# When is $A \leq B$ with fixed probability $p> \frac{1}{2}$ a partial order?

Surprisingly, (to me) the relation defined as $$(A,B) \in \leq_q$$ if $$p(A \leq B) \geq q$$ is not generally a partial order.

I would ideally like to know the weakest conditions sufficient for it to be true. I don't really mind whether the definitions use $$\lt$$ instead, or the desired conditions secure a strict partial order. For just one example, is it true when all underlying variables are jointly Gaussian distributed with arbitrary covariance? Obviously the relation I have in mind is based on the (possibly dependent) joint distribution.

For a quite trivial example that induces a strict partial order, take an independent trivariate Gaussian with means $$0,1,2$$ and unit covariance has $$p(A 0.5$$, $$p(B0.5$$ and $$p(A 0.5$$. Also obviously $$p(A is greater than either $$p(A or $$p(B. This is a transitive, irreflexive, anti-symmetric relation. This is a strict partial order.

• Do you have a non-trivial example where this is true? Oct 13, 2023 at 16:08
• For the case of variables with joint Gaussian distribution, it's true, if $A < B$ is defined as $P( A < B) > 1/2$. Then $P(A < B) > 1/2$ means that $P(B-A > 0) > 1/2$ and $B-A$ is normal with mean $E[B] - E[A]$. For a Gaussian $X$, $P(X >c) > 1/2$ iff $c < E[X]$. Thus $A<B$ in this ordering if and only if $E[A] < E[B]$, which gives a partial order. Oct 24, 2023 at 19:08
• Thanks @JairTaylor, so this holds irrespective of the marginal variances of A,B,C, and their covariance? Does this generalize to some larger class of (marginally symmetric?) distributions?
– JRC
Nov 7, 2023 at 14:32
• @JRC Yes, it holds regardless of covariance considerations. I think it should generalize to any family of continuous distributions with equal mean and median. Nov 7, 2023 at 19:36
• @JairTaylor Ah yes. I see. Thanks! I'm wonder what leverage there would be to generalize further by requiring p to be large. The case of p = 1 appears to be called strict "almost sure" dominance?
– JRC
Nov 8, 2023 at 12:22

Here are some conditions under which we can form a poset in this way, although there may be a more general solution.

Let $$\mathcal{F}$$ be a set of random variables. For any $$X,Y \in \mathcal{F}$$, $$X \neq Y$$, set $$Z= Y-X$$; assume that

(i) $$P(Z > E[Z]) = 0.5$$, and

(ii) the cdf $$f(x) = P(Z < x)$$ is strictly increasing for $$x \in (E[Z] - \epsilon, E[Z] + \epsilon)$$ for some $$\epsilon$$.

Again setting $$Z = Y-X$$, we claim that $$P(Z > 0) > 0.5$$ if and only if $$E[Z] > 0$$. For suppose that $$E[Z] > 0$$; then $$P(Z > 0) > P(Z >E[Z]) = 0.5$$, by (i) and (ii); and similarly if $$E[Z] \leq 0$$ then $$P(Z >0) \leq P(Z > E[Z]) = 0.5$$ by (i).

Defining $$X<_PY$$ if $$P(X 0.5$$, it follows that $$X<_PY$$ if and only if $$E[X] < E[Y]$$, and so $$S$$ is a poset isomorphic to the poset of the multiset $$\{E[X] | X \in S\}$$ in the usual ordering in $$\mathbb{R}$$.

In particular, any set of Gaussian random variables, regardless of covariance, satisfies the criteria. It's possible that (ii) could be removed or replaced; I'm not sure how to proceed in that case.

Regarding your other comment, if we define $$X to mean $$P(X p$$ for some $$p > 0.5$$, this forms a poset if we add some more restrictions, for example that they are independent Gaussians with the same variance $$\sigma^2$$. If that's the case, then $$P(X p$$ is equivalent to $$E[Y] - E[X] > f(p)$$ where $$f(p)$$ is defined by $$P(N >f(p)) = p$$ where $$N \sim N(0,\sigma\sqrt{2})$$. Probably there are some less restrictive hypotheses we could give.