About testing the hypothesis of equality of two bernoulli

Mathematical Theory and Modeling www.iiste.org
ISSN 2224-5804 (Paper) ISSN 2225-0522 (Online)
Vol.4, No.9, 2014
About Testing the Hypothesis of Equality of Two Bernoulli
Regression Curves
Petre Babilua (Corresponding author)
Faculty of Exact and Natural Sciences of I. Javakhishvili Tbilisi State University
2 University St., Tbilisi 0186, Georgia
Tel: +995599408383 E-mail: petre.babilua@tsu.ge
Elizbar Nadaraya
Faculty of Exact and Natural Sciences of I. Javakhishvili Tbilisi State University
Tel: +995599570555 E-mail: elizabar.nadaraya@tsu.ge
Grigol Sokhadze
I. Vekua Institute of Applied Mathematics of I. Javakhishvili Tbilisi State University
Tel: +995591313197 E-mail: grigol.sokhadze@tsu.ge
Abstract
The limiting distribution of an integral square deviation between two kernel type estimators of Bernoulli
regression functions is established in the case of two independent samples. The criterion of testing is constructed
for both simple and composite hypotheses of equality of two Bernoulli regression functions. The question of
consistency is studied. The asymptotics of behavior of the power of test is investigated for some close
alternatives.
Keywords: Bernoulli Regression Function, Power of Test, Consistency, Composite Hypothesis

t j n
142
1. Introduction
Let random variables
i  Y , i =1,2, take two values 1 and 0 with probabilities i p (succes) and i 1 p ,
i =1,2 (failure), respectively. Assume that the probability of success i p is the function of an independent
variable x0,1 , i.e.   p p x Y   x i
i i = = P =1| i 1,2 (see [1]-[3]). Let j t , j =1,,n , be the
devision points of the interval 0,1:
j
2 1
= , =1, , .
2 j
n
Let further
1
i Y and
2
i Y , i =1,,n , be mutually independent random Bernoulli variables with
      i k i
k
i P Y =1| t = p t ,
      i k i
k
i P Y = 0 | t = 1 p t , i =1,,n , k =1,2 . Using the samples
1 1
1 , , n Y  Y
and
2 2
1 , , n Y  Y we want to chek the hypothesis
:  =  =  , 0,1 0 1 2 H p x p x p x x ,
against the sequence of “close” alternatives of the form
:  =      , =1,2. 1 H p x p x u x o k n k n k n   

Vol.4, No.9, 2014
where 0 n  relevantly, u x u x 1 2  , x0,1 and   n o uniformly in x0,1.
The problem of comparing two Bernoulli regression functions arises in some applications, for example in
quantal biossays in pharmacology. There x denotes the dose of a drung and p x the probability of response
1 垐2 1 2 2
,
2 2
T nb p x p x p x dx nb p x p x dx
       
 
n n 1 n 2 n n n 1 n 2
n
 
 
n n
b b
       


ˆ ,
p x p x p x
x t
p x K Y i
in j
nb b
n j n
x t

nb b
1
= , = , = 1, , ,
T nb p x p x dx p x p x Ep x i n
   
n n n n in in in

     
   
   
Q  t t 
u v K K dx
n

d d Q d d t p t p t
1
n
 x  t   y  t   x  u   y  u
  
          
         
1 1 1
     
 
K K p t K K p u du O
nb b b b b b nb
n i n n n n n n
143
to the dose x .
We consider the crietrion of testing the hypothesis 0 H based on the statistic
     
 
   
 
   
= , 1 , 0,
n n n
where
     
in in n
n
   
 
1
1
n
p x K
1
1
j i
= , = 1,2,
1
i
= ,
n
n i 
n

  
 
 
  
 
 


Kx is some distribution density and 0 n b is a sequence of positive numbers, p x in ˆ is the kernel
estimator of the regression function (see [4], [5]).
2. Assumptions and Notation
We assume that a kernel Kx 0 is chosen so that it is a function of bounded variation and satisfies the
conditions: Kx= K x , Kx= 0 for x  > 0 ,  Kxdx 1 . The class of such functions is denoted by
H .
We also introduce the notation:
     
 
     
