Dear all,
I would like to compare the R-squared of a log log model and a linear model to find out which has the better fit. Is there a tool in Stata with which I can compare the R-square of the log log model with the R-square obtained from OLS estimation of the linear model? Natalie * * 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/ |
--- On Thu, 17/6/10, Natalie Trapp wrote:
> I would like to compare the R-squared of a log log model > and a linear model to find out which has the better fit. Is > there a tool in Stata with which I can compare the R-square > of the log log model with the R-square obtained from OLS > estimation of the linear model? Comparing R-squares only makes sense when you don't change the dependent variable: the proportion of variance explained depends both the how much you explain and on how much variance you had to begin with. A non-linear transformation like taking the logarithm will influence the variance of your dependent variable, making the R-squares of the linear model and the log-log model incomparable. Hope this helps, Maarten -------------------------- Maarten L. Buis Institut fuer Soziologie Universitaet Tuebingen Wilhelmstrasse 36 72074 Tuebingen Germany http://www.maartenbuis.nl -------------------------- * * 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/ |
Thank you Maarten.
That's right, an R-square comparison is meaningful only if the dependent variable is the same for both models. 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. On 6/17/2010 11:49 AM, Maarten buis wrote: > --- On Thu, 17/6/10, Natalie Trapp wrote: > >> I would like to compare the R-squared of a log log model >> and a linear model to find out which has the better fit. Is >> there a tool in Stata with which I can compare the R-square >> of the log log model with the R-square obtained from OLS >> estimation of the linear model? >> > Comparing R-squares only makes sense when you don't change > the dependent variable: the proportion of variance explained > depends both the how much you explain and on how much variance > you had to begin with. A non-linear transformation like taking > the logarithm will influence the variance of your dependent > variable, making the R-squares of the linear model and the > log-log model incomparable. > > Hope this helps, > Maarten > > -------------------------- > Maarten L. Buis > Institut fuer Soziologie > Universitaet Tuebingen > Wilhelmstrasse 36 > 72074 Tuebingen > Germany > > http://www.maartenbuis.nl > -------------------------- > > > > > * > * 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/ > > * * 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/ |
there have been attempts in Stata; in my opinion the best of these is
-brsq- from an old STB (type -findit brsq-); of course, as one of the authors, I'm undoubtedly somewhat biased; look carefully at the STB article to ensure it does what you want and to see some references to other attempts Rich On 6/17/10 6:01 AM, Natalie Trapp wrote: > Thank you Maarten. > > That's right, an R-square comparison is meaningful only if the dependent > variable is the same for both models. > > 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. > > On 6/17/2010 11:49 AM, Maarten buis wrote: >> --- On Thu, 17/6/10, Natalie Trapp wrote: >> >>> I would like to compare the R-squared of a log log model >>> and a linear model to find out which has the better fit. Is >>> there a tool in Stata with which I can compare the R-square >>> of the log log model with the R-square obtained from OLS >>> estimation of the linear model? >>> >> Comparing R-squares only makes sense when you don't change >> the dependent variable: the proportion of variance explained >> depends both the how much you explain and on how much variance >> you had to begin with. A non-linear transformation like taking >> the logarithm will influence the variance of your dependent >> variable, making the R-squares of the linear model and the >> log-log model incomparable. >> >> Hope this helps, >> Maarten >> >> -------------------------- >> Maarten L. Buis >> Institut fuer Soziologie >> Universitaet Tuebingen >> Wilhelmstrasse 36 >> 72074 Tuebingen >> Germany >> >> http://www.maartenbuis.nl >> -------------------------- * 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/ |
Thank you very much!
On 6/17/2010 12:10 PM, Richard Goldstein wrote: > there have been attempts in Stata; in my opinion the best of these is > -brsq- from an old STB (type -findit brsq-); of course, as one of the > authors, I'm undoubtedly somewhat biased; look carefully at the STB > article to ensure it does what you want and to see some references to > other attempts > > Rich > > On 6/17/10 6:01 AM, Natalie Trapp wrote: > >> Thank you Maarten. >> >> That's right, an R-square comparison is meaningful only if the dependent >> variable is the same for both models. >> >> 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. >> >> On 6/17/2010 11:49 AM, Maarten buis wrote: >> >>> --- On Thu, 17/6/10, Natalie Trapp wrote: >>> >>> >>>> I would like to compare the R-squared of a log log model >>>> and a linear model to find out which has the better fit. Is >>>> there a tool in Stata with which I can compare the R-square >>>> of the log log model with the R-square obtained from OLS >>>> estimation of the linear model? >>>> >>>> >>> Comparing R-squares only makes sense when you don't change >>> the dependent variable: the proportion of variance explained >>> depends both the how much you explain and on how much variance >>> you had to begin with. A non-linear transformation like taking >>> the logarithm will influence the variance of your dependent >>> variable, making the R-squares of the linear model and the >>> log-log model incomparable. >>> >>> Hope this helps, >>> Maarten >>> >>> -------------------------- >>> Maarten L. Buis >>> Institut fuer Soziologie >>> Universitaet Tuebingen >>> Wilhelmstrasse 36 >>> 72074 Tuebingen >>> Germany >>> >>> http://www.maartenbuis.nl >>> -------------------------- >>> > * > * 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/ > > * * 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/ |
In reply to this post by Natalie Trapp
