Partial sums of analytic functions defined by binomial distribution and negative binomial distribution

ABSTRACT

The study of statistical distributions in a complex variable is one of the most vibrant areas of research. The complex analogue of many distributions has been studied. This article introduces the complex analogue of Binomial distribution and Negative Binomial distribution and studies certain geometrical properties of these analogues. It comprises the study of analytic functions defined by using the complex versions of Binomial distribution and Negative Binomial distribution functions. It includes the problems of finding the lower bounds of real parts of certain ratios of partial sums to the infinite series sums for these analytic functions. Such lower bounds are determined for the said analytic functions, for their first derivatives and for the Alexander transformation of these analytic functions.

KEYWORDS:

• Analytic functions
• statistical distribution
• binomial distribution
• negative binomial distributions
• partial sums

MATHEMATICS SUBJECT CLASSIFICATIONS:

• 30A10
• 30B10
• 30C50
• 30C55
• 60E05

1. Introduction

The probability distributions are widely applied in the field of mathematics, ecology, physics, machine learning, economics, artificial intelligence, biomedical, game theory and many more. Their strong relationship with the other fields attracts the researchers to develop new dimensions of research. Certain discrete distributions play an important role in mathematics and other fields. Many scientists and researchers have studied these distributions due to their noteworthy properties and their involvements in many areas of emerging sciences.

The initial studies about statistical distributions in complex variables are relatively a new area of research. For the first time, Wooding defined the analogue of the complex Gaussian distribution in 1956 that was multivariate in nature. That was carried by Goodman, who in 1963 discussed similar aspects of the normal distribution but again in complex multivariate. He also considered the analogue of the multiple and partial correlation along with the Wishart distribution. Afterwards, this area became a well-known area of research and many mathematician and statisticians including Brillinger, Dubman, and Young contributed to it by giving some practical uses of the complex analogue. The other remarkable contributions include Capon and Goodman, Giri, Kabe, Khatri, Saxena, Srivastava, Tan, and Young, who discussed different complex analogues of the classical real variable multivariate analysis. For more details and for some recent references on complex random variables, see [1] and references therein. This study of complex probability distributions was extended for Poisson distribution as well. The geometrical properties of these distributions have also been extensively studied by researchers and this has been a core area of interest for geometric function theorists. For references, see [2–10] in which the derivations of certain geometric properties were carried out to establish the connections between Geometric Function Theory and Statistics by means of generalized discrete probability distributions. To find the partial sum and certain ratios of partial sums to their derivatives is one of the important problems associated with the series of analytic functions. Several researchers like Silvia [11], Silverman [12], Brickman et al. [13], Liu and Owa [14], Orhan and Yagmur [15] have investigated bounds of the real of these ratios of partial sums of Bessel functions. Recently, partial sums of Wright functions were studied and discussed in [16]. Motivated by the above-mentioned work, this paper includes the investigation of the ratios of analytic functions related to binomial distribution and negative binomial distribution functions to their sequence of partial sums.

In 1936, Ronald Fisher, a British statistician, had made use of binomial distribution in studying the possibility of scientific chicanery in the famous experiments on pea genetics. The term ‘negative binomial distribution’ is due to the fact that a binomial with a negative exponent generates this distribution. Negative binomial distribution is one of the most important discrete distribution which is useful in fitting data, obtained in biological and medical research, see [17]. This distribution is also known as Pascal distribution and it also appears in the study of randomization of parameters of a distribution, see [18].

