Hi R users
I'm trying to fit a model y=ax^b.
I know if I made ln(y)=ln(a)+bln(x) this is a linear regression.
But I obtain differente results with nls() and lm()
My commands are: nls(CV ~a*Est^b, data=limiares, start =list(a=100,b=0),
trace = TRUE) for nonlinear regression
and : lm(ln_CV~ln_Est, data=limiares) for linear
regression
Nonlinear regression model: a=738.2238151 and b=-0.3951013
Linear regression: Coefficients:
Estimate Std. Error t value
Pr(>|t|)
(Intercept) 7.8570224 0.0103680 757.8 <2e-16 ***
ln_Est -0.5279412 0.0008658 -609.8 <2e-16 ***
I think it should be a=exp("(Intercept) ") = exp(7.8570224) =
2583.815
and b=ln_Est
Probably I'm wrong, but why??
Thanks in advance.
Ana Quiterio
[[alternative HTML version deleted]]
> -----Original Message----- > From: r-help-bounces at stat.math.ethz.ch [SMTP:r-help-bounces at stat.math.ethz.ch] On Behalf Of Ana Quit??rio > Sent: Wednesday, January 25, 2006 3:33 PM > To: r-help at stat.math.ethz.ch > Subject: [R] Question about fitting power > >From: Ana Quit??rio > > > > Hi R users > > > > I'm trying to fit a model y=ax^b. > > > > I know if I made ln(y)=ln(a)+bln(x) this is a linear regression. > > > > But I obtain differente results with nls() and lm() > > > > My commands are: > nls(CV ~a*Est^b, data=limiares, start =list(a=100,b=0), trace = TRUE) > for nonlinear regression, and : > lm(ln_CV~ln_Est, data=limiares) > for linear regression > > > > Nonlinear regression model: a=738.2238151 and b=-0.3951013 > > Linear regression: Coefficients: > > Estimate Std. Error t value Pr(>|t|) > > (Intercept) 7.8570224 0.0103680 757.8 <2e-16 *** > ln_Est -0.5279412 0.0008658 -609.8 <2e-16 *** > > > > I think it should be a=exp("(Intercept) ") = > > exp(7.8570224) = 2583.815 and b=ln_Est > > > > Probably I'm wrong, but why?? > > > > Thanks in advance. > > Ana QuiterioIn addition to what Andy said, maybe your nls() model is not natural because a power statistical model should produce multiplicative outcomes. An alternative is to try the nonlinear power model with multiplicative deviates assuming lognormality (which is what the linear model does but without the need for back-transformation), with nlm(), for instance: fn<-function(p){ + CV_mod=p[1]*Est^p[2]; + squdiff=(log(limiares$CV)-log(CV_mod))^2; + lik=(length(limiares$CV)/2)*log(sum(squdiff)/length(limiares$CV)) + } CV.lik<-nlm(fn,p=c(100,0),hessian=TRUE) fec.lik covmat<-solve(CV.lik$hessian) covmat Ruben
The two methods are fitting different models. With lm(), the model is y = a * x^b * error or, equivalently, ln(y) = ln(a) + b * ln(x) + ln(error) With nls(), the model is y = a * x^b + error Thus you will get two different estimates. Andy From: Ana Quit??rio> > Hi R users > > > > I'm trying to fit a model y=ax^b. > > I know if I made ln(y)=ln(a)+bln(x) this is a linear regression. > > > > But I obtain differente results with nls() and lm() > > > > My commands are: nls(CV ~a*Est^b, data=limiares, start > =list(a=100,b=0), > trace = TRUE) for nonlinear regression > > and : lm(ln_CV~ln_Est, > data=limiares) for linear > regression > > > > > > Nonlinear regression model: a=738.2238151 and b=-0.3951013 > > > > Linear regression: Coefficients: > > Estimate Std. Error t value > Pr(>|t|) > > (Intercept) 7.8570224 0.0103680 757.8 <2e-16 *** > > ln_Est -0.5279412 0.0008658 -609.8 <2e-16 *** > > > > > > I think it should be a=exp("(Intercept) ") = > exp(7.8570224) = 2583.815 > and b=ln_Est > > > > Probably I'm wrong, but why?? > > > > > > Thanks in advance. > > > > Ana Quiterio > > > > > > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help at stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! > http://www.R-project.org/posting-guide.html > >