Re: Re: st: nl command - error#130 expression too long

[Carlotta Schuster <schuster.carlotta@gmail.com>]

Thank you Maarten. I have been exploring the ml command, with which I
think I should be able to solve my problem. I am following Gould,
Pitblado and Poi's Maximum Likelihood estimation with Stata, but I am
unable to implement my problem correctly. I will try to describe the
problem here.

My model is the following (I omit subscripts hoping it is clear enough
without them):

Y = alpha + beta1*A(lambda)+beta2*X2+...+beta15*X15+ error

Where y is a discrete choice variable and I want to estimate alpha,
beta1,..., beta15 and LAMBDA. Here is where problem comes from, since
A(lambda) is a non-linear function and I want to estimate lambda jointly
with the rest of the parameters. If I set a particular value for lambda
the model becomes a regular probit which I can easily estimate with
Stata's probit command, but as I said, I also want to estimate lambda.
In particular A(lambda) takes the following form:

A(lambda) = sumatory(k=1 to k=age-1) weight(k,lambda)*Return_T-k

And weight(k, lambda) is a non-linear function of lambda which also
includes a sumatory and the age variable).

Any ideas on how this could be program using the ml command?

My tries so far have led me to the following program, where I create the
A(lambda) variable inside the program using one of the parameters to be
estimated.  But this does not work.


capture program drop nlprobitlf
program nlprobitlf
  version 11
  args todo b lnfj
  tempvar w w_den N a

local N = _N
gen w_den = 0
gen w = 0
forvalues i = 1(1)`N' {
local a = age-1 in `i'
forvalues y = 1(1)`a'{
replace w_den = w_den + (age-`y')^`b'[1,4] in `i'
}
forvalues z = 1(1)`a'{
replace w = w + ((age-`z')^`b'[1,4])*rets`z'/w_den in `i'
}
}


quietly generate double `xb' = `b'[1,1]+`b'[1,2]*w+`b'[1,3]*income

quietly replace `lnfj' = lnnormal(`xb') if ($ML_y1 == 1)
quietly replace `lnfj' = lnnormal(-1*`xb') if ($ML_y1 == 0)


end


Best,

Carlotta
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