Timeline for Expected value of squared least squares estimator
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 13, 2018 at 21:25 | vote | accept | cabeer | ||
| Oct 13, 2018 at 20:29 | comment | added | whuber♦ | When $K \gt 1,$ it is never the case that $\sum \lambda_k^{-1} = 1/\operatorname{Tr}(X^\prime X).$ | |
| Oct 13, 2018 at 17:35 | answer | added | user158565 | timeline score: 1 | |
| Oct 13, 2018 at 8:39 | comment | added | cabeer | Thank you very much for the answer and explanation! If you post this as an answer I happily accept it. :-) | |
| Oct 13, 2018 at 0:21 | comment | added | user158565 | In $\beta'(X'X)^{-1}(X'\epsilon)$, only $\epsilon$ is random, $X$ and $\beta$ are constant. So $E(\beta'(X'X)^{-1}(X'\epsilon)) = \beta'(X'X)^{-1}X'E(\epsilon) = \beta'(X'X)^{-1}X'0 = 0$ | |
| Oct 12, 2018 at 23:31 | history | edited | cabeer | CC BY-SA 4.0 |
deleted 10 characters in body
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| Oct 12, 2018 at 23:03 | comment | added | cabeer | @a_statistician Oh damn, that does seem to make a lot of sense... But how come we can pull $\epsilon$ out of $E(.)$? Isn't this only allowed for scalar values? | |
| Oct 12, 2018 at 22:56 | comment | added | user158565 | $E(\beta'(X'X)^{-1}(X'\epsilon)) = 0$, because $E(\epsilon) = 0$ | |
| Oct 12, 2018 at 22:37 | comment | added | cabeer | What do you mean exactly by that? I mean it's not about finding the estimate, but about the expected value of squared estimator? | |
| Oct 12, 2018 at 22:10 | comment | added | Maxtron | If $\hat{\beta}$ is an estimate of $\beta$, why are you trying to find the its estimate again? Once you estimate $\beta$, it become deterministic, so there won't be any covariance term. | |
| Oct 12, 2018 at 21:30 | review | First posts | |||
| Oct 12, 2018 at 22:16 | |||||
| Oct 12, 2018 at 21:29 | history | asked | cabeer | CC BY-SA 4.0 |