Pipeline-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Euler–Maruyama Strong and Weak Convergence — Examples

1 · Prerequisites

2 · Summary

These examples check the rate contracts on the Ornstein–Uhlenbeck equation, including a terminal strong error with an explicit $O (h)$ root-mean-square bound and a cosine payoff with verified backward regularity for weak order one. A constant learned-drift error remains even when the Euler solver is exact for the learned equation. The final example couples fine and coarse Euler paths with summed Brownian increments and verifies the $O (h)$ correction variance, $O (h^{1 / 2})$ strong-route bias, and $O (h^{- 1})$ work per path pair.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

Additive-noise Ornstein–Uhlenbeck Euler strong error

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 $F_{s}$ . On $[0, 1]$ , solve the additive-noise Ornstein–Uhlenbeck equation $d X_{t} = - X_{t} d t + d W_{t}, X_{0} = 0.$ For $h = 1 / N \leq 1 / 2$ , couple the Euler recursion to the same Brownian path. The finite- $L^{2}$ theorem gives the general strong supremum bound $E \sup_{t \leq 1} ∣ X_{t} - Y_{t}^{h} ∣^{2} = O (h)$ ; in this additive-noise example the terminal root-mean-square error is in fact $O (h)$ .

Facts & Assumptions

Given: AC; the complete usual filtered Brownian setup above; $T = 1$ ; the $F_{0}$ -measurable initial value $X_{0} = 0$ ; $h = 1 / N \leq 1 / 2$ ; and the SDE and Euler recursion displayed in the Example.

[F1]

The Euler recursion is $Y_{k + 1} = Y_{k} + b (t_{k}, Y_{k}) h + σ (t_{k}, Y_{k}) Δ W_{k}$ , and the continuous interpolation agrees with the recursion at grid points. Euler–Maruyama step process and continuous interpolation

[F2]

For a one-dimensional continuous Brownian Itô process and $F \in C^{1, 2}$ , Itô's formula has drift $F_{t} + b F_{x} + \frac{1}{2} σ^{2} F_{x x}$ and stochastic term $σ F_{x} d W$ . One-dimensional Ito formula

[F3]

For a deterministic finite-energy integrand $q$ , $E (\int_{0}^{T} q_{s} d W_{s})^{2} = \int_{0}^{T} q_{s}^{2} d s$ . Ito isometry and linearity in predictable L2

[F4]

For a deterministic step integrand, its Itô integral is the corresponding finite sum of Brownian increments. Ito integral of an elementary predictable process

[F5]

Under global state-Lipschitz, linear-growth, half-Hölder time, and square-integrable initial-value assumptions, the Euler strong supremum error satisfies $E \sup_{t \leq T} ∣ X_{t} - Y_{t}^{h} ∣^{2} \leq C h$ . Euler–Maruyama strong rate for square-integrable initial data

[A1]

AC is included because the Brownian, Itô-formula and Itô-integral suppliers in [F2]–[F4] assume it. The Axiom of Choice

Verification

[F1] The Euler recursion is $Y_{k + 1} = Y_{k} + b (t_{k}, Y_{k}) h + σ (t_{k}, Y_{k}) Δ W_{k}$ , and the continuous interpolation agrees with the recursion at grid points. Euler–Maruyama step process and continuous interpolation

[F2] For a one-dimensional continuous Brownian Itô process and $F \in C^{1, 2}$ , Itô's formula has drift $F_{t} + b F_{x} + \frac{1}{2} σ^{2} F_{x x}$ and stochastic term $σ F_{x} d W$ . One-dimensional Ito formula

[F3] For a deterministic finite-energy integrand $q$ , $E (\int_{0}^{T} q_{s} d W_{s})^{2} = \int_{0}^{T} q_{s}^{2} d s$ . Ito isometry and linearity in predictable L2

[F4] For a deterministic step integrand, its Itô integral is the corresponding finite sum of Brownian increments. Ito integral of an elementary predictable process

[F5] Under global state-Lipschitz, linear-growth, half-Hölder time, and square-integrable initial-value assumptions, the Euler strong supremum error satisfies $E \sup_{t \leq T} ∣ X_{t} - Y_{t}^{h} ∣^{2} \leq C h$ . Euler–Maruyama strong rate for square-integrable initial data

[A1] AC is included because the Brownian, Itô-formula and Itô-integral suppliers in [F2]–[F4] assume it. The Axiom of Choice

1.1F5givenalgebra

Here $b (x) = - x$ and $σ (x) = 1$ . Thus $∣ b (x) - b (y) ∣ = ∣ x - y ∣$ , $∣ b (x) ∣^{2} + ∣ σ (x) ∣^{2} = x^{2} + 1$ , and both coefficients are time independent. The initial value is square-integrable and independent of future Brownian increments, so the hypotheses of [F5] hold.

