[R] Query on R-squared correlation coefficient for linear regression through origin
Eric Berger
er|cjberger @end|ng |rom gm@||@com
Thu Sep 27 14:50:23 CEST 2018
See also this thread in stats.stackexchange
https://stats.stackexchange.com/questions/26176/removal-of-statistically-significant-intercept-term-increases-r2-in-linear-mo
On Thu, Sep 27, 2018 at 3:43 PM, J C Nash <profjcnash using gmail.com> wrote:
> This issue that traces back to the very unfortunate use
> of R-squared as the name of a tool to simply compare a model to the model
> that
> is a single number (the mean). The mean can be shown to be the optimal
> choice
> for a model that is a single number, so it makes sense to try to do better.
>
> The OP has the correct form -- and I find no matter what the software, when
> working with models that do NOT have a constant in them (i.e., nonlinear
> models, regression through the origin) it pays to do the calculation
> "manually". In R it is really easy to write the necessary function, so
> why take a chance that a software developer has tried to expand the concept
> using a personal choice that is beyond a clear definition.
>
> I've commented elsewhere that I use this statistic even for nonlinear
> models in my own software, since I think one should do better than the
> mean for a model, but other workers shy away from using it for nonlinear
> models because there may be false interpretation based on its use for
> linear models.
>
> JN
>
>
> On 2018-09-27 06:56 AM, Patrick Barrie wrote:
> > I have a query on the R-squared correlation coefficient for linear
> > regression through the origin.
> >
> > The general expression for R-squared in regression (whether linear or
> > non-linear) is
> > R-squared = 1 - sum(y-ypredicted)^2 / sum(y-ybar)^2
> >
> > However, the lm function within R does not seem to use this expression
> > when the intercept is constrained to be zero. It gives results different
> > to Excel and other data analysis packages.
> >
> > As an example (using built-in cars dataframe):
> >> cars.lm=lm(dist ~ 0+speed, data=cars) # linear regression through
> > origin
> >> summary(cars.lm)$r.squared # report R-squared [1] 0.8962893 >
> > 1-deviance(cars.lm)/sum((cars$dist-mean(cars$dist))^2) # calculates
> > R-squared directly [1] 0.6018997 > # The latter corresponds to the value
> > reported by Excel (and other data analysis packages) > > # Note that we
> > expect R-squared to be smaller for linear regression through the origin
> > > # than for linear regression without a constraint (which is 0.6511 in
> > this example)
> >
> > Does anyone know what R is doing in this case? Is there an option to get
> > R to return what I termed the "general" expression for R-squared? The
> > adjusted R-squared value is also affected. [Other parameters all seem
> > correct.]
> >
> > Thanks for any help on this issue,
> >
> > Patrick
> >
> > P.S. I believe old versions of Excel (before 2003) also had this issue.
> >
>
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