EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 14, No. 4, 2021, 1517-1529
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

The Bivariate Extended Poisson Distribution of type 1

Rufin Bidounga1,3,∗, Michel Koukouatikissa Diafouka1,3, Réolie Foxie Mizélé
Kitoti2,3, Dominique Mizère2,3

1 Exact Science Department, Normal School Higher, Marien Ngouabi University,
Brazzaville, Congo
2 Department of Mathematics, Faculty of Science and Technology,
Marien Ngouabi University, Brazzaville, Congo
3 Laboratory of statistics and analysis of data (Labsad), Faculty of Science and Technology,
Marien Ngouabi University, Brazzaville, Congo

Abstract. In this paper, we will construct the bivariate extended Poisson distribution which
generalizes the univariate extended Poisson distribution. This law will be obtained by the method
of the product of its marginal laws by a factor. This method was demonstrated in [7]. Thus we call
the bivariate extended Poisson distribution of type 1 the bivariate extended Poisson distribution
obtained by the method of the product of its marginal distributions by a factor. We will show that
this distribution belongs to the family of bivariate Poisson distributions and and will highlight the
conditions relating to the independence of the marginal variables. A simulation study was realised.

2020 Mathematics Subject Classifications: 62E15, 62H10, 60E05

Key Words and Phrases: Extended Poisson distribution, bivariate Poisson distribution, esti-
mation and statistical testing

1. Introduction

Several authors have studied bivariate Poisson laws, in particular, Berkhout and Plug[2]
and Lakshminarayna et al.[7]. Then, [3] has highlighted the weighted bivariate Poisson law
having as a basic law, the bivariate Poisson law according to Berkhout and Plug[2]; a law
that allows to generate all the bivariate Poisson laws. The bivariate Poisson distribution
according to Berkhout and Plug[2] is rightly considered the standard distribution in N2

as is the Poisson distribution in N. In this paper, we will construct the bivariate extended
Poisson distribution which generalizes the univariate extended Poisson distribution. This
law is obtained by the method of the product of its marginal laws by a factor. This method
was demonstrated in [7], thus we call the bivariate extended Poisson distribution of type

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v14i4.4151

Email addresses: rufbid@yahoo.fr (R. Bidounga), michel.koukouatikissa@umng.cg (M. Koukouatikissa
Diafouka), foxie reolie2000@yahoo.fr (R. F. Mizélé Kitoti), domizere@gmail.com (D. Mizère)

http://www.ejpam.com 1517 © 2021 EJPAM All rights reserved.



R. Bidounga et al. / Eur. J. Pure Appl. Math, 14 (4) (2021), 1517-1529 1518

1 the bivariate extended Poisson distribution obtained by the method of the product of
its marginal distributions by a factor. We have shown that this distribution belongs to
the family of bivariate Poisson distributions and have highlighted the conditions relating
to the independence of the marginal variables. A simulation study was realised.

2. A review of laws

2.1. The Univariate Extended Poisson distribution

Definition 1. The laws following probability mass function

P (Y = y) =


1− e−θ

β
, y = 0,

(
θy

y!
e−θ

)
β−1

(
β

θ
y − 1

)
, y = 1, 2, . . .

∀ θ > 0, and β ≥ θ, (1)

is renamed as the univariate extended Poisson distribution (see [5]).

It can be written in the form [8]

P (Y = y) =
θy

y!
e−θ

(
β

θ
y − 1

)
β−1


1− e−θ(

β

θ
y − 1

)
e−θ


δ0(y)

, y ∈ N, ∀ θ > 0 and β ≥ θ,

(2)
where δ0 is the Dirac function in 0. θ is the canonic parameter. This distribution has the
following characteristics

Eθ(Y ) = 1 +
β − 1

β
θ, (3)

var(Y ) =
β − 1

β2
θ2 +

β + 1

β
θ. (4)

Under- or over-dispersed distribution

The following facts are immediate.

