Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

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

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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