# Create an exchangeable sequence from a non-exchangeable sequence

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.

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 $$c = P(X_i = a_0) = P(T(i) = 0) = P(T^{-1}(0) = i).$$ However, due to countable additivity $$\sum_{i=0}^{\infty} P(T^{-1}(0) = i) = 1,$$ 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.
• 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. – Ben - Reinstate Monica May 22 '18 at 23:55
• 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 $\mathbb{N}=\{0,1,2,\ldots\}$) – Juho Kokkala May 23 '18 at 4:59
• 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. – Ben - Reinstate Monica May 23 '18 at 5:33