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From |
Hitomi Amaya <tommy.amaya@gmail.com> |

To |
statalist@hsphsun2.harvard.edu |

Subject |
st: Maximum Likelihood |

Date |
Tue, 1 Apr 2014 11:52:39 -0700 (PDT) |

Hello, I defined the following maximum likelihood program for bivariate probit model. STATA gave me an error "unknown function 1-()". I found William Gould of Stata corp mentioning: "Never code 1-normprob(z). Code instead normprob(-z), which is far more accurate." to Deborah who raised a similar question on this STATA list. However, as you can see in my code below, I have 1-(normprob(z)^131) instead of 1-normprob(z). I cannot easily convert it into a form without using 1-(...). I read "Maximum Likelihood Estimation with Stata", but I couldn't find a solution. I would very much appreciate it if anyone could come up with a solution to this problem. There is my code (which is a modification of the code written by Antoine, also found on the STATA list): program define mybiprobt args lnf theta1 theta2 theta3 tempvar TH quietly gen double `TH'=((exp(`theta3')-exp(-`theta3'))/(exp(`theta3')+exp(-`theta3'))) quietly replace `lnf'=(ln(binorm(-`theta1',-`theta2',`TH'))-ln((1–(normprob(-‘theta2'))^131)^(1/48))) if $ML_y1==0 & $ML_y2==0 quietly replace `lnf'=(ln(binorm(-`theta1',`theta2',-`TH'))-ln((1–(normprob(-‘theta2'))^131)^(1/48))) if $ML_y1==0 & $ML_y2==1 quietly replace `lnf'=(ln(binorm(`theta1',-`theta2',-`TH'))-ln((1–(normprob(-‘theta2'))^131)^(1/48))) if $ML_y1==1 & $ML_y2==0 quietly replace `lnf'=(ln(normprob(`theta2')-binorm(-`theta1',`theta2',-`TH'))-ln((1–(normprob(-‘theta2'))^131)^(1/48))) if $ML_y1==1 & $ML_y2==1 end Thank you. Hitomi Amaya -- View this message in context: http://statalist.1588530.n2.nabble.com/Maximum-Likelihood-tp7580502.html Sent from the Statalist mailing list archive at Nabble.com. * * 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: RE: Maximum Likelihood***From:*Timothy Mak <tshmak@hku.hk>

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