# Is the relation "not FOSD" transitive?

$$X:\Omega\to [0,1]$$ 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?

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.
• 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? Commented Dec 3, 2023 at 12:50
• 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. Commented Dec 4, 2023 at 2:47