Product/conditional filtration Suppose I have a finite list of random variables $X_i$, $1\leq i\leq n$, which reveal their values one after the other, but in random order, meaning that for a random (unknown) permutation $\sigma\in S_n$, $X_{\sigma(1)},\ldots,X_{\sigma(n)}$ is the list observed. Without the permutation one usually models the flow of information with the natural filtration $\mathcal{F}_i=\sigma(X_1,\ldots,X_i)$ ($\sigma$ here is not the permutation, but the sigma-algebra)for all $0\leq i\leq n$. Consider the event $\Sigma=\{\sigma=\pi\}$ for some fixed $\pi\in S_n$. In the product probability space of the common probability space $\Omega$ on which the $X_i$'s are defined and $\Omega'$ on which the random permutation is defined, does it make sense to consider $\mathcal{F}_{i}^\pi=\sigma(X_{\pi(1)},\ldots,X_{\pi(i)})\times\{\sigma=\pi\}$ as a filtration on $\Omega\times\Sigma$? Is there a standard way to define, alternatively, a filtration conditionally on $\Sigma$?
 A: I would model this as follows, start to finish.
Let $(\Omega,\Sigma,\mathbb P)$ be a probability space.
Further, let $(X,\sigma)\in\mathbb R^n\times S_n$, that is $\mathbb R^n\times S_n$ equipped with the product $\sigma$-algebra, further $\mathbb R^n$ equipped with the Borel algebra induced by the canonical topology (the ultimate default), and $S_n$ equipped with the discrete $\sigma$-algebra (which can also be considered a sub-$\sigma$-algebra for $S_n$ embedded in $\mathbb R$). This gives you a joint distribution of $X=(X_1,\dots,X_n):\Omega\rightarrow\mathbb R^n$ and $\sigma:\Omega\rightarrow S_n$ on the same underlying probability space $\Omega$.
With respect to the filtration and the flow of information, one has to actively incorporate $\sigma$ in this flow of information. Say, there are two very intuitive options. The first option is that $\sigma$ is revealed first, and then $X_1,\dots,X_n$ are revealed one by one, in this order. The corresponding filtration is
$$\{\emptyset,\Omega\},\sigma(\sigma),\sigma(\sigma,X_1),\sigma(\sigma,X_1,X_2),\dots,\sigma(\sigma,X).$$
The other intuitive option is that the $X_i$'s are revealed in the order prescribed by $\sigma$, but we don't know the order. Well, which $X_i$ does appear first? Well, it's $X_{\sigma(1)}$ because $\sigma(1)$ is the first position revealed under $\sigma$. It is easy to see that $X_{\sigma(1)}$ is also measurable, since $\{X_{\sigma(1)}\in\mathcal E\}=\bigcap_{i=1}^n\{\sigma(1)=i,X_i\in\mathcal E\}$ is, by definition, for measurable $\mathcal E$ (in $\mathbb R$). Clearly, this reasoning extends to $(X_{\sigma(1)},\dots,X_{\sigma(n)})$ and so we may consider the filtration
$$\{\emptyset,\Omega\},\sigma(X_{\sigma(1)}),\sigma(X_{\sigma(1)},X_{\sigma(2)}),\dots,\sigma(X_{\sigma(1)},\dots,X_{\sigma(n)})
$$
where we are only shown the permuted values. Answers to follow-up questions, like what information is revealed by the last $\sigma$-algebra, are subject to the joint distribution of $X$ and $\sigma$, where only the values (with multiplicities) are revealed if $\sigma$ is independent of $X$, and all of $(X,\sigma)$ is revealed if $\sigma$ is determined by the values (with multiplicities) - intuitively speaking.
