Weak Euler Bias for Ornstein-Uhlenbeck Moments

Moment Bias for the Ornstein–Uhlenbeck Process

The Ornstein–Uhlenbeck (OU) process is a useful test case for an SDE simulation scheme: its mean and variance can be calculated exactly, while Euler–Maruyama produces explicit discrete-time formulas. Comparing these formulas isolates the time-discretization bias in two basic weak quantities. Here, “bias” means the numerical moment minus the corresponding exact moment; it is a deterministic difference, not the random error from a finite Monte Carlo sample.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. Fix parameters $\kappa>0$, $\theta\in\mathbb{R}$, $\sigma\in\mathbb{R}$, and a deterministic initial value $x_0$. The OU SDE is

dX_t=\kappa(\theta-X_t)\,dt+\sigma\,dB_t,\qquad X_0=x_0.

The drift is $\mu(x)=\kappa(\theta-x)$, and the diffusion coefficient is the constant $\sigma$. For a finite horizon $T$, the Itô integrals below have progressively measurable integrands in $\mathcal{L}^2$, the class of processes $H_s$ satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. In particular, deterministic kernels such as $e^{-\kappa(t-s)}$ meet this condition on every finite time interval.

Exact OU Moments

Proposition: For $t\ge0$, the OU process with deterministic initial value $x_0$ has the representation $$ X_t=\theta+(x_0-\theta)e^{-\kappa t} +\sigma\int_0^t e^{-\kappa(t-s)}\,dB_s. $$ Consequently, $$ \mathbb{E}[X_t]=\theta+(x_0-\theta)e^{-\kappa t},\qquad \operatorname{Var}(X_t)=\frac{\sigma^2}{2\kappa}(1-e^{-2\kappa t}). $$

Proof. Multiply the SDE for $X_t-\theta$ by the deterministic integrating factor $e^{\kappa t}$. The product rule gives $d(e^{\kappa t}(X_t-\theta))=\sigma e^{\kappa t}\,dB_t$. Integrating from $0$ to $t$, then multiplying by $e^{-\kappa t}$, yields the stated representation. The integrand $s\mapsto e^{-\kappa(t-s)}$ is deterministic and satisfies $\int_0^t e^{-2\kappa(t-s)}ds<\infty$; thus the Itô integral is defined in $\mathcal{L}^2$. Its expectation is zero, and the Itô isometry gives its variance as $\int_0^t e^{-2\kappa(t-s)}ds=(1-e^{-2\kappa t})/(2\kappa)$. Multiplication by $\sigma$ scales that variance by $\sigma^2$, proving both moment formulas. $\square$

The zero-noise case $\sigma=0$ is included: the integral and variance are then zero. If $x_0=\theta$, the mean remains $\theta$. These cases will also be covered by the discrete formulas.

One Euler Step: Bias in Mean and Variance

For a step of length $h>0$, Euler–Maruyama started at $X_0=x$ gives $X_1^{\mathrm{EM}}=x+\kappa(\theta-x)h+\sigma(B_h-B_0)$. The increment $B_h-B_0$ has mean zero and variance $h$, so the numerical mean and variance are $\theta+(x-\theta)(1-\kappa h)$ and $\sigma^2h$, respectively. The exact OU process started at $x$ has mean $\theta+(x-\theta)e^{-\kappa h}$ and variance $\sigma^2(1-e^{-2\kappa h})/(2\kappa)$.

Theorem: Define one-step bias as Euler–Maruyama moment minus exact moment. Then $$ b_{\mathrm{mean}}=(x-\theta)(1-\kappa h-e^{-\kappa h}),\qquad b_{\mathrm{var}}=\sigma^2\left(h-\frac{1-e^{-2\kappa h}}{2\kappa}\right). $$ For $h>0$, the variance bias is nonnegative, and it is strictly positive if $\sigma\ne0$.

