Parameter Estimation and Hypothesis Testing of Two ... - Science Direct

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Physics Procedia 33 (2012) 1475 – 1480

2012 International Conference on Medical Physics and Biomedical Engineering

Parameter Estimation and Hypothesis Testing of Two Negative Binomial Distribution Population with Missing Data Zhiwen Zhao College of Mathematics,Jilin Normal University,Siping, China [email protected]

Abstract Our purpose is to deal with the parameter estimation and hypothesis testing on the equality of two negative binomial distribution populations with missing data. The consistency and asymptotic normality of the estimations are proved. In addition statistic on testing equality of two negative distributions and its limiting distribution are obtained.

©2012 2011Published Published by Elsevier Ltd. Selection and/or peer-review under responsibility of [name Committee. organizer] © by Elsevier B.V. Selection and/or peer review under responsibility of ICMPBE International Open access under CC BY-NC-ND license. Keywords:missing data, maximum likelihood estimation, hypothesis testing

1 Introduction The problem of missing data is very common in many studies and field experiments (see Lepkowski, Landis & Stehouwer[1], Kim & Curry[2]). Missing data can bias parameter estimate. A variety of methods have been developed to estimate the unknown parameters of different models when there exists missing data. Reference [3] deals with the parameter estimation and hypothesis testing on the equality of censoring sample with missing data. Reference [4] two exponential distribution under Type considered the estimation for two binomial distributions with partially missing data. Exact Likelihood Inference for Two Exponential Populations under Joint Type -II Censoring was studied by Balakrishnan and Rasouli (see [5]). Moreover, Reference [6] investigated the estimation and test for two normal populations with partially missing data. In this paper, Our purpose is to deal with the parameter estimation and hypothesis testing on the equality of two negative binomial distribution populations with missing data. The consistency and asymptotic normality of the estimations are proved. In addition statistic on testing equality of two negative distributions and its limiting distribution are given. The rest of this paper is organized as follows. In Section II, we introduce obtain the parameter estimator. In Section III, we get the limiting distribution of estimator. The hypothesis testing and

1875-3892 © 2012 Published by Elsevier B.V. Selection and/or peer review under responsibility of ICMPBE International Committee. Open access under CC BY-NC-ND license. doi:10.1016/j.phpro.2012.05.241

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Zhiwen Zhao / Physics Procedia 33 (2012) 1475 – 1480

confidence interval about the difference of parameters in two populations are discussed in Section IV.

2 Maximum likelihood estimation In this section, we consider the maximum likelihood estimation for the parameter in negative binomial distribution populations with missing data. Consider the following two negative binomial distribution populations whose probability functions are

f ( x; i )

1 r 1 i

C xr

(1

) x - r , for x

i

0,

r

r 1

otherwise,

where i ,i 1,2, are unknown parameters. For this distribution, we focus on the case where some X values in a sample of size n may be missing. That is, we obtain the following incomplete observations ( X i , i ), i 1,2, , n and (Yi , i ), i 1,2, , n from two above populations. When i 1 , X i is missing. When i 1 , Yi is missing. Furthermore, we assume that P ( i 1) P ( i 1) p . In what follows, we consider the estimator of likelihood function can be written n

L1 ( 1 )

i 1

1

. Based on the observations ( X i , i ), i 1,2,

[C Xr i 1 1

r i

i

(1

i

) X i -r ] .

Hence, the logarithm of the likelihood function is given by ln L1 ( 1 )

n i 1

i

[ln C Xr i 1 1 r ln

( X i - r)ln(1

1

Note that (ln L1 ( 1 ))

n

r

[

i

i 1

1

i 1

Solving the log-likelihood equation, we obtain r

ˆ

1

n i 1

n i 1

i

Xi

. i

In a similar way we have ˆ

2

r

n i 1

n i 1

Yi

i

i

.

Xi - r ]. 1 1

1

)] .

, n , the

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3 The limiting property of estimators In this section, we consider the limiting property of estimators. a. s. " denotes convergence almost surely. Theorem 1: ˆ1 a.s. 1 , where " Proof. Note that X i i ,1 i n are independently identical distributed variable, we have

1 n Xi ni 1

a .s .

E( X1 1 ) .

i

(1)

After simple calculation, we get E( X1 1 )

E( X1 )E( 1 )

p

r

,

which, combing with (1) , gives 1 n Xi ni 1

a .s .

r

p

i

.

