In linear regression model, the means of the errors are assumed to be zero. Furthermore, we can assume either that the errors are uncorrelated and have the same variance, or even that the errors are iid. Note normality isn't assumed on the errors.

What extra properties does assuming errors are iid bring to the OLS estimates, compared to assuming errors are uncorrelated and common variance? Thanks!

  • $\begingroup$ Are you treating the regressors as stochastic or as deterministic? $\endgroup$ Aug 2, 2013 at 21:54
  • $\begingroup$ @AlecosPapadopoulos: determinisitic. $\endgroup$
    – Tim
    Aug 2, 2013 at 23:22

1 Answer 1


The zero-mean, constant and equal variance, and the uncorrelatedness assumptions, are the "white noise" assumption. The i.i.d assumption is stronger and implies the white-noise assumption.

The finite-sample OLS properties do not gain anything if we strengthen our assumptions from white noise to i.i.d.(it is normality that would make a difference). This can be verified if one goes through the proofs of the OLS properties (like unbiasedness and efficiency) without normal error: what is invoked for these properties to be proved is the "white noise" properties only.

Regarding the asymptotic properties of OLS (consistency and asymptotic normality), note that when the regressors $\mathbf X$ are deterministic, the regressors-error sequence $\left(\mathbf X, \varepsilon\right)$ is in all cases a heterogeneous sequence -because the variability of $\mathbf X$ cannot now be treated as the natural variability of a sequence of identical random variables (since $\mathbf X$ does not contain random variables). And this holds irrespective of whether the errors are identically distributed or just white noise. So the "identically distributed errors" assumption does not bring anything new to the asymptotic properties of OLS, nor does it help make these properties more easy to obtain (compared to the "white noise" assumption). The "independent errors" assumption is rather more influential. Without it: for OLS to be consistent and asymptotically normal, we must assume in addition stronger regularity conditions on the higher moments of the distribution, and in all cases, we cannot let the dependence assumption rage unchecked: $\left(\mathbf X, \varepsilon\right)$ must be a "mixing process" (avoiding technicalities, a mixing process consists of dependent variables, but "rather mildly" dependent, which are eventually "asymptotically independent"). Otherwise the asymptotic OLS properties do not survive. So by assuming independent errors, we essentially avoid stating rather technically advanced requirements for the same OLS asymptotic properties to hold.

  • 1
    $\begingroup$ +1 Welcome to our site and thank you for a clear and informative explanation! $\endgroup$
    – whuber
    Aug 4, 2013 at 15:53
  • 1
    $\begingroup$ @whuber Thanks for the welcome. I sure hope to contribute something, and also learn here. $\endgroup$ Aug 4, 2013 at 16:25

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.