Ornstein–Uhlenbeck Process Theory

1 Project aims

This notebook develops the Ornstein-Uhlenbeck (OU) process as a model for mean-reverting spreads in pairs trading. We introduce the stochastic differential equation; derive its basic properties; simulate paths in discrete time; connect the model to empirical data.

2 Motivation

Pairs trading relies on the idea that a linear combination of two asset prices behaves like a mean-reverting spread. If this spread deviates sufficiently from some typical level, then one may bet on reversion.

A mean-reverting process has the deterministic pull back to the long-run mean and random shocks which disrupt this reversion. Hence we may model such behaviour using an OU process.

3 Definition of the OU process

Consider the stochastic differential equation:

\[ dX_t = \theta(\mu - X_t)\,dt + \sigma\,dW_t, \]

where:

• \(X_t\) is the state at time \(t\)
• \(\mu\) is the long-run mean
• \(\theta > 0\) is the mean reversion speed
• \(\sigma \ge 0\) is the volatility parameter
• \((W_t)_{t \ge 0}\) is a standard Wiener process

Larger \(\theta\) means faster pull toward \(\mu\). Larger \(\sigma\) means noisier paths.

Examining the drift term, \(\theta(\mu - X_t)\): if \(X_t > \mu\), then the drift is negative, so the process pulls downwards; if \(X_t < \mu\), then the drift is positive, so the process pulls upwards.

4 Explicit solution

The OU process admits an explicit solution. We can do this by solving the stochastic differential equation via an integrating factor. We use some standard properties of stochastic integrals with deterministic integrands.

We can rewrite the OU SDE as:

\[ dX_t + \theta X_t\,dt = \theta \mu\,dt + \sigma\,dW_t. \]

Notice that this resembles a linear ODE. Hence we multiply through by the usual integrating factor \(e^{\theta t}\).

\[ e^{\theta t} dX_t + \theta e^{\theta t} X_t\,dt = \theta \mu e^{\theta t}\,dt + \sigma e^{\theta t}\,dW_t. \]

Hence by the product rule,

\[ d(e^{\theta t} X_t) = \theta \mu e^{\theta t}\,dt + \sigma e^{\theta t}\,dW_t. \]

Integrating from 0 to t gives

\[ e^{\theta t} X_t = X_0 + \theta \mu \int_{0}^{t} e^{\theta s}\,ds + \sigma \int_{0}^{t} e^{\theta s}\,dW_s. \]

Solving the deterministic integral gives \[ \int_{0}^{t} e^{\theta s}\,ds = \frac{e^{\theta t} - 1}{\theta}. \]

Therefore the solution to the OU SDE is

\[ X_t = X_0 e^{-\theta t} + \mu (1 - e^{-\theta t}) + \sigma \int_{0}^{t} e^{-\theta(t-s)}\,dW_s. \]

This then gives the more standard form

\[ X_t = \mu + (X_0 - \mu) e^{-\theta t} + \sigma \int_{0}^{t} e^{-\theta(t-s)}\,dW_s. \]

Note that this derivation avoids using Itô calculus hence is not fully rigorous. However this manipulation gives the correct structure.

We may now take the expectation and use linearity. Note that the integrand is deterministic so the expectation of the integral is \(0\).

\[ \mathbb{E}[X_t] = \mu + (X_0 - \mu) e^{-\theta t} \]

This is a deterministic exponential decay which describes how fast deviations from the long-run mean disappear on average. We have \(\mathbb{E}[X_t] \to \mu\) as \(t \to \infty\).

As for the variance, assuming \(X_0\) is fixed, only the stochastic integral contributes to the variance. Using the variance property of stochastic integrals,

\[ \operatorname{Var}[X_t] = \operatorname{Var}[\sigma \int_{0}^{t} e^{-\theta(t-s)}\,dW_s] = \sigma^2 \int_{0}^{t} e^{-2\theta(t-s)}\,ds. \]

Let \(u = t - s\). Then

\[ \int_{0}^{t} e^{-2\theta(t-s)}\,ds = \int_{0}^{t} e^{-2\theta u}\,du = \frac{1 - e^{-2\theta t}}{2\theta}. \]

The final variance function is therefore

\[ \operatorname{Var}[X_t] = \frac{\sigma^2}{2\theta}(1-e^{-2\theta t}). \]

For the long-run variance, we let \(t \to \infty\), \(\operatorname{Var}[X_t] \to \frac{\sigma^2}{2\theta}\).

As \(t \to \infty\), the distribution of \(X_t\) converges to the stationary law \[ X_t \xrightarrow{d} \mathcal{N}\!\left(\mu,\frac{\sigma^2}{2\theta}\right). \]

5 Discrete-time simulation formula

If we consider the same OU SDE problem and apply the explicit solution over the interval \([t, t+\Delta t]\), we get

\[ X_{t+\Delta t} = \mu + (X_t - \mu) e^{-\theta\Delta t} + \sigma \int_{t}^{t+\Delta t} e^{-\theta(t+\Delta t-s)}\,dW_s. \]

Conditioning on \(X_t\), the only random term is the stochastic integral, and since it is a centred Gaussian integral over future Brownian increments, its conditional expectation is 0. Hence

\[ \mathbb{E}[X_{t+\Delta t} \mid X_t] = \mu + (X_t - \mu) e^{-\theta\Delta t}. \]

Given \(X_t\) similarly to before, only the integral contributes to the variance:

\[ \operatorname{Var}[X_{t+\Delta t} \mid X_t] = \operatorname{Var}[\sigma \int_{t}^{t+\Delta t} e^{-\theta(t+\Delta t-s)}\,dW_s]. \]

Then using the same variance property of stochastic integrals,

\[ \operatorname{Var}[X_{t+\Delta t} \mid X_t] = \sigma^2 \int_{t}^{t+\Delta t} e^{-2\theta(t+\Delta t-s)}\,ds. \]

Set \(u=t+\Delta t-s\) then

\[ \int_{t}^{t+\Delta t} e^{-2\theta(t+\Delta t-s)}\,ds = \int_{0}^{\Delta t} e^{-2\theta u}\,du = \frac{1-e^{-2\theta\Delta t}}{2\theta}. \]

Therefore we have the conditional variance

\[ \operatorname{Var}[X_{t+\Delta t} \mid X_t] = \frac{\sigma^2}{2\theta}(1-e^{-2\theta\Delta t}). \]

The stochastic integral of a deterministic function against Brownian motion is Gaussian, so the whole conditional law is Gaussian. Hence we have the conditional distribution of \(X_{t+\Delta t}\) given \(X_t\).

\[ X_{t+\Delta t} \mid X_t \sim \mathcal{N}\!\left(\mu + (X_t - \mu) e^{-\theta\Delta t}, \frac{\sigma^2}{2\theta}(1-e^{-2\theta\Delta t})\right). \]

We now have the exact simulation formula. If \(Z \sim \mathcal{N}\!(0, 1)\), then we have the formula

\[ X_{t+\Delta t} = \mu + (X_t - \mu)e^{-\theta\Delta t} + \sigma\sqrt{\frac{1-e^{-2\theta\Delta t}}{2\theta}}\,Z. \]