EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 17, No. 3, 2024, 2336-2348
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Probabilistic Type 2 Poly-Bernoulli Polynomials

Si Hyeon Lee1 ,∗, Li Chen2, Wonjoo Kim3

1 Kwangwoon University, Seoul 139-701, Republic of Korea
2 School of Mathematics, Xi’an University of Finance and Economics, Xi’an 710100, China
3 Department of Applied Mathematics, Kyunghee University, Seoul, Republic of Korea

Abstract. The main purpose of this article is to introduce the probabilistic type 2 poly-Bernoulli
polynomials under the condition that Y is a random variable. This means that we will consider
the probabilistic extension of the type 2 poly-Bernoulli polynomials and study to obtain some new
results. Furthermore, we also define the probabilistic unipoly-Bernoulli polynomials and numbers
attached to p, and investigate their interesting basic properties. Based on these new definition,
we derive some meaningful formulae of probabilistic type 2 poly-Bernoulli polynomials and prob-
abilistic unipoly-Bernoulli polynomials and numbers attached to p.

2020 Mathematics Subject Classifications: 11B68

Key Words and Phrases: Bernoulli polynomials, Stirling numbers, Probabilistic type 2 poly-
Bernoulli polynomials, Probabilistic unipoly-Bernoulli polynomials.

1. Introduction

The Bernoulli polynomials are defined by

t

et − 1
ext =

∞∑
n=0

Bn(x)
tn

n!
, (see[1, 2, 7, 15, 27, 30], [12, 19, 20, 28]). (1)

For k ∈ Z, the polylogarithm function is defined by

Lik(x) =

∞∑
n=1

xn

nk
, (|x| < 1), (see[4, 5, 24], [23]). (2)

For k ∈ Z, Kim defined the polyexponential function ek(x), which is given by

ek(x) =
∞∑
n=1

xn

(n− 1)!nk
, (see[6]). (3)

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v17i3.5275

Email addresses: ugug11@naver.com (S. H. Lee),
chenli 0928@xaufe.edu.cn (L. Chen), ugug11@naver.com (W. Kim)

https://www.ejpam.com 2336 © 2024 EJPAM All rights reserved.



S. H. Lee, L. Chen, W. Kim / Eur. J. Pure Appl. Math, 17 (3) (2024), 2336-2348 2337

When k = 1, we note that

e1(x) =
∞∑
n=1

xn

n!
= ex − 1. (4)

As we all know, the poly-Bernoulli polynomials are defined by Kaneko. It is given by

Lik(1− e−t)

1− e−t
ext =

∞∑
n=0

PB(k)
n (x)

tn

n!
, (see[5]). (5)

When x = 0, we note that PB
(k)
n = PB

(k)
n (0) are called the poly-Bernoulli numbers.

In 2019, Kim considered the definition of type 2 poly-Bernoulli polynomials. It is given
by

ek(log(1 + t))

et − 1
ext =

∞∑
n=0

β(k)
n (x)

tn

n!
, (see[6, 22]). (6)

When x = 0, we note that β
(k)
n = β

(k)
n (0) are called the type 2 poly-Bernoulli numbers.

Kim also studied the unipoly function attached to p. Its definition as follows.

uk(x|p) =
∞∑
n=1

p(n)

nk
xn, (k ∈ Z), (see[6]). (7)

Later, he defined the unipoly-Bernoulli polynomials attached to p by

1

1− e−t
uk(1− e−t|p)ext =

∞∑
n=0

B(k)
n,p(x)

tn

n!
, (see[6]). (8)

Recently, Kim studied the probabilistic poly-Bernoulli polynomials associated with Y .
Assume that Y is a random variable such that the moment generating function of Y given
by

E[eY t] =

∞∑
n=0

E[Y n]
tn

n!
, (|t| < r), ([6, 14, 16]). (9)

exist for some r ≥ 0. Then the definition of the probabilistic poly-Bernoulli polynomials
are given by

Lik(1− e−t)

1− E[e−Y t]
(E[e−Y t])x =

∞∑
n=0

B(k,Y )
n (x)

tn

n!
, (see[3, 8, 9, 18, 31, 32]). (10)

When k = 1, it is obvious that B
(1,Y )
n = (−1)nBY

n (x). This type of polynomials is a
new extension. Inspired by this, the aim of our paper is to explore the probabilistic type
2 poly-Bernoulli polynomials and obtain some new results. Meanwhile, the probabilistic
unipoly-Bernoulli polynomials are also another research.



