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(ISSN
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VOLUME
03
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SSUE
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Pages:
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5.
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(2023:
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ABSTRACT
The results of this work were published in the journal Reports of the Academy of Sciences of the Republic of
Uzbekistan without proof. In this paper, all the results obtained with complete proofs. Several theorems for a linear
combination of functions of order statistics are proved in this work.
KEYWORDS
Superpositions of functions, inverse function, Kolmogorov-Smirnov transformation, random variables, estimation of
concentration functions, Chebyshev's inequalities, linear combination of order statistics composed of uniform
distribution, beta distribution.
INTRODUCTION
Research Article
ESTIMATES OF THE CONCENTRATION FUNCTION FOR STATISTICS
n
T
Submission Date:
February 11, 2023,
Accepted Date:
February 16, 2023,
Published Date:
February 21, 2023
Crossref doi:
https://doi.org/10.37547/ajast/Volume03Issue02-03
Madrakhimov Askarali
Fergana State University Of An Associate Professor, Uzbekistan
Khonkulov Ulugbek Khursanalievich
Associate Professor Of Fergana State University, Uzbekistan
Akhmedov Olimjon Ulugbek Ugli
Fergana State University At Is A Lecturer, Uzbekistan
Journal
Website:
https://theusajournals.
com/index.php/ajast
Copyright:
Original
content from this work
may be used under the
terms of the creative
commons
attributes
4.0 licence.
Volume 03 Issue 02-2023
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VOLUME
03
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SSUE
02
Pages:
18-25
SJIF
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FACTOR
(2021:
5.
705
)
(2022:
5.
705
)
(2023:
7.063
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CONSIDER THE STATISTICS
,
1
(
)
n
n
in
i n
i
T
c h x
=
=
. Define the function
( )
H
using the superposition of functions
( )
h
and
1
( )
F
−
,. by equality
1
( )
(
( )).
H
h F
−
=
Here
1
( )
F
−
- the inverse function to
1
( ), ( )
(
)
F
F x
P x
x
=
. Using the Kolmogorov-Smirnov transformation, we can make sure that.
n
T
и
1
(
)
n
in
in
i
c H u
=
equally distributed[8].
In fact, if
F(x)
is continuous, then according to the Kolmogorov-Smirnov transformation
( )
i
F x
represents a uniform
S.V.
i
u
on [0,1]. Therefore
,
,
(
)
.
i n
i n
F x
u
=
Say, by virtue of continuity
1
,
,
( ),
(
)
i n
i n
F x
F
u
x
−
=
. Therefore
1
,
,
,
(
)
(
(
))
(
)
i n
i n
i n
h x
h F
u
H u
−
=
=
. Thus
,
1
(
).
n
n
in
i n
i
T
c H u
=
=
Suppose that
H
has a continuous bounded
derivative of the second order.
Analysis of literature on the topic (Literature review). As in [9], we use the representation
,
,
1
1
(
)
(
)
n
n
n
in
i n
in
i n
n
n
n
i
i
T
c H x
c H u
U
R
=
=
=
=
=
+
+
(1)
where
,
1
1
,
1
1
1
n
n
n
in
n
in
i n
i
i
i
i
i
c H
u
c H
u
n
n
n
=
=
=
=
−
+
+
+
,
2
,
1
1
(
)
2
1
n
n
in
in
i n
i
i
R
c H
u
n
=
=
−
+
,,
,
1
1
in
i n
i
i
u
n
n
=
+
−
+
+
,
,
| | 1,
1
i n
i
Mu
n
=
+
.
We investigate the evaluation of the concentration function of S.V.
n
T
, т.е.
( ; )
sup
{
}
n
n
x
Q T
P x T
x
−
=
+
at any
0
. Using (1) and the fact that
(
; )
( ; )
Q
const
Q
+
=
we have
( ; )
(
; )
n
n
n
n
Q T
Q
U
R
=
+
+
(
; )
n
n
n
Q
U
R
=
+
+
(2)
(Research Methodology).
Let 's first make sure of the validity of the following statements[10].
Lemma 1.