   
 
n
ij n i j n
 
x u x v
b b
n n
     
1 2
1 2
1 2
n k
2 2
2
2 1 1
2
= , , , = ,
1
= , = = 1 .
n k i ik i i k i k i
nb
n k i k




  



  
3. Auxiliary Assertions
Lemma 1 ([6]). Let K  xH   and p x , x0,1 , be a function of bounded variation. If n nb   ,
then
1 i 2 i
3   =
1 2 3  
i

1 0
uniformly in x, y0,1, where N 0 i  , i =1,2,3.
Lemma 2. Let KxH  ,   0,1 1 p x C and u x 1 , u x 2 be continuous functions on 0,1 . If

Vol.4, No.9, 2014
 2
n nb and 0 1/2  
n n  b , then for the hypothesis n H1
       
b p p x p x dx K x dx
2 1 n n
     (1)
    
      
ET p p x p x dx K u du
  
= , = 1 ,

K K K is the convolution operator
n n
 
    
 
1 1 1
    2 2 2 2
 d d Q  d Q A n A n
(3)
n i k ik i ii
n
 
    
1 1
2
  . (5)
      
   
 
      
      
    
1
= 2 1 ,
2
n n n n n
x n n
b
144
1
1 2 2 2 2 2
0
x
0 2

  

and
       1/2 1/2 1/2
1
3/2
= n n n n n
n
b p O b O b O
nb
 
, (2)
where
1
 1        2
 
x
0 | |
n n
0
=  , 
.
 
Proof. We have
 
2 1 2
, 1 1
=
2 2
nb
n i k i

 
where
          1 1 2 2 1 1 , 1, , , k k k k k k d  d t  p t  p t  p t  p t k  n
= 2  1    , 1,2 k k k n d p t  p t O  k  , (4)
uniformly in [0,1] k t .
It can be easily established that
 
 
2
1 2 3 2 2
2 1 2
1

i n
n n i
i n n n
n
x t
b A n n b d K dx c c
b nb nb

  


From the definition of ik Q and (4) follows
          
 

     
2
2
1
1
= .
n
n
i i
n i i n
i n n
n n n
x t y t
A n nb p t p t O K K dx dy
b b 
  

 
   
 
Further, using Lemma 1 and also taking into account that
1

 
x
  0,1 1 p x C and ,   , 
  

b
n n b
x
for all
  n x , it is easy to show that
     
 
      
x
b
n
2 2 1 2 2 1/2
1 0 2
1
1
= 2 1 .
n
n
x y
b A n p x p x K dx dy O b O b O O
b nb



 

 
 

    
        
   
 
Thus

Vol.4, No.9, 2014
       
b A n p x p x dx K x dx
2 1 n
    . (6)
x t
ET nb K d t dx
b

K u p x b u p x b u du dx O O
p x p x dx K x dx O b O O
    
      
L nb p x p x Ep x Ep x dx
      

n n n n n n

L nb Ep x Ep x dx
   
n n n n
145
1
1 2 2 2
1 0
x
0 2



From (5) and (6) follows statement (1).
Further, using the above-mentioned method, we can write
 1    1 2
 
 
1
n
     
 
 
2
n
1
         
1
2
0
1
= = =
2
1
= 1 =
nb
1
1 .
n
n
n
i
n n n i
i n
x
b
n n n
x n
b
n n
nb
x n





 
 

  
  
 
 
   
        
   
 
 
      
 
 
 
 
Thus
       1/2 1/2 1/2
1
3/2
= n n n n n
n
b p O b O b O
nb
 
.
The lemma is proved.
Asymptotical Normality of the Statistic n T
We have the following assertion.
Theorem 1. Let KxH  and
 ,  ,   0,1 1
1 2 p x u x u x C .
If  2
n nb , 0 1/2  
n n  b and 0
1/2 2 nb c n n   , 0 < <  0 c , then for the hypothesis n H1
       1/2 1 ,1 d
n n b T p  p N a      ,
where p and p 2  are defined in Lemma 2 and d  denotes convergence in distribution and
Na,1 is a random variable having the standard normal distribution with parameters a,1 ,
 
    
1
0 2
1 2
0
=
2
c
a u x u x dx
 p
  .
Proof. We have
1 1 2 = n n n n T T  L  L ,
where
         
 
     
