# A note on negative λ-binomial distribution

## Abstract

In this paper, we introduce one discrete random variable, namely the negative λ-binomial random variable. We deduce the expectation of the negative λ-binomial random variable. We also get the variance and explicit expression for the moments of the negative λ-binomial random variable.

## Introduction

In a sequence of independent Bernoulli trials, let the random variable X denote the trial at which the rth success occurs, where r is a fixed nonnegative integer. Then

$$P(X=x)={\binom{x-1}{r-1}}p^{r}(1-p)^{x-r}, \quad x=r,r+1,r+2,\ldots ,$$

and we say that X has a negative binomial distribution with parameters $$(r,p)$$ (see [13, 12, 13]).

The negative binomial distribution is sometimes defined in terms of the random variable Y, the number of failures before the rth success. This formulation is statistically equivalent to one given above in terms of X denoting the trial at which the rth success occurs, since $$Y=X-r$$. The alternative form of the negative binomial distribution is

$$p(k)=P(Y=k)={\binom{r+k-1}{k}}p^{r}(1-p)^{k},\quad k=0,1,2,\ldots ,$$

where p is the probability of success in the trial (see [1, 3, 12, 13]).

It is known that the degenerate exponential function is defined by

$$e_{\lambda }^{x}(t)=(1+\lambda t)^{\frac{x}{\lambda }}=\sum_{n=0}^{ \infty }(x)_{n,\lambda } \frac{t^{n}}{n!}, \quad \lambda \in \mathbb{R},$$
(1)

where

$$(x)_{0,\lambda }=1,\qquad (x)_{n,\lambda }=x(x-\lambda ) \cdots \bigl(x-(n-1) \lambda \bigr)\quad (n\ge 1)\ (\text{see [5--7, 10, 11]}).$$
(2)

Recently, λ-analogue of binomial coefficients was considered by Kim to be

$${\binom{x}{0}}_{\lambda }=1,\qquad { \binom{x}{n}}_{\lambda }= \frac{(x)_{n,\lambda }}{n!}= \frac{x(x-\lambda )\cdots (x-(n-1)\lambda )}{n!}\quad (n\ge 1)\ ( \text{see [6, 8, 9]}).$$
(3)

In this paper, we consider the negative λ-binomial distribution and obtain expressions for its moments.

## Negative λ-binomial distribution

### Definition 2.1

$$Y_{\lambda }$$ is the negative λ-binomial random variable if the probability mass function of $$Y_{\lambda }$$ with parameters $$(r,p)$$ is given by

$$p_{\lambda }(k)=P_{\lambda }(Y_{\lambda }=k)={ \binom{r+(k-1)\lambda }{k}}_{\lambda }e_{\lambda }^{r}(p-1) (1-p)^{k},$$
(4)

where λ (0,1) and p is the probability of success in the trials.

Note that

$${\binom{r+(k-1)\lambda }{k}}_{\lambda }=(-1)^{k}{ \binom{-r}{k}}_{\lambda },\quad k\ge 0\ (\text{see })$$
(5)

and

\begin{aligned} \sum_{n=0}^{\infty }p_{\lambda }(k) =& \sum_{n=0}^{\infty }{ \binom{r+(k-1)\lambda }{k}}_{\lambda }(1-p)^{k} e_{\lambda }^{r}(p-1) \\ =&e_{\lambda }^{r}(p-1)e_{\lambda }^{-r}(p-1)=1. \end{aligned}
(6)

From (4), we note that

$$\lim_{\lambda \rightarrow 1}p_{\lambda }(k)$$
(7)

is the probability mass function of negative binomial random variable with parameters $$(r,p)$$, and

$$\lim_{\lambda \rightarrow 0}p_{\lambda }(k)$$
(8)

is the probability mass function of Poisson random variable with parameters $$r(1-p)$$.

Let X be a discrete random variable, and let $$f(x)$$ be a real-valued function. Then we have

$$E\bigl(f(X)\bigr)=\sum_{x}f(x)p(x),$$
(9)

where $$p(x)$$ is the probability mass function.