Proof. Subtract the exact mean from the numerical mean to obtain the displayed mean bias. Subtracting the exact variance from $\sigma^2h$ gives the displayed variance bias. To establish its sign, set $y=2\kappa h>0$. The function $g(y)=e^{-y}-1+y$ satisfies $g(0)=0$ and $g'(y)=1-e^{-y}>0$ for $y>0$. Hence $1-e^{-y}<y$, so $(1-e^{-2\kappa h})/(2\kappa)<h$. Multiplication by $\sigma^2$ proves the claim, including equality when $\sigma=0$. $\square$

Worked Example: One-Step Bias from a Nonstationary Start

Take $\kappa=1$, $\theta=2$, $x=0$, $\sigma=1$, and $h=\tfrac12$. Compute the one-step numerical and exact means and variances, then compare them.

The Euler mean is $2+(0-2)(1-\tfrac12)=2-1=1$. The exact mean is $2+(0-2)e^{-1/2}=2-2e^{-1/2}$. Thus the signed mean bias, numerical minus exact, is $1-(2-2e^{-1/2})=2e^{-1/2}-1$.

The Euler variance is $\sigma^2h=1\cdot\tfrac12=\tfrac12$. The exact variance is $\frac{1}{2}(1-e^{-2(1)(1/2)})=\frac{1-e^{-1}}{2}$. Their difference is $\tfrac12-\frac{1-e^{-1}}2=\frac{e^{-1}}2>0$.

Check: the mean bias also equals $(x-\theta)(1-\kappa h-e^{-\kappa h})=(-2)(\tfrac12-e^{-1/2})=2e^{-1/2}-1$, as calculated directly. The variance difference is positive because the one-step Euler variance is $\tfrac12$, while the exact variance is $(1-e^{-1})/2<\tfrac12$.

Finite-Time Moments on a Uniform Grid

Let $T=Nh$ for an integer $N\ge1$, with grid points $t_n=nh$. The Euler–Maruyama recursion is $X_{n+1}^{\mathrm{EM}}=X_n^{\mathrm{EM}}+\kappa(\theta-X_n^{\mathrm{EM}})h+\sigma\Delta B_n$, where $\Delta B_n=B_{t_{n+1}}-B_{t_n}$. Each increment is independent of the past, has mean zero, and has variance $h$. Write $q=1-\kappa h$. Then the recursion becomes $X_{n+1}^{\mathrm{EM}}-\theta=q(X_n^{\mathrm{EM}}-\theta)+\sigma\Delta B_n$.

Proposition: For $0<\kappa h<2$, the finite-time Euler moments are $$ \mathbb{E}[X_N^{\mathrm{EM}}]=\theta+(x_0-\theta)q^N,\qquad \operatorname{Var}(X_N^{\mathrm{EM}}) =\sigma^2h\sum_{j=0}^{N-1}q^{2j} =\frac{\sigma^2(1-q^{2N})}{2\kappa-\kappa^2h}. $$ The exact moments at $T=Nh$ are $$ \mathbb{E}[X_T]=\theta+(x_0-\theta)e^{-\kappa T},\qquad \operatorname{Var}(X_T)=\frac{\sigma^2}{2\kappa}(1-e^{-2\kappa T}). $$

Proof. Iterating the recursion gives $X_N^{\mathrm{EM}}-\theta=q^N(x_0-\theta)+\sigma\sum_{j=0}^{N-1}q^{N-1-j}\Delta B_j$. The sum has mean zero, so taking expectations gives the stated Euler mean. Its increments are independent and centered; consequently, its variance is the sum of the squared coefficients times $h$, namely $\sigma^2h\sum_{j=0}^{N-1}q^{2j}$. Since $0<\kappa h<2$, $|q|<1$, so $1-q^2\ne0$ and the finite geometric sum is $(1-q^{2N})/(1-q^2)$. Also $1-q^2=1-(1-\kappa h)^2=2\kappa h-\kappa^2h^2$. Substitution gives the closed form. The exact moments follow from the preceding proposition at $t=T$. If $\sigma=0$, both variance formulas give zero; if $q=0$, which occurs at $\kappa h=1$, the finite sum formula still applies and the closed form has denominator $1-q^2=1$. $\square$

