pdf of the rth order statistics

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Transcript pdf of the rth order statistics

‫اعداد ‪ :‬د‪ .‬زكية الصيعري‬
‫كلية العلوم للبنات‬
‫‪2014 -1435‬‬
‫‪1‬‬
‫‪Dr. Zakeia A.Alsaiary‬‬
Refrences: ‫المراجع‬
1- First course in order
statistics
Arnold, Balakrishnan and Nagaraja
2- Order statistics
H. A. David
3- ‫اإلحصاءات الترتيبية‬
‫ مكتبة المتنبي‬-‫للدكتورة ثروت محمد عبدالمنعم‬
Dr. Zakeia
2 A.Alsaiary
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Dr. Zakeia A.Alsaiary
Definition
The order statistics of a random sample
X1, . . . ,Xn are the sample values placed in
ascending order. They are denoted by X(1), . .
. ,X(n).
The smallest of the Xi,s is denoted by X1:n,
the second smallest is denoted by X2:n,…,
and finally the largest is denoted by Xn:n thus
order statistics are random variables that
satisfy X(1) < X(2) <· · · · < X(n). The
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‫اإلحصاءات المرتبة ‪ :‬هي عناصر عينة عشوائية مرتبة من األصغر إلى‬
‫األكبر‪ .‬وفي أغلب مناقشاتنا لإلحصاءات المرتبة سوف نعتبر العينة‬
‫العشوائية تتبع توزيعات متصلة‪.‬‬
‫‪Example:‬‬
‫‪Let x1 = 0.62 , x2 = 0.98, x3= 0.31 , x4 = 0.81 ,‬‬
‫‪x5 = 0.53 are observation for independent‬‬
‫‪expeirement , the order statistics of it are:‬‬
‫=‪Y1 = 0.31 , y2 = 0.53 , y3 = 0.62 , y4 = 0.81 , y5‬‬
‫‪0.98‬‬
‫‪Y3 = 0.62 is the median and‬‬
‫‪the range = y5 - y1= 0.98 – 0.31 = 0.67‬‬
‫‪Dr. Zakeia A.Alsaiary‬‬
‫‪5‬‬
ORDER STATISTICS
• If X1, X2,…,Xn be a r.s. of size n from a population
with continuous pdf f(x), then the joint pdf of the
order statistics X(1), X(2),…,X(n) is
g  y 1 , y  2 ,
, y  n    n !f  y 1  f  y  2 
f  y n  
for   y (1)  ...  y ( n )  
=0
g  y 1 , y  2 ,
(elsewhere)
, y  n    n ! f  y i  
n
i 1
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ORDER STATISTICS
Example 1:4 ‫كتاب اإلحصاءات الترتيبية صفحة‬
Find the joint pdf of the order statistics for the
uniform distribution , the standard exponential
distribution and normal distribution?
Solution: p.d.f for the uniform is:
f ( x )  1, 0  x 1
 g ( y 1 , y 1,..., y n )  n !, 0  y 1  y 1 ...,  y n 1,
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ORDER STATISTICS
Solution: p.d.f for the standard Exponential
distribution is:
f (x )  e
 g ( y 1 , y 1 ,..., y n )  n !e

x
, x 0
n
yi
i 1
, 0  y 1  y 1  ...,  y n  ,
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ORDER STATISTICS
Solution: p.d.f for the standard normal distribution
is:
x2
f (x ) 

1
e
2
2
,  x  
n
y i2

2
i 1
n! 
 g ( y 1 , y 1 ,..., y n ) 
e
,  y 1  y 1  ...,  y n  ,
2
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The marginal distributions for
the Oder Statistics
p.d.f of the rth order statistics:
Theorem:
If X1, X2,…,Xn be a r.s. of size n from a population
with continuous pdf f(x), then the p.d.f. of the rth
order statistics X(r) is given as:
n!
gr ( y r ) 
f (yr)
( r 1)!( n  r )!
r 1
n 1
 [ F ( y r ) ] [1  F ( y r ) ] ,   y r 
PROOF:
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ORDER STATISTICS
• r-th Order Statistic
y1 y2 … yr-1 yr yr+1 …
P(X<yr)
y
yn
P(X>yr)
# of possible orderings
n!/{(r1)!1!(n  r)!}
f(yr)
gr ( y r ) 
n!
f (yr)
( r 1)!( n  r )!
 [ F ( y r ) ]r 1 [1  F ( y r ) ]n 1 ,    y r  
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The marginal distributions for the oder
statistics
p.d.f of the largest order statistics:
Theorem:
If X1, X2,…,Xn be a r.s. of size n from a population
with continuous pdf f(x), then the p.d.f. of the
Largest order statistics Y(n) is given as:
n 1
g n ( y n )  n f ( y n )[ F ( y n )] ,   y n 
PROOF:
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The marginal distributions for
the oder statistics
p.d.f of the smallest order statistics:
Theorem:
If X1, X2,…,Xn be a r.s. of size n from a population
with continuous pdf f(x), then the p.d.f. of the
smallest order statistics X(1) is given as:
n 1
g1 ( y 1 )  n f ( y 1 )[1 F ( y 1 )] ,   y n 
PROOF:
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Example
Let
Y 1  Y 2  ...  Y 6
are an O. S. of sample size n = 6
and the p.d.f. of this sample is
1
f (x ) 
, 0 x 2
2
Find: g r ( y r ) , g1 ( y 1 ) , g 6 ( y 6 )
Solution:
6!
r 1
61
gr ( y r ) 
y r [2 y r ] , 0  y r 2
6
( r 1)!( 6  r )!2
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6
5
g1 ( y 1 )  6 [ 2  y 1 ] , 0  y 1  2
2
6 5
g6 ( y 6 )  6 y 6 , 0  y 6 2
2
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Example
Let X
,X 2 ,X 3 ,X 4 ,X 5
is a random sample from beta
population with parameters  1 ,  2 .
Y 1  Y 2  Y 3  Y 4 Y 5 Let the order statistics of the
sample,
1
f ( x )  2 x , 0  x 1
Find:
g n ( y n ) , g 1 ( y 1 ) then find P (Y  0.2 ),
g 3 ( y 3 ) then
find the var iance
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Solution:
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Solution
f (x ) 
1
x  1 (1  x )  1 , 0  x 1 ,   1,   0
 ( ,  )
f ( x )  2 x , 0  x 1
0 , x  0
 2
F ( x )  x , 0  x 1
1 , x  1

