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?

  • 2
    $\begingroup$ 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. $\endgroup$
    – whuber
    Jul 26, 2014 at 17:35

2 Answers 2


From @whuber,

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

  • $\begingroup$ Is the end result then: $$f_{X_{(1)},X_{(2)},..., X_{(r)}}=n! \frac{(1-F(x_{(r)}))^{n-r-1}}{(n-r-1)!} \prod^r_{i=1}f(x_{(i)}) $$ ? $\endgroup$
    – axk
    Jan 29, 2018 at 19:59
  • $\begingroup$ @axk Yes, I think so. $\endgroup$ Jan 29, 2018 at 21:14
  • $\begingroup$ But then for $r=n$ how do we get rid of the fraction? $\endgroup$
    – axk
    Jan 29, 2018 at 21:20
  • $\begingroup$ @axk It's written on the original question. The derivation is for when $r\leq n-1$. For $r=n$ it's already known. $\endgroup$ Jan 30, 2018 at 10:21
  • $\begingroup$ The question sais $r\le n$. This is what I don't understand: For the case $r=n-1$ the formula gives what is known for the case $r=n$, and for the case $r=n$ it doesn't. $\endgroup$
    – axk
    Jan 31, 2018 at 1:29

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))


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