Consider the series form of complex valued functions as, (1) $f\left(z\right)=z+\sum _{j=2}^{\mathrm{\infty }}{a}_j{z}^j,$(1) which are analytic in $E=\{z:z\in \mathbb{C},|z|<1\}$ and hold the normalization condition $f\left(0\right)=f^{\mathrm{\prime }}\left(0\right)-1=0$. For more about analytic functions, the interested readers are referred to [19–25] and the references therein. Alexander transform of an analytic function $f\left(z\right)$ is defined by $A\left[f\right]\left(z\right)={\int }_0^z\frac{f\left(t\right)}{t}\mathrm{d}t=z+\sum _{j=2}^{\mathrm{\infty }}\frac{a_j}{j}{z}^j.$ The core aim of this research is to provide some results of certain analytic functions connected with binomial and negative Binomial distribution and use the concept of partial sums to obtain the lower bound of the probability mass function of these distributions. This research will show the strong correlation of the discrete probability distributions and Geometric Functions Theory and it opens up an interesting and new direction of research in the theory of analytic functions.

2. Main results

Silvia [11] was the first who gave the concept of finding the lower bound of the real part of the ratio of the partial sum of analytic functions to its infinite series sum. Silverman [12] developed more useful techniques to find these bounds. He found the partial sums of convex and starlike functions. After that, several researchers investigated such partial sums for differential subclasses of analytic functions. For more work on partial sums, the interested readers are referred to [13–16, 26–31]. This article intends to find certain bounds related to partial sums of analytic functions which are defined by using the Poisson and Geometric distributions.

2.1. Partial sums of binomial distribution:

One of the most essential discrete probability distributions is Binomial distribution. When there are two possible outcomes, then Binomial distribution model is used which is an important probability model. In a Bernoulli trial, the random experiment has two hypothetical results that are success or failure. If the number of trials m = 1, then it is called Bernoulli distribution that is special case for the Binomial distribution. Binomial distribution determines the probability of successful outcomes. It has two parameters, m and p where m denotes the number of trial and p denotes the success outcomes.

Binomial distribution has the following properties.

1. For Binomial experiment, the probability of success trial must be constant from trial to trial.

2. Successive trials must be independent.

3. The experiment is repeated for a fixed number of times.

4. For this distribution, the sample from a finite or infinite population by using the replacement method is drawn.

Mathematically, the expression of Binomial distribution is $p(X=x)={\displaystyle (\genfrac{}{}{0ex}{}{m}{x})}{p}^x{q}^{m-x},x=0,1,2,3\dots ,m,$ where m denotes the repetition of the experiment and q denotes the probability of failure with $q=1-p.$ The binomial distribution can be used in test of the goodness. This distribution was discovered early in the eighteenth century formulated by Swiss Mathematician Takob Bernoulli, see [32]. Here, the partial sums of Binomial distribution are discussed. For this, consider the following series form of Binomial distribution function. (2) $f(z,m;p)=z+\sum _{j=2}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{m}{j-1})p^{j-1}{q}^{m-j+1}{z}^j,z\in E.$(2) The partial sum of the above series is defined as (3) $f(z,m;p{)}_n=z+\sum _{j=2}^n(\genfrac{}{}{0ex}{}{m}{j-1})p^{j-1}{q}^{m-j+1}{z}^j.$(3) Consider the Alexander transform of the function $f(z,m;p)$ as defined by (4) $Af(z,m;p)=z+\sum _{j=2}^{\mathrm{\infty }}\frac{1}{j}(\genfrac{}{}{0ex}{}{m}{j-1})p^{j-1}{q}^{m-j+1}{z}^j.$(4) Now, compute the lower bounds of certain ratios of these partial sums like $\mathrm{\Re }\left\{\frac{f(z,m;p)}{f(z,m;p{)}_n}\right\},$ $\mathrm{\Re }\left\{\frac{f(z,m;p{)}_n}{f(z,m;p)}\right\},$ $\mathrm{\Re }\left\{\frac{f^{\mathrm{\prime }}(z,m;p)}{f^{\mathrm{\prime }}(z,m;p{)}_n}\right\},$ $\mathrm{\Re }\left\{\frac{f^{\mathrm{\prime }}(z,m;p{)}_n}{f^{\mathrm{\prime }}(z,m;p)}\right\},$ $\mathrm{\Re }\left\{\frac{Af(z,m;p)}{Af(z,m;p{)}_n}\right\},$ $\mathrm{\Re }\left\{\frac{Af(z,m;p{)}_n}{Af(z,m;p)}\right\}.$ For this, we need to establish the following lemmas.

