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I'm pretty new to the concepts of stationarity/cointegration. I am using the "urca" package in "Rstudio" to run my tests.

I have been trying to run cointegration tests, but the frustrating thing is that I haven't been able to find two series that are non-stationary, even when I try using examples cited by cointegration tutorials. My $p$-value is always too big such that I have to reject the null straight away. However, if I look at the $t$-values and compare them to the critical values, they seem to suggest otherwise.

Should I then ignore the $p$-value in the ADF test? Here are my test results. My two price series are XLE US Equity and CO1 Comdty (Brent 1st futures) from 01/01/2010 - today (5/11/2015).

Any help/elaboration will be very much appreciated, thank you!

> testXLE<-ur.df(XLE,type="drift",selectlags="AIC")
> summary(testXLE)

############################################### 
# Augmented Dickey-Fuller Test Unit Root Test # 
############################################### 

Test regression drift 


Call:
lm(formula = z.diff ~ z.lag.1 + 1 + z.diff.lag)

Residuals:
     Min       1Q   Median       3Q      Max 
-10.3948  -2.5809   0.6846   2.7908  10.1940 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)  
(Intercept)  6.58864    3.43524   1.918   0.0596 .
z.lag.1     -0.08584    0.04533  -1.894   0.0628 .
z.diff.lag   0.05529    0.12544   0.441   0.6609  
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 4.162 on 64 degrees of freedom
Multiple R-squared:  0.05337,   Adjusted R-squared:  0.02379 
F-statistic: 1.804 on 2 and 64 DF,  p-value: 0.1729


Value of test-statistic is: -1.8936 1.8395 

Critical values for test statistics: 
      1pct  5pct 10pct
tau2 -3.51 -2.89 -2.58
phi1  6.70  4.71  3.86

My interpretation of the results:

  • according to p-value (0.1729>0.05) do not reject null; series is stationary
  • t-value = (-1.8936>-2.89) --> do not reject null hypothesis; series is not stationary
  • t-value = (1.8395<4.71) --> do not reject a0=0 --> there is no drift

Conclusion: The series is non-stationary: Random Walk with no drift.

ElizaTYX
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  • In concluding "do not reject null hypothesis; series is not stationary" for the second test, your reading of the results as "not stationary" would hold even when the observed t-value was negative and near zero. Shouldn't the comparison be to the *absolute* values of the t-statistics? – Mike Hunter Nov 05 '15 at 10:34
  • Hi @DJohnson, my understanding is that by the comparison of absolute values, if the |t-value|>|critical value|, you reject the null hypothesis (because the probability of wrongly rejecting the null is very low). So in this case, as |-1.8936| – ElizaTYX Nov 06 '15 at 03:43
  • I assumed the reverse wrt critical vs observed t. – Mike Hunter Nov 06 '15 at 11:11
  • Hi @DJohnson, I'm not sure if I am right, but I arrived at my conclusion taking reference to this explanation: http://stats.stackexchange.com/questions/24072/interpreting-rs-ur-df-dickey-fuller-unit-root-test-results – ElizaTYX Nov 09 '15 at 04:03
  • @ElizaTYX, you are correct. In the ADF test, a test statistic *to the left* of the critical value yields rejection of $H_0$: "the process has a unit root". – Richard Hardy Nov 09 '15 at 06:37
  • I don't think you should change your question. It showed your confusion that was addressed properly. Once you start changing it, some comments and the answer may become irrelevant. But it's up to you. – Richard Hardy Nov 09 '15 at 06:40
  • @RichardHardy Ah I get what you mean, I have added back the last comment on p-value that I deleted, thanks. – ElizaTYX Nov 09 '15 at 06:54

1 Answers1

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The $p$-value in the output above corresponds to the $F$-statistic for the overall significance of the test regression. However, this is not what the ADF test is about. When interested in presence/absence of a unit root, you should look at the test statistic and compare it to the critical values of tau2. So the results are not conflicting; the two statistics are for answering two different questions.

Richard Hardy
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  • Hi @Richard, thanks for the response, but could you in layman terms explain then what it means when the p-value<0.05 in an ADF test? Normally it would mean that we are rejecting the null hypothesis, but in the scenario where the t-value indicates that the null hypothesis of a unit root existing is not rejected, what are we rejecting then? – ElizaTYX Nov 05 '15 at 09:25
  • The null hypothesis associated with the $p$-value in the output above is that all regressors (except for the intercept) have zero coefficients in population. If $p$-value<0.05, then at least one of the regressors has a non-zero coefficient in population. – Richard Hardy Nov 05 '15 at 09:35
  • By regressors, are we talking about the different lag terms here, since this is a unit root test and involves only one time series? Thanks again, @Richard. – ElizaTYX Nov 05 '15 at 09:42
  • Yes. If you include trend in the ADF specification, it will also be one of the regressors, and I guess it should be included in the $F$-test, too (deterministic time trends are a little peculiar, so I am not completely sure). – Richard Hardy Nov 05 '15 at 10:45
  • Hi @Richard, sorry to trouble again but could you verify if my interpretation of the ADF results above are correct? Another user raised that it is wrong (in the comments) and we're not too sure about it. – ElizaTYX Nov 09 '15 at 05:22
  • @ElizaTYX You shouldnt be looking at the pvalue in the regression at all. This would be valid if you'd have asymptotic normal distributions, but asymptotically you have non-standard distributions (Dickey-Fuller distribution). All you can say is that -1.89>-2.58 so the pvalue is >0.10. In other packages you get more detailed pvalues. – hejseb Nov 09 '15 at 06:23
  • Hej @hejseb, yes I understand now that I should not look at the p-value to determine stationarity of my series (perhaps I should remove that portion in my question), but as to my interpretation that because the t-value -1.8936>-2.58, therefore we cannot reject the null hypothesis that a unit root exists -- is this right? [edit -- Richard Hardy has just answered to my question in the comment thread above, thanks!] – ElizaTYX Nov 09 '15 at 06:38
  • Hi @RichardHardy, another related issue that I have is that I do not understand why my test result can change when I change the type of test. E.g. when I set the ADF type to "drift", my result is to reject the null. However, when I perform the test again with type = "none", the result is to not reject the null. Shouldn't the results be the same for both types, just that I should get the result of alpha = 0 (i.e. no drift in the second t-statistic) in the type="drift" test? – ElizaTYX Nov 11 '15 at 03:45
  • @RichardHardy Additionally, may I know how I can test if the series is I(1) or I(2) etc. after having successfully tested that it has a unit root? – ElizaTYX Nov 11 '15 at 06:02
  • If the drift is not there in reality but you allow for it in the test, a time trend will be (mistakenly) subtracted from the trajectory and the residual tested for presence of unit root. The residual will look more stationary than it should, and you might therefore get a false result. If you do not allow for a drift when it is actually there in reality, your residual will look less stationary than it should. Also, critical values for the unit root parameter will differ depending on the absence/presence of drift. Actually, this should be posted as a separate question (if not answered before). – Richard Hardy Nov 11 '15 at 16:48
  • Once you have found a unit root, you difference the data and conduct the test again. If you find a unit root in the differenced data, that means there are at least two unit roots, thus you have an $I(d)$ process with $d \geqslant 2$. You keep differencing and testing until you no longer find a unit root. If the process does not have a unit root after having differenced $d$ times but it has a unit root after having differenced $d-1$ times, then it is $I(d)$ (that's the definition). Again, this should be a separate question. – Richard Hardy Nov 11 '15 at 16:50