ON THE ESTIMATION OF DISTRIBUTION FUNCTION ON INDIRECT SAMPLE
Elizbar Nadaraya1, Petre Babilua2, Grigol Sokhadze3
Iv. Javakhishvili Tbilisi State University, Tbilisi, Georgia
elizbar.nadaraya@tsu.ge, 2petre.babilua@tsu.ge, 3grigol.sokhadze@tsu.ge
1
1. Let X 1 , X 2 ,, X n be a sample of independent observations of a non-negative random
value X with a distribution function F x  . In problems of the theory of censored observations
the sample values are pairs of random values Yi   X i  ti  and Z i  I Yi  X i  , i  1, n ,
where t i are given numbers ( t i  t j for i  j ) or random values independent of X i , i  1, n .
Throughout the paper I   denotes the indicator of the set A.
We will consider here several different cases: the observer has an access only to random
2i  1
, i  1, n , cF  inf x : F x  1   .
2n
The problem consists in estimating distribution functions F x  by the sample
1 , 2 ,, n . Such a problem arises, for example, in corrosion investigations (see [1] where an
values  i  I  X i  ti  , t i  c F
experiment connected with corrosion is described).
We will consider estimates for F x  that are analogous to regression curve estimates of
Nadaraya-Watson type and have the form
1
 n  x  ti  
 x  ti 


Fˆn x   Fn1 x Fn 2 x , Fn1 x    K 

,
F
x

  K 
 i
  ,
n2
 h 
i 1
 i1  h  
where K x  is some weight function (kernel), h  hn is a sequence of positive numbers
n
converging to zero.
Lemma. Assume that
10. K x  is some distribution density with bounded variation and K x  K  x ,
x  R   ,  . If nh   , then
c
1 n m11  x  t j  m2 1
1 F m11  x  u  m2 1
 1 




K
F
t

K 
 F u  du  O 

j



nh j 1
cF h 0
 h 
 nh 
 h 
(2)
uniformly with respect to x  0, cF  , m1 , m2 are natural.
Without loss of generality we assume below that the interval 0, cF   0,1.
Theorem 1. Let F x  be continuous and the conditions of the lemma be fulfilled. Then
the estimate Fˆn x  is asymptotically unbiased and consistent at all points x  0,1. Moreover,
Fˆn x  is distributed asymptotically normally, i.e.
d
nh Fˆ x   EFˆ x   1 x  

N 0,1,

n
n

 x   F x 1  F x  K 2 u du,
2
,
where d denotes convergence in distribution, and N 0,1 a random value having a normal
distribution with mean 0 and variance 1.
1
2. Uniform consistency. We define the conditions under which the estimate Fˆn x 
converges uniformly in probability (almost surely) to a true F x  .
Following E. Parzen [2], we introduce the Fourie transform of K x 

 t    eitx K  x  dx

and assume that
20.  t  is absolutely integrable. Then Fn1  x  can be written in the form
Fn1 x  
1
2

e
iu
x
h
 u 

t
j
iu
1 n
h

e
 j du .
nh j 1
Thus
1
Fn1 x   EFn1 x  
2
Denote

e
iu

x
h
t
j
iu
1 n
 u    j  F t j e h du .
nh j 1


d n  sup Fˆn x   EFˆn x , n  h ,1  h , 0    1 .
xn
Theorem 2. Let K x  satisfy conditions 10 and 20.
1
(a) Let F x  be continuous and n 4 hn    , then
p
Dn  sup Fˆn x   F x  

0;
xn

(b) If

p
 n 4 hn   , p  4 , then Dn  0 almost surely.
n1
Proof. We have
 1

 1 1  x  u   h
sup 1   K 
 du    K u  du   K u  du  0
x n
 h 0  h   
h 1
This and this lemma imply
sup Fn 2 x   1  0
(1)
(2)
x n
i.e., due to uniform convergence, for any  0  0 , 0   0  1, and sufficiently large n  n0 we
have Fn 2 x   1   0 uniformly with respect to x   n .
Denote
t
j
iu
1 n
h
An  sup

e
 j ,  j   j  F t j  .
uR nh j 1
Then
d
p
n
1   0  p

2  p

p
n
A
  u  du

Note that
2
p
, p  4.
(3)
p
n
  tk  t j   2
1
2
 
u  
EAnp 
E
sup




cos

j
k j

h
nh p uR 
j 1
k j

 

p
n
1
k j 2
2

E
sup




cos
u 


j
k j
nh p uR 
 nh 
j 1
k j

p
2
n m
n
n1
1
m 
2
E
sup


 j j m cos u  .



j
p
nh  uR j1
 nh 
m n1 j 1
m0
From this, by the inequality
q
n
n
 a j  n q1  a j , q  1 ,
j 1
q
j 1
we have
EAnp 
p
1
2
p
1
2
p
2
n m
n 1
2
 n 2
2
m 
E


