International Mathematical Forum, 4, 2009, no. 22, 1091 - 1103
Joint Distribution of any Record
Value and an Order Statistics
Cihan Aksop
Gazi University, Department of Statistics
06500 Teknikokullar, Ankara, Turkey
[email protected]
Salih Çelebioğlu
Gazi University, Department of Statistics
06500 Teknikokullar, Ankara, Turkey
[email protected]
Abstract
Let X1 , X2 , . . . , Xn be a random sample drawn from an absolutely
continuous distribution function F . Let X1:n < X2:n < · · · < Xn:n be
the order statistics and Yp , p ≥ 1 be the record values of this random
sample. In this study, the joint distribution of a record value Yp and an
order statistic Xr:n is given.
Mathematics Subject Classification: 62H10
Keywords: Order statistics, record values, record times
1
Introduction
Let X1 , X2 , . . . , Xn be a random sample drawn from an absolutely continuous
distribution function F and let (a, b) be the support of F with −∞ ≤ a < b ≤
∞. Then record times and the record values are defined respectively by
L (1) = 1, Y1 = X1
L (n + 1) = min j : j > L (n) , Xj > XL(n)
Yn = XL(n) (n ≥ 1) .
The record statistics and the order statistcs are frequently used so as to
characterize some distribution functions. Nagaraja and Nevzorov (1997) [8]
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Cihan Aksop and Salih Çelebioğlu
are interested in the distribution of Y2 given X2:2 and thus obtained a characterization of the exponential distribution. Su et al. (2008) [10] dealt with
the distribution of Y1 given Xn:n while Balakrishnan and Stepanov (2004) [3]
are interested in the distribution of Y2 given Xn:n . The results by Ahsanullah (1979) [1], Balakrishnan and Balasubramanian (1995) [2], Gupta (1978)
[4], Huang (1975) [5], Kirmani (1984) [6], Nagaraja (1977) [7] and Nevzorov
(2000) [9] are some other studies obtained in similar context.
In this study, the joint distribution of Xr:n and Yp , p ≥ 1 are examined.
2
The joint distribution of Xr:n and Yp
Let b > yp > yp−1 > · · · > y1 > y0 = a, b ≤ x ≤ x and examine the probability
P {Yp = yp , Xr:n = x} for the cases yp < x, y = x and yp > x separately.
1) yp < x ⇒ L (p) < r. Hence we have
P {Yp = yp , Xr:n = x, L (p) = kp , Yp−1 = yp−1 , L (p − 1) = kp−1 ,
. . . , Y2 = y2 , L (2) = k2 , Y1 = y1 }
= P {X1 = y1 , X2 < y1 , . . . , Xk2 −1 < y1 , Xk2 = y2 , Xk2 +1 < y2 , . . .
. . . , Xkp−1 −1 < yp−2 , Xkp−1 = yp−1 , Xkp−1 +1 < yp−1 , . . . , Xkp −1 < yp−1 ,
Xkp = yp , Xr−kp :n−kp = x
=
p−1
f (yi ) F ki+1 −ki −1 (yi ) f (yp ) P Xr−kp :n−kp = x
i=1
p−1
=
f (yi ) F ki+1 −ki −1 (yi ) f (yp )
i=1
(n − kp )!
F r−kp −1 (x)
(r − kp − 1)! (n − r)!
× [1 − F (x)]n−r f (x)
⇒ P {Yp = yp , Xr:n = x, L (p) = kp , L (p − 1) = kp−1 , . . . , L (2) = k2 }
yp yp−1
y2
(n − kp )!
F r−kp −1 (x) [1 − F (x)]n−r f (x)
=
···
(r
−
k
−
1)!
(n
−
r)!
p
a
a
a
p−1
f (yi) F ki+1 −ki −1 (yi) f (yp ) dy1 · · · dyp−2 dyp−1
×
i=1
=
(n − kp )!
F r−kp −1 (x) [1 − F (x)]n−r f (x)
(r − kp − 1)! (n − r)!
p
1
kp −1
(yp ) f (yp )
×F
k −1
i=2 i
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Record values and order statistic
⇒ P {Yp = yp , Xr:n = x} =
kp −1
r−1
···
kp =p kp−1 =p−1
×F
r−kp −1
n−r
(x) [1 − F (x)]
f (x) F
k
3 −1
k2 =2
kp −1
(n − kp )!