These formulas make the finite-time biases explicit. The mean bias is $(x_0-\theta)(q^N-e^{-\kappa T})$; its sign depends on the starting point relative to $\theta$, so it need not be positive. The variance bias is $\sigma^2(1-q^{2N})/(2\kappa-\kappa^2h)-\sigma^2(1-e^{-2\kappa T})/(2\kappa)$. Unlike the one-step sign argument, its sign should be determined from this finite-time expression or from a justified comparison, not inferred from positivity of only one term.

Worked Example: Three Euler Steps and the Finite-Time Variance Bias

Take $\kappa=1$, $\theta=0$, $x_0=0$, $\sigma=2$, $h=\tfrac13$, and $N=3$, so $T=1$. Compute the finite-time variance bias exactly.

Here $q=1-\tfrac13=\tfrac23$. The Euler variance is $4\cdot\tfrac13[1+(\tfrac23)^2+(\tfrac23)^4]$. The squared powers are $1,\tfrac49,\tfrac{16}{81}$, whose sum is $\tfrac{81+36+16}{81}=\tfrac{133}{81}$. Therefore the Euler variance is $\tfrac43\cdot\tfrac{133}{81}=\tfrac{532}{243}$.

The exact variance at $T=1$ is $\frac{4}{2}(1-e^{-2})=2(1-e^{-2})$. Thus the signed variance bias is $$ \frac{532}{243}-2(1-e^{-2}) =\frac{532-486}{243}+2e^{-2} =\frac{46}{243}+2e^{-2}>0. $$

Check: substituting $q=\tfrac23$ into the closed geometric-sum formula gives $\frac{4(1-(\tfrac23)^6)}{2-\tfrac13} =\frac{4(1-\tfrac{64}{729})}{5/3} =\frac{4(665/729)}{5/3} =\frac{532}{243}$, agreeing with the term-by-term sum. The bias is positive because both $\tfrac{46}{243}$ and $2e^{-2}$ are positive; positivity does not follow from $e^{-2}>0$ alone without simplifying the other term.

Bias as the Step Size Decreases

The finite-time formulas also show the scale of the bias. Fix $\kappa>0$ and $T>0$, and take $h=T/N$ with $N$ increasing. In the following expansions, $O(h^2)$ denotes a quantity whose absolute value is bounded by a constant times $h^2$ for all sufficiently small $h$, with $\kappa,T$ fixed.

Proposition: As $h=T/N\to0$, the finite-time Euler mean and variance biases satisfy $$ \mathbb{E}[X_N^{\mathrm{EM}}]-\mathbb{E}[X_T] =-(x_0-\theta)e^{-\kappa T}\frac{\kappa^2T}{2}h+O(h^2), $$ $$ \operatorname{Var}(X_N^{\mathrm{EM}})-\operatorname{Var}(X_T) =\frac{\sigma^2}{2\kappa} \left(\kappa^2T e^{-2\kappa T} +\frac{\kappa}{2}(1-e^{-2\kappa T})\right)h+O(h^2). $$ In particular, both moment biases are of order at most $h$, and the leading variance-bias coefficient is positive when $\sigma\ne0$.

Proof. Taylor expansion of the logarithm at zero gives $\log(1-\kappa h)=-\kappa h-\kappa^2h^2/2+O(h^3)$. Since $N=T/h$, it follows that $q^N=e^{-\kappa T}(1-\kappa^2Th/2+O(h^2))$. Subtracting $e^{-\kappa T}$ and multiplying by $x_0-\theta$ proves the mean expansion.

