# Why does the degree of freedom of SSR equals 1 in simple linear regression

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.
• 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). – Glen_b Apr 3 '16 at 2:16