1.2F2givenalgebra

Apply [F2] to $F (t, x) = e^{t} x$ along $X$ . Since $F_{t} + (- x) F_{x} + \frac{1}{2} (1)^{2} F_{x x} = 0$ and $F_{x} = e^{t}$ , it follows that $e^{t} X_{t} = \int_{0}^{t} e^{s} d W_{s}$ . At $t = 1$ this gives $X_{1} = \int_{0}^{1} e^{- (1 - s)} d W_{s} .$

1.3F1F4givenalgebra

The recursion is $Y_{k + 1} = (1 - h) Y_{k} + Δ W_{k}$ . Induction gives $Y_{N} = \sum_{k = 0}^{N - 1} (1 - h)^{N - 1 - k} Δ W_{k} = \int_{0}^{1} (1 - h)^{N - 1 - ⌊ s / h ⌋} d W_{s},$ where the second equality is [F4] on the grid partition.

2.1F3step 1.2step 1.3algebra

By [F3] applied to the difference of the deterministic integrands, $E ∣ X_{1} - Y_{N} ∣^{2} = \int_{0}^{1} {∣ e^{- (1 - s)} - (1 - h)^{N - 1 - ⌊ s / h ⌋} ∣}^{2} d s .$ For $s \in [k h, (k + 1) h]$ , set $j = N - 1 - k$ and $u = s - k h$ . Then $(1 - s) = j h + (h - u)$ , so $∣ e^{- (1 - s)} - e^{- j h} ∣ \leq h$ . Also $0 \leq e^{- j h} - (1 - h)^{j} = e^{- j h} (1 - e^{- j (- \log (1 - h) - h)}) \leq j (- \log (1 - h) - h) .$ For $h \leq 1 / 2$ , the integral formula gives $0 \leq - \log (1 - h) - h = \int_{0}^{h} \frac{x}{1 - x} d x \leq h^{2}$ ; since $j < 1 / h$ , the last bound is at most $h$ . Hence the integrand difference is at most $2 h$ on every grid interval.

3.1A1F5step 2.1algebra∎

Integrating the bound of step 2.1 over $[0, 1]$ gives $E ∣ X_{1} - Y_{N} ∣^{2} \leq 4 h^{2}$ , hence $(E ∣ X_{1} - Y_{N} ∣^{2})^{1 / 2} \leq 2 h$ . This is a special additive-noise terminal rate; it is stronger than the general $O (h^{1 / 2})$ root-mean-square guarantee in [F5]. AC is inherited through [A1].

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

A verified weak-order contract for the Ornstein–Uhlenbeck process

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 $F_{s}$ . On $[0, 1]$ , take $d X_{t} = - X_{t} d t + d W_{t}$ with $X_{0} = 0$ and bounded Lipschitz payoff $φ (x) = \cos x$ . For $0 \leq t \leq 1$ , put $r (t) = e^{t - 1}$ and $u (t, x) = \exp ⁣ (- \frac{1 - r (t)^{2}}{4}) \cos (r (t) x) .$ This value function satisfies the backward equation, the mixed derivative growth requirement with $q = 2$ , and the moment threshold $p \geq 12$ . Thus the conditional weak theorem applies and gives Euler weak error $O (h)$ .

Facts & Assumptions

Given: AC; the complete usual filtered Brownian setup above; $T = 1$ ; the $F_{0}$ -measurable initial value $X_{0} = 0$ ; the Ornstein–Uhlenbeck SDE and payoff displayed in the Example; and Euler step size $0 < h \leq 1$ with $1 / h \in N$ .

