expected value of error term equals zero, formal proof

Question

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?

Answer 1

So we suppose that $E(u)\not=0$, say =3

Now take your original model then add and subtract 3, so that it doesn't change:

$y=\beta_0+3+\beta_1x+(u-3)$

We do this `trick' because we want to manipulate the original equation without changing it.

Now , since $E(u)=3$, we know $E(u-3)=0$

Take our constant $\beta_0$ and transform it into $\beta^*_0=\beta_0+3$

We can do this because the constant just absorbs anything in the regression equation.

Do the same thing to u and we get

$y=\beta^*_0+\beta_1x+u^* $

Estimating this model is identical to estimating the old one, we've just shifted that 3 around between the error and constant term.

The point of the exercise is that we can re-specify it so that is has zero mean, which we want to be able to estimate the constant.