Re: st: constant in nonlinear squares

["Brian P. Poi" <bpoi@stata.com>]

On Tue, 3 Apr 2007, Maritza Sotomayor wrote:

> Thank you Brian,
> It worked fine with the option hasconstant(xb_constante). But when I
> compared the results with my other alternative (just include
> "constante" as variable without hasconstant option) I got the same
> result.
> I'd like to be sure if I am using the substitutable expression option
> correctly. What I have to specify is a logistic function like this:
> y = 1/1+exp(-bX)
> where b is a parameter vector with X vector of independent
> variables(eight). So I am not sure if using linear combination xb
> -{eqname:varlist}- was correct.
> This is what I used:
> nl (iitt = 1/(1+exp(-1*{xb: lavrgdp ldpcgdp ltc lroyalsale latarif
> lagi lavgest ldcap})))  nolog
> What do you think my mistake was?
> Maritza Sotomayor
Maritza,

When you do not specify the hasconstant() option, -nl- will try to 
determine whether your function has a constant term automatically.  Since 
your constante variable is equal to one for all observations, -nl- 
correctly identified its parameter as being a constant term.  You 
typically don't need to use the hasconstant() option; -nl- will find the 
constant for you.  The only time -nl- has trouble finding a constant term 
is when the data is badly scaled and a variable has a very small (but 
nonzero) variance.

  -- Brian Poi
  -- bpoi@stata.com
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