(r − kp − 1)! (n − r)!
(yp ) f (yp )
p
i=2
=
1
ki − 1
r−1
kp
(n − kp )!
1
F r−kp −1 (x) [1 − F (x)]n−r f (x)
kp − 1 (r − kp − 1)! (n − r)!
=p
×F kp −1 (yp ) f (yp ) ζ (p, kp )
where ζ (p, kp ) is a function independent of the distribution function and defined by
kp −1
ζ (p, kp ) =
···
kp−1 =p−1
p−1
k
3 −1 k2
i=2
1
.
ki − 1
The following Table 1 gives the values of the function ζ (p, kp ) corresponding
to some p and kp values. These values can be obtained by the Octave program
code given in Appendix-1.
4
3 1,5000
4
5
6
7
8
9
10
5
6
1,8333 2,0833
1,0000 1,4583
0,4167
7
2,2833
1,8750
0,7083
0,1250
8
2,4500
2,2556
1,0208
0,2431
0,0292
Table 1. Some values of ζ (p, kp )
2) The case yp = x. We have
9
2,5929
2,6056
1,3431
0,3889
0,0639
0,0056
10
2,7179
2,9297
1,6687
0,5568
0,1125
0,0135
0,0009
11
2,8290
3,2316
1,9943
0,7422
0,1744
0,0260
0,0024
0,0001
12
2,9290
3,5145
2,3174
0,9416
0,2486
0,0435
0,0050
0,0004
13
3,0199
3,7808
2,6369
1,1523
0,3342
0,0661
0,0090
0,0008
14
3,1032
4,0325
2,9520
1,3720
0,4302
0,0939
0,0145
0,0016
15
3,1801
4,2712
3,2622
1,5991
0,5357
0,1270
0,0217
0,0027
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Cihan Aksop and Salih Çelebioğlu
P {Yp = x, Xr:n = x, L (p) = kp , Yp−1 = yp−1 , L (p − 1) = kp−1 ,
. . . , Y2 = y2 , L (2) = k2 , Y1 = y1 }
= P {X1 = y1 , X2 < y1 , . . . , Xk2 −1 < y1 , Xk2 = y2 , Xk2 +1 < y2 , . . .
. . . Xkp−1 −1 < yp−2 , Xkp−1 = yp−1 , Xkp−1 +1 < yp−1 , . . . , Xkp −1 < yp−1
Xkp = x, Xr−kp :n−kp = x
=
p−1
f (yi ) F ki+1 −ki −1 (yi ) P Xr−(kp −1):n−(kp −1) = x
i=1
p−1
=
f (yi ) F ki+1 −ki −1 (yi )
i=1
(n − kp + 1)!
[1 − F (x)]r−kp F n−r (x) f (x)
(r − kp )! (n − r)!
⇒ P {Yp = x, Xr:n = x, L (p) = kp , L (p − 1) = kp−1 , . . . , L (2) = k2 }
y2
x yp−1
(n − kp + 1)!
···
[1 − F (x)]r−kp F n−r (x) f (x)
=
(r − kp ) 1 (n − r)!
a
a
a
p−1
×
f (yi ) F ki+1 −ki −1 (yi ) dy1 · · · dyp−2 dyy−1
i=1
p
1
(n − kp + 1)!
r−kp
n−r+kp −1
[1 − F (x)]
=
F
(x) f (x)
(r − kp )! (n − r)!
k −1
i=2 i
⇒ P {Yp = x, Xr:n = x} =
r
kp −1
···
kp =p kp−1 =p−1
× [1 − F (x)]r−kp F n−r+kp −1 (x) f (x)
=
r
kp
k
3 −1
k2
p
i=2
(n − kp + 1)!
(r − kp )! (n − r)!
=2
1
ki − 1
(n − kp + 1)!
1
[1 − F (x)]r−kp F n−r+kp −1 (x) f (x) ζ (p, kp ) .
kp − 1 (r − kp )! (n − r)!