S. H. Lee, L. Chen, W. Kim / Eur. J. Pure Appl. Math, 17 (3) (2024), 2336-2348 2338

The Stirling number of the first kind are defined by

(x)n =
n∑

k=0

S1(n, k)x
k, (see[10, 28, 29]). (11)

Where (x)0 = 1, (x)n = x(x− 1) · · · (x− n+ 1), (n ≥ 1).
From (11), we can easily know

1

k!
(log(1 + t))k =

∞∑
n=k

S1(n, k)
tn

n!
, (see[10, 11, 29]). (12)

The Stirling number of the second kind are defined by

xn =

n∑
k=0

S2(n, k)(x)k, (see[17, 21, 26]). (13)

From (13), we also derive the generating function as follows.

1

k!
(et − 1)k =

∞∑
n=k

S2(n, k)
tn

n!
, (see[21, 26]). (14)

In 2024, Kim defined the probabilistic Stirling number of the second kind associated
with Y are given by

1

k!
(E[eY t]− 1)k =

∞∑
n=k

{
n
k

}
Y

tn

n!
, (see[3, 9, 18], [14]). (15)

The Bell polynomials are defined by

ex(e
t−1) =

∞∑
n=0

Beln(x)
tn

n!
, (see[13, 16, 22, 23, 25]). (16)

2. probabilistic type 2 poly-Bernoulli polynomials

Let (Yj)j≥1 be a sequence of mutually independent copies of the random variable Y ,
and let

S0 = 0, Sk = Y1 + Y2 + · · ·+ Yk, (k ∈ N). (17)

In this section we consider probabilistic type 2 poly-Bernoulli polynomials.

ek(log(1 + t))

E[eY t]− 1
(E[eY t])x =

∞∑
n=0

β(k,Y )
n (x)

tn

n!
. (18)



S. H. Lee, L. Chen, W. Kim / Eur. J. Pure Appl. Math, 17 (3) (2024), 2336-2348 2339

When x = 0, β
(k,Y )
n (0) = β

(k,Y )
n are called probabilistic type 2 poly-Bernoulli numbers.

From (18), we get

∞∑
n=0

β(k,Y )
n (x)

tn

n!
=

ek(log(1 + t))

E[eY t]− 1
(E[eY t])x (19)

=
∞∑
j=0

β
(k,Y )
j

tj

j!

∞∑
k=0

(
x

k

)
k!

∞∑
m=k

{
m
k

}
Y

tm

m!

=

∞∑
n=0

n∑
m=0

m∑
k=0

(
n

m

)
β
(k,Y )
n−m (x)k

{
m
k

}
Y

tn

n!
.

Therefore, by comparing the coefficients on both sides of (19), we have the following
theorem.

Theorem 1. For n, k ≥ 0, we have

β(k,Y )
n =

n∑
m=0

m∑
k=0

(
n

m

)
β
(k,Y )
n−m (x)k

{
m
k

}
Y

.

From (18), we have

∞∑
n=0

β(k,Y )
n (x)

tn

n!
=

ek(log(1 + t))

t

t

E[eY t]− 1
(E[eY t])x (20)

=

∞∑
l=0

BY
l (x)

tl

l!

∞∑
i=1

(log(1 + t))i

(i− 1)!ik

=

∞∑
l=0

BY
l (x)

tl

l!

1

t

∞∑
i=1

1

ik−1

∞∑
j=i

S1(j, i)
tj

j!

=

∞∑
l=0

BY
l (x)

tl

l!

∞∑
j=0

j+1∑
i=1

1

ik−1

S1(j + 1, i)

j + 1

tj

j!

=

∞∑
n=0

 n∑
j=0

j+1∑
i=1

(
n

j

)
S1(j + 1, i)

ik−1(j + 1)
BY

n−j(x)

 tn

n!
.

Thus, by comparing the coefficients on both sides of (20), we have the following theorem.

Theorem 2. For n, j ≥ 0, we have

β(k,Y )
n (x) =

n∑
j=0

j+1∑
i=1

(
n

j

)
S1(j + 1, i)

ik−1(j + 1)
BY

n−j(x).



S. H. Lee, L. Chen, W. Kim / Eur. J. Pure Appl. Math, 17 (3) (2024), 2336-2348 2340

Now, we observe that

n∑
m=0

(
E[eY t]

)m
=

E[eY t]n+1 − 1

E[eY t]− 1
. (21)

From (21), we have

n∑
m=0

E[eY t] =
1

e1(log(1 + t))

e1(log(1 + t))

E[eY t]− 1
(E[eY t]n+1 − 1) (22)

=
1

t

t

E[EY t]− 1

(
E[eY t]n+1 − 1

)
=

1

t

( ∞∑
l=0

β
(1,Y )
l −

∞∑
l=0

β
(1,Y )
l

tl

l!

