I Googled this, but some answers still leave me confused. One of the answers says:

"I just think of it in terms of the fitted line having 2 parameters, the intercept and the slope. But the intercept doesn't count after we adjust for the mean (which we do in obtaining SStotal), so that leaves just one df for regression."

How can you drop the intercept after adjust for the mean?


If you have a model $$y=x\beta_x+\beta_0$$ No matter what you do this holds for the linear model $$E[y]=E[x]\beta_x+\beta_0$$ Hence $$E[y]-E[x]\beta_x=\beta_0$$

For a given data set $x,y$ when you pick any given $\beta_x$, it constrains $\beta_0$ to be $\bar y-\bar x \beta_x$. That's why you really have only one degree of freedom, the other coefficient is not free. That's why in econometrics sometimes non-intercept variables are called bona fide regressors to contrast them to the intercept.

  • $\begingroup$ Well, it's only not free in the sense that it's necessarily related to $\bar{y}$; if you condition on $\bar{y}$, it's not free. But regression doesn't condition on $\bar{y}$, which is a random variable, so it seems an odd thing to insist on (at least bolding it makes it sound insistent). $\endgroup$ – Glen_b Apr 3 '16 at 2:16

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