3) The case yp > x. In addition, let yt < x and yt+1 ≥ x be satisfied for a
t = 0, 1, . . . , p − 1. Let us call any nonrecord observation as an ordinary value.
In the case we have
Record values and order statistic
1095
P {Yp = yp , Xr:n = x, L (p) = kp , Yp−1 = yp−1 , L (p − 1) = kp−1 ,
. . . , Y2 = y2 , L (2) = k2 , Y1 = y1 }
= P {X1 = y1 , X2 < y1 , . . . , Xk2 −1 < y1 , Xk2 = y2 , Xk2 +1 < y2 ,
. . . , Xkt −1 < yt−1 , Xkt = yt , Xkt +1 < yt , . . . , Xkt+1 −1 < yt ,
Xkt+1 = yt+1 − Xkt+1 +1 < yt+1 , . . . , Xkp −1 < yp−1 , Xkp = yp , Xr:n = x
t
f (yi ) F ki+1 −ki −1 (yi )
=
i=1
×P Xkt+1 = yt+1 , Xkt+1 +1 < yt+1 , . . . , Xkp −1 < yp−1 , Xkp = yp ,
Xr−(kt+1 −1):n−(kt+1 −1) = x .
Now, let us consider the probability in the last equation closely. If we
call the ordinary values between two successive record values as ”the set of
ordinary values”, then we have p −t−1 sets of ordinary values in the examined
probability. Let the numbers of values greater than x in these sets of ordinary
values be i1 , i2 , . . . , ip−t−1 , respectively. For the sake of simple notation, let
I = (i1 , i2 , . . . , ip−t−1 ) and J be the set of indices for the random variables
with the ordinary values. Thus the examined probability can be partitioned
into three parts as follows
P Xkt+1 = yt+1 , Xkt+1 +1 < yt+1 , . . . , Xkp −1 < yp−1 , Xkp = yp ,
Xr−(kt+1 −1):n−(kt+1 −1) = x
P Xkt +1 = yt+1 , Xkt+1 +1 < yt+1 , . . . , Xkp −1 < yp−1 , Xkp = yp ,
=
I∈I1
Xr−(kt+1 −1):n−(kt+1 −1) = x
P Xkt+1 = yt+1 , Xkt+1 +1 < yt+1 , . . . , Xkp −1 < yp−1 ,
+
I∈I2 j∈J
Xkp = yp , Xkp = yp , Xr−(kt+1 −1):n−(kt+1 −1) = x, Xj = x
P Xkt+1 = x, Xkt+1 +1 < x, . . . , Xkp −1 < yp−1 , Xkp = yp ,
+
I∈I3
Xr−(kt+1 −1):n−(kt+1 −1) = x
where
(1)
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Cihan Aksop and Salih Çelebioğlu
p−t−1
(i1 , i2 , . . . , ip−t−1 ) | max {0, kp − r − p + t} ≤
I1 =
ij
j=1
≤ min {kp − kt+1 + 1, n − r} − (p − t)}
p−t−1
(i1 , i2 , . . . , ip−t−1 ) | max {0, kp − r − p + t − 1} ≤
I2 =
ij
j=1
I3 =
≤ min {kp − kt+1 , n − r} − (p − t)}
p−t−1
(i1 , i2 , . . . , ip−t−1 ) | max {0, kp − r − p + t} ≤
ij
j=2
≤ min {kp − kt+2 + 1, n − r} − (p − t − 1)} .
In this partition, the first sum represents the case where none of the ordinary and record values are not equal to x while the second sum points out
the case random variable indexed by j of ordinary values is equal to x. The
last sum denotes the case where one of therecord values is equal to x, and is
p−t−1
ij and examine these sums,
only possible for t < p − 1. Define Ikt = j=k
respectively:
the first sum gives
P Xkt+1 = yt+1 , Xkt+1 +1 < yt+1 , . . . , Xkp −1 < yp−1 , Xkp = yp ,
I∈I1
Xr−(kt+1 −1):n−(kt+1 −1) = x
p−1
kj+1 − kj − 1
f (yj )
[F (yj ) − F (x)]ij−t F kj+1 −kj −1−ij−t (x)
=
ij−t
I∈I1 j=t+1
×f (yp ) P Xr−kp +p−t+I1t :n−kp = x
p−1 kj+1 − kj − 1
f (yj ) [F (yj ) − F (x)]ij−t f (yp )
=
ij−t
I∈I1 j=t+1
×
(n − kp )!
