Dickey-Fuller stationarity test

The Dickey-Fuller test checks the stationarity of a scalar time series using one time series. It assumes that the X: \Omega \times \cD \rightarrow \Rset process with \cD \in \Rset, discretized on the time grid (t_0, \dots, t_{N-1}) writes:

(1)X_t = a + bt + \rho X_{t-1} + \varepsilon_{t}

where \rho > 0 and where a or b or both (a,b) can be assumed to be equal to 0.

The Dickey-Fuller test checks whether the random perturbation at time t vanishes with time.

When a \neq 0 and b=0, the model (1) is said to have a drift. When a = 0 and b \neq 0, the model (1) is said to have a linear trend.

In the model (1), the only way to have stochastic non stationarity is to have \rho = 1 (if \rho > 1, then the process diverges with time which is readily seen in the data). In the general case, the Dickey-Fuller test is a unit root test to detect whether \rho=1 against \rho < 1:

The test statistics and its limit distribution depend on the a priori knowledge we have on a and b. In case of absence of a priori knowledge on the structure of the model, several authors have proposed a global strategy to cover all the sub-cases of the model (1), depending on the possible values on a and b.

The strategy implemented is recommended by Enders (Applied Econometric Times Series, Enders, W., second edition, John Wiley & sons editions, 2004.).

We note (X_1, \hdots, X_n) the data, by W(r) the Wiener process, and W^{a}(r) = W(r) - \int_{0}^{1} W(r)\di{r}, W^{b}(r) = W^{a}(r) - 12 \left(r - \frac{1}{2} \right) \int_{0}^{1} \left(s - \frac{1}{2} \right) W(s)\di{s}.

1. We assume the model (2):

(2)\boldsymbol{X_t = a + bt + \rho X_{t-1} + \varepsilon_{t}}

The coefficients (a,b,\rho) are estimated by (\Hat{a}_n, \Hat{b}_n, \Hat{\rho}_n) using ordinary least-squares fitting, which leads to:

(3)\underbrace{\left( \begin{array}{lll} \displaystyle n-1 &\sum_{i=1}^n t_{i} &\sum_{i=2}^n y_{i-1}\\ \displaystyle \sum_{i=1}^n t_{i} &\sum_{i=1}^n t_{i}^2 &\sum_{i=2}^n t_{i} y_{i-1}\\ \displaystyle \sum_{i=2}^n y_{i-1}& \sum_{i=2}^n t_{i}y_{i-1} &\sum_{i=2}^n y_{i-1}^2 \end{array} \right)}_{\mat{M}} \left( \begin{array}{c} \hat{a}_n\\ \hat{b}_n\\ \hat{\rho}_n \end{array} \right)= \left( \begin{array}{l} \displaystyle \sum_{i=1}^n y_{i} \\ \displaystyle \sum_{i=1}^n t_{i} y_{i}\\ \displaystyle \sum_{i=2}^n y_{i-1} y_{i} \end{array} \right)

We first test:

(4)\left\{ \begin{array}{lr} \cH_0: & \rho = 1 \\ \cH_1: & \rho < 1 \end{array} \right.

thanks to the Student statistics:

t_{\rho=1} = \frac{\rho_n-1}{\hat{\sigma}_{\rho_n}}

where \sigma_{\rho_n} is the least square estimate of the standard deviation of \Hat{\rho}_n, given by:

\sigma_{\rho_n}=\mat{M}^{-1}_{33}\sqrt{\frac{1}{n-1}\sum_{i=2}^n\left(y_{i}-(\hat{a}_n+\hat{b}_nt_i+\hat{\rho}_ny_{i-1})\right)^2}

which converges in distribution to the Dickey-Fuller distribution associated to the model with drift and trend:

t_{\rho = 1} \stackrel{\mathcal{L}}{\longrightarrow} \frac{\int_{0}^{1}W^{b}(r) \di{W(r)}}{\int_{1}^{0} W^{b}(r)^2 \di{r}}

The null hypothesis \cH_0 from (4) is accepted when t_{\rho=1} > C_{\alpha} where C_{\alpha} is the test threshold of level \alpha.

The quantiles of the Dickey-Fuller statistics for the model with drift and linear trend are:

\left\{ \begin{array}{ll} \alpha = 0.01, & C_{\alpha} = -3.96 \\ \alpha = 0.05, & C_{\alpha} = -3.41 \\ \alpha = 0.10, & C_{\alpha} = -3.13 \end{array} \right.

