Dear all,
Departing from the following data
v1 v2 v3 z
0.0375815 0.0480932 -1 -0.3227311
-0.0192965 0.0261546 -0.1428161 -0.0480932
-0.097088 0.0096482 0 0.0187907
I am using NLS to estimate the following function
z = v1*rho+v2*(rho^2)+v3*(sigma^2)
where you have two parameters (rho and sigma^2). The problem is I
tried two different ways to estimate the parameters that lead me to
different values but this shouldnt be the case. Originally, I was
using the following programme:
program define nlequ
version 9
if "`1'"== "?" {
global S_1 " rho sigma2 "
global rho=1
global sigma2=1
exit
}
replace `1'=v1*$rho +v2*$rho^2 +v3*$sigma2
end
and typing in the command line something like:
nl equ z, init(rho=0.7, sigma2=1)
The regression output is
Iteration 0: residual SS = .409458
Iteration 1: residual SS = .0018054
Iteration 2: residual SS = .0000357
Iteration 3: residual SS = .0000355
Iteration 4: residual SS = .0000355
Source | SS df MS Number of obs = 3
-------------+------------------------------ F( 2, 1) = 1503.70
Model | .106785885 2 .053392943 Prob > F = 0.0182
Residual | .000035508 1 .000035508 R-squared = 0.9997
-------------+------------------------------ Adj R-squared = 0.9990
Total | .106821393 3 .035607131 Root MSE = .0059588
Res. dev. = -25.51948
(equ)
------------------------------------------------------------------------------
z1 | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
rho | -.1857942 .0626628 -2.96 0.207 -.9820006 .6104122
sigma2 | .3149169 .0067697 46.52 0.014 .2288992 .4009345
------------------------------------------------------------------------------
Likewise, we should be able to get to the same result if we type in
the command line
nl (z1 = v1*{rho=1}+v2*{rho}^2+v3*{sigma2=1}) ,
init(rho -.1857942 sigma2 .3149169)
the regression output is
Iteration 0: residual SS = .0026024
Iteration 1: residual SS = .0000491
Iteration 2: residual SS = .0000487
Iteration 3: residual SS = .0000487
Iteration 4: residual SS = .0000487
Source | SS df MS
-------------+------------------------------ Number of obs = 3
Model | .106772687 2 .053386344 R-squared = 0.9995
Residual | .000048706 1 .000048706 Adj R-squared = 0.9986
-------------+------------------------------ Root MSE = .006979
Total | .106821393 3 .035607131 Res. dev. = -24.57133
------------------------------------------------------------------------------
z | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+----------------------------------------------------------------
/rho | -.169541 .0664627 -2.55 0.238 -1.01403 .6749475
/sigma2 | .3186849 .0069996 45.53 0.014 .229746 .4076239
------------------------------------------------------------------------------
Notice that I even introduced the previous estimates as initial values
but even then Stata finds an alternative minimum. Further, the
residual SS is higher.
I would appreciate any comment that help me to understand why Stata
provides different values depending on what strategy I use for NLS.Is
there any mistake in my command lines?
Cheers
Paulo
SP: I am creating and ado file and I have some additional questions.
1- The reason i ask about nl is because I had some problems
introducing a nl programme in an ado file. is it illegal to use an nl
programme? The structure of my ado file is as follows:
program define estim1
...
nl equ z
...
end
program define nl equ
...
end
2- Is illegal to name a programme if the name start with numbers?
(i.e.: a programme named 2sls)
*
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