<>
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. Kit Kit Baum | Boston College Economics & DIW Berlin | http://ideas.repec.org/e/pba1.html An Introduction to Stata Programming | http://www.stata-press.com/books/isp.html An Introduction to Modern Econometrics Using Stata | http://www.stata-press.com/books/imeus.html * * 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/ |
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_xb=(totexp-xb)^2/1e6 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 <[hidden email]> 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/ |
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 <[hidden email]>: > 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 <[hidden email]> 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/ |
<> Shorthand for a "million", I would say: ************* di 1e6 ************* HTH Martin -----Ursprüngliche Nachricht----- Von: [hidden email] [mailto:[hidden email]] Im Auftrag von Joao Ricardo F. Lima Gesendet: Samstag, 19. Juni 2010 19:11 An: [hidden email] 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 <[hidden email]>: > 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 <[hidden email]> 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 >> 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/ * * 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/ |
In reply to this post by Joao Ricardo F. Lima
Martin,
exactly! Thx! JL 2010/6/19 Martin Weiss <[hidden email]>: > > <> > > Shorthand for a "million", I would say: > > ************* > di 1e6 > ************* > > > > HTH > Martin > > > -----Ursprüngliche Nachricht----- > Von: [hidden email] > [mailto:[hidden email]] Im Auftrag von Joao Ricardo F. > Lima > Gesendet: Samstag, 19. Juni 2010 19:11 > An: [hidden email] > 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 <[hidden email]>: >> 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 <[hidden email]> 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/ > > > * > * 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/ |
In reply to this post by Natalie Trapp
It has been awhile since I looked at this but recall a technique from grad school using the geometric mean. For your dependent variable y, create a log variable
gen lny = log(y) get the average of this su lny create a transform for y gen ty = y/exp(r(mean)) where r(mean) is the mean of the log y's and exp converts it to the geometric mean create the log of the ty gen ln_ty = log(ty) now regress ty and ln_ty separately and compare the standard errors of the regressions; they are directly comparable on the transformed dependent variable For references on this see S. Wiesberg Applied Linear Regression, 2nd ed, 1985, sec 6.4 especially p. 148 and for a more advanced explanation...Cook and Weisberg, Residuals and Influence in Regression, 1982, sec 2.4 Regards...Jay Tuthill -----Original Message----- From: [hidden email] [mailto:[hidden email]] On Behalf Of Natalie Trapp Sent: Thursday, June 17, 2010 7:28 AM To: [hidden email] Subject: Re: st: Comparison of the R-squared in a loglog and linear model Thank you very much! On 6/17/2010 12:10 PM, Richard Goldstein wrote: > there have been attempts in Stata; in my opinion the best of these is > -brsq- from an old STB (type -findit brsq-); of course, as one of the > authors, I'm undoubtedly somewhat biased; look carefully at the STB > article to ensure it does what you want and to see some references to > other attempts > > Rich > > On 6/17/10 6:01 AM, Natalie Trapp wrote: > >> Thank you Maarten. >> >> That's right, an R-square comparison is meaningful only if the dependent >> variable is the same for both models. >> >> 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. >> >> On 6/17/2010 11:49 AM, Maarten buis wrote: >> >>> --- On Thu, 17/6/10, Natalie Trapp wrote: >>> >>> >>>> I would like to compare the R-squared of a log log model >>>> and a linear model to find out which has the better fit. Is >>>> there a tool in Stata with which I can compare the R-square >>>> of the log log model with the R-square obtained from OLS >>>> estimation of the linear model? >>>> >>>> >>> Comparing R-squares only makes sense when you don't change >>> the dependent variable: the proportion of variance explained >>> depends both the how much you explain and on how much variance >>> you had to begin with. A non-linear transformation like taking >>> the logarithm will influence the variance of your dependent >>> variable, making the R-squares of the linear model and the >>> log-log model incomparable. >>> >>> Hope this helps, >>> Maarten >>> >>> -------------------------- >>> Maarten L. Buis >>> Institut fuer Soziologie >>> Universitaet Tuebingen >>> Wilhelmstrasse 36 >>> 72074 Tuebingen >>> Germany >>> >>> http://www.maartenbuis.nl >>> -------------------------- >>> > * > * 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/ > > * * 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/ * * 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/ |
--- On Mon, 21/6/10, Jay Tuthill wrote:
> It has been awhile since I looked at > this but recall a technique from grad school using the > geometric mean. For your dependent variable y, create a log > variable Another reference is: Roger Newson (2003) "Stata tip 1: The eform() option of regress" The Stata Journal, 3(4):445. http://www.stata-journal.com/article.html?article=st0054 The real question is whether you really want to know how the geometric mean changes when an explanatory variable changes, or whether you want to know how the arithmatic (regular) mean changes when an explanatory variable changes... If you care about the latter then your best choice is -glm- with the -line(log)- option, see the article by Nick et al. I refered to earlier today: <http://www.stata.com/statalist/archive/2010-06/msg01168.html>. Hope this helps, Maarten -------------------------- Maarten L. Buis Institut fuer Soziologie Universitaet Tuebingen Wilhelmstrasse 36 72074 Tuebingen Germany http://www.maartenbuis.nl -------------------------- * * 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/ |
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