Timeline for Orthogonal projection $X X^+$ from random Gaussian matrix $X$
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 7 at 12:57 | comment | added | Iosif Pinelis | @JoãoF.Doriguello : I am glad this was of help. In such a case, these guidelines may be relevant. | |
| Jun 7 at 9:09 | comment | added | João F. Doriguello | Many thanks for the answer, it isn't as hard as I anticipated, it is quite simple actually. | |
| Jun 7 at 9:00 | vote | accept | João F. Doriguello | ||
| Jun 6 at 21:59 | comment | added | Iosif Pinelis | @MichaelHardy : I agree with your latter comments. | |
| Jun 6 at 21:49 | comment | added | Michael Hardy | $\ldots X_i \sim \operatorname N(a + bw_i, \sigma^2)$ and $\widehat a,\widehat {b\,}$ are the least-squares estimators of $a,b$ then $\sum_{i=1}^n \left(X_i - (\widehat a + \widehat {b\,} w_i)\right)^2 \sim \sigma^2 \chi_{n-2}^2$ and that that sum is probabilistically independent of the least-squares estimators $\widehat a,\widehat{b\,}.$ | |
| Jun 6 at 21:47 | comment | added | Michael Hardy | Noted. One should also note that the case $\det \Sigma = 0$ arises very naturally in some contexts, perhaps the simplest of which is that if $X_1,\ldots,X_n\sim\text{i.i.d.} \operatorname N_1(\mu,\sigma^2)$ so that $\overline X = (X_1 + \cdots + X_n)/n \sim \operatorname N(\mu, \sigma^2/n),$ then the vector whose $i$th component is $X_i - \overline X$ for $i=1,\ldots,n$ has a singular covariance matrix, and one uses that to show that $\sum_{i=1}^n (X_i-\overline X)^2 \sim \sigma^2 \chi_{n-1}^2,$ and if$\,\ldots\qquad$ | |
| Jun 6 at 21:29 | comment | added | Iosif Pinelis | @MichaelHardy : I have added a detail on this. | |
| Jun 6 at 21:28 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jun 6 at 20:42 | comment | added | Michael Hardy | What if $\det\Sigma=0 \text{?} \qquad$ | |
| Jun 6 at 20:26 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jun 6 at 20:09 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |