0
$\begingroup$

I have difficulty in solving unit root test by using both adf.test and pp.test in R.

The result p-value using adf.test is 0.2677 (mean that it has unit root), and the result p-value using pp.test is 0.01696 (mean that is has no unit root).

What means that? I wonder whether I have to use diff function to remove unit root or not.

(k argument in adf.test, How do I use this? what means exactly for k?)

Raw data is below.

2935833
2622529
2719635
2625179
2311187
2101758
2552638
2883423
3128904
2959348
2759000
2233755
2560858
2548821
2625675
2326076
1662956
1772409
1797275
2639852
2799990
3133285
2438296
2583766
2610157
2493415
2094163
2174301
2283420
2505128
2873785
2339727
2985829
3037351
1828265
1038562
1474727
1523331
2122667
2571006
2252161
2422347
2155973
2294976
2809652
2436293
2561852
2199544
2674423
2551363
3110508
3177925
3046952
2850904
3002830
2910913
2809172
3136842
3355368
3604565
3013310
3125751
2548605
2646575
2231458
1962095
1958019
2143073
2305966
2620302
2356447
2427571
$\endgroup$
1
  • 3
    $\begingroup$ Your interpretation of p-values is suspect; 0.27 does not mean there is no unit root and 0.017 does not mean there is; these are measures of evidence (in one interpretation, anyway), where a smaller p-value means that there is more evidence for a unit root. $\endgroup$ Commented Nov 30, 2011 at 3:45

2 Answers 2

1
$\begingroup$

Your finding is interesting. I could give you couple of reason for this.

1) pp test is non parametric, it just need the residual to be stationary. But in adf test, the residual should be independent and identically distributed. ADF test is not robust when the series is autocorrelated. I guess this is the main reason.

2) Rejection is stronger than acception. Also from the plot also you can see there is no evidence of stochstic trend.

$\endgroup$
0
$\begingroup$

I am not familiar with adf.test or pp.test. In any case it looks like you want to use a 5% (or 10%?) criterion in a kind of unit root decision rule and doing so with 2 different approaches means you get different outcomes. Well the approaches do different things so this is quite possible.

The ADF test uses lagged difference terms to address serial correlation. Lagged differences are added with the aim of removing serial correlation as an issue ( Aside: I am not sure why vinux says "ADF test is not robust when the series is autocorrelated". The augmentation is explicitly designed to remove serial correlation, if adding lagged difference term(s) removes serial correlation as an issue why would ADF not be robust to serial correlation?).

The Phillips Perron approach applies a nonparametric correction to the standard ADF test statistic, allowing for more general dependence in the errors, including conditional heteroskedasticity. If there were strong concerns over heteroskedasticity in the ADF residuals this might influence lead an analyst to go for PP. If the addition of lagged differences in ADF did not remove serial correlation then this again might suggest PP as an alternative.

The fact that your results are sensitive to the different approaches suggest further exploration/explanation is needed - do your ADF residuals still indicate serial correlation or heteroskedasticitiy, what does the graph look like, what are your theoretical expectations for stationarity or serial correlation or heteroskedasticity, what would the interpretation of a regression using differenced values mean etc.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.