AW: st: Comparison of the R-squared in a loglog and linear model

["Martin Weiss" <martin.weiss1@gmx.de>]

<> 

Shorthand for a "million", I would say:

*************
di 1e6
*************



HTH
Martin


-----Ursprüngliche Nachricht-----
Von: owner-statalist@hsphsun2.harvard.edu
[mailto:owner-statalist@hsphsun2.harvard.edu] Im Auftrag von Joao Ricardo F.
Lima
Gesendet: Samstag, 19. Juni 2010 19:11
An: statalist@hsphsun2.harvard.edu
Betreff: Re: st: Comparison of the R-squared in a loglog and linear model

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.
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>



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Joao Ricardo Lima, D.Sc.
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