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From |
"Joao Ricardo F. Lima" <jricardofl@gmail.com> |

To |
statalist@hsphsun2.harvard.edu |

Subject |
Re: st: Comparison of the R-squared in a loglog and linear model |

Date |
Sat, 19 Jun 2010 14:11:25 -0300 |

Austin, the question is not my opinion to the thread. I only don't understand this part of the code: g mse_xb=(totexp-xb)^2/1e6 What's -1e6-?? Thx a lot, Joao Lima 2010/6/18 Austin Nichols <austinnichols@gmail.com>: > Kit et al.-- > Duan's smearing method is one approach to dealing with a logged > depvar; a better approach is to use a regression technique that > respects the functional form, like -poisson- (or another member of the > -glm- family). But you still cannot compare the R-squared across > non-nested models and hope to conclude anything about which model is > better from that information alone. Mean squared prediction error in > levels for the nonzero outcomes seems a reasonable criterion for > rejecting the log(y) regression model below. > > use http://fmwww.bc.edu/ec-p/data/mus/mus03data, clear > qui reg totexp suppins phylim actlim totchr age female income > predict xb > qui reg ltotexp suppins phylim actlim totchr age female income > levpredict tenorm > levpredict teduan, duan print > qui poisson totexp suppins phylim actlim totchr age female income > predict tepois > qui nbreg totexp suppins phylim actlim totchr age female income > predict tenbreg > su totexp xb te* > su totexp xb te* if totexp>0 > corr totexp xb te* > > g mse_norm=(totexp-tenorm)^2/1e6 > g mse_duan=(totexp-teduan)^2/1e6 > g mse_pois=(totexp-tepois)^2/1e6 > g mse_nbreg=(totexp-tenbreg)^2/1e6 > su mse* > su mse* if totexp>0 > > Variable | Obs Mean Std. Dev. Min Max > -------------+-------------------------------------------------------- > mse_xb | 2955 127.0504 642.6503 .00005 12779.11 > mse_norm | 2955 142.4353 641.0374 3.32e-06 11744.09 > mse_duan | 2955 140.7604 644.1605 .0000549 11842.16 > mse_pois | 2955 128.3255 648.1356 4.52e-06 12841.78 > mse_nbreg | 2955 131.8694 642.3027 2.48e-06 12432.65 > > For those enamored of scatter plots for this kind of comparison, much > more work is required to get a good picture of fit. This is one > approach: > > g cr_te=totexp^(1/3) > g cr_xb=sign(xb)*abs(xb)^(1/3) > g cr_norm=tenorm^(1/3) > g cr_duan=teduan^(1/3) > g cr_pois=tepois^(1/3) > g cr_nbreg=tenbreg^(1/3) > sc cr_* cr_te if totexp>0, msize(1 1 1 1 1 1) > > On Fri, Jun 18, 2010 at 9:47 AM, Christopher Baum <kit.baum@bc.edu> wrote: >> <> >> On Jun 18, 2010, at 2:33 AM, Natalie wrote: >> >>> Can I not maybe obtain the antilog predicted values for the log log >>> model and compute the R-squared between the antilog of the observed and >>> predicted values. And then compare this R-square with the R-square >>> obtained from OLS estimation of the linear model? >>> >>> There are other statistical programs that can do this automatically, but >>> as I work with Stata, I'd rather do it with this program. >> >> >> findit levpredict >> >> Generate the level form of the dependent variable (correctly, using this routine) and then >> compute the squared correlation between that and the original level variable. That will be the >> R^2 of the log form of the regression. > * > * 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/ > -- ---------------------------------------- Joao Ricardo Lima, D.Sc. Professor UFPB-CCA-DCFS Fone: +558387264913 Skype: joao_ricardo_lima ---------------------------------------- * * 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**:**AW: st: Comparison of the R-squared in a loglog and linear model***From:*"Martin Weiss" <martin.weiss1@gmx.de>

**References**:**Re: st: Comparison of the R-squared in a loglog and linear model***From:*Christopher Baum <kit.baum@bc.edu>

**Re: st: Comparison of the R-squared in a loglog and linear model***From:*Austin Nichols <austinnichols@gmail.com>

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