Use of binomial coefficient in order statistics Appreciate all your help with the following:
Let $U_{1}, ... U_{n}$ be n independent and uniformly distributed on (0,1) random variables. Let $U_{(1)} \geq U_{(2)} \geq ... \geq U_{(n)}$ be an ordering of the $U_{i}$'s in descending order. Find P($U_{(2)} \lt$ u).
My solution: 
P(($U_{(1)} \geq u) \cap (U_{(2)} < u)) = u^{n} + u^{n-1}(1-u)$ 
The correct solution is: $u^{n} + \binom{n}{1} u^{n-1}(1-u)$
What I do not understand is why would you need the $\binom{n}{1}$ part if the set of RV is already ordered and that you would have already known that which is the largest element already?
Many thanks in advance.
 A: I believe the point that addresses your question is that even though you have ordered your random variables as $U_{(1)}\geq U_{(2)} \geq \cdots \geq U_{(n)}$, you do not know which random variable among the $U_i$ that is largest. The order statistics are still random variables: You only declare that $U_{(1)} = \max\{U_1,U_2,\ldots,U_n\}$ and so on, but you do not know which of the $U_i$ will be $U_{(1)}$, which will be $U_{(2)}$ etc.
I would do the calculation somewhat along the following lines:
$$P(U_{(2)} < u) = P((U_{(1)} < u)\cup(U_{(1)}>u,U_{(2)}<u))$$
$$=P(U_{(1)} < u) + P(U_{(1)}>u,U_{(2)}<u)$$
$$=u^n + P(U_{(1)}>u,U_{(2)}<u)$$
$$=u^n + P(U_i>u\text{ for some }i, U_j<u\text{ for all }j\neq i).$$
Here I have used that $P(U_{(1)} < u) = P(\max\{U_1,U_2,\ldots,U_n\} <u) = P(U_i<u\text{ for all }i)=u^n$.
Also, the probability $P(U_i>u\text{ for some }i, U_j<u\text{ for all }j\neq i)$ can be calculated by noting that you need to choose one $U_i$ that is larger that $u$. You can do this in $\binom{n}{1}$ ways. Then this chosen $U_i$ is larger than $u$ with probability $1-u$, and the rest $n-1$ of the $U_i$ is smaller than $u$ with probability $u^{n-1}$ (by independence).
I hope this makes sense.
