Stochastic dominance is a partial order between random variables. The concept arises in decision theory and decision analysis in situations where one gamble (a probability distribution over possible outcomes) can be ranked as superior to another gamble for a broad class of decision-makers.
Statewise dominance
The simplest case of stochastic dominance is statewise dominance (also known as state-by-state dominance), defined as follows:
$A$ is statewise dominant over $B$ if $A$ gives at least as good a result in every state (every possible set of outcomes), and a strictly better result in at least one state.
Anyone who prefers more to less (in the standard terminology, anyone who has monotonically increasing preferences) will always prefer a statewise dominant gamble.
First-order
First-order stochastic dominance is defined as:
$A$ has first-order stochastic dominance over $B$ if for any outcome $x$, $A$ gives at least as high a probability of receiving at least $x$ as does $B$, and for some $x$, A gives a higher probability of receiving at least $x$.
In notation form, $\mathrm{P}(A \geq x) \geq \mathrm{P}(B \geq x)$ for all $x$, and for some $x, \mathrm{P}(A \geq x) > \mathrm{P}(B \geq x)$.
In terms of the cumulative distribution functions, $A$ first-order stochastically dominates $B$ means that ${\displaystyle F_{A}(x)\leq F_{B}(x)}$ for all $x$, with strict inequality at some $x$.
For example $$ \begin{array}{lllllll} \text { State (die result) } & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text { Gamble A wins } \$ & 1 & 1 & 2 & 2 & 2 & 2 \\ \text { Gamble B wins } \$ & 1 & 1 & 1 & 2 & 2 & 2 \\ \text { Gamble C wins } \$ & 3 & 3 & 3 & 1 & 1 & 1 \\ \hline \end{array} $$ Here gamble $A$ statewise dominates gamble $B$, and $C$ first-order stochastically dominates $B$. Gambles $A$ and $C$ cannot be ordered relative to each other.
As long as the decision maker prefers having more wealth to less (his utility function is increasing), he would prefer $A$ to $B$ if $A$ first-order stochastically dominates $B$. For every weakly increasing utility function $u$, we have $$ \int u(x) \mathrm{d} F_A(x) \geq \int u(x) \mathrm{d} F_B(x) $$
图示：
Second-order
Now assume that $A$ and $B$ have the same mean, so that one does not dominate the other in the sense of first-order stochastic dominance. Can we still say that a risk-averse decision maker prefers $A$ to $B$ without knowing his utility function $u$ ?
$A$ second-order stochastically dominates $B$ if and only if the decision maker weakly prefers $A$ to $B$ under every weakly increasing concave utility function $u$. That means $$ \int u(x) \mathrm{d} F_A(x) \geq \int u(x) \mathrm{d} F_B(x) \quad \text{for concave }\, u $$ Equivalently, if and only if $$ \int_{-\infty}^{x}\left[F_{B}(t)-F_{A}(t)\right] \mathrm{d} t \geq 0 \;\;\text { for all }\, x $$
First-order stochastic dominance of $A$ over $B$ is a sufficient condition for second-order dominance of $A$ over $B$.
图示：
