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I've got a dataset which clearly shows a trend. However, I want to assess whether this trend is deterministic or stochastic. If I understood it right, I would need to use differences if the trend is stochastic and I could just detrend it if the trend is deterministic.

Hence, I used the kpss.test() from the tseries package in R with the null hypothesis that my data is trend-stationary which gave me a p-value of 0.01, i.e. I have evidence that it is not trend-stationary and the trend is stochastic, right?

On the other hand, I made a regression to identify the (possibly present) deterministic trend and made an additional kpss.test() on the residuals with the null of level-stationarity. This gave me a p-value of 0.1, i.e. I can't reject the null of my data being stationary after detrending, right?

Am I missing something or is there a possible further test I should use?

Here is my data:

y <- ts(c(12.9860268, 12.5362944, 10.9379455, 10.7029227, 9.6421311, 
  8.168712, 7.0846535, 6.7134053, 6.5685634, 5.6701865, 4.2352191, 
  4.3919294, 3.1960928, 2.8841746, 2.1974112, 0.5650275, -0.5647561, 
  -1.7419743, -2.9294583, -4.456346, -4.9608364, -5.3176373, -7.8000258, 
  -8.4957238, -10.1346795, -10.9322896, -11.491641, -12.1036813, 
  -13.022572, -14.5290742), start = 1982)

The regression and its fitted values and residuals are given by

m <- lm(y ~ time(y))
f <- ts(fitted(m), start = 1982)
r <- ts(residuals(m), start = 1982)

The KPSS test results are

library("tseries")
kpss.test(y, null = "Trend")
##         KPSS Test for Trend Stationarity
## 
## data:  y
## KPSS Trend = 0.30727, Truncation lag parameter = 1, p-value = 0.01    
kpss.test(r, null = "Level")
##         KPSS Test for Level Stationarity
## 
## data:  r
## KPSS Level = 0.30727, Truncation lag parameter = 1, p-value = 0.1

And a plot of the data and the residuals from the regression:

plot(y, type = "o", main = "data", xlab = "years", ylab = "")
lines(f, col = 2)
plot(r, type = "o", main = "residuals", xlab = "years", ylab = "")

enter image description here

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1 Answer 1

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Note that the test statistic is in the two versions of your test is exactly identical. The reason for this is that for the null = "Trend" case exactly the same computation is used as you did: computing the residuals from a simple linear regression and then compute the test statistic "as usual".

However, this changes the distribution of the test statistic so that the computation of critical values and p-values has to be adapted. This is also easy to see in the source code of kpss.test() which is based on the simulated critical values at 1%, 2.5%, 5%, and 10% level from the original paper. In the case of trend stationarity the critical values are: 0.216, 0.176, 0.146, 0.119 (i.e., 0.307 is significant). In contrast, in the case of level stationarity the critical values are: 0.739, 0.574, 0.463, 0.347 (where 0.307 would not be significant).

In short: In your case the first test of trend stationarity is the appropriate one while the second test of the residuals is not.

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  • $\begingroup$ Great explanation and editing, many thanks. By the way, is the kpss test appropriate for the sample size? Would you recommend further tests? $\endgroup$
    – Stats_L
    Commented Jul 30, 2015 at 0:12
  • 1
    $\begingroup$ With this sample size, I would mainly be concerned about a potential lack of power. However, given that the test is significant, this does not appear to be an issue. Also, adf.test(y) cannot reject the null hypothesis of a unit root which is consistent with the kpss.test() result. My take would be that the series looks like a random walk with drift which is also what auto.arima(y, d = 1, seasonal = FALSE, approximation = FALSE, stepwise = FALSE) from forecast selects. $\endgroup$ Commented Jul 30, 2015 at 7:47

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