Estimating Ornstein-Uhlenbeck Parameters from Discrete Data

From Continuous-Time Dynamics to a Regression

Suppose observations of a mean-reverting process are available at equally spaced times. The exact transition law turns each consecutive pair into a linear regression observation with a Gaussian error. The regression is useful because it fits the discrete-time law directly; the fitted coefficients can then be translated back to the continuous-time parameters. That translation has constraints, so a numerical regression output is not automatically a valid Ornstein–Uhlenbeck fit.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). Consider the Ornstein–Uhlenbeck SDE \(dX_t=\kappa(\theta-X_t)\,dt+\sigma\,dB_t\), where \(\kappa>0\) is the mean-reversion rate, \(\theta\in\mathbb{R}\) is the long-run level, and \(\sigma>0\) is the diffusion coefficient. Observe \(X_{t_k}\) at \(t_k=kh\), where \(h>0\), for \(k=0,\ldots,n\). Set \(U_k=X_{t_k}\) and \(Y_k=X_{t_{k+1}}\).

For each finite \(T\), the stochastic integral below has integrand \(s\mapsto e^{-\kappa(t-s)}\), which is deterministic, progressively measurable, and satisfies \(\mathbb{E}\int_0^t e^{-2\kappa(t-s)}\,ds<\infty\). Thus it belongs to the Ito-integrand class \(\mathcal{L}^2\).

Proof. Apply Ito's formula to \(f(t,x)=e^{\kappa t}x\). Its partial derivatives are \(f_t=\kappa e^{\kappa t}x\), \(f_x=e^{\kappa t}\), and \(f_{xx}=0\), so the second-order correction is \(\tfrac12 f_{xx}\sigma^2\,dt=0\). The drift terms cancel, giving \(d(e^{\kappa t}X_t)=\kappa\theta e^{\kappa t}\,dt+\sigma e^{\kappa t}\,dB_t\). Integrating from \(s\) to \(t\) and multiplying by \(e^{-\kappa t}\) yields \(X_t=\theta+(X_s-\theta)e^{-\kappa(t-s)}+\sigma\int_s^t e^{-\kappa(t-u)}\,dB_u\). The integral is centered normal, independent of \(\mathcal{F}_s\), with variance \(\sigma^2\int_s^t e^{-2\kappa(t-u)}\,du=\frac{\sigma^2}{2\kappa}(1-e^{-2\kappa(t-s)})\). Brownian increments on disjoint intervals are independent, so the corresponding integrals are independent. Taking \(s=kh\) and \(t=(k+1)h\) gives the stated coefficients and variance. \(\square\)

Least-Squares Fit and Its Justification

Given the \(n\) pairs \((U_k,Y_k)\), fit an intercept and slope by minimizing the residual sum of squares \(S(a,b)=\sum_{k=0}^{n-1}(Y_k-a-bU_k)^2\). Define \(\bar U=n^{-1}\sum_k U_k\), \(\bar Y=n^{-1}\sum_k Y_k\), and \(S_{UU}=\sum_k(U_k-\bar U)^2\). If \(S_{UU}>0\), the predictors are not all equal, and the unconstrained least-squares coefficients are uniquely determined.

Proof. For any fixed \(b\), expand the sum around the response mean: \(\sum_k(Y_k-a-bU_k)^2=\sum_k[(Y_k-\bar Y)-b(U_k-\bar U)]^2+n(\bar Y-a-b\bar U)^2\). The cross term vanishes because both centered sums are zero. The second term is minimized uniquely at \(a=\bar Y-b\bar U\). Substitution leaves a quadratic in \(b\), namely \(\sum_k(Y_k-\bar Y)^2-2b\sum_k(U_k-\bar U)(Y_k-\bar Y)+b^2S_{UU}\). Since \(S_{UU}>0\), completing the square gives its unique minimizer as the stated \(\hat b\). Substitution into \(a=\bar Y-b\bar U\) gives \(\hat a\). The minimized value is exactly the sum of squared residuals \(S_{\min}\), by its definition. \(\square\)

For the Gaussian transition model, conditional on the observed starting value \(X_0\), the likelihood is proportional to \(q^{-n/2}\exp[-S(a,b)/(2q)]\), with \(q>0\). Thus, for fixed \(q\), maximizing the likelihood over unrestricted \(a,b\) is equivalent to least squares. For the fitted coefficients, if \(S_{\min}>0\), the maximum-likelihood estimate of the innovation variance is \(\hat q=S_{\min}/n\): differentiating the log likelihood \(-\frac n2\log q-\frac{S_{\min}}{2q}\) gives derivative \(-n/(2q)+S_{\min}/(2q^2)\), which vanishes at \(S_{\min}/n\), increases before that point, and decreases after it. If \(S_{\min}=0\), the likelihood is proportional to \(q^{-n/2}\), which diverges as \(q\downarrow0\). Since \(q=0\) is excluded, there is no finite maximum-likelihood estimate in that case.

To recover continuous-time parameters from an admissible fit, require \(0<\hat b<1\). Then the inverse formulas are \(\hat\kappa=-\frac{\log\hat b}{h}\), \(\hat\theta=\frac{\hat a}{1-\hat b}\), and \(\hat\sigma=\sqrt{\frac{2\hat\kappa\hat q}{1-\hat b^2}}\), provided \(\hat q>0\). These follow by solving \(b=e^{-\kappa h}\), \(a=\theta(1-b)\), and \(q=\sigma^2(1-b^2)/(2\kappa)\). The conditions ensure \(\hat\kappa>0\), a nonzero denominator for \(\hat\theta\), and a positive real volatility.

Worked Fits and Parameter Recovery

Admissibility and a Likelihood Pitfall

A related variance pitfall occurs even when the predictors vary. For example, take \(n=2\), \(U=(0,1)\), and \(Y=(1,3)\). The line \(Y=1+2U\) has residual sum of squares \(S_{\min}=0\). The conditional likelihood at that line is proportional to \(q^{-2/2}=q^{-1}\), so it tends to infinity as \(q\downarrow0\). Reporting \(\hat q=S_{\min}/n=0\) as an MLE is incorrect: zero is only a boundary limit, and no maximum exists over \(q>0\). Also, the slope \(2\) fails the OU admissibility condition. The example demonstrates two separate checks, not one issue.

Do not silently clip an inadmissible slope to \(0\) or \(1\). Neither endpoint corresponds to a finite positive \(\kappa\): \(b=1\) would give \(\kappa=0\), while \(b=0\) is only approached as \(\kappa\) grows without bound. If the scientific model requires \(\kappa>0\), estimate within that parameter space and report the constraint used. Finally, the fitted intercept and slope summarize the conditional transition law; a finite data set does not guarantee that either coefficient, or the resulting volatility, equals its simulation value.