Timeline for Variance of a weighted linear regression
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 6 at 20:35 | comment | added | Iosif Pinelis | @loqueelviento : I have commented on your reasons for unequal weights. As for $A$ and $B$, I did not say that you chose them. Instead, I said you chose (weighted) estimators of $A$ and $B$. These estimators will depend on your choice of the weights. I also said that there is no reason to tie estimation of $V$ to your choice of estimators for $A$ and $B$. | |
| Sep 6 at 20:24 | comment | added | loqueelviento | Yes, I could answer my own question. 1. There are reasons for unequal weights (e.g. filter out distant data if law is linear only locally). 2. I don't understand your comment about A and B, I do not chose them, I try to fit them. | |
| Sep 6 at 14:25 | comment | added | Iosif Pinelis | @loqueelviento : Do you a further response to my answer and comment? | |
| Sep 5 at 14:17 | comment | added | Iosif Pinelis | What do you mean, precisely, by "in the weighted fit"? Why (and how) should an unbiased estimator of $V$ depend on your choice of estimators of $A$ and $B$? Also, using unequal weights given the same variance for all $i$ does not seem reasonable. | |
| Sep 5 at 13:36 | comment | added | loqueelviento | Yes, it is indeed standard in the unweighted case. However, in the weighted fit, it's no longer a n / (n-2) correction. | |
| Sep 5 at 12:57 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |