Prove or disprove the existence of rv's X, Y and Z such that:

Yes, it is possible.There are 6 possibilities which are mutually disjoint: three cyclic: X > Y > Z ; Y > Z > X ; Z > X > Y and the 3 counter-cyclic combinations.Now suppose I have a sample space of 6 (X,Y,Z) events such that there is exactly 1 in each disjoint set. Also suppose that all cyclic events have probability = a which is > 1/6 and the counter-cyclic events have probability = b (< 1/6).Then, 3a + 3b = 1 => a + b = 1/3.Now, P(X > Y) = P(X > Y > Z) + P(X > Z > Y) + P(Z > X > Y) = 2a + b = a + 1/3 > 1/2 because a > 1/6Similarly, P(Y > Z) and P(Z > X) are also > 1/2 because each is composed of 2 cyclic and only 1 counter-cyclic permutation.

Maybe this is the same with what needaclue already posted.Let X, Y, Z live on [0,1].Define,X=2, Y=1, Z=0 on [0, 1/3);X=0, Y=2, Z=1 on [1/3, 2/3);X=1, Y=0, Z=2 on [2/3, 1]Then P[X>Y]=P[Y>Z]=P[Z>X]=2/3