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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.

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3 Answers 3

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.

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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.

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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.)

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