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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?

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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.

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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 $$\frac{k}{n+1} = \frac{n!}{(k-1)! (n-k)!} \int_{x=0}^1 x^k (1-x)^{n-k} \, dx.$$

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