From (9), we note that

\begin{aligned} E(Y_{\lambda }) =&\sum_{k=0}^{\infty }kp_{\lambda }(k)= \sum_{k=0}^{ \infty }k{\binom{r+(k-1)\lambda }{k}}_{\lambda }(1-p)^{k} e_{\lambda }^{r}(p-1) \\ =&\frac{r}{e_{\lambda }^{\lambda }(p-1)}\sum_{k=1}^{\infty } \frac{(r+(k-1)\lambda )\cdots (r+\lambda )}{(k-1)!}(1-p)^{k} e_{\lambda }^{r+\lambda }(p-1) \\ =&\frac{r}{e_{\lambda }^{\lambda }(p-1)}\sum_{k=0}^{\infty } \frac{(r+k\lambda )\cdots (r+\lambda )}{k!}(1-p)^{k+1} e_{\lambda }^{r+ \lambda }(p-1) \\ =&\frac{r(1-p)}{e_{\lambda }^{\lambda }(p-1)}\sum_{k=0}^{\infty }{ \binom{r+\lambda +(k-1)\lambda }{k}}_{\lambda }(1-p)^{k} e_{\lambda }^{r+ \lambda }(p-1) \\ =&\frac{r(1-p)}{e_{\lambda }^{\lambda }(p-1)}e_{\lambda }^{-(r+\lambda )}(p-1)e_{\lambda }^{r+\lambda }(p-1) \\ =&\frac{r(1-p)}{e_{\lambda }^{\lambda }(p-1)}. \end{aligned}
(10)

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

### Theorem 2.1

Let $$Y_{\lambda }$$ be a negative λ-binomial random variable with parameters $$(r,p)$$. Then we have

$$E(Y_{\lambda })=\frac{r(1-p)}{e_{\lambda }^{\lambda }(p-1)}.$$

### Note 2.1

$$\lim_{\lambda \rightarrow 1}E(Y_{\lambda })=\frac{r(1-p)}{p}=E(Y),$$

where Y is the negative binomial random variable with parameters $$(r,p)$$.

### Note 2.2

$$\lim_{\lambda \rightarrow 0}E(Y_{\lambda })=r(1-p)=E(Y),$$

where Y is the Poisson random variable with parameter $$r(1-p)$$.

Now, we observe that

\begin{aligned} E\bigl(Y^{2}_{\lambda }\bigr) =&\sum _{k=0}^{\infty }k^{2}p_{\lambda }(k)= \sum_{k=0}^{ \infty }k(k+1-1)p_{\lambda }(k) \\ =&\sum_{k=0}^{\infty }k(k-1)p_{\lambda }(k)+ \sum_{k=0}^{\infty }kp_{\lambda }(k) \\ =&\sum_{k=0}^{\infty }k(k-1){ \binom{r+(k-1)\lambda }{k}}_{\lambda }(1-p)^{k} e_{\lambda }^{r}(p-1)+E(Y_{\lambda }) \\ =&\frac{r(r+\lambda )}{e_{\lambda }^{2\lambda }(p-1)}\sum_{k=2}^{ \infty } \frac{(r+(k-1)\lambda )\cdots (r+2\lambda )}{(k-2)!}(1-p)^{k} e_{\lambda }^{r+2\lambda }(p-1)+E(Y_{\lambda }) \\ =&\frac{r(r+\lambda )}{e_{\lambda }^{2\lambda }(p-1)}\sum_{k=0}^{ \infty }{ \binom{r+(k+1)\lambda }{k}}_{\lambda }(1-p)^{k+2} e_{\lambda }^{r+2 \lambda }(p-1)+E(Y_{\lambda }) \\ =&\frac{r(r+\lambda )(1-p)^{2}}{e_{\lambda }^{2\lambda }(p-1)}\sum_{k=0}^{ \infty }{ \binom{r+2\lambda +(k-1)\lambda }{k}}_{\lambda }(1-p)^{k} e_{\lambda }^{r+2\lambda }(p-1)+E(Y_{\lambda }) \\ =&\frac{r(r+\lambda )(1-p)^{2}}{e_{\lambda }^{2\lambda }(p-1)}e_{\lambda }^{-(r+2\lambda )}(p-1)e_{\lambda }^{r+2\lambda }(p-1)+E(Y_{\lambda }) \\ =&\frac{r(r+\lambda )(1-p)^{2}}{e_{\lambda }^{2\lambda }(p-1)}+ \frac{r(1-p)}{e_{\lambda }^{\lambda }(p-1)}. \end{aligned}
(11)

The variance of random variable X is defined by

$$\operatorname{Var}(X)=E\bigl(X^{2}\bigr)-\bigl[E(X) \bigr]^{2}\quad (\text{see [1, 3]}).$$
(12)

From Theorem 2.1, (11), and (12), we note that

\begin{aligned} \operatorname{Var}(Y_{\lambda }) =&E\bigl(Y_{\lambda }^{2} \bigr)-\bigl[E(Y_{\lambda })\bigr]^{2} \\ =&\frac{r(r+\lambda )(1-p)^{2}}{e_{\lambda }^{2\lambda }(p-1)}+ \frac{r(1-p)}{e_{\lambda }^{\lambda }(p-1)}- \frac{r^{2}(1-p)^{2}}{e_{\lambda }^{2\lambda }(p-1)} \\ =&\frac{r(1-p)^{2}}{e_{\lambda }^{2\lambda }(p-1)}(r+\lambda -r)+ \frac{r(1-p)}{e_{\lambda }^{\lambda }(p-1)} \\ =&\lambda \frac{r(1-p)^{2}}{e_{\lambda }^{2\lambda }(p-1)}+ \frac{r(1-p)}{e_{\lambda }^{\lambda }(p-1)}. \end{aligned}

Therefore, we obtain the following theorem.

