Page 22 - Contributed Paper Session (CPS) - Volume 5
P. 22

CPS649 A-Hadi N. Ahmed et al.
               Proof.
               We have
                                                    2
                                           ()   (1 +  )(1 −  ) 
                                   () =    =  1      2       1  .
                                                    2
                                           ()   (1 +  )(1 −  ) 
                                           
                                                                   2
                                                    2
                                                           1

               Clearly, one can see that ( + 1) ≤ () ∀   >  .
                                                              2
                                                         1
               Theorem  2  shows  that  the  NDL  distribution  is  ordered  according  to  the
               strongest stochastic order (v).

               Corollary  2:  Based  on  the  chain  of  stochastic  orders  in  the  definition  (4),
                ≤ ℎ  ,  ≤ ℎ  ,  ≤      ≤   .

               Definition 6: The discrete random variable  is said to be smaller than  in
               weak likelihood ratio ordering (denoted by  ≤   )  if    (+1)  ≤    (0)  ∀  ≥
                                                                                  (0)
                                                                        (+1)
                                                                                 
               0 (see, Khider et al., 2002).

               Theorem 3: Let ~( ) and ~( ). Then,  is said to be smaller than
                                       1
                                                       2
                in weak likelihood ratio ordering, denoted by  ≤   , for all   >  .
                                                                              1
                                                                                   2

               Proof.
               According to Definition 2.5 of Khider et al. (2002), we can prove that

                                             ( + 1)   (0) .
                                                         
                                             
                                             ( + 1)  ≤   (0)
                                             
                                                         
               Then, we obtain

                     2
                    (1 +  )[(1 −  ) +1  − (1 −  ) +1 ]  ≤ 0, ∀  <  . Hence  ≤  .
                                    1
                                                  2
                            2
                    1
                            2
                            (1 +  )(1 −  ) +1           1    2           
                            2
                                           2
                                   1

               The mean residual life (MRL) function of the NDL distribution is defined by
                                                    1 −       (1 − )(2 − )
                          () = ( = | ≥ ) =  () +            ,
                                                       2       (1 +  + )
               where ()is the hrf of the NDL distribution.

               4.  Moments
               The first four raw moments of the NDL distribution are, respectively, given by

                                                             3
                                                                  2
                                      (1 − )(2 + )      +  − 8 + 6
                                                        ′
                          ′
                          = () =   (1 + )  ,     =   (1 + )  ,
                                                        2
                          1
                                                                2

                                                     2
                                               3
                                     (1 − )( + 2 − 24 + 24)
                                 ′
                                 =            (1 + )
                                 3
                                                3
               and
                                                  2
                                          3
                                    4
                         (1 − )(− − 2 + 78 − 192 + 120)
                     ′
                                                                           ̅
                     =                  (1 + )              , ℎ  = (1 − ).
                     4
                                          4

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