Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

4.4. Order Statistics 255

The pdf ofY 2 is then

h(y 2 )=6f(y 2 )

∫b

y 2

∫y 2

a

f(y 1 )f(y 3 )dy 1 dy 3

=

{

6 f(y 2 )F(y 2 )[1−F(y 2 )] a<y 2 <b

0elsewhere.

Accordingly,

P(Y 2 ≤m)=6

∫m

a

{F(y 2 )f(y 2 )−[F(y 2 )]^2 f(y 2 )}dy 2

=6

{

[F(y 2 )]^2

2

−

[F(y 2 )]^3

3

}m

a

=

1

2

.

Hence, for this situation, the median of the sample medianY 2 is the population
medianm.

Once it is observed that

∫x

a

[F(w)]α−^1 f(w)dw=

[F(x)]α

α

,α> 0 ,

and that ∫
b

y

[1−F(w)]β−^1 f(w)dw=

[1−F(y)]β

β

,β> 0 ,

it is easy to express the marginal pdf of any order statistic, sayYk,intermsofF(x)

andf(x). This is done by evaluating the integral

gk(yk)=

∫yk

a

···

∫y 2

a

∫b

yk

···

∫b

yn− 1

n!f(y 1 )f(y 2 )···f(yn)dyn···dyk+1dy 1 ···dyk− 1.

The result is

gk(yk)=

{ n!

(k−1)!(n−k)![F(yk)]

k− (^1) [1−F(yk)]n−kf(yk) a<yk<b
0elsewhere.
(4.4.2)
Example 4.4.2.LetY 1 <Y 2 <Y 3 <Y 4 denote the order statistics of a random
sample of size 4 from a distribution having pdf
f(x)=
{
2 x 0 <x< 1
0elsewhere.
We express the pdf ofY 3 in terms off(x)andF(x) and then computeP(^12 <Y 3 ).
HereF(x)=x^2 , provided that 0<x<1, so that
g 3 (y 3 )=
{ 4!
2! 1!(y
2
3 )
(^2) (1−y 2
3 )(2y^3 )0<y^3 <^1
0elsewhere.