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Consider the probability space $(\mathcal{S}, \mathcal{A}, \mu)$. Suppose that the random variables $X_1,..., X_n$ are exchangeable and take values in the Borel space $(\mathcal{X}, \mathcal{B})$. We can prove that the empirical probability measure $P_n$ is a measurable function from $(\mathcal{X}^n, \mathcal{B}^n)$ to $(\mathcal{P}, \mathcal{C}_{\mathcal{P}})$, where $\mathcal{P}$ is the set of all probability measures on $(\mathcal{X}, \mathcal{B})$ (Schervish, Theory of Statistics, Sec. 1.4 Prob. 24.).

I have a problem in understanding where do we use exchangeability.


Indeed, suppose for simplicity $\mathcal{X}=\{0,1\}$.

$\forall x:=(x_1,..., x_n)\in \mathcal{X}^n$, we have

$$ \begin{cases} P_n(x)(\{1\}):=\frac{1}{n}\sum_{i=1}^n 1(x_i=1)\\ P_n(x)(\{0\}):=\frac{1}{n}\sum_{i=1}^n 1(x_i=0)\\ P_n(x)(\{1,0\}):=\frac{1}{n}\sum_{i=1}^n 1(x_i\in \{0,1\})\\ P_n(x)(\emptyset):=\frac{1}{n}\sum_{i=1}^n 1(x_i\in \emptyset)\\ \end{cases} $$ and hence $P_n(x):\mathcal{B}\rightarrow[0,1]$ is a probability measure on $(\mathcal{X}, \mathcal{B})$ assigning $P(x)(B)\in [0,1]$ to each $B\in \mathcal{B}$.

Hence, $P_n: \mathcal{X}^n\rightarrow \mathcal{P}$.


Or (equivalently?), suppose for simplicity $\mathcal{S}:=\{s_1,s_2,s_3\}$, $\mathcal{X}:=\{0,1\}$, $n=2$, $X_1(s_1)=1,X_1(s_2)=1 , X_1(s_1)=0$, and $X_2(s_1)=0,X_2(s_2)=1, X_2(s_1)=0$.

$\forall s:=(s_1,s_2)\in \mathcal{S}^2$, we have

$$ \begin{cases} P_n(s)(\{1\}):=\frac{1}{2}\sum_{i=1}^2 1(X_i(s_i)=1)\\ P_n(s)(\{0\}):=\frac{1}{2}\sum_{i=1}^2 1(X_i(s_i)=0)\\ P_n(s)(\{1,0\}):=\frac{1}{2}\sum_{i=1}^2 1(X_i(s_i)\in \{0,1\})\\ P_n(s)(\emptyset):=\frac{1}{2}\sum_{i=1}^2 1(X_i(s_i)\in \emptyset)\\ \end{cases} $$ and hence $P_n(s):\mathcal{B}\rightarrow[0,1]$ is a probability measure on $(\mathcal{X}, \mathcal{B})$ assigning $P(s)(B)\in [0,1]$ to each $B\in \mathcal{B}$.

Hence, $P_n: \mathcal{S}^n\rightarrow \mathcal{P}$.


Where do we use exchangeability?

Similar question here

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We don't. Fix $B\in\mathcal B$ and consider the map $f_B(P):=P(B)$. You have, for a Borel subset $A$ of $\Bbb R$, $$ \eqalign{ \{x\in\mathcal X^n:P_n(x)\in f_B^{-1}(A)\}&=\{x\in\mathcal X^n:P_n(x)(B)\in A\}\cr &=\{x\in\mathcal X^n:{1\over n}\sum_{k=1}^n 1_B(x_k)\in A\},\cr } $$ and each of the maps $x\mapsto 1_B(x_k)$ is $\mathcal B^n/\mathcal B(\Bbb R)$-measurable, hence so is their average. It follows that $\{x\in\mathcal X^n:P_n(x)\in f_B^{-1}(A)\}\in\mathcal B^n$ for each $A\in\mathcal B(\Bbb R)$ and $B\in\mathcal B$, and since $\mathcal C_{\mathcal P}$ is generated by $\{f_B^{-1}(A): A\in\mathcal B(\Bbb R), B\in\mathcal B\}$, the asserted measurability of $P_n$ follows.

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