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I quote Mörters/Peres (2010)

Definition Let $X_1, X_2,\ldots$ be a sequence of random variables on a probability space $\left(\Omega,\mathcal{F},\mathbb{P}\right)$ and consider a set $A$ of sequences such that \begin{equation} \left\{X_1, X_2,\ldots\in A\right\}\in\mathcal{F} \end{equation} The event $\left\{X_1, X_2,\ldots\in A\right\}$ is called exchangeable if \begin{equation} \left\{X_1, X_2,\ldots\in A\right\}\subset\left\{X_{\sigma_1}, X_{\sigma_2},\ldots\in A\right\} \end{equation} for all finite permutations $\sigma:\mathbb{N}\mapsto\mathbb{N}$. Here finite permutation means that $\sigma$ is a bijection with $\sigma_n=n$ for all sufficiently large $n$.



What I cannot understand is why definition is: \begin{equation} \left\{X_1, X_2,\ldots\in A\right\}\color{red}{\subset}\left\{X_{\sigma_1}, X_{\sigma_2},\ldots\in A\right\}\tag{1} \end{equation} and not: \begin{equation} \left\{X_1, X_2,\ldots\in A\right\}\color{red}{=}\left\{X_{\sigma_1}, X_{\sigma_2},\ldots\in A\right\}\tag{2} \end{equation} Looking at definition of exchangeable event from other references, it seems to me that $(2)$ is the "good" definition and not $(1)$.




Am I wrong? If so, why - in the spirit of Mörters/Peres definition - doesn't it hold true that: \begin{equation} \left\{X_1, X_2,\ldots\in A\right\}\supset\left\{X_{\sigma_1}, X_{\sigma_2},\ldots\in A\right\} \end{equation}?

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1 Answer 1

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Those definitions are equivalent. Note that $\sigma^{-1}$ is also a finite permutation. If $A$ is an exchangeable event by the Mortërs/Peres definition, since we know $\{X_{\sigma_1}, X_{\sigma_2}, \dots \in A\} \in \mathcal{F}$ we can apply the definition using the finite permutation $\sigma^{-1}$ to obtain $\{X_1,X_2, \cdots \in A\} \supseteq \{X_{\sigma_1}, X_{\sigma_2}, \dots \in A\}$.

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  • $\begingroup$ So, what you are saying is that they just focus on the "direction" $\subseteq$, but the other way round is true as well. And it is easily showable following your approach, by means of $\sigma^{-1}$. Correct? $\endgroup$ Commented Oct 22, 2020 at 13:42
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    $\begingroup$ Yes. Basically, if you exchange a finite numbers of rv',s you stay on $A$. If you reverse them back, you also stay in $A$. $\endgroup$ Commented Oct 22, 2020 at 14:10
  • $\begingroup$ Why do we know that $\{X_{\sigma_1},X_{\sigma_2},\ldots\in A\}$ is an element of $\mathcal{F}$? It seems to me that all we know is that the unpermuted set belongs to $\mathcal{F}$. Without any additional assumptions on $A$ or the probability triplet, I don't see how we can conclude that. $\endgroup$
    – Simon SMN
    Commented May 9 at 7:00
  • $\begingroup$ @SimonSMN The definitions are equivalent if we take $A$ to be a borelian of $\mathbb{R}^{\mathbb{N}}$ (what we need to do in most cases). $\endgroup$ Commented May 9 at 20:02
  • $\begingroup$ @CélioAugusto Ok, that makes sense. I still don't see how we conclude $\supseteq$. For $\sigma$ fixed, we do not know that the event $\{X_{\sigma_1},X_{\sigma_2},\ldots\in A\}$ is exchangeable. $\endgroup$
    – Simon SMN
    Commented May 10 at 7:36

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