Is the assumption that:

$E(e_i|x_i) = 0$

And the two assumptions that:

$cov(x_i, e_i)=0$ $E(e_i)=0$

Equivalent? (I have seen both formulation of the assumptions).

The two are certainly not mathematically equivalent, since covariance can be zero even if there are conditional errors. Such process would lead to OLS being inefficient, but not biased (since the process would increase the variance). So is it not the case that the first assumption is in fact the correct one?

But then, if $cov(x, e)=0$ and $E(e_i|x_i) != 0$, I presume that would lead to autocorrelation. Thus, the assumption that $E(e_i|x_i) = 0$ should replace the autocorrelation assumption as well. That means that the assumptions are more coherent using the 2nd formulation as no condition is stated twice. Adding the autocorrelation assumption to the bottom assumptions would also make the two mathematically equivalent, since there is no way I can think of that $E(e_i|x_i)$ could be unequal to zero without one of the listed assumptions also being unequal to zero. Am I correct in this?

So according to the above we should have:


$E(e_i|x_i) = 0$


$cov(x_i, e_i)=0,$ $E(e_i)=0$, $cov(e_i, e_j)= E(e_i, e_j) = 0$.

And as such the Markov assumptions that use the $E(e_i|x_i)$ formulation should not have the autocorrelation assumption as it is redundant.

  • $\begingroup$ Related: stats.stackexchange.com/questions/190703/non-linear-endogeneity/… $\endgroup$ Feb 26, 2016 at 5:21
  • 1
    $\begingroup$ Note that in the classical Gauss-Markov theorem (en.wikipedia.org/wiki/Gauss%E2%80%93Markov_theorem) the $x_i$ are not stochastic. So $E(e_i|x_i)=E(e_i)$. If we move to stochastic $x_i$, then $E(e_i|x_i)=0$ is essential, because it guarantees unbiasedness of the OLS estimate, and the Gauss-Markov theorem states that OLS is BLUE, i.e. it is efficient among unbiased estimators. The autocorrelation in this case is irrelevant, as there is a variant of Gauss-Markov theorem in the general case when covariance matrix of regression disturbances is any positive-definite matrix. $\endgroup$
    – mpiktas
    Feb 26, 2016 at 9:38
  • $\begingroup$ @mpktas Indeed, these are the modified assumptions (two different ways of presenting them). The $cov(x_i,e_i)=0$ relaxes the stochasticity assumption just the same. What I want to know is whether the assumption $E(e_i|x_i)$ can relax the autocorrelation assumption as well (as it seems to me it's included in the assumption). $\endgroup$
    – Dole
    Feb 26, 2016 at 10:41

1 Answer 1


No, they are not equivalent. $E(disturbance|x)$ {not residual($e$) but actual error($\epsilon$), check your question} is conditional expectation of the error term and does not need to assume the deterministic nature of matrix $X$. It is one of Gauss Markov's assumption that makes OLS estimates BLUE (Best Linear Unbiased Estimator). Co-variance of $X$ with residual($e$) is a property of OLS, which does not need any assumption. And can be easily proven by the following:

Let the slope estimate be $\hat{\beta}$; After taking derivative of SSE, $e'e$, and equating it to zero we get:

$$\hat{\beta} = inv[(X'X)]X'Y$$

$$(X'X)\hat{\beta} = X'Y$$

Putting $Y = (X)\hat{\beta} + (e)$ in above equation;

$$(X'X)\hat{\beta} = x'[(X)\hat{\beta} + e]$$

$$(X'X)\hat{\beta} = (X'X)\hat{\beta} + X'e$$

Till here we haven't made any assumptions (except $X'X$ is invertible, which might arise due to perfect multicollenearity or trivially because number of observations is less than number of variables).

Therefore from the above equation we get that $X'e = 0$ This will always be true for OLS. Now this property leads to derivation of several other properties of OLS.

$E(e)$ is one of them, which is true if the regression contains a constant.

Proof: If the regression contains a constant, first column of $X$ will be vector of $1$'s and $X'e = 0$, $\therefore 1*e(1) + 1*e(2)\dots 1*e(n) = 0$ Therefore, meaning $Sum(e) = 0$, dividing by non-zero $n$, we get $E(e) = 0$

Therefore these two are properties and the above one is a Gauss Markov assumption for BLUE.

The no-autocorrelation and homoscedasticity comes from another Gauss Markov assumption that;

$$E(\epsilon \epsilon'|X) = (\sigma^2) \text{Identity}$$

$E(\epsilon \epsilon'|X)$ gives the covariance matrix of disturbances terms conditional on $X$. This matrix is reduced to the term on right hand side by applying two assumptions,

  1. No autocorrelation hence Covariance between errors is zero, therefore making non diagonal terms zero.

  2. Variance of error term is constant (homoscedasticity), $\therefore \epsilon^2= \sigma^2$.

An excellent source can be found at: https://web.stanford.edu/~mrosenfe/soc_meth_proj3/matrix_OLS_NYU_notes.pdf

Thanks Anurag

  • $\begingroup$ Thanks for an answer to a years old question! Indeed, they are not equivalent, the errors can be correlated even if the expectation is zero, which breaks the equivalence. $\endgroup$
    – Dole
    Feb 26, 2017 at 22:20

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