Re: st: Maximum likelihood estimation

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

On 02/13/2013 03:27 PM, Joseph Monte wrote:
> Dear Statalisters,
>
> I'll try again with a little more info since I did not get any
> responses. Here is the code I have so far (based on previous Statalist
> posts and "Maximum Likelihood estimation with Stata" by
> Gould, Pitbaldo and Poi, 4th ed.). The paper I cited in my first email
> below models the log of the variance of the regression error in
> equation 2 while I believe I have modelled the log of sigma. I would
> preferably like to model the log of the variance as in the paper cited
> but am not sure how.
>
> cscript
>    program mynormal_lf1
>            version 12
>                    args todo b lnfj g1 g2
>            tempvar mu lnsigma sigma
>                    mleval `mu' = `b', eq(1)
>                    mleval `lnsigma' = `b', eq(2)
>            quietly {
>                                  gen double `sigma' = exp(`lnsigma')
>                                  replace `lnfj' =
> ln(normalden($ML_y1,`mu',`sigma'))
>                                  if (`todo'==0) exit
>                                  tempvar z
>                                  tempname dmu dlnsigma
>                                  gen double `z' = ($ML_y1-`mu')/`sigma'
>                                  replace `g1' = `z'/`sigma'
>                                  replace `g2' = `z'*`z'-1
>                  }
>    end
>
>
> ml model lf1 mynormal_lf1 (mu: y = x1 x2 x3 x4 x5 x6 x7 x8 x9)
> (lnsigma: y = x1 x2 x3 x4 x5 x6 x7 x8 x9)
> ml max
>
> Thanks,
>
> Joe

Joe,

Letting Stata do the work of computing numerical derivatives, here is the equivalent program that models the log variance rather than the log standard deviation:

-------------
cscript

program mynormal_lf0
	version 12
	args todo b lnfj
	tempvar mu lnvar var
	mleval `mu' = `b', eq(1)
	mleval `lnvar' = `b', eq(2)
	quietly {
		gen double `var' = exp(`lnvar')
		replace `lnfj' =  ln(normalden($ML_y1,`mu',sqrt(`var')))
	}
end

ml model lf0 mynormal_lf0 (mu: y = x1 x2 x3 x4 x5 x6 x7 x8 x9)		///
	(lnvar: y = x1 x2 x3 x4 x5 x6 x7 x8 x9)
ml max
-------------

The only real change is that since the normalden() function's third argument is the standard deviation, not the variance, we need to take the square root of the variance.  By construction, the variance variable `var' will be non-negative, so taking the square root is not a problem here.

   -- Brian Poi
   -- bpoi@stata.com


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