Let $(X_n)_n$ be a sequence of iid real-valued random variables with a distribution absolutely continuous with respect to the Lebesgue measure (hence any two variables are almost surely distinct). For $n\in \mathbb N^*$ let $R_n$ be the position of $X_n$ "so far", that is, among $(X_1, \dots, X_n)$. In other words $R_n=1+\Sigma_{k=1}^n 1_{X_k>X_n}$. I need to prove that the $R_n$'s are mutually independent. At first glance it seems a little counter-intuitive if you ask me.
All I can think of is that $\lbrace R_n=k \rbrace = \cup_{\sigma \in S_{n-1}} \lbrace X_{\sigma(n-1)} < \dots < X_{\sigma(k)} < X_n < X_{\sigma(k-1)} < \dots < X_{\sigma(1)} \rbrace$.
It seems like a handy formulation but I don't know where to take it next.
Sketch of the proof: Three basic independence arguments, all based on the random set $$\mathcal X_n=\{X_k;1\leqslant k\leqslant n\}$$ which collects the $n$ distinct values of the sequence up to time $n$ but "forgets" their time labels.
By exchangeability, conditionally on the set $\mathcal X_n$, the random variable $R_n$ is uniformly distributed on $\{1,2,\ldots,n\}$, thus $\mathcal X_n$ and $R_n$ are independent.
The random variable $X_{n+1}$ is independent on $R_n$ and $\mathcal X_n$.
The random variable $R_{n+1}$ is measurable with respect to $X_{n+1}$ and $\mathcal X_n$.
Thus, by 1. and 2., $X_{n+1}$ and $\mathcal X_n$ are jointly independent on $R_n$. By 3., this implies that $R_{n+1}$ is independent on $R_n$.