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 $p(A<B) > 0.5$, $p(B<C)>0.5$ and $p(A<C) > 0.5$. Also obviously $p(A<C)$ is greater than either $p(A<B)$ or $p(B<C)$. This is a transitive, irreflexive, anti-symmetric relation. This is a strict partial order.

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    $\begingroup$ Do you have a non-trivial example where this is true? $\endgroup$
    – GEdgar
    Oct 13, 2023 at 16:08
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    $\begingroup$ 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. $\endgroup$ Oct 24, 2023 at 19:08
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    $\begingroup$ 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? $\endgroup$
    – JRC
    Nov 7, 2023 at 14:32
  • $\begingroup$ @JRC Yes, it holds regardless of covariance considerations. I think it should generalize to any family of continuous distributions with equal mean and median. $\endgroup$ Nov 7, 2023 at 19:36
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    $\begingroup$ @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? $\endgroup$
    – JRC
    Nov 8, 2023 at 12:22

1 Answer 1


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 $P(X <Y) > 0.5$, 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 $X<Y$ to mean $P(X<Y) > p$ 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 $P(X<Y) > p$ 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.


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