Suppose you have an arbitrary sequence of real values $\{ a_i | i \in \mathbb{N} \}$. Now, suppose you want to randomise the order of this sequence so that it is now exchangeable. To do this, you choose a random permutation $T: \mathbb{N} \rightarrow \mathbb{N}$ and form the new sequence $\{ X_i | i \in \mathbb{N} \}$, where $X_i \equiv a_{T(i)}$.

How do you form a random permutation $T$ to achieve exchangeability for a sequence? If this is possible, please show me how; if not, please explain why.


This is not possible in the general case.

Assume the $a_i$s are unique constants. Exchangeability of the $X_i$s implies that $P(X_i = a_0)$ does not depend on $i$. That is, $P(X_i = a_0)=c$ where $c$ does not depend on $i$. With the construction defined in the question, we have \begin{equation} c = P(X_i = a_0) = P(T(i) = 0) = P(T^{-1}(0) = i). \end{equation} However, due to countable additivity \begin{equation} \sum_{i=0}^{\infty} P(T^{-1}(0) = i) = 1, \end{equation} which is a contradiction with $P(T^{-1}(0) = i)$ being a constant.

Intuitively, the random variable $T^{-1}(0)$ would need to follow a "uniform distribution over the natural numbers," which does not exist.

  • $\begingroup$ Wonderful! (+1) Can I please request a favour: can you please edit this to make it consistent with the range of $a$ values in the question - i.e., exclude $a_0$, and take $i$ over $\mathbb{N}$, excluding $i=0$. Thanks for this answer. $\endgroup$ – Ben May 22 '18 at 23:55
  • $\begingroup$ Why not, although the question does not specify whether $0 \in \mathbb{N}$ (math.stackexchange.com/questions/283). (And am slightly worried that this change in the sum may confuse someone who assumes $\mathbb{N}=\{0,1,2,\ldots\}$) $\endgroup$ – Juho Kokkala May 23 '18 at 4:59
  • $\begingroup$ Okay, fair enough. I guess I should have clarified my use of the convention in the question. $\endgroup$ – Ben May 23 '18 at 5:12
  • $\begingroup$ You can still do that if you think it's useful (feel free to edit this answer accordingly) $\endgroup$ – Juho Kokkala May 23 '18 at 5:29
  • $\begingroup$ That's alright - I'll leave it as you have answered, since there may indeed be readers who are reading $\mathbb{N}$ in the way you read it, rather than the way I did. Your comment already links to the ambiguous convention here, so I'd rather leave all this up. $\endgroup$ – Ben May 23 '18 at 5:33

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