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The Mann–Whitney U test has the following interpretation reported on wikipedia:

[...] is a nonparametric test of the null hypothesis that, for randomly selected values X and Y from two populations, the probability of X being greater than Y is equal to the probability of Y being greater than X.

I had intuitively taken this to mean "one variable tends to be larger than the other", but I recently encountered different notions of stochastic dominance including statewise dominance, first order dominance, second-order dominance, third-order dominance, and a less-specific notion of higher-order stochastic dominance.

Statewise stochastic dominance:

Random variable A is statewise dominant over random variable 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.

First-order stochastic dominance:

Random variable A has first-order stochastic dominance over random variable 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, $ P[A\geq x]\geq P[B\geq x]P[A\geq x]\geq P[B\geq x]$ for all $x$, and for some $x$, $P[A>x]>P[B>x] P[A>x]>P[B>x]$

Second-order stochastic dominance:

In terms of cumulative distribution functions $F_{A}$ and $F_{B}$, $A$ is second-order stochastically dominant over $B$ if and only if the area under $F_{A}$ from minus infinity to $x$ is less than or equal to that under $F_{B}$ from minus infinity to $x$ for all real numbers $x$, with strict inequality at some $x$; that is, $\int _{-\infty }^{x}[F_{B}(t)-F_{A}(t)]\,dt\geq 0$ for all $x$, with strict inequality at some $x$. Equivalently, $A$ dominates $B$ in the second order if and only if ${E} [u(A)]\geq \operatorname {E} [u(B)]$ for all nondecreasing and concave utility functions $u(x)$.

Third-order stochastic dominance:

Let $F_{A}$ and $F_{B}$ be the cumulative distribution functions of two distinct investments $A$ and $B$. $A$ dominates $B$ in the third order if and only if $\int _{-\infty }^{x}\int _{-\infty }^{z}[F_{B}(t)-F_{A}(t)]\,dt\,dz\geq 0{\text{ for all }}x$ [AND] ${E} _{A}(x)\geq \operatorname {E} _{B}(x)$, and there is at least one strict inequality. Equivalently, $A$ dominates $B$ in the third order if and only if $\operatorname {E} _{A}U(x)\geq \operatorname {E} _{B}U(x)$ for all nondecreasing, concave utility functions $U$ that are positively skewed (that is, have a positive third derivative throughout).

Higher-order stochastic dominance:

[Yet to be defined]

Which form of stochastic dominance is tested for using the Mann–Whitney U hypothesis test?

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According to this paper, the alternative hypothesis is,

$$H_1: P(X_i > Y_j) \neq 0.5. $$

It implies the usual stochastic ordering which is the first order stochastic dominance.

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    $\begingroup$ And you can verify that but just examining how you estimate the Wilcoxon-Mann-Whitney concordance probability: take all possible pairs of observations from different groups and compute the proportion of pairs for which the group 1 value exceeds the group 2 value. $\endgroup$ Jul 19 '21 at 11:30
  • $\begingroup$ This is not 1st order stochastic dominance, because it is possible to reject the null for a rank sum test, but, for example have CDFs which cross (even more than once). $\endgroup$
    – Alexis
    Oct 30 '21 at 23:31

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