### Theorem 2.2

Let $$Y_{\lambda }$$ be a negative λ-binomial random variable with parameters $$(r,p)$$. Then we have

$$\operatorname{Var}(Y_{\lambda })=\lambda \frac{r(1-p)^{2}}{e_{\lambda }^{2\lambda }(p-1)}+ \frac{r(1-p)}{e_{\lambda }^{\lambda }(p-1)}.$$

### Note 2.3

$$\lim_{\lambda \rightarrow 1}\operatorname{Var}(Y_{\lambda })= \frac{r(1-p)}{p^{2}}=\operatorname{Var}(Y),$$

where Y is the negative binomial random variable with parameters $$(r,p)$$.

### Note 2.4

$$\lim_{\lambda \rightarrow 0}\operatorname{Var}(Y_{\lambda })=r(1-p)= \operatorname{Var}(Y),$$

where Y is the Poisson random variable with parameter $$r(1-p)$$.

Note that

$$k^{n}=\sum_{l=0}^{n}S_{2}(n,l) (k)_{l},$$
(13)

where $$S_{2}(n,l)$$ is the Stirling number of the second kind, and

$$(k)_{0}=1,\qquad (k)_{l}=k(k-1)\cdots (k-l+1)\quad (l\ge 1)\ (\text{see [14, 15]}).$$

From (13), we note that

\begin{aligned} E\bigl(Y^{n}_{\lambda }\bigr) =&\sum _{k=0}^{\infty }k^{n}p_{\lambda }(k)= \sum_{l=0}^{n}S_{2}(n,l) \sum_{k=l}^{\infty }(k)_{l} p_{\lambda }(k) \\ =&\sum_{l=0}^{n}S_{2}(n,l) \sum_{k=l}^{\infty }(k)_{l} { \binom{r+(k-1)\lambda }{k}}_{\lambda }(1-p)^{k} e_{\lambda }^{r}(p-1) \\ =&\sum_{l=0}^{n}S_{2}(n,l) \frac{r(r+\lambda )\cdots (r+(l-1)\lambda )}{e_{\lambda }^{l\lambda }(p-1)} \\ &{}\times\sum_{k=l}^{\infty } \frac{(r+(k-1)\lambda )\cdots (r+l\lambda )}{(k-l)!}(1-p)^{k} e_{\lambda }^{r+l\lambda }(p-1) \\ =&\sum_{l=0}^{n}S_{2}(n,l) \frac{r(r+\lambda )\cdots (r+(l-1)\lambda )}{e_{\lambda }^{l\lambda }(p-1)} \\ &{}\times\sum_{k=0}^{\infty } \frac{(r+(k+l-1)\lambda )\cdots (r+l\lambda )}{k!}(1-p)^{k+l} e_{\lambda }^{r+l\lambda }(p-1) \\ =&\sum_{l=0}^{n}S_{2}(n,l) \frac{r(r+\lambda )\cdots (r+(l-1)\lambda )}{e_{\lambda }^{l\lambda }(p-1)} \\ &{}\times\sum_{k=0}^{\infty }{ \binom{r+(k+l-1)\lambda }{k}}_{\lambda }(1-p)^{k+l}e_{\lambda }^{r+l\lambda }(p-1) \\ =&\sum_{l=0}^{n}S_{2}(n,l) \frac{r(r+\lambda )\cdots (r+(l-1)\lambda )(1-p)^{l}}{e_{\lambda }^{l\lambda }(p-1)} \\ &{}\times\sum_{k=0}^{\infty }{ \binom{r+l\lambda + (k-1)\lambda }{k}}_{\lambda }(1-p)^{k} e_{\lambda }^{r+l\lambda }(p-1) \\ =&\sum_{l=0}^{n}S_{2}(n,l) \frac{r(r+\lambda )\cdots (r+(l-1)\lambda )(1-p)^{l}}{e_{\lambda }^{l\lambda }(p-1)} e_{\lambda }^{-r-l\lambda }(p-1) e_{\lambda }^{r+l\lambda }(p-1) \\ =&\sum_{l=0}^{n}S_{2}(n,l) \frac{r(r+\lambda )\cdots (r+(l-1)\lambda )(1-p)^{l}}{e_{\lambda }^{l\lambda }(p-1)} \\ =&\sum_{l=0}^{n}S_{2}(n,l) \frac{(r+(l-1)\lambda )_{l,_{\lambda }}(1-p)^{l}}{e_{\lambda }^{l\lambda }(p-1)}. \end{aligned}

Therefore, we obtain the following theorem.