Let x and y be arbitrary S.V. We put
( )
(
),
( )
(
)
F x
P x
x
G x
P x
y
x
=
=
+
.
Then
(
)
(| |
)
( )
(
)
(| |
)
F x
P y
G x
F x
P y
−
−
+
+
for any
0
и
х
.
The proof of this lemma is given in [6].
Lemma 2.
For any
0
и
0
inequality is fair
Volume 03 Issue 02-2023
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American Journal Of Applied Science And Technology
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VOLUME
03
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Pages:
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SJIF
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FACTOR
(2021:
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705
)
(2022:
5.
705
)
(2023:
7.063
)
OCLC
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( ; )
(
; )
2 (
; )
2 (|
|
)
n
n
n
n
Q T
Q U
Q U
P R
+
+
.
Proof. Applying Lemma 1 to
(
)
n
n
P x U
R
x
+
+
and by virtue of (2) we find that
( ; )
sup
{
}
n
n
n
x
Q T
P x U
R
x
−
=
+
+
=
sup { (
)
(
)}
n
n
n
n
x
P U
R
x
P U
R
x
−
=
+
+
−
+
sup { (
)
(
)}
n
n
n
n
x
P U
R
x
P U
R
x
−
=
+
+
−
+
(3)
Since
(
)
(
)
n
n
P U
x
P U
x
+ +
−
−
=
(
)
(
)
n
n
P x
U
x
P x U
x
=
−
+
+
+
(
)
(
)
n
n
P x
U
x
P x U
x
+
+
+ +
=
+
+
(
)
(
)
n
n
P x U
x
P x U
x
+
+
+
+
=
(
)
2 (
)
n
n
P x U
x
P x U
x
=
+
+
+
,
then from the relation (3) we get the inequality
( ; )
sup
(
)
2 (
)
n
n
n
x
Q T
P x U
x
P x U
x
−
+
+
+
2 (|
|
)
n
P R
+
sup
(
)
2 sup
(
)
n
n
x
x
P x U
x
P x U
x
−
−
+
+
+ +
2 (|
|
)
n
P R
=
(
; )
2 (
; )
2 (|
|
)
n
n
n
Q U
Q U
P R
=
+
+
.
Lemma 2 is proved. Let 's put
.
1
in
in
i
c
c H
n
=
+
Takes place
Theorem 1.
Let
0
in
in
c
c
(
1,2,..., ;
1,2,...)
i
n n
=
=
and
max |
( ) |
x
H x
C
. Then
1
2
3/4
( ; )
n
n
C
CB
Q T
n
n
+
(4)
where
1
n
n
in
i
B
c
=
=
.
Proof.
Based on Theorem II.2.1
(
; )
n
C
Q U
n
(5)
Volume 03 Issue 02-2023
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VOLUME
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Pages:
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(2023:
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(
; )
n
C
Q U
n
(6)
Now let's estimate the probability
(|
|
)
n
P R
. By virtue of the definition
n
R
and according to the conditions of
Theorem 1, we obtain that
2
,
1
1
(|
|
)
(
)
2
1
n
n
in
in
i n
i
i
P R
P
c H
u
n
=
=
−
+
2
,
1
2
1
n
in
i n
i
i
P
c
u
n
C
=
−
+
Applying Chebyshev's inequality to the latter, we have
2
,
1
(|
|
)
2
1
n
n
in
i n
i
C
i
P R
M
c
u
n
=
−
+
2
,
,
1
1
2
1
2
n
n
in
i n
in
i n
i
i
C
i
C
c M u
c Du
n
=
=
−
=
+
(7)
Since (sm. [7])
,
2
(
1)
(
1) (
2)
i n
i n i
Du
n
n
− +
=
+
+
,
then it follows from (7) that
2
1
(
1)
(|
|
)
2
(
1) (
2)
n
n
in
i
C
i n i
P R
c
n
n
=
− +
+
+
(8)
Нетрудно убедиться в том, что при
1
i
n
2
(
1)
1
(
1) (
2)
4(
2)
i n i
n
n
n
− +
+
+
+
(9)
Consequently, from the relations (8) and (9) we obtain the following inequality
1
(|
|
)
8
8
n
n
in
n
i
C
C
P R
c
B
n
n
=
=
(10)
In turn, from the relations (5), (6) and (10) according to Lemma 2 we have
2
( ; )
4
n
n
C
C
CB
Q T
n
n
n
+
+
(11)
Minimizing the last two terms on the right side of the relation
Volume 03 Issue 02-2023
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VOLUME
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(10) regarding
, we find, what
1/4
.
