Statsmodels: Calculate fitted values and R squared

Question

I am running a regression as follows (df is a pandas dataframe):

import statsmodels.api as sm
est = sm.OLS(df[\'p\'], df[[\'e\', \'         


        

Answer 1

If you do not include an intercept (constant explanatory variable) in your model, statsmodels computes R-squared based on un-centred total sum of squares, ie.

tss = (ys ** 2).sum()  # un-centred total sum of squares

as opposed to

tss = ((ys - ys.mean())**2).sum()  # centred total sum of squares

as a result, R-squared would be much higher.

This is mathematically correct. Because, R-squared should indicate how much of the variation is explained by the full-model comparing to the reduced model. If you define your model as:

ys = beta1 . xs + beta0 + noise

then the reduced model can be: ys = beta0 + noise, where the estimate for beta0 is the sample average, thus we have: noise = ys - ys.mean(). That is where de-meaning comes from in a model with intercept.

But from a model like:

ys = beta . xs + noise

you may only reduce to: ys = noise. Since noise is assumed zero-mean, you may not de-mean ys. Therefore, unexplained variation in the reduced model is the un-centred total sum of squares.

This is documented here under rsquared item. Set yBar equal to zero, and I would expect you will get the same number.