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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].

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Sources

Desmond J. Higham, Xuerong Mao and Andrew M. Stuart, Strong convergence of Euler-type methods for nonlinear stochastic differential equations, §5 (standard reference, not scraped)