Proposition 1. The Fisher dispersion index of the variable Y which follows the extended
Poisson distribution of parameters (θ, β) noted I(Y ) is such that

• I(Y ) > 1 if
β

1 +
√
β

< θ ≤ β, i.e. the extended Poisson distribution is overdispersed

;



R. Bidounga et al. / Eur. J. Pure Appl. Math, 14 (4) (2021), 1517-1529 1519

• I(Y ) < 1 if 0 < θ <
β

1 +
√
β
, i.e. the extended Poisson distribution is underdispersed

;

• I(Y ) = 1 if θ =
β

1 +
√
β
, i.e. the extended Poisson distribution is equiderdispersed.

Proof. Indeed,

var(Y )− E(Y ) =
β − 1

β2
θ2 +

β + 1

β
θ − 1− β − 1

β
θ,

=
θ2β − (β − θ)2

β2
,

=
(θ
√
β + β − θ)[θ(1 +

√
β)− β]

β2
.

Since β ≥ θ > 0 then the sign of var(Y )−E(Y ) depends only on θ(1+
√
β)−β. Then

I(Y ) is greater, smaller or equal to 1 depending on whether var(Y ) − E(Y ) is positive,
negative or null respectively. We are assured of the answer

Let us recall the result of [5].

Proposition 2. The moments generating function of the extended Poisson distribution is
equal to

MY (t) =
1−

(
1− βet

)
eθ
(
et − 1

)
β

, t ∈ [−1, 1]. (5)

Now, we have the following result.

Proposition 3.

Eθ

[
e−Y

]
=

1−
(
1− βe−1

)
eθ
(
e−1 − 1

)
β

, (6)

Eθ

[(
e−Y

)2]
=

1−
(
1− βe−2

)
eθ
(
e−2 − 1

)
β

, (7)

Eθ

[
Y e−Y

]
=

e−1eθ
(
e−1 − 1

) [
β − θ

(
1− βe−1

)]
β

. (8)

Proof. Indeed, expression (6) is obvious because it is equal to MY (−1). Ditto for
expression (7) which is equal to MY (−2). And for expression (8), we have Eθ(Y etY ) =

d

dt
MY (t). Since

d

dt
MY (t) =

eteθ
(
et − 1

) [
β − θ

(
1− βet

)]
β

, by setting t=-1, we are as-

sured of the answer.



R. Bidounga et al. / Eur. J. Pure Appl. Math, 14 (4) (2021), 1517-1529 1520

2.2. The Bivariate Poisson distribution according to Berkhout and Plug[2]

Definition 2. Let Yj (j = 1, 2) be a random variable that follows the Poisson distribution
of parameter θj (j = 1, 2). The vector (Y1, Y2) follows the bivariate Poisson distribution
according to Berkhout and Plug [2] if its mass function fBP is equal to

fBP (y1, y2; θ1, θ2) =

(
θ
y1
1

y1!
e−θ1

)(
θ
y2
2

y2!
e−θ2

)
, (y1, y2) ∈ N2, (θ1, θ2) ∈ R∗2

+ , (9)

under conditions

ln θ1 = x′ρ1, (10)

ln θ2 = x′ρ2 + ηy1, (11)

where ρ1, ρ2 and η are parameters, x = (x1, . . . , xp) is a vector of deterministic variables
or factors.

The generalized model (10) has the response variable Y1 and the model (11) the variable

Y2. The expression (10) induces that P (Y1 = y1; θ1) =
θ
y1
1
y1!

e−θ1 is a marginal law while

the model (11) induces that

P (Y2 = y2; θ2) = P (Y2 = y2/Y1 = y1),

=
θ
y2
2

y2!
e−θ2 ,

=
ey2(x

′ρ2 + ηy1)

y2!
e−(x

′ρ2 + ηy1),

is a conditional law.

When η = 0 then the conditional probability P (Y2 = y2/Y1 = y1) is not depend of
observation y1 and the variables Y1 and Y2 are independent.

The bivariate Poisson distribution according to Berkhout and Plug[2] has the charac-
teristics (see [1]).

Eθ1(Y1) = var(Y1) = θ1, (12)

Eθ2(Y2) = ex
′ρ2+a2+θ1(eη−1), (13)



R. Bidounga et al. / Eur. J. Pure Appl. Math, 14 (4) (2021), 1517-1529 1521

var(Y2) = Eθ2(Y2) + [Eθ2(Y2)]
2
(
eθ1(e

η−1) − 1
)
, (14)

cov(Y1, Y2) = θ1Eθ2(Y2) (e
η − 1) . (15)

Expression (14) shows that the variable Y2 is overdispersed. And the covariance is neg-
ative, null or positive depending on whether the parameter η is negative, null or positive.

