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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].

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