 
1
1 2 1 2
2 2
1 2
,
1
.
2
n
n




By the Lemma 1, it is clear

Vol.4, No.9, 2014
  1 1 1
   
n n 2 n n
  . (7)
n n n n

  
1 1
2 n n n n n n
b L nb K t u x b t u x b t dt O dx
       
  . (8)
  
1
c
1/2 2 0 2
b L u t u t dt     . (9)
b L nb p x Ep x Ep x dx
E I E I nb E p x Ep x Ep x dx

n n n n n n
n


nb p x p x Ep x Ep x Ep x Ep x dx dx
n n n n n n n
t
 x  t   x  t


p x p x K K p t p t
     
nb  b b
   
  
p x p x n b
n n n
       
 K   K   p u  p u du  O
 
 b   b    nb
 
146
 
2
1
1/2 2 1/2 2
1 2
0
x t
b L nb K u t u t dt O dx
b b nb 
 

     
         
    
1

 
x
Since ,   , 
  

b
n n b
x
for all   n x , then from (7) we find
       
 
2
1/2 2 1/2 2
1 2
nb
n  
n
 
 
Further, since  ,   0,1 1
1 2 u x u x C , then from (8) we have
      
1 2
0 2 n n
Now, we show that 1/2 1 0 n n b L   . We have
1
=
2 n n n n n n
       
 
1/2 1 1/2
1 1 2
n



1
2 n n n n
        
nb p x Ep x Ep x dx
   
 
1/2
2 1 2
 
n
1 2
n n  I  I . (10)
It is clear that
1
           
 
2
            
 
     
1/2
2
1/2
2
1 1 1/2
1 1 2
1/2
1/2
=
1 1 1 2 1 1 2 1 1 2 2 2 1 2
1
cov , ,
2
= .
n
n n n

 

                        
 
    
 
   

Easily verify, that
    
 
    1 2
n
i i
1 1 1 2 2 1 1
1
1
cov , 1
n n i i
n i n n
and by Lemma 2 we can now write
    
   
 
1 2
1 1 1 2
1
x u x u
1 2
1 1 2
0
cov , =
1
1 .
n n n

Thus

Vol.4, No.9, 2014

        
1 1 1
     
       1
  
   
      

        
nb
 
      
 .
      
   
147
1
1 1/2 1 2
2 1 1 2
0
1 2
1/2 2
1/2
1 1 2 1 1 2 2 2 1 2 3 3
2
= 0,
n
n n
n n n n
n n
n n n n n n
n
x u x u
E I nb K K p u p u du
nb b b nb
nb
Ep x Ep x Ep x Ep x dx dx c n b c
n






     

 
1/2 2 nb c n n   , 0 < <  0 c and
since, by condition, 0
1/2 2
1/4 = n n
n
n
n
b

   .
So, 1 0 n I  . Analogously we can show that
2 0 n I  .
Hence
1 0 n L  . (11)
Further, to prove the theorem it remains to show
 
 
1
0,1 n n d
n
T
N

 . (12)
Since the proof of (12) is similar to that of Theorem 1 from [7], we omit it.
Using the representation
1 1 2 = n n n n T T  L  L , Lemma 2, (9), (11) and (12), we find that
 
 
    
1
1/2 0 2
1 2
0
,1
2 ( )
n d
n
T p c
b N u x u x dx
 p  p
The theorem is proved.
The conditions of Theorem 1 for bn and n are fulfilled if we assume  b b n n 0 = and 1 2 4
n 0n       for
0  1 2 .
Corollary. Let KuH  and   0,1 1 p x C . If  2
n nb , then for the hypothesis 0 H
       1/2 1 0,1 d
n n b T p  p N      . (13)
4. Application of the Statistic n T for the Hypothesis Testing
As an important application of the result of the corollary let us construct the criterion of testing the simple
hypothesis 0 H :       1 2 p x = p x = p x (this is the case with given p x ); the critical domain is defined by
the inequality
      1/2 = n n n T d p b p        ,

Vol.4, No.9, 2014
And from Theorem 1 we establish that the local behavior of the power    H1n n n
T d
1 H n n n
 
    
p   

A u   u x u x dx u  u u ,
x dx K x dx

 
 
x dx K x dx
 
 

b E p
   
n n
 
c b E p x Ep x dx E p x Ep x dx
      
 
n n n n n
 
 
n n
b Ep x p x dx b Ep x p x dx
   
 
 
