Soppose $\left(X_{(1)}，X_{(2)},\dots,X_{(n)}\right)$ is the ordered statistics of random variables $X_k,k=1,\dots n$. The density fucntions of $X_k,$ for $k=1,\dots,n$ are the same, denoted by $f(x)$. The distribution of $X_k$ is $F(x)$. And $X_k$ are independent.
I want to know whether it is true that $$E(\sum_{k=1}^{n}{X_{(k)}})=E(\sum_{k=1}^{n}{X_{k}})$$
notice that the density function of $X_{(k)}$ is in the form of $\frac{n!}{(k-1)!(n-k)!}F^{k-1}(x)[1-F(x)]^{n-k}\cdot f(x)$ so I think that $$\begin{split}E(\sum_{k=1}^{n}{X_{(k)}})&=\sum_{k=1}^{n}E{X_{(k)}}) \\ &=\sum_{k=1}^{n}\int x\cdot \frac{n!}{(k-1)!(n-k)!}F^{k-1}(x)[1-F(x)]^{n-k}\cdot f(x)dx\end{split}\\$$
From above is it true that $$E(\sum_{k=1}^{n}{X_{(k)}})=E(\sum_{k=1}^{n}{X_{k}})$$