Consider the following problem:
Suppose $f: [0; \infty) \to [0; \infty)$ is a continuous bijection. Suppose two people are given two envelopes with $x \$ $ and $f(x) \$ $ respectively. They know the value of $f$ and the amount of money in their envelope (but do not know, whether this is $x$ or $f(x)$). They consider getting $x \$ $ and $f(x) \$ $ to be equally likely. Then they are allowed to request exchange of envelopes (and if they both do, the envelopes are exchanged between them). Is the situation, where both of them request the exchange, always a Bayes-Nash equilibrium?
First, note that the game is symmetric.
Then, suppose, a player got $y$ in their envelope. Then, they consider the amount of money in another players envelope to be either $f(y)$ (in case, when $y = x$) or $f^{-1}(y)$ in case ($y = f(x)$) with equal probability. Thus the expected gain in case when both request exchange is $\frac{f(y) + f^{-1}(y)}{2}$ and $y$, if at least some of them refuses. Thus our question is equivalent to the following problem:
Suppose $f: [0; \infty) \to [0; \infty)$ is a continuous bijection and $y \in [0; \infty)$. Is it always true, that $f(y) + f^{-1}(y) \geq 2y$?
I managed to solve this problem (with positive answer) only for the case when $f$ is linear. Indeed, if $f(x) \equiv cx$ for $c \in (0 ; \infty)$, then $f(y) + f^{-1}(y) - 2y = cy + c^{-1}y - 2y = c^{-1}y(c - 1)^2 \geq 0$. However, I do not know, whether this method can be generalised to an arbitrary $f$...
It's not always true. If $f(y)=\sqrt y$, then $f(.5)+f^{-1}(.5)-2\times .5=\sqrt{\frac 12}+\frac 1 4 - 1<0.$