# Exchangeability of random variables and conditional expectation

Let $(\Omega, \mathcal{F}, \mathbb{P} )$ be a product space ${\mathbb{R}}^{\mathbb{N}}$ equipped with the product $\sigma$-algebra $\mathcal{B} ({\mathbb{R}}^{\mathbb{N}})$. Let $(X_n)_{n \geq 1}$ be the coordinate process on $\Omega$, i.e. \begin{equation} \omega = (X_1 (\omega), X_2 (\omega), \ldots). \end{equation} Moreover, let $(X_n)_{n \geq 1}$ be exchangeable, i.e. the process $(X_ {\tau(n)})_{n \geq 1}$ has the same law as $(X_ {n})_{n \geq 1}$ under $\mathbb{P}$ for any permutation $\tau$ of $\mathbb{N}$ which fixes all but finitely many elements of $\mathbb{N}$.

Let $S$ be the $\sigma$-algebra on $\Omega$ made up of events $A \in \mathcal{B} ({\mathbb{R}}^{\mathbb{N}})$ such that \begin{equation} (X_1 (\omega), X_2 (\omega), \ldots ) \in A \quad \text{iff} \quad (X_{\tau (1)} (\omega), X_{\tau(2)} (\omega), \ldots ) \in A. \end{equation}

Show that for any $A \in S$, \begin{equation} \mathbb{E} ( {\mathbf{1}}_A | \, \sigma ( X_1, X_2, \ldots, X_n ) ) = \mathbb{E} ( {\mathbf{1}}_A | \, \sigma ( X_{n+1}, X_{n+2}, \ldots, X_{2n} ) ). \end{equation} (I only know that we need to use the permutation $\tau$ that interchanges $1$ and $n+1$, $2$ and $n+2$, etc, but I don't know how to prove this fact rigorously.)

## 1 Answer

A starting point could be the following: if $\tau$ is a permutation with finite support, then for each bounded random variable $Y$ and each $\sigma$-algebra $\mathcal M$, we have $$\mathbb E[Y\mid\mathcal M](\tau(\omega))=\mathbb E[Y^\tau\mid \tau^{-1}\mathcal M](\omega),$$ where $Y^\tau$ denotes the random variable defined by $Y^\tau:=Y(\tau(\omega))$.

• I don't really understand what is meant by ${\tau}^{-1} \mathcal{M}$ and how to prove this statement. – Richard Sep 23 '14 at 3:32
• Can I argue that \begin{equation}\mathbb{E} ( {\mathbf{1}}_A (X_1, X_2, \ldots) | \sigma (X_1, X_2, \ldots, X_n )) = \mathbb{E} ( {\mathbf{1}}_A (X_{n+1}, X_{n+2}, \ldots, X_{2n}, X_1, \ldots , X_n, X_{2n+1}, \ldots ) | \sigma (X_1, \ldots, X_n )) = \mathbb{E} ( {\mathbf{1}}_A (X_1, \ldots , X_n, X_{n+1}, X_{n+2}, \ldots, X_{2n}, X_{2n+1}, \ldots ) | \sigma (X_{n+1}, X_{n+2}, \ldots, X_{2n} )) \end{equation} , with the last step replacing $X_1$ by $X_{n+1}$, $X_2$ by $X_{n+2}$, etc ? – Richard Sep 23 '14 at 10:20
• $\tau^{-1}\mathcal M$ is the collection of all sets of the form $\tau^{-1}A$ where $A\in\mathcal M$. You have to use the fact that $(X_n)$ is exchangeable. Then use this with $Y=\chi_A$ and $\tau$ the permutation you mentioned. – Davide Giraudo Sep 28 '14 at 12:34