13A.7 The Ornstein–Uhlenbeck spread: estimation, half-life and precision

Motivation

Once a spread is judged stationary, a trader needs three numbers: the level it reverts to, how fast, and how noisy. The OU process supplies all three, and its estimation errors are larger than most backtests admit.

13A.7.1Model and exact discretisation#

dX_t = \kappa(\theta - X_t)\,dt + \sigma\,dW_t, \qquad \kappa>0. \tag{13A.22}

Apply Itô to $e^{\kappa t}X_t$ : $d(e^{\kappa t}X_t) = \kappa\theta e^{\kappa t}dt + \sigma e^{\kappa t}dW_t$ . Integrate over $[t, t+\Delta]$ :

X_{t+\Delta} = \theta + (X_t-\theta)e^{-\kappa\Delta} + \sigma\int_t^{t+\Delta}e^{-\kappa(t+\Delta-s)}dW_s.

The stochastic integral is Gaussian with mean zero and variance $\sigma^2(1-e^{-2\kappa\Delta})/(2\kappa)$ by the Itô isometry. This is M10 (10.18). So the sampled series is an exact Gaussian AR(1):

X_{t+\Delta} = a + b\,X_t + e_{t+\Delta},\qquad b = e^{-\kappa\Delta},\quad a = \theta(1-b),\quad \sigma_e^2 = \frac{\sigma^2(1-b^2)}{2\kappa}. \tag{13A.23}

The stationary law is $N(\theta, \sigma^2/(2\kappa))$ , and the stationary standard deviation equals $\sigma_e/\sqrt{1-b^2}$ . The half-life is $\ln 2/\kappa$ (M10 (10.19)).

13A.7.2Maximum likelihood#

Proposition 13A.7exact conditional MLE

Given $X_0,\dots,X_n$ at spacing $\Delta$ , the maximum-likelihood estimates conditional on $X_0$ are

\hat\kappa = -\frac{\ln\hat b}{\Delta},\qquad \hat\theta = \frac{\hat a}{1-\hat b},\qquad \hat\sigma = \hat\sigma_e\sqrt{\frac{2\hat\kappa}{1-\hat b^2}}, \tag{13A.24}

where $(\hat a,\hat b)$ are OLS estimates from regressing $X_{t+\Delta}$ on $(1, X_t)$ and $\hat\sigma_e^2 = \text{SSR}/n$ , provided $0<\hat b<1$ .

Proof

By (13A.23) the conditional log-likelihood is $-\tfrac n2\ln(2\pi\sigma_e^2) - \tfrac{1}{2\sigma_e^2}\sum(X_{t+\Delta}-a-bX_t)^2$ , maximised over $(a,b)$ by OLS and over $\sigma_e^2$ by $\text{SSR}/n$ . The map $(\kappa,\theta,\sigma)\mapsto(a,b,\sigma_e)$ of (13A.23) is a bijection from $(0,\infty)\times\mathbb R\times(0,\infty)$ onto $\{0<b<1\}\times\mathbb R\times(0,\infty)$ , and maximum likelihood is invariant under reparametrisation.∎

If $\hat b\ge1$ there is no OU fit: the data are more consistent with a random walk or an explosive process. fit_ou raises an error in that case rather than return a negative $\kappa$ . The exact (unconditional) likelihood adds the stationary density of $X_0$ ; the difference is negligible for long samples. The Euler approximation $\kappa\approx(1-\hat b)/\Delta$ is the first-order expansion of $-\ln\hat b/\Delta$ ; it understates $\kappa$ by a relative amount of about $\kappa\Delta/2$ , which is tiny for daily data and slow spreads.

13A.7.3How precisely is $\kappa$ known?#

Proposition 13A.8precision and bias of $\hat\kappa$ , first order

With $n$ observations spanning $S = n\Delta$ and $\kappa\Delta$ small,

\frac{\operatorname{se}(\hat\kappa)}{\kappa}\approx\sqrt{\frac{2}{\kappa S}},\qquad \frac{\mathbb E\hat\kappa - \kappa}{\kappa}\approx \frac{4}{\kappa S}. \tag{13A.25}

Proof

For a stationary AR(1), $\operatorname{Var}(\hat b)\approx(1-b^2)/n$ . The delta method on $\hat\kappa = -\ln\hat b/\Delta$ gives $\operatorname{se}(\hat\kappa)\approx\operatorname{se}(\hat b)/(b\Delta)$ . With $1-b^2 = 1 - e^{-2\kappa\Delta}\approx2\kappa\Delta$ and $b\approx1$ : $\operatorname{se}(\hat\kappa)^2\approx 2\kappa\Delta/(n\Delta^2) = 2\kappa/S$ . For the bias, Kendall's approximation (M10 · 10.7.2) is $\mathbb E\hat b - b\approx-(1+3b)/n\approx -4/n$ , so $\mathbb E\hat\kappa - \kappa\approx (4/n)/(b\Delta)\approx 4/S$ .∎

So the precision depends on $\kappa S$ , the number of mean-reversion time units in the sample (about 1.44 times the number of half-lives observed), and not on the sampling frequency.

Worked example

A daily spread with $\hat b = 0.97$ and $\hat\sigma_e = 0.01$ (log units), $\Delta = 1/250$ year:

$\kappa = -250\ln 0.97 = 7.615$ per year; half-life $\ln2/\kappa = 0.0910$ years $= 22.76$ trading days (the approximation $-\ln 2/(b-1)$ gives 23.10);
$\sigma = 0.01\sqrt{2\times7.615/(1-0.97^2)} = 0.1605$ per $\sqrt{\text{year}}$ ; stationary standard deviation $0.01/\sqrt{1-0.9409} = 0.0411$ , i.e. about 4.1%.
With two years of data ( $n = 500$ ), $\kappa S = 0.0305\times500 = 15.2$ . By (13A.25) the relative standard error is about 36% and the upward bias about 26%. A simulation of 2,000 such samples (m13a_ou.py, computed for this course) gives a mean $\hat\kappa/\kappa$ of 1.27, a median of 1.20 and a standard deviation of 0.44. Inverting that sampling distribution, the point estimate "half-life 23 days" from two years of data is consistent with a true half-life from about 19 to 38 days (16th to 84th percentiles of the simulated $\hat\kappa/\kappa$ , 4,000 draws) and from about 14 to 52 days at the 2.5th and 97.5th percentiles.

Practitioner notes

Report the half-life with an interval. Bias-correct $\hat b$ (M10 · 10.7.2) before converting to $\kappa$ .
A half-life of a few days estimated on intraday data from one month is not evidence of anything: $\kappa S$ is small whatever the number of bars.
If the half-life exceeds the period you are willing to hold a losing trade, the spread is not tradeable for you, whatever the test says.

13A.7.1Model and exact discretisation#

13A.7.2Maximum likelihood#

13A.7.3How precisely is κ\kappaκ known?#

13A.7.3How precisely is $\kappa$ known?#