2.3. The Bivariate Poisson distribution according to Lakshminarayna et
al.[7]

In [7], the authors defined the bivariate Poisson law as the product of its marginal laws
by a multiplier factor.

Definition 3. Let Y1 and Y2 be two Poisson random variables with respective parameters
θ1 and θ2. The bivariate distribution of the couple (Y1, Y2), denoted fLPS, has the mass
function

fLPS(y1, y2; θ1, θ2, α) =

(
θy11
y1!

e−θ1

)(
θ
y2
2

y2!
e−θ2

)[
1 + α

(
e−y1 − e−dθ1

)(
e−y2 − edθ2

)]
,

(16)
with e−dθi = Eθi

(
e−Yi

)
, yi ∈ N, θi ∈ R∗

+ (i = 1, 2), α ∈ R∗
+ and d = 1− e−1.

This distribution has the following characteristics

Eθ1(Y1) = var(Y1) = θ1, (17)

cov(Y1, Y2) = θ1θ2d
2e−d(θ1 + θ2). (18)

In [4], the authors showed that the bivariate Poisson distribution according to Laksh-
minarayna et al.[7] is a distribution of the bivariate Poisson family and that it converges
to the bivariate Poisson distribution according to Berkhout and Plug[2].

3. The Bivariate Extended Poisson distribution of type 1

Based on the work [7], we define the bivariate extended Poisson of type 1 distribution
as follows.

Definition 4. Let us consider Y1 and Y2 two univariate extended Poisson variables with
respective parameters (θ1, β1) and (θ2, β2). The bivariate Poisson distribution of the pair
(Y1, Y2), denoted fBEP,1, has the mass function (see [7])

fBEP,1(y1, y2; θ1, θ2, β1, β2, α) =

 2∏
j=1

(θ
yj
j

yj !
e−θj

)(
βj
θj

yj − 1

)
β−1
j

 1− e−θj(
βj

θj
yj − 1

)
e−θj


δ0(yj)


×



R. Bidounga et al. / Eur. J. Pure Appl. Math, 14 (4) (2021), 1517-1529 1522

g(y1, y2; θ1, θ2, α), (19)

where g(y1, y2; θ1, θ2, α) = [1 + α (e−y1 − c1) (e
−y2 − c2)], with cj = Eθj

(
e−Yj

)
, yj ∈ N,

θj ∈ R∗
+, βj ≥ θj (j = 1, 2) and α ∈ R.

The initials ”BEP,1” stand for Bivariate Extended Poisson of type 1. We have the
following result.

Proposition 4.

(i) The marginal laws of Y1 and Y2 are extended Poisson laws of respective parameters
(θ1, β1) and (θ2, β2).

(ii)

cov(Y1, Y2) = αcov
(
Y1, e

−Y1
)
cov

(
Y2, e

−Y2
)
. (20)

Let be P (Yj = yj) =

(
βj
θj

yj − 1

)
β−1

 1− e−θj(
βj

θ yj − 1
)
e−θj

δ0(yj)

∀ θj > 0, βj ≥ θj ,

j = 1, 2, the marginal law of variable Yj j = 1, 2. It follows the result.

Corollary 1.

fBEP,1(y1, y2; θ1, θ2, β1, β2, α) = P (Y1 = y1)P (Y2 = y2)
[
1 + α

(
e−y1 − c1

) (
e−y2 − c2

)]
,

(21)

cj = Eθj

(
e−Yj

)
, yj ∈ N, θj ∈ R∗

+, βj ≥ θj (j = 1, 2) and α ∈ R.

This result confirms that the definition 4 is rigorously correct.

Corollary 2. When α = 0, the variables Y1 and Y2 are independent.

