# Intuitive understanding of joint probability of order statistics

I am struggling to understand why for the order statistics of $$\{X_1,..,X_n\}$$ iid for values $$x_1 < x_2 <... my joint density is $$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.

• 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 Nov 19, 2022 at 17:53
• @jlammy oh it doesn't matter which variable gets a certain value, it just depends on the values observed. Nov 19, 2022 at 17:54

(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

(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 $$\{x_i\}$$
= $$\sum_{\text{all } \sigma(\{x_i\})} \mathbb{P}(\sigma(\{x_i\}))$$
= $$n! \times \displaystyle \prod_{i=1}^nf_X(x_i)$$