F r−kt+1 (x)
(r − kp + p − t + I1t − 1)! (n − r − p + t − I1t )!
t
× [1 − F (x)]n−r−p+t−I1 f (x)
and the second sum gives
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Record values and order statistic
P Xkt+1 = yt+1 , Xkt+1 +1 < yt+1 , . . . , Xkp −1 < yp−1 , Xkp = yp ,
I∈I2 j∈J
Xr−(kt+1 −1):n−(kt+1 −1) = x, Xj = x
p
(kt+2 − kt+1 − 1)!
=
f (yz )
(kt+2 − kt+1 − i1 − 2)!i1 !
I∈I z=t+1
2
i1
kt+2 −kt+1 −i1 −2
(x) f (x)
× [F (yt+1 ) − F (x)] F
kt+3 − kt+2 − 1
i2
kt+3 −kt+2 −1−i2
[F (yt+2 ) − F (x)] F
(x) · · ·
×
i2
kp − kp−1 − 1
[F (yp−1 ) − F (x)]ip−t−1
×
ip−t−1
× F kp −kp−1 −1−ip−t−1 (x)
kt+2 − kt+1 − 1
i1
kt+2 −kt+1 −i1 −1
+
(x)
[F (yt+1 ) − F (x)] F
i1
(kt+3 − kt+2 − 1)!
×
[F (yt+2 ) − F (x)]i2
(kt+3 − kt+2 − i2 − 2)!i2 !
× F kt+3 −kt+2 −i2 −2 (x) f (x) × · · ·
kp − kp−1 − 1
×
[F (yp−1 ) − F (x)]ip−t−1
ip−t−1
× F kp −kp−1 −1−ip−t−1 (x) + · · ·
t
t
n − kp
×
F r−kp +p−t+I1 (x) [1 − F (x)]n−r−p+t−I1 .
t
r − kp + p − t + I1
Defining
Bul =
1 ,u = 1
0 , otherwise
the second sum can be written in the following form
=
p−t−1
p−t−1
I∈I2 u=1
l=1
(kt+l+1 − kt+1 − 1)!
f (yt+l )
(kt+l+1 − kt+l − il − 1 − Bul )!il !
× [F (yt+l ) − F (x)]il f (yp )
t
n − kp
F r−kt+1 (x) [1 − F (x)]n−r−p+t−I1 f (x) .
×
t
r − kp + p − t + I1
For the last sum, it is obvious that
1098
Cihan Aksop and Salih Çelebioğlu
p−1
f (yj )
I∈I3 j=t+2
×
kj+1 − kj − 1
ij−t
[F (yj ) − F (x)]ij−t f (yp )
(n − kp + 1)!
F r−kt+1 −1 (x)
(r − kp + p − t + I2t − 1)! (n − r − p + t − I2t + 1)!
t
× [1 − F (x)]n−r−p+t−I2+1 f (x) .
On the other hand, it is also satisfied that
P {Yp = yp , Xr:n = x, L (p) = kp , L (p − 1) , kp−1 , . . . , L (2) = k2 }
yt+2 x yt
y2 p−1 yp
t
=
···
···
f (yi ) F ki+1 −ki −1 (yi )
t=0
x
x
a
a
a
i=1
×P Xkt+1 = yt+1 , Xkt+1 +1 < yt+1 , . . . , Xkp −1 < yp−1 , Xkp = yp ,
Xr−(kt+1 −1):n−(kt+1 −1) = x dy1 dy2 · · · dyt dyt+1 · · · dyp−1
yt+2 p−1 yp
t+1
1
=
···
F kt+1 −1 (x)
k
−
1
i
x
t=0 x
i=2
×P Xkt+1 = yt+1 , Xkt+1 +1 < yt+1 , . . . , Xkp −1 < yp−1 , Xkp = yp ,
Xr−(kt+1 −1):n−(kt+1 −1) = x dyt+1 · · · dyp−1 .
