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I've been looking for methods like Mann Whitney, but without homogeneity of variance. So far, I found that I suppose to test for stochastic dominance instead - i.e. modifying the null hypothesis. Could anybody refer books or any literature for me to study further? Thanks.

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    $\begingroup$ like Mann Whitney, but without homogeneity of variance. M-W tests for the stochastic dominance. It does not require homogeneity of variance. $\endgroup$ – ttnphns Jun 26 '16 at 13:05
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    $\begingroup$ There have been many, many good discussions of Mann-Whitney and its direct extension Kruskal-Wallis tests on this site. May I recommend you to search and browse though them first? $\endgroup$ – ttnphns Jun 26 '16 at 13:08
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    $\begingroup$ Moreover, M-W knows nothing about specifically variances. It is rank-based nonparametric test. $\endgroup$ – ttnphns Jun 26 '16 at 13:21
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    $\begingroup$ @AndyW Looking around, since "P(X>Y)=P(Y>X)" is often called "stochastic equality", it looks like some people have occasionally taken to calling "P(X>Y)>P(Y>X)" stochastic dominance. It appears not to be very widespread though. Mann and Whitney's original 1947 paper uses the term "stochastically larger", which would seem to be a better term. $\endgroup$ – Glen_b -Reinstate Monica Jun 26 '16 at 16:40
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    $\begingroup$ However, the Mann-Whitney can be considered a test for the more strict case of stochastic dominance (as given at the link AndyW gave) -- one could accompany the minimal assumptions of the test with say the assumption of equality under the null and a dominance alternative -- as long as you can add those assumptions it will work fine. The test is somewhat sensitive to other kinds of differences, so if you can't make those additional assumptions it wouldn't be suitable. $\endgroup$ – Glen_b -Reinstate Monica Jun 26 '16 at 17:30
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First, let me point out that Mann-Whitney does not test stochastic dominance. It tests the null hypothesis $P(X > Y) = P(X < Y)$. If $X$ is uniform on $[0,1]$ and $Y$ is 1 with probability 0.9 and 0 with probability 0.1, then the null is false but neither variable stochastically dominates the other. Stochastic dominance is a stronger condition.

There's plenty of literature on testing for dominance, though it is highly technical and requires some understanding of empirical processes. A Google Scholar search for "test stochastic dominance" will work. A famous reference is, e.g., Barrett and Donald (2003), Econometrica, 71, 1, "Consistent Tests for Stochastic Dominance".

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  • $\begingroup$ (+1) But see Glen_b's comment. The necessary assumptions can often be reasonable ones - more so than a location-shift alternative - & gross violations are apparent from examination of the empirical distribution functions. $\endgroup$ – Scortchi - Reinstate Monica Aug 8 '19 at 9:34

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