Further, by 1 n r n i1

a. s .

pr ,

i

we have r

ˆ

1

n i 1

n i 1

Xi

i

a .s . 1

.

i

a .s . Theorem 2: ˆ2 a.s. " denotes convergence almost surely. 2 , where " By using the same method as we used to prove Theorem 1, we can prove Theorem 2, so we omit the proof here. Lemma 1: Let Tn (T1n , T2 n , , Tkn )T and ( 1 , 2 , , k )T . g (t1 , , t k ) has continuous partial derivation . If

n (Tn

)

N (0, ),

L

then n [ g (T1n , T2 n ,

where

(

ij

2

)k k ,

(

, Tkn ) g ( 1 ,

g

g

i

Theorem 3:

n ( ˆ1

1

)

2 1

Proof.

Let Wi

d

p

N (0, 2 1

p

2 2 1

)

ij

and

2

,

,

k

)]

N (0,

d

2

),

denote convergence in distribution.

d

j

), where

r

p2

2 1

( i , i X i ) T . We have {Wi , i

2 p 2 12 r p p 2r 2

3 1

r

p

r

p2

2 2 1

r

.

1} is independently identical distributed variable and

E (W1 )

( p, p

r

)T .

1

Let E (W1

2 2 1

EW1 )(W1

EW1 ) T .

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By multivariate central limit theorem, we have 1 n n( Wi ni 1 Write

N (0, ) .

EW1 )

d

a11 a21

a12 , a22

where a11

a12

E

E(

a21

2 1

2

(E 1 )2

1

p (1 p ) ,

X 1 ) E 1E( 1 X 1 )

r

p(1 p )

1

and 2

E(

a 22

1

t1 , T1n t2

Let g (t1 , t 2 ) and

2

p

r

2

X 1 ) (E 1 X 1 )2

p

r r

1 2 1

r2

1 n i Xi, ni 1

r n i , T2 n ni 1

r2

p2

.

2 1

1

rp

. We have

1

1 n r i n i1 1 n i Xi ni 1

g (T1n ,T2 n )

1

and g( 1,

2

)

1

.

By simple calculation, we have g

1

rp

1

and 2 1

g

rp

2

.

By Lemma 1, we have n(

1

1

)

n [ g (T1n , T2 n ) g ( 1 ,

2

d

)]

N (0,

2

),

where g

2

p n ( ˆ2

Theorem 4: p

p

2 2

r

p

2

)

2 2

1

1

r

p2

2 1

p

d

N (0,

2p

2

2 2 2 2

2 2

a11 2

g 1

2 1

g

g

a12

2

2 p 2 12 r p p2r 2

g

2 3 1

r

p

a 22

2

2 2 1

r

p2

2 2 1

r

.

), where

r p r p 22 r 2 p 2 22 r 2 By using the same method as we used . p r to prove Theorem 3, we can prove Theorem 4, so we omit the proof here. 2 1

2 2

2

2 1

g

3 2

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4 Testing the equality of two populations and confidence interval for In this section, we consider the following hypotheses: H 0 : 1 2 0 H1 : First we establish test statistics and discuss the limiting distribution of test statistics. Let n

pˆ pˆ ˆ12

ˆ 12

pˆ ˆ12 r

1

2

2

0.

n i

i 1

i

i 1

2n 2 ˆ2 2 pˆ r

pˆ 2 ˆ12

1

,

pˆ ˆ13 r

1 2 2

pˆ ˆ12 r 2

pˆ 2 ˆ12 r 2

pˆ r

and pˆ ˆ22

2 pˆ 2 ˆ22 r pˆ ˆ23 r pˆ 2 r 2 By the strong large number law, we have ˆ 22

pˆ ˆ22 r

pˆ 2 ˆ22

pˆ ˆ22 r 2

pˆ 2 ˆ22 r 2

pˆ

a.s.

p,

ˆ12

a.s .

2 1

ˆ 22

a.s.

2 2

.

and . Therefore we can obtain the following result: n [ 1 2 ( 1 2 )] d N (0,1). Particularly under the-null hypothesis H 0 , we have Theorem 5: ˆ12 ˆ 2 2

n[ ˆ1

1 2

2

ˆ2

]

N (0,1) .

d

2

Proof. By Slutsky theorem, Theorem 2 and Theorem 3, we can prove Theorem 4. Let in what follows we discuss the asymptotic confidence interval of 1 2 For 0