Lemma 2.1

Let $m>0,$ $0\le p\le 1$. Then, the function $f(z,m;p):E\to \mathbb{C}$ defined by (2(2) $f(z,m;p)=z+\sum _{j=2}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{m}{j-1})p^{j-1}{q}^{m-j+1}{z}^j,z\in E.$(2) ) satisfies the following inequalities. (5) $\begin{array}{rl}& \left(1\right)\left|f\right(z,m;p\left)\right|\le 1+\mathrm{\Phi }\frac{p{q}^m}{1-2p},\end{array}$(5) (6) $\begin{array}{rl}& \left(2\right)\left|f^{\mathrm{\prime }}\right(z,m;p\left)\right|\le 1+\frac{\mathrm{\Phi }p{q}^m(2q-p)}{{(p-q)}^2},\end{array}$(6) (7) $\begin{array}{rl}& \left(3\right)\left|Af\right(z,m;p\left)\right|\le 1+\xi \frac{p{q}^m}{1-2p},\end{array}$(7) where $\mathrm{\Phi }=\frac{m!}{2^{m-2}}$ and $\xi =\frac{m!}{2^{m-1}}.$

Proof.

$\left(1\right)$ From (2(2) $f(z,m;p)=z+\sum _{j=2}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{m}{j-1})p^{j-1}{q}^{m-j+1}{z}^j,z\in E.$(2) ), one can have $\begin{array}{rl}\left|f\right(z,m;p\left)\right|& \le 1+\sum _{j=2}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{m}{j-1})p^{j-1}{q}^{m-j+1}\\ & =1+\sum _{j=2}^{\mathrm{\infty }}\frac{m!}{(j-1)!(m-j+1)!}{p}^{j-1}{q}^{m-j+1}.\end{array}$ It is noted that $(j-1)!\ge 2^{j-2},$ which gives $\frac{1}{(j-1)!}\le \frac{1}{2^{j-2}}.$ This implies that $\frac{1}{(m-j+1)!}\le \frac{1}{2^{m-j}}.$ Using these, one can have $\begin{array}{rl}\left|f\right(z,m;p\left)\right|& \le 1+\frac{m!}{2^{m-2}}\sum _{j=2}^{\mathrm{\infty }}{p}^{j-1}{q}^{m-j+1}\\ & =1+\frac{m!}{2^{m-2}}\frac{p{q}^m}{1-2p}.\end{array}$ $\left(2\right)$ From (2(2) $f(z,m;p)=z+\sum _{j=2}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{m}{j-1})p^{j-1}{q}^{m-j+1}{z}^j,z\in E.$(2) ), one can have $\begin{array}{rl}\left|f^{\mathrm{\prime }}\right(z,m;p\left)\right|& \le 1+\sum _{j=2}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{m}{j-1})j{p}^{j-1}{q}^{m-j+1}\\ & =1+\frac{m!}{2^{m-2}}\frac{q^{m+1}}{p}\sum _{j=2}^{\mathrm{\infty }}j{\left(\frac{p}{q}\right)}^j.\end{array}$ Now using the series (8) $\frac{1}{{(1-x)}^2}=1+2x+3x+\dots ,$(8) from which, one can have $\frac{\frac{p}{q}}{{(1-\frac{p}{q})}^2}=\frac{p}{q}+2{\left(\frac{p}{q}\right)}^2+3{\left(\frac{p}{q}\right)}^3+\dots ,$ which reduces to $\begin{array}{rl}\frac{-p^2(p-2q)}{q{(q-p)}^2}& =2{\left(\frac{p}{q}\right)}^2+3{\left(\frac{p}{q}\right)}^3+\dots \\ & =\sum _{j=2}^{\mathrm{\infty }}j{\left(\frac{p}{q}\right)}^j.\end{array}$ Thus, we have $\begin{array}{rl}\left|f^{\mathrm{\prime }}\right(z,m;p\left)\right|& \le 1+\frac{m!}{2^{m-2}}\frac{q^{m+1}}{p}\left(\frac{-p^2(p-2q)}{q{(q-p)}^2}\right)\\ & =1+\frac{m!}{2^{m-2}}\left(\frac{p{q}^m(2q-p)}{{(q-p)}^2}\right).\end{array}$ $\left(3\right)$ From (4(4) $Af(z,m;p)=z+\sum _{j=2}^{\mathrm{\infty }}\frac{1}{j}(\genfrac{}{}{0ex}{}{m}{j-1})p^{j-1}{q}^{m-j+1}{z}^j.$(4) ), one can have $\begin{array}{rl}\left|Af\right(z,m;p\left)\right|& \le 1+\sum _{j=2}^{\mathrm{\infty }}\frac{1}{j}(\genfrac{}{}{0ex}{}{m}{j-1})p^{j-1}{q}^{m-j+1}\\ & =1+\frac{m!}{2^{m-1}}\sum _{j=2}^{\mathrm{\infty }}{p}^{j-1}{q}^{m-j+1}\\ & =1+\frac{m!}{2^{m-1}}\frac{p{q}^m}{q-p}\\ & 1+\frac{m!}{2^{m-1}}\frac{p{q}^m}{1-2p}.\text{■}\end{array}$

