Expected value of the k-th order statistic from uniform random variables I am trying to find the expected value of $X_{(k)}$ 
Here is my work so far:
$$f_{X_{(k)}}(x)=\frac{n!}{(k-1)!(n-k)!}f_X(x)[F_{X}(x)]^{k-1}[1-F_X(x)]^{n-k}$$
by $X_i \sim U(0,1)$ this becomes
$$E(X_{(k)})=\frac{n!}{(k-1)!(n-k)!}\int_0^1 x^{k}[1-x]^{n-k}dx$$
This almost looks like a beta
$$B(a,b) = \int_0^1\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}x^{a-1}[1-x]^{b-1}dx=1$$
$$E(X_{(k)})=\frac{n!}{(k-1)!(n-k)!}\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\int_0^1
x^{k}[1-x]^{n-k}dx$$
So we can integrate out the beta pdf
$$a =k+1,b=n-k+1$$ 
$$E(X_{(k)})=\frac{n!}{(k-1)!(n-k)!}\frac{(k+1)!(n-k+1)!}{(n+2)!}$$
I cannot seem to simplify this to the right answer given in the solutions. Did I make a mistake somewhere? 
 A: I think you're messing up in this way:  $\Gamma(m) = (m-1)$! for a positive integer $m$. I'm getting 
\begin{align*}
E(X_{(k)}) &=\frac{n!}{(k-1)!(n-k)!}\int_0^1 x^{k}[1-x]^{n-k}dx \\[5pt]
&= \frac{ \Gamma(n+1)}{\Gamma(k) \Gamma(n-k+1)}\int_0^1 x^{k}[1-x]^{n-k}dx \\[5pt]
&=\frac{ \Gamma(n+1)}{\Gamma(k) \Gamma(n-k+1)} \cdot \frac{\Gamma(k+1) \Gamma(n-k+1) }{ \Gamma(n+2)} \int_0^1 \frac{ \Gamma(n+2)}{\Gamma(k+1) \Gamma(n-k+1) } x^{k}[1-x]^{n-k}dx \\[5pt]
&= \frac{ \Gamma(n+1)}{\Gamma(k) \Gamma(n-k+1)} \cdot \frac{\Gamma(k+1) \Gamma(n-k+1) }{ \Gamma(n+2)} \\[5pt]
&= \frac{\Gamma(k+1) \Gamma(n+1)}{\Gamma(k) \Gamma(n+2)} =  \frac{k}{n+1}.
\end{align*}
Final comment: $X_{(k)} \sim \text{Beta}(k,n+1-k)$, which agrees with above.
A: The PDF of a beta distribution with parameters $a = k+1$, $b = n-k+1$, is $$\frac{\Gamma(n+2)}{\Gamma(k+1)\Gamma(n-k+1)} x^k (1-x)^{n-k}, \quad x \in (0,1).$$  This gives you the desired integrand you want to evaluate, so $$1 = \frac{\Gamma(n+2)}{\Gamma(k+1)\Gamma(n-k+1)} \int_{x=0}^1 x^k (1-x)^{n-k} \, dx = \frac{(n+1)!}{k! (n-k)!} \int_{x=0}^1 x^k (1-x)^{n-k} \, dx.$$  And now it is evident that there is an extra factor of $(n+1)/k$ on the RHS, so we have $$\frac{k}{n+1} = \frac{n!}{(k-1)! (n-k)!} \int_{x=0}^1 x^k (1-x)^{n-k} \, dx.$$