2 2
n
n
B
n
=
=
Putting the found expression for
and (11) we finally conclude
that
1
2
3/4
( ; )
n
n
C
CB
Q T
n
n
+
.
Theorem 1 has been fully proved.
Remark 1. Note that the right side of relation (4) tends to zero if
3/2
(
)
n
B
o n
=
.
As the following theorem shows, in the case when the extreme terms of the variation series are removed from
the considered linear combination, the conditions on the function
( )
H
may be weakened.
Theorem
2. Let
0
in
c
=
, if
i
n
and
i
n
,
а
for the rest
,
0
in
i
c
c
.
Пусть
H
has continuous
derivatives
,
H H
в
1
1
[
( )
,
( )
]
F
F
−
−
−
+
(
τ
–
arbitrary number). Then
1
2
3/4
( ; )
n
n
C
CB
Q T
n
n
+
где
1
n
n
in
i
B
c
=
=
.
Proof.
By the condition of the theorem
0
in
c
only in the case when the inequality holds
i
n
. Let
0
and
min( ,1
)
−
. Let 's put
,
1
: max
1
i n
i n
i
A
u
n
=
−
+
. We show that
1
( )
.
P A
O
n
=
.
It is known that (sm. ratio (2.1.1))
1
2
,
1
1
...
...
i
i n
n
z
z
z
u
z
z
+
+ + +
=
+ +
,
где
1
2
1
, ,...,
n
z z
z
+
- independent S.V. with general
F.R.
( )
max(0,1
).
x
G x
e
−
=
−
According to the latter
,
1
: max
1
i n
i n
i
A
u
n
=
−
=
+
1
2
1
1
1
1
1
(
1)(
...
)
(
...
)
max
(
1)(
...
)
i
n
i n
n
n
z
z
z
i z
z
n
z
z
+
+
+
+
+ +
−
+ +
=
=
+
+ +
1
1
1
2
1
1
2
1
(
1)(
...
)
(
...
)
1
max
(
1)
(
...
)
i
n
i n
n
n
z
z
i z
z
n
n
z
z
z
+
+
+
+ +
−
+ +
+
=
+
+
+ +
1
1
1
2
1
(
1)(
...
)
(
...
)
max
(
1)
2
i
n
i n
n
z
z
i z
z
n
+
+
+ +
−
+ +
+
Volume 03 Issue 02-2023
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American Journal Of Applied Science And Technology
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VOLUME
03
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Pages:
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SJIF
I
MPACT
FACTOR
(2021:
5.
705
)
(2022:
5.
705
)
(2023:
7.063
)
OCLC
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Publisher:
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Servi
1
1
1
:
2
...
n
n
z
z
+
+
+ +
(12)
By Chebyshev's inequality, we have
1
2
1
1
2
...
n
n
P
z
z
z
+
+
+ + +
1
2
1
...
1
1
(
1)
2
n
z
z
z
P
O
n
n
+
+
+ +
=
+
.
Therefore
1
1
1
2
1
(
1)(
...
)
(
...
)
1
( )
max
(
1)
2
i
n
i n
n
z
z
i z
z
P A
P
O
n
n
+
+
+ +
−
+ +
+
+
(13)
Let's introduce an event into consideration
1
1
1
2
(
1)(
...
)
(
...
)
:
(
1)
2
i
n
i
n
z
z
i z
z
A
n
+
+
+ +
−
+ +
=
+
.
Since the event
1
1
1
2
| (
1)(
...
)
(
...
(
1)) |
(
1)
2
i
n
n
z
z
i z
z
n
n
+
+
+ +
−
+ +
− +
+
1
1
2
2
| (
1)(
...
) |
| (
...
