I am trying to solve the next exercise but I don´t know how to start.

Let $X\in\mathbb{R}^n$ be a random vector which is exchangeable in the sense that $X$ has the same distribution as $(X_{\pi(i)})_{i=1}^n$ for any fixed permutation $\pi$ of $\{1,2,...,n\}$.

a) Show that the covariance matrix of $X$ equals $$\Sigma=\alpha I_n+\beta 1_n 1_n^T$$ for certain real numbers $\alpha,\beta$ where $1_n:=(1)_{i=1}^n$.

Actually, I don't understand how to start, I have the next things:

The covariance is given by $$Cov(X,X)=\begin{bmatrix}{Var(X_1)}&{Cov(X_1,X_2)}&{\ldots Cov(X_1,X_n)}\\{Cov(X_2,X_1)}&{Var(X_2)}&{\ldots Cov(X_2,X_n)}\\{\vdots}&{\vdots}&{\vdots}\\{cov(X_n,X_1)}&{Cov(X_n,X_2)}&{Var(X_n)}\end{bmatrix}$$ Here I suppose that because the vector $X$ can be interchangeable $Cov(X_1,X_2)=Cov(X_i,X_k)$ for any pair of value $i,k$, also I think that $Var(X_1)=Var(X_2)=...Var(X_n)$, however I don´t know how to prove that I am only supposing that because of the statement of the exercise, I ddon´t know how to proceed in a formal way.

Then I need to answer the next questions: b) Under which conditions on $\alpha$ and $\beta$ is $\Sigma$ non singular? Show that, under these conditions, the inverse of $\Sigma$ is also of the type: $$\Sigma^{-1}=\tilde\alpha I_n+\tilde\beta 1_n 1_n^T$$ for some $\tilde \alpha ,\tilde\beta$

  • 1
    $\begingroup$ Hint: let $\alpha=\operatorname{Var}(X_1) - \operatorname{Cov}(X_1,X_2)$ and $\beta=\operatorname{Cov}(X_1,X_2)$. Everything else is straightforward algebra, so just do the calculations. $\endgroup$
    – whuber
    Sep 24, 2017 at 17:16
  • 2
    $\begingroup$ Intuitively: $\mbox{Cov}(X_i,X_j)=\mbox{Covariance}(X_1,X_2)$ from exchangeability. $\endgroup$
    – Alex R.
    Sep 24, 2017 at 17:36

1 Answer 1


From exchangeability of $X=(X_1, \dotsc,X_n)$ it follows that all the $X_i$ has the same marginal distributions, so in particular, the same variance (assuming existence). So the diagonal of the covariance matrix must be constant.

Likewise, from exchangeability all pairwise joint distributios $(X_i,X_j), i\not=j$ must be equal, so the off-diagonal elements must all be equal.


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