For the variance, the same expansion gives $q^{2N}=e^{-2\kappa T}(1-\kappa^2Th+O(h^2))$, while $(2\kappa-\kappa^2h)^{-1}=(2\kappa)^{-1}(1+\kappa h/2+O(h^2))$. Insert these into the finite-time Euler variance. Multiplying the numerator and reciprocal-denominator expansions yields $$ \operatorname{Var}(X_N^{\mathrm{EM}}) =\frac{\sigma^2}{2\kappa} \left(1-e^{-2\kappa T} +h\left(\kappa^2T e^{-2\kappa T} +\frac{\kappa}{2}(1-e^{-2\kappa T})\right) +O(h^2)\right). $$ Subtracting the exact variance proves the claimed formula. For $T>0$ and $\kappa>0$, both terms in the displayed leading coefficient are positive; multiplication by $\sigma^2$ makes it strictly positive if $\sigma\ne0$. $\square$

The expansions concern exact moment bias, not the variance of a Monte Carlo estimator. In a simulation, sampling variability is an additional source of error. The finite-time formulas here allow the discretization component to be studied without that sampling uncertainty.

Worked Example: Finite-Time Mean and Variance from Two Steps

Take $\kappa=2$, $\theta=3$, $x_0=1$, $\sigma=2$, $h=\tfrac14$, and $N=2$, so $T=\tfrac12$. Compute both Euler moments, their exact counterparts, and the signed biases.

The step multiplier is $q=1-2(\tfrac14)=\tfrac12$, hence $q^2=\tfrac14$. The Euler mean is $3+(1-3)\tfrac14=3-\tfrac12=\tfrac52$. The exact mean is $3+(1-3)e^{-2(1/2)}=3-2e^{-1}$. The mean bias is therefore $\tfrac52-(3-2e^{-1})=2e^{-1}-\tfrac12$.

The Euler variance is $4\cdot\tfrac14[1+(\tfrac12)^2]=1(1+\tfrac14)=\tfrac54$. The exact variance is $\frac{4}{2(2)}(1-e^{-2(2)(1/2)})=1-e^{-2}$. The variance bias is $\tfrac54-(1-e^{-2})=\tfrac14+e^{-2}>0$.

Check: the closed form gives $\frac{4(1-(\tfrac12)^4)}{4-4(\tfrac14)} =\frac{4(15/16)}{3}=\tfrac54$, matching the direct sum. The mean bias agrees with $(x_0-\theta)(q^N-e^{-\kappa T})=(-2)(\tfrac14-e^{-1})=2e^{-1}-\tfrac12$. For the variance bias, both terms $\tfrac14$ and $e^{-2}$ are positive.

Common Pitfall: Mixing Up the Bias and Its Sign

A frequent error is to compare the numerical and exact moments but then report the difference with the opposite sign. The convention here is always numerical minus exact. For the three-step example, a wrong calculation would write $\text{bias}=2(1-e^{-2})-\tfrac{532}{243}$, which is the exact-minus-numerical difference, not the defined bias. The correction is to retain the order in the subtraction: $\tfrac{532}{243}-2(1-e^{-2})=\tfrac{46}{243}+2e^{-2}>0$. The sign requires simplifying the entire expression; observing only that $e^{-2}>0$ would not determine the sign of an expression containing other terms.

Also distinguish the drift’s effect on the mean from the diffusion’s effect on the variance. The mean bias depends on $x_0-\theta$, so it can vanish when the process starts at its mean-reversion level, even when the variance bias does not. Conversely, when $\sigma=0$, both variance formulas vanish, while the mean can still have a time-discretization bias.

Check Your Understanding

Use numerical-minus-exact as the signed-bias convention in each question.

For one Euler step with $\kappa=2$, $h=\tfrac14$, and $\sigma=1$, calculate the exact and Euler variances and their difference.
For a uniform grid with $N$ steps, write the finite-time Euler mean in terms of $q=1-\kappa h$, and state the exact mean at $T=Nh$.
Explain why the finite-time mean bias can be zero when the variance bias is nonzero.
For $\sigma=0$, what do the exact and Euler variance formulas give? Can the mean bias still be nonzero?
In the three-step worked example, verify the Euler variance by evaluating the geometric sum rather than adding its three terms.