Proof. [Proof of the proposition 5]

(i)

P (Y1 = y1) =
∑
y2

fBEP,1(y1, y2; θ1, θ2, β1, β2, α),

=

(
θ
y1
1

y1!
e−θ1

)(
β1
θ
y1 − 1

)
β−1
1


1− e−θ1(

β1
θ1

y1 − 1

)
e−θ1


δ0(y1)

×

∑
y2

(
θ
y2
2

y2!
e−θ2

)(
β2
θ
y2 − 1

)
β−1
2


1− e−θ2(

β2
θ2

y2 − 1

)
e−θ2


δ0(y2)

+



R. Bidounga et al. / Eur. J. Pure Appl. Math, 14 (4) (2021), 1517-1529 1523

α
(
e−y1 − c1

)∑
y2

(
θ
y2
2

y2!
e−θ2

)(
β2
θ
y2 − 1

)
β−1
2


1− e−θ2(

β2
θ2

y2 − 1

)
e−θ2


δ0(y2)

×

(
e−y2 − c2

)
,

=

(
θ
y1
1

y1!
e−θ1

)(
β1
θ
y1 − 1

)
β−1
1


1− e−θ1(

β1
θ1

y1 − 1

)
e−θ1


δ0(y1)

+

(
e−y1 − c1

)
Eθ2

(
e−Y2 − c2

)
.

Since Eθ2

(
e−Y2 − c2

)
= 0, we are sure of the answer.

By symmetry, we have

P (Y2 = y2) =

(
θ
y2
2

y2!
e−θ2

)(
β2
θ
y2 − 1

)
β−1
2


1− e−θ2(

β2
θ2

y2 − 1

)
e−θ2


δ0(y2)

.

(ii) cov(Y1, Y2) = Eθ1,θ2(Y1Y2)− Eθ1(Y1)Eθ2(Y2).

We have

Eθ1,θ2(Y1Y2) =
∑
y1

∑
y2

y1y2fBEP,1(y1, y2; θ1, θ2, α),

=
∑
y1

∑
y2

y1y2

2∏
j=1


(
θ
yj
j

yj !
e−θj

)(
βj
θj

yj − 1

)
β−1
j


1− e−θj(

βj
θj

y1 − 1

)
e−θj


δ0(yj)

+

∑
y1

∑
y2

y1y2

2∏
j=1


(
θ
yj
j

yj !
e−θj

)(
βj
θj

yj − 1

)
β−1
j


1− e−θj(

βj
θj

yj − 1

)
e−θj


δ0(yj)

×

(
e−y1 − c1

)(
e−y2 − c2

)
,

=Eθ1(Y1)Eθ2(Y2) + αEθ1

[
Y1

(
e−Y1 − c1

)]
Eθ2

[
Y2

(
e−Y2 − c2

)]
,

=Eθ1(Y1)Eθ2(Y2) + α
[
Eθ1

(
Y1e

−Y1
)
− Eθ1(Y1)Eθ1

(
e−Y1

)]
×[

Eθ2

(
Y2e

−Y2
)
− Eθ2(Y2)Eθ2

(
e−Y2

)]
.



R. Bidounga et al. / Eur. J. Pure Appl. Math, 14 (4) (2021), 1517-1529 1524

And

cov(Y1, Y2) = α
[
Eθ1

(
Y1e

−Y1
)
− Eθ1(Y1)Eθ1

(
e−Y1

)]
×[

Eθ2

(
Y2e

−Y2
)
− Eθ2(Y2)Eθ2

(
e−Y2

)]
.

Ultimately, we have cov(Y1, Y2) = αcov
(
Y1, e

−Y1
)
cov

(
Y2, e

−Y2
)
. We are sure of

the answer.

Proposition 5. Under conditions (10) and (11) we have

fBEP,1(y1, y2; θ1, θ2, β1, β2, α) =

2∏
j=1


(
βj
θj

yj − 1

)
β−1
j


1− e−θj(

βj
θj

yj − 1

)
e−θj


δ0(yj)

×

g(y1, y2; θ1, θ2, α)× fBP (y1, y2, θ1, θ2). (22)

Expression (22) confirms that the bivariate Poisson extended distribution is a member
of the family of bivariate Poisson distributions (see [1]).