Partition the integral under the last sum into three parts exactly as in (1)
1
kt+1 −1
without considering the coefficient t+1
(x) for one moment and
i=2 ki −1 F
evaluate each part separately:
yp
x
···
x
yt+2
p−1 kj+1 − kj − 1
f (yj ) [F (yj ) − F (x)]ij−t
ij−t
I∈I1 j=t+1
×f (yp ) dyt+1 · · · dyp−1
(n − kp )!
F r−kt+1 (x)
×
t
t
(r − kp + p − t + I1 − 1)! (n − r − p + t − I1 )!
t
× [1 − F (x)]n−r−p+t−I1 f (x) .
Making the transformation F (yj ) − F (x) = uj , j = t + 1, . . . , p gives
y +1
F (y
)−F (x)
u
f (yj ) dyj = duj and |xj → |0 j+1
= |0 j+1 , and results
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Record values and order statistic
p−t−1
p−1 t
1
kj+1 − kj − 1
=
[F (yp ) − F (x)]I1 +p−t−1 f (yp )
k
ij−t
l=1 il + k
I∈I1 j=t+1
k=1
×
(n − kp )!
F r−kt+1 (x)
(r − kp + p − t + I1t − 1)! (n − r − p + t − I1t )!
t
× [1 − F (x)]n−r−p+t−I1 f (x)
p−1 t
1
kj+1 − kj − 1
=
[F (yp ) − F (x)]I1 +p−t−1 f (yp )
j−t
ij−t
il + j − t
l=1
I∈I1 j=t+1
×
(n − kp )!
F r−kt+1 (x)
t
t
(r − kp + p − t + I1 − 1)! (n − r − p + t − I1 )!
t
× [1 − F (x)]n−r−p+t−I1 f (x) .
The evaluation of second integral is as follows:
yp
x
···
yt+2
x
p−t−1
p−t−1
I∈I2 u=1
l=1
il
(kt+l+1 − kt+l − 1)!
f (yt+l )
(kt+l+1 − kt+l − il − 1 − Bul )!il !
× [F (yt+l ) − F (x)] f (yp ) dyt+1 · · · dyp−1
t
n − kp
×
F r−kt+1 (x) [1 − F (x)]n−r−p+t−I1 f (x)
t
r − kp + p − t + I1
=
p−t−1
p−t−1
I∈I2 u=1
l=1
p−t−1
×
k=1
×
=
k
t
(kt+l+1 − kt+l − 1)!
[F (yp ) − F (x)]I1 +p−t−1
(kt+l+1 − kt+l − il − 1 − Bul )!il !
1
j=1 ij
+k
f (yp )
n − kp
r − kp + p − t + I1t
p−t−1
p−t−1
I∈I2 u=1
l=1
t
F r−kt+1 (x) [1 − F (x)]n−r−p+t−I1
1
(kt+l+1 − kt+l − 1)!
l
(kt+l+1 − kt+l − il − 1 − Bul )!il ! j=1 ij + l
t
× [F (yp ) − F (x)]I1 +p−t−1 f (yp )
t
n − kp
F r−kt+1 (x) [1 − F (x)]n−r−p+t−I1 −1 f (x)
×
t
r − kp + p − t + I1
p−t−1
p−t−1
t
[F (yp ) − F (x)]I1 +p−t−1
(kt+l+1 − kt+l − 1)!
=
l
(kt+l+1 − kt+l − il − 1 − Bul )!il !
j=1 ij + l
I∈I2 u=1
l=1
t
n − kp
×f (yp )
F r−kt+1 (x) [1 − F (x)]n−r−p+t−I1 f (x) .
t
r − kp + p − t + I1
1100
Cihan Aksop and Salih Çelebioğlu
The evaluation of last integral is similar to the first integral which results
p−1 t
1
kj+1 − kj − 1
[F (yp ) − F (x)]I2 +p−t−1 f (yp )
j−t
ij−t
il + j − t
l=2
I∈I3 j=t+2
×
(n − kp + 1)!
F r−kt+2 (x)
t
t
(r − kp + p − t + I2 − 1)! (n − r − p + t − I2 + 1)!
t
× [1 − F (x)]n−r−p+t−I2 f (x) .
