#  Some results on $r$-truncated degenerate Poisson Random Variables

Taekyun Kim  Department of Mathematics, Kwangwoon University, Seoul 139-701,
Republic of Korea<EMAIL_ADDRESS>,  DAE SAN KIM  Department of Mathematics,
Sogang University, Seoul 121-742, Republic of Korea<EMAIL_ADDRESS>,
Si-Hyeon Lee  Department of Mathematics, Kwangwoon University, Seoul 139-701,
Republic of Korea<EMAIL_ADDRESS>,  Seong-Ho Park  Department of
Mathematics, Kwangwoon University, Seoul 139-701, Republic of Korea
<EMAIL_ADDRESS>and  Lee-Chae Jang  Graduate School of Education, Konkuk
University, Seoul 143-701, Republic of Korea<EMAIL_ADDRESS>
###### Abstract.

The zero-truncated Poisson distributions are certain discrete probability
distributions whose supports are the set of positive integers, which are also
known as the conditional Poisson distributions or the positive Poisson
distributions. Recently, as a natural extension of those distributions, Kim-
Kim studied the zero-truncated degenerate Poisson distributions. In this
paper, we introduce the $r$-truncated degenerate Poisson random variable with
parameter $\alpha>0$, whose probability mass function is given by
$p_{\lambda,r}(i)=\cfrac{(1)_{i,\lambda}}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\frac{\alpha^{i}}{i!},\quad(i=r+1,r+2,r+3,\dots)$,
and investigate various properties of this random variable.

###### Key words and phrases:

$r$-truncated degenerate Poisson random variables; $r$-truncated degenerate
Stirling numbers of the second kind

###### 2010 Mathematics Subject Classification:

11B73; 60G50

##  1\. Introduction

It is well known that a random variable $X$, taking on one of the values
$0,1,2,\dots$, is said to be the Poisson random variable with parameter
$\alpha>0$, if the probability mass function of $X$ is given by

$p(i)=P\\{X=i\\}=e^{-\alpha}\frac{\alpha^{i}}{i!},\quad(i=0,1,2,\dots),\quad(\mathrm{see}\
[11,13]).$

We note that a Poisson random variable indicates how many events occured in a
given period of time. Further, a random variable $X_{r}$, taking on one of the
values $r+1,r+2,\dots$, is called the $r$-truncated Poisson random variable
with parameter $\alpha>0$, if the probability mass function of $X_{r}$ is
given by

$p_{r}(i)=P\\{X_{r}=i\\}=\frac{1}{e^{\alpha}-\sum_{l=0}^{r}\frac{\alpha^{l}}{l!}}\frac{\alpha^{i}}{i!},$

where $i=r+1,r+2,r+3,\dots$, for $r\geq 0$. In the special case of $r=0$, we
obtain the zero-truncated Poisson distribution.

On the other hand, the degenerate Poisson random variables with parameter
$\alpha>0$, whose probability mass function is given by (see (1))

$p_{\lambda}(i)=e_{\lambda}^{-1}(\alpha)\frac{\alpha^{i}}{i!}(1)_{i,\lambda}\quad(i=0,1,2,\dots),$

were studied by Kim-Kim-Jang-Kim in [10]. In addition, the zero-truncated
degenerate Poisson random variables with parameter $\alpha>0$, whose
probability mass function is given by

$p_{\lambda,0}(i)=\frac{1}{e_{\lambda}(\alpha)-1}\frac{\alpha^{i}}{i!}(1)_{i,\lambda}\quad(i=1,2,3,\dots),$

were studied in [9].

The aim of this paper is to generalize the results on the zero-truncated
degenerate Poisson distribution in [9] to the case of the $r$-truncated
degenerate Poisson distribution as a natural extension of the $r$-truncated
Poission distribution. We will obtain, among other things, its expectation,
its variance, its $n$-th moment, and its cumulative distribution function.

One motivation for this research is its potential applications to the
cornavirus pandemic. It has been spreading unpredictably around the world,
terrorizing many people. Although several vaccines for the coronavirus have
been developed and many people are getting vaccinated, they have many
unexpected side effects as well. We would like to predict stability of the
coronavirus vaccines after $r$-days the vaccines were shot. For this purpose,
we study the $r$-truncated degenerate Poisson random variables (see (6)),
which has the ‘degenerate factor’ $\lambda$ reflecting abnormal situations.
Indeed, we think that Theorem 5 is useful in predicting the probability of the
coronavirus vaccines becoming stable after the $r$-days of getting vaccinated.
Another motivation is applications of various probabilistic methods in
studying special numbers and polynomials arising from combinatorics and number
theory. For this, we let the reader refer to [5-10,14 and the references
therein]. For the rest of this section, we recall the necessary facts that are
needed throughout this paper.

