0
$\begingroup$

Let $y = X\beta + u$ be a regression model. If we assume $\mathbb V[u|X] = \sigma^2$ then does this imply $\mathbb E[u|X] = c$? Clearly, $\mathbb V[u|X] = \mathbb E[uu'|X] - \mathbb E[u|X]\mathbb E[u|X]'$, so $\mathbb E[uu'|X] = \sigma^2 + \mathbb E[u|X]\mathbb E[u|X]'$. Since both, $\mathbb E[uu'|X]$ and $\mathbb E[u|X]\mathbb E[u|X]'$ can vary, $\mathbb E[u|X]$ need not be constant. But what would be a counter example?

If we assume iid observations and only a single regressor, it suffices to regard a single $u_i$. So, $E[u_i^2|x_i] = \sigma^2 + \mathbb E[u_i|x_i]^2$. If we take the derivative wrt to $x_i$, we find $D\mathbb E[u_i^2|x_i] = 2D\mathbb E[u_i|x_i]$ so I thought about assuming $\mathbb E[u_i|x_i] = 0.5x_i$ and $\mathbb E[u^2_i|x_i] = x_i$. Is this reasoning correct or do I miss a point?

$\endgroup$

1 Answer 1

1
$\begingroup$

$\mathbb E[u|X]$ can be whatever you want. The variance assumption alone does not imply it must be anything. Most commonly, $\mathbb E[u|X]$ is assumed to be $0$. If this was not done, and instead you assumed it was equal to some parameter to be estimated, like $c$, then all of the parameters would probably not be identifiable, and they would be less interpretable.

If you assume $\mathbb E[u_i|x_i] = 0.5x_i$ then you can write your model as $$ y_i = \beta_0 + \beta_1 x_i + (.5x_i + \sigma z_i) $$ where $z_i$ is a standard normal variate. The parameters $(\beta_0, \beta_1, .5)$ yield the same likelihood as $(\beta_0, \beta_1 + .5, 0)$. You are not estimating $.5$, so technically your model is still identifiable; but I don't see any advantage to be gained from not following the convention of assuming your noise is mean zero.

$\endgroup$
2
  • $\begingroup$ Thanks for you reply. Maybe I was not precise enough. My question concerns whether $\mathbb E[u^2|X] = const$ implies $\mathbb E[u|X] = const$. Certainly this is not true but I wanted to provide a counter example (that is why I chose the above functional form) $\endgroup$ Commented Sep 22, 2019 at 20:58
  • $\begingroup$ @SydAmerikaner if you want a counterexample, just pick any (full rank) normal distribution. This family is parameterized by a mean and a variance. The parameter space is rectangular, so the mean can be anything while the variance can be anything positive. $\endgroup$
    – Taylor
    Commented Sep 23, 2019 at 1:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.