Timeline for Unbiasedness of Covariance Matrix Estimator in OLS
Current License: CC BY-SA 4.0
11 events
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| Nov 17, 2021 at 8:22 | history | edited | Christoph Hanck | CC BY-SA 4.0 |
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| Nov 17, 2021 at 7:49 | history | edited | Christoph Hanck | CC BY-SA 4.0 |
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| Nov 17, 2021 at 6:40 | history | edited | Christoph Hanck | CC BY-SA 4.0 |
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| Nov 16, 2021 at 19:39 | comment | added | whuber♦ | The part I did not catch--and now I see you asked for clarification--is whether the model is heteroscedastic or not. If, at the outset, you were to stipulate that you are discussing the heteroscedastic model, there would be less chance of misunderstanding. (+1) | |
| Nov 16, 2021 at 18:30 | comment | added | Christoph Hanck | Yeah, $tr(H)=tr(X(X'X)^{-1}X')tr((X'X)^{-1}X'X)=k$ only depends on $k$, which gives rise to the unbiasedness of the standard estimator $\sum_i\hat e_i^2/(n-k)$ under homoskedasticity, eg stats.stackexchange.com/questions/76738/…. I am not sure how that applies here, though, since the heteroskedasticity robust estimator operates on the product $x_i\hat e_i$, so that we need to work on individual squared residuals, not their sum. See also econstor.eu/bitstream/10419/189084/1/qed_wp_0537.pdf | |
| Nov 16, 2021 at 17:31 | comment | added | whuber♦ | I believe that when you sum the squared residuals, though, you obtain a quantity closely related to the trace of $H,$ which depends only on $n$ and $p,$ regardless of the design, thereby leading to the claim in the question. | |
| Nov 16, 2021 at 17:25 | history | edited | Christoph Hanck | CC BY-SA 4.0 |
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| Nov 16, 2021 at 13:48 | history | edited | Christoph Hanck | CC BY-SA 4.0 |
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| Nov 16, 2021 at 10:38 | history | edited | Christoph Hanck | CC BY-SA 4.0 |
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| Nov 16, 2021 at 10:30 | history | edited | Christoph Hanck | CC BY-SA 4.0 |
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| Nov 16, 2021 at 10:21 | history | answered | Christoph Hanck | CC BY-SA 4.0 |