## Unit Root Tests

### Test Simulated Data for a Unit Root

This example shows how to test univariate time series models for stationarity. It shows how to simulate data from four types of models: trend stationary, difference stationary, stationary (AR(1)), and a heteroscedastic, random walk model. It also shows that the tests yield expected results.

Simulate four time series.

T = 1e3; % Sample size t = (1:T)'; % Time multiple rng(142857); % For reproducibility y1 = randn(T,1) + .2*t; % Trend stationary Mdl2 = arima(D=1,Constant=0.2,Variance=1); y2 = simulate(Mdl2,T,Y0=0); % Difference stationary Mdl3 = arima(AR=0.99,Constant=0.2,Variance=1); y3 = simulate(Mdl3,T,Y0=0); % AR(1) Mdl4 = arima(D=1,Constant=0.2,Variance=1); sigma = (sin(t/200) + 1.5)/2; % Std deviation e = randn(T,1).*sigma; % Innovations y4 = filter(Mdl4,e,Y0=0); % Heteroscedastic

Plot the first 100 points in each series.

y = [y1 y2 y3 y4]; figure; plot1 = plot(y(1:100,:)); plot1(1).LineWidth = 2; plot1(3).LineStyle = ":"; plot1(3).LineWidth = 2; plot1(4).LineStyle = ":"; plot1(4).LineWidth = 2; title("\bf First 100 Periods of Each Series"); legend("Trend Stationary','Difference Stationary","AR(1)", ... "Heteroscedastic",Location="NorthWest");

All of the models appear nonstationary and behave similarly. Therefore, you might find it difficult to distinguish which series comes from which model simply by looking at their initial segments.

Plot the entire data set.

plot2 = plot(y); plot2(1).LineWidth = 2; plot2(3).LineStyle = ":"; plot2(3).LineWidth = 2; plot2(4).LineStyle = ":"; plot2(4).LineWidth = 2; title("\bf Each Entire Series"); legend("Trend Stationary','Difference Stationary","AR(1)", ... "Heteroscedastic",Location="NorthWest");

The differences between the series are clearer here:

The trend stationary series has little deviation from its mean trend.

The difference stationary and heteroscedastic series have persistent deviations away from the trend line.

The AR(1) series exhibits long-run stationary behavior; the others grow linearly.

The difference stationary and heteroscedastic series appear similar. However, that the heteroscedastic series has much more local variability near period 300, and much less near period 900. The model variance is maximal when $$\mathrm{sin}(t/200)=1$$, at time $$100\pi \approx 314$$. The model variance is minimal when $$\mathrm{sin}(t/200)=-1$$, at time $$300\pi \approx 942$$. Therefore, the visual variability matches the model.

Use the Augmented Dicky-Fuller test on the three growing series (`y1`

, `y2`

, and `y4`

) to assess whether the series have a unit root. Since the series are growing, specify that there is a trend. In this case, the null hypothesis is $${H}_{0}:{y}_{t}={y}_{t-1}+c+{b}_{1}\Delta {y}_{t-1}+{b}_{2}\Delta {y}_{t-2}+{\epsilon}_{t}$$ and the alternative hypothesis is $${H}_{1}:{y}_{t}=a{y}_{t-1}+c+\delta t+{b}_{1}\Delta {y}_{t-1}+{b}_{2}\Delta {y}_{t-2}+{\epsilon}_{t}$$. Set the number of lags to 2 for demonstration purposes.

`hY1 = adftest(y1,Model="ts",Lags=2)`

`hY1 = `*logical*
1

`hY2 = adftest(y2,Model="ts",Lags=2)`

`hY2 = `*logical*
0

`hY4 = adftest(y4,Model="ts",Lags=2)`

`hY4 = `*logical*
0

`hY1 = 1`

indicates that there is sufficient evidence to suggest that`y1`

is trend stationary. This is the correct decision because`y1`

is trend stationary by construction.`hY2 = 0`

indicates that there is not enough evidence to suggest that`y2`

is trend stationary. This is the correct decision since`y2`

is difference stationary by construction.`hY4 = 0`

indicates that there is not enough evidence to suggest that`y4`

is trend stationary. This is the correct decision, however, the Dickey-Fuller test is not appropriate for a heteroscedastic series.

