Expected value, first-order dominance and super modularity

Question

Suppose a function $f:X\times Y\to\mathbb{R}$ - with $X\subseteq\mathbb{R}$ and $Y\subseteq \mathbb{R}$ - satisfies increasing differences and is increasing in $X$.

Consider another function $g:\Delta(X) \times Y\to\mathbb{R}$ given by $g(\mu,y)=\mathbb{E}_{\mu}[f(x,y)]$. That is, $g(\mu,y)$ is the expected value of $f(x,y)$ with respect to a measure $\mu$, for a fixed $y$.

Endow $\Delta(X)$ with the first order stochastic dominance relation, so that $(\Delta(X), \succsim_{FSD})$ is a lattice.

Question: can we say that $g(\mu,y)$ is super modular in $(\Delta(X), \succsim_{FSD})$ for every fixed y?

Relevant definitions:

1. $f$ satisfies increasing differences if $x'\ge x$ and $y'\ge y$ implies $f(x',y')-f(x',y)\ge f(x,y')-f(x,y)$;
2. $\mu'\succsim_{FSD}\mu$ if $\mathbb{E}_{\mu'}[h]\ge\mathbb{E}_{\mu}[h]$ for every increasing, bounded function $h$;
3. g is supermodular if $g(\mu\vee \mu',y)+g(\mu\wedge\mu',y)\ge g(\mu,y)+g(\mu',y)$ for any $\mu, \mu'$ and any $y$.

Answer 1

This is an elaboration of your comment that actually $$ g(\mu\vee \mu',y)+g(\mu\wedge\mu',y)= g(\mu,y)+g(\mu',y) $$ (the variable $y$ is irrelevant as well as the assumptions about $f$). As you have mentioned, the equality boils down to the linearity of expectation (w.r.t. the measure) and the equality (of measures) $\mu\vee\mu'+\mu\wedge\mu=\mu+\mu'$.

To understand why this is true, recall that the stochastic dominance $X\precsim_{FSD} Y$ is equivalent to the pointwise comparison $F_X(x)\ge F_Y(x)$, $x\in \mathbb{R}$, of cdfs. In other words, $(\Delta(X), \precsim_{FSD})$ as a lattice is isomorphic to the set of cdfs with pointwise (reversed) order $\ge$. Therefore, for any $f$, which is integrable with respect to both $\mu$ and $\mu'$, $$ \mathbb{E}_{\mu\wedge \mu'}[f] + \mathbb{E}_{\mu\vee \mu'}[f] = \int_{\mathbb R} f(x) dF_{\mu\wedge \mu'}(x) + \int_{\mathbb R} f(x) dF_{\mu\vee \mu'}(x)\\ = \int_{\mathbb R} f(x) d\big(F_{\mu\wedge \mu'}(x) + F_{\mu\vee \mu'}(x)\big) = \int_{\mathbb R} f(x) d\big(F_{\mu}(x)\vee F_{\mu'}(x) + F_{\mu}(x)\wedge F_{\mu'}(x)\big)\\ = \int_{\mathbb R} f(x) d\big(F_{\mu}(x) + F_{\mu'}(x)\big) = \mathbb{E}_{\mu}[f] + \mathbb{E}_{\mu'}[f], $$ as claimed.

---

You might also be interested in a slightly less trivial property, which can be proved similarly under the "increasing differences" assumption: $$ \mathbb{E}_{\mu\wedge \mu'}[f(x,y\wedge y')] + \mathbb{E}_{\mu\vee \mu'}[f(x,y\vee y')]\ge \mathbb{E}_{\mu}[f(x,y)] + \mathbb{E}_{\mu'}[f(x,y')]. $$