My question is much more specific than the title:

Given a symmetric distribution $\Xi$ on $\mathbb R^2$, when is it possible to construct a sequence $\xi_1,\xi_2,\dots$ of random variables such that the joint distribution of any two of them is $\Xi$?

For example, if the pdf of $\Xi$ is decomposable: $p_\Xi(x,y) = p(x) p(y)$, then one can just take a sequence of independent r.v.'s.

(To construct certain counterexample related to fractional Brownian motion) I am particularly interested in the pdf $p_\Xi(x,y) = \frac{(a+1)(a+2)}2 |x-y|^{a}1_{[0,1]}(x)1_{[0,1]}(y)$, $a\in(-1,0)$.

  • $\begingroup$ Do you have an example of such a sequence for a non-decomposable joint distribution? $\endgroup$ – ohai Oct 27 '10 at 16:08

I recently came upon this question in the context of distributions taking values in a finite set, but since yours take values in the compact interval $[0,1]$ I don't think much will go wrong applying the answer to your setting.

Certainly a sufficient condition is that you can construct an exchangeable sequence $\chi_1,\chi_2,\ldots$ for which the marginal of $\chi_1$ and $\chi_2$ is $\Xi$, or equivalently that $\Xi$ is a mixture of i.i.d. distributions per de Finetti's theorem. It turns out this is also necessary.

To see this suppose that $\Xi$ satisfies the condition you give on its two-variable marginals. Then for any finite $n$ there is a finite sequence of random variables $\xi_1,\ldots,\xi_n$ (just take the first $n$ variables of your given sequence) all of whose two-variable marginals are $\Xi$. We can construct a new sequence of random variables $\chi_1,\ldots,\chi_n$ by randomly permuting the $\xi_1,\ldots,\xi_n$. By linearity the marginal of any two of these will still be $\Xi$.

But the distribution of the $\chi_1,\ldots,\chi_n$ is invariant under arbitrary permutations, by symmetry. Using a diagonalization and compactness argument, we get from the existence of such a sequence $\chi_1,\ldots,\chi_n$ for all finite $n$ the existence of an exchangeable sequence $\chi_1,\chi_2,\ldots$ whose two-variable marginal distribution is $\Xi$.

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  • $\begingroup$ Neat! And it should generalise easily to n dimensional marginals. $\endgroup$ – George Lowther Oct 27 '10 at 16:27

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