Proof. Indeed, we have

fBEP,1(y1, y2; θ1, θ2, β1, β2, α) =

(
θ
y1
1

y1!
e−θ1

)(
θ
y2
2

y2!
e−θ2

)
×

2∏
j=1


(
βj
θj

yj − 1

)
β−1
j


1− e−θj(

βj
θj

yj − 1

)
e−θj


δ0(yj)


×

g(y1, y2; θ1, θ2, α),

And under conditions (10) and (11) we are assured of the result.

3.1. Estimation of parameters θ1, θ2, β1, β2, α

The parameters θ1, θ2, β1, β2 and α will be estimated by the maximum likelihood
method. Let us consider an n-sample (y1,1, y2,1), (y1,2, y2,2), . . . , (y1,n, y2,n) of the couple
of random variables (Y1, Y2) of density fBEP,1(y1, y2; θ1, θ2, β1, β2, α). The log-likelihood
function L((y1, y2), θ1, θ2, β1, β2, α) is given by



R. Bidounga et al. / Eur. J. Pure Appl. Math, 14 (4) (2021), 1517-1529 1525

L((y1, y2), θ1, θ2, β1, β2, α) =
n∑

i=1

ln fBEP,1(y1,j , y2,j ; θ1, θ2, β1, β2, α).

The following system of normal equations

∂

∂θj
L((y1, y2), θ1, θ2, β1, β2, α) =0,

∂

∂βj
L((y1, y2), θ1, θ2, β1, β2, β, α) =0,

∂

∂α
L((y1, y2), θ1, θ2, β1, β2, α) =0,

is used to calculate the estimators θ̂j , β̂j , (j = 1, 2) and α̂ using the package maxLik for
the statistical environment R (see [6]).

Student’s t test to test α=0

To ensure the independance of variables Y1 and Y2, we must perform a statistical test
that allow discriminate between the following hypotheses [9]

• Null hypothesis H0 : α = 0, vs

• alternative hypothesis H1 : α ̸= 0.

Let α̂ = α̂n the maximum likelihood estimator of α. The variable

U =
√
n

α̂n − α√
I−1(α; θ1, θ2, β1, β2)

, (23)

follows, when n is large, the normal distribution N (0, 1), with I(α; θ1, θ2, β1, β2) the
amount of information provided by the pair (Y1, Y2) at parameter α.

The result is as follows.

Proposition 6.

(i)

I(α; θ1, θ2, β1, β2) = Eθ1,θ2

[ (
e−Y1 − c1

)2 (
e−Y2 − c2

)2[
1 + α

(
e−Y1 − c1

) (
e−Y2 − c2

)]2
]
. (24)

(ii) And under the null hypothesis

I(0; θ1, θ2, β1, β2) = var
(
e−Y1

)
var

(
e−Y2

)
. (25)

The estimator of I(0; θ1, θ2, β1, β2) which we denote Î(0) will be calculated following
two approaches:



R. Bidounga et al. / Eur. J. Pure Appl. Math, 14 (4) (2021), 1517-1529 1526

First approach: the substitution method

Knowing the estimators θ̂1n, θ̂2n, β̂1n and β̂2n we can estimate I(0; θ1, θ2, β1, β2) by Î(0) =
I(0; θ̂1n, θ̂2n, β̂1n, β̂2n).

Second approach: basic statistics

Let yjn =
1

n

n∑
i=1

e−yji and s
′2
jn =

1

n− 1

n∑
i=1

(
e−yji − yjn

)2
(j = 1, 2) the empirical mean

and the empirical unbiased variance of the sample(
e−Yj1 , . . . , e−Yji , . . . , e−Yjn

)
(j = 1, 2) of size n of the variable e−Yj (j = 1, 2).

Since the variances var
(
e−Y1

)
and var

(
e−Y2

)
can be estimated by the respective

empirical unbiased variances s
′2
1n and s

′2
2n, then we can estimate I(0; θ1, θ2, β1, β2) by

Î(0) = s
′2
1n × s

′2
2n.

Test statistics:
The test statistic

T =
√
n

α̂n√
Î−1(0)

, (26)

follows under the null hypothesis when n is large, the Student’s law of degree of freedom
n.

Decision: let x = P (> |T |) the p-value. Given a first order risk α = 5%, if x < α
then H0 is rejected, otherwise it is accepted.

