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From |
"Miguel Angel Duran" <maduran@uma.es> |

To |
<statalist@hsphsun2.harvard.edu> |

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 - $ML_y2^`theta4'), `sigma')) Then what would be ml model command? "ml model lf datos4mean (b:vdmean=) (c:vlagmean=) (theta4:) (sigma:)? Thanks in advance. Miguel. -----Mensaje original----- De: owner-statalist@hsphsun2.harvard.edu [mailto:owner-statalist@hsphsun2.harvard.edu] En nombre de Aljar Meesters Enviado el: viernes, 26 de abril de 2013 17:16 Para: statalist@hsphsun2.harvard.edu Asunto: Re: st: Does ml requires a non-linear function to have a linear part? Dear Miquel, You can solve this issue by not passing v(/m)lagmean as an independent but dependent variable and using the constants as parameters. This would give: program datos4mean version 10.1 args lnf theta2 theta3 sigma quietly replace `lnf' = ln(normalden($ML_y1, `theta2' * $ML_y2^`theta3' * (1 - $ML_y2), `sigma')) end ml model lf datos4mean (b:vdmean=) (c:vlagmean=) (sigma:) ml check ml maximize However, since you are estimating a nonlinear least squares model you can also estimate nl (vdmean = {b}*vlagmean^{c}*(1-vlagmean)) Which is less restrictive on the distribution assumption of the residuals. Best, Aljar 2013/4/26 Miguel Angel Duran <maduran@uma.es>: > In all the examples that I have been able to find about how to use ml > to estimate a non-linear equation, there is always a linear part that > makes it possible to specify the dependent variable. Nevertheless, the > equation I am trying to estimate does not have that linear part. Can > anyone help me to know whether I can use ml (and how if it were possible)? > > Just to explain myself beter, this is my equation, > > vdmean = b*vlagmean^c*(1-vlagmean) > > And this is one of the things what I have tried to do, > > . program datos4mean > 1. version 10.1 > 2. args lnf theta2 theta3 sigma > 3. quietly replace `lnf' = ln(normalden($ML_y1, `theta2' * > vlagmean^`theta3' * (1-vlagmean), `sigma')) > 4. end > > . ml model lf datos4mean (vdmean=mlagmean, nocons) (theta2:) (theta3:) > (sigma:), vce(robust) > > . ml check > > RESULT: datos3mean HAS PASSED ALL TESTS > > . 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: > * http://www.stata.com/help.cgi?search > * http://www.stata.com/support/faqs/resources/statalist-faq/ > * http://www.ats.ucla.edu/stat/stata/ * * For searches and help try: * http://www.stata.com/help.cgi?search * http://www.stata.com/support/faqs/resources/statalist-faq/ * http://www.ats.ucla.edu/stat/stata/ * * For searches and help try: * http://www.stata.com/help.cgi?search * http://www.stata.com/support/faqs/resources/statalist-faq/ * http://www.ats.ucla.edu/stat/stata/

**Follow-Ups**:**st: Non-linear model + More parameters than variables + -ml-***From:*"Miguel Angel Duran" <maduran@uma.es>

**References**:**st: Does ml requires a non-linear function to have a linear part?***From:*"Miguel Angel Duran" <maduran@uma.es>

**Re: st: Does ml requires a non-linear function to have a linear part?***From:*Aljar Meesters <aljar.meesters@gmail.com>

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