1 1
b E p O O b O
n b nb
148
T  d  is as follows
  
 
A u
 
 
1  
,
where
1
2
c
  0
      
1 2 1 2
0
, ,
2
  =1     ,   is a standard normal distribution.
Note note that in (13) the statistic function n T is normalized by the values  p and   2  p which depend
on p x . If p x is not defined by a hypothesis, then the parameters  p and   2  p should be replaced
respectively by
 
n n
 
 
 
 
 
2
n
2 2 2
0
2
,
2 ,
n
x
n n
x
 
 

 
 
              1 1 2 2 = n n n n n n n  x p x p x  p x  p x p x  p x
and we show that
     1/2 2 2 0, n n n b p   p       . (14)
Let us prove (14). Since  
 
1
1 n
   
n
p x O
nb
 
uniformly in   n x  and   in 4 p x  c , x0,1 , i 1,2 ,
we obtain
 
      
 
      
 
   
 
   
 
1/2
1/2 1/2
1/2 2 2
5 1 1 2 2
1/2 1/2
1 2 .
n n n n
n n

 
 
 
 x  x

   
 b b

Further, using Lemma 1 and and also taking into account that p  x  C 1   1
0,1 and ,  ,
 n n
 
for all
  n x  , it is easy to see that
    1/2 1/2
3/2
n n n
n

   
               
.

Vol.4, No.9, 2014
Hence    1/2 0 n n b p      . Analogously, it can be shown that   2 2
 , j 1 , n, ,
H x   h u du , hu is some known continuous distribution density o 0,1 . In this case, by a
149
n   p .
Theorem 2. Let K  xH   and
      1
1 2 p x = p x C 0,1 .
If  2
n nb , then for n
    1/2 1 0,1 . d
n n n n b T  N     
Proof. Follows from (13) and (14).
Theorem 2 enables us to construct an asymptotical criterion of testing the composite hypothesis
    0 1 2 H : p x = p x , x0,1 . The critical domain for testing this hypothesis is defined by the inequality
    1/2 , 1 n n n n n T d b                . (15)
Theorem 3. Let K  xH   ,       1
1 2 p x , p x C 0,1 . If 2
n nb   , then for n  
   1
1 H n n T  d  
Here the alternative hypothesis 1 H is any pair      1 2 p x , p x ,       1
1 2 p x , p x C 0,1 , 0   1
 p x  ,
i
i 1,2 , such that     1 2 p x  p x on the set of positive measure.
Proof. Is similar to the proof of Theorem 3 from [7].
Remark. Let i t be the division points of the interval 0,1 , which are chosen so that
j
  2 1
2 j
H t

n
x
where    
0
similar reasoning to the above one we can be generalize the results obtained in this paper.
Acknowledgement. The work is supported by Shota Rustaveli National Scientific Foundation, Project
No. PG/18/5-104/13

Vol.4, No.9, 2014
150
References
1. Efromovich, S. (1999), Nonparametric curve estimation. Methods, theory, and applications. Springer Series
in Statistics. New York: Springer-Verlag.
2. Copas, J. B. (1983), Plotting p gainst x . Appl. Statist. 32:1, 25-31
3. Okumura, H. & Naito, K. (2004), Weighted kernel estimators in nonparametric binomial regression. The
International Conference on Recent Trends and Directions in Nonparametric Statistics. J. Nonparametr. Stat.,
16:1-2, 39-62. DOI:10.1080/10485250310001624828,
http://www.tandfonline.com/doi/abs/10.1080/10485250310001624828
4. Nadaraya, E. A. (1964), On estimating regression. (Russian) Teor. Veroyatn. Primen., 9, 157-159
5. Watson, G. S. (1964), Smooth regression analysis. Sankhyā Ser. A, 26, 359-372
6. Nadaraya, E., Babilua, P., & Sokhadze, G. (2010), Estimation of a distribution function by an indirect
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7. Nadaraya, E. Babilua, P., & Sokhadze, G. (2013), On the integral square measure of deviation of a
nonparametric estimator of the Bernoulli regression. Teor. Veroyatn. Primen., 57:2, 322-336; translation in
Theory Probab. Appl., 57:2, 265-278.

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