Timeline for Under what conditions is a multivariate normal distribution exchangeable?
Current License: CC BY-SA 4.0
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| when toggle format | what | by | license | comment | |
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| Jun 14, 2023 at 14:59 | vote | accept | user449277 | ||
| Jun 14, 2023 at 14:42 | answer | added | Iosif Pinelis | timeline score: 3 | |
| Jun 14, 2023 at 13:46 | comment | added | Augusto Santos | Assume it is zero mean or centered (otherwise one needs to account for the mean): we have $p(x) \propto e^{-\mathbf{x} \Sigma^{-1} \mathbf{x}} = \Pi_{i\neq j} e^{-\beta x_ix_j}\times \Pi_{i=1}^{N}e^{-(\alpha+\beta) x_i^2}$. Any heterogeneity that you introduce in the underlying "interaction network" given by the precision matrix can produce a counter-example for exchangeability. Remark that, the underlying "network" given by the support of the precision matrix should be complete (even if it is regular, but not complete, exchangeability is not granted). | |
| Jun 14, 2023 at 13:21 | comment | added | user449277 | Thank you! I had a similar hunch, $\Sigma = \alpha I + \beta (\mathbf{1}\mathbf{1}^T - I)$, i.e. all diagonal elements are identical and all off-diagonal elements are identical. Can you explain why the inverse? Does it make a difference? | |
| Jun 14, 2023 at 13:11 | comment | added | Augusto Santos | If the inverse of the covariance is of the form $\Sigma^{-1}=(\alpha I+ \beta \mathbf{1}\mathbf{1}^{\top})$, for some $\alpha>0$ and $\beta$, then it is exchangeable. I believe it is also a necessary condition (I might be missing something). | |
| S Jun 14, 2023 at 12:44 | review | First questions | |||
| Jun 14, 2023 at 14:40 | |||||
| S Jun 14, 2023 at 12:44 | history | asked | user449277 | CC BY-SA 4.0 |