Use the Augmented Dickey-Fuller test on the AR(1) series (`y3`

) to assess whether the series has a unit root. Since the series is not growing, specify that the series is autoregressive with a drift term. In this case, the null hypothesis is $${H}_{0}:{y}_{t}={y}_{t-1}+{b}_{1}\Delta {y}_{t-1}+{b}_{2}\Delta {y}_{t-2}+{\epsilon}_{t}$$ and the alternative hypothesis is $${H}_{1}:{y}_{t}=a{y}_{t-1}+{b}_{1}\Delta {y}_{t-1}+{b}_{2}\Delta {y}_{t-2}+{\epsilon}_{t}$$. Set the number of lags to 2 for demonstration purposes.

`hY3 = adftest(y3,Model="ard",Lags=2)`

`hY3 = `*logical*
1

`hY3 = 1`

indicates that there is enough evidence to suggest that `y3`

is a stationary, autoregressive process with a drift term. This is the correct decision because `y3`

is an autoregressive process with a drift term by construction.

Use the KPSS test to assess whether the series are unit root nonstationary. Specify that there is a trend in the growing series (`y1`

, `y2`

, and `y4`

). The KPSS test assumes the following model:

$${y}_{y}={c}_{t}+\delta t+{u}_{t}$$

$${c}_{t}={c}_{t-1}+{\epsilon}_{t},$$

where $${u}_{t}$$ is a stationary process and $${\epsilon}_{t}$$ is an independent and identically distributed process with mean 0 and variance $${\sigma}^{2}$$. Whether there is a trend in the model, the null hypothesis is $${H}_{0}:{\sigma}^{2}=0$$ (the series is trend stationary) and the alternative hypothesis is $${H}_{1}:{\sigma}^{2}>0$$ (not trend stationary). Set the number of lags to 2 for demonstration purposes.

hY1 = kpsstest(y1,Lags=2,Trend=true)

`hY1 = `*logical*
0

hY2 = kpsstest(y2,Lags=2,Trend=true)

`hY2 = `*logical*
1

hY3 = kpsstest(y3,Lags=2)

`hY3 = `*logical*
1

hY4 = kpsstest(y4,Lags=2,Trend=true)

`hY4 = `*logical*
1

All is tests result in the correct decision.

Use the variance ratio test on al four series to assess whether the series are random walks. The null hypothesis is $${H}_{0}$$: $$Var(\Delta {y}_{t})$$ is constant, and the alternative hypothesis is $${H}_{1}$$: $$Var(\Delta {y}_{t})$$ is not constant. Specify that the innovations are independent and identically distributed for all but `y1`

. Test `y4`

both ways.

hY1 = vratiotest(y1)

`hY1 = `*logical*
1

hY2 = vratiotest(y2,IID=true)

`hY2 = `*logical*
0

hY3 = vratiotest(y3,IID=true)

`hY3 = `*logical*
0

hY4NotIID = vratiotest(y4)

`hY4NotIID = `*logical*
0

hY4IID = vratiotest(y4,IID=true)

`hY4IID = `*logical*
0

All tests result in the correct decisions, except for `hY4_2 = 0`

. This test does not reject the hypothesis that the heteroscedastic process is an IID random walk. This inconsistency might be associated with the random seed.

Alternatively, you can assess stationarity using `pptest`

### Test Time Series Data for Unit Root

This example shows how to test a univariate time series for a unit root. It uses wages data (1900-1970) in the manufacturing sector. The series is in the Nelson-Plosser data set.

Load the Nelson-Plosser data. Extract the nominal wages data.

```
load Data_NelsonPlosser
wages = DataTable.WN;
```

Trim the `NaN`

values from the series and the corresponding dates (this step is optional because the test ignores `NaN`

values).

wDates = dates(isfinite(wages)); wages = wages(isfinite(wages));

Plot the data to look for trends.

```
plot(wDates,wages)
title('Wages')
```

The plot suggests exponential growth.

