Probabilistic ordering

Question

I want to characterize the probabilistic ordering of some (random) variables without going into a parametric from of the variables themselves. I couldn't easily find any theory for this and I am not sure what it is called, but I can try to explain through an example what I mean. Consider two random variables, $X$ and $Y$. There are only two possible orderings and I can characterize it as $P(X\le Y) = p_0$ and $P(X>Y) = 1-p_0$. What if I now have a third variable $Z$. Obviously, there now $3!=6$ orderings and we can give 6 parameters summing to 1, but I want to give more "separable" parameters, utilizing $p_0$ that we already have. Is it sufficient to also give $P(Y\le Z)$, $P(X\le Z)$? Once I give these parameters I should be able to compute the probability of observing any ordering. It seems to me that there must be a theory of such probabilistic orderings but I am unable to find it anywhere, or come up with a solution on first principles.

Answer 1

Once I give these parameters I should be able to compute the probability of observing any ordering.

No, one cannot reconstruct quantities like $P(X\le Y\le Z)$ from $P(X\le Y)$, $P(Y\le Z)$ and $P(Z\le X)$ only. Note that there are 5 free parameters on the side of the 3-dimensional ordering and only 3 on the side of the 2-dimensional orderings hence this would be surprising if one could. Here is an explicit counterexample.

First consider $X$, $Y$ and $Z$ i.i.d. uniform on $(0,1)$. Then, by symmetry, $P(X\le Y\le Z)=\frac16$ and $P(X\le Y)=P(Y\le Z)=P(Z\le X)=\frac12$.

Consider on the other hand $X$ and $Y$ i.i.d. uniform on $(0,1)$ and $Z=X+Y$ mod $1$. Then $Z$ is uniform on $(0,1)$ and each pair $(X,Y)$, $(Y,Z)$ and $(Z,X)$ is uniform on $(0,1)\times(0,1)$ hence $P(X\le Y)=P(Y\le Z)=P(Z\le X)=\frac12$. But $[X\le Y\le Z]=[X\le Y\le 1-X]$ hence $P(X\le Y\le Z)=\frac14$.