)

=

∞∑
l=0

β
(1,Y )
l+1 (n+ 1)− β

(1,Y )
l+1

l + 1

tl

l!
.

On the other hand,

n∑
m=0

(
E[eY t]

)m
=

n∑
m=0

E[e(Y1+Y2+···+Ym)t] (23)

=
n∑

m=0

∞∑
l=0

E[Sl
m]

tl

l!

=
∞∑
l=0

n∑
m=0

E[Sl
m]

tl

l!
.

Hence, comparing the coefficients on both sides of (22) and (23), we have the following
theorem.

Theorem 3. For n ≥ 0, we have

n∑
m=0

E[Sm] =
β
(1,Y )
l+1 (n+ 1)− β

(1,Y )
l+1

l + 1
.

From (3), we have

em(log(1 + t)) =

∞∑
k=1

(log(1 + t))k

(k − 1)!km
(24)

=

∞∑
k=0

(log(1 + t))k+1

k!(k + 1)m



S. H. Lee, L. Chen, W. Kim / Eur. J. Pure Appl. Math, 17 (3) (2024), 2336-2348 2341

=

∞∑
k=0

1

(k + 1)m−1

∞∑
n=k+1

S1(n, k + 1)
tn

n!

=
∞∑

n=k+1

n−1∑
k=0

S1(n, k + 1)

(k + 1)m−1

tn

n!
.

On the other hand,

em(log(1 + t)) =

∞∑
l=0

β
(m,Y )
l

tl

l!

(
E[eY t − 1]

)
(25)

=
∞∑
l=0

β
(m,Y )
l

tl

l!

 ∞∑
j=0

E[Y j ]
tj

j!
− 1


=

∞∑
n=0

(
n∑

l=0

(
n

l

)
β
(m,Y )
l E[Y n−l]− β(m,Y )

n

)
tn

n!
.

Therefore, by comparing the coefficients on both sides of (24) and (25), we have the
following theorem.

Theorem 4. For n, k ≥ 0, we have

n−1∑
k=0

S1(n, k + 1)

(k + 1)m−1
=

{∑n
l=0

((
n
l

)
β
(m,Y )
l E[Y n−l]− β

(m,Y )
n

)
, if n ≥ k + 1,

0, if n < k + 1.

Let Y be the Poisson random variable with parameter α > 0, then we have

ek(log(1 + t))

E[eY t]− 1

(
E[eY t]

)x
=

ek(log(1 + t))

eα(et−1) − 1
eαx(e

t−1) (26)

=
α(et − 1)

α(et − 1)

ek(log(1 + t))

eα(et−1)−1
eαx(e

t−1)

=
1

α

∞∑
j=0

β
(k)
j

tj

j!

α(et − 1)

eα(et−1) − 1
eαx(e

x−1)

=
1

α

∞∑
j=0

β
(k)
j

tj

j!

∞∑
l=0

αlBl(x)
(ex − 1)l

l!

=

∞∑
j=0

β
(k)
j

tj

j!

∞∑
m=0

m∑
l=0

αl−1Bl(x)S2(m, l)
tm

m!

=
∞∑
n=0

n∑
m=0

m∑
l=0

(
n

m

)
β
(k)
n−mαl−1Bl(x)S2(m, l)

tn

m
.



S. H. Lee, L. Chen, W. Kim / Eur. J. Pure Appl. Math, 17 (3) (2024), 2336-2348 2342

From (18) and (26), we have the following theorem.

Theorem 5. Let Y be the Poisson random variable with parameter α, we have

β(k,Y )
n (x) =

n∑
m=0

m∑
l=0

(
n

m

)
β
(k)
n−mαl−1Bl(x)S2(m, l).

From (18), we have

∞∑
n=0

B(k,Y )
n (α+ 1) =

ek(log(1 + t))

E[eY t]− 1

(
E[eY t]

)α
E[eY t] (27)

=
∞∑
l=0

B
(k,Y )
l (α)

tl

l!

∞∑
m=0

E[Y m]
tm

m!

=
∞∑
n=0

n∑
l=0

(
n

l

)
B

(k,Y )
l (α)E[Y n−l]

tn

n!
.

