Jonathan Zaozao Zhang writes,

For the dataset in my research, I am currently trying to compare the fit between a linear (y=a+bx) and a nonlinear model y=(a0+a1*x)/(1-a2*x).

The question is: For the goodness of fit, can I compare R-squared values?(I doubt it… Also, the nls command in R does not give R-squared value for the nonlinear regression) If not, why not? and what would be a common goodness of fit measure that can be used for such comparsion?

My response: first off, you can compare the models using the residual standard deviations. R^2 is ok too, since that’s just based on the residual sd divided by the data sd. Data sd is same in 2 models (since you’re using the same dataset), so comparing R^2 is no different than comparing residual sd.

Even simpler, I think, is to note that model 2 includes model 1 as a special case. If a2=0 in model 2, you get model 1. So you can just fit model 2 and look at the confidence interval for a2 to get a sense of how close you are to model 1.

Continuing on this theme, I’d graph the fitted model 2 as a curve of E(y) vs x, showing a bunch of lines indicating inferential uncertainty in the fitted regression curve. Thien you can see the fitted model and related possibilities, and see how close it is to linear.

You want to find if your data need a more complex model or not.

What I'd try is bayesian Model comparison:

ie computing P(Model1 | Data) / P(Model2 | Data)

assuming flat prior P(M1)=P(M2) and a, say, a gaussian noise P( data_i | M1 ) = gauss( yi ; a xi +b; sigma) and P(di | M2 )= gauss( yi ; (a0+a1*xi)/(1-a2*xi) ; sigma).

This will require marginalising on parameter space, which is a 2D space for M1 and a 3D for M2.

This will favor the simpler model in case of equal fit.