My data rejects unit root, but shows structural break, is this possible?

  • 4
    Your data doesn't reject anything, it's just numbers. I assume you applied some test. Wouldn't a structural break invalidate the test assumptions? But yes, you can write models that have a unit root and a structural break, so data could have both. – Glen_b May 11 '13 at 6:12

Others have already answered about the wrong idea of hypothesis testing. Regarding your question : yes it is possible. Take for example this time series. It is the price series of a stock of the S&P 500 (therefore, real data) in a time span of more than 10 years. As you can see the series presents what seems to be a structural change. The ADF, with constant and constant plus trend, gives a p-value respectively of 0.003629 and 0.01257, therefore we can reject the null hypothesis of a unit root (if we assume a level of significance of 5%). Since we suspect the presence of a break, I run also a unit root test with an unknown breakpoint, which computed p-value is equal to 0.0239, once again we can reject the null of a unit root.

The second figure shows an OLS-based CUSUM test with alternative boundaries and alfa equals to 0.05 plotted with the R package "strucchange", which shows you the contemporaneous precence of a break.

price series

CUSUM

The two older answers already give you a good sense, yet I would like to add some points about the nature of unit-roots and unit root testing.

Please refer to the following question in order

What is the difference between a stationary test and a unit root test?

Here is the part of my answer which matters for you:

How unit-root test and stationarity-test complement each other If you have a time series data set how it usually appears in econometric time series I propose you should apply both a Unit root test: (Augmented) Dickey Fuller or Phillips-Perron depending on the structure of the underlying data and a KPSS test.

Case 1 Unit root test: you can’t reject $H_0$; KPSS test: reject $H_0$. Both imply that series has unit root.

Case 2 Unit root test: Reject $H_0$. KPSS test: don`t reject $H_0$. Both imply that series is stationary.

Case 3 If we can’t reject both test: data give not enough observations.

Case 4 Reject unit root, reject stationarity: both hypothesis are component hypothesis – heteroskedasticity in series may make a big difference; if there is structural break it will affect inference.

Power problem: if there is small random walk component (small variance $σ^2$ u ), we can’t reject unit root and can’t reject stationarity.

Your data does not reject a unit root or reject stationarity. Certain hypothesis tests reject a unit root or reject stationarity. Maybe you can go deeper into hypothesis testing for unit roots and stationarity and there won't be any contradiction between unit roots and stationarity anymore.




If your data has a structural break it means that a pattern in the data "breaks" exactly one time. If your data has a unit root it means that a pattern in the data "breaks" in every time period. However there are various possibilities between "one break" and a "break in every time period". The paper Multiple Trend Breaks And The Unit-Root Hypothesis by Lumsdaine and Papell examines the possibility of having more than one structural break in the data. They also show a test for testing for more than one structural break.

Yes, it can. Nothing restricts the following DGP to exist:

$$ y_t = \begin{cases} c,& \text{if } t = 0 \\ y_{t-1} + e_t,& \text{if } 0 < t < t_1\\ -y_{t-1} + e_t,& \text{if } t_1 \leq t \end{cases} $$

where $e_t \sim N(0,\sigma^2)$.

An implementation in R:

remove(list = ls())

set.seed(1)
i <- 1
n1 <- 50
n2 <- 100
y <- seq(10, 10, length.out=n2)
e <- rnorm(n2,0,6)

for(i in 2:n1){
  y[i] <- y[i - 1] + e[i]
}

for(j in n1:n2){
  y[j] <- -y[j - 1] + e[j]
}

plot(y, type="o")

library(tseries)
adf.test(y)

The graph is

enter image description here

and the test result is:

    Augmented Dickey-Fuller Test

data:  y
Dickey-Fuller = -2.564, Lag order = 4, p-value = 0.343
alternative hypothesis: stationary

Partially answered in comments:

Your data doesn't reject anything, it's just numbers. I assume you applied some test. Wouldn't a structural break invalidate the test assumptions? But yes, you can write models that have a unit root and a structural break, so data could have both. – Glen_b

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