Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Say for two independent distributions $X,Y$, I define that $X \triangleleft Y$ means $P(X > Y) < P(X < Y)$.

Is this relationship transitive? That is if $X \triangleleft Y$ and $Y \triangleleft Z$ then $X \triangleleft Z$?

My instinct is that it isn't and there's a fairly obvious counter example of discrete distributions. But I can't seem to come up with it.

share|cite|improve this question
up vote 3 down vote accepted

Let $X=0$ with probability $1$. Let $Y=1$ with probability $p$ and $-2$ with probability $1-p$. Let $Z=2$ with probability $q$ and $-1$ with probability $1-q$.

Then $X\triangleleft Y\,$ iff $p>\frac12$, $Y\triangleleft Z$ iff $q+(1-q)(1-p)>\frac12$, and $Z\triangleleft X$ iff $q<\frac12$, so to get a counterexample you need only choose $p,q\in[0,1]$ so that the system

$$\begin{cases} p>\frac12&\\ q+(1-q)(1-p)>\frac12&\\ q<\frac12& \end{cases}$$

of inequalities is satisfied. This is not hard; you can even have $p+q=1$, if you want.

share|cite|improve this answer
FYI ... To avoid circularity in the definition, the question has been edited to use "$\triangleleft$" instead of "$<$". – r.e.s. Dec 10 '11 at 13:17
I do not understand $X<Y$ iff $p>\frac12$ (nor two other similar statements later on). – Did Dec 10 '11 at 13:38
@DidierPiau: $X \triangleleft Y$ $\Leftrightarrow$ $P(X > Y) < P(X < Y)$ $\Leftrightarrow$ $P(0 > Y) < P(0 < Y)$ $\Leftrightarrow$ $1-p < p$ – r.e.s. Dec 10 '11 at 13:55
@r.e.s.: Notation changed to match; thanks for the heads-up. – Brian M. Scott Dec 10 '11 at 18:28
Now I understand... :-) Well done Brian. – Did Dec 10 '11 at 21:37

Let $X$ be uniform on the set $\{1, 6, 8\}$, let $Y$ be uniform on the set $\{2, 4, 9\}$, and let $Z$ be uniform on the set $\{3, 5, 7\}$, where these three random variables are independent. Then $P(Y>X) = P(Z>Y) = P(X>Z) = 5/9$, so $X \triangleleft Y \triangleleft Z \triangleleft X$. This is an example of nontransitive dice.

share|cite|improve this answer

These are nontransitive coins such that each coin has the same pair of probabilities $p, 1-p$ for its two sides, which are labelled from the set $\{1,2,3,4,5,6\}$:

        Side with probability 
Coin          p      1-p
----         ---     ---
  X           2       6
  Y           3       4
  Z           5       1

Because $P(X < Y) = P(Y < Z) = p$ and $P(Z < X) = 1-p^2$, it follows from $p = 1-p^2$ that $p$ must be the "little" golden ratio $\frac{\sqrt{5} - 1}{2} \approx 0.6180$.

NB: This can be seen as an embellishment of the answer by Brian Scott, modified to use a set of consecutive positive integers, and to make the coins equally "biased". (It was merely coincidental that I happened to assign prime numbers to the sides with probability $p$ and non-primes to the opposite sides, and also that the golden ratio happens to be involved.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.