# Density of the $k^{th}$ smallest of $X_1,X_2,…,X_n$

Show that if $(X_1,X_2,...,X_n)$ are i.i.d. with common density $f$ and distribution function $F$, then $X_{(k)}$ has density $$f_{(k)}=k\binom{n}{k}f(y)(1-F(y))^{n-k}F(y)^{k-1}$$

where $X_{(k)}$ is the $k^{th}$ smallest of $X_1,X_2,...,X_n$

for $n=2$ it is true;

let $X_{(1)}=\min\{X_1,X_2\}$ then

$F_{X_{(1)}}(y)=\Pr(X_{(1)}\le y)=1-\Pr(X_{(1)}>y)=1-\Pr(X_1>y)\Pr(X_2>y)=1-(1-F)^2$

Hence $\displaystyle f_{X_{(1)}}(y)=\frac{d}{dy}F_{X_{(1)}}(y)=2f(1-F)=1\binom{2}{1}f(1-F)^{2-1}F^{1-1}$

So it is OK, but how can I prove it for $n\ge 2$ ?