The Ornstein-Uhlenbeck Process

A spread between cointegrated assets does not wander freely; it is pulled toward a long-run level at a rate proportional to its displacement. The diffusion that captures exactly this behaviour is the Ornstein-Uhlenbeck process, and it is tractable enough that its transition law, stationary law, and discrete-time estimation all follow in closed form.

#The mean-reverting equation

Definition1

The Ornstein-Uhlenbeck process solves the linear stochastic differential equation

dX_t=\kappa\,(\theta-X_t)\,dt+\sigma\,dW_t, \tag{1}

with mean-reversion speed $\kappa>0$ , long-run level $\theta\in\R$ , volatility $\sigma>0$ , and $W$ a Brownian motion.

The affine drift $\kappa(\theta-X_t)$ is globally Lipschitz with linear growth, so Equation (1) has a unique strong solution for every square-integrable initial condition [1].

#Closed form

Proposition2

The solution of Equation (1) is

X_t=\theta+(X_0-\theta)\,e^{-\kappa t}+\sigma\int_0^t e^{-\kappa(t-s)}\,dW_s. \tag{2}

Proof

Apply Ito's formula to $Y_t=e^{\kappa t}X_t$ . Since $Y$ has no second-order contribution from the finite-variation factor,

dY_t=\kappa e^{\kappa t}X_t\,dt+e^{\kappa t}\,dX_t=\kappa e^{\kappa t}\theta\,dt+\sigma e^{\kappa t}\,dW_t, \tag{3}

after substituting Equation (1) and cancelling the $\kappa e^{\kappa t}X_t$ terms. Integrating gives $Y_t=Y_0+\theta(e^{\kappa t}-1)+\sigma\int_0^t e^{\kappa s}\,dW_s$ , and multiplying by $e^{-\kappa t}$ returns Equation (2).

#Transition and stationary laws

Proposition3

Conditional on $X_0$ , the value $X_t$ is Gaussian with

\E[X_t\mid X_0]=\theta+(X_0-\theta)\,e^{-\kappa t}, \qquad \Var(X_t\mid X_0)=\frac{\sigma^2}{2\kappa}\big(1-e^{-2\kappa t}\big). \tag{4}

As $t$ grows the law converges to the stationary distribution $\mathcal N\!\big(\theta,\sigma^2/2\kappa\big)$ , which is invariant under Equation (1).

Proof

The stochastic integral in Equation (2) is a Wiener integral of a deterministic function, hence Gaussian with zero mean, so the conditional mean is the deterministic part of Equation (2). The Ito isometry gives the variance,

\Var(X_t\mid X_0)=\sigma^2\int_0^t e^{-2\kappa(t-s)}\,ds=\frac{\sigma^2}{2\kappa}\big(1-e^{-2\kappa t}\big). \tag{5}

Letting $t$ grow, the mean tends to $\theta$ and the variance to $\sigma^2/2\kappa$ . The solution is a time-homogeneous Markov process, so the law of $X_t$ given $X_0$ is governed by a transition semigroup $P_t$ , and $\mu P_t$ is the marginal of $X_t$ when $X_0\sim\mu$ . If $X_0$ is drawn from $\mu=\mathcal N(\theta,\sigma^2/2\kappa)$ independently of $W$ , the summands in Equation (2) are independent Gaussians, so $X_t$ has mean $\theta$ and variance $e^{-2\kappa t}(\sigma^2/2\kappa)+(\sigma^2/2\kappa)(1-e^{-2\kappa t})=\sigma^2/2\kappa$ , hence $X_t\sim\mathcal N(\theta,\sigma^2/2\kappa)=\mu$ for every $t$ . Thus $\mu P_t=\mu$ for all $t$ , so $\mu$ is invariant.

A deviation from the mean decays in conditional expectation as $e^{-\kappa t}$ , so it loses half its size over the half-life

t_{1/2}=\frac{\ln 2}{\kappa}, \tag{6}

a single scalar summarizing the reversion timescale.

#Exact reduction to an autoregression

Proposition4

Sampled at a fixed spacing $\Delta$ , the process obeys the exact recursion

X_{(k+1)\Delta}=\theta(1-\varphi)+\varphi\,X_{k\Delta}+\varepsilon_k, \qquad \varphi=e^{-\kappa\Delta}, \tag{7}

where the $\varepsilon_k$ are independent and $\mathcal N\!\big(0,\sigma^2(1-\varphi^2)/2\kappa\big)$ .

