Timeline for Are Gauss-Markov assumptions: $E(e_i|x_i) = 0$ and $cov(x_i, e_j)=0$ equivalent?
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| Feb 26, 2017 at 22:03 | vote | accept | Dole | ||
| Feb 26, 2017 at 5:39 | answer | added | anuragsodhi | timeline score: 2 | |
| Feb 26, 2016 at 10:41 | comment | added | Dole | @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). | |
| Feb 26, 2016 at 9:38 | comment | added | mpiktas | 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. | |
| Feb 26, 2016 at 8:30 | history | edited | mpiktas | CC BY-SA 3.0 |
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| Feb 26, 2016 at 5:46 | history | edited | Dole | CC BY-SA 3.0 |
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| Feb 26, 2016 at 5:21 | comment | added | Christoph Hanck | Related: stats.stackexchange.com/questions/190703/non-linear-endogeneity/… | |
| Feb 26, 2016 at 5:20 | history | edited | Dole | CC BY-SA 3.0 |
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| Feb 26, 2016 at 5:04 | history | edited | Dole | CC BY-SA 3.0 |
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| Feb 26, 2016 at 4:56 | history | edited | Dole | CC BY-SA 3.0 |
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| Feb 26, 2016 at 4:41 | history | edited | Dole | CC BY-SA 3.0 |
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| Feb 26, 2016 at 4:22 | history | asked | Dole | CC BY-SA 3.0 |