For any $\lambda\in\mathbb{R}$, the degenerate exponential functions are
defined by

(1) $e_{\lambda}^{x}(t)=(1+\lambda
t)^{\frac{x}{\lambda}}=\sum_{k=0}^{\infty}(x)_{k,\lambda}\frac{t^{k}}{k!},\quad(\mathrm{see}\
[1-3,5-10,14]),$

where $(x)_{0,\lambda}=1$,
$(x)_{n,\lambda}=x(x-\lambda)\cdots(x-(n-1)\lambda)$, $(n\geq 1)$.

In particular, for $x=1$, we denote $e_{\lambda}^{1}(t)$ by $e_{\lambda}(t)$.

In addition, it is convenient to introduce $e_{\lambda,r}(t)$, for any integer
$r\geq 0$, given by

(2) $e_{\lambda,r}(t)=\sum_{k=0}^{r}(1)_{k,\lambda}\frac{t^{k}}{k!}.$

The Bell polynomials are given by

(3)
$e^{x(e^{t}-1)}=\sum_{n=0}^{\infty}\mathrm{Bel}_{n}(x)\frac{t^{n}}{n!},\quad(\mathrm{see}\
[4,12]).$

In light of (3), the degenerate Bell polynomials are defined in [9] by

(4)
$e_{\lambda}^{-1}(x)e_{\lambda}(xe^{t})=\sum_{n=0}^{\infty}\mathrm{Bel}_{n,\lambda}(x)\frac{t^{n}}{n!}.$

In particular, for $x=1$,
$\mathrm{Bel}_{n,\lambda}=\mathrm{Bel}_{n,\lambda}(1)$, $(n\geq 0)$, are
called the degenerate Bell numbers. Note that

$\lim_{\lambda\rightarrow
1}\mathrm{Bel}_{n,\lambda}(x)=\mathrm{Bel}_{n}(x),\quad(n\geq 0).$

Let $X$ be a Poisson random variable with parameter $\alpha>0$. Then the
moment of $X$ is given by

$E[X^{n}]=\mathrm{Bel}_{n}(\alpha),\quad(n\geq 0),\quad(\mathrm{see}\
[11,13]).$

Let $Y$ be a discrete random variable taking values in the nonnegative
integers. Then the probability generating function of $Y$ is given by

$F(t)=E[t^{Y}]=\sum_{i=0}^{\infty}p(i)t^{i},\quad(\mathrm{see}\ [11,13]),$

where $p(i)=P\\{Y=i\\}$ is the probability mass function of $Y$.

The degenerate Stirling numbers of the second kind $S_{2,\lambda}(n,k)$ are
defined by

$(x)_{n,\lambda}=\sum_{l=0}^{n}S_{2,\lambda}(n.l)(x)_{l},\quad(n\geq
0),\quad(\mathrm{see}\ [5]),$

or equivalently by

$\frac{1}{k!}\bigg{(}e_{\lambda}(t)-1\bigg{)}^{k}=\sum_{n=k}^{\infty}S_{2,\lambda}(n,k)\frac{t^{n}}{n!}.$

Here $(x)_{0}=1,\ (x)_{n}=x(x-1)(x-2)\cdots(x-n+1)$, $(n\geq 1)$.

Now, we consider the $r$-truncated degenerate Stirling numbers of the second
kind $S_{2,\lambda}^{[r]}(n,kr+k)$ given by

(5)
$\frac{1}{k!}\bigg{(}e_{\lambda}(t)-e_{\lambda,r}(t)\bigg{)}^{k}=\sum_{n=kr+k}^{\infty}S_{2,\lambda}^{[r]}(n,kr+k)\frac{t^{n}}{n!}.$

Note that

$S_{2,\lambda}^{[0]}(n,k)=S_{2,\lambda}(n,k),\quad(\mathrm{see}\ [5]).$

##  2\. $r$-truncated degenerate Poisson random variables

The random variable $X_{\lambda,r}$ is called the $r$-truncated degenerate
Poisson random variable with parameter $\alpha>0$, if the probability mass
function of $X_{\lambda,r}$ is given by

(6)
$p_{\lambda,r}(k)=\cfrac{(1)_{k,\lambda}}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\frac{\alpha^{k}}{k!},$

where $k=r+1,r+2,\dots$, with $r\geq 0$.