Lemma 2.2

Let $m>0,0\le p\le 1$. Then the function $f(z,m;p):E\to \mathbb{C}$ defined by (2(2) $f(z,m;p)=z+\sum _{j=2}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{m}{j-1})p^{j-1}{q}^{m-j+1}{z}^j,z\in E.$(2) ) satisfies the following inequalities. $\begin{array}{rl}& \left(1\right)\frac{1-2p}{\mathrm{\Phi }p{q}^m}\sum _{j=2}^{\mathrm{\infty }}\left|a_j\right|\le 1,\\ & \left(2\right)\frac{{(p-q)}^2}{\mathrm{\Phi }p{q}^m(2q-p)}\sum _{j=2}^{\mathrm{\infty }}j\left|a_j\right|\le 1,\\ & \left(3\right)\frac{1-2p}{\xi p{q}^m}\sum _{j=2}^{\mathrm{\infty }}\frac{\left|a_j\right|}{j}\le 1,\end{array}$ where $\mathrm{\Phi }=\frac{m!}{2^{m-2}}$ and $\xi =\frac{m!}{2^{m-1}}.$

Proof.

$\left(1\right)$ From (5(5) $\begin{array}{rl}& \left(1\right)\left|f\right(z,m;p\left)\right|\le 1+\mathrm{\Phi }\frac{p{q}^m}{1-2p},\end{array}$(5) ), one can have $1+\sum _{j=2}^{\mathrm{\infty }}\left|a_j\right|\le 1+\mathrm{\Phi }\frac{p{q}^m}{1-2p},$ where $a_j=(\genfrac{}{}{0ex}{}{m}{j-1})p^{j-1}{q}^{m-j+1}.$ This implies that $\frac{1-2p}{\mathrm{\Phi }p{q}^m}\sum _{j=2}^{\mathrm{\infty }}\left|a_j\right|\le 1.$ $\left(2\right)$ From (6(6) $\begin{array}{rl}& \left(2\right)\left|f^{\mathrm{\prime }}\right(z,m;p\left)\right|\le 1+\frac{\mathrm{\Phi }p{q}^m(2q-p)}{{(p-q)}^2},\end{array}$(6) ), one can have $1+\sum _{j=2}^{\mathrm{\infty }}j\left|a_j\right|\le 1+\mathrm{\Phi }\left(\frac{p{q}^m(2q-p)}{{(q-p)}^2}\right),$ where $a_j=(\genfrac{}{}{0ex}{}{m}{j-1})p^{j-1}{q}^{m-j+1}.$ This implies that $\frac{{(q-p)}^2}{\mathrm{\Phi }p{q}^m(2q-p)}\sum _{j=2}^{\mathrm{\infty }}j\left|a_j\right|\le 1.$ $\left(3\right)$ From (6(6) $\begin{array}{rl}& \left(2\right)\left|f^{\mathrm{\prime }}\right(z,m;p\left)\right|\le 1+\frac{\mathrm{\Phi }p{q}^m(2q-p)}{{(p-q)}^2},\end{array}$(6) ), one can have $1+\sum _{j=2}^{\mathrm{\infty }}\frac{\left|a_j\right|}{j}\le 1+\xi \frac{p{q}^m}{1-2p},$ where $a_j=(\genfrac{}{}{0ex}{}{m}{j-1})p^{j-1}{q}^{m-j+1}.$ This implies that $\frac{1-2p}{\xi p{q}^m}\sum _{j=2}^{\mathrm{\infty }}\frac{\left|a_j\right|}{j}\le 1.\text{■}$

Theorem 2.3

Let $m\in \mathbb{N}$ with m>0 and $0\le p\le 1$. Then, $\begin{array}{rl}& \left(1\right)\mathrm{\Re }\left\{\frac{f(z,m;p)}{f(z,m;p{)}_n}\right\}\ge \frac{1-2p-\mathrm{\Phi }p{q}^m}{1-2p},\\ & \left(2\right)\mathrm{\Re }\left\{\frac{f(z,m;p{)}_n}{f(z,m;p)}\right\}\ge \frac{1-2p}{1-2p+\mathrm{\Phi }p{q}^m},\end{array}$ where $\mathrm{\Phi }=\frac{m!}{2^{m-2}}.$

Proof.

(1) Consider the function $\begin{array}{rl}h\left(z\right)& =\frac{1-2p}{\mathrm{\Phi }p{q}^m}\{\frac{f(z,m;p)}{f(z,m;p{)}_n}-(1-\frac{\mathrm{\Phi }p{q}^m}{1-2p})\}\\ & =\frac{1-2p}{\mathrm{\Phi }p{q}^m}\{\frac{1+\sum _{j=2}^{\mathrm{\infty }}{a}_j{z}^{j-1}}{1+\sum _{j=2}^n{a}_j{z}^{j-1}}-(1-\frac{\mathrm{\Phi }p{q}^m}{1-2p})\},\end{array}$ where $a_j=(\genfrac{}{}{0ex}{}{m}{j-1})p^{j-1}{q}^{m-j+1}.$ This implies that $\begin{array}{rl}h\left(z\right)& =\frac{\frac{1-2p}{\mathrm{\Phi }p{q}^m}\sum _{j=n+1}^{\mathrm{\infty }}{a}_j{z}^{j-1}+1+\sum _{j=2}^n{a}_j{z}^{j-1}}{1+\sum _{j=2}^n{a}_j{z}^{j-1}}\\ & =\frac{1+w\left(z\right)}{1-w\left(z\right)},\left(\mathrm{s}\mathrm{a}\mathrm{y}\right).\end{array}$ This implies that $\begin{array}{rl}w\left(z\right)& =\frac{h\left(z\right)-1}{h\left(z\right)+1}\\ & =\frac{\frac{1-2p}{\mathrm{\Phi }p{q}^m}\sum _{j=n+1}^{\mathrm{\infty }}{a}_j{z}^{j-1}}{2+2\sum _{j=2}^n{a}_j{z}^{j-1}+\frac{1-2p}{\mathrm{\Phi }p{q}^m}\sum _{j=n+1}^{\mathrm{\infty }}{a}_j{z}^{j-1}}.\end{array}$ Now applying $\left|w\right(z\left)\right|\le 1$, we have $\left|\frac{\frac{1-2p}{\mathrm{\Phi }p{q}^m}\sum _{j=n+1}^{\mathrm{\infty }}{a}_j{z}^{j-1}}{2+2\sum _{j=2}^n{a}_j{z}^{j-1}+\frac{1-2p}{\mathrm{\Phi }p{q}^m}\sum _{j=n+1}^{\mathrm{\infty }}{a}_j{z}^{j-1}}\right|\le 1,$ which implies that $\frac{1-2p}{\mathrm{\Phi }p{q}^m}\sum _{j=n+1}^{\mathrm{\infty }}\left|a_j\right|\le 2-2\sum _{j=2}^n\left|a_j\right|-\frac{1-2p}{\mathrm{\Phi }p{q}^m}\sum _{j=n+1}^{\mathrm{\infty }}\left|a_j\right|.$ That is, we get $\frac{1-2p}{\mathrm{\Phi }p{q}^m}\sum _{j=n+1}^{\mathrm{\infty }}\left|a_j\right|+\sum _{j=2}^n\left|a_j\right|\le 1.$ Since the left-hand side of the above expression is bounded above by $\frac{1-2p}{\mathrm{\Phi }p{q}^m}\sum _{j=2}^{\mathrm{\infty }}\left|a_j\right|\le 1,$by Lemma 2.2$\left(1\right)$. This completes the proof.