Thus in the case yp > x , we have
P {Yp = yp , Xr:n = x, L (p) = kp , L (p − 1) = kp−1 , . . . , L (2) = k2 }
p−1 t+1
1
=
F kt+1 −1 (x)
k −1
t=0 i=2 i
p−1 I t +p−t−1
kj+1 − kj − 1 [F (yp ) − F (x)] 1
f (yp )
×
j−t
ij−t
l=1 il + j − t
I∈I1 j=t+1
×
(n − kp )!
F r−kt+1 (x)
(r − kp + p − t + I1t − 1)! (n − r − p + t − I1t )!
t
+
× [1 − F (x)]n−r−p+t−I1 f (x)
p−t−1
p−t−1
(kt+l+1 − kt+l − 1)!
I∈I2 u=1
l=1
(kt+l+1 − kt+l − il − 1 − Bul )!il !
t
[F (yp ) − F (x)]I1 +p−t−1
f (yp )
×
l
j=1 ij + l
n − kp
n−r−p+t−I1t
r−kt+1
(x) [1 − F (x)]
f (x)
×
F
r − kp + p − t + I1t
+
p−2 t+1
1
F kt+1 −1 (x)
k
−
1
i
t=0 i=2
kj+1 − kj − 1 [F (yp ) − F (x)]I2t +p−t−1
p−1
f (yp )
×
j−t
ij−t
il + j − t
l=2
I∈I3 j=t+2
×
(n − kp + 1)!
F r−kt+2 (x)
t
t
(r − kp + p − t + I2 − 1)! (n − r − p + t − I2 + 1)!
t
× [1 − F (x)]n−r−p+t−I2 f (x) .
Hence in the case yp > x ,we have the joint distribution of pth record value
1101
Record values and order statistic
Yp , and rth order statistic Xr:n as follows:
P {Yp = yp , Xr:n = x}
=
kp
n
···
kp =p kp−1 =p−1
p−1 t+1
k3 k2 =2 t=0 i=2
1
F kt+1 −1 (x)
ki − 1
p−1 I t +p−t−1
kj+1 − kj − 1 [F (yp ) − F (x)] 1
×
f (yp )
j−t
ij−t
il + j − t
l=1
I∈I1 j=t+1
×
(n − kp )!
F r−kt+1 (x)
(r − kp + p − t + I1t − 1)! (n − r − p + t − I1t )!
t
× [1 − F (x)]n−r−p+t−I1 f (x)
p−t−1
p−t−1
(kt+l+1 − kt+l − 1)!
t
[F (yp ) − F (x)]I1 +p−t−1
+
f (yp )
l
(k
−
k
−
i
−
1
−
B
)!i
!
t+l+1
t+l
l
ul
l
i
+
l
j
j=1
I∈I2 u=1
l=1
n − kp
n−r−p+t−I1t
r−kt+1 (x)
[1 − F (x)]
f (x)
×
F
r − kp + p − t + I1t
+
n
kp
kp =p kp−1 =p−1
···
p−2 t+1
k3 k2
t=0 i=2
1
F kt+1 −1 (x)
ki − 1
kj+1 − kj − 1 [F (yp ) − F (x)]I2t +p−t−1
×
f (yp )
j−t
ij−t
il + j − t
p−1
l=2
I∈I3 j=t+2
×
(n − kp + 1)!
F r−kt+2 (x)
(r − kp + p − t + I2t − 1)! (n − r − p + t − I2t + 1)!
t
× [1 − F (x)]n−r−p+t−I2 f (x) .
3
Appendix 1. Octave Code for Calculating
ζ (p, kp)
Define the values of p and k (p) for which ζ (p, k (p)) is to be calculated. toplam
in the last line is the value of ζ (p, k (p)).
# p=
# k(p)=
for i=2:p-1
k(i)=i;
end
toplam=0;
1102
Cihan Aksop and Salih Çelebioğlu
devam=1;
i=1;
while (i<p)
carpim=1;
for j=2:p-1
carpim=carpim*1/(k(j)-1);
end
toplam=toplam+carpim;
i=2;
while (i<p)
k(i)=k(i)+1;
if (k(i)>k(i+1)-1)
k(i)=i;
i=i+1;
else
break
end
end
end
toplam
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1103
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Received: September, 2008