Here we must observe that

$\displaystyle\sum_{k=r+1}^{\infty}p_{\lambda,r}(k)$
$\displaystyle=\frac{1}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\sum_{k=r+1}^{\infty}\frac{\alpha^{k}}{k!}(1)_{k,\lambda}$
$\displaystyle=\frac{1}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}(e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha))$
$\displaystyle=1.$

In addition, we note that

$\lim_{\lambda\rightarrow
0}p_{\lambda,r}(k)=\frac{1}{e^{\alpha}-\sum_{l=0}^{r}\frac{\alpha^{l}}{l!}}\frac{\alpha^{k}}{k!}$

is the probability mass function of the $r$-truncated Poisson random variable
with parameter $\alpha>0$.

Let us assume that $X_{\lambda,r}$ is the $r$-truncated degenerate Poisson
random variable with parameter $\alpha>0$. Then we note that the expectation
of $X_{\lambda,r}$ is given by

(7) $\displaystyle E[X_{\lambda,r}]\ $ $\displaystyle=\
\sum_{n=r+1}^{\infty}np_{\lambda,r}(n)=\frac{1}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\sum_{n=r+1}^{\infty}n\frac{\alpha^{n}}{n!}(1)_{n,\lambda}$
$\displaystyle=\
\frac{\alpha}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\sum_{n=r}^{\infty}\frac{\alpha^{n}}{n!}(1)_{n+1,\lambda}$
$\displaystyle=\
\frac{\alpha}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\bigg{(}\sum_{n=r+1}^{\infty}\frac{\alpha^{n}}{n!}(1)_{n+1,\lambda}+\frac{\alpha^{r}}{r!}(1)_{r+1,\lambda}\bigg{)}$
$\displaystyle=\
\frac{\alpha}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\bigg{(}\sum_{n=r+1}^{\infty}\frac{\alpha^{n}}{n!}(1)_{n,\lambda}(1-n\lambda)+\frac{\alpha^{r}}{r!}(1)_{r+1,\lambda}\bigg{)}$
$\displaystyle=\
\alpha-\frac{\alpha\lambda}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\sum_{n=r+1}^{\infty}n\frac{\alpha^{n}}{n!}(1)_{n,\lambda}+\frac{\alpha}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\frac{\alpha^{r}}{r!}(1)_{r+1,\lambda}$
$\displaystyle=\ \alpha-\alpha\lambda
E[X_{\lambda,r}]+\frac{\alpha}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\frac{\alpha^{r}}{r!}(1)_{r+1,\lambda}.$

From (7), we obtain the following theorem.

######  Theorem 1.

Let $X_{\lambda,r}$ be the $r$-truncated degenerate Poisson random with
parameter $\alpha>0$. Then, for $\lambda\neq-\frac{1}{\alpha}$, the
expectation of $X_{\lambda,r}$ is given by

$E[X_{\lambda,r}]=\frac{1}{1+\alpha\lambda}\bigg{[}\alpha+\frac{\alpha^{r+1}}{r!}\frac{(1)_{r+1,\lambda}}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\bigg{]}.$

