# Order of independent random variables

Let $(p_\pi)_{\pi\in S_3}$ be given nonnegative reals such that $\sum_{\pi \in S_3} p_\pi = 1$. What are necessary and sufficient conditions for there to exist independent random variables $X_1,X_2,X_3$ such that, for each $\pi$, $p_\pi$ is the probability of $X_{\pi(1)} < X_{\pi(2)} < X_{\pi(3)}$?

• Do you already know anything about the structure of the set of attainable point of the 5-dimensional simplex, $\{p\colon \exists \text{independent} X_1,X_2,X_3 \text{s.t.} p_\pi=\mathbb P(X_{\pi(1)}<X_{\pi(2)}<X_{\pi(3)})\}$? – Anthony Quas May 1 '17 at 14:17
• @AnthonyQuas I do know that not all points are attainable, because I have convinced myself that the three orders $123$, $231$, and $312$ can't each have positive probability without other orders also having positive probability. No doubt this is just the qualitative tip of the quantitative iceberg, but I was hoping somebody would instantly recall the answer from somewhere. – Sean Eberhard May 1 '17 at 14:51
• I see. For what it's worth, I just used mathematica to prove that there is an open set of attainable probability vectors. – Anthony Quas May 1 '17 at 15:06
• @SeanEberhard Is there more background you can provide for this specific question? – Henry.L May 2 '17 at 18:24

This is too long to be a comment. But I felt it serves as a pointer.

Without loss of generality we assume $X_{1},X_{2},X_{3}$ share the same image domain $[0,1]$.

For a specific $\pi\in S_{3}$, the probability of $X_{\pi(1)}<X_{\pi(2)}<X_{\pi(3)}$ is $$\int_{0}^{1}dF_{X_{\pi(3)}}(\nu)\int_{0}^{\nu}dF_{X_{\pi(2)}}(\omega)\int_{0}^{\omega}dF_{X_{\pi(1)}}(\eta)=\int_{0}^{1}f_{X_{\pi(3)}}(\nu)d\nu\int_{0}^{\nu}f_{X_{\pi(2)}}(\omega)d\omega\int_{0}^{\omega}f_{X_{\pi(3)}}(\eta)d\eta$$where $F_{X_{i}}$ are probability measures on $[0,1]$; $f_{X_{i}}$ are probability densities w.r.t. Lebesgue measure on $[0,1]$, therefore it suffices to make the 3! integral equations have a consistent solution $f_{X_{1}},f_{X_{2}},f_{X_{3}}$. But since they are independent, $f_{X_{1}},f_{X_{2}},f_{X_{3}}$ are at the same time marginal probability measures, so the problem becomes finding a joint distribution $g$ of $(X_{1},X_{2},X_{3})$ subject to 3! conditions $$\int_{0}^{1}f_{X_{\pi(3)}}(\nu)d\nu\int_{0}^{\nu}f_{X_{\pi(2)}}(\omega)d\omega\int_{0}^{\omega}f_{X_{\pi(3)}}(\eta)d\eta=\int_{0}^{1}\int_{0}^{\nu}\int_{0}^{\omega}f_{X_{\pi(3)}}(\nu)f_{X_{\pi(2)}}(\omega)f_{X_{\pi(3)}}(\eta)d\nu d\omega d\eta=\int_{0}^{1}\int_{0}^{\nu}\int_{0}^{\omega}g_{X_{\pi(3)},X_{\pi(2)},X_{\pi(1)}}(\nu,\omega,\eta)d\nu d\omega d\eta=p_\pi$$.

If you are willing to assume that $p_{\pi}\equiv\frac{1}{3!}$ then a special case that suffices to work is $X_{i}$ are exchangeable, which deserves an extended discussion like the one in Chap 5,7 in [Kallenberg].

[Kallenberg]Kallenberg, Olav. Probabilistic symmetries and invariance principles. Springer Science & Business Media, 2006.

• Surely using copulas is exactly what you're not allowed to do? The OP is rather explicit that he's looking for independent $X$'s, whereas if I understand correctly, the point of copulae is to produce non-independent joint distributions with specified marginals? – Anthony Quas May 1 '17 at 13:37
• @AnthonyQuas You are right, just noticed that and remove that line. So it finally reduces to a system of integral equations? – Henry.L May 1 '17 at 13:38
• For what it's worth, the problem would be trivial if you were allowed dependent variables: just choose $3!$ points in $\mathbb R^3$, one point $x_\pi$ for each ordering $\pi$ with the correct ordering of the coordinates, and set $(X_1,X_2,X_3)$ to be $x_\pi$ with probability $p_\pi$. – Anthony Quas May 1 '17 at 13:51
• @AnthonyQuas But OP does not say anything special about the domain, so I do not think it can be reduced further than integral equations. – Henry.L May 1 '17 at 13:57