Summation of binomials I'm struggling to see how the distribution function of an order statistic is obtained. So far, from here, I understand that the distribution function $F_k$ of $X_{(k)}$ is given by:
\begin{equation}
\sum_{j=k}^n \binom{n}{j} \left[F(x)\right]^j \left[1 - F(x)\right]^{n - j}
\end{equation}
Now, I want to find the CDF of $F_1(x)$, that is, I choose $k$ to be equal to 1. I'm striving to obtain the same solution shown in the link above, which is
\begin{equation}
F_1(x) = 1 - \left[1 - F(x)\right]^n.
\end{equation}
I tried to use the transformations and change of variables that are used here. Mainly this one: $\dbinom r l = \dfrac r l \dbinom {r - 1} {l - 1}$. The problem is that, in our case, $l$ depends on the summation so I'm not able to take it out to do the change of variables.
Could anyone please shed some light on this? Is there a better way to solve this? Is there a closed expression for any $k$?
Thanks in advance for the time you take in reviewing this question.
 A: Ok, so finally I think I have solved it. I leave here the solution for anyone who is curious about it:
Let $F(x)$ be the CDF fo the random variable from which we sample points:
We know that the CDF of the $k$-th order statistic is
\begin{equation}\label{eq:original}
F_k(x) = \sum_{j=k}^n \binom{n}{j} F(x)^j [1 - F(x)]^{n-j}
\end{equation}
Additionally, note that, by the Binomial Theorem:
\begin{equation}\label{eq:binomialTh}
\sum_{j=0}^n \binom{n}{j} F(x)^j [1 - F(x)]^{n-j} = [ F(x) + (1 - F(x))]^n = 1^n
\end{equation}
By using eq.~\ref{eq:original}, we may expand eq.~\ref{eq:binomialTh} as follows:
\begin{equation}\label{eq:combination}
1^n = \sum_{j=0}^{k-1} \binom{n}{j} F(x)^j [1 - F(x)]^{n-j} + \sum_{j=k}^{n} \binom{n}{j} F(x)^j [1 - F(x)]^{n-j}
\end{equation}
Note, in eq.~\ref{eq:combination}, that the second summation is the expression of $F_k(x)$. Therefore, we may rewrite~\ref{eq:combination} as follows and derive the expression for $F_k(x)$:
\begin{equation}
1 = \sum_{j=0}^{k-1} \binom{n}{j} F(x)^j [1 - F(x)]^{n-j} + F_k(x)
\end{equation}
\begin{equation}\label{eq:final}
F_k(x) = 1 - \sum_{j=0}^{k-1} \binom{n}{j} F(x)^j [1 - F(x)]^{n-j}
\end{equation}
Now, particularising eq.~\ref{eq:final} for both $k = 1$ and $k = n$, we obtain the following expressions for the $1$st and $n$-th order statistics.
\begin{equation}
F_1(x) = 1 - \sum_{j=0}^{0} \binom{n}{0} F(x)^0 [1 - F(x)]^{n} = 1 - [1 - F(x)]^{n}
\end{equation}
\begin{equation}
\begin{aligned}
   F_n(x) &=1 - \sum_{j=0}^{n-1} \binom{n}{j} F(x)^j [1 - F(x)]^{n} = \\
    &= 1 - [1 - \sum_{j=n}^{n} \binom{n}{n} F(x)^n [1 - F(x)]^{n-n}] = \\
    &= 1 - [1 - F(x)^n] = F(x)^n
  \end{aligned}
\end{equation}
I would appreciate if anyone could help me with labeling equations in this forum, since it does not appear to work as in standard latex.
Regards.
A: You ask if there's a "better way to solve this" so I assume you're open to the idea of solutions which don't involve messing with binomials, and that you're only looking to solve this for the first order statistic. If I've misinterpreted this I apologize. 
We have:
$$
Pr(X_{(1)} \leq x) = 1 - Pr(X_{(1)} \geq x)
$$
$$
Pr(X_{(1)} \geq x) = Pr(X_{1} \geq x, X_{2} \geq x, ... , X_n \geq x) 
$$
$$
= Pr(X_1 \geq x) \times Pr(X_2 \geq x) \times ... \times Pr(X_n \geq x)
$$
$$
= [Pr(X_1 \geq x)]^n = [1-Pr(X_1 \leq x)]^n = [1-F_x(x)]^n
$$
$$
\therefore Pr(X_{(1)} \leq x) = 1 - [1-F_x(x)]^n
$$
