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 $X_{\pi(1)} < X_{\pi(2)} < X_{\pi(3)}$?

  • $\begingroup$ Do you already know anything about the structure of the set of attainable point of the 5-dimensional simplex, $\{p\colon \exists \text{independent} X_1,X_2,X_3 \text{s.t.} p_\pi=\mathbb P(X_{\pi(1)}<X_{\pi(2)}<X_{\pi(3)})\}$? $\endgroup$ – Anthony Quas May 1 '17 at 14:17
  • $\begingroup$ @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. $\endgroup$ – Sean Eberhard May 1 '17 at 14:51
  • $\begingroup$ I see. For what it's worth, I just used mathematica to prove that there is an open set of attainable probability vectors. $\endgroup$ – Anthony Quas May 1 '17 at 15:06
  • $\begingroup$ @SeanEberhard Is there more background you can provide for this specific question? $\endgroup$ – 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.

  • 1
    $\begingroup$ 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? $\endgroup$ – Anthony Quas May 1 '17 at 13:37
  • $\begingroup$ @AnthonyQuas You are right, just noticed that and remove that line. So it finally reduces to a system of integral equations? $\endgroup$ – Henry.L May 1 '17 at 13:38
  • 1
    $\begingroup$ 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 $p_\pi$. $\endgroup$ – Anthony Quas May 1 '17 at 13:51
  • $\begingroup$ @AnthonyQuas But OP does not say anything special about the domain, so I do not think it can be reduced further than integral equations. $\endgroup$ – Henry.L May 1 '17 at 13:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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