Consider a multivariate normal distribution. What are necessary and sufficient conditions (esp. on the covariance matrix) under which this distribution is exchangeable?



Under what conditions is a multivariate normal distribution exchangeable?
$\newcommand\Si\Sigma\newcommand\si\sigma\newcommand\1{\mathbf1}$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. $\quad\Box$


