# Probability that $\max(U_1,U_2) > U_3$ for independent uniform random variables $U_i$

Suppose $U_1$, $U_2$ and $U_3$ are independent uniform $(0,1)$. I am supposed to find $P(\max(U_1,U_2) > U_3)$.

What I rewrote the question as was this is equal to:

$$2P(U_1>U_3) - P(U_1 \mathrm{ and }\,\, U_2 > 3) = 2(1/2) - 1/3 = 2/3.$$

Can someone check to see if $2/3$ is also what they got? Thanks

-
Please accept answers to your questions. Accepting answers is a simple way of thanking the strangers who are helping you out. You can accept an answer by ticking the check-mark next to the one you found most helpful. – Qiaochu Yuan Apr 15 '11 at 4:30

## 3 Answers

This question is an excellent example of a setting where symmetry considerations yield the answer with almost no effort. Symmetry is of common use in statistical mechanics and, in the more general setting of mathematics as a whole, one could go back (at least) to Hermann Weyl for thoughts about its role. As regards probability, some recent discussion is in Symmetry and Probability by Jill North, with some Comments on the preceding, by Branden Fitelson.

The key remark in the question at hand is that the distribution of the random vector $(U_1,U_2,U_3)$ is invariant by the permutations of its coordinates, for example $(U_1,U_2,U_3)$ and $(U_3,U_1,U_2)$ share the same statistical properties, as well as $(U_{s(1)},U_{s(2)},U_{s(3)})$ for any permutation $s$ in the symmetric group $\mathfrak{S}_3$.

Now, one is interested in the probability of the event $[\max(U_1,U_2) > U_3]$. The complementary event is $$A_3=[\max(U_1,U_2) < U_3]$$ and $A_3$ simply means that $U_3$ is the largest of the three values $U_1$, $U_2$, $U_3$ (there is almost surely no tie here because the common distribution has no atom). By symmetry, the events $A_1$, $A_2$ and $A_3$ have the same probability where $$A_1=[\max(U_2,U_3) < U_1]\quad\mbox{and}\quad A_2=[\max(U_1,U_3) < U_2].$$ Since these events are disjoint and their union is the universe, their common probability is $1/3$ and $$P(\max(U_1,U_2) > U_3)=1-P(A_3)=2/3.$$ One sees that everything above works for an i.i.d. sample of size $n$ based on any atomless distribution on the real line (just replace $P(A_i)=1/3$ by $P(A_i)=1/n$), and even in the more general setting of exchangeable random variables, as soon as the ties have probability zero.

-

So I believe I know the setting of the problem that you're asking about here. (This is review exercise 5.14(e) in Pitman's text, which I know because I assigned it to my students this week.) In particular, given the context, you might be tempted to try to find the density of $\max(U_1, U_2)$, state that it's independent of $U_3$, write down the joint density of $\max(U_1, U_2)$ and $U_3$, and integrate over the appropriate region. These are probably all things you should be able to do.

But don't do them! Symmetry is the right approach to this problem. Actually, the simplest way to solve this problem is to note that the event $\max(U_1, U_2) < U_3$ is just the event that $U_3$ is the largest of the three random variables.

-
Using \max will render $\max$ correctly. Cheers. – cardinal Apr 21 '11 at 22:10

Yes, the answer $2/3$ is correct.

Hints for a solution. Work with the complement of the event $\max\{U_1,U_2\}>U_3$, and use the law of total probability conditioning on $U_3$. More generally, try showing the following (using the hints): ${\rm P}(\max\{U_1,\ldots,U_n\}>U_{n+1})=n/(n+1)$, where the $U_i$ are independent uniform$(0,1)$ random variables.

-