$\left(2\right)$ Consider the function $\begin{array}{rl}h\left(z\right)& =\frac{1-2p+\mathrm{\Phi }p{q}^m}{\mathrm{\Phi }p{q}^m}\{\frac{f(z,m;p{)}_n}{f(z,m;p)}-\frac{1-2p}{1-2p+\mathrm{\Phi }p{q}^m}\}\\ & =\frac{1-2p+\mathrm{\Phi }p{q}^m}{\mathrm{\Phi }p{q}^m}\{\frac{1+\sum _{j=2}^n{a}_j{z}^{j-1}}{1+\sum _{j=2}^{\mathrm{\infty }}{a}_j{z}^{j-1}}-\frac{1-2p}{1-2p+\mathrm{\Phi }p{q}^m}\}\\ & =\frac{1+\sum _{j=2}^n{a}_j{z}^{j-1}-\frac{1-2p}{\mathrm{\Phi }p{q}^m}\sum _{j=n+1}^{\mathrm{\infty }}{a}_j{z}^{j-1}}{1+\sum _{j=2}^{\mathrm{\infty }}{a}_j{z}^{j-1}}\\ & =\frac{1+w\left(z\right)}{1-w\left(z\right)},\left(\mathrm{s}\mathrm{a}\mathrm{y}\right).\end{array}$ This implies that $\begin{array}{rl}w\left(z\right)& =\frac{h\left(z\right)-1}{h\left(z\right)+1}\\ & =\frac{(-\frac{1-2p}{\mathrm{\Phi }p{q}^m}-1)\sum _{j=n+1}^{\mathrm{\infty }}{a}_j{z}^{j-1}}{2+2\sum _{j=2}^n{a}_j{z}^{j-1}-(\frac{1-2p}{\mathrm{\Phi }p{q}^m}-1)\sum _{j=n+1}^{\mathrm{\infty }}{a}_j{z}^{j-1}}.\end{array}$ Now applying $\left|w\right(z\left)\right|\le 1$, we have $\left|\frac{(-\frac{1-2p}{\mathrm{\Phi }p{q}^m}-1)\sum _{j=n+1}^{\mathrm{\infty }}{a}_j{z}^{j-1}}{2+2\sum _{j=2}^n{a}_j{z}^{j-1}-(\frac{1-2p}{\mathrm{\Phi }p{q}^m}-1)\sum _{j=n+1}^{\mathrm{\infty }}{a}_j{z}^{j-1}}\right|\le 1,$ which reduces to $\frac{1-2p}{\mathrm{\Phi }p{q}^m}\sum _{j=n+1}^{\mathrm{\infty }}\left|a_j\right|+\sum _{j=2}^n\left|a_j\right|\le 1.$ Since the left-hand side of above expression is bounded above by $\frac{1-2p}{\mathrm{\Phi }p{q}^m}\sum _{j=2}^{\mathrm{\infty }}\left|a_j\right|\le 1,$by Lemma 2.2(1). This completes the proof.

