Timeline for If $e_0$ are the OLS residuals, what is random in $\hat{\beta}_{OLS}|f(e_0) < \hat{\beta} < f^*(e_0)$?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| S Apr 3, 2021 at 18:14 | history | bounty ended | Adrian | ||
| S Apr 3, 2021 at 18:14 | history | notice removed | Adrian | ||
| Apr 3, 2021 at 18:14 | vote | accept | Adrian | ||
| Apr 3, 2021 at 16:11 | comment | added | Sextus Empiricus | What is the meaning of $$\hat{\beta}_1 | f(\hat{e}) < \hat{\beta}_1 < f^*(\hat{e}), \hat{e} = e_0$$ | |
| Apr 2, 2021 at 20:43 | comment | added | psboonstra | Ok. I understand a unit vector to be any vector with length 1, and I was confused by your stating that $u_1 \in \mathbb{R}^p$ instead of just saying $u_1=\{1, 0, \ldots, 0\}$. | |
| Apr 2, 2021 at 20:41 | answer | added | psboonstra | timeline score: 2 | |
| Apr 2, 2021 at 20:21 | comment | added | Adrian | $\beta_1$ is the coefficient corresponding to the first covariate. $u_1$ is a unit vector with 1 in the first position and 0 everywhere else. So $\beta_1 = u_1^T\beta$, where $\beta_1$ is a scalar, and $\beta$ is the entire coefficient vector of length $p$. | |
| Apr 2, 2021 at 18:42 | comment | added | psboonstra | Thanks. another question. I see now that my first comment may have misled you, notationally speaking, because you are using $\beta$ in a different sense than I assumed. I had interpreted $\beta_1$ as the regression coefficient corresponding to the first predictor, but you are writing $\beta_1$ as the fitted value of the outcome given $X=u_1$ for some covariate pattern $u_1$. In other words, what you are calling $\hat\beta_1$, I would write as $\hat Y|X=u_1$, or $\hat Y(u_1)$. Is my understanding correct? | |
| Apr 2, 2021 at 3:48 | history | edited | Adrian | CC BY-SA 4.0 |
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| Apr 2, 2021 at 3:47 | comment | added | Adrian | Yes, technically $y|X \sim N(X\beta, \sigma^2I)$. $f$ is arbitrary and maps from $\mathbb{R}^n$ to $\mathbb{R}$. | |
| Apr 1, 2021 at 20:33 | comment | added | psboonstra | I'm a little confused by your notation. How does your notation $y\sim N(\mu,\sigma^2 I)$ relate to the rest of the problem. Based on the rest of the problem, I would have guessed you need something like $Y|X\sim N(X\beta,\sigma^2 I)$? Also, what is $f$? Is it assumed known but arbitrary? Does it need to be invertible? And it sounds like it maps from $\mathbb{R}^n$ to $\mathbb{R}$? | |
| S Apr 1, 2021 at 17:57 | history | bounty started | Adrian | ||
| S Apr 1, 2021 at 17:57 | history | notice added | Adrian | Draw attention | |
| Mar 31, 2021 at 0:00 | history | tweeted | twitter.com/StackStats/status/1377048055736897538 | ||
| Mar 30, 2021 at 15:31 | history | edited | Adrian | CC BY-SA 4.0 |
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| Mar 30, 2021 at 15:26 | history | asked | Adrian | CC BY-SA 4.0 |