Proof

Write Equation (2) from time $k\Delta$ to $(k+1)\Delta$ ,

X_{(k+1)\Delta}=\theta+(X_{k\Delta}-\theta)\,e^{-\kappa\Delta}+\sigma\int_{k\Delta}^{(k+1)\Delta} e^{-\kappa((k+1)\Delta-s)}\,dW_s. \tag{8}

The deterministic part rearranges to $\theta(1-\varphi)+\varphi X_{k\Delta}$ . The stochastic term $\varepsilon_k$ is a Wiener integral over $(k\Delta,(k+1)\Delta]$ , hence Gaussian with mean zero, independent of $\mathcal F_{k\Delta}$ and so of the past increments, and its variance is $\sigma^2\int_0^\Delta e^{-2\kappa u}\,du=\sigma^2(1-\varphi^2)/2\kappa$ by the Ito isometry.

Equation Equation (7) is a first-order autoregression. Because the innovations are i.i.d. $\mathcal N(0,s^2)$ with $s^2=\sigma^2(1-\varphi^2)/2\kappa$ constant in $k$ (homoskedastic), the conditional log-likelihood is $\sum_k\big[-\tfrac12\log(2\pi s^2)- (X_{(k+1)\Delta}-\theta(1-\varphi)-\varphi X_{k\Delta})^2/(2s^2)\big]$ . For fixed $s^2$ the maximiser over $(\theta(1-\varphi),\varphi)$ minimises the residual sum of squares, independent of the value of $s^2$ , and so is exactly the ordinary least-squares regression of $X_{(k+1)\Delta}$ on $X_{k\Delta}$ (the profile over $s^2$ being the residual variance). The slope estimates $\varphi$ , from which $\kappa=-\ln\varphi/\Delta$ and the half-life of Equation (6) follow.

#Illustration

Simulating the exact recursion Equation (7), regressing successive samples, and recovering the half-life shows the estimator at work.

import numpy as np
from numpy.random import Generator


def simulate_ou(
    kappa: float, theta: float, sigma: float, dt: float, n: int, rng: Generator
) -> np.ndarray:
    """Simulate an Ornstein-Uhlenbeck path by its exact discretization.

    Args:
        kappa: Mean-reversion speed.
        theta: Long-run level.
        sigma: Volatility.
        dt: Sampling spacing.
        n: Number of steps.
        rng: Seeded generator for reproducibility.

    Returns:
        The sampled path of length n + 1. It starts at path[0] = theta, the
        stationary mean, and converges to the stationary law N(theta,
        sigma^2 / (2 kappa)) after a transient initial segment.
    """
    phi = np.exp(-kappa * dt)
    innovation_sd = sigma * np.sqrt((1.0 - phi**2) / (2.0 * kappa))
    noise = rng.normal(0.0, innovation_sd, size=n)
    path = np.empty(n + 1)
    path[0] = theta
    for k in range(n):
        path[k + 1] = theta * (1.0 - phi) + phi * path[k] + noise[k]
    return path


def estimate_half_life(path: np.ndarray, dt: float) -> float:
    """Estimate the mean-reversion half-life by least squares on the AR(1) form.

    Args:
        path: A sampled Ornstein-Uhlenbeck trajectory.
        dt: The sampling spacing used to generate the path.

    Returns:
        The estimated half-life ln(2) / kappa.
    """
    lagged = path[:-1]
    nextval = path[1:]
    design = np.vstack([np.ones_like(lagged), lagged]).T
    slope = np.linalg.lstsq(design, nextval, rcond=None)[0][1]
    kappa = -np.log(slope) / dt
    return float(np.log(2.0) / kappa)


rng = np.random.default_rng(0)
path = simulate_ou(kappa=1.5, theta=0.0, sigma=0.3, dt=1.0 / 252.0, n=50_000, rng=rng)
half_life = estimate_half_life(path, dt=1.0 / 252.0)

The closed form Equation (2), the stationary law of Proposition 3, and the autoregressive reduction Equation (7) give the Ornstein-Uhlenbeck process closed-form transition, stationary, and estimation theory.

[1]

B. Øksendal, Stochastic Differential Equations: An Introduction with Applications, 6th ed. Springer, 2003.

Explore connections

see in the atlas

referenced by (2)

cite

@misc{ornstein-uhlenbeck,
  author = {Zac Kienzle},
  title  = {The Ornstein-Uhlenbeck Process},
  year   = {2026},
  month  = {05},
  url    = {https://zackienzle.com/blog/ornstein-uhlenbeck}
}

#The mean-reverting equation

Definition1

The Ornstein-Uhlenbeck process solves the linear stochastic differential equation

dX_t=\kappa\,(\theta-X_t)\,dt+\sigma\,dW_t, \tag{1}

with mean-reversion speed $\kappa>0$ , long-run level $\theta\in\R$ , volatility $\sigma>0$ , and $W$ a Brownian motion.

The affine drift $\kappa(\theta-X_t)$ is globally Lipschitz with linear growth, so Equation (1) has a unique strong solution for every square-integrable initial condition [1].

#Closed form

Proposition2

The solution of Equation (1) is

X_t=\theta+(X_0-\theta)\,e^{-\kappa t}+\sigma\int_0^t e^{-\kappa(t-s)}\,dW_s. \tag{2}

Proof

Apply Ito's formula to $Y_t=e^{\kappa t}X_t$ . Since $Y$ has no second-order contribution from the finite-variation factor,

dY_t=\kappa e^{\kappa t}X_t\,dt+e^{\kappa t}\,dX_t=\kappa e^{\kappa t}\theta\,dt+\sigma e^{\kappa t}\,dW_t, \tag{3}

#Transition and stationary laws

Proposition3

Conditional on $X_0$ , the value $X_t$ is Gaussian with

\E[X_t\mid X_0]=\theta+(X_0-\theta)\,e^{-\kappa t}, \qquad \Var(X_t\mid X_0)=\frac{\sigma^2}{2\kappa}\big(1-e^{-2\kappa t}\big). \tag{4}

As $t$ grows the law converges to the stationary distribution $\mathcal N\!\big(\theta,\sigma^2/2\kappa\big)$ , which is invariant under Equation (1).

Proof

\Var(X_t\mid X_0)=\sigma^2\int_0^t e^{-2\kappa(t-s)}\,ds=\frac{\sigma^2}{2\kappa}\big(1-e^{-2\kappa t}\big). \tag{5}

A deviation from the mean decays in conditional expectation as $e^{-\kappa t}$ , so it loses half its size over the half-life

t_{1/2}=\frac{\ln 2}{\kappa}, \tag{6}

a single scalar summarizing the reversion timescale.

#Exact reduction to an autoregression

Proposition4

Sampled at a fixed spacing $\Delta$ , the process obeys the exact recursion

X_{(k+1)\Delta}=\theta(1-\varphi)+\varphi\,X_{k\Delta}+\varepsilon_k, \qquad \varphi=e^{-\kappa\Delta}, \tag{7}

where the $\varepsilon_k$ are independent and $\mathcal N\!\big(0,\sigma^2(1-\varphi^2)/2\kappa\big)$ .

Proof

Write Equation (2) from time $k\Delta$ to $(k+1)\Delta$ ,

X_{(k+1)\Delta}=\theta+(X_{k\Delta}-\theta)\,e^{-\kappa\Delta}+\sigma\int_{k\Delta}^{(k+1)\Delta} e^{-\kappa((k+1)\Delta-s)}\,dW_s. \tag{8}

#Illustration

Simulating the exact recursion Equation (7), regressing successive samples, and recovering the half-life shows the estimator at work.

import numpy as np
from numpy.random import Generator


def simulate_ou(
    kappa: float, theta: float, sigma: float, dt: float, n: int, rng: Generator
) -> np.ndarray:
    """Simulate an Ornstein-Uhlenbeck path by its exact discretization.

    Args:
        kappa: Mean-reversion speed.
        theta: Long-run level.
        sigma: Volatility.
        dt: Sampling spacing.
        n: Number of steps.
        rng: Seeded generator for reproducibility.

    Returns:
        The sampled path of length n + 1. It starts at path[0] = theta, the
        stationary mean, and converges to the stationary law N(theta,
        sigma^2 / (2 kappa)) after a transient initial segment.
    """
    phi = np.exp(-kappa * dt)
    innovation_sd = sigma * np.sqrt((1.0 - phi**2) / (2.0 * kappa))
    noise = rng.normal(0.0, innovation_sd, size=n)
    path = np.empty(n + 1)
    path[0] = theta
    for k in range(n):
        path[k + 1] = theta * (1.0 - phi) + phi * path[k] + noise[k]
    return path


def estimate_half_life(path: np.ndarray, dt: float) -> float:
    """Estimate the mean-reversion half-life by least squares on the AR(1) form.

    Args:
        path: A sampled Ornstein-Uhlenbeck trajectory.
        dt: The sampling spacing used to generate the path.

    Returns:
        The estimated half-life ln(2) / kappa.
    """
    lagged = path[:-1]
    nextval = path[1:]
    design = np.vstack([np.ones_like(lagged), lagged]).T
    slope = np.linalg.lstsq(design, nextval, rcond=None)[0][1]
    kappa = -np.log(slope) / dt
    return float(np.log(2.0) / kappa)


rng = np.random.default_rng(0)
path = simulate_ou(kappa=1.5, theta=0.0, sigma=0.3, dt=1.0 / 252.0, n=50_000, rng=rng)
half_life = estimate_half_life(path, dt=1.0 / 252.0)

[1]

B. Øksendal, Stochastic Differential Equations: An Introduction with Applications, 6th ed. Springer, 2003.

Explore connections

see in the atlas

referenced by (2)

cite

@misc{ornstein-uhlenbeck,
  author = {Zac Kienzle},
  title  = {The Ornstein-Uhlenbeck Process},
  year   = {2026},
  month  = {05},
  url    = {https://zackienzle.com/blog/ornstein-uhlenbeck}
}