From G. Casella & R. Berger "Statistical Inference":

Definition 5.4.1 The order statistics of a random sample $X_1,\dots,X_n$ are the sample values placed in ascending order. They are denoted by $X_{(1)},\dots, X_{(n)}$.

The order statistics are random variables that satisfy $X_{(1)}\leq\dots\leq X_{(n)}$. In particular, $X_{(1)}=\min\limits_{1 \leq i \leq n} X_i,\dots,X_{(n)}=\max\limits_{1 \leq i \leq n} X_i$.

My question: what is the mathematical meaning of $ \min\limits_{1 \leq i \leq n}X_i $ and $\max\limits_{1 \leq i \leq n} X_i$ ?

My understanding: I know that $X<Y$ means $X(\omega)<Y(\omega) $ for all $\omega \in \Omega$. Here, $X_i$'s are independent and identically distributed so $X_i(\omega)=X_j(\omega) $ for all $\omega \in \Omega$. So how can we choose or define the $\min$ and $\max$ from those identical random variables?

  • 3
    $\begingroup$ Being iid does bot imply equality! Only equal distribution and independence. $\endgroup$
    – Michael M
    Commented Sep 13, 2016 at 6:37
  • $\begingroup$ @MichaelM could you give a more detailed explanation ? thanks $\endgroup$
    – Bungbu
    Commented Sep 13, 2016 at 8:53
  • $\begingroup$ If the second paragraph in the high-lighted text is how the Casella and Berger book actually reads, then I think that it is a deplorable use of $\ldots$ since the descriptions of $X_{(2)}$, $X_{(3)}$, etc up to $X_{(n-1)}$ are nowhere so simple as those of $X_{(1)}$ and $X_{(n)}$ as to be hidden in the elision. Why couldn't they have written $$X_{(1)} = \min_{1 \leq i \leq n} X_i \ \mathbf{and} \ X_{(n)} = \max_{1\leq i \leq n} X_i ~~??$$ $\endgroup$ Commented Sep 13, 2016 at 17:04
  • 1
    $\begingroup$ @Bungbu: please correct your question about the incorrect assertion that $X_i(.)=X_j(,)$ when the variates are iid. $\endgroup$
    – Xi'an
    Commented Sep 15, 2016 at 7:18

1 Answer 1


Given one realisation of the iid random sample $(X_1,\ldots,X_n)$, $(x_1,\ldots,x_n)$, corresponding to one element $\omega$ of $\Omega$, the random variable $$X_{(1)}=\min_{1\le i\le n} X_i$$ takes the value $$\min_{1\le i\le n} x_i=\min_{1\le i\le n} X_i(\omega)$$ The index $i$ of the smallest realisation among $(x_1,\ldots,x_n)$ is itself the realisation of a random variable that varies with $\omega\in\Omega$.

For instance,if I simulate ten realisations of a triplet of Normal variables, $X_1,X_2,X_3$,

> matrix(rnorm(30),10,3)
             [,1]       [,2]        [,3]
 [1,] -1.02746612  1.4859813 -0.27463682
 [2,]  1.71744642 -1.0629540 -0.91810440
 [3,] -0.02366358  1.2791298 -1.03245628
 [4,]  0.19509761  0.6689736  2.38240372
 [5,] -0.27112138  0.3897314  2.00995758
 [6,]  0.72376531 -0.2788061  0.06265955
 [7,] -1.99991699  1.0521950 -1.06698400
 [8,]  0.96063353 -0.1083878  0.01145838
 [9,] -1.79045784 -1.6397322  0.92556376
[10,]  0.64114091  0.9712841 -0.18193148

the ten realisations of $X_{(1)}$ are

 [1,] -1.02746612
 [2,] -1.0629540
 [3,] -1.03245628
 [4,]  0.19509761  
 [5,] -0.27112138  
 [6,] -0.2788061 
 [7,] -1.99991699
 [8,] -0.1083878
 [9,] -1.7904578
[10,] -0.18193148

with respective indices 1, 2, 3, 1, 1, 2, 1, 2, 1, 3

  • $\begingroup$ Thank you. It makes sense to me now. By the way $\min\limits_{1\leq i\leq n}X_i$ is just a notation, isn't it ? I mean the "operator" $\min$ in $\min\limits_{1\leq i\leq n}X_i$ and in $\min\limits_{1\leq i\leq n}x_i$ do not work the same way. $\endgroup$
    – Bungbu
    Commented Sep 13, 2016 at 9:03
  • 4
    $\begingroup$ @Bungbu: The operator is the same in the sense that you consider the minimum of the random variables, for each realisation of the sample. It is no different from considering $X_1^2$ or $X_3/X_7$, which are also functions applied to one or several random variables. $\endgroup$
    – Xi'an
    Commented Sep 13, 2016 at 9:21
  • $\begingroup$ I appreciate your help but i am still very confused. First of all, since $X_i$'s are iid then $X_1(\omega)=\dots=X_n(\omega)\in R,$ for all $\omega \in \Omega$. So it does not make sense to me to take $\min\limits_{1\leq i\leq n} X_i(\omega)$. According to the example you gave, I sense how $X_{(1)}$ works, but i cannot figure out what exactly $X_{(1)}$ is, like, what is the domain and range of these statistics. $\endgroup$
    – Bungbu
    Commented Sep 15, 2016 at 3:20
  • 1
    $\begingroup$ You should read back your probability notes or book as iid does not mean that $X_1(\omega)=\dots=X_n(\omega)$ at each realisation! If you think about it, how can independent variates take the same value at each realisation? All the $X_i$'s are different functions, albeit with the same distribution. $\endgroup$
    – Xi'an
    Commented Sep 15, 2016 at 7:16

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