I am struggling to understand why for the order statistics of $\{X_1,..,X_n\}$ iid for values $x_1 < x_2 <... <x_n$ my joint density is $f_{Y_1,...,Y_n}(x_1,...,x_n) = n!\Pi_{i=1}^nf_X(x_i)$.

My best interpretation is that there are $n!$ ways to permute the order of the values $x_i$ and each $x_i$ has a probability $f_X(x_i)$. But this doesn't make sense to me as I know the data is already ordered, so surely there are not $n!$ ways to rearrange the data as there is only $1$ order allowed.

  • 3
    $\begingroup$ You multiply by $n!$ because eg in the case $n=3$ it doesn't matter if $X_1=x_1$, $X_2=x_2$, $X_3=x_3$ or alternatively $X_1=x_3$, $X_2=x_1$, $X_3=x_2$, both give rise to the same set of order statistics $\endgroup$
    – jlammy
    Nov 19, 2022 at 17:53
  • $\begingroup$ @jlammy oh it doesn't matter which variable gets a certain value, it just depends on the values observed. $\endgroup$
    – Governor
    Nov 19, 2022 at 17:54

2 Answers 2


(Too long for a comment)

IMHO, a good way to get an intuition about the interaction between the different order statistics (in the case of uniform distribution on [0,1]) is to build their correlation matrix $C$ which is the symmetric matrix with the following entries of its lower part:

$$\text{If} \ j<i \ : \ C_{i,j}=\operatorname{Cov} (U_{(i)},U_{(j)})=\frac {j(n-i+1)}{\sqrt{ij(n+1-i)(n+1-j)}}$$

(see a proof here for the variance matrix).

What kind of values do we have ; in the case $n=5$, here is the correlation matrix :

$$C=\pmatrix{1&0.6325&0.4472&0.3162&0.2\\ 0.6325&1&0.7071&0.5&0.3162\\ 0.4472&0.7071&1&0.7071&0.4472\\ 0.3162&0.5&0.7071&1&0.6325\\ 0.2&0.3162&0.4472&0.6325&1}$$

These different entries quantify the degree of interaction of the different $U_{(k)}$s. For example the interaction "degree" between $U_{(1)}$ and $U_{(5)}$, measured by $0.2$ is low, whereas the interaction degree between $U_{(2)}$ and $U_{(3)}$ is much higher.


The probability of any permutation of $\{x_i\}$ is $\prod_{i=1}^nf_X(x_i)$

Whenever the experiment is any permutation of $\{x_i\}$, the order statistics will be $\{x_i\}$.

Probability that the order statistics is $\{x_i\}$
= Probability that the un-ordered statistics is one of the permutations of $\{x_i\}$
= $\sum_{\text{all } \sigma(\{x_i\})} \mathbb{P}(\sigma(\{x_i\})) $
= $n! \times \displaystyle \prod_{i=1}^nf_X(x_i)$


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