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de Finetti's theorem roughly states that infinite sequence of exchangeable random variables are conditionally independent. I am looking for tight bounds for de Finetti's theorem in the following scenario.

Suppose the random variable $X_i$ is drawn from $[n] = \{1, \cdots, n\}$ for all $1 \le i \le m$ (not necessarily i.i.d). Further suppose that the sequence $X_1, \cdots, X_m$ is exchangeable meaning that $$ \mathbb{P}((X_1, \cdots, X_m)) = \mathbb{P}((X_{\sigma(1)}, \cdots, X_{\sigma(m)})) $$ for any permutation $\sigma$.

Are there tight bounds known on the distance (in total variation) between the distribution of the sequence $(X_1, \cdots, X_m)$ and the closest mixture of product distributions? In particular, I am interested in bounds that are tight on the size of $n$ (in the Theorem below it is $|S|$, where the variables take values in $S$) .

I have found only one paper that deals with this issue which is this paper by Diaconis and Freedman. Theorem 3 in this paper gives a distance between the distribution of such a sequence mentioned above and the closest product distribution but it is not mentioned if the dependence on $|S|$ in their result is necessary. I would appreciate any references that deal with my situation.

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  • $\begingroup$ Is $S=[n]$? And do you have an application in mind? The extreme points of the set of exchangeable measures on $[n]^m$ are easily described. $\endgroup$
    – Yuval Peres
    Commented May 22, 2019 at 16:51
  • $\begingroup$ I think if the correlation between any two of them is nonnegative then the sequence is a mixture of i.i.d. sequences. Even for one who has never heard of de Finetti's theorem, it is routine to show that a finite exchangeable sequence with negative correlation cannot be extended beyond a finite length that depends on the correlation. Also, I seem to recall that if instead of mixing by taking a convex combination, one allows affine combintations (thus integrating with respect to a signed measure with total measure $1$) then the conclusion of de Finetti's theorem holds without the$\,\ldots$ $\endgroup$
    – Michael Hardy
    Commented May 22, 2019 at 20:05
  • $\begingroup$ $\ldots\,$hypothesis that the sequence is infinitely long. $\qquad$ $\endgroup$
    – Michael Hardy
    Commented May 22, 2019 at 20:05
  • $\begingroup$ You refer to "the size of $|S|$" without having said what $S$ is. Do you mean the length $m$ of the sequence? $\qquad$ $\endgroup$
    – Michael Hardy
    Commented May 22, 2019 at 20:46
  • $\begingroup$ oh: ${}\qquad{}$ The set $S$ is defined in the abstract of the paper by Diaconis and Freedman, mentioned in the posting some lines below where $S$ is referred to in the posting. $\endgroup$
    – Michael Hardy
    Commented May 22, 2019 at 20:48

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