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Let $Y_t = \rho Y_{t-1} + \epsilon_t$ and $Y_0$ be some constant.

I generated a time series data for the above model like this

y=rnorm(1,5,1)
for(i in 1:100)
  y=append(y,tail(y,1)+rnorm(1,1,1))

Plot of the time series data y_t

The above is a plot of the same. Now to test the stationarity of this series I used the adf.test function from the package tseries in R. For the above model a Dickey Fuller Test (i.e adf test with 0 lag) should be enough (and appropriate too) to test for stationarity. However the output of the following code :

> adf.test(y,k = 0)

    Augmented Dickey-Fuller Test

data:  y
Dickey-Fuller = -3.4732, Lag order = 0, p-value = 0.04793
alternative hypothesis: stationary

Tells that the null hypothesis is rejected at %5 level and the series is stationary. While from the plot the series is clearly NOT stationary.

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If run your code many times and check how often you get a pvalue less than 5%:

S <- 10000
rejections <- rep(NA, S)
for (s in 1:S){
  y=rnorm(1,5,1)
  for(i in 1:100){
      y=append(y,tail(y,1)+rnorm(1,1,1))
  }
  reject <- (adf.test(y,k = 0)$p.value <= 0.05)
  rejections[s] <- reject
}
mean(rejections, na.rm=TRUE)

You see that the adf test does its job, it only incorrectly rejects the null of not stationary about 5% of the time.

The data you posted is obviously non-stationary because it seems to have a linear trend - but the ADF implemented in adf.test uses a model that allows for a constant and a linear trend term. So if these are taken out, it can happen that the data then looks stationary.

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  • $\begingroup$ thanks for pointing out that tseries::adf.test implementation forces a constant and a linear term. I used the urca::df after reading your answer and now things are working fine :) $\endgroup$ – Vishaal Sudarsan Sep 3 '19 at 13:24

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