ubreak for a single regressor

[feporrua1]

Hi,

I am trying to test for a break in one of the independent variables in several ols regressions.
However, the tests results that I get are for all equation variables (i.e., Varying regressors: All equation variables) and not only for the variable I want.
Please help.

The relevant part of the code is:

Code: Select all

freeze(aqtable) temp{!x}.ubreak 15 @ @trend

Note that for this example the variable of interest is @trend but the same happens if I choose any other of the independent variables

Many thanks,

Francisco

[EViews Gareth]

Looks like that should work

[feporrua1]

But it does not... the results you get are not for @trend but for all the variables in your regression. I really need the results only for one of the variables, not for all of them.
I am using Eviews 9.5 version. Is this a known bug? Is there a way to get results for only one variable in a program?

Please help

[feporrua1]

Dear Gareth,
Here is a simple example of the problem with this function. Please tell me if I am doing something wrong or if it is a bug (and hopefully if there is a way to solve it):

Code: Select all

genr x=nrnd genr y=nrnd equation eq1.ls y c @trend x freeze(qa) eq1.ubreak 15 @ @trend

As you can see if you run the program, the results are:

Quandt-Andrews unknown breakpoint test
Null Hypothesis: No breakpoints within 15% trimmed data
Varying regressors: All equation variables
Equation Sample: 1 100
Test Sample: 16 86
Number of breaks compared: 71

And indeed the results are different when i do the test using the "equation views" way by clicking directly in the regression window. The problem is that I cannot do this by hand for the 1000's of series I am analyzing.

Many thanks,

Francisco

[EViews Gareth]

ok, that looks like a bug. We'll fix.

[feporrua1]

Dear Gareth,

Many thanks. Do you have an idea about when this bug will be fixed?

Best regards,

Francisco

[EViews Gareth]

Within a couple of weeks or so, at a guess. It will be in the next patch.

[EViews Gareth]

You could use multibreak(maxbreaks=1) in the meantime.

[feporrua1]

Many thanks.
I have a doubt about what is assumed in the multibreak;
The multibreak (Bai Perron) is for a disjoint trend, right? The trend is not joined at the breakpoint (it assumes a jump as opposed to the Perron-Yabu, for example). In this way, the multibreak (maxbreaks=1) is not a close replacement for the ubreak, I t

Is this correct?

Thanks again.