Exchangeability with random effects?

Question

Consider a $N\times N$ random matrix $$ \epsilon:= \begin{bmatrix} \epsilon_{11} & \epsilon_{12} & \dots &\epsilon_{1N} \\ \epsilon_{21} & \epsilon_{22} & \dots & \epsilon_{2N} \\ \vdots & \vdots & \ddots & \vdots \\ \epsilon_{N1} & \dots & \dots & \epsilon_{NN}\\ \end{bmatrix} $$ $\epsilon$ is exchangeable if $$ \begin{bmatrix} \epsilon_{11} & \epsilon_{12} & \dots &\epsilon_{1N} \\ \epsilon_{21} & \epsilon_{22} & \dots & \epsilon_{2N} \\ \vdots & \vdots & \ddots & \vdots \\ \epsilon_{N1} & \dots & \dots & \epsilon_{NN}\\ \end{bmatrix} $$ is distributed as $$ \begin{bmatrix} \epsilon_{\varphi{(1)}\varphi{(1)}} & \epsilon_{\varphi{(1)}\varphi{(2)}} & \dots &\epsilon_{\varphi{(1)}\varphi{(N)}} \\ \epsilon_{\varphi{(2)}\varphi{(1)}} & \epsilon_{\varphi{(2)}\varphi{(2)}} & \dots & \epsilon_{\varphi{(2)}\varphi{(N)}} \\ \vdots & \vdots & \ddots & \vdots \\ \epsilon_{\varphi{(N)}\varphi{(1)}} & \dots & \dots & \epsilon_{\varphi{(N)}\varphi{(N)}}\\ \end{bmatrix} $$ for any permutation $\varphi$.

I know that if $\epsilon_{ij}$ are i.i.d. across $i,j$, then $\epsilon$ is exchangeable.

My question is: suppose $\epsilon_{ij}:=u_{ij}+\alpha$ where $u_{ij}$ are i.i.d. across $i,j$, $\alpha_i$ are i.i.d across $i$ and $\alpha_i$ is independent of $u_i$, i.e. $$ \epsilon:= \begin{bmatrix} u_{11}+ \alpha_1 & u_{12}+ \alpha_1 & \dots &u_{1N}+ \alpha_1 \\ u_{21}+ \alpha_2 & u_{22}+ \alpha_2 & \dots &u_{2N}+ \alpha_2\\ \vdots & \vdots & \ddots & \vdots \\ u_{N1}+ \alpha_N & \dots & \dots & u_{NN}+ \alpha_N\\ \end{bmatrix} $$ Is $\epsilon$ defined in this way exchangeable?

Answer 1

Yes, it is.

In fact, for any two exchangeable random matrices $A$ and $B$ independent of each other, $A+B$ will also be exchangeable.

The matrix $U=[u_{ij}]$ is exchangeable since it is assumed to be i.i.d., and $A=[a_{ij}]$ where $a_{ij}=\alpha_i$ is exchangeable since the permutation on $A$ simply acts as a permutation on the $\alpha_i$ (also exchangeable since it is i.i.d.).

In fact, it would be sufficient to assume that $U$ and $[\alpha_i]$ are exchangeable and independent.

The proof of the exchangeability of $A+B$ is a mere technicality, I suppose. If $\psi$ is a permutation, and we for any matrix $M=[m_{ij}]$ denote the permutation as $\psi(M)=[m_{\psi(i)\psi(j)}]$, then $A$ and $\psi(A)$ have the same distribution, and $B$ and $\psi(B)$ have the same distribution. Since they are independent, that makes $A+B$ and $\psi(A+B)=\psi(A)+\psi(B)$ also have identical distributions.