First order stochastic dominance

Key idea

Higher expected value from the dominant distribution

Details

We say $F(x)$ F.O.S.D $G(x)$ if $F(x) \leq G(x), \forall x$

Higher expected value (1)

Suppose the support of the distribution is some interval $[\alpha, \beta]$

$\int^{\beta}_{\alpha} x f(x) dx = \int^{\beta}_{\alpha} x dF(x) = \left[ xF(x) \right]^{\beta}_{\alpha} - \int^{\beta}_{\alpha} F(x) dx$

As $F(x) \leq G(x) \forall x$, then $\int^{\beta}_{\alpha} F(x) dx \leq \int^{\beta}_{\alpha} G(x) dx$, minus a smaller quantity in turn implies that $E(F(x)) \geq E(G(x))$.

Higher expected value (2)

Consider some non-decreasing function $u(x)$:

$\int^{\beta}_{\alpha} u(x) [f(x) - g(x)] dx = \int^{\beta}_{\alpha} u(x) d[F(x) - G(x)]$

Using integration by parts again, this equals to:

$\left[ u(x)[F(x) - G(x)] \right]^{\beta}_{\alpha} - \int^{\beta}_{\alpha} [F(x) - G(x)] u'(x) dx = - \int^{\beta}_{\alpha} [F(x) - G(x)] u'(x) dx$

Again as $F(x) \leq G(x) \forall x$ and $u'(x) \geq 0$, then we have:

$E_{F(x)}(u(x)) \geq E_{G(x)}(u(x))$

Conditional stochastic dominance

Key idea

Truncate from right, left-tail probability mass higher

Details

Consider two differentiable distribution functions $F(\theta)$ and $G(\theta)$ with common support $[\alpha, \beta]$.

Define $F_y(\theta)$ and $G_y(\theta)$ to be the two distribution functions right truncated at point $y$. That is,

$F_y(\theta) = \frac{F(\theta)}{F(y)}; \forall \theta \in [\alpha, y]$ $G_y(\theta) = \frac{G(\theta)}{G(y)}; \forall \theta \in [\alpha, y]$

Then we say $F(.)$ conditionally stochastic dominate $G(.)$ if

$F_y(\theta) = \frac{F(\theta)}{F(y)} < \frac{G(\theta)}{G(y)} = G_y(\theta), \forall y \in [\alpha, \beta]; \forall \theta < y$

In words, we say $F(.)$ conditionally stochastic dominate $G(.)$ if for every right truncation $y$, the resultant probability mass in the left-tail is less in $F(.)$ than in $G(.)$.

Useful links

Conditional stochastic dominance: http://www.econ.ucla.edu/riley/201C/2017/MathematicalNote.pdf