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.