This question is followup to the previous similar question. There I was trying to find good sufficient condition for abstract preorder to be symmetric, but now, as I have found good formalization of my specific preorder, I'll better ask about it directly.

Let's say we have arbitrary finite measure space $\langle \Omega, \mathfrak{I}, \mathbb{P} \rangle$. For any functions $\rho : Y \rightarrow X$ and measurable $f : \Omega \rightarrow X$ with any finite $X$ and $Y$ we define $H_{\rho}(f) = \{\mathbb{P} \circ g^{-1} \mid f = \rho \circ g, g \text{ is measurable}\}$ – set of functions $Y \rightarrow \mathbb{R}_{\ge 0}$ characterizing internal structure of $f$ induced by $\rho$. Preorder $\precsim$ is defined on set of measurable functions $\{ f : \Omega \rightarrow X \}$ as follows – $f_1 \precsim f_2 \iff H_{\rho}(f_1) \subseteq H_{\rho}(f_2), \forall \rho$.

I believe that such preorder is symmetric ($f_1 \precsim f_2 \Leftrightarrow f_2 \precsim f_1, \forall f_1, f_2$) due to external reasons (it has respective game-theoretical interpretation), but I'm unable to prove it as defined. Can anyone help me, please?

UPDATE: Please, take note, $\forall \rho$ here means for all $\rho$ across every possible finite $Y$, not just one externally fixed.

UPDATE: To keep you guys interested and maybe bootstrap thoughts, I'll add some facts.

  1. $f_1 \precsim f_2 \Rightarrow \xi \circ f_1 \precsim \xi \circ f_2, \forall f_1, f_2, \xi$ – this follows trivially from definition of $H_{\rho}$.
  2. Consequentially, symmetry of $\precsim$ is easily provable for $f_2 = \pi \circ f_1, \pi : X \leftrightarrow X$. Because of $f_1 \precsim f_2 \Rightarrow \pi \circ f_1 \precsim \pi \circ f_2$ we can build infinite chain of $f_1 \precsim \pi \circ f_1 \precsim \pi \circ \pi \circ f_1 \precsim \ldots$. Combinatorics says that there must be finite cycle $\pi \circ ... \circ \pi$ synonymous to identity permutation, which leads to $f_2 = \pi \circ f_1 \precsim \pi \circ ... \circ \pi \circ f_1 = f_1$.
  3. Consequentially, symmetry of $\precsim$ is easily provable for measure spaces with finite algebra. In such case any measurable function $f_i$ can be derived from singular function $g$, that transforms each atom to its own element of codomain, as $f_i = \xi_i \circ g$. For such $g$ and its permutations $g \precsim \pi \circ g \Leftrightarrow \pi \circ g \precsim g \Leftrightarrow \mathbb{P} \circ g^{-1} = \mathbb{P} \circ (\pi \circ g)^{-1}$ which leads to $f_1 \precsim f_2 \Leftrightarrow f_2 \precsim f_1$ by definition of $H_{\rho}$.

Unfortunately, here my progress bumps into wall – going from this subcases to generic case seems to be beyond my knowlegde. And this is sad, because symmetry of preorder in question is a cornerstone of formal argumentation for very interesting new result in game theory. I would be very grateful even for slightest hints.

  • $\begingroup$ You should add measurability of $g$ in the definition of $H_\rho(f)$ (and perhaps delete or rethink the explanation after that definition). $\endgroup$ – Jochen Wengenroth May 8 at 7:16
  • $\begingroup$ If $f_1$ and $f_2$ have different distributions $\mathbb P^{f_j}=\mathbb P\circ f_j^{-1}$ the sets $H_\rho(f_j)$ are disjoint becaus any $Q\in H_\rho(f)$ satisfies $Q^\rho=\mathbb P^f$. Does this help? $\endgroup$ – Jochen Wengenroth May 8 at 7:19
  • $\begingroup$ @JochenWengenroth, added measurability of $g$, but I'm not sure what to do with explanation of $H_\rho$, as proper explanation would require bringing here pretty esoteric game-theoretical stuff, and I really want to avoid it. $\mathbb{P}^{f_1} = \mathbb{P}^{f_2}$ is a requirement for comparability of $f_1$ and $f_2$, as you well noted, but unfortunately it doesn't really help because roadblock here is a bit more complex - it's about quantified $\rho$ in both parts of $H_{\rho}(f_1) \subseteq H_{\rho}(f_2), \forall \rho \Leftrightarrow H_{\rho}(f_1) \supseteq H_{\rho}(f_2), \forall \rho$. $\endgroup$ – Doktor Diagoras May 8 at 7:51

I think I found a counterexample.

Let $Y=\{1\}$, $X=\Omega=\{1,2\}$, $f_1(\omega)=\omega$, $f_2(\omega)=1$.

Then for all $\rho:Y\to X$ there exists no $g$ such that $f=\rho \circ g$, thus $H_\rho(f_1)$ is empty and $f_1\precsim f_2$.

On the other hand, for $\rho = x\mapsto 1$ we have $H_\rho(f_2)=\{\mathbb P(\Omega)\}$. Thus $f_2\not\precsim f_1$.

  • $\begingroup$ I believe, we have a misunderstanding here. When I write $\forall \rho$, I mean for all $\rho$ across every possible finite $Y$, not just one externally fixed. I'll update my question. $\endgroup$ – Doktor Diagoras May 8 at 13:33

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