If U1, U2, U3 are IID Uniform[0,1] random variables. How can we find the P(U1 < U2 < U3)?

Note: the possible orderings of U1 U2 and U3 are equally likely

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    $\begingroup$ This appears to be routine bookwork. Please add the self-study tag and read its tag wiki, modifying your question as needed to follow the guidelines there. $\endgroup$ – Glen_b -Reinstate Monica Apr 7 '15 at 3:34
  • $\begingroup$ What are the possible orderings of U1,U2,U3? Are they equally likely? $\endgroup$ – Glen_b -Reinstate Monica Apr 7 '15 at 3:34
  • $\begingroup$ U1, U2 and U3 are equally likely $\endgroup$ – Ali Mubashir Apr 7 '15 at 3:37
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    $\begingroup$ So you’ve edited the question to add an answer to @Glen_b comment. Now, note that this (almost) answers the original question too, and you’re done. $\endgroup$ – Elvis Apr 7 '15 at 3:56
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    $\begingroup$ If all orderings are equally likely, you need only know how many there are to compute the probability of observing a particular one. Even if you don't know how to compute the number, there are so few you could simply list them. $\endgroup$ – Glen_b -Reinstate Monica Apr 7 '15 at 4:43

The sample space of triple $\Omega=\{U_1, U_2, U_3\}$ is a unit 3D cube. Consider events $$\mathcal{E}_{i,j,k} =\{U_i<U_j<U_k\}$$ Clearly $$ \Omega = \cup_{\sigma \in S_3} \mathcal{E}_{\sigma(1),\sigma(2),\sigma(3)} $$ Moreover $\mathcal{E}_{\sigma(1),\sigma(2),\sigma(3)}$ and $\mathcal{E}_{\tau(1),\tau(2),\tau(3)}$ are disjoint for any distinct permutations $\sigma$ and $\tau$.

Since all orderings are equally likely: $$ \Pr\left(\mathcal{E}_{\sigma(1),\sigma(2),\sigma(3)}\right) = \Pr\left(\mathcal{E}_{1,2,3}\right) $$ and $$ 1 = \Pr(\Omega) = \sum_{\sigma \in S_3} \Pr\left(\mathcal{E}_{\sigma(1),\sigma(2),\sigma(3)}\right) = \vert S_3 \vert \cdot \Pr\left(\mathcal{E}_{1,2,3}\right) $$ Hence $$ \Pr\left(\mathcal{E}_{1,2,3}\right)= \frac{1}{\vert S_3 \vert} = \frac{1}{3!} = \frac{1}{6} $$

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    $\begingroup$ Please read the guidelines to answering self-study questions. This goes well beyond 'hints and guidance', and appears to stray into 'doing someone's assignment'. Also see here. $\endgroup$ – Glen_b -Reinstate Monica Apr 7 '15 at 4:29
  • $\begingroup$ And consider moreover that it is very likely that this answer is written in too technical manner for the OP... $\endgroup$ – Elvis Apr 7 '15 at 7:24
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    $\begingroup$ I think the latter (being too technical for the OP) is, in fact, a very neat resolution to the concerns raised by the first comment. $\endgroup$ – wolfies Apr 7 '15 at 8:02
  • $\begingroup$ There is something logically missing from this argument (and therefore it does not appear to be correct). It might not be obvious that there is anything wrong, because it requires only a trivial fix, but please notice that anyone trying to apply this argument to a discrete variable would come up with an incorrect answer. (E.g., if "uniform" is replaced by "Bernoulli" the answer changes from $1/6$ to $0$.) It is therefore worthwhile looking over this answer to see exactly where the uniformity of the variables was used: I would like to suggest explicitly pointing out that place. $\endgroup$ – whuber Apr 7 '15 at 14:42

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