13A.2 Unit roots and the Dickey–Fuller test

13A.2.1The regression form#

Start from

Subtract $y_{t-1}$:

$H_0: \gamma = 0$ (unit root; shocks are permanent; no mean reversion). $H_1: \gamma < 0$ (stationary; shocks decay at rate $\rho$ per period). The test is one-sided: $\gamma > 0$ is an explosive process, not a stationary one. OLS gives

13A.2.2The limiting distribution#

Proof

Under $H_0$, $y_t = \sum_{s\le t} e_s$.

Numerator. Square $y_t = y_{t-1} + e_t$: $y_t^2 = y_{t-1}^2 + 2y_{t-1}e_t + e_t^2$. Sum from 1 to $T$; the $y^2$ terms telescope:

$$\sum_{t=1}^T y_{t-1}e_t = \tfrac12\Big(y_T^2 - \sum_{t=1}^T e_t^2\Big).$$

Divide by $T$. By the CLT $y_T/\sqrt T \Rightarrow \sigma W(1)$, and by the LLN $T^{-1}\sum e_t^2 \to \sigma^2$. So $T^{-1}\sum y_{t-1}e_t \Rightarrow \tfrac{\sigma^2}{2}(W(1)^2 - 1)$.

Denominator. Define the step function $X_T(r) = T^{-1/2}y_{\lfloor Tr\rfloor}$ on $[0,1]$. The functional CLT (Donsker) gives $X_T \Rightarrow \sigma W$ in $D[0,1]$. Since $X_T$ is constant on $[(t-1)/T, t/T)$, $\int_0^1 X_T(r)^2dr = T^{-2}\sum_{t=1}^T y_{t-1}^2$. The map $f \mapsto \int_0^1 f^2$ is continuous, so $T^{-2}\sum y_{t-1}^2 \Rightarrow \sigma^2\int_0^1 W^2$.

Jointly. Both limits are functionals of the same $X_T$ (the numerator is $\tfrac12(X_T(1)^2 - T^{-1}\sum e_t^2)$), so they converge jointly and the continuous mapping theorem applies to the ratio:

$$T(\hat\rho-1) = \frac{T^{-1}\sum y_{t-1}e_t}{T^{-2}\sum y_{t-1}^2} \Rightarrow \frac{\tfrac12(W(1)^2-1)}{\int W^2}.$$

The $t$-ratio. $\operatorname{se}(\hat\rho)^2 = s^2/\sum y_{t-1}^2$ with $s^2 \to \sigma^2$, so $\hat\tau = T(\hat\rho-1)\,\big(T^{-2}\sum y_{t-1}^2\big)^{1/2}/s$, which converges to the second expression.∎

Three consequences a trader should know.

• The rate is $T$, not $\sqrt T$. $\hat\rho$ is "superconsistent" under a unit root. The statistic $\sqrt T(\hat\rho - 1)$ collapses to zero, so the usual normal approximation is useless.
• The distribution is skewed left. $P(\hat\rho < 1) \to P(W(1)^2 < 1) = P(|Z|<1) = 0.6827$. Even for a pure random walk, about two thirds of samples show "mean reversion" ($\hat\rho<1$). (Simulated with $T = 1{,}000$: 0.686.) This is the same downward bias that M10 · 10.7.2 corrected with Kendall's formula.
• Using $-1.645$ over-rejects. The 5% quantile of $\hat\tau$ is about $-1.95$ with no constant and $-2.86$ with a constant, not $-1.645$.

Deterministic terms With a constant in (13A.2) the regressor is effectively demeaned and $W$ is replaced by the demeaned Brownian motion $\tilde W(r) = W(r) - \int_0^1 W$ (M10 · 10.7.2). With a constant and a linear trend, $W$ is replaced by the residual of its projection on $(1, r)$. The proofs are identical after projecting out the deterministic regressors (Hamilton 1994, 17.4). The distribution changes each time:

Regression	1%	5%	10%	Simulated, $T=500$, 20,000 draws (1%, 5%, 10%)
No constant (n)	−2.57	−1.94	−1.62	−2.55, −1.95, −1.62
Constant (c)	−3.43	−2.86	−2.57	−3.46, −2.89, −2.57
Constant and trend (ct)	−3.96	−3.41	−3.13	−3.96, −3.42, −3.14

The first three columns are MacKinnon's asymptotic values, as used by statsmodels.tsa.stattools.adfuller; the last column is computed by simulate_df_quantiles in m13a_adf.py.

Which case for a spread? A spread has an unknown non-zero mean, so use a constant. The no-constant case is only for series whose mean is zero by construction (the Engle–Granger residual, 13A.5). The trend case answers a different question: "is the spread stationary around a trend?" A trend-stationary spread is not something you can trade by betting on a return to a fixed level.

13A.2.3Augmentation#

Daily spreads rarely follow an AR(1) exactly. If the true model is AR($p$), the residual in (13A.2) is serially correlated and the DF distribution no longer applies. The fix is to add lagged differences.

Proof

For $i\ge2$, $y_{t-i} = y_{t-1} - \sum_{j=1}^{i-1}\Delta y_{t-j}$ (telescoping). Substitute:

$$y_t = \Big(\sum_{i=1}^p\phi_i\Big)y_{t-1} - \sum_{i=2}^p\phi_i\sum_{j=1}^{i-1}\Delta y_{t-j} + e_t.$$

Swap the order of summation over the triangle $1\le j < i \le p$: $\sum_{i=2}^p\sum_{j=1}^{i-1} = \sum_{j=1}^{p-1}\sum_{i=j+1}^p$. Subtract $y_{t-1}$ from both sides. The coefficient of $y_{t-1}$ is $\sum\phi_i - 1 = -\Phi(1)$, and $z=1$ is a root of $\Phi$ iff this is zero.∎

With deterministic terms, the ADF regression actually run is

dropping $d$ or $c$ as the case requires. Under $H_0$ and with the other roots of $\Phi$ outside the unit circle, $\Delta y$ is a stationary AR($p-1$), the coefficients $\delta_j$ are asymptotically normal, and the $t$-ratio on $\gamma$ has the same DF limit as before. Said and Dickey (1984) extend this to ARMA errors by letting $p$ grow with $T$ (at rate $T^{1/3}$ or slower). In practice choose $p$ by AIC or BIC, or by testing down from a generous maximum; adfuller(autolag="AIC") does the former.

Exercise link m13a_adf.py builds (13A.7) by hand and must match adfuller(maxlag=p, autolag=None) to $10^{-9}$ for all three trend cases and several $p$.

13A.2.4Power, and why more data is not always more information#

Unit-root tests have low power against near-unit-root alternatives. The useful way to see it: under an OU alternative sampled every $\Delta$, $\rho = e^{-\kappa\Delta}$, and the information about $\kappa$ grows with the span $S = T\Delta$ of the data, not with $T$. Sampling a five-year spread every minute instead of every day gives many more observations with $\rho$ closer to one and almost no extra power (Shiller and Perron 1985). 13A.7 makes this precise: the relative standard error of $\hat\kappa$ is about $\sqrt{2/(\kappa S)}$.

A complementary test reverses the null. KPSS (Kwiatkowski, Phillips, Schmidt and Shin 1992) tests $H_0$: stationary. Reporting both is good practice: "ADF does not reject a unit root and KPSS rejects stationarity" is a much stronger statement than either alone.