Consider a multivariate normal distribution. What are necessary and sufficient conditions (esp. on the covariance matrix) under which this distribution is exchangeable?
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1$\begingroup$ 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). $\endgroup$– Augusto SantosCommented Jun 14, 2023 at 13:11
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$\begingroup$ 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? $\endgroup$– user449277Commented Jun 14, 2023 at 13:21
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$\begingroup$ 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). $\endgroup$– Augusto SantosCommented Jun 14, 2023 at 13:46
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$\newcommand\Si\Sigma\newcommand\si\sigma\newcommand\1{\mathbf1}$Let $D:=N(\mu,\Si)$ be the $n$-variate normal distribution with mean vector $\mu=[\mu_1,\dots,\mu_n]^\top$ and covariance matrix $\Si=[\si_{ij}\colon i,j=1,\dots,n]$.
Suppose that $D$ is the distribution of an exchangeable random vector $X=[X_1,\dots,X_n]^\top$. Then (i) $\mu_i=EX_i=EX_j=\mu_j=:a$ and $\si_{ii}=Var\,X_i=Var\,X_j=\si_{jj}=:b^2$ for all $i$ and $j$ and (ii) $\si_{ij}=Cov(X_i,X_j)=Cov(X_k,X_l)=\si_{kl}=:c$ for all $i,j,k,l$ such that $i\ne j$ and $k\ne l$. So, $$\mu=a\1\quad\text{and}\quad\Si=c\1\1^\top+(b^2-c)I,\tag{1}\label{1}$$ where $\1:=[1,\dots,1]^\top$ and $I$ is the identity matrix. For such a covariance matrix $\Si$, $\1$ is an eigenvector belonging to eigenvalue $b^2+(n-1)c$, and any vector orthogonal to $\1$ is an eigenvector belonging to eigenvalue $b^2-c$. Since $\Si$ is positive definite, we conclude that $$b^2+(n-1)c>0\quad\text{and}\quad b^2-c>0. \tag{2}\label{2}$$
Vice versa, suppose that conditions \eqref{1} and \eqref{2} hold. Then $$\Si^{1/2}=u\1\1^\top+vI,$$ where $$v:=\sqrt{b^2-c}\quad\text{and}\quad u:=\frac{\sqrt{b^2+(n-1)c}-\sqrt{b^2-c}}n.$$ So, letting $$X:=a\1+\Si^{1/2}Z=a\1+u(\1^\top Z)\1+vZ, \tag{3}\label{3}$$ where $Z$ is a standard normal random vector in $\mathbb R^n$, we see that $X$ is an exchangeable normal random vector in $\mathbb R^n$ with the mean vector $\mu$ and the covariance matrix $\Si$ as in \eqref{1}.
Thus, the conjunction of conditions \eqref{1} and \eqref{2} is necessary and sufficient for the $n$-variate normal distribution $N(\mu,\Si)$ to be exchangeable. $\quad\Box$
answered Jun 14, 2023 at 14:42
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$\begingroup$ fantastic, thank you so much! $\endgroup$ Commented Jun 14, 2023 at 15:08
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$\begingroup$ @IosifPinelis: incidentally, the inverse of this covariance necessarily obeys $\alpha I+\beta \mathbf{1}\mathbf{1}^{\top}$ for some $\alpha$ and $\beta$ if I am not missing anything? To me, it was easier to reason over the inverse as it appears explicitly in the distribution (and exchangeability becomes apparent there)... $\endgroup$ Commented Jun 14, 2023 at 15:24
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1$\begingroup$ @AugustoSantos : Yes, of course, the inverse (and actually any function) of the covariance matrix will be of the same form. The more difficult part of the problem here is the "vice versa" part, that is, the sufficiency of (1)&(2), where $X$ as in (3) is constructed. $\endgroup$ Commented Jun 14, 2023 at 15:31
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