability space, the square-integrable vectors, the recurrence, orthogonality, and numerical bounds in the Statement.

[F1]

Filtration and filtered probability space: $F_{n} \subseteq F_{n + 1}$ and each $F_{n}$ is a sigma algebra on the underlying probability space.

Proof

Given: The assumptions above.

1.1givenF1algebra

Since $∣ e_{n} + h u_{n} ∣ \leq (1 + L h) ∣ e_{n} ∣$ and $∣ r_{n} ∣ \leq C_{0} h^{2}$ , the elementary inequality $2 a b \leq h a^{2} + h^{- 1} b^{2}$ and $h \leq 1$ give $E ∣ e_{n} + h u_{n} + r_{n} ∣^{2} \leq (1 + C h) E ∣ e_{n} ∣^{2} + C h^{3} .$ Indeed, the square of $e_{n} + h u_{n}$ contributes at most $(1 + L h)^{2} ∣ e_{n} ∣^{2} \leq (1 + C h) ∣ e_{n} ∣^{2}$ , its cross term with $r_{n}$ is bounded by $C h^{2} ∣ e_{n} ∣ \leq C h ∣ e_{n} ∣^{2} + C h^{3}$ , and $∣ r_{n} ∣^{2} \leq C h^{4} \leq C h^{3}$ .

2.1givenstep 1.1algebra

Orthogonality makes the cross expectation between $e_{n} + h u_{n} + r_{n}$ and $M_{n + 1} + Q_{n + 1}$ zero. To bound the remaining cross term, put $a = ∣ M_{n + 1} ∣$ and $b = ∣ Q_{n + 1} ∣$ ; nonnegativity of $E (a - λ b)^{2}$ for every real $λ$ implies $E [a b] \leq (E a^{2})^{1 / 2} (E b^{2})^{1 / 2}$ . Hence $2 ∣ E ⟨ M_{n + 1}, Q_{n + 1} ⟩ ∣ \leq 2 (E ∣ M_{n + 1} ∣^{2})^{1 / 2} (E ∣ Q_{n + 1} ∣^{2})^{1 / 2} \leq C h^{2} (E ∣ e_{n} ∣^{2})^{1 / 2} \leq C h E ∣ e_{n} ∣^{2} + C h^{3} .$ The final inequality is $2 a b \leq a^{2} + b^{2}$ with factors proportional to $h^{1 / 2} (E ∣ e_{n} ∣^{2})^{1 / 2}$ and $h^{3 / 2}$ . Therefore the recurrence and step 1.1 imply $E ∣ e_{n + 1} ∣^{2} \leq (1 + C h) E ∣ e_{n} ∣^{2} + C h^{3} .$

3.1givenstep 2.1algebra∎

Iterating the inequality in step 2.1 from $e_{0} = 0$ gives $E ∣ e_{N} ∣^{2} \leq C h^{3} \sum_{j = 0}^{N - 1} (1 + C h)^{j} \leq C h^{3} N e^{C N h} \leq C_{T} h^{2},$ because $(1 + C h)^{j} \leq e^{C j h} \leq e^{C T}$ and $N h \leq T$ . Taking square roots proves the stated RMS estimate. This recurrence argument uses only finite algebraic estimates and the given orthogonality conditions.

Depends on

Filtration and filtered probability space

Used by

Strong order one for scalar and commutative-noise Milstein schemes Theorem

Dependency tree · two levels

3 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Kloeden and Platen, Numerical Solution of Stochastic Differential Equations, Theorem 10.6.3 and proof, printed pp. 361-365 (standard reference, not scraped)
M. Tretyakov, Lecture Notes for a Mini-Course on Numerical Methods for SDEs, Theorem 6.1 and proof, PDF pp. 32-35 (standard reference, not scraped)