1.1. Case 1: The null hypothesis \cH_0 from (4) is rejected

We test whether b=0:

(5)\left\{ \begin{array}{lr} \cH_0: & b = 0 \\ \cH_1: & b \neq 0 \end{array} \right.

where the statistics t_n = \frac{|\hat{b}_n|}{\sigma_{b_n}} converges in distribution to the Student distribution Student with \nu=n-4, where \sigma_{b_n} is the least square estimate of the standard deviation of \Hat{b}_n, given by:

\sigma_{b_n}=\mat{M}^{-1}_{22}\sqrt{\frac{1}{n-1}\sum_{i=2}^n\left(y_{i}-(\hat{a}_n+\hat{b}_nt_i+\hat{\rho}_ny_{i-1})\right)^2}

The decision to be taken is:

  • If \cH_0 from (5) is rejected, then the model 1 (2) is confirmed. And the test (4) proved that the unit root is rejected : \rho < 1. We then conclude that the final model is : \boldsymbol{X_t = a + bt + \rho X_{t-1} + \varepsilon_{t}} with \boldsymbol{\rho < 1} which is a trend stationary model.

  • If \cH_0 from (5) is accepted, then the model 1 (2) is not confirmed, since the trend presence is rejected and the test (4) is not conclusive (since based on a wrong model). We then have to test the second model (7).

1.2. Case 2: The null hypothesis \cH_0 from (4) is accepted

We test whether (\rho, b) = (1,0):

(6)\left\{ \begin{array}{lr} \cH_0: & (\rho, b) = (1,0) \\ \cH_1: & (\rho, b) \neq (1,0) \end{array} \right.

with the Fisher statistics:

\displaystyle \hat{F}_1 = \frac{(S_{1,0} - S_{1,b})/2}{S_{1,b}/(n-3)}

where S_{1,0}=\sum_{i=2}^n\left(y_i-(\hat{a}_n+y_{i-1})\right)^2 is the sum of the square errors of the model 1 (2) assuming \cH_0 from (6) and S_{1,b}=\sum_{i=2}^n\left(y_i-(\hat{a}_n+\hat{b}_nt_i+\hat{\rho}_ny_{i-1})\right)^2 is the same sum when we make no assumption on \rho and b.

The statistics \hat{F}_1 converges in distribution to the Fisher-Snedecor distribution FisherSnedecor with d_1=2, d_2=n-3. The null hypothesis \cH_0 from (4) is accepted when \hat{F}_1 < \Phi_{\alpha} where \Phi_{\alpha} is the test threshold of level \alpha.

The decision to be taken is:

  • If \cH_0 from (6) is rejected, then the model 1 (2) is confirmed since the presence of linear trend is confirmed. And the test (4) proved that the unit root is accepted: \rho = 1. We then conclude that the model is: \boldsymbol{X_t = a + bt + X_{t-1} + \varepsilon_{t}} which is a non stationary model.

  • If \cH_0 from (6) is accepted, then the model 1 (2) is not confirmed, since the presence of the linear trend is rejected and the test (4) is not conclusive (since based on a wrong model). We then have to test the second model (7).

2. We assume the model (7):

(7)\boldsymbol{X_t = a + \rho X_{t-1} + \varepsilon_{t}}

The coefficients (a,\rho) are estimated as follows:

(8)\underbrace{\left(\begin{array}{lll} \displaystyle n-1 &\sum_{i=2}^n y_{i-1}\\ \displaystyle \sum_{i=2}^n y_{i-1} &\sum_{i=2}^n y_{i-1}^2 \end{array} \right)}_{\mat{N}} \left( \begin{array}{c} \hat{a}_n\\ \hat{\rho}_n \end{array} \right)= \left( \begin{array}{l} \displaystyle \sum_{i=1}^n y_{i} \\ \displaystyle \sum_{i=2}^n y_{i-1} y_{i} \end{array} \right)

We first test:

(9)\left\{ \begin{array}{lr} \mathcal{H}_0: & \rho = 1 \\ \mathcal{H}_1: & \rho < 1 \end{array} \right.

thanks to the Student statistics:

t_{\rho=1} = \frac{\rho_n-1}{\sigma_{\rho_n}}

where \sigma_{\rho_n} is the least square estimate of the standard deviation of \Hat{\rho}_n, given by:

\sigma_{\rho_n}=\mat{N}^{-1}_{22}\sqrt{\frac{1}{n-1}\sum_{i=2}^n\left(y_{i}-(\hat{a}_n+\hat{\rho}_ny_{i-1})\right)^2}

which converges in distribution to the Dickey-Fuller distribution associated to the model with drift and no linear trend:

t_{\rho = 1} \stackrel{\mathcal{L}}{\longrightarrow} \frac{\int_{0}^{1}W^{a}(r) \di{W(r)}}{\int_{1}^{0} W^{a}(r)^2 \di{r}}

The null hypothesis \cH_0 from (9) is accepted when t_{\rho=1} > C_{\alpha} where C_{\alpha} is the test threshold of level \alpha.

The quantiles of the Dickey-Fuller statistics for the model with drift are:

\left\{ \begin{array}{ll} \alpha = 0.01, & C_{\alpha} = -3.43 \\ \alpha = 0.05, & C_{\alpha} = -2.86 \\ \alpha = 0.10, & C_{\alpha} = -2.57 \end{array} \right.