From (18), we also have

∞∑
n=0

B(k,Y )
n (α)

tn

n!
=

∞∑
n=0

BY
n

tn

n!
E[e(Y1+Y2+···+Yα)t] (28)

=

∞∑
l=0

B
(k,Y )
l

tn

n!

∞∑
m=0

E[Sm
α ]

tm

m!

=

∞∑
n=0

n∑
l=0

(
n

l

)
B

(k,Y )
l E[Sm

α ]
tn

n!
.

Therefore, by (27) and (28), we have the following theorem.

Theorem 6. For any α ∈ Z and n, α ≥ 0, we have

B(k,Y )
n (α+ 1)−B(k,Y )

n (α) =
n∑

l=0

(
n

l

)(
B

(k,Y )
l (α)E[Y n−l]−B

(k,Y )
l E[Sm

α ]
)
.

3. The probabilistic unipoly-Bernoulli polynomials

In this section, we give the definition of the probabilistic unipoly-Bernoulli polynomials
attached to p as follows.

1

1− E[e−Y t]
uk(1− e−t|p)(E[e−Y t])x =

∞∑
n=0

B(k,Y )
n,p (x)

tn

n!
. (29)



S. H. Lee, L. Chen, W. Kim / Eur. J. Pure Appl. Math, 17 (3) (2024), 2336-2348 2343

If x = 0, B
(k,Y )
n,p = B

(k,Y )
n,p (0) are called the probabilistic unipoly-Bernoulli numbers.

Particularly, if p(n) = 1, then B
(k,Y )
n,1 = B

(k,Y )
n (x).

From (29)

1

1− E[e−Y t]
uk(1− e−t|p) = 1

1− E[e−Y t]

∞∑
m=1

P (m)(1− e−t)m

mk
(30)

=
t

1− E[e−Y t]

1

t

∞∑
m=1

p(m)

mk

(1− e−t)m

m!
m!

=

∞∑
j=0

BY
j (−1)j

tj

j!

1

t

∞∑
m=1

p(m)m!

mk

∞∑
l=m

S2(l,m)(−1)l−m tl

l!

=

∞∑
j=0

BY
j (−1)j

tj

j!

∞∑
l=0

l+1∑
m=1

p(m)m!

mk

S2(l + 1,m)(−1)l+1−m

l + 1

tl

l!

=

∞∑
n=0

n∑
l=0

l+1∑
m=1

(
n

l

)
p(m)(m− 1)!

mk−1
(−1)n−m+1S2(l + 1,m)

l + 1
BY

n−l

tn

n!
.

Therefore, by compring the coefficients on both sides of (29) and (30), we have the following
theorem.

Theorem 7. For n, k ≥ 0, we have

B(k,Y )
n,p =

n∑
l=0

l+1∑
m=1

(
n

l

)
p(m)(m− 1)!

mk−1
(−1)n−m+1S2(l + 1,m)

l + 1
BY

n−l.

Let Y be the Poisson random variable with parameter α > 0. Then we have

uk(1− e−t|p)exα(et−1) =

∞∑
n=0

B(k,Y )
n,p (x)

tn

n!
(1− eα(e

−t−1)) (31)

=
∞∑
n=0

B(k,Y )
n,p (x)

tn

n!
−

∞∑
m=0

B(k,Y )
m (x)

tm

m!

∞∑
l=0

αl(e−t − 1)l

l!

=
∞∑
n=0

B(k,Y )
n,p (x)

tn

n!
−

∞∑
m=0

B(k,Y )
m (x)

tm

m!

∞∑
l=0

αl
∞∑
i=l

S2(i, l)(−1)i
ti

i!

=

∞∑
n=0

B(k,Y )
n,p (x)

tn

n!
−

∞∑
n=0

n∑
i=0

i∑
l=0

(
n

i

)
(−1)iαlS2(i, l)B

(k,Y )
n−i (x)

tn

n!

=

∞∑
n=0

(
B(k,Y )

n,p (x)−
n∑

i=0

i∑
l=0

(
n

i

)
(−1)iαlS2(i, l)B

(k,Y )
n−i (x)

)
tn

n!
.



S. H. Lee, L. Chen, W. Kim / Eur. J. Pure Appl. Math, 17 (3) (2024), 2336-2348 2344

On the other hand

uk(1− e−t|p)exα(et−1) =
∞∑

m=1

p(m)

mk
(1− e−t)m

∞∑
i=0

Beli(x)
αi(e−t − 1)i

i!
(32)

=

∞∑
j=1

j∑
i=0

p(j − i)

(j − i)k
Beli(x)

αi

i!
(−1)j−i (e

−t − 1)j

j!
j!

