1
$\begingroup$

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 $\tilde \alpha ,\tilde\beta$

$\endgroup$
2
  • 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

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.