E
sup
cos
u

  j j  m
i 
nh  p 
nh  p uR m
 nh  j 1
i 1
 n 1

p
2
 C n1  C n 2 .
(4)
m0
Let us estimate C n1 and C n 2 :
p
22
Cn1 
n
1
p
1
2
p
n
 E j
h
p
22

p j 1
n
1
p
1
2
h
 1  F t  F t   F t 1  F t  c
n
p
p j 1
1
p
j
j
j
j
2
p
2
n h
.
(5)
p
Further, using Whittle’s inequality [3] for moments of quadratic form, we obtain
p
Cn 2 
22
1
p
2n  3 2 1 n1 E n m  2 ,
 
j j m
nhp
m n1
j 1
p
m0
thus
p
2
n m
E
 
j 1
j
where c p  depends only on p and E  j
 c p n  m 4 ,
p
j m
p
 1.
Thus
n m
n 1

m   n 1
m0
E
 j j m
j 1
p
2
p
n 1
 p 1 
 2c p  m 4  O n 4 
m 1


and

 1
C n 2  O p
 4 p
n h


 .

After combining the relations (3), (4), (5) and (6), we obtain
3
(6)

 1
Ed  O p
 4 p
n h
p
n


, p  4 .

Therefore


P sup Fˆn x   EFˆn x     
 xn

c3
p
4
 n h
p
.
(7)
p
Further we obtain
sup EFˆn x   F x  
x n
1 

 sup EFn1 x   F x   sup 1  Fn 2 x   .
1   0  xn
x n

(8)
The second summand in the right-hand part of (8) tends, by virtue of (2), to zero, while the first
summand is estimated as follows:
 1 
sup EFn1  x   F  x   S n1  S n 2  O  ,
x n
 nh 
(9)
1
 x y
S n1  sup  F  y   F x K 
 dy ,
0 x 1 h 0
 h 
1
Sn2
 1 1  x y 
 sup 1   K 
 dy ,
h
h
x n

 
0

and, by virtue of (1),
Sn2  0 .
(10)
Now let us consider S n1 . Note that
1  x y
1 u
Sn1  sup  F  y   F x  K 
 dy  sup  F x  u   F x  K   du 
h  h 
h h
0 x1 0
0 x1 x1
1
x

 sup
u
 F x  u   F x  h K  h  du .
1
(11)
0 x 1 
Assume that   0 and divide the integration domain in (11) into two domains u  
and u   . Then
S n1  sup
1 u
1 u
 F x  u   F x h K  h  du  sup  F x  u   F x h K  h  du 
0 x1 u 
0 x1 u 
 sup sup F x  u   F x   2
xR u 
K u du .
u
h
(12)
By a choice of   0 the first summand in the right-hand part of (12) can be made
arbitrarily small. After choosing   0 and making n tend to infinity, we obtain that the second
summand tends to zero.
Thus
lim S n1  0 .
(13)
n 
Finally, the proof of the theorem follows from the relations (7)-(10) and (13).
Remarks.
1) If K x   0 , x  1 and   1, i.e.,  n  h,1  h , then S n 2  0 .
2) Under the conditions of Theorem 2,
4
sup Fˆn  x   F  x   0
xa ,b 
in probability (almost surely) for any fixed interval a, b  0,1 since there exists n0 such that
a, b   n , n  n0 .
Let us assume that h  n  ,   0 . The conditions of Theorem 2 are fulfilled:
1
n 4 hn   if  
1
4
and

n

n 1
p
4
hn p   if  
1 1
 , p  4.
4 p
3. Estimation of moments. In the considered problem there naturally arises the question
of estimation of integral functional of F x  , for example, of moments  m , m  1 :
1
 m  m t m1 1  F t dt .
0
As estimates for  m we will consider the statistics
ˆ nm  1 
m n
1
j

n j 1 h
1h
t
h
m1
 t tj
K 
 h
 1
 Fn 2 t dt .

Theorem 3. Let K x  satisfy condition 10 and, in addition to this, K x   0 outside the
interval  1,1 . If nh   as n   , then ̂ nm is an asymptotically unbiased, consistent
estimate for  m and, moreover,
n ˆ nm  Eˆ nm 

1
d


N 0,1,  2  m 2  t 2 m2 F t 1  F t dt .
0
References
1. Manjgaladze, K.V. On one estimate of a distribution function and its moments. (Russian)
Bull. Acad. Sci. Georgian SSR 124 (1980), No. 2, pp. 261-268.
2. Parzen, E. On estimation of a probability density function and mode. Ann. Math. Statist. 33
(1962), pp. 1065-1076.
3. Whittle, P. Bounds for the moments of linear and quadratic forms in independent variables.
Teor. Verojatnost. i Primenen. 5 (1960), pp. 331-335.
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