Sum of CDF of order statistics

Question

Suppose I have $n$ i.i.d. random variables $X_1,\dots,X_n$ with CDF $F$, and I order and relabel them such that $Y_1>\dots>Y_n$. The distribution of $Y_k$, the $k$-th highest value, is then $$G_k(x) = \sum_{i=0}^{k-1} {n \choose i} F(x)^{n-i} (1-F(x))^i.$$ How can I show that $$\sum_{k=1}^n G_k (x) = n F(x) \quad ?$$ And what is the intuition here? Is it simply that $X_i,\dots,X_n$ is a reordering of $Y_1,\dots,Y_n$? I have a completely unrelated reason why I believe the above is true, but I was not able to show it algebraically in general.

Answer 1

Partial answer (without the intuition part):

$$\sum_{k=1}^n G_k (x) = \sum_{i=0}^{n-1} (n-i){n \choose i} F(x)^{n-i} (1-F(x))^i=\sum_{i=0}^{n-1} n{n-1 \choose i} F(x)^{n-i} (1-F(x))^i=n F(x)\sum_{i=0}^{n-1} {n-1 \choose i} F(x)^{n-1-i} (1-F(x))^i=nF(x)(F(x)+1-F(x))^{n-1}=nF(x)$$