Theorem 2.4

Let $m\in \mathbb{N}$ with m>0 and $0\le p\le 1$. Then $\begin{array}{rl}& \left(1\right)\mathrm{\Re }\left\{\frac{f^{\mathrm{\prime }}(z,m;p)}{f^{\mathrm{\prime }}(z,m;p{)}_n}\right\}\ge 1+\frac{\mathrm{\Phi }p{q}^m(p-2q)}{{(q-p)}^2},\\ & \left(2\right)\mathrm{\Re }\left\{\frac{f^{\mathrm{\prime }}(z,m;p{)}_n}{f^{\mathrm{\prime }}(z,m;p)}\right\}\ge \frac{{(q-p)}^2}{{(q-p)}^2-\mathrm{\Phi }p{q}^m(p-2q)},\end{array}$ where $\mathrm{\Phi }=\frac{m!}{2^{m-2}}.$

Proof.

$\left(1\right)$ Consider the function $\begin{array}{rl}h\left(z\right)& =\frac{{(q-p)}^2}{\mathrm{\Phi }p{q}^m(2q-p)}\{\frac{f^{\mathrm{\prime }}(z,m;p)}{f^{\mathrm{\prime }}(z,m;p{)}_n}-(1+\frac{\mathrm{\Phi }p{q}^m(p-2q)}{{(q-p)}^2})\}\\ & =\frac{{(q-p)}^2}{\mathrm{\Phi }p{q}^m(2q-p)}\{\frac{1+\sum _{j=2}^{\mathrm{\infty }}j{a}_j{z}^{j-1}}{1+\sum _{j=2}^{n}j{a}_j{z}^{j-1}}-(1+\frac{\mathrm{\Phi }p{q}^m(p-2q)}{{(q-p)}^2})\}\\ & =\frac{1+\sum _{j=2}^{n}j{a}_j{z}^{j-1}-\frac{{(q-p)}^2}{\mathrm{\Phi }p{q}^m(2q-p)}\sum _{j=n+1}^{\mathrm{\infty }}j{a}_j{z}^{j-1}}{1+\sum _{j=2}^{n}j{a}_j{z}^{j-1}}\\ & =\frac{1+w\left(z\right)}{1-w\left(z\right)},\left(\mathrm{s}\mathrm{a}\mathrm{y}\right).\end{array}$ This gives that $\begin{array}{rl}w\left(z\right)& =\frac{h\left(z\right)-1}{h\left(z\right)+1}\\ & =\frac{\frac{{(q-p)}^2}{\mathrm{\Phi }p{q}^m(2q-p)}\left(\sum _{j=n+1}^{\mathrm{\infty }}j{a}_j{z}^{j-1}\right)}{2+2\left(\sum _{j=2}^{n}j{a}_j{z}^{j-1}\right)-\frac{{(q-p)}^2}{\mathrm{\Phi }p{q}^m(2q-p)}\left(\sum _{j=n+1}^{\mathrm{\infty }}j{a}_j{z}^{j-1}\right)}.\end{array}$ Now applying the fact that $w\left(z\right)\le 1,$ we have $\frac{{(q-p)}^2}{\mathrm{\Phi }p{q}^m(2q-p)}\left(\sum _{j=n+1}^{\mathrm{\infty }}j\left|a_j\right|\right)+\left(\sum _{j=2}^{n}j\left|a_j\right|\right)\le 1.$ Since the left-hand side of the above expression is bounded above by$\frac{(q-p{)}^2}{\mathrm{\Phi }p{q}^m(2q-p)}\sum _{j=2}^{\mathrm{\infty }}j\left|a_j\right|\le 1,$ by virtue of Lemma 2.2(2). This completes the proof.

