FitResult.CorrelationCoefficient returning R^2 rather than R?

May 22, 2013 at 9:53 PM
So I have the following data:

-0.054357662 150
-0.04830468 170
-0.043465694 190


I used the BivariateSample class to find a linear regression line. The FitResult I got back is returning 0.99584198247835987 as the correlation coefficient between the two variables. If I try to get the correlation coefficient from Excel 2007, I get 0.997935928. Well, 0.997935928 squared is 0.99584198247835987.

It appears that the FitResult.CorrelationCoefficient is actually returning the correlation coefficient squared, rather than the actual value.

Am I missing something? I admit I am not a statistician.
Coordinator
May 24, 2013 at 6:41 AM
If you want to compute the correlation coefficient between X and Y in a BivariateSample s, call s.PearsonRTest().Statistic. (The Pearson R test is a test for correlation and the test statistic is R.) Alternatively, compute s.Covariance / Math.Sqrt(s.X.Variance * s.Y.Variance), which is the definition of R. Doing so on the data you give returns the same value as you quote from Excel. (Yes, we should probably add a property to BivariateSample that gives this directly.)

When you compute s.LinearRegression().CorrelationCoefficient(0, 1), you are computing the correlation between the fitted slope and the fitted intercept. For a small change in the intercept da/a, this number tells you how much you should change the slope db/b in order to maintain a good fit. It just happens for a linear fit to be related to R^2. Notice that you had to give parameter indexes to compute this number. For a more complex fit involving D parameters, there would be D(D-1)/2 of these numbers, but there will always be just one correlation coefficient for a bivariate sample.
May 24, 2013 at 4:28 PM
That makes sense. Now I'm getting the right results. Thanks!