Proving the PDF of the $r$th order statistic If $X_1,...,X_n$ are iid random variables with $r$th order statistic $X_{(r)}$ I'm trying to prove it's pdf is
$$f_{(r)}(x)=\frac{n!}{(r-1)!(n-r)!}F(x)^{r-1}[1-F(x)]^{n-r}f(x)$$
Is is true for $r=1$. I then assume true for $r$. The only step in this proof is I don't understand why this assumption implies that the cdf for $X_{(r)}$ is:
$$F_{(r)}(x)=\sum_{j=r}^n {n\choose j} F(x)^{j}[1-F(x)]^{n-j}$$
Is someone could explain this I would be very grateful
 A: The formula 
$$F_{(r)}(x)=\sum_{j=r}^n {n\choose j} F(x)^{j}[1-F(x)]^{n-j}\qquad\qquad(\ast)$$
comes from basic probability considerations.  
Fix $x$. If $X_i \le x$, call the result on trial $i$ a "success."  By the definition of $F(x)$, the probability of success on the $i$-th trial is $p=F(x)$, and the probability of failure is $1-p$.
The $r$-th order statistic $X_{(r)}$ is $\le x$ precisely if there are at least $r$ successes.  The number of successes has binomial distribution. The probability of exactly $j$ successes is $\binom{n}{j}p^j(1-p)^{n-j}$. So the probability of $r$ or more successes is
$$\sum_{j=r}^n \binom{n}{j}p^j(1-p)^{n-j}.$$
 Since $p=F(x)$, this is exactly your formula $(\ast)$.
A: Since it seems it's a proof by induction, we can also show this intermediate result by induction. We have 
\begin{align*}
F_{(r)}(x)&=\int_{-\infty}^xf_{(r)}(t)dt\\
&=\frac{n!}{(r-1)!(n-r)!}\left[-\frac{(1-F(t))^{n-r+1}F(t)^{r-1}}{n-r+1}
\right]_{-\infty}^x\\
&+\frac{n!}{(r-1)!(n-r)!}\cdot\int_a^x\frac{(1-F(t))^{n-r+1}}{n-r+1}F(t)^{r-2}(r-1)f(t)dt\\
&=-\frac{n!}{(r-1)!(n-(r-1))!}(1-F(x))^{n-r+1}F(x)^{r-1}+\sum_{j=r-1}^n\binom njF(x)^j(1-F(x))^{n-j}\\
&=\sum_{j=r}^n\binom njF(x)^j(1-F(x))^{n-j}
\end{align*}
and applying the result for $r-1$ we get the wanted result.
