# ordering three random variables

Let's consider three random variables $\xi_1, \xi_2, \xi_3$ distributed over $\{1, 2, 3, \ldots, M\}$ values. Is it possible that $P(\xi_1 < \xi_3) > 0.5, P(\xi_3 < \xi_2) > 0.5, P(\xi_2 < \xi_1) > 0.5$? So we cannot order them using probabilities. Thank you.

-

## 1 Answer

Let $\Omega := [0,1]$ and $\mathbb{P} := \lambda|_{[0,1]}$ the lebesgue measure on $[0,1]$. Define

$$\xi_1(x) := \begin{cases}3 & x \in \left[0,\frac{1}{3} \right] \\ 1 & x \in \left(\frac{1}{3},\frac{2}{3} \right) \\ 2 & x \in \left[\frac{2}{3},1 \right] \end{cases} \\ \xi_2(x) := \begin{cases}2 & x \in \left[0,\frac{1}{3} \right] \\ 3& x \in \left(\frac{1}{3},\frac{2}{3} \right) \\ 1 & x \in \left[\frac{2}{3},1 \right] \end{cases} \\ \xi_3(x) := \begin{cases}1 & x \in \left[0,\frac{1}{3} \right] \\ 2 & x \in \left(\frac{1}{3},\frac{2}{3} \right) \\ 3 & x \in \left[\frac{2}{3},1 \right] \end{cases}$$

Then $\mathbb{P}[\xi_1 < \xi_3] = \mathbb{P}[\xi_3<\xi_2] = \mathbb{P}[\xi_2<\xi_1] = \frac{2}{3}$.

-
what is x? I thought that x should be 1,2,3 – ashim Dec 6 '12 at 19:30
I thought that the random variables should take its values in $\{1,\ldots,M\}$ (i.e. $\xi_j: (\Omega,\mathcal{A},\mathbb{P}) \to \{1,\ldots,M\}$ where $(\Omega,\mathcal{A},\mathbb{P})$ is a probability space). So you mean random variables $\xi_j: \{1,\ldots,M\} \to \mathbb{R}$? – saz Dec 6 '12 at 19:33
you are right. Could you explain why $\mathbb{P}[\xi_2<\xi_1] = \frac{2}{3}$ – ashim Dec 6 '12 at 19:39
I see it why this is a case – ashim Dec 6 '12 at 19:40