Now, we observe that

(8) $\displaystyle E[X_{\lambda,r}^{2}]\ $ $\displaystyle=\
\sum_{n=k+1}^{\infty}n^{2}p_{\lambda,r}(n)\ =\
\frac{1}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\sum_{n=r+1}^{\infty}n^{2}\frac{\alpha^{n}}{n!}(1)_{n,\lambda}$
$\displaystyle=\
\frac{1}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\bigg{(}\sum_{n=r+1}^{\infty}\frac{\alpha^{n}}{(n-2)!}(1)_{n,\lambda}+\sum_{n=r+1}^{\infty}\frac{\alpha^{n}}{(n-1)!}(1)_{n,\lambda}\bigg{)}$
$\displaystyle=\
\frac{1}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\bigg{(}\sum_{n=r+2}^{\infty}\frac{\alpha^{n}}{(n-2)!}(1)_{n,\lambda}+\sum_{n=r+1}^{\infty}\frac{\alpha^{n}}{(n-1)!}(1)_{n,\lambda}\bigg{)}$
$\displaystyle\quad+\frac{1}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\frac{\alpha^{r+1}}{(r-1)!}(1)_{r+1,\lambda}$
$\displaystyle=\
\frac{1}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\bigg{(}\alpha\sum_{n=r+1}^{\infty}\frac{\alpha^{n}}{(n-1)!}(1)_{n+1,\lambda}+\sum_{n=r+1}^{\infty}\frac{n\alpha^{n}}{n!}(1)_{n,\lambda}\bigg{)}$
$\displaystyle\quad+\frac{1}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}r\frac{\alpha^{r+1}}{r!}(1)_{r+1,\lambda}$
$\displaystyle=\
\frac{\alpha}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\sum_{n=r+1}^{\infty}\frac{n\alpha^{n}}{n!}(1)_{n,\lambda}(1-n\lambda)+E[X_{\lambda,r}]$
$\displaystyle\quad+\frac{r\alpha^{r+1}(1)_{r+1,\lambda}}{r!\big{(}e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)\big{)}}$
$\displaystyle=\ \alpha E[X_{\lambda,r}]-\lambda\alpha
E[X_{\lambda,r}^{2}]+E[X_{\lambda,r}]+\frac{\alpha^{r+1}}{r!}\frac{r(1)_{r+1,\lambda}}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}.$

From (8), we note that

(9)
$E[X_{\lambda,r}^{2}]=\frac{\alpha+1}{1+\lambda\alpha}E[X_{\lambda,r}]+\frac{1}{1+\lambda\alpha}\frac{r(1)_{r+1,\lambda}}{e_{\lambda}(t)-e_{\lambda,r}(\alpha)}\frac{\alpha^{r+1}}{r!},$

where $\lambda\neq-\frac{1}{\alpha}$.

For $\lambda\neq-\frac{1}{\alpha}$, we note that the variance of
$X_{\lambda,r}$ is given by

(10) $\displaystyle\mathrm{Var}(X_{\lambda,r})\ =\
E[X_{\lambda,r}^{2}]-\big{(}E[X_{\lambda,r}]\big{)}^{2}$ $\displaystyle=\
E[X_{\lambda,r}]\bigg{(}\frac{\alpha+1}{1+\lambda\alpha}-E[X_{\lambda,r}]\bigg{)}+\frac{1}{1+\lambda\alpha}\frac{r(1)_{r+1,\lambda}}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\frac{\alpha^{r+1}}{r!}$
$\displaystyle=\
\frac{1}{1+\alpha\lambda}\bigg{(}\alpha+\frac{\alpha^{r+1}}{r!}\frac{(1)_{r+1,\lambda}}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\bigg{)}\bigg{(}\frac{\alpha+1}{1+\alpha\lambda}-\frac{1}{1+\alpha\lambda}\bigg{(}\alpha+\frac{\alpha^{r+1}}{r!}\frac{(1)_{r+1,\lambda}}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\bigg{)}\bigg{)}$
$\displaystyle\quad+\frac{1}{1+\alpha\lambda}\frac{r(1)_{r+1,\lambda}}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\frac{\alpha^{r+1}}{r!}$
$\displaystyle=\
\frac{1}{(1+\alpha\lambda)^{2}}\bigg{(}\alpha+\frac{\alpha^{r+1}}{r!}\frac{(1)_{r+1,\lambda}}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\bigg{)}\bigg{(}1-\frac{\alpha^{r+1}}{r!}\frac{(1)_{r+1,\lambda}}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\bigg{)}$
$\displaystyle\quad+\frac{1}{1+\alpha\lambda}\frac{r(1)_{r+1,\lambda}}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\frac{\alpha^{r+1}}{r!}.$

Therefore, by (10), we obtain the following theorem.

######  Theorem 2.

Let $X_{\lambda,r}$ be the $r$-truncated degenerate Poisson random variable
with parameter $\alpha>0$. For $\lambda\neq-\frac{1}{\alpha}$, we have

$\displaystyle\mathrm{Var}(X_{\lambda,r})$
$\displaystyle=\frac{1}{(1+\alpha\lambda)^{2}}\bigg{(}\alpha+\frac{\alpha^{r+1}}{r!}\frac{(1)_{r+1,\lambda}}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\bigg{)}\bigg{(}1-\frac{\alpha^{r+1}}{r!}\frac{(1)_{r+1,\lambda}}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\bigg{)}$
$\displaystyle+\frac{1}{1+\alpha\lambda}\frac{r(1)_{r+1,\lambda}}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\frac{\alpha^{r+1}}{r!}.$

