ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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A learned-drift bias that persists after solver error vanishes

Example

Work on a complete filtered probability space whose filtration is right-continuous and contains all $P$ -null subsets of $F$ in $F_{0}$ . Let $W$ be an adapted standard one-dimensional Brownian motion whose increments after time $s$ are independent of $F_{s}$ . Compare the target equation $d X_{t} = d W_{t}$ , $X_{0} = 0$ , on $[0, 1]$ with a learned equation $d {\hat{X}}_{t} = (π / 2) d t + d W_{t}$ , ${\hat{X}}_{0} = 0$ . For the bounded Lipschitz payoff $φ (x) = \cos x$ , the Euler solver is exact at the terminal grid point for the learned equation, while the learned drift creates a nonzero bias relative to the target law that is independent of $h$ .

Facts & Assumptions

Given: AC; the complete usual filtered Brownian setup above; $T = 1$ ; the $F_{0}$ -measurable initial values $X_{0} = {\hat{X}}_{0} = 0$ ; $h = 1 / N$ ; the target and learned equations in the Example; and $φ (x) = \cos x$ .

[F1]

A one-dimensional standard Brownian motion has $W_{1} \sim N (0, 1)$ . $d$ -dimensional Brownian motion

[F2]

If $Z \sim N (m, v)$ , its characteristic function is $E e^{i t Z} = \exp (i m t - v t^{2} / 2)$ . Characteristic function of a normal law

[F3]

The Euler–Maruyama recursion uses $Y_{k + 1} = Y_{k} + b (t_{k}, Y_{k}) h + σ (t_{k}, Y_{k}) Δ W_{k}$ . Euler–Maruyama recursion

[F4]

Under global Lipschitz, linear-growth, half-Hölder time and square-integrable initial-value hypotheses, the Euler strong supremum error tends to zero at root-mean-square rate $O (h^{1 / 2})$ . Euler–Maruyama strong rate for square-integrable initial data

[A1]

AC is stated because the Brownian and normal characteristic-function interfaces in [F1]–[F2] assume it. The Axiom of Choice

Verification

1.1F1F3givenalgebra

The target equation has terminal value $X_{1} = W_{1}$ . By [F1]–[F2], $E \cos (X_{1}) = Re E e^{i W_{1}} = e^{- 1 / 2} .$ The learned Euler recursion is ${\hat{Y}}_{k + 1} = {\hat{Y}}_{k} + (π / 2) h + Δ W_{k}$ . Summing its $N$ increments gives ${\hat{Y}}_{N} = \frac{π}{2} N h + \sum_{k = 0}^{N - 1} Δ W_{k} = \frac{π}{2} + W_{1},$ which is exactly ${\hat{X}}_{1}$ for every $h = 1 / N$ . Thus the discretization error relative to the learned equation is zero.

2.1A1F2F4step 1.1algebra∎

By [F2] and the addition formula, $E \cos ({\hat{X}}_{1}) = Re (e^{i π / 2} E e^{i W_{1}}) = 0.$ Therefore the absolute expectation bias relative to the target is $∣ 0 - e^{- 1 / 2} ∣ = e^{- 1 / 2} > 0$ for every step size, so it persists as $h \to 0$ . The solver theorem [F4] controls error to the learned SDE solution; it supplies no bound on the difference between the learned and target drifts. AC is inherited through [A1].

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