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st: maximum likelihood


From   "Abhimanyu Arora" <abhimanyu.arora@student.kuleuven.be>
To   <statalist@hsphsun2.harvard.edu>
Subject   st: maximum likelihood
Date   Tue, 13 Apr 2010 13:06:59 +0200

Dear statalist
I have been trying to find the maximum likelihood estimates of my model, but
I do not obtain standard errors (likelihood function non-concave and just a
couple of iterations achieved). Following the section "Importance of
generating temporary variables as doubles" pg 58-60 of the book by Gould et
al, I generate temporary variables which are functions of my parameters to
be estimated, only to obtain totally different coefficient values but still
no standard errors+no concavity. What is the reason behind this puzzling
phenomenon and its solution? Is my ml model command ok even though I may not
have any _dependent_ variable as such?

Please find below the initial and final likelihood evaluators. I have two
variables "i" (0/1) and "age"(1-5) and a constant matrix V(1X5) and need to
find parameters theta_1 and R which determine the likelihood of each
observation as below.  

Thanks and regards
Abhimanyu Arora
KU Leuven


*******************
Initial ml commands
*******************

cap program drop maxl

program define maxl
qui{
args lnf theta_1 R

replace
`lnf'=ln((exp(`R'+0.9*V[1,1]))/exp(age*(`theta_1')+0.9*V[1,2])+exp(`R'+0.9*V
[1,1])) if i==1 & age ==1 

replace
`lnf'=ln((exp(`R'+0.9*V[1,1]))/exp(age*(`theta_1')+0.9*V[1,3])+exp(`R'+0.9*V
[1,1])) if i==1 & age ==2 

replace
`lnf'=ln((exp(`R'+0.9*V[1,1]))/exp(age*(`theta_1')+0.9*V[1,4])+exp(`R'+0.9*V
[1,1])) if i==1 & age ==3 

replace
`lnf'=ln((exp(`R'+0.9*V[1,1]))/exp(age*(`theta_1')+0.9*V[1,5])+exp(`R'+0.9*V
[1,1])) if i==1 & age ==4 

replace
`lnf'=ln((exp(`R'+0.9*V[1,1]))/exp(age*(`theta_1')+0.9*V[1,5])+exp(`R'+0.9*V
[1,1])) if i==1 & age ==5 


replace
`lnf'=1-ln((exp(`R'+0.9*V[1,1]))/exp(age*(`theta_1')+0.9*V[1,2])+exp(`R'+0.9
*V[1,1])) if i==0 & age ==1
replace
`lnf'=1-ln((exp(`R'+0.9*V[1,1]))/exp(age*(`theta_1')+0.9*V[1,3])+exp(`R'+0.9
*V[1,1])) if i==0 & age ==2
replace
`lnf'=1-ln((exp(`R'+0.9*V[1,1]))/exp(age*(`theta_1')+0.9*V[1,4])+exp(`R'+0.9
*V[1,1])) if i==0 & age ==3
replace
`lnf'=1-ln((exp(`R'+0.9*V[1,1]))/exp(age*(`theta_1')+0.9*V[1,5])+exp(`R'+0.9
*V[1,1])) if i==0 & age ==4
replace
`lnf'=1-ln((exp(`R'+0.9*V[1,1]))/exp(age*(`theta_1')+0.9*V[1,5])+exp(`R'+0.9
*V[1,1])) if i==0 & age ==5

}

end
ml model lf maxl (theta_1:) (R:)
ml check
ml init 0 0,copy
ml maximize,difficult

*****************************************
Introducing double temporary variables
*****************************************
cap program drop maxl

program define maxl
qui{
args lnf theta_1 R
tempvar a b 
gen double `a' =age*(`theta_1')
gen double `b'=exp(`R'+0.9*V[1,1])

replace `lnf'=ln(`b'/(exp(`a'+0.9*V[1,2])+`b'))  if i==1 & age ==1 

replace `lnf'=ln(`b'/(exp(`a'+0.9*V[1,3])+`b'))  if i==1 & age ==2 

replace `lnf'=ln(`b'/(exp(`a'+0.9*V[1,4])+`b'))  if i==1 & age ==3 

replace `lnf'=ln(`b'/(exp(`a'+0.9*V[1,5])+`b'))  if i==1 & age ==4 

replace `lnf'=ln(`b'/(exp(`a'+0.9*V[1,5])+`b'))  if i==1 & age ==5 


replace `lnf'=1-ln(`b'/(exp(`a'+0.9*V[1,2])+`b'))  if i==0 & age ==1
replace `lnf'=1-ln(`b'/(exp(`a'+0.9*V[1,3])+`b'))  if i==0 & age ==2
replace `lnf'=1-ln(`b'/(exp(`a'+0.9*V[1,4])+`b'))  if i==0 & age ==3
replace `lnf'=1-ln(`b'/(exp(`a'+0.9*V[1,5])+`b'))  if i==0 & age ==4
replace `lnf'=1-ln(`b'/(exp(`a'+0.9*V[1,5])+`b'))  if i==0 & age ==5

}

end
ml model lf maxl (theta_1:) (R:)
ml check
ml init 0 0,copy
ml maximize,difficult

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