Transform the data using the log function to linearize the series.

```
logWages = log(wages);
plot(wDates,logWages)
title('Log Wages')
```

The plot suggests that time series has a linear trend.

Test the null hypothesis that there is no unit root (trend stationary) against the alternative hypothesis that the series is a unit root process with a trend (difference stationary). Set `'Lags',7:2:11`

, as suggested in Kwiatkowski et al., 1992.

`[h1,pValue1] = kpsstest(logWages,'Lags',7:2:11)`

`h1 = `*1x3 logical array*
0 0 0

`pValue1 = `*1×3*
0.1000 0.1000 0.1000

`kpsstest`

fails to reject the null hypothesis that the wage series is trend stationary.

Test the null hypothesis that the series is a unit root process (difference stationary) against the alternative hypothesis that the series is trend stationary.

[h2,pValue2] = adftest(logWages,'Model','ts')

`h2 = `*logical*
0

pValue2 = 0.8327

`adftest`

fails to reject the null hypothesis that the wage series is a unit root process.

Because the results of the two tests are inconsistent, it is unclear that the wage series has a unit root. This is a typical result of tests on many macroeconomic series.

`kpsstest`

has a limited set of calculated critical values. When it calculates a test statistic that is outside this range, the test reports the p-value at the appropriate endpoint. So, in this case, `pValue`

reflects the closest tabulated value. When a test statistic lies inside the span of tabulated values, `kpsstest`

linearly interpolates the p-value.

### Test Stock Data for a Random Walk

This example shows how to assess whether a time series is a random walk. It uses market data for daily returns of stocks and cash (money market) from the period January 1, 2000 to November 7, 2005.

Load the data.

`load CAPMuniverse`

Extract two series to test. The first column of data is the daily return of a technology stock. The last (14th) column is the daily return for cash (the daily money market rate).

tech1 = Data(:,1); money = Data(:,14);

The returns are the logs of the ratios of values at the end of a day over the values at the beginning of the day.

Convert the data to prices (values) instead of returns. `vratiotest`

takes prices as inputs, as opposed to returns.

tech1 = cumsum(tech1); money = cumsum(money);

Plot the data to see whether they appear to be stationary.

subplot(2,1,1) plot(Dates,tech1); title('Log(relative stock value)') datetick('x') hold on subplot(2,1,2); plot(Dates,money) title('Log(accumulated cash)') datetick('x') hold off

Cash has a small variability, and appears to have long-term trends. The stock series has a good deal of variability, and no definite trend, though it appears to increase towards the end.

Test whether the stock series matches a random walk.

[h,pValue,stat,cValue,ratio] = vratiotest(tech1)

`h = `*logical*
0

pValue = 0.1646

stat = -1.3899

cValue = 1.9600

ratio = 0.9436

`vratiotest`

does not reject the hypothesis that a random walk is a reasonable model for the stock series.

Test whether an i.i.d. random walk is a reasonable model for the stock series.

`[h,pValue,stat,cValue,ratio] = vratiotest(tech1,'IID',true)`

`h = `*logical*
1

pValue = 0.0304

stat = -2.1642

cValue = 1.9600

ratio = 0.9436

`vratiotest`

rejects the hypothesis that an i.i.d. random walk is a reasonable model for the `tech1`

stock series at the 5% level. Thus, `vratiotest`

indicates that the most appropriate model of the `tech1`

series is a heteroscedastic random walk.

Test whether the cash series matches a random walk.

[h,pValue,stat,cValue,ratio] = vratiotest(money)

`h = `*logical*
1

pValue = 4.6093e-145

stat = 25.6466

cValue = 1.9600

ratio = 2.0006

`vratiotest`

emphatically rejects the hypothesis that a random walk is a reasonable model for the cash series (`pValue = 4.6093e-145`

). The removal of a trend from the series does not affect the resulting statistics.

## References

[1] Kwiatkowski, D., P. C. B. Phillips, P. Schmidt and Y. Shin. “Testing the Null Hypothesis of Stationarity against the Alternative of a Unit Root.”
*Journal of Econometrics*. Vol. 54, 1992, pp. 159–178.

## See Also

`adftest`

| `kpsstest`

| `pptest`

| `vratiotest`