Joint density of first r order statistics

Question

Let $X_i \sim^{iid} F$ for $i=1,...,n$, where $F$ is a continuous distribution.

I want to find the pdf for $X_{(1)},X_{(2)},..., X_{(r)}$, with $r\leq n$.

We know that $f_{X_{(1)},X_{(2)},..., X_{(n)}}=n!\prod^n_{i=1}f(x_{(i)})$, when $x_{(1)}\leq x_{(2)}\leq \cdots\leq x_{(n)}$; $f_{X_{(1)},X_{(2)},..., X_{(n)}}=0$ otherwise.

Well when integrating

$$f_{X_{(1)},X_{(2)},..., X_{(r)}}=n! \prod^r_{i=1}f(x_{(i)}) \int^{\infty}_{x_{(r)}}\cdots \int^{\infty}_{x_{(n-1)}}f(x_{(n)})\cdots f(x_{(r+1)})dx_{(n)}\cdots dx_{(r+1)} $$

$$=n! \prod^r_{i=1}f(x_{(i)}) \int^{\infty}_{x_{(r)}}\cdots \int^{\infty}_{x_{(n-2)}}(1-F(x_{(n-1)})) f(x_{(n-1)}) \cdots f(x_{(r+1)})dx_{(n-1)}\cdots dx_{(r+1)}$$

$$=n! \prod^r_{i=1}f(x_{(i)}) \int^{\infty}_{x_{(r)}}\cdots \int^{\infty}_{x_{(n-3)}}\frac{-(1-F(x_{(n-2)}))^2}{2} f(x_{(n-2)}) \cdots f(x_{(r+1)})dx_{(n-2)}\cdots dx_{(r+1)}$$

and reaching at the end $$f_{X_{(1)},X_{(2)},..., X_{(r)}}=n! (-1)^{n-r}\frac{(1-F(x_{(r)}))^{n-r-1}}{(n-r-1)!} \prod^r_{i=1}f(x_{(i)}) $$

This doesn't seem to be correct, since we have that annoying -1 factor. Where did I go wrong?

Answer 1

From @whuber,

$\int^{\infty}_{x_{(n-2)}}(1-F(x_{(n-1)})) f(x_{(n-1)})dx_{(n-1)}=\left[-\frac{(1-F(x_{(n-1)}))^2}{2}\right]_{x_{(n-2)}}^{\infty}=0+\frac{(1-F(x_{(n-2)}))^2}{2}$

Answer 2

The end result should be $$f_{X_{(1)},X_{(2)},..., X_{(r)}}=n! \frac{(1-F(x_{(r)}))^{n-r}}{(n-r)!} \prod^r_{i=1}f(x_{(i)}) $$

...by the same approach as on page 9 of this book https://books.google.co.uk/books/about/Order_Statistics_and_Inference.html?id=chvvAAAAMAAJ by N. Balakrishnan, A. Clifford Cohen (we get it directly by using only the third of the integrals in their derivation of joint density of 2 order statistics, i.e. using eq. (2.3.4))