Suppose $\left(X_{(1)}，X_{(2)},\dots,X_{(n)}\right)$ is the ordered statistics of random variables $X_k,k=1,\dots n$. The density functions 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 $$\mathbb E\left[\sum_{k=1}^nX_{(k)}\right]=\mathbb E\left[\sum_{k=1}^nX_k\right]$$
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 $$\begin{split}\mathbb E\left[\sum_{k=1}^nX_{(k)}\right]&=\sum_{k=1}^n\mathbb E\left[X_{(k)}\right]\\ &=\sum_{k=1}^n\int x\cdot\frac{n!}{(k-1)!(n-k)!}F^{k-1}(x)[1-F(x)]^{n-k}\cdot f(x)\ dx\end{split}$$
From above is it true that $$\mathbb E\left[\sum_{k=1}^nX_{(k)}\right]=\mathbb E\left[\sum_{k=1}^nX_{k}\right]$$
