Hi everybody
'scuse my sixpennyworth cos I'm still confused and don't really want to
get involved but I'm thinking that always that unit roots need to be outside the unit circle.
The question here is, Are they ?.
Stationarity about a trendline means there is information in the data not yet described.
If the trend information is removed then any inferences that follow are not about the original data.
Gretl can handle all of this I'd be certain.
Long before Gretl when I was researching,there was a lack of great proggies such as Gretl.
This compelled me to write my own programs. I'm thinking that researchers who haven't learned
Wayne Fuller's and David Dickey's stuff off by heart from Fuller's book or their joint work,
coupled with Solo, Evans & Savin,Banajee and Hendry, Said and Dickey, etc etc might like to give 
them a good read. I'm merely minding that its tough to be valid with Unit root stuff.  
   Hudson
 
Dr RJF Hudson Qld Australia
rjfhud@powerup.com.au
----- Original Message -----
From: Muheed Jamaldeen
To: Gretl list
Sent: Tuesday, December 13, 2011 12:41 PM
Subject: Re: [Gretl-users] Deterministic trend in VAR

PS: JMULTI computes  eigenvalues of the reverse characteristic polynomial which must be greater than one for stability, while Gretl computes inverse roots which must be less than 1.

On Tue, Dec 13, 2011 at 1:33 PM, Muheed Jamaldeen <mj.myworld@gmail.com> wrote:
Hi Peter,

Yes. USGDP is part of the 11 variables. I didn't want to clutter the list with the rest of the tests and other information.

I am estimating the reduced form (near) VAR and SVAR on JMULTI whilst using Gretl for preliminary testing. JMULTI computes modulus of the eigenvalues of the reverse characteristic polynomial which are all greater than one (which is the stability condition, implying stationarity).

Let me explain what I mean by stability implying stationarity using Lutekepohl's text:

" The process is stable if det(IK − A1z − ·· ·− Apz p) = 0 for |z| ≤ 1, (3.2) that is, the polynomial defined by the determinant of the autoregressive operator has no roots in and on the complex unit circle. On the assumption that the process has been initiated in the infinite past (t = 0,±1,±2, . . .), it generates stationary time series that have time-invariant means, variances, and covariance structure."

He makes a similar argument here: http://books.google.com.au/books?id=muorJ6FHIiEC&q=Stable#v=snippet&q=Stable&f=false

See the proposition in page 25.

That makes sense to me because regardless of what the univariate Data Generating Process (DGP) suggests, the description of the process by each VAR equation is stable, implying stationarity. If this is  WITHOUT a trend term, then is there a case for a trend term? That is the crux of the question. The distinction is that the DGP may not be stationary, but the VAR process is. So there is no problem if the VAR process is stable (and thereby stationary) without a trend term.

I agree with your small sample point. But my hands are tied there! lol. 

Also, for consistency I computed the VAR inverse roots on Gretl. They are all inside the unit circle. Implying stability of the VAR process.

roots (real, imaginary, modulus, frequency)

 1: (-0.6179,  0.4963,  0.7925,  0.3923)
 2: (-0.6179, -0.4963,  0.7925, -0.3923)
 3: (-0.7550,  0.0000,  0.7550,  0.5000)
 4: (-0.7032,  0.1521,  0.7195,  0.4661)
 5: (-0.7032, -0.1521,  0.7195, -0.4661)
 6: (-0.4947,  0.6105,  0.7857,  0.3584)
 7: (-0.4947, -0.6105,  0.7857, -0.3584)
 8: (-0.5062,  0.3974,  0.6435,  0.3941)
 9: (-0.5062, -0.3974,  0.6435, -0.3941)
10: (-0.5245,  0.2641,  0.5872,  0.4258)
11: (-0.5245, -0.2641,  0.5872, -0.4258)
12: (-0.2243,  0.7003,  0.7354,  0.2993)
13: (-0.2243, -0.7003,  0.7354, -0.2993)
14: (-0.3287,  0.5380,  0.6305,  0.3373)
15: (-0.3287, -0.5380,  0.6305, -0.3373)
16: ( 0.3527,  0.8513,  0.9215,  0.1875)
17: ( 0.3527, -0.8513,  0.9215, -0.1875)
18: (-0.0961,  0.5961,  0.6038,  0.2754)
19: (-0.0961, -0.5961,  0.6038, -0.2754)
20: ( 0.2624,  0.7219,  0.7681,  0.1945)
21: ( 0.2624, -0.7219,  0.7681, -0.1945)
22: ( 0.2109,  0.6545,  0.6876,  0.2004)
23: ( 0.2109, -0.6545,  0.6876, -0.2004)
24: (-0.2329,  0.0000,  0.2329,  0.5000)
25: (-0.1000,  0.0000,  0.1000,  0.5000)
26: ( 0.6966,  0.6594,  0.9592,  0.1206)
27: ( 0.6966, -0.6594,  0.9592, -0.1206)
28: ( 0.4453,  0.5594,  0.7149,  0.1430)
29: ( 0.4453, -0.5594,  0.7149, -0.1430)
30: ( 0.7258,  0.4395,  0.8485,  0.0866)
31: ( 0.7258, -0.4395,  0.8485, -0.0866)
32: ( 0.8118,  0.3423,  0.8810,  0.0635)
33: ( 0.8118, -0.3423,  0.8810, -0.0635)
34: ( 0.9206,  0.2748,  0.9608,  0.0462)
35: ( 0.9206, -0.2748,  0.9608, -0.0462)
36: ( 0.9928,  0.0788,  0.9959,  0.0126)
37: ( 0.9928, -0.0788,  0.9959, -0.0126)
38: ( 0.9958,  0.0000,  0.9958,  0.0000)
39: ( 0.8962,  0.1124,  0.9032,  0.0198)
40: ( 0.8962, -0.1124,  0.9032, -0.0198)
41: ( 0.5908,  0.2922,  0.6591,  0.0731)
42: ( 0.5908, -0.2922,  0.6591, -0.0731)
43: ( 0.5367,  0.2025,  0.5736,  0.0574)
44: ( 0.5367, -0.2025,  0.5736, -0.0574)

Hope that helps.

Cheers,

Mj


 


On Tue, Dec 13, 2011 at 12:51 PM, Summers, Peter <psummers@highpoint.edu> wrote:
I'm confused too.

MJ, is the (log) level of GDP one of the 11 series in your VAR? If so, then based on the unit root tests you showed earlier, it is not "stable (and stationary) without a trend." On the contrary, it has a unit root -- and is therefore non-stationary -- whether or not a deterministic trend is included in the dgp.

In other words, including the level of GDP in your reduced-form VAR renders it non-stationary.

Back to the potential small-sample issue: a VAR with 11 variables and 4 lags has 44 parameters per equation, not counting a constant (or trend?!). There are also 55 parameters in the covariance matrix. With 100 observations per series, you're asking quite a lot of your data set. Even if you knew the covariance matrix for sure, you'd have just over 2 obs/parameter for estimating the dynamics. I don't think that's asymptotic yet, but I could be wrong ;-)


________________________________
From: gretl-users-bounces@lists.wfu.edu [gretl-users-bounces@lists.wfu.edu] on behalf of Dr RJF Hudson [rjfhud@powerup.com.au]
Sent: Monday, December 12, 2011 8:28 PM
To: Gretl list
Subject: Re: [Gretl-users] Deterministic trend in VAR

Greetings all
Have to say I'm getting confused, here.
I'd be appreciative please if somebody would tell me please
what this means  "the reduced form".... of what?
Also if a set is stable as you say, and to produce its stationarity you are confident that you haven't
squelched out important information from the data by differencing etc, what's the reason to introduce trend information and then trust inferences from the results ?
Trend in their Unit Roots?
I'm cool
rest easy
Richard Hudson

Dr RJF Hudson Qld Australia
rjfhud@powerup.com.au<mailto:rjfhud@powerup.com.au>
----- Original Message -----
From: Muheed Jamaldeen<mailto:mj.myworld@gmail.com>
To: Gretl list<mailto:gretl-users@lists.wfu.edu>
Sent: Tuesday, December 13, 2011 10:59 AM
Subject: Re: [Gretl-users] Deterministic trend in VAR

You're right about the VAR not being stable if USGDP were the only series in the model. Well, the VAR is a 11 variable VAR (4). The 11 variables are GDP and macroeconomic variables.

I am testing the impact of cash rate innovations on GDP. The question is, if the reduced form is stable (and stationary) WITHOUT a trend, should one include a trend when the univariate tests suggest that SOME of the series may have trend in their unit roots.

Hope that makes sense?


On Tue, Dec 13, 2011 at 11:46 AM, Summers, Peter <psummers@highpoint.edu<mailto:psummers@highpoint.edu>> wrote:
MJ,

You're right that the unit root tests are telling you that you have a unit root in at least one series.

I'm confused about what your VAR looks like though (and maybe the rest of the list is too). If this is one of the series in your VAR, then it's not stable/stationary, by definition. That is, the lag operator polynomial will have at least one root on the unit circle. My earlier answer assumed that your unit root & cointegration tests ruled out both, but now it seems that's not the case.

Relating to ths, how many series do you have in your VAR? My feeling is that 100 obs per series isn't really a lot, especially if you're trying to sort out issues related to deterministic vs stochastic trends, cointegration vs none, etc.

At this point I'd suggest a) reading the gretl manual and/or your favorite reference on VARs & VECMs, and/or b) providing some more detail about what you're trying to do.

PS
________________________________
From: gretl-users-bounces@lists.wfu.edu<mailto:gretl-users-bounces@lists.wfu.edu> [gretl-users-bounces@lists.wfu.edu<mailto:gretl-users-bounces@lists.wfu.edu>] on behalf of Muheed Jamaldeen [mj.myworld@gmail.com<mailto:mj.myworld@gmail.com>]
Sent: Monday, December 12, 2011 6:59 PM
To: Gretl list
Subject: Re: [Gretl-users] Deterministic trend in VAR

Peter,

I have 100 observations in the model. So small samples may or may not be an issue. I am wondering if the deterministic trend is an issue at all because the VAR is stable implying stationarity of the described process in each equation WITHOUT the trend (i.e. the polynomial defined by the determinant of the autoregressive operator has no roots in and on the complex unit circle without the time trend term).

The ADF tests suggest that we cannot reject the trend term. Let me show you an example. Following is the ADF tests for logged US GDP.

Monte Carlo studies suggest that choosing the lag order (p) of the unit root tests according to the formula: Int {12(T /100)1/ 4} so the lag order is 12 with 100 observations.

test without constant
test statistic: tau_nc(1) = 2.13551
asymptotic p-value 0.9927

test with constant
test statistic: tau_c(1) = -1.28148
asymptotic p-value 0.6405

with constant and trend
test statistic: tau_ct(1) = -0.728436
 asymptotic p-value 0.9702

Following is the estimate for the trend term in the last ADF regression.

                      coefficient    std. error          t-ratio   p-value
 -------------------------------------------------------------
time            0.000200838   0.000317669    0.6322   0.5292

So all three tests are saying that I cannot reject the null of unit root. Including I(1) variables in an unrestricted VAR is fine as Lutekepohl and Toda and Yammoto have demonstrated. It's a question of whether a trend term is to be included. I am inclined to think not because the VAR is stable WITHOUT a trend.

Thoughts?

Cheers,

Mj

On Tue, Dec 13, 2011 at 1:17 AM, Summers, Peter <psummers@highpoint.edu<mailto:psummers@highpoint.edu><mailto:psummers@highpoint.edu<mailto:psummers@highpoint.edu>>> wrote:
MJ,

If your data have deterministic trends, then unit root tests should pick that up (though there may be a problem in small samples). If you include a trend but the dgp is stationary, then a t-test should conclude that the trend coefficient is zero. Presumably your unit root tests reject the null, right?

From: gretl-users-bounces@lists.wfu.edu<mailto:gretl-users-bounces@lists.wfu.edu><mailto:gretl-users-bounces@lists.wfu.edu<mailto:gretl-users-bounces@lists.wfu.edu>> [mailto:gretl-users-bounces@lists.wfu.edu<mailto:gretl-users-bounces@lists.wfu.edu><mailto:gretl-users-bounces@lists.wfu.edu<mailto:gretl-users-bounces@lists.wfu.edu>>] On Behalf Of Muheed Jamaldeen
Sent: Monday, December 12, 2011 5:52 AM
To: Gretl list
Subject: [Gretl-users] Deterministic trend in VAR

Hi all,
Just a general VAR related question. When is it appropriate to include a deterministic time trend in the reduced form VAR? Visually some of the data series (not all) look like they have trending properties. In any case, does the inclusion of the time trend matter if the process is stable and therefore stationary (i.e. the polynomial defined by the determinant of the autoregressive operator has no roots in and on the complex unit circle) without the time trend term. Other than unit root tests, is there a better way to test whether the underlying data generating process has a stochastic or deterministic process?

I am mainly interested in the impulse responses.

Cheers,

Mj


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