2.1. Case 1: The null hypothesis \cH_0 from (9) is rejected

We test whether a=0:

(10)\left\{ \begin{array}{lr} \mathcal{H}_0: & a = 0 \\ \mathcal{H}_1: & a \neq 0 \end{array} \right.

where the statistics t_n = \frac{|\hat{a}_n|}{\sigma_{a_n}} converges in distribution to the Student distribution Student with \nu=n-3, where \sigma_{a_n} is the least square estimate of the standard deviation of \Hat{a}_n, given by:

\sigma_{a_n}=\mat{N}^{-1}_{11}\sqrt{\frac{1}{n-1}\sum_{i=2}^n\left(y_{i}-(\hat{a}_n+\hat{\rho}_ny_{i-1})\right)^2}

The decision to be taken is:

  • If \cH_0 from (10) is rejected, then the model 2 (7) is confirmed. And the test (9) proved that the unit root is rejected: \rho < 1. We then conclude that the final model is: \boldsymbol{X_t = a + \rho X_{t-1} + \varepsilon_{t}} with \boldsymbol{\rho < 1} which is a stationary model.

  • If \cH_0 from (10) is accepted, then the model 2 (7) is not confirmed, since the drift presence is rejected and the test (4) is not conclusive (since based on a wrong model). We then have to test the third model (12).

2.2. Case 2: The null hypothesis \cH_0 from (9) is accepted

We test whether (\rho, a) = (1,0):

(11)\left\{ \begin{array}{lr} \mathcal{H}_0: & (\rho, a) = (1,0) \\ \mathcal{H}_1: & (\rho, a) \neq (1,0) \end{array} \right.

with a Fisher test. The statistics is:

\displaystyle \hat{F}_2 = \frac{(SCR_{2,c} - SCR_{2})/2}{SCR_{2}/(n-2)}

where SCR_{2,c} is the sum of the square errors of the model 2 (7) assuming \cH_0 from (11) and SCR_{2} is the same sum when we make no assumption on \rho and a.

The statistics \hat{F}_2 converges in distribution to the Fisher-Snedecor distribution FisherSnedecor with d_1=2, d_2=n-2. The null hypothesis \cH_0 from (4) is accepted if when \hat{F}_2 < \Phi_{\alpha} where \Phi_{\alpha} is the test threshold of level \alpha.

The decision to be taken is:

  • If \cH_0 from (11) is rejected, then the model 2 (7) is confirmed since the presence of the drift is confirmed. And the test (9) proved that the unit root is accepted: \rho =1. We then conclude that the model is: \boldsymbol{X_t = a + X_{t-1} + \varepsilon_{t}} which is a non stationary model.

  • If \cH_0 from (11) is accepted, then the model 2 (7) is not confirmed, since the drift presence is rejected and the test (9) is not conclusive (since based on a wrong model). We then have to test the third model (12).

3. We assume the model (12):

(12)\boldsymbol{X_t = \rho X_{t-1} + \varepsilon_{t}}

The coefficients \rho are estimated as follows:

(13)\hat{\rho}_n=\frac{\sum_{i=2}^ny_{i-1}y_i}{\sum_{i=2}^ny_{i-1}^2}

We first test:

(14)\left\{ \begin{array}{lr} \mathcal{H}_0: & \rho = 1 \\ \mathcal{H}_1: & \rho < 1 \end{array} \right.

thanks to the Student statistics:

t_{\rho=1} = \frac{\hat{\rho}_n-1}{\sigma_{\rho_n}}

where \sigma_{\rho_n} is the least square estimate of the standard deviation of \Hat{\rho}_n, given by:

\sigma_{\rho_n}=\sqrt{\frac{1}{n-1}\sum_{i=2}^n\left(y_{i}-\hat{\rho}_ny_{i-1}\right)^2}/\sqrt{\sum_{i=2}^ny_{i-1}^2}

which converges in distribution to the Dickey-Fuller distribution associated to the random walk model:

t_{\rho = 1} \stackrel{\mathcal{L}}{\longrightarrow} \frac{\int_{0}^{1}W(r) \di{W(r)}}{\int_{1}^{0} W(r)^2 \di{r}}

The null hypothesis \cH_0 from (14) is accepted when t_{\rho=1} > C_{\alpha} where C_{\alpha} is the test threshold of level \alpha.

The quantiles of the Dickey-Fuller statistics for the random walk model are:

\left\{ \begin{array}{ll} \alpha = 0.01, & C_{\alpha} = -2.57 \\ \alpha = 0.05, & C_{\alpha} = -1.94 \\ \alpha = 0.10, & C_{\alpha} = -1.62 \end{array} \right.

The decision to be taken is:

  • If \cH_0 from (14) is rejected, we then conclude that the model is : \boldsymbol{X_t = \rho X_{t-1} + \varepsilon_{t}} where \rho < 1 which is a stationary model.

  • If \cH_0 from (14) is accepted, we then conclude that the model is: \boldsymbol{X_t = X_{t-1} + \varepsilon_{t}} which is a non stationary model.

API: