Retrieving inverse roots of the characteristic equation

Questions and discussions on Time Series Analysis
holyw160
Posts: 10
Joined: Thu Nov 09, 2006 3:07 am

Retrieving inverse roots of the characteristic equation

Unread post by holyw160 »

Hi, I'm trying to implement the ARDL/bounds testing methodology of Pesaran, Shin & Smith (2001) and was trying to retrieve the inverse roots of the characteristic equation from the unrestricted ECMs. Does anyone have an example of the commands/code I should use, please? Thanks.
TomDoan
Posts: 7814
Joined: Wed Nov 01, 2006 4:36 pm

Re: Retrieving inverse roots of the characteristic equation

Unread post by TomDoan »

You can pull the lag polynomial out with %EQNLAGPOLY. %EQNLAGPOLY(0,depvar) will get the lag polynomial for the dependent variable with 1 as the 0 lag coefficient and the other lags with - signs attached. See, for instance, the DISTRIBLAG.RPF example. You can then use %POLYCXROOTS to get the roots. See, for instance, the TSAYP042.RPF example.

However, it's not clear that those have any value here. If you're estimating an ECM, then the dependent variable (in lagged level form) appears on the right side. Extracting the lag polynomial of the difference won't include that. And there is no way to include the effect of the omitted EC term without the remainder of the VECM. The roots of the "characteristic polynomial" of the differences thus tell you nothing about the dynamics of the process. And what's the point if they don't?

I would note that the paper is one of the sloppiest pieces of empirical work I've run across. They post two versions of the data file, one of which is supposed to be derived from the other. However, several of the derived series don't match with what they're supposed to be from the original data. And neither version can generate the published results. Given that this is just applied least squares, that shouldn't happen.
holyw160
Posts: 10
Joined: Thu Nov 09, 2006 3:07 am

Re: Retrieving inverse roots of the characteristic equation

Unread post by holyw160 »

Thanks Tom
Post Reply