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Stable sde schemes for superlinear drift — Examples

1 · Prerequisites

2 · Summary

The cubic dissipative equation shows that explicit Euler moments can diverge even when the exact SDE has finite moments. The CIR example checks positivity of the implicit Lamperti step from the boundary input, including a negative Brownian increment. The linear multiplicative-noise example separates deterministic Euler stability from stochastic mean-square stability and compares explicit and drift-implicit updates quantitatively.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Explicit Euler moments diverge for a cubic dissipative SDE

Statement

Fix $T > 0$ and consider $d X_{t} = - X_{t}^{3} d t + d W_{t}, X_{0} = 0.$ For $N \geq 1$ , let $h = T / N$ and let $Y_{0} = 0$ , $Y_{n + 1} = Y_{n} - h Y_{n}^{3} + (W_{t_{n + 1}} - W_{t_{n}}), t_{n} = n h .$ The exact solution has finite moments of every order at time $T$ , but for every $p \geq 1$ , $\lim_{N \to \infty} E ∣ Y_{N} ∣^{p} = + \infty .$

Facts & Assumptions

Given: AC, a one-dimensional standard Brownian motion, fixed $T > 0$ , and the SDE and Euler–Maruyama scheme in the Statement.

[F1]

Euler–Maruyama has grid update $Y_{n + 1} = Y_{n} + h b (Y_{n}) + σ (Y_{n}) Δ W_{n}$ . Autonomous Euler–Maruyama scheme and interpolation

[F2]

Brownian increments on disjoint deterministic intervals are independent, and $Δ W_{n}$ has law $N (0, h)$ with density $(2 π h)^{- 1 / 2} e^{- x^{2} / (2 h)}$ and second moment $h$ . $d$ -dimensional Brownian motion Standard normal and normal laws

[F3]

If $b$ is one-sided Lipschitz with polynomially bounded derivative, $σ$ is globally Lipschitz, and the initial state has all finite moments, the local solution is global and has finite moments of every order on finite horizons. Global existence and finite-horizon moments for one-sided Lipschitz SDEs Locally Lipschitz Itô SDE on an open state domain

[A1]

AC is inherited through the Brownian and SDE interfaces. The Axiom of Choice

Proof

Given: The assumptions and supplier statements above.

1.1F3givenalgebra

The drift $b (x) = - x^{3}$ satisfies $(x - y) (b (x) - b (y)) = - (x - y)^{2} (x^{2} + x y + y^{2}) \leq 0$ , since $x^{2} + x y + y^{2} = (x + y / 2)^{2} + 3 y^{2} / 4 \geq 0$ . Also $∣ b^{'} (x) ∣ = 3 x^{2} \leq 3 (1 + ∣ x ∣^{2})$ , and $σ (x) = 1$ is globally Lipschitz. Thus [F3] applies with deterministic $X_{0} = 0$ , so $E ∣ X_{T} ∣^{p} < \infty$ for every finite $p \geq 1$ .

1.2F1F2givenalgebra

The algorithm [F1] gives the displayed recursion. For $h \leq 1 / 4$ , set $E_{N} = {h^{- 1} \leq Δ W_{0} \leq 2 h^{- 1}} \cap ⋂_{n = 1}^{N - 1} {∣ Δ W_{n} ∣ \leq 1} .$ The increments are independent by [F2]. The normal density, bounded below on $[h^{- 1}, 2 h^{- 1}]$ by its value at $2 h^{- 1}$ , gives $P (h^{- 1} \leq Δ W_{0} \leq 2 h^{- 1}) \geq (2 π)^{- 1 / 2} h^{- 3 / 2} e^{- 2 h^{- 3}} .$ Also $P (∣ Δ W_{n} ∣ > 1) \leq E ∣ Δ W_{n} ∣^{2} = h$ , since $1_{{∣ z ∣ > 1}} \leq ∣ z ∣^{2}$ . For $h \leq 1 / 4$ , $(1 - h)^{N - 1} \geq e^{- 2 h (N - 1)} \geq e^{- 2 T}$ . Hence $P (E_{N}) \geq (2 π)^{- 1 / 2} e^{- 2 T} h^{- 3 / 2} e^{- 2 h^{- 3}} .$

2.1F1F2step 1.2givenalgebra

On $E_{N}$ , $Y_{1} = Δ W_{0} \geq h^{- 1}$ . If $∣ y ∣ \geq h^{- 1}$ and $∣ δ ∣ \leq 1$ , then $∣ y - h y^{3} + δ ∣ \geq h ∣ y ∣^{3} - ∣ y ∣ - 1 = h ∣ y ∣^{3} (1 - \frac{1}{h ∣ y ∣^{2}} - \frac{1}{h ∣ y ∣^{3}}) \geq \frac{h}{2} ∣ y ∣^{3},$ because the two fractions are at most $h$ and $h^{2}$ , whose sum is at most $5 / 16 < 1 / 2$ . The lower bound is at least $h^{- 1}$ when $h \leq 1 / 2$ and $∣ y ∣ \geq h^{- 1}$ . Induction therefore gives $∣ Y_{n} ∣ \geq h^{- 1}$ for all $1 \leq n \leq N$ on $E_{N}$ . With $z_{n} = \sqrt{h / 2} ∣ Y_{n} ∣$ , the same inequality gives $z_{n + 1} \geq z_{n}^{3}$ , while $z_{1} \geq (2 h)^{- 1 / 2} > 1$ . Thus $∣ Y_{N} ∣ \geq \sqrt{2 / h} (2 h)^{- 3^{N - 1} / 2} on E_{N} .$

3.1A1F3step 1.2step 2.1givenalgebra∎

Combining steps 1.2 and 2.1 yields $E ∣ Y_{N} ∣^{p} \geq (2 π)^{- 1 / 2} e^{- 2 T} h^{- 3 / 2} e^{- 2 h^{- 3}} {(\sqrt{2 / h} (2 h)^{- 3^{N - 1} / 2})}^{p} .$ Here $h = T / N$ . The positive logarithmic term $\frac{p}{2} 3^{N - 1} \log ((2 h)^{- 1})$ grows faster than the negative term $2 h^{- 3} = 2 N^{3} / T^{3}$ , so this lower bound tends to $+ \infty$ . AC is exactly [A1].

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The CIR Lamperti step remains positive from the boundary value

Statement

Let $κ, θ, σ, h > 0$ satisfy $2 κ θ \geq σ^{2}$ . Starting the numerical Lamperti state at the boundary value $z_{n} = 0$ , even a negative Brownian increment $Δ W_{n} = - r$ with arbitrary $r \geq 0$ gives a strictly positive next state under the CIR scheme.

Facts & Assumptions

Given: AC, positive parameters satisfying the Feller condition, $z_{n} = 0$ , and $Δ W_{n} = - r$ for $r \geq 0$ .

[F1]

Put $a = (4 κ θ - σ^{2}) / 8$ , $A = 1 + κ h / 2$ , and $c = z_{n} + σ Δ W_{n} / 2$ . The scheme returns $(c + \sqrt{c^{2} + 4 A a h}) / (2 A)$ . Positive Lamperti drift-implicit scheme for the CIR process

[F2]

Under $κ, θ, σ > 0$ and $2 κ θ \geq σ^{2}$ , $a > 0$ and the displayed root is the unique positive solution of the drift-implicit Lamperti equation for every $z_{n} \geq 0$ and finite increment. CIR Lamperti equation and unique positive implicit root

[A1]

AC is inherited from the numerical Brownian and algebraic interfaces. The Axiom of Choice

Proof

Given: The hypotheses and supplier statements above.

1.1F1F2givenalgebra

From the parameter assumptions, $4 κ θ - σ^{2} \geq σ^{2} > 0$ , so $a > 0$ ; also $A = 1 + κ h / 2 > 0$ . With $z_{n} = 0$ and $Δ W_{n} = - r$ , $c = - σ r / 2 \leq 0$ .

2.1A1F1F2step 1.1givenalgebra∎

Since $4 A a h > 0$ , $\sqrt{c^{2} + 4 A a h} > ∣ c ∣ = σ r / 2.$ Therefore the numerator $c + \sqrt{c^{2} + 4 A a h}$ and denominator $2 A$ are both strictly positive, so the algorithm's next Lamperti state is positive. Its square is also strictly positive. The uniqueness and equation interpretation follow from [F2], including the case $r = 0$ . AC is exactly [A1].

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Noise destabilizes explicit Euler while drift-implicit Euler contracts

Statement

Consider the mean-square stable scalar SDE $d X_{t} = - 2 X_{t} d t + \sqrt{3} X_{t} d W_{t}, X_{0} = 1.$ Its exact second moment is $E ∣ X_{t} ∣^{2} = e^{- t}$ . On a grid with $h = 1 / 2$ , explicit Euler–Maruyama has one-step mean-square amplification $3 / 2 > 1$ , even though the deterministic Euler multiplier is zero. The drift-implicit Euler–Maruyama update has one-step mean-square factor $5 / 8 < 1$ .

