Bookmark and Share

Notice: On March 31, it was announced that Statalist is moving from an email list to a forum. The old list will shut down on April 23, and its replacement, is already up and running.

[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

RE: st: Does ml require a non-linear function to have a linear part?

From   "Miguel Angel Duran" <>
To   <>
Subject   RE: st: Does ml require a non-linear function to have a linear part?
Date   Mon, 29 Apr 2013 11:24:20 +0200

Dear Aljar,

Thank you very much for your help. Your suggestion works. Indeed, I obtain
the same results that from nl. By the way, the reason to use ml is because I
have to estimate different versions of the same basic model. To estimate
some of them, nl requires introducing starting values, but I have the
feeling that results are too dependent on the starting values used.

One more question if you do not mind. Suppose I would have a third parameter
(theta3), so that the "quietly replace" line of the program becomes:

quietly replace `lnf' = ln(normalden($ML_y1, `theta2' * $ML_y2^`theta3' * (1
- $ML_y2^`theta4'), `sigma'))

Then what would be ml model command? "ml model lf datos4mean (b:vdmean=)
(c:vlagmean=) (theta4:) (sigma:)?

Thanks in advance.



-----Mensaje original-----
[] En nombre de Aljar Meesters
Enviado el: viernes, 26 de abril de 2013 17:16
Asunto: Re: st: Does ml requires a non-linear function to have a linear

Dear Miquel,

You can solve this issue by not passing v(/m)lagmean as an independent but
dependent variable and using the constants as parameters. This would give:
program datos4mean
    version 10.1
    args lnf theta2 theta3 sigma
    quietly replace `lnf' = ln(normalden($ML_y1, `theta2' * $ML_y2^`theta3'
* (1 - $ML_y2), `sigma')) end

ml model lf datos4mean (b:vdmean=) (c:vlagmean=) (sigma:) ml check ml

However, since you are estimating a nonlinear least squares model you can
also estimate

nl (vdmean = {b}*vlagmean^{c}*(1-vlagmean))

Which is less restrictive on the distribution assumption of the residuals.


2013/4/26 Miguel Angel Duran <>:
> In all the examples that I have been able to find about how to use ml 
> to estimate a non-linear equation, there is always a linear part that 
> makes it possible to specify the dependent variable. Nevertheless, the 
> equation I am trying to estimate does not have that linear part. Can 
> anyone help me to know whether I can use ml (and how if it were possible)?
> Just to explain myself beter, this is my equation,
> vdmean = b*vlagmean^c*(1-vlagmean)
> And this is one of the things what I have tried to do,
> . program datos4mean
>   1. version 10.1
>   2. args lnf theta2 theta3 sigma
>   3. quietly replace `lnf' = ln(normalden($ML_y1, `theta2' * 
> vlagmean^`theta3' * (1-vlagmean), `sigma'))
>   4. end
> . ml model lf datos4mean (vdmean=mlagmean, nocons) (theta2:) (theta3:) 
> (sigma:), vce(robust)
> . ml check
> . ml maximize
> And I get this message,
> initial:       log pseudolikelihood = -72.946848
> rescale:       log pseudolikelihood =  219.01781
> rescale eq:    log pseudolikelihood =  219.52686
> could not calculate numerical derivatives flat or discontinuous region 
> encountered r(430);
> *
> *   For searches and help try:
> *
> *
> *
*   For searches and help try:

*   For searches and help try:

© Copyright 1996–2015 StataCorp LP   |   Terms of use   |   Privacy   |   Contact us   |   Site index