Let us consider the generating function of the moments of the $r$-truncated
degenerate Poisson random variable with parameter $\alpha>0$. Then, from (4),
we have

(11) $\displaystyle\sum_{n=0}^{\infty}E[X_{\lambda,r}^{n}]\frac{t^{n}}{n!}\ $
$\displaystyle=\ E[e^{X_{\lambda,r}t}]\ =\
\sum_{m=r+1}^{\infty}e^{mt}p_{\lambda,r}(m)$ $\displaystyle=\
\frac{1}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\sum_{m=r+1}^{\infty}\frac{\alpha^{m}}{m!}(1)_{m,\lambda}e^{mt}$
$\displaystyle=\
\frac{1}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\bigg{(}e_{\lambda}(\alpha
e^{t})-\sum_{m=0}^{r}\frac{\alpha^{m}}{m!}(1)_{m,\lambda}e^{mt}\bigg{)}$
$\displaystyle=\
\frac{e_{\lambda}(\alpha)}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\bigg{(}e_{\lambda}^{-1}(\alpha)e_{\lambda}(\alpha
e^{t})-e_{\lambda}^{-1}(\alpha)\sum_{m=0}^{r}\frac{\alpha^{m}}{m!}(1)_{m,\lambda}e^{mt}\bigg{)}$
$\displaystyle=\
\frac{e_{\lambda}(\alpha)}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\bigg{(}\sum_{n=0}^{\infty}\mathrm{Bel}_{n,\lambda}(\alpha)\frac{t^{n}}{n!}-\sum_{n=0}^{\infty}e_{\lambda}^{-1}(\alpha)\sum_{m=0}^{r}\frac{\alpha^{m}}{m!}(1)_{m,\lambda}m^{n}\frac{t^{n}}{n!}\bigg{)}$
$\displaystyle=\
\sum_{n=0}^{\infty}\frac{e_{\lambda}(\alpha)}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\bigg{(}\mathrm{Bel}_{n,\lambda}(\alpha)-e_{\lambda}^{-1}(\alpha)\sum_{m=0}^{r}\frac{\alpha^{m}}{m!}(1)_{m,\lambda}m^{n}\bigg{)}\frac{t^{n}}{n!}.$

Thus, by (11), we get the next theorem.

######  Theorem 3.

Let $X_{\lambda,r}$ be the $r$-truncated degenerate Poisson random variable
with parameter $\alpha>0$. For $n\geq 0$, we have

$E[X_{\lambda,r}^{n}]=\frac{e_{\lambda}(\alpha)}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\bigg{(}\mathrm{Bel}_{n,\lambda}(\alpha)-e_{\lambda}^{-1}(\alpha)\sum_{m=0}^{r}\frac{\alpha^{m}}{m!}(1)_{m,\lambda}m^{n}\bigg{)}.$

For $x\geq r+1$, we note that the cumulative distribution function is given by

(12) $\displaystyle F_{X_{\lambda,r}}(x)\ $ $\displaystyle=\
P\\{X_{\lambda,r}\leq x\\}\ =\ \sum_{k=r+1}^{[x]}p_{\lambda,r}(k)$
$\displaystyle=\
\frac{1}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\sum_{k=r+1}^{[x]}\frac{\alpha^{k}}{k!}(1)_{k,\lambda}$
$\displaystyle=\
\frac{1}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\bigg{(}\sum_{k=0}^{[x]}\frac{\alpha^{k}}{k!}(1)_{k,\lambda}-\sum_{k=0}^{r}\frac{\alpha^{k}}{k!}(1)_{k,\lambda}\bigg{)}.$

From (12) and (2), we can derive the following equation.