Facts & Assumptions

Given: AC; a one-dimensional Brownian motion; the scalar SDE and $h = 1 / 2$ grid in the Statement.

[F1]

Under globally Lipschitz coefficients and square-integrable initial state, the stated filtered SDE has a unique strong solution. Itô SDE data and strong solution Global strong well-posedness for Lipschitz Itô SDEs

[F2]

Itô's formula applies to the exponential candidate and to the square of a continuous Brownian Itô process. Multidimensional Ito formula for Brownian-driven processes

[F3]

Brownian increments are independent, centered normal with variance $h$ ; in particular $E e^{u W_{t}} = e^{u^{2} t / 2}$ . $d$ -dimensional Brownian motion Standard normal and normal laws

[F4]

Explicit Euler uses $Y_{n + 1} = Y_{n} + h b (Y_{n}) + σ (Y_{n}) Δ W_{n}$ . Autonomous Euler–Maruyama scheme and interpolation

[F5]

Drift-implicit Euler applies the backward resolvent to $Y_{n} + σ (Y_{n}) Δ W_{n}$ and is well-defined for the dissipative linear drift $b (x) = - 2 x$ . Drift-implicit Euler–Maruyama with explicit solve-failure status

[A1]

AC is inherited through the Brownian, SDE, and algorithm interfaces. The Axiom of Choice

Proof

Given: The hypotheses and supplier statements above.

1.1F1F2F3givenalgebra

Define $X_{t} = \exp (- 7 t / 2 + \sqrt{3} W_{t})$ . Itô's formula [F2] gives $d X_{t} = (- 7 / 2 + 3 / 2) X_{t} d t + \sqrt{3} X_{t} d W_{t} = - 2 X_{t} d t + \sqrt{3} X_{t} d W_{t}$ and $X_{0} = 1$ . The coefficients are globally Lipschitz, so [F1] identifies this with the unique strong solution. For $t > 0$ , completing the square in the $N (0, t)$ density gives $E e^{u W_{t}} = e^{u^{2} t / 2}$ ; at $t = 0$ both sides are $1$ . Hence $E ∣ X_{t} ∣^{2} = e^{- 7 t} E e^{2 \sqrt{3} W_{t}} = e^{- 7 t} e^{6 t} = e^{- t}$ .

1.2F3F4givenalgebra

Let $Y_{n}$ be explicit Euler with $h = 1 / 2$ . Since $b (x) = - 2 x$ and $σ (x) = \sqrt{3} x$ , [F4] gives $Y_{n + 1} = \sqrt{3} Y_{n} Δ W_{n}$ . Conditional on $F_{t_{n}}$ , independence and variance $h$ imply $E (∣ Y_{n + 1} ∣^{2} ∣ F_{t_{n}}) = 3 h ∣ Y_{n} ∣^{2} = \frac{3}{2} ∣ Y_{n} ∣^{2} .$ Without the noise term, the deterministic multiplier is $1 - 2 h = 0$ , so the factor greater than one here is caused by the stochastic increment.

2.1A1F3F5givenalgebra∎

For drift-implicit Euler, the resolvent equation is $Y_{n + 1} + 2 h Y_{n + 1} = Y_{n} + \sqrt{3} Y_{n} Δ W_{n}$ . At $h = 1 / 2$ this gives $Y_{n + 1} = \frac{1}{2} (1 + \sqrt{3} Δ W_{n}) Y_{n}$ . Hence $E (∣ Y_{n + 1} ∣^{2} ∣ F_{t_{n}}) = \frac{1}{4} (1 + 3 h) ∣ Y_{n} ∣^{2} = \frac{5}{8} ∣ Y_{n} ∣^{2} .$ The factor is strictly below one, so iteration gives $E ∣ Y_{n} ∣^{2} \leq (5 / 8)^{n} ∣ Y_{0} ∣^{2}$ . The implicit resolvent exists because $x \mapsto x + 2 h x$ is bijective. AC is exactly [A1].

Stable sde schemes for superlinear drift — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Explicit Euler moments diverge for a cubic dissipative SDE

Statement

Facts & Assumptions

Proof

The CIR Lamperti step remains positive from the boundary value

Statement

Facts & Assumptions

Proof

Noise destabilizes explicit Euler while drift-implicit Euler contracts

Statement

Facts & Assumptions

Proof

Sources