I have the following question.
If I have the following setup: $X_i=F(\xi_i)$ for $i\leq N$, i.i.d. RVs $(\xi_i)_{i\leq N}$ and some measurable $F:\mathbb{R}\to \mathbb{R}$, and $(Y_1,...,Y_N)=G^N(\xi_1,...,\xi_N)$ for some measurable $G^N:\mathbb{R}^N\to \mathbb{R}^N$. If I know that the $(Y_i)_{i\leq N}$ are exchangeable in the way that $(Y_{\sigma(1)},...,Y_{\sigma(N)})=G^N(\xi_{\sigma(1)},...,\xi_{\sigma(N)})$ for any perturbation $\sigma$, can I conclude that the tuple processes $(X_i,Y_i)_{i\leq N}$ are also exchangeable? If not, are they at least identically distributed?
Thank you for any help
Yes, and it holds more generally whenever $\{\xi_1, ..., \xi_n\}$ are exchangeable, meaning that the distribution of $(\xi_1, ..., \xi_n)$ is the same as that of $(\xi_{\sigma(1)}, ..., \xi_{\sigma(n)})$ for any permutation $\sigma$ of $\{1, ..., n\}$.
Fix $n$ as a positive integer. Let $\{\xi_i\}_{i=1}^n$ be exchangeable random variables. Suppose $g:\mathbb{R}^n\rightarrow \mathbb{R}^n$ is measurable and has the desired permutation property: For all $(x_1, ..., x_n) \in \mathbb{R}^n$ and all permutations $\sigma$ of $\{1, ..., n\}$ we have $$ \left((y_1, ..., y_n) = g(x_1, ..., x_n)\right)\implies \left( (y_{\sigma(1)}, ...., y_{\sigma(n)}) = g(x_{\sigma(1)}, ..., x_{\sigma(n)})\right)$$
Define $(Y_1, ..., Y_n)=g(\xi_1, ..., \xi_n)$. Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be measurable and define $X_i=f(\xi_i)$ for $i \in \{1, ..., n\}$.
Define $W_i=(X_i, Y_i)$ for $i \in \{1, ..., n\}$.
Claim: $\{W_1, ..., W_n\}$ are exchangeable.
Proof: Let $\sigma$ be a permutation of $\{1, ..., n\}$. By the permutation property of $g$ we have
\begin{align} &Y_{\sigma(1)} = g_1(\xi_{\sigma(1)}, \xi_{\sigma(2)}, ..., \xi_{\sigma(n)})\\ &Y_{\sigma(2)} = g_1(\xi_{\sigma(2)}, \xi_{\sigma(3)}, ..., \xi_{\sigma(n)}, \xi_{\sigma(1)})\\ &Y_{\sigma(3)} = g_1(\xi_{\sigma(3)}, \xi_{\sigma(4)}, ..., \xi_{\sigma(n)}, \xi_{\sigma(1)}, \xi_{\sigma(2)})\\ &\cdots\\ &Y_{\sigma(n)} = g_1(\xi_{\sigma(n)}, \xi_{\sigma(1)}, ..., \xi_{\sigma(n-1)})\\ \end{align}
Then:
(Equation A) \begin{align} W_{\sigma(1)} &= (F(\xi_{\sigma(1)}), g_1(\xi_{\sigma(1)}, \xi_{\sigma(2)}, \ldots, \xi_{\sigma(n)}))\\ W_{\sigma(2)} &= (F(\xi_{\sigma(2)}), g_1(\xi_{\sigma(2)}, \xi_{\sigma(3)}, \ldots, \xi_{\sigma(n)}, \xi_{\sigma(1)}))\\ \cdots\\ W_{\sigma(n)} &= (F(\xi_{\sigma(n)}), g_1(\xi_{\sigma(n)}, x_{\sigma(1)}, \ldots, \xi_{\sigma(n-1)})) \end{align} So
(Equation B) \begin{align} W_{1} &= (F(\xi_1), g_1(\xi_1, \xi_2, ..., \xi_n))\\ W_{2} &= (F(\xi_2), g_1(\xi_2, \xi_3, ..., \xi_n, \xi_1))\\ W_{3} &= (F(\xi_3), g_1(\xi_3, \xi_4, ..., \xi_n, \xi_1, \xi_2))\\ \cdots\\ W_{n} &= (F(\xi_n), g_1(\xi_n, \xi_1, ...., \xi_{n-1})) \end{align}
Define $(Z_1, ..., Z_n) = (\xi_{\sigma(1)}, \ldots, \xi_{\sigma(n)})$. Substituting this definition into (Equation A) gives
(Equation C) \begin{align} W_{\sigma(1)} &= (F(Z_1), g_1(Z_1, Z_2, ..., Z_n))\\ W_{\sigma(2)} &= (F(Z_2), g_1(Z_2, Z_3, ..., Z_n, Z_1))\\ W_{\sigma(3)} &= (F(Z_3), g_1(Z_3, Z_4, ..., Z_n, Z_1, Z_2))\\ \cdots\\ W_{\sigma(n)} &= (F(Z_n), g_1(Z_n, Z_1, ...., Z_{n-1})) \end{align} Now compare the labels in (Equation C) and (Equation B). Since $(Z_1, ..., Z_n)$ and $(\xi_1, ..., \xi_n)$ have the same distribution, it is clear that $(W_1, ..., W_n)$ and $(W_{\sigma(1)}, ..., W_{\sigma(n)})$ have the same distribution. $\Box$