(
1)) |
(
1)
2
(
1)
4
i
i
n
z
z
i
i z
z
n
n
n
+
+ + −
+ + − +
+
+
.
Then, it follows that
1
1
1
2
| (
1)(
...
)
(
...
(
1)) |
( )
(
1)
2
i
n
i
n
z
z
i
i z
z
n
P A
P
n
+
+
+ + − −
+ +
− +
=
+
1
1
1
2
|
...
|
|
...
(
1) |
(
1)
4
(
1)
4
i
n
z
z
i
i z
z
n
P
P
n
n
+
+ + −
+ +
− +
+
+
+
(14)
1
1
1
2
|
...
|
|
...
(
1) |
(
1)
4
(
1)
4
i
n
z
z
i
z
z
n
P
P
n
n
+
+ + −
+ +
− +
+
+
+
Further, again by virtue of Chebyshev's inequality, we have
1
1
|
...
(
...
) |
( )
(
1)
4
i
i
i
z
z
M z
z
P A
P
n
+ + −
+ +
+
+
1
1
1
1
|
...
(
...
) |
(
1)
4
n
n
z
z
M z
z
P
n
+
+
+ +
−
+ +
+
+
Volume 03 Issue 02-2023
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VOLUME
03
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Pages:
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FACTOR
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705
)
(2022:
5.
705
)
(2023:
7.063
)
OCLC
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Publisher:
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4
4
4
4
1
1
(
)
(
)
(
1)
(
1)
i
i
i
i
i
i
j
j
C
C
M
z
Mz
M
z
Mz
n
n
=
=
−
+
−
+
+
(15)
Applying inequalities for moments of sums of independent S.V. given in V.V.Petrov's monograph ([5] page. 79) from
(15) we get that
2
( )
(
1)
i
C
P A
n
+
.
By virtue of the latter, from the relation (13) we find that
1
1
1
1
1
( )
( )
n
n
i
i
i
i
P A
P
A
O
P A
O
O
n
n
n
=
=
+
+
=
.
Say, we have shown that
1
( )
P A
O
n
=
. Note that if an event occurs
A
, that is , the values
,
1
1
in
i n
i
i
u
n
n
=
−
+
+
, included in expression (1) satisfy the obvious inequality
0
1
in
−
+
.
From this, according to the conditions of Theorem 2, if an event occurs
A
, it follows that
|
(
) |
in
H
C
. Further
reasoning coincides with the reasoning of Theorem 1 Theorem 2 is proved.
Remark 2.
Suppose
( )
h x
x
=
. Then
1
1
1
3
1
1
(
( ))
( )
( ),
( )
,
( )
(
( ))
(
( ))
p F
x
H x
F
x
H x
H x
p F
x
p F
x
−
−
−
−
=
=
= −
,
где
( )
( )
d
p x
F x
dx
=
. If
( )
( )
p x
F x
=
exists
( )
0
p x
, with
р
(
х
)
,
( )
p x
- continuous and
|
( ) |
p x
C
, then
the conditions of Theorem 1 are fulfilled[2, 3].
REFERENCES
1.
Bjerve S. Error bounda for linear combinations of order statictics. Ann. Statist, v.5, 1977, No 2, pp.357-369.
2.
Мадрахимов, А. Э. (1981). Оценки функции концентрации для линейной комбинации порядковых
статистик.
Изв. АН Узб., Серия физ.
-
мат. наук
, (5), 12-17.
3.
Мадрахимов, А., & Кукиева, С. ПРЕДЕЛЬНЫЕ СВОЙСТВА ПОРЯДКОВЫХ СТАТИСТИК.
FarDU. ILMIY XABARLAR
,
5.
4.
Madrahimov, A. E. (2019). Estimation of the function of consentration for an ordered statistics.
Scientific journal of
the Fergana State University
,
2
(4), 6-12.
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SSUE
02
Pages:
18-25
SJIF
I
MPACT
FACTOR
(2021:
5.
705
)
(2022:
5.
705
)
(2023:
7.063
)
OCLC
–
1121105677
Publisher:
Oscar Publishing Services
Servi
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