How is this possible: A time series with a stationary level but non-stationary first differences ???

[Nora Kaimann]

It's the following time series that has this strange property:

* Example generated by -dataex-. To install: ssc install dataex
clear
input float ln_CPI_AUT
 4.388754
 4.388754
 4.388754
4.3907385
 4.388754
 4.388754
4.3896227
4.3896227
 4.393708
 4.394449
4.4012165
 4.398269
4.4050107
 4.407938
 4.407207
 4.407938
4.4157033
4.4165487
4.4165487
4.4194427
4.4213676
4.4249663
4.4268804
4.4279556
4.4318876
 4.433789
 4.438761
 4.440649
4.4426513
 4.443592
4.4415917
 4.444532
 4.445588
 4.444532
 4.446526
4.4494514
 4.450386
 4.452252
  4.45609
4.4589877
4.4599133
4.4589877
 4.460953
 4.460953
   4.4628
 4.461877
 4.464758
4.4665976
 4.467631
 4.470381
 4.469465
 4.469465
4.4714103
 4.470381
4.4732375
4.4751754
4.4732375
4.4751754
 4.476996
 4.477905
4.4816456
4.4853725
4.4853725
 4.490096
4.4937916
4.4918895
4.4955783
4.4955783
4.4992537
 4.501142
4.5056815
4.5065646
4.5096498
4.5146985
 4.512616
 4.513603
 4.516667
 4.515683
 4.515683
4.5176497
4.5186315
 4.515683
4.5196123
4.5186315
 4.521571
4.5246105
4.5295844
4.5305543
4.5315237
4.5305543
 4.533459
4.5315237
4.5305543
4.5315237
4.5344257
4.5344257
4.5374265
4.5422306
 4.547117
4.5499744
end

When I apply an augmented Dickey-Fuller test the level is stationary but the first difference of the stationary level is non-stationary:

dfuller   ln_CPI_AUT, lags(12) trend 

dfuller d.ln_CPI_AUT, lags(12) trend 

I should admit, if I apply a Phillips-Perron test to the same time series, the result is consistent, in the sense that the level is non-stationary and the first difference is stationary.

How would you interpret this ADF result ???

Thanks for any comments!
Nora

[Phil Bromiley]

When I run this after your data
g t=_n
tsset t
dfuller ln_CPI_AUT, lags(12) trend reg
dfuller d.ln_CPI_AUT, lags(12) trend reg

I get negative Z(t) on both and it is almost significant on the first and insignificant on the second.

[Sebastian Kripfganz]

By the way: The two specifications are not really consistent. If you have a trend in levels, this becomes an intercept in first differences; hence no need for the trend option in the second specification. On top of that, if the model is nonstationary in levels and you difference it, the number of lags gets reduced by one; hence lags(11) in the second specification. Given the data extract, almost all of the lags are statistically insignificant anyway; hence lags(12) does not seem to be the most appropriate choice to begin with.

[Nora Kaimann]

Thanks a lot for your replies! But is it really true the "If you have a trend in levels, this becomes an intercept in first differences; hence no need for the trend option in the second specification." ? Take e.g. the following unit root process: ln⁡(p(t) )= α1 + (t+1) α2 + ln⁡(p(t-1)) + ε(t). Here, subtracting ln⁡(p(t-1)) would not eliminate the linear trend - or am I confused now? Thanks for a reply!

[Eric de Souza]

You are substracting ln(p(t-1)) from both sides. What you should do is subtract the equation for ln(p(t-1)) from the equation for ln(p(t)). Write down the equation for ln(p(t-1)) and substract it from the equation for ln(p(t)).
You get (t+1)a2 - t.a2 and t disappears.