[F1]

[F2]

An Itô integral with deterministic $h \in L^{2} [0, T]$ has centered normal law $N (0, \int_{0}^{T} h (s)^{2} d s)$ . Deterministic Ito integrals are Gaussian

[F3]

A $N (m, v)$ variable has characteristic function $t \mapsto \exp (i m t - v t^{2} / 2)$ . Characteristic function of a normal law

[F4]

The Euler interpolation on a grid interval is $Y_{t} = Y_{k} + b (t_{k}, Y_{k}) (t - t_{k}) + σ (t_{k}, Y_{k}) (W_{t} - W_{t_{k}})$ . Euler–Maruyama step process and continuous interpolation

[F5]

Under the Euler coefficient and initial-moment hypotheses, for every finite $p \geq 2$ the step and continuous interpolation have uniformly bounded pointwise $p$ th moments. Euler–Maruyama uniform moments and interpolation increments

[F6]

For a scalar standard normal $Z$ , $E ∣ Z ∣^{12} = 11!! < \infty$ by the Gaussian even-moment lemma at $m = 6$ . The normal scaling convention gives $N (0, v)$ as the law of $\sqrt{v} Z$ , including $v = 0$ . Gaussian even moments for Brownian increments Standard normal and normal laws

[F7]

If the bounded test payoff's backward value function has the stated mixed derivatives and $p \geq 2 q + 8$ moments, the weak Euler error is at most $C h$ . Euler–Maruyama weak order one under backward regularity

[A1]

AC is included because the Brownian, deterministic-integral Gaussian, normal-characteristic-function and normal-moment suppliers in [F1]–[F3], [F6] adopt it. The Axiom of Choice

Verification

Given: The hypotheses in the Statement or Example and the facts above.

1.1A1F1F2given

Apply [F1] to $e^{t} X_{t}$ . The drift terms cancel, so $X_{t} = \int_{0}^{t} e^{- (t - s)} d W_{s} .$ For a solution started at $(t, x)$ , the same calculation gives $X_{1}^{t, x} = r (t) x + \int_{t}^{1} e^{- (1 - s)} d W_{s} .$ By [F2], the last integral is $N (0, (1 - r (t)^{2}) / 2)$ .

1.2givenalgebra

Since $r^{'} = r$ and $A (t) : = e^{- (1 - r (t)^{2}) / 4}$ satisfies $A^{'} / A = r^{2} / 2$ , direct differentiation gives $u_{t} = A (\frac{r^{2}}{2} \cos (r x) - r x \sin (r x)), u_{x} = - A r \sin (r x), u_{x x} = - A r^{2} \cos (r x) .$ For the generator $L = - x \partial_{x} + \frac{1}{2} \partial_{x x}$ , these formulas imply $(\partial_{t} + L) u = 0$ . Every mixed derivative $\partial_{t}^{j} \partial_{x}^{k} u$ with $j \leq 2$ , $k \leq 4$ is a finite sum of bounded smooth coefficient functions times $x^{a} \sin (r x)$ or $x^{a} \cos (r x)$ with $a \leq 2$ ; hence it is bounded by $C (1 + ∣ x ∣^{2})$ . The backward contract holds with $q = 2$ .

2.1F2F3step 1.1algebra

By [F3], for that normal variable $Z$ , $E e^{i Z} = e^{- (1 - r^{2}) / 4}$ . Taking the real part after multiplying by $e^{i r x}$ gives $E \cos (X_{1}^{t, x}) = e^{- (1 - r (t)^{2}) / 4} \cos (r (t) x) = u (t, x) .$ Thus $u$ is the stated backward value function and $u (1, x) = \cos x$ .

2.2F2F5F6givenstep 1.1

The exact process is centered normal at each time with variance $(1 - e^{- 2 t}) / 2 \leq 1 / 2$ by [F2], so [F6] at order $12$ gives $E ∣ X_{t} ∣^{12} = 11!! {(1 - e^{- 2 t}) / 2}^{6} \leq 11!! / 64$ uniformly in $t \in [0, 1]$ , including $t = 0$ . For the Euler process, $b (x) = - x$ is globally Lipschitz with linear growth, $σ (x) = 1$ is globally Lipschitz with linear growth, the time modulus is zero, and $X_{0} = 0$ has every moment. Applying [F5] at $p = 12$ therefore gives uniform pointwise 12th moments for $Y^{h}$ and ${\overset{ˉ}{Y}}^{h}$ . The threshold is $p = 12 = 2 q + 8$ .

3.1A1F4F7step 2.1step 1.2step 2.2∎

The payoff is bounded and $1$ -Lipschitz. By [F4], the interpolation agrees with the Euler recursion at $T$ , so the conclusion of [F7] applies to $Y_{N}^{h}$ . Steps 2.1–2.2 verify its backward regularity and moment premises, giving $∣ E \cos (X_{1}) - E \cos (Y_{N}^{h}) ∣ \leq C h$ . AC is inherited through [A1].

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

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].

ExampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

Shared-increment Euler coupling and payoff rates

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 $F_{s}$ . For $d X_{t} = - X_{t} d t + d W_{t}$ , $X_{0} = 0$ , and $T = 1$ , couple the Euler paths with fine step $h = 1 / (2 n)$ and coarse step $2 h = 1 / n$ by summing each pair of fine Brownian increments for the coarse update. For $φ (x) = \cos x$ , the shared-Brownian theorem gives correction variance $O (h)$ , strong bias $O (h^{1 / 2})$ , and coupled path work $O (h^{- 1})$ .

Facts & Assumptions

Given: AC; the complete usual filtered Brownian setup above on $[0, 1]$ ; the $F_{0}$ -measurable initial value $X_{0} = 0$ ; $h = 1 / (2 n)$ for $n \geq 1$ ; the fine/coarse coupling described in the Example; and $φ (x) = \cos x$ .

[F1]

Under the stated nested-grid coupling and a globally Lipschitz payoff, the Euler correction has second moment $O (h)$ , its variance is $O (h)$ , the strong-route bias is $O (h^{1 / 2})$ , and fixed-refinement path work is $O (h^{- 1})$ . Shared-Brownian Euler level correction rates

[F2]

For a real continuous Brownian Itô process $X_{t} = X_{0} + \int_{0}^{t} b_{s} d s + \int_{0}^{t} σ_{s} d W_{s}$ , applying Itô's formula to $e^{t} X_{t}$ is valid and gives the displayed drift and diffusion terms. One-dimensional Ito formula

[F3]

For a square-integrable Brownian Itô integral $M_{t} = \int_{0}^{t} H_{s} d W_{s}$ , $E \sup_{t \leq T} ∣ M_{t} ∣^{2} \leq 4 E \int_{0}^{T} H_{s}^{2} d s$ . Doob maximal bound for the Ito integral

[F4]

A standard Brownian motion has independent normal increments of variance equal to the elapsed time. $d$ -dimensional Brownian motion

[A1]

The refinement factor is $M = 2$ ; each coarse increment is the sum of its two fine increments, and both paths start from the same $X_{0}$ and use the same Brownian path.

[A2]

The payoff is globally Lipschitz with constant $1$ , since $∣ \cos x - \cos y ∣ = ∣ \int_{y}^{x} - \sin s d s ∣ \leq ∣ x - y ∣$ .

[A3]

AC is inherited through the Brownian and Itô interfaces in [F2]–[F4]. The Axiom of Choice

Verification

1.1A3F2F3givenalgebra

The coefficients are $b (x) = - x$ and $σ (x) = 1$ . They are globally Lipschitz with constant $1$ , satisfy $∣ b (x) ∣^{2} + ∣ σ (x) ∣^{2} = x^{2} + 1$ , and are time independent, so the shared coefficient and time-modulus conditions hold. Applying [F2] to $e^{t} X_{t}$ gives $X_{t} = e^{- t} \int_{0}^{t} e^{s} d W_{s}$ . Applying [F2] to $e^{t} W_{t}$ then gives $X_{t} = \int_{0}^{t} e^{- (t - s)} d W_{s} = W_{t} - \int_{0}^{t} e^{- (t - s)} W_{s} d s .$ Hence $\sup_{t \leq 1} ∣ X_{t} ∣ \leq 2 \sup_{t \leq 1} ∣ W_{t} ∣$ , and [F3] gives $E \sup_{t \leq 1} ∣ X_{t} ∣^{2} \leq 16 < \infty$ .

1.2A1A2F4given

The payoff satisfies [A2]. The fine and coarse meshes obey $h \leq 1 / 2$ , $1 / h = 2 n \in N$ , and $1 / (2 h) = n \in N$ ; [A1] makes every coarse increment the corresponding Brownian increment over its full interval, [F4] gives its centered normal law with variance $2 h$ , and the given filtration-relative Brownian hypothesis supplies independence from the sigma-algebra at the start of that coarse interval. Thus the shared-rate theorem hypotheses hold.

2.1F1step 1.2∎

Applying [F1] yields $Var (\cos (Y_{1}^{h}) - \cos (Y_{1}^{2 h})) \leq C h, ∣ E \cos (X_{1}) - E \cos (Y_{1}^{h}) ∣ \leq C \sqrt{h},$ and work $O (h^{- 1})$ per coupled path pair. These rates correspond to $(α, β) = (1 / 2, 1)$ for this payoff.