Stochastic order

Question

I need help with the following exercise:

$X$ and $Y$ are random variables in $\mathbb{R}$ with the distribution functions $F_X$ and $F_Y$, so $X$ is stochastically smaller or equal to $Y$, i.e. $(X\leq_{st}Y)$ if $F_X(t)\geq F_Y(t)$ holds for all $t\in\mathbb{R}$.

How can I show the following statement?

$X\leq_{st}Y$ implies $E[f(X)]\leq E[f(Y)]$ for every monotone increasing function $f:\mathbb{R}\rightarrow \mathbb{R}$, so that the given expected values exist.

Answer 1

Let $\Phi_{X}$ and $\Phi_{Y}$ be functions $\left(0,1\right)\to\mathbb{R}$ prescribed by $u\mapsto\inf\left\{ z\in\mathbb{R}\mid F_{X}\left(z\right)\geq u\right\} $ and $u\mapsto\inf\left\{ z\in\mathbb{R}\mid F_{Y}\left(z\right)\geq u\right\} $ respectively.

These are the so-called "inverses" of $F_X$ and $F_Y$ and it is well known that $\phi_{X}\left(U\right)\stackrel{d}{=}X$ and $\phi_{Y}\left(U\right)\stackrel{d}{=}Y$ if $U$ has uniform distribution on $(0,1)$.

Eventually have a look at this answer about that fact.

Further $X\leq_{st}Y$ implies that $\Phi_{X}\left(u\right)\leq\Phi_{Y}\left(u\right)$ for every $u\in\left(0,1\right)$ and consequently: $$f\left(\Phi_{X}\left(u\right)\right)\leq f\left(\Phi_{Y}\left(u\right)\right)\text{ for every }u\in\left(0,1\right)$$

Then we find:

$$\mathbb{E}f\left(\phi_{X}\left(U\right)\right)=\int_{0}^{1}f\left(\phi_{X}\left(u\right)\right)du\leq\int_{0}^{1}f\left(\phi_{Y}\left(u\right)\right)du=\mathbb{E}f\left(\phi_{Y}\left(U\right)\right)$$

This with $\mathbb Ef(X)=\mathbb{E}f(\phi_{X}(U)$ and $\mathbb Ef(Y)=\mathbb{E}f(\phi_{Y}(U)$ so we are ready.

This all under assumption that the expectations exist of course.

Answer 2

This is a fundamental result in stochastic dominance. See theorem 1' on page 29 of Hadar & Russel 1969.

TL;DR: integrate by parts the difference of the expectations $\int f(x) [F'(x)-G'(x)]dx$

Answer 3

The typical argument of the first-order stochastic dominance property is that $$ \int_{a}^{b} f_x(t) u(t) dt = [(1-F_x(t))u(t)]_{a}^b + \int_{a}^b u'(t) (1-F_x(t))dt $$ so that $$ \int_{a}^{b} f_x(t) u(t) dt -\int_{a}^{b} f_y(t) u(t) dt = \int_{a}^b u'(t) (F_y(t)-F_x(t))dt, $$ and since $u'(t) \ge 0$ and $F_x(t) \ge F_y(t)$ for all $t$, the right-hand side is non-positive, so $\int_{a}^{b} f_y(t) u(t) dt \ge \int_{a}^{b} f_x(t) u(t) dt$.

I don't think you can use this kind of argument to prove the expectations exist? If $(1-F_x(b))u(b)$ diverges as $b\rightarrow \infty$, so that $u(t)$ is growing faster than $1-F_x(t)$ is tending to zero, the expectation might fail to exist and the calculation on the first line would fail.

If the distributions have compact support or if $f$ is not just continuous but also bounded, you could prove existence, but it wouldn't really follow from the stochastic dominance argument, just standard results like the dominated convergence theorem.