I've been trying to figure out why the expected value of the error term equals zero when the intercept is included. I don't understand the formal proof. In my book the following proof is given:
$y = \beta_0 +\beta_1x + u$
Suppose the expectation of $u$ is $3$ instead of $0$, then $E(u-3)=0$. If we add $3$ to the constant term and subtract it from the error term, we obtain:
$y = (\beta_0+3) +\beta_1x+ (u-3)$
Since both equations are equivalent, and since $E(u-3)=0$, then the latter equation can be written in a form that has a zero expectation for the error term:
$y = \beta_0^* +\beta_1x+ u^*$
where $\beta_0^*=\beta_0+3$ and $u^*=u-3$
I have the following questions:
Why do we subtract $3$ from the error term ($E(u-3)=0$)? Why can't we just set $E(u)=3$? I don't understand this specification.
And why is the $3$ added to the intercept?