Suppose that $X_1,\ldots, X_n$ are i.i.d. with common distribution $\mu$ on $\mathbb{R}$, and assume $X_i \neq X_j$ almost surely for $i\neq j.$
Consider now the random variable $Y$ which satisfies that $Y \mid (X_1,\ldots,X_n) \sim \mu_n$, where $\mu_n= \frac{1}{n} \sum_{j=1}^n \delta_{X_j}$. In other words $Y$ is drawn from the sample $(X_1,\ldots,X_n)$ uniformly (i.e. $P(Y=X_j \mid (X_1,\ldots,X_n)) = \frac{1}{n}$ for all $j=1,\ldots,n$).
My question is: What is the distribution of $Y$?
Intuitively the distribution would definitely depend on $\mu$. My guess would be that $Y \sim \mu$ but I am not sure.
I have tried the following for $y \in \mathbb{R}$:
$$ P(Y=y) = P(Y=X_i =y \mid X_i=y) P(X_i=y) = P(Y=X_i \mid X_i = y) \mu(y). $$ If this holds, then one would need to compute $P(Y=X_i \mid X_i = y)$. Here I was thinking maybe Bayes' rule might come in handy but I can't quite make it work.
Can anyone help me?
\begin{align} P\big(Y\in A\big)&=\int_{\mathbb{R}^n}P\big(Y\in A\,\big|\,X=x\big)d\mu^n\\ &=\int_{\mathbb{R}^n}\sum_{j=1}^nP\big(Y\in A,Y=X_j\,\big|\,X=x\big)d\mu^n\\ &=\int_{\mathbb{R}^n}\sum_{j=1}^nP\big(X_j\in A,Y=x_j\,\big|\,X=x\big)d\mu^n\\ &=\int_{\mathbb{R}^n}\sum_{j=1}^nI_A(x_j)P\big(Y=x_j\,\big|\,X=x\big)d\mu^n\\ &=\int_{\mathbb{R}^n}\frac{1}{n}\sum_{j=1}^nI_A(x_j)d\mu^n\\ &=\mu(A) \end{align}