Thread Subject: glmfit

I am working with glmfit to perform probit regressions. I want to regress my y on just a costant but the glmfit points out that the limit is reached.

My syntax is;
[beta dev stat]=glmfit(x,[y n],'binomial','probit')

if x has not regressors, what's the function's input in place of x? and if the regression contains already a (default) constant, in such a case, how can I exclude the constant?

thanks for your help

Luca

Luca Di Simone wrote:
> I am working with glmfit to perform probit regressions. I want to regress my y on just a costant but the glmfit points out that the limit is reached.
>
> My syntax is;
> [beta dev stat]=glmfit(x,[y n],'binomial','probit')
>
> if x has not regressors, what's the function's input in place of x? and if the regression contains already a (default) constant, in such a case, how can I exclude the constant?

As the help says,

>> help glmfit
 GLMFIT Fit a generalized linear model.
[snip]
       'constant' - specify as 'on' (the default) to include a constant
          term in the model, or 'off' to omit it. The coefficient of the
          constant term is the first element of B.

Or you can do this:

   x = zeros(size(y,1),0)
   glmfit(x,[y n],'binomial','link','probit')

But why are you using GLMFIT if you don't have any predictor variables? The only reason I can think of would be to compute some version of r-squared.

I am computing the mcfadden r^2 and I have to calculate the log likelihood of a model with only intercept. Now I used the vector of zeros as only regressor but the output of the function is that the matrix is singular to working precision.
While if I perform the regression on an specific regressor the results have not this problem.
How can I do?
thanks
Luca






Peter Perkins <Peter.Perkins@MathRemoveThisWorks.com> wrote in message <hcs88q$qgf$1@fred.mathworks.com>...
> Luca Di Simone wrote:
> > I am working with glmfit to perform probit regressions. I want to regress my y on just a costant but the glmfit points out that the limit is reached.
> >
> > My syntax is;
> > [beta dev stat]=glmfit(x,[y n],'binomial','probit')
> >
> > if x has not regressors, what's the function's input in place of x? and if the regression contains already a (default) constant, in such a case, how can I exclude the constant?
>
> As the help says,
>
> >> help glmfit
> GLMFIT Fit a generalized linear model.
> [snip]
> 'constant' - specify as 'on' (the default) to include a constant
> term in the model, or 'off' to omit it. The coefficient of the
> constant term is the first element of B.
>
> Or you can do this:
>
> x = zeros(size(y,1),0)
> glmfit(x,[y n],'binomial','link','probit')
>
> But why are you using GLMFIT if you don't have any predictor variables? The only reason I can think of would be to compute some version of r-squared.

> Now I used the vector of zeros as only regressor but the output of the
> function is that the matrix is singular to working precision.
...
> Peter Perkins <Peter.Perkins@MathRemoveThisWorks.com> wrote in message
>> x = zeros(size(y,1),0)
>> glmfit(x,[y n],'binomial','link','probit')

Did you do it the way Peter suggested? He has created not a vector of zeros,
but an array with the right number of rows and zero columns.

-- Tom

Yes, I used the way suggusted by peter and the output is the matrix is singular. Can you help?
Luca

"Tom Lane" <tlane@mathworks.com> wrote in message <hcsevl$2o4$1@fred.mathworks.com>...
> > Now I used the vector of zeros as only regressor but the output of the
> > function is that the matrix is singular to working precision.
> ...
> > Peter Perkins <Peter.Perkins@MathRemoveThisWorks.com> wrote in message
> >> x = zeros(size(y,1),0)
> >> glmfit(x,[y n],'binomial','link','probit')
>
> Did you do it the way Peter suggested? He has created not a vector of zeros,
> but an array with the right number of rows and zero columns.
>
> -- Tom
>

Luca Di Simone wrote:
> Yes, I used the way suggusted by peter and the output is the matrix is singular. Can you help?

It may be that you are using an old version of this function. It definitely does work in the current version.

If that's the case, you might try passing in a column of ones for x, and turning off the constant, as I suggested in an earlier post.