=

∞∑
j=1

j∑
i=0

p(j − i)

(j − i)k
Beli(x)

αij!

i!
(−1)j−i

∞∑
n=j

S2(n, j)(−1)n
tn

n!

=
∞∑
n=1

n∑
j=1

j∑
i=0

p(j − i)

(j − i)k
Beli(x)

αij!

i!
(−1)j−i+nS2(n, j)

tn

n!
.

Therefore, by comparing the coefficients on both sides of (31) and (32), we have the
following theorem.

Theorem 8. Let Y be the Poisson random variable with parameter α(> 0). Then we have

B(k,Y )
n,p (x) =

n∑
j=1

j∑
i=0

p(j − i)

(j − i)k
Beli(x)

αij!

i!
(−1)j−i+nS2(n, j)+

n∑
i=0

i∑
l=0

(
n

i

)
(−1)iαlS2(i, l)B

(k,Y )
n−i (x).

From (29), we have

∞∑
n=0

B(k,Y )
n,p (α)

tn

n!
=

∞∑
n=0

B
(k,Y )
l,p

tl

l!

(
E[e−Y t]

)α
(33)

=

∞∑
l=0

B
(k,Y )
l,p

tl

l!

∞∑
m=0

(−1)mE[Y1 + Y2 + · · ·+ Yα]
tm

m!

=

∞∑
n=0

n∑
m=0

(−1)m
(
n

m

)
B

(k,Y )
n−m,pE[Sm

α ]
tn

n!
.

Therefore, by compring the coefficients on bosides (33), we have the following theorem.

Theorem 9. For α, n ≥ 0 and α ∈ Z, we have

B(k,Y )
n,p (α) =

n∑
m=0

(−1)m
(
n

m

)
B

(k,Y )
n−m,pE[Sm

α ].

Let Y be the Bernoulli random variable with probability of success A. Then we have

∞∑
n=0

B(k,Y )
n,p (x)

tn

n!
=

1

A(e−t − 1)
uk(1− e−t|p)

(
A(e−t − 1) + 1)

)x
(34)



REFERENCES 2345

=
1

A(e−t − 1)

∞∑
l=1

p(l)

lk
(1− e−t)l

∞∑
m=0

(
x

m

)
Am(e−t − 1)m

=
1

A(e−t − 1)

∞∑
i=1

i∑
l=1

p(l)

lk

(
x

i− l

)
Ai−l(e−t − 1)i

=
∞∑
i=1

i∑
l=1

(−1)l
p(l)

lk

(
x

i− l

)
Ai−l−1(e−t − 1)i−1

=
∞∑
i=0

i+1∑
l=1

(−1)l
p(l)

lk

(
x

i− l + 1

)
Ai−l(e−t − 1)i

=

∞∑
i=0

i+1∑
l=1

(−1)l
p(l)

lk
i!

(
x

i− l + 1

)
Ai−l

∞∑
n=i

(−1)nS2(n, i)
tn

n!

=
∞∑
n=0

n∑
i=0

i+1∑
l=1

(−1)l+n p(l)

lk
(i)l−1(x)i−lA

i−lS2(n, i)
tn

n!
.

Therefore, by compring the coefficients on both sides of (34), we have the following theo-
rem.

Theorem 10. Let Y be the Bernoulli random variable with probability of success A, then
we have

B(k,Y )
n,p (x) =

n∑
i=0

i+1∑
l=1

(−1)l+n p(l)

lk
(i)l−1(x)i−lA

i−lS2(n, i).

4. Conclusion

In this paper, we present a probabilistic version of the type 2 poly-Bernoulli polynomi-
als associated with a random variable Y satisfying suitable moment conditions. We call it
probabilistic type 2 poly-Bernoulli polynomials. We study some properties of such poly-
nomials and obtain relevant results. More specifically, we derived an exact expression for
βk,Y
n (x), and establish a relation between the type 2 poly-Bernoulli numbers and the Stir-

ling number of the first kind, and obtain a explicit formula of β
(k,Y )
n (x), In the case where

Y is the Poisson variable with parameter α. Similarly, we define the unipoly-Bernoulli
polynomials attached to p. Then we show the explicit expression of Bk,Y

n,p (x) and other
results by skilful calculations. As a next step in our research, we will study this probability
type of polynomials more deeply so that give better and generalizable results.

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