On Tue, 6 Jan 2015, Daniel Bencik wrote: > I am struggling with my variable lists once again. I expected the following > two ML estimations to give the same results, but they don't. Any ideas? > > #estimation 1 > scalar c_ = 0.1 > scalar rng = 0.1 > scalar err = 0.2 > > mle ll = -ln(lambda) - sqrtPark/lambda > > series lambda = mean(sqrtPark) > series lambda = c_ + rng * sqrtPark(-1) + err * lambda(-1) > > params c_ rng err > end mle --robust > > > #estimation 2 > list xDepVars = sqrtPark > list xExpVars = sqrtPark(-1) > > c_ = 0.1 > rng = 0.1 > err = 0.2 > > series xDepVar = xDepVars[1] > series xExpVar = xExpVars[1] > > mle ll2 = -ln(lambda2) - xDepVar/lambda2 > > series lambda2 = mean(xDepVar) > series lambda2 = c_ + rng * xExpVar + err * lambda2(-1) > > params c_ rng err > end mle --robust
Sorry, I realize that the answer I gave yesterday in http://lists.wfu.edu/pipermail/gretl-users/2015-January/010564.html will not do! My explanation was that an expression such as "xDepVars[1]" gives the integer ID number of the series at position 1 in the named list, not the series itself. Well, that was true, but it's not consistent with the documentation I added recently on indexing into a list, and it's surely not what users will expect. So today I've modified the result of indexing into a list to match the doc and the most likely expectation. You now get a series, and your ML script should behave as you intended. That's in CVS and snapshots. Note that if you wish to retrieve the ID number of the series at a certain position in a list this can be done by converting the list into a matrix. The following little script illustrates this point as well as the effect of indexing into a list directly. <hansl> open data4-1 list L = sqft price series x = L[1] print sqft x --byobs ols L[2] 0 L[1] matrix m = L printf "The series at position 1 in L has ID %d\n", m[1] </hansl> Allin
