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For the given model $y = x\beta + \varepsilon$, the four assumptions would be

  1. The model is linear in $x$ and $\beta$, also additive in $\varepsilon$
  2. The conditional mean of the error terms are zero (i.e. $E[\varepsilon|x] = 0)$
  3. $Var[\varepsilon|x] = \sigma^2I_n$
  4. Exogeneity of $x$ (i.e. either $x$ is fixed or independent of $\varepsilon$)

My questions is that in the case of the following situations which assumptions have been violated.

Case 1. $E[x'\varepsilon] \neq 0$ [I am not sure whether this implies the second or fourth assumption has been violated]

Case 2. $Cov(x_i,\varepsilon_i) \neq 0$ [I suppose the third one has been violated but could it imply the violation of other assumptions?]

Any help will be greatly appreciated.

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Case 1: $$\mathbb{E}(x'\epsilon) = \mathbb{E}(\mathbb{E}(x'\epsilon|x)) = \mathbb{E}(x'\mathbb{E}(\epsilon|x))$$

Therefore, $$\mathbb{E}(\epsilon|x) = 0 \implies \mathbb{E}(x'\epsilon) = 0 $$ Equivalently, $$\mathbb{E}(x'\epsilon) \neq 0 \implies \mathbb{E}(\epsilon|x) \neq 0 $$

Thus, violating assumption (2).

Case 2: If two random variables are independent, then they are uncorrelated i.e. their covariance is 0. Equivalently, if two random variables are correlated, then they are not independent. Therefore, $Cov(x_i, \epsilon)\neq 0$ implies that $x$ is not independent of $\epsilon$ violating assumption (4).

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I assume that x is a scalar.( but argument is similar if it isn't).

They both violate condition 4) because they both say that x, the regressor, is not independent of $\epsilon$. Case 1 a theoretical statement about the regressor and the noise term. Case 2 is an empirical statement about some of the actual regressor data and the noise term. If you want a nice intuitive explanation of econometrics ( not a standard textbook exposition ), I highly recommend Peter Kennedy's text. I think it's called "econometrics"but I'm pretty sure he only wrote one book so you can google for him on amazon.

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  • $\begingroup$ @amit: nice on case 1. but I think case 1 also violates 4) because $E(x^\prime \epsilon)$ is the definition of the covariance of $x$ and $\epsilon$. ( assuming mean of $\epsilon$ = 0 ). thanks for fix on case 1. $\endgroup$ – mark leeds Jun 6 '18 at 5:56

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