I was reading about how to compute the joint distribution of an order statistics but I don't really follow the beginning. How do we arrive to the formula?
$f_{X_{(k)}}(x) = \frac{n!}{(k-1)!(n-k)!}[F_X(x)]^{k-1}[1-F_X(x)]^{n-k}f_x(x)$
(it is not very detailed and I am bad in probabilities).
This is strictly meant to give you some intuition behind the following equation: $f_{X_{(k)}}(x) = \frac{n!}{(k-1)!(n-k)!}[F_X(x)]^{k-1}[1-F_X(x)]^{n-k}f_X(x).$
Let's break it down piece-by-piece. $f_{X_{(k)}}(x) $ is the PDF of the $k$th order statistic, i.e., if you draw $n$ independent identically distributed random variables from some distribution, arrange them in ascending order by their size, we are looking at how the $k$th one is distributed ($k$th smallest). Informally, it is the "probability" that the $k$th order statistic is $x$. Let's come back to $\frac{n!}{(k-1)!(n-k)!}$ in a bit. Next we have $[F_X(x)]^{k-1}$, which is the CDF of the original distribution raised to the $(k-1)$th power---the idea here is that in order for the $k$th order statistic to be $x$, then we need $k-1$ of the random variables to be less than or equal to $x$. The way we measure this probability is with the CDF and because our random variables are independent, we can just multiply these probabilities. The next term, $[1-F_X(x)]^{n-k}$, is similar; it gives the probability that $n-k$ of the random variables are greater than $x$. At the end, we have $f_X(x)$ which is just the "probability" that one of the random variables is $x$. Returning to $\frac{n!}{(k-1)!(n-k)!}$, this is combinatorial bookkeeping. The $n!$ gives the number of ways that the $n$ random variables might be ordered, and the denominator says we aren't concerned about the actual order of the first $k-1$ or the last $n-k$ random variables.