Dear All,
I am analyzing a DCE to estimate WTP values for environmental protection and I am trying different approaches mixed logit gmnl and WTP space.
1) Question
I run a mixed logit model with lognormal price coefficient (p)
firstly I transformed p into a non negative variable

gen mprice=-1*p

Then I run the mixed logit

mixlogit d , group(id) rand(f r mprice) id(Id) ln(1) nrep(50)

with the following results

d Coef. Std. Err. z P>z [95% Conf. Interval]

Mean
f .8806705 .055364 15.91 0.000 .772159 .989182
r 1.739759 .0919498 18.92 0.000 1.559541 1.919978
mprice .8917631 .0717043 12.44 0.000 .7512252 1.032301

SD
f .7638781 .0943571 8.10 0.000 .5789416 .9488145
r .8516495 .1039393 8.19 0.000 .6479323 1.055367
mprice .9619734 .0943854 10.19 0.000 .7769815 1.146965


As seen the mprice sign is positive is this correct?
Then I transform the price coefficients and estimate the mean median and standard deivation
nlcom (mean_price: -1*exp([Mean]_b[mprice]+0.5*[SD]_b[mprice]^2)) (med_price: -1*exp([Mean]_b[mprice]))(sd_price: exp([Mean]_b[mprice]+0.5*[SD]_b[mprice]^2) * sqrt(exp([SD]_b[mprice]^2)-1))

when I obtain these values how can I estimate the WTP values ?

d Coef. Std. Err. z P>z [95% Conf. Interval]

mean_price -3.874666 .4394025 -8.82 0.000 -4.735879 -3.013453
med_price -2.439427 .1749175 -13.95 0.000 -2.782259 -2.096595
sd_price 4.7815 1.189924 4.02 0.000 2.449292 7.113707

and now the second question I want to run a second regression with correlated coefficients using the starting values from the mixed logit: how can I do that? this is what I have done but I am not sure is correct….
mixlogit d p, group(id) rand(f r) id(Id)nrep(1000)

Log likelihood = -1753.3229 Prob > chi2 = 0.0000


d Coef. Std. Err. z P>z [95% Conf. Interval]

Mean
p -1.961931 .0822945 -23.84 0.000 -2.123226 -1.800637
f .7863766 .0577565 13.62 0.000 .6731759 .8995773
r 1.53036 .068974 22.19 0.000 1.395174 1.665547

SD
f .8402785 .0635261 13.23 0.000 .7157697 .9647874
r .6378639 .0602608 10.59 0.000 .5197549 .7559729

matrix b = e(b)


matrix b = b[1,5],0,0,0,0,0

results:



LR chi2(3) = 464.32
Log likelihood = -1746.8525 Prob > chi2 = 0.0000


d Coef. Std. Err. z P>z [95% Conf. Interval]

p 1.954885 .0825494 -23.68 0.000 -2.116679 -1.793091
f .8327778 .0608246 13.69 0.000 .7135638 .9519918
r 1.536822 .0693505 22.16 0.000 1.400898 1.672747

/l11 .8628636 .0669893 12.88 0.000 .7315669 .9941603
/l21 .2042594 .0559398 3.65 0.000 .0946194 .3138994
/l22 -.599216 .0619962 -9.67 0.000 -.7207263 -.4777058


Thank you very much for your help
Kind Regards
Carla