Timeline for Is there a formula for the determinant of the covariance matrix $\mathbf{X_n}^T \mathbf{X_n}$ in the case of multiple regression?
Current License: CC BY-SA 4.0
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| Mar 22, 2021 at 23:01 | answer | added | Good Luck | timeline score: 1 | |
| Mar 22, 2021 at 21:00 | history | tweeted | twitter.com/StackStats/status/1374103614634012677 | ||
| Mar 22, 2021 at 20:45 | history | edited | sonicboom | CC BY-SA 4.0 |
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| Mar 22, 2021 at 20:23 | history | edited | sonicboom | CC BY-SA 4.0 |
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| Mar 22, 2021 at 20:15 | history | edited | sonicboom | CC BY-SA 4.0 |
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| Mar 22, 2021 at 20:13 | comment | added | sonicboom | While each row corresponds to an iid observation of the covariates, the random variables in a given row can be dependent on one another. | |
| Mar 22, 2021 at 20:07 | comment | added | sonicboom | @whuber Each row is a vector of covariate observations of iid random variables. E.g. the $n$ rows in the second column correspond to $n$ observations of some random variable. So | |
| Mar 22, 2021 at 20:04 | history | edited | sonicboom | CC BY-SA 4.0 |
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| Mar 22, 2021 at 20:02 | comment | added | sonicboom | @jld What does it converge to in expectation if we scale it appropriately, i.e. what does $\frac{1}{n^p} E[\text{det} X_n^T X_n]$ converge to? I will edit the post to put that scaling in. | |
| Mar 22, 2021 at 19:17 | comment | added | jld | Under reasonable regularity conditions $\frac 1n X_n^TX_n \to_p \Sigma$ with $\Sigma$ being the covariance matrix of the new observations, so, since $\det$ is continuous, $\det \frac 1n X^T_nX_n \to_p \det \Sigma$. Then $\det X_n^TX_n = n^p \det \frac 1n X_n^T X_n$ so this will generally blow up | |
| Mar 22, 2021 at 19:10 | comment | added | whuber♦ | Yes, you do need conditions. The expectation of the determinant depends on the specifics of the process that creates a sequence of rows of $X.$ (My earlier comment referred to a now-deleted comment wherein another user referred to the determinants of $X$ and $X^\prime,$ btw). | |
| Mar 22, 2021 at 18:55 | comment | added | sonicboom | @whuber A matrix times its transpose is a square matrix. I am interested in the classical regression setting, the elements of the random vectors $X_i = [1, X_{i1},\dots,X_{ip}]$ are i.i.d. continuous random variables. I am not sure if we need additional conditions for the expectation of the determinant, and its limit, to exist in the multiple regression case? | |
| Mar 22, 2021 at 18:39 | comment | added | whuber♦ | @Erik I wondered, because there are methods to construct determinants from rectangular matrices. I describe one at stats.stackexchange.com/a/512862/919. | |
| Mar 22, 2021 at 18:37 | comment | added | Eric Perkerson | Oh yes, good point. | |
| Mar 22, 2021 at 18:37 | comment | added | whuber♦ | In (3), what are you assuming about the distribution of the $X_n$? @Erik How are you computing the determinants of non-square matrices?? | |
| Mar 22, 2021 at 18:25 | history | edited | sonicboom | CC BY-SA 4.0 |
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| Mar 22, 2021 at 18:17 | comment | added | jcken | $det(X^T X)$ is non negative, but will be zero iff there exists a linear dependence among columns of $X$ so not guaranteed to be positive | |
| Mar 22, 2021 at 18:10 | history | asked | sonicboom | CC BY-SA 4.0 |