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

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

Subject |
st: Non-linear model + More parameters than variables + -ml- |

Date |
Tue, 30 Apr 2013 12:11:52 +0200 |

Hi, Statalisters. I am stuck with a problem. I have been able to solve part of it (with your help), but not the following. One of the versions of my model is not linear and quite complex. This is it (x and y are variables, and thetas are parameters), y=[theta1+theta2*(x/theta3)^theta4]*[1-(x/theta3)] To estimate it using -ml- I have used this program (assuming that residuals are normally distributed): program datos6mean version 10.1 args lnf theta1 theta2 theta3 theta4 sigma quietly replace `lnf' = ln(normalden($ML_y1, (`theta1'+`theta2'*($ML_y2/`theta3')^`theta4')*(1 - ($ML_y2/`theta3')), `sigma')) end Nevertheless, I do not know how to write the command -ml model lf datos6mean- in order to indicate Stata that there are just two variables (that I have introduced as if both of them were dependent variables), that there is no linear part in the model, but there are four parameters to be estimated. Will anyone out there please help me? Thanks. Miguel. * * 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/

**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>

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

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