Timeline for Zero Covariance vs Independence of Slope and Intercept Estimators in Linear Models with Least Squares
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 10, 2021 at 15:45 | vote | accept | Adrian Keister | ||
| Sep 10, 2021 at 2:51 | history | became hot network question | |||
| Sep 9, 2021 at 21:04 | comment | added | StubbornAtom | Joint distribution of OLS estimators is discussed at several posts, including stats.stackexchange.com/q/347628/119261. | |
| Sep 9, 2021 at 21:00 | history | tweeted | twitter.com/StackStats/status/1436071919690919944 | ||
| Sep 9, 2021 at 20:54 | comment | added | Adrian Keister | Well, actually the book did say that $\operatorname{Var}(Y_i)=\operatorname{Var}(\varepsilon)=\sigma^2.$ But it didn't say anything at all about showing the bivariate distribution. | |
| Sep 9, 2021 at 20:52 | comment | added | COOLSerdash | Yeah I agree: The question probably shouldn't mention independence without making some further comments about the assumptions of the errors. | |
| Sep 9, 2021 at 20:47 | comment | added | Adrian Keister | @COOLSerdash Right, that's evidently where I'm meant to go. And that's very straight-forward (I've shown how to do that in my question, here). So apparently, the question was poorly worded (for being in its location in the book)? | |
| Sep 9, 2021 at 20:45 | comment | added | COOLSerdash | Ah! According to the solutions manual, the book just want's you to calculate the covariance (which you did) and show that it's zero under some circumstances. Yes, "jointly normal" extends to multiple coefficients not just two. It's probably better to say that they have a multivariate normal distribution. | |
| Sep 9, 2021 at 20:40 | comment | added | Adrian Keister | @COOLSerdash Yes, but I don't understand jld's post, I'm afraid. This result is, apparently, supposed to be able to be proven without the linear algebra approach at all, since that comes later in this book. Question: does "jointly normally distributed" mean the same thing as "bivariate normally distributed"? | |
| Sep 9, 2021 at 20:31 | comment | added | COOLSerdash | It seems to me that you have all ingredients together to answer your question: i) if $X$ and $Y$ have a bivariate normal distribution, zero covariance means independence, ii) if we assume normal errors, the least square coefficient estimates are jointly normally distributed (@jld mentions this in their answer). So the book is correct if the errors are assumed to be normal. | |
| Sep 9, 2021 at 19:59 | history | edited | Adrian Keister | CC BY-SA 4.0 |
added 128 characters in body
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| Sep 9, 2021 at 19:30 | answer | added | jld | timeline score: 6 | |
| Sep 9, 2021 at 18:57 | history | edited | Adrian Keister | CC BY-SA 4.0 |
edited title
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| Sep 9, 2021 at 18:50 | history | asked | Adrian Keister | CC BY-SA 4.0 |