Timeline for Show that the estimator of GMM weigh matrix is consistent
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 20, 2021 at 9:51 | comment | added | Bertrand | Thanks for your clarification. | |
| Jul 19, 2021 at 15:59 | vote | accept | shk910 | ||
| Jul 19, 2021 at 12:34 | answer | added | chan1142 | timeline score: 1 | |
| Jul 19, 2021 at 12:10 | comment | added | chan1142 | In my previous comment, "add up" is suppose to be "average". | |
| Jul 19, 2021 at 10:49 | comment | added | Q9y5 | @Bertrand Well, $\tilde{e}_{i}$ does converge in probability to $e_{i}$, it's just from consistency of $\tilde{\beta}$ and slutsky lemma. But this method won't work because the infinite sum of little op is not necessary a little op, $\tilde{e}_{i}=e_{i}+o_{p}\left(1\right)$ doesn't yield $\frac{1}{n}\sum_{i=1}^{n}Z_{i}Z_{i}^{\top}\tilde{e}_{i}=\frac{1}{n}\sum_{i=1}^{n}Z_{i}Z_{i}^{\top}e_{i}+o_{p}\left(1\right)$. | |
| Jul 19, 2021 at 10:13 | comment | added | chan1142 | Note $\tilde{e}_i = e_i - X_i' (\tilde{\beta}-\beta)$ so $\tilde{e}_i^2 = e_i^2 - 2e_i x_i' (\tilde\beta - \beta) + (\tilde\beta - \beta)' X_i X_i' (\tilde\beta - \beta)$. Multiply $Z_iZ_i'$, add up, and then you will see that the second term and the third term converge in probability to zero due to the consistency of $\tilde\beta$. The first term converges in probability to $E[Z_i Z_i' e_i^2] = \Omega$ by LLN. Note that it's not $E[Z_iZ_i' \tilde{e}_i^2]$. | |
| Jul 18, 2021 at 21:31 | comment | added | Bertrand | At the level of a single random term, $\tilde e_i$ will not converge in probability to $e_i$. Averaging is necessary (LLN). | |
| Jul 18, 2021 at 16:24 | history | asked | shk910 | CC BY-SA 4.0 |