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?


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:


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.

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  • $\begingroup$ lets say the expected error isn't zero (0>), then the intercept absorbs the non-zero E(u) and the best fit line shifts up to make E(u)=0. Is this correct? $\endgroup$ – S. Ming Aug 26 '16 at 22:16
  • $\begingroup$ Sounds right to me. $\endgroup$ – VCG Aug 26 '16 at 22:23

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