I am trying to fit a polynomial function to some multi-dimensional data. However, I am getting a negative $R_{adj}^2$ as you can see:

datat = {{2.0000, 35.229, 2.61922, 1.26667, 5.9160*10^6, 14217., 
   0.04422}, {2.0000, 40.522, 2.70379, 9.76264, 2.6300*10^6, 5202.0, 
   0.039375}, {1.0000, 24.3, 2.62143, 5.0000, 3.4141*10^6, 8163.0, 
   0.042542}, {2.0000, 48.87, 2.66604, 15.000, 4.8240*10^6, 10408., 
   0.040983}, {2.0000, 41.59, 2.67677, 2.47553, 2.3944*10^6, 5040.0, 
   0.042063}, {2.0000, 39.629, 2.99468, 1.18333, 5.8163*10^6, 5888.0, 
   0.039648}, {2.0000, 39.618, 2.6657, 1.8, 2.3768*10^6, 5132.0, 
   0.038273}, {2.0000, 42.924, 2.68517, 1.2557, 2.4082*10^6, 4972.0, 
   0.041902}, {1.0000, 32.8, 2.48408, 5.0000, 1.8193*10^7, 59681., 
lmf = LinearModelFit[datat, {1}~Join~{o, g, p, d, a, t}, {o, g, p, d, a, t}];
(*out: -0.511164*)

By definition, in 2-dimensional space an Adjusted $R^2<0$ indicates that the predicted line is worse at fitting the data than an horizontal line (if I'm not mistaken). In this 6-dimensional case, I believe a negative $R^2_{adj}$ indicates that the hyperplane is worse at describing the data than a random hyperplane.

So, my questions are:

  1. How is this possible? I would expect that this could happen if I didn't have a constant term in the fitting function (indicated by the {1} term in lmf);
  2. Are there other algorithms that I can use instead of the default regression one that maybe could solve this problem? (i.e. to get $R^2_{adj} = 0$ as lowest coefficient of determination)


  • $\begingroup$ ehm, now that I look at both questions at the bottom, I am wondering if the title is the most appropriate $\endgroup$
    – Sos
    Commented Oct 5, 2015 at 17:45
  • $\begingroup$ Are you really trying to fit "data to a model (polynomial)" - an intriguing endeavor? I do not quite understand, why you joining the {1} to the list of functions; it does not change a thing and by definition an intercept is automatically added (e.g. beta0)? $\endgroup$
    – gwr
    Commented Oct 5, 2015 at 18:06
  • $\begingroup$ @gwr - I meant the other way around. I apologize and will correct that right away. Indeed, I notice that the documentation adds an intercept by default, but having that {1} changes nothing. The main issue remains $\endgroup$
    – Sos
    Commented Oct 5, 2015 at 18:10
  • $\begingroup$ Adjusted $R^2$ might be negative if there are terms in your model that do not help to predict the response, as this page about goodness of fit tells us. If I look at the proposed linear model equation, then $a$ and $t$ have coefficients of approximately zero. $\endgroup$
    – gwr
    Commented Oct 5, 2015 at 18:18
  • 1
    $\begingroup$ And fitting a model with 8 parameters (1 intercept, 6 covariates, and a variance) with 9 data points is, well, kinda silly. $\endgroup$
    – JimB
    Commented Oct 5, 2015 at 23:21

1 Answer 1


Please look at https://stats.stackexchange.com/questions/48703/what-is-the-adjusted-r-squared-formula-in-lm-in-r-and-how-should-it-be-interpret for a fuller explanation of a variety of estimators for “adjusted R2”.

The formula used by Mathematica is $$R_{adj}^2={1-(1-R^2){{n-1}\over{n-p-1}}}$$ where $n$ is the sample size and $p$ is the number of parameters (excluding the intercept and assume for this example we are fitting an intercept). This means that $R_{adj}^2$ will be negative whenever $R^2<p/(n-1)$. So as the ratio of the number of parameters relative to the sample size gets large, one will need larger and larger values of $R^2$ to result in non-negative values of $R_{adj}^2$.

I understand from your comment above that you provided an example with a small sample size to show or question that one can obtain negative values. However, it is the use of a small sample size relative to the number of parameters to be fit that results in such extreme negative values of $R_{adj}^2$.

To explicitly answer your questions:

(1) By definition $R_{adj}^2$ will be negative whenever $R^2<p/(n-1)$ (when there is an intercept in the model).

(2) Negative $R_{adj}^2$ values are not a problem. Overfitting is. Further, one should consider looking less at $R^2$’s and more at residual plots and the root mean square error to judge the quality of a fit.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.