(13)
$F_{X_{\lambda,r}}(x)=\frac{1}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\big{(}e_{\lambda,[x]}(\alpha)-e_{\lambda,r}(\alpha)\big{)}.$

######  Theorem 4.

Assume that $X_{\lambda,r}$ is the $r$-truncated degenerate Poisson random
variable with parameter $\alpha>0$. For $x\geq r+1$, the cumulative
distribution function of $X_{\lambda,r}$ is given by

$F_{X_{\lambda,r}}(x)=\frac{1}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\big{(}e_{\lambda,[x]}(\alpha)-e_{\lambda,r}(\alpha)\big{)}.$

Let us assume that
$X_{\lambda,r}^{(1)},X_{\lambda,r}^{(2)},\dots,X_{\lambda,r}^{(k)}$ are
identically independent $r$-truncated degenerate Poisson random variables with
parameter $\alpha>0$, and let

$X_{\lambda,r}=\sum_{i=1}^{k}X_{\lambda,r}^{(i)},\quad(k\in\mathbb{N}).$

From the probability generating function of random variable, we note that

(14) $\displaystyle E[t^{X_{\lambda,r}^{(i)}}]\ $ $\displaystyle=\
\sum_{n=r+1}^{\infty}t^{n}P[X_{\lambda,r}^{(i)}=n]$ $\displaystyle=\
\frac{1}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\sum_{n=r+1}^{\infty}\frac{\alpha^{n}}{n!}(1)_{n,\lambda}t^{n}$
$\displaystyle=\ \frac{e_{\lambda}(\alpha t)-e_{\lambda,r}(\alpha
t)}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}.$

By (14), we get

(15) $\displaystyle E[t^{X_{\lambda,r}}]\ $ $\displaystyle=\
\prod_{i=1}^{k}E[t^{X_{\lambda,r}^{(i)}}]$ $\displaystyle=\
\bigg{(}\frac{1}{e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha)}\bigg{)}^{k}k!\frac{1}{k!}\big{(}e_{\lambda}(\alpha
t)-e_{\lambda,r}(\alpha t)\big{)}^{k}$ $\displaystyle=\
\frac{k!}{(e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha))^{k}}\sum_{n=kr+k}^{\infty}S_{2,\lambda}^{[r]}(n,kr+k)\frac{\alpha^{n}t^{n}}{n!}.$

On the other hand,

(16) $\displaystyle E[t^{X_{\lambda,r}}]\ $ $\displaystyle=\
E[t^{X_{\lambda,r}^{(1)}+X_{\lambda,r}^{(2)}+\cdots+X_{\lambda,r}^{(k)}}]$
$\displaystyle=\
\sum_{n=kr+r}^{\infty}t^{n}P\bigg{[}\sum_{i=1}^{k}X_{\lambda,r}^{(i)}=n\bigg{]}.$

Therefore, by (15) and (16), we obtain the following theorem.

######  Theorem 5.

Let $X_{\lambda,r}^{(1)},X_{\lambda,r}^{(2)},\dots,X_{\lambda,r}^{(k)}$ be
identically independent $r$-truncated degenerate Poisson random variables with
parameter $\alpha>0$, and let $\displaystyle
X_{\lambda,r}=\sum_{i=1}^{k}X_{\lambda,r}^{(i)}\displaystyle$. Then the
probability for $X_{\lambda,r}$ is given by

$P[X_{\lambda,r}=n]=\left\\{\begin{array}[]{ccc}\cfrac{k!}{(e_{\lambda}(\alpha)-e_{\lambda,r}(\alpha))^{k}}\frac{\alpha^{n}}{n!}S_{2,\lambda}^{[r]}(n,kr+k),&\textrm{if
$n\geq kr+r$},\\\ 0,&\textrm{otherwise.}\end{array}\right.$

##  3\. Conclusion

We generalized the results on the zero-truncated degenerate Poisson
distribution to the case of the $r$-truncated degenerate Poisson distribution,
with its potential applications to the cornavirus pandemic and applications of
probabilistic methods to the study of some special numbers and polynomials in
mind.

Let $X_{\lambda,r}$ be the $r$-truncated degenerate Poisson random variable
with parameter $\alpha$. Then, for the random variable $X_{\lambda,r}$, we
derived its expectation, its variance, its $n$-th moment, and its cumulative
distribution function. In addition, we obtained two different expressions for
the probability generating function of a finite sum of independent
$r$-truncated degenerate Poisson random variables with equal parameters.
As one of our future projects, we would like to continue this line of
research, namely to explore applications of various methods of probability
theory to science, engineering and social science, and to the study of some
special polynomials and numbers.

Acknowledgements: Not applicable.

Funding: Not applicable.

Availability of data and material: Not applicable.

Competing interests: The authors declare no conflict of interest.

Author Contributions: D.S.K. and T.K. wrote the paper; L.-C.J. and S.-H. L.
and S.-H. P. checked the results of the paper and typed the paper. All authors
have read and agreed to the published version of the manuscript.

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