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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?

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The $-1$ was never there. Go back to the second integral and notice it had to be evaluated between $x_{(n-2)}$ and $\infty$. It's zero at the upper limit and the value at the lower limit has to be subtracted, yielding $(1-F(x_{(n-2)})^2/2$ rather than its negative. – whuber Jul 26 '14 at 17:35
Thanks @whuber. – An old man in the sea. Jul 26 '14 at 17:42
up vote 4 down vote accepted

From @whuber,

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

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