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I am having trouble in solving the following problem;

If $X_1, X_2, . . . \in \mathbb{R}$ are exchangeable with $EX_i^2 < \infty$ then $E(X_1X_2) ≥ 0.$

What I know is that the definition of exchangeable sequence;

A sequence $X_1, X_2, . . . $is said to be exchangeable if for each $n$ and permutation $\pi$ of $\{1, . . . , n\}, (X_1, . . . , X_n)$ and $ (X_{\pi(1)}, . . . , X_{\pi(n)})$ have the same distribution.

How could I develop ideas from the above simple definition? Are there any hints available? Thanks in advance.

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Here is a naive way of proving this.

Define the matrix $A=(E(X_iE_j))$; by exchangeability all the diagonal elements are equal (to $a$, say) and all the off diagonal elements are equal (to $b$, say). This matrix is positive semidefinite. Now evaluate the quadratic form $q_n=x'Ax$, where the first $n$ elements of $x$ are $1$ and the rest are $0$. You get $q_n = na+(n^2-n)b$. But $q_n\ge0$ for all $n$. Hence $b\ge0$.

Equivalently, $q_n=E(X_1+\cdots +X_n)^2 = nE(X_1^2)+(n^2-n)E(X_1X_2)=na+(n^2-n)b\ge0$ for all $n>0$, so $b\ge0$.

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  • $\begingroup$ Thank you K.L. You really saved me. $\endgroup$ – Euduardo Mar 22 at 1:32

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