### Theorem 2.3

Let $$Y_{\lambda }$$ be a negative λ-binomial random variable with parameters $$(r,p)$$. Then we have

$$E\bigl(Y^{n}_{\lambda }\bigr)=\sum _{l=0}^{n}S_{2}(n,l) \frac{(r+(l-1)\lambda )_{l,_{\lambda }}(1-p)^{l}}{e_{\lambda }^{l\lambda }(p-1)}.$$

### Note 2.5

$$\lim_{\lambda \rightarrow 1}E\bigl(Y^{n}_{\lambda }\bigr)= \sum_{l=0}^{n}S_{2}(n,l) \frac{(r+(l-1))_{l}(1-p)^{l}}{p^{l}}=E\bigl(Y^{n}\bigr),$$

where Y is the negative binomial random variable with parameters $$(r,p)$$ (see [4, 12]).

### Note 2.6

$$\lim_{\lambda \rightarrow 0}E\bigl(Y^{n}_{\lambda }\bigr)= \sum_{l=0}^{n}S_{2}(n,l) \bigl(r(1-p)\bigr)^{l}=E\bigl(Y^{n}\bigr),$$

where Y is the Poisson random variable with parameter $$r(1-p)$$ (see ).

Note that

\begin{aligned} E\bigl(Y^{n}_{\lambda }\bigr) =&\sum _{k=0}^{\infty }k^{n} p_{\lambda }(k) \\ =&\sum_{k=0}^{\infty }k^{n}{ \binom{r+(k-1)\lambda }{k}}_{\lambda }(1-p)^{k} e_{\lambda }^{r}(p-1) \\ =&\sum_{k=1}^{\infty }k^{n-1} \frac{(r+(k-1)\lambda )\cdots (r+\lambda )r}{(k-1)!}(1-p)^{k} e_{\lambda }^{r}(p-1) \\ =&\sum_{k=0}^{\infty }(k+1)^{n-1} \frac{(r+k\lambda )\cdots (r+\lambda )r}{k!}(1-p)^{k+1} e_{\lambda }^{r}(p-1) \\ =&r(1-p)\sum_{k=0}^{\infty }\sum _{i=0}^{n-1}{\binom{n-1}{i}}k^{i} \frac{(r+k\lambda )\cdots (r+\lambda )}{k!}(1-p)^{k} e_{\lambda }^{r}(p-1) \\ =&\frac{r(1-p)}{e_{\lambda }^{\lambda }(p-1)}\sum_{i=0}^{n-1}{ \binom{n-1}{i}}\sum_{k=0}^{\infty }k^{i}{ \binom{r+\lambda +(k-1)\lambda }{k}}_{\lambda }(1-p)^{k} e_{\lambda }^{r+ \lambda }(p-1) \\ =&\frac{r(1-p)}{e_{\lambda }^{\lambda }(p-1)}\sum_{i=0}^{n-1}{ \binom{n-1}{i}}E\bigl(Z^{i}_{\lambda }\bigr), \end{aligned}

where $$Z_{\lambda }$$ is the negative λ-binomial random variable with parameters $$(r+\lambda ,p)$$.

Therefore, we obtain the following theorem.

### Theorem 2.4

Let $$Y_{\lambda }$$, $$Z_{\lambda }$$ be two negative λ-binomial random variables with parameters $$(r,p)$$, $$(r+\lambda ,p)$$ respectively. Then we have

$$E\bigl(Y^{n}_{\lambda }\bigr)=\frac{r(1-p)}{e_{\lambda }^{\lambda }(p-1)}\sum _{i=0}^{n-1}{ \binom{n-1}{i}}E \bigl(Z^{i}_{\lambda }\bigr).$$

## Conclusion

In this paper, we introduced one discrete random variable, namely the negative λ-binomial random variable. The details and results are as follows. We defined the negative λ-binomial random variable with parameter $$(r,p)$$ in (4) and deduced its expectation in Theorem 2.1. We also obtained its variance in Theorem 2.2 and derived explicit expression for the moment of the negative λ-binomial random variable in Theorem 2.3.

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### Acknowledgements

The authors thank Jangjeon Institute for Mathematical Science for the support of this research.

Not applicable.

## Funding

This research was funded by the National Natural Science Foundation of China (No. 11871317, 11926325, 11926321).

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All authors contributed equally to the manuscript and typed, read, and approved the final manuscript.

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Correspondence to Taekyun Kim.

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