International Journal of Probability and Statistics
p-ISSN: 2168-4871 e-ISSN: 2168-4863
2020; 9(2): 42-44
doi:10.5923/j.ijps.20200902.03
Received: Aug. 21, 2020; Accepted: Sep. 9, 2020; Published: Sep. 17, 2020

G. G. Hamedani
Department of Mathematical and Statistical Sciences, Marquette University, Milwaukee, WI, USA
Correspondence to: G. G. Hamedani, Department of Mathematical and Statistical Sciences, Marquette University, Milwaukee, WI, USA.
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Copyright © 2020 The Author(s). Published by Scientific & Academic Publishing.
This work is licensed under the Creative Commons Attribution International License (CC BY).
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The problem of characterizing a distribution is an important problem in applied sciences, where an investigator is vitally interested to know if their model follows the right distribution. To this end the investigator relies on conditions under which their model would follow specifically the chosen distribution. In this work, we present certain characterizations of the new two-parameter Poisson-Sujatha distribution introduced by Shanker et al. (2020) with the intention of completing, in some way, their work. These characterizations are based on the conditional expectation of certain function of the random variable and in terms of the inverse hazard function.
Keywords: Poisson, Sujatha
Cite this paper: G. G. Hamedani, Characterizations of the New Two-Parameter Poisson-Sujatha Distribution, International Journal of Probability and Statistics , Vol. 9 No. 2, 2020, pp. 42-44. doi: 10.5923/j.ijps.20200902.03.
, based on: (i) the conditional expectation of certain function of the random variable and (ii) in terms of the reverse hazard function. The main goal here is to complete, in some way, the work of Shanker et al. (2020).The cumulative distribution function (cdf),
the corresponding probability mass function (pmf),
and the reversed hazard function,
of the NTPPS distribution are given, respectively, by ![]() | (1) |
![]() | (2) |
![]() | (3) |
are parameters,
is the normalizing constant and, for simplicity, 
be a random variable. The pmf of
is (2) if and only if ![]() | (4) |
has pmf (2), then the left-hand side of (4), using the formula for the finite geometric series, will be
Conversely, if (4) holds, then ![]() | (5) |
![]() | (6) |
or
or
After some rearranging the terms and simplifying, we arrive at
From the last equality, we have
which, in view of (3), implies that
has pmf (2).Proposition 2. Let
be a random variable. The pmf of
is (2) if and only if its reverse hazard function,
satisfies the following difference equation![]() | (7) |
Proof. Clearly, if
has pmf (2), then (7) holds. Now, if (7) holds, then
or
or, in view of the initial condition
we have
which is the reverse hazard function corresponding to the pmf (2).