Timeline for Limiting distribution of "scatter matrix" $\frac{1}{n}XX^T:=\frac{1}{n}\sum_{i=1}^nx_ix_i^T$ for iid $x_1,\ldots,x_n \in \mathbb R^p$
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Jan 15, 2022 at 20:11 | history | edited | Michael Hardy | CC BY-SA 4.0 |
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| Jul 2, 2019 at 19:30 | vote | accept | dohmatob | ||
| Jul 2, 2019 at 19:30 | comment | added | dohmatob | OK, makes sense. Thanks! | |
| Jul 2, 2019 at 18:20 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jul 2, 2019 at 12:52 | comment | added | Iosif Pinelis | @dohmatob : Concerning your latest comment: That will never be the case if $Var(x^i x^j)\ne0$ for some $i,j$ such that $i\ne j$. Indeed, then $(i,j)\ne(j,i)$, whereas $Cov(x^ix^j,x^jx^i)=Var(x^i x^j)\ne0$, and so, the covariance operator $R$ is not a multiple of the identity operator and hence cannot be the covariance operator of a Gaussian matrix with iid entries. | |
| Jul 2, 2019 at 12:38 | comment | added | dohmatob | Any obvious conditions which would imply the convergence is to a Ginibre (i.e matrices with iid Gaussian entries) ? | |
| Jul 2, 2019 at 12:35 | comment | added | dohmatob | Oops, that was an oversight on my part. Indeed, the difference is $\sqrt{n}\mu$ which diverges (except when $\mu=0$). Thanks! | |
| Jul 2, 2019 at 12:20 | comment | added | Iosif Pinelis | If the $x_i$'s are non-degenerate Gaussian, then $\mu$ is a nonzero (positive-semidefinte) matrix, and so, the convergence of $(s-n\mu)/\sqrt n$ implies that $s/\sqrt n$ does not converge. | |
| Jul 2, 2019 at 12:11 | comment | added | dohmatob | Thanks for the response. A (perhaps) naive question though, in case the $x_i$'s are draw from a centered multivariate Gaussian, we know that $s$ has a Wishart distribution. Reasoning along your lines, is it ok that the translate $(s-n\mu)/\sqrt{n}$ converges to a Gaussian, as this would immediately imply $s/\sqrt{n}$ also converges to a Gaussian ? This sounds weird. | |
| Jul 2, 2019 at 12:03 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jul 2, 2019 at 11:44 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |