$X:\Omega\to [0,1]$ is a random variable. It is known that first order stochastic dominance FOSD is a partial order that is transitive:

$X$ FOSD $Y$, $Y$ FOSD $Z$ implies $X$ FOSD $Z$.

Now consider the relation not FOSD. $Z$ not FOSD $Y$ and $Y$ not FOSD $X$.

Is it true that $X$ FOSD $Z$ is not possible?


1 Answer 1



Suppose $Y$ is Bernoulli(0.5) and $X$ and $Z$ are point masses at 1/4 and 3/4. Then $Z\succ X$ in the partial ordering, but $(X,Y)$ and $(Y,Z)$ are both incomparable, so $X\not\prec Y$ and $X\not\succ Y$ and so on. You'd need a total order or preorder rather than a partial order to get transitivity both ways.

  • $\begingroup$ Many thanks for your elegant answer! Regarding your last statement: is it universally true that if $\succ$ is a nontotal partial order, then $\not\succ$ must always violate transitivity? $\endgroup$
    – High GPA
    Commented Dec 3, 2023 at 12:50
  • 1
    $\begingroup$ There are some edge cases that are transitive, such as the partial order where no elements are comparable, or the order where one element $A$ has $A\succ x$ for all other $x$ in the set and there are no other comparable pairs. $\endgroup$ Commented Dec 4, 2023 at 2:47

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