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Example
Let
is a random sample from
standard uniform distribution.
Find: p.d.f. for the median?
________________________________________
Y m 1
m
m
X 1 , X 2 ,..., X 2m 1
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Joint p.d.f. of i-th and j-th Order
Statistic (for i<j)
Theorem:
If X1, X2,…,Xn be a r.s. of size n from a population
with continuous pdf f(x), and Y1< Y2<…<Yn are
the order statistics of that sample, then the p.d.f.
of the two order statistics Yi< Yj , i<j and i,j = 1,2,
…,n is given as
g ij ( y i , y j ) 
n!
[F ( y i )]i 1 f ( y i )[F ( y j )  F ( y i )] j i 1 f ( y j )[1  F ( y j )]n  j
(i  1)!( j  i  1)!(n  j )!
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ORDER STATISTICS
• Joint p.d.f. of i-th and j-th Order Statistic (for i<j)
i-1 items
y1 y2 … yi-1
1 item
n-j items
1 item
# of possible orderings
n!/{(i1)!1!(j-i-1)!1!(n  j)!}
y
yi
P(X<yi)
yi+1
…
yj-1 yj yj+1
P(yi<X<yj)
f(yi)
g ij ( y i , y j ) 
j-i-1 items
yn
P(X>yj)
f(yj)
n!
[F ( y i )]i 1 f ( y i )[F ( y j )  F ( y i )] j i 1 f ( y j )[1  F ( y j )]n  j
(i  1)!( j  i  1)!(n  j )!
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Example
Let
is a random sample from beta
population with parameters  1 ,  2 .
Let Y 1  Y 2  Y 3  Y 4 Y 5 the order statistics of the sample,
X 1 ,X 2 ,X 3 ,X 4 ,X 5
f ( x )  2 x , 0  x 1
Find: 1- Joint p.d.f. for y 2 , y 4 ( g 2,4 ( y 2 , y 4 ) )
1
1
2
2- Let n = 4 then find P ( y 3  ) , P (  y 3  )
2
3
3
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Example
Let Y
 Y 2  Y 3  Y 4 Y 5
the order statistics of random
sample with size n = 5, from distribution with
p.d.f.
1
f ( x )  e x , 0  x
Prove that the two O.S. Z 2 Y 4 Y 2 , Z 1 Y 2
Are independent.
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Solution
f ( x )  e x , 0  x
0 , x  0

F ( x )  1  e  x , 0  x  1
1 , x  1

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g ( y 2 , y 4 )  120e  y 2e  2 y 4 (1e  y 2 )(e  y 2 e  y 4 ) , 0  y 2  y 4 
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c.d.f of the rth order statistics:
Theorem:
If X1, X2,…,Xn be an independent and identical r.s.
of size n from a population with pdf f(x) and cdf
F(x), then the c.d.f. of the rth order statistics X(r)
is given as:
r j
n r n r
n!
[
F
(
x
)]
 
j
Fr :n ( x ) 
(

1)



j
( r 1)!( n  r )! j  0  
rj
PROOF:
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p.d.f of the rth order statistics:
Special cases:
1  Fr :n (   )  0
n r n r
n!
1
 
j
2  Fr :n (  ) 
( 1)



( r 1)!( n  r )! j  0  j 
rj
 1 , prove that ??
3  Fn :n ( x )  [ F ( x ) ]n why ??
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c.d.f of the rth order statistics:
Theorem:
If X1, X2,…,Xn be an independent and identical r.s.
of size n from a population with pdf f(x) and cdf
F(x), then the c.d.f. of the rth order statistics X(r)
is given as:
n r  j 1
r 1 r 1
n!
1

[1

F
(
x
)
]
 
j
Fr :n ( x ) 
(

1)



j
( r 1)!( n  r )! j  0  
n  r  j 1
PROOF:
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p.d.f of the rth order statistics:
Special case:
F1:n ( x ) 1 [1 F ( x )]
n
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why ??
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p.d.f of the rth order
statistics:
‫صور أخرى لدالة التوزيع التراكمية لإلحصاء المرتب الرائي‬
:‫مثبتة في الكتب‬
1
Fr :n ( x ) 
 ( r , n  r  1)
F (x )

y r 1 [1  y ]n r dy
0
 I F ( x ) ( r , n  r  1)
n 
j
nj
Fr :n ( x )     [ F ( x ) ] [1  F ( x ) ]
j
j  
n
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Example
Let Y 1  Y 2  Y 3  Y 4 Y 5 the order statistics of the sample
with n = 5, from the distribution
f ( x )  2 x , 0  x 1
1
Find: P ( y 4  )
2
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Example
Let Y 1  Y 2  Y 3  Y 4 Y 5  Y 6 Y 7 the order statistics of the
sample with n = 5, from the distribution
f ( x )  3(1 x )2 , 0  x 1
Find: P ( y 4  1 3 0.6)
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