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At the start, please forgive me if my question is too elementary.

I am fitting a non-linear curve. Say a parabola. The data points I have are close to a parabola and the best output I get is a parabola. I want to quantify the quality if fit. Something like an R-Squared metric. I was wondering if the R-Squared metric, like in the case of a linear OLS, makes sense since one of the inputs for R-squared is the average of the input values, which I'm not sure makes sense for a parabola.

Can someone please help?

nimbus3000
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2 Answers2

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OLS quadratic model: $y = \beta_0 + \beta_1 X + \beta_2 X^2$

Your model is still a linear function of the unknown parameters $\beta$ with the features $X$ and $X^2$. Hence $R^2$ is still applicable.

Kane Chua
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  • The equation "y = B0 + B1 * sin(x) + B2 * log(x)" is linear in the coefficients and can be fit using linear algebra just as can be done with quadratic polynomials. Would you consider R-squared applicable in this case? Both this example equation and a quadratic polynomial can also be fit using non-linear regression, in both of those cases would R-squared be applicable? My understanding is yes for these questions. – James Phillips May 25 '19 at 15:52
  • So (non-)linearity is about the coefficients $ \beta_i $ and not about $ X $ ? – Ben Jul 31 '20 at 09:18
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I calculate R-squared (R2) as "R2 = 1.0 - (regression_error_variance / dependent_data_variance)" and use it to tell me the fraction of the dependent data variance that is explained by the regression model. For non-linear equations this is both approximate and useful. For example if the R-squared value from an exponential regression is 0.75, I interpret this as meaning that the fitted equation explains 75 percent of the dependent data variance. In the case of your parabola example, the model is not a straight line and so the R-squared value is also both approximate and useful. My understanding is that R-squared is only exact for straight lines.

James Phillips
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