# Why such a definiton for "exchangeable" event?

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 $$$$\left\{X_1, X_2,\ldots\in A\right\}\in\mathcal{F}$$$$ The event $$\left\{X_1, X_2,\ldots\in A\right\}$$ is called exchangeable if $$$$\left\{X_1, X_2,\ldots\in A\right\}\subset\left\{X_{\sigma_1}, X_{\sigma_2},\ldots\in A\right\}$$$$ for all finite permutations $$\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: $$$$\left\{X_1, X_2,\ldots\in A\right\}\color{red}{\subset}\left\{X_{\sigma_1}, X_{\sigma_2},\ldots\in A\right\}\tag{1}$$$$ and not: $$$$\left\{X_1, X_2,\ldots\in A\right\}\color{red}{=}\left\{X_{\sigma_1}, X_{\sigma_2},\ldots\in A\right\}\tag{2}$$$$ Looking at definition of exchangeable event from other references, it seems to me that $$(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: $$$$\left\{X_1, X_2,\ldots\in A\right\}\supset\left\{X_{\sigma_1}, X_{\sigma_2},\ldots\in A\right\}$$$$?

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 $$\{X_1,X_2, \cdots \in A\} \supseteq \{X_{\sigma_1}, X_{\sigma_2}, \dots \in A\}$$.
• 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? Commented Oct 22, 2020 at 13:42
• Yes. Basically, if you exchange a finite numbers of rv',s you stay on $A$. If you reverse them back, you also stay in $A$. Commented Oct 22, 2020 at 14:10
• 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. Commented May 9 at 7:00
• @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). Commented May 9 at 20:02
• @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. Commented May 10 at 7:36