[Nora Kaimann]

I think the algebra is correct. I can simply substract ln(p(t-1)) or do it your way, what would yield:

ln(p(t) ) - ln(p(t-1) ) = α2 + ln(p(t-1)) - ln(p(t-2)) ε(t) - ε(t-1)

Here I can now again substitute the equation for ln(p(t-1)) and we are back to:

ln(p(t) ) - ln(p(t-1) ) = α2 + α1 + (t) α2 + ln(p(t-2)) + ε(t-1) - ln(p(t-2)) ε(t) - ε(t-1)
<=>
ln(p(t) ) - ln(p(t-1) ) = α1 + (t+1) α2 + ε(t)

[Eric de Souza]

No.
(1) ln⁡(p(t) )= α1 + (t+1) α2 + ln⁡(p(t-1)) + ε(t)
(2) ln⁡(p(t-1) )= α1 + (t) α2 + ln⁡(p(t-2)) + ε(t-1)

Now substract (2) from (1):
ln(p(t) ) - ln(p(t-1) ) = (1) α2 + ln(p(t-1)) - ln(p(t-2)) + ε(t) - ε(t-1)

[Nora Kaimann]

Sorry, but your last line in post #7 is identical with the first line of my post #6. As I have demonstrated in this post #6, if you substitute now ln⁡(p(t-1) )= α1 + (t) α2 + ln⁡(p(t-2)) + ε(t-1) this yields

ln(p(t) ) - ln(p(t-1) ) = α1 + (t+1) α2 + ε(t)

which clearly proves that it is possible that the level-process my contain a linear trend and the first difference of the level my also contain a linear trend. Your last line in post #7 aka my first line in post #6 "hides" this aspect only.

And that's the crucial point here. We cannot exclude by apriori arguments that a linear trend disappears always, if we take the first difference. Therefore it makes sense to test empirically for a linear trend in the level and also in the first difference. So Sebastian's argument "The two specifications are not really consistent. If you have a trend in levels, this becomes an intercept in first differences; hence no need for the trend option in the second specification." is not correct.

Nevertheless thanks a lot ! To me this discussion was very usefull. I'm not sure, if I would have found the argument fast enough in an oral presentation of my results...

[Sebastian Kripfganz]

Both calculations are mathematically correct. The problem with Eric's calculation is that he is creating an I(-1) error term ε(t) - ε(t-1) in his first differenced model while Nora's error term ε(t) is still I(0).

Another aspect here is that both calculations are done under the null hypothesis of a unit root in levels. If there is a linear trend term (t+1) α2 on the right hand side of such a random-walk model, it essentially characterizes a quadratic trend. This becomes apparent if you backward iterate the model in levels. This quadratic time trend then correctly becomes a linear time trend in the trend stationary first-differenced model following Nora's derivations.

The crucial point to notice is that the trend option of the dfuller command assumes that under the null hypothesis the process is only a random-walk with drift but not an unrestricted time trend. Such a drift term (the intercept) becomes a linear (not quadratic) time trend by backward iteration. The model you have written down is not consistent with the trend option of the dfuller command. The assumed model rather is
ln⁡(p(t)) = α1 + (t+1) α2 (α3 - 1) + α3 ln⁡(p(t-1)) + ε(t)

If α3 = 1 (unit root), the trend term disappears.

[Eric de Souza]

To summarise from my point of view:

If ln(p(t)) is generated by ln(p(t)) = α1 + (t+1) α2 + ln(p(t-1)) + ε(t) = (α1+ α2) + t.α2 + ln(p(t-1)) + ε(t), then Δln(p(t)) has an MA1 error term. That is clear. If ε(t) is stationary (I(0)), then so is ε(t) - ε(t-1). I don’t know what Sebastian means by I(-1).

It is also clear that ln(p(t)) has a unit root and a deterministic quadratic trend.

The null hypothesis for the dfuller test with the trend option is that the series (ln(p(t)) has a unit root and may or may not have a drift but not a quadratic deterministic trend (and that the error term is white noise).

To satisfy these assumptions, we need to reformulate the model which is what Sebastian does at the end of #9