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  • 1.  Use R-squared for nonlinear regression?

    Posted Mon January 31, 2022 07:27 PM
    Edited by System Test Fri January 20, 2023 04:13 PM
    I'm a novice trying to learn statistics.

    I want to measure the accuracy of an exponential regression (the model is of the form y=a+b*exp(cx) ).

    For nonlinear regression, such as exponential regression, is R-squared a valid measure of accuracy?
    • For what it's worth, I did find a blog that says the answer is noR-squared isn't valid for nonlinear regression. With that said, I'm hoping to get some more  input on this subject...rather than base my analysis on a single blog posting...I'm looking for additional guidance.

    Thanks!
    #SPSSStatistics


  • 2.  RE: Use R-squared for nonlinear regression?

    Posted Mon January 31, 2022 07:57 PM
    Well, yes and no.  In general, with nonlinear regression, the residual mean at the minimum may not be zero, which is assumed by the usual calculation, so this can throw things off a lot.

    However, with your example, I wonder whether you really meant a to be both the intercept and the coefficient of the exponent.  That would be very unusual and might cause estimation to fail.

    If those two should really be different parameters, then the model isn't essentially nonlinear.  You could take the log of both sides and estimate it as an ordinary linear model.  In its current form, you could either use Curve Fit or nonlinear regression to estimate it.

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  • 3.  RE: Use R-squared for nonlinear regression?

    Posted Mon January 31, 2022 08:12 PM

    Thanks. Sorry, there was something funky going on with the equation that I wrote in the original post. I tried to update/correct it just now.

    Also, it occurred to me that my original post might not have had enough information.
    Here's some background info, from back when I was using Excel: Tune Exponential Regression.




  • 4.  RE: Use R-squared for nonlinear regression?

    Posted Mon January 31, 2022 09:00 PM
    So
    21 - y = e**ax
    ln(21-y) = ax
    so this is linear in parameters, which is what regression cares about. 

    compute yy = 21 - y.
    REGRESSION
      /STATISTICS COEFF OUTS R ANOVA
      /ORIGIN 
      /DEPENDENT yy
      /METHOD=ENTER x.


    image.png
    image.png

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