[FernandoRios]

Hi Joro
Just one point on the Example that you posted.
This code
nl (y = {a} * x^max(3, {c}))

only constrains that the max(3, {c}) can have a value of 3 or above. It make any constraints on {c}.
Also, when {c} is less than 3, the function doesn't change with respect to {c}, which may be why Stata is assuming {c} is constrained to its initial value of 0.

Fernando

[Viridiana Corona]

Nick Cox Joro Kolev



Thank you for all your comments, they have been very helpful to me. Regarding this question, today I tried to estimate the same nonlinear equation by maximum likelihood. In fact, the I wanted the nl estimation to be sure that the maximum likelihood performs correctly. Â The code I use is the following:

capture program drop mymodel_gf0

program mymodel_gf0




    args todo b lnfj

    tempvar xb

    quietly generate double `xb' =`b’[1,1]*((x)^(`b'[1,2]-1)) 

    quietly replace `lnfj' = ln(normalden((y - `b')/`b'[1,3])) - ln(`b'[1,3])

end




ml model gf0 mymodel_gf0 (y = x) /sigma




ml maximize

However I can't get the program to run. Is it because stata assumes this is a linear equation? How can I correct this?







Thank you very much for your help!




Viridiana

[Nick Cox]

That's really a new question which should start a new thread. More crucially, you don't supply a reproducible example. However, the code looks very confused to me.

If local macro b contains the name of a matrix you supply, then you can refer to particular elements but you can't in the same code subtract the matrix from your outcome variable.

Further, I don't get any sense that this code would iterate as every time you call it within ml he result would be the same.

I think you're confusing nl programming and ml programming. There is scope for initializing parameters in ml, but the mechanism is different.

But, but, but: The biggest deal of all here is the idea. I don't see why maximum likelihood should give a different answer here any way if you're assuming normal errors.

[Joro Kolev]

Here is the same model with the data Nick generated estimated by maximum likelihood. The recently introduced command -mlexp- makes estimation of maximum likelihood models a lot easier:

. nl (y = {a} * x^{c}), nolog
(obs = 100)


      Source |      SS            df       MS
-------------+----------------------------------    Number of obs =        100
       Model |  237913.27          2  118956.634    R-squared     =     0.9996
    Residual |  94.467737         98  .963956498    Adj R-squared =     0.9996
-------------+----------------------------------    Root MSE      =   .9818129
       Total |  238007.74        100  2380.07735    Res. dev.     =   278.0965

------------------------------------------------------------------------------
           y |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
          /a |   3.016182   .0613011    49.20   0.000     2.894532    3.137831
          /c |   .6979471   .0048235   144.70   0.000     .6883749    .7075192
------------------------------------------------------------------------------

. mlexp (ln(normalden(y, {b}*x^{c}, {sigma}))), nolog
initial:       log likelihood =     -<inf>  (could not be evaluated)
feasible:      log likelihood = -16142.302
rescale:       log likelihood = -16142.302
rescale eq:    log likelihood = -392.11126

Maximum likelihood estimation

Log likelihood = -139.04826                     Number of obs     =        100

------------------------------------------------------------------------------
             |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
          /b |   3.016181   .0609311    49.50   0.000     2.896759    3.135604
          /c |   .6979471   .0047947   145.57   0.000     .6885497    .7073444
      /sigma |   .9719452   .0687269    14.14   0.000     .8372429    1.106647
------------------------------------------------------------------------------

The results are the same as from the NLS.

[Joro Kolev]

Thank you, Fernando ! You resolved the mystery of what Stata is doing, and why Stata thinks that c is a constant.

Unfortunately the two methods of imposing the inequality constraint that c>1 do not lead to the same results. Here are the results with Nick's data:

. nl (y = {a} * x^max(1,{c=3})), nolog
(obs = 100)


      Source |      SS            df       MS
-------------+----------------------------------    Number of obs =        100
       Model |  36030.414          0           .    R-squared     =     0.9219
    Residual |  3052.1515         99  30.8298129    Adj R-squared =     0.9219
-------------+----------------------------------    Root MSE      =    5.55246
       Total |  39082.566         99  394.773392    Res. dev.     =   625.6309

------------------------------------------------------------------------------
           y |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
          /a |   .8333161   .0095456    87.30   0.000     .8143756    .8522566
          /c |  -.0433347          .        .       .            .           .
------------------------------------------------------------------------------
  Parameter c taken as constant term in model & ANOVA table

. nl (y = {a} * x^(1+exp({lnc}))), nolog
(obs = 100)


      Source |      SS            df       MS
-------------+----------------------------------    Number of obs =        100
       Model |  35808.757          1  35808.7575    R-squared     =     0.9162
    Residual |  3273.8083         98  33.4062071    Adj R-squared =     0.9154
-------------+----------------------------------    Root MSE      =    5.77981
       Total |  39082.566         99  394.773392    Res. dev.     =   632.6416

------------------------------------------------------------------------------
           y |      Coef.   Std. Err.      t    P>|t|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
          /a |   .8589113   .0389093    22.07   0.000     .7816971    .9361254
        /lnc |  -25.45258   4.80e+08    -0.00   1.000    -9.52e+08    9.52e+08
------------------------------------------------------------------------------
  Parameter lnc taken as constant term in model & ANOVA table

. nlcom 1+exp( _b[lnc:_cons])

       _nl_1:  1+exp( _b[lnc:_cons])

------------------------------------------------------------------------------
           y |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
       _nl_1 |          1          .        .       .            .           .
------------------------------------------------------------------------------

.

Do you see me doing anything wrong? Or these are numerical problems?

Hi Joro
Just one point on the Example that you posted.
This code
nl (y = {a} * x^max(3, {c}))

only constrains that the max(3, {c}) can have a value of 3 or above. It make any constraints on {c}.
Also, when {c} is less than 3, the function doesn't change with respect to {c}, which may be why Stata is assuming {c} is constrained to its initial value of 0.

Fernando

[FernandoRios]

Hi Joro
I would say that the difference comes from numerical differences.
The way I see it, for example, is that neither is imposing an specific constraint on {c}. But on the power of "x".
In the first example, (with a minor variable change)
x^{d}
{d}=max(1,{c=3})
the constrain is that one is imposing the restriction that {d} >=1. if the true solution was that d<1, (based on how the data was created), it implies that a solution is for c<=1.
Because the initial value being used is c=3, NL algorithm is looking for the closest numerical solution towards c<1. And It seems that the last solution was to use c=-.04.

In the second case, the restriction is more explicit, since for any c> -infinity. then d>>1
In this case, the closest numerical solution is c=-25. And at that value d=1 (as you show)

I would also say that at the boundaries of the solution, neither function is continuous, which is why both approaches give slightly different results.
F