$x_{1}...x_{n}$ are independent continuous random variables with common distribution function $F(x)$,compute $E(F(x_{(n)})-F(x_{(1)}))$ $x_{1}...x_{n}$ are independent continuous random variables with common distribution function $F(x)$,consider the order statistics $(x_{(1)},...,x_{(n)})$, compute $E(F(x_{(n)})-F(x_{(1)}))$
I have no idea to this problem, anyone could help me? Thanks!
 A: Let's do this the roundabout way (the direct way is what @JohnK's answer remarked).  
To consider the expected value, we need to treat the variables involved as random variables. To stress this we write
$$E[F(X_{(n)})-F(X_{(1)})]$$
and we set $F(X_{(n)}) \equiv Z$, $F(X_{(1)}) \equiv Y$,
so we want to calculate
$$E[F(X_{(n)})-F(X_{(1)})] = E(Z) - E(Y)$$
The cumulative distribution function of $X_{(n)}$ is
$F_{X_{(n)}}(x_{(n)}) = [F(x_{(n)})]^n $ and $[F(X_{(n)})]^n$, viewed as a random variable, follows a uniform $U(0,1)$ by the Probability Integral Transform.
So
$$[F(X_{(n)})]^n = U \Rightarrow Z^n = U$$
Applying the change-of-variable formula
$$f_Z(z) = \left|\frac{\partial U}{\partial Z}\right|\cdot f_U(u) = nz^{n-1} \cdot 1= nz^{n-1}, z\in [0,1]$$
Therefore
$$E(Z) = \int_0^1nz^{n-1}zdz = \frac {n}{n+1} \tag{1}$$
The cumulative distribution function of $X_{(1)}$ is
$F_{X_{(1)}}(x_{(1)}) =1- [1-F(x_{(1)})]^n $ and, $1- [1-F(X_{(1)})]^n$, viewed as a random variable, follows too a uniform $U(0,1)$.
So
$$1-[1-F(X_{(1)})]^n = U \Rightarrow 1-[1-Y]^n = U$$
Applying the change-of-variable formula
$$f_Y(y) = \left|\frac{\partial U}{\partial Y}\right|\cdot f_U(u) = n(1-y)^{n-1} , y\in [0,1]$$
Therefore
$$E(Y) = \int_0^1n(1-y)^{n-1}ydy = nB(2,n) = \frac {1}{n+1} \tag{2}$$
where $B(2,n)$ is the beta function. See also this derivation, since, indeed, as mentioned in another answer, $Y$ is the minimum order statistic of an i.i.d. sample of standard uniform random variables (and $Z$ is the corresponding maximum).
So
$$E[F(X_{(n)})-F(X_{(1)})] = E(Z) - E(Y) = \frac {n}{n+1} - \frac {1}{n+1} = \frac {n-1}{n+1}$$
For the direct way, it is a very good suggestion to study the Probability Integral Transform.
A: You need to consider the Probability Integral Transform. For ease of notation denote the order statistics from smallest to largest by $Y_1, \ldots, Y_n $
You can afterwards see that 
$$E \left[ F \left(Y_n \right)-F \left( Y_1 \right)  \right] $$
is basically the expected value of the difference between the maximum and the minimum of a uniform $(0,1)$ distribution.
A: You can distribute the expectation function. Then you would have the E[x_n]-E[x_1]. Then you need to find the CDF of the max and min order statistic. From there you can find the expectation of each. To find the CDF of the max and min order statistic use the CDF substitution method and use the fact that x_1,...,x_n are independent. 
