Errata corrige ;-)
Constrained maximum:
matrix ac = atanh((2*c-c_max-c_min)./(c_max-c_min))
matrix lg = ln(gamma)
mle logl = - 0.5*ln(2*pi) - ln(sigma) - 0.5*(e/sigma)^2
series e = pstr_cres(y,X,q,gamma,c,m,Z)
matrix gamma = exp(lg)
matrix c = c_min + 0.5*(tanh(ac) + 1).*(c_max-c_min)
params lg ac sigma
end mle
Bye
Giuseppe
On Thu, 2011-10-27 at 14:54 +0200, Giuseppe Vittucci wrote:
> Don't care about my previous email.
> I have eventually modified the function and it gives me now the series
> of the residuals. Hence, I can use the following commands:
>
> Unconstrained maximum:
>
> mle logl = - 0.5*ln(2*pi) - ln(sigma) - 0.5*(e/sigma)^2
> series e = pstr_cres(y,X,q,gamma,c,m,Z)
> params gamma c sigma
> end mle
>
> Constrained maximum:
>
> matrix ac = atanh((2*c-c_max-c_min)./(c_max-c_min))
> matrix lg = ln(gamma)
> mle logl = - 0.5*ln(2*pi) - ln(sigma) - 0.5*(e/sigma)^2
> series e = pstr_cres(y,X,q,gamma,c,m,Z)
> matrix gamma = exp(lg)
> matrix c = c_min + 0.5*(tanh(ac) + 1)*(c_max-c_min)
> params lg ac sigma
> end mle
>
> Thanks
> Giuseppe
>
>
> On Thu, 2011-10-27 at 09:57 +0200, Giuseppe Vittucci wrote:
> > Still on the mle command.
> > The argument of the mle command in gretl is actually the log-L
> > contributions and not the Log-L.
> > Clearly every combination of parameters that maximize the latter also
> > maximize the former.
> > So, as far as the point estimates of the parameters are concerned, it
> > does not really matters which one is used.
> >
> > So far I have used mle simply as a maximization BFGS method, and I was
> > not looking at the covariance or the other ancillary statistics.
> >
> > In my case working directly with the log-likelihood is much easier cause
> > I have a quite complex function that returns the SSR and I use it
> > directly in the command.
> > In case of homoscedastic normally distributed residuals, the Log-L is
> > indeed just:
> >
> > Log-L = -n/2*(ln (ssr/n) + 1 + ln 2pi)
> >
> > and I can use my function directly in the formula.
> > On the contrary, working directly with the log-L contributions is not
> > straightforward in my case.
> >
> > I would like to know if I could use the covariance matrix and the other
> > statistics generated by the program (in particular the information
> > criteria), if, instead of using the Log-L contributions, I simply divide
> > the Log-L by n.
> >
> > As far as the covariance is concerned, likely I cannot use the matrix
> > calculated from the outer product of the gradient, but can I use the
> > Hessian?
> >
> > Thanks a lot
> > Giuseppe