Proof. [Proof of the Proposition 6] We have

∂

∂α
ln fBEP,1 =

(
e−Y1 − c1

) (
e−Y2 − c2

)
1 + α

(
e−Y1 − c1

) (
e−Y2 − c2

) ,
and

∂2

∂α2
ln fBEP,1 = −

(
e−Y1 − c1

)2 (
e−Y2 − c2

)2[
1 + α

(
e−Y1 − c1

) (
e−Y2 − c2

)]2 .
Therefore

I(α; θ1, θ2, β1, β2) = Eθ1,θ2

[ (
e−Y1 − c1

)2 (
e−Y2 − c2

)2[
1 + α

(
e−Y1 − c1

) (
e−Y2 − c2

)]2
]
,

and under the null hypothesis the variables Y1 and Y2 are independent. This leads to

I(0; θ1, θ2, β1, β2) = var
(
e−Y1

)
var

(
e−Y2

)
.

And we are sure of the answer. Expressions (6) and (7) allow us to calculate the

variances of the variables e−Y1 and e−Y2 .



R. Bidounga et al. / Eur. J. Pure Appl. Math, 14 (4) (2021), 1517-1529 1527

4. Simulation study

4.1. Simulation and basic statistics

In this section, we realize a simulation study. On this, we consider two random variables
Y1 and Y2 following the extended Poisson distribution of respective parameters (θ1, β1) and
(θ2, β2). The table 1 contains the simulations of the variables Y1 and Y2 and the table 2
the basic statistics. We have the presumption that according to the Fisher indices of table
2, the variables are overdispersed. We have simulated samples of size n = 150.

Table 1: Simulated data, θ1 = 1, β1 = 2 for Y1 and θ2 = 3, β2 = 5 for Y2

Count 0 1 2 3 4 5 6 7 8 9 10
n = 150NY1 42 23 48 24 12 0 1

NY2 25 3 22 23 29 24 10 5 5 3 1

Table 2: Basic statistics

Variable Mean Variance Fisher index

Y1 1.6333 1.7371 1.0635

Y2 3.4933 5.3657 1.5359

4.2. Estimation of model parameters and remark

Using the package maxLik for the statistical environment R (see [6]), we have the
output R in table 3.

The parameter estimates are θ̂1 = 1.14132, θ̂2 = 2.70274, β̂1 = 2.64952, β̂2 = 4.58362
and α̂ = 1.45859. The table 3 shows that α is different to 0. Indeed, the corresponding
p-value is equal to 0.00096, lower than at the usual significance level 0.05, so we reject the
hypothesis that α = 0 at the risk of significance 5%. For this set of simulated data, the
variables Y1 and Y2 are dependent.

5. Conclusion

We constructed the bivariate extended Poisson distribution as a generalization of the
univariate extended Poisson distribution by the method of the product of its marginal
laws by a factor. This method was demonstrated by Lakshminarayna et al.[7]. We have
shown that this distribution belongs to the family of bivariate Poisson distributions. The
Student’s statistical test allows to highlight the independence between the variables Y1
and Y2.



REFERENCES 1528

Table 3: Output R

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
Maximum Like l ihood es t imat i on
Newton−Raphson maximisation , 5 i t e r a t i o n s
Return code 2 : s u c c e s s i v e function va lue s with in t o l e r an c e l im i t
Log−Like l ihood : −494.9254
5 f r e e parameters
Est imates :

Estimate Std . e r r o r t va lue Pr(> t )
[ 1 , ] 1 .14132 0.09977 11 .439 < 2e−16 ∗∗∗
[ 2 , ] 2 .70274 0.16122 16 .764 < 2e−16 ∗∗∗
[ 3 , ] 2 .64952 0.43437 6 .100 1 .06 e−09 ∗∗∗
[ 4 , ] 4 .58362 0.81856 5 .600 2 .15 e−08 ∗∗∗
[ 5 , ] 1 .45859 0.44173 3 .302 0.00096 ∗∗∗
−−−
S i g n i f . codes : 0 ’ ∗∗∗ ’ 0 .001 ’ ∗∗ ’ 0 .01 ’ ∗ ’ 0 .05 ’ . ’ 0 . 1 ’ ’ 1

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

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