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From |
"Paulo Regis" <pauloregis.ar@googlemail.com> |

To |
statalist@hsphsun2.harvard.edu |

Subject |
st: Alternative uses of -nl |

Date |
Tue, 18 Nov 2008 23:54:34 +0800 |

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) * * For searches and help try: * http://www.stata.com/help.cgi?search * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/

**Follow-Ups**:**Re: st: Alternative uses of -nl***From:*"Brian P. Poi" <bpoi@stata.com>

**st: RE: Alternative uses of -nl***From:*"Martin Weiss" <martin.weiss1@gmx.de>

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