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Let $t_1,\ldots,t_n$ be a set of $n$ intervals, in the form $[l_i,u_i]$.

I define the precedence $t_i \succ t_j$ as the event in which a sample drawn from $t_i$ is greater than a sample drawn from $t_j$.

I have this running example: $$ \left \{ \begin{array}{l} t_1 = [1, 3] \\ t_2 = [2, 4] \\ t_3 = [5, 9] \\ t_4 = [7, 12] \end{array} \right . $$ and I find that: $$ \Pr(t_4 \succ t_3) = \Pr(t_4 \succ t_3 \succ t_1 \succ t_2)+\Pr(t_4 \succ t_3 \succ t_2 \succ t_1) $$ It is always true that $\Pr(t_i \succ t_j) = \Pr(t_i\succ t_j \succ t_k \succ t_m) + \Pr(t_i \succ t_j \succ t_m \succ t_k)$?

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  • $\begingroup$ How are you defining $t_i \succ t_j \succ t_k \succ t_m$? Can $t_i \succ t_j$ be interpreted as an interval? $\endgroup$ – Patrick Oct 9 '13 at 15:10
  • $\begingroup$ It is an ordering: the samples drawn from the four distributions are presented in this order: [i - j - k - m]. $\endgroup$ – Eleanore Oct 9 '13 at 15:11
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In your example, the statement would be true if $t_i = t_2$ and $t_j = t_1$ since:

$$\mathbb{Pr}(t_2 \succ t_1) = \mathbb{Pr}(t_4 \succ t_3 \succ t_2 \succ t_1) + \mathbb{Pr}(t_3 \succ t_4 \succ t_2 \succ t_1)$$

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