Given $n$ iid nonnegative r.v.'s $\{X_i\}$ distributed as an absolutely continuous r.v. $X$ with density $f$, defined on a common $(\Omega,\mathcal{F},\mathbb{P})$ and a stopping time $T$ (which stops at the first largest seen value, after a fixed waiting time), in a paper I was reading the author seeks to estimate $\mathbb{E}(X_T)$.
In order to do this, the iid sequence $\{X_i\}$ is claimed to be equivalent to see $X$ generating $n$ values $\{x_i\}$ independently, which are then permuted uniformly at random, independently. This is true since the stopping rule $T$ is defined in terms of the relative ordering in which the samples appear. Thus just take for granted that $\mathbb{E}X_T=\mathbb{E}X_{\pi(T)}$, where $\pi$ is a uniform random permutation.
My problem is when the following formula appears: $\mathbb{E}X_T=\int\ldots\int \prod_{i=1}^n f(x_i)\mathbb{E}_{\pi}x_{\pi(T)}dx_1 \ldots dx_n$. My main question is what the subscript $\pi$ means here. I understand that the integrals are an integration with respect to the joint density of repeatedly sampling from $X$, independently, but how is the expectation $\mathbb{E}_{\pi}$, which refers to the space $(\Omega,\mathcal{F},\mathbb{P})$ directly (if I understand correctly, the subscript refers to the law with respect to which the expectation is taken, not to a conditioning, but then where is the randomness of the stopping time? That should also be appearing as a subscript...), compatible, within the same formula, with the other expectation, taken as an integral, that is expressed in the state space?
The only conclusion that possibly makes sense to me, should be that the law of total expectation has been implicitly used here, but somehow the conditioning on the $X_1=x_1,\ldots,X_n=x_n$ has been omitted in $\mathbb{E}_{\pi}$, which should really be $\mathbb{E}(X_{\pi(T)}|X_1=x_1,\ldots, X_n=x_n)$. Is this the correct interpretation of such formalism? Thanks for any help.