Is $A \leq B$ with probability $p>1/2$ a partial order on random variables?

Question

It seems reflexive, antisymmetric and transitive. My mathematics training is limited though and I would love proof or contradiction. A quick search turns up many references cataloging on stochastic orderings - e.g. stochastic dominance, hazard rate order, etc. None so far seems to mention this most basic one though.

Is the relation defined as $(A,B) \in \leq_q$ if $p(A \leq B) \geq q$ a partial order too? I feel like it should be - at least for $q>1/2$, and maybe assuming something about the dependence between any pair $A,B$? - but need help to prove it.

Answer 1

No, as it is not anti-symmetric.

Anti-symmetry requires that $a≤b\,\,\&\,\,b≤a\implies a=b$.

To get a counter example in your case, suppose that $A\in \{0,1\}$ with probabilities $\{.99,\, .01\}$ and $B\in \{0,2\}$ also with probabilities $\{.99,\, .01\}$ Then, following your definition we see that $A≤B$ and $B≤A$ yet they are not the same variable.

More broadly, you get $A≤B$ and $B≤A$ any time $A=B$ with probability $>\frac 12$, yet that relation does not imply equality of the variables, even if you interpret "equality" to mean "identical distribution".

Answer 2

I don't believe your definition will work.

Here's a counter-example: Efron's non-transitive dice.

Consider four six-sided non-standard dice with the following markings on each side:

• A: {1,2,3,9,10,11}
• B: {0,1,7,8,8,9}
• C: {5,5,6,6,7,7}
• D: {3,4,4,5,11,12}

If you roll A versus B, you'll find that more than 50% of the time A beats B. And B beats C. And C beats D. And D beats A.

You might wish to look into an extension of this idea, my Lake Wobegon Dice.