Stochastic dominance characterization

Question

Consider two probability measures on $\Bbb R$ given by $\mu$ and $\nu$. We write $\mu\leq \nu$ if there exists a joint distribution $P$ with the latter marginals such that $P(x\leq y) = 1$. In particular, that implies that $\mu f \leq \nu f$ for any non-decreasing function $f$. I guess that there does not exist $f^*$ such that $\mu f^* \leq \nu f^*$ implies that $\mu\leq \nu$. How can I prove that?

Answer 1

Suppose such nondecreasing $f^*$ exists.

If $f^*$ is strictly increasing then choose $x < y <z\in\mathbb R$ and we have $f^*(x)<f^*(y)<f^*(z)$. now choose $\varepsilon > 0 $ such that $f^*(y) > \varepsilon f^*(x) + (1-\varepsilon)f^*(z)$.

Now let $\mu$ be the measure concentrated on $y$ and let $\nu$ be the measure concentrated on $\{x,z\}$ with $\nu\{x\} = \varepsilon$ and $\nu\{z\} = 1-\varepsilon$.

Now $\mu f^* = f^*(y) > \varepsilon f^*(x) + (1-\varepsilon)f^*(z) = \nu f^*$ but it is clearly not the case that $\mu\geq\nu$.

If $f^*$ is not strictly increasing then there must exist $x < y$ with $f^*(x) = f^*(y)$, so let $\mu$ be the probability measure concentrated on $y$ and $\nu$ the probability measure concentrated on $x$. Clearly we do not have $\mu\leq \nu$ but $\mu f^* = f^*(y) \leq f^*(x) = \nu f^*$.