$\left(2\right)$ Consider the function $\begin{array}{rl}h\left(z\right)& =\frac{\mathrm{\Phi }p{q}^m(p-2q)-{(q-p)}^2}{\mathrm{\Phi }p{q}^m(p-2q)}\{\frac{f^{\mathrm{\prime }}(z,m;p{)}_n}{f^{\mathrm{\prime }}(z,m;p)}-\frac{{(q-p)}^2}{{(q-p)}^2-\mathrm{\Phi }p{q}^m(p-2q)}\}\\ & =\frac{\mathrm{\Phi }p{q}^m(p-2q)-{(q-p)}^2}{\mathrm{\Phi }p{q}^m(p-2q)}\{\frac{1+\sum _{j=2}^{n}j{a}_j{z}^{j-1}}{1+\sum _{j=2}^{\mathrm{\infty }}j{a}_j{z}^{j-1}}-\frac{{(q-p)}^2}{{(q-p)}^2-\mathrm{\Phi }p{q}^m(p-2q)}\}\\ & =\frac{1+\sum _{j=2}^{n}j{a}_j{z}^{j-1}-\frac{{(q-p)}^2}{\mathrm{\Phi }p{q}^m(2q-p)}\sum _{j=n+1}^{\mathrm{\infty }}j{a}_j{z}^{j-1}}{\sum _{j=2}^{\mathrm{\infty }}j{a}_j{z}^{j-1}}\\ & =\frac{1+w\left(z\right)}{1-w\left(z\right)},\left(\mathrm{s}\mathrm{a}\mathrm{y}\right).\end{array}$ This gives that $\begin{array}{rl}w\left(z\right)& =\frac{h\left(z\right)-1}{h\left(z\right)+1}\\ & =\frac{-\frac{\mathrm{\Phi }p{q}^m(p-2q)-{(q-p)}^2}{\mathrm{\Phi }p{q}^m(p-2q)}\left(\sum _{j=n+1}^{\mathrm{\infty }}j{a}_j{z}^{j-1}\right)}{2+2\left(\sum _{j=2}^{n}j{a}_j{z}^{j-1}\right)-\frac{\mathrm{\Phi }p{q}^m(p-2q)-{(q-p)}^2}{\mathrm{\Phi }p{q}^m(p-2q)}\left(\sum _{j=n+1}^{\mathrm{\infty }}j{a}_j{z}^{j-1}\right)}.\end{array}$ Now applying the fact that $w\left(z\right)\le 1,$ we have $\frac{{(q-p)}^2}{\mathrm{\Phi }p{q}^m(2q-p)}\sum _{j=n+1}^{\mathrm{\infty }}j{a}_j+\sum _{j=2}^{n}j{a}_j\le 1.$ Since the left-hand side of the above expression is bounded above by$\frac{(q-p{)}^2}{\mathrm{\Phi }p{q}^m(p-2q)}\sum _{j=2}^{\mathrm{\infty }}j\left|a_j\right|\le 1,$ by virtue of Lemma 2.2(2). This completes the proof.

Theorem 2.5

Let $m\in \mathbb{N}$ with m>0 and $0\le p\le 1$. Then $\begin{array}{rl}& \left(1\right)\mathrm{\Re }\left\{\frac{Af(z,m;p)}{Af(z,m;p{)}_n}\right\}\ge \frac{1-2p-\xi p{q}^m}{1-2p},\\ & \left(2\right)\mathrm{\Re }\left\{\frac{Af(z,m;p{)}_n}{Af(z,m;p)}\right\}\ge \frac{1-2p}{1-2p+\xi p{q}^m},\end{array}$ where $\xi =\frac{m!}{2^{m-1}}.$

Proof.

$\left(1\right)$ Consider the function $h\left(z\right)=\frac{1-2p}{\xi p{q}^m}\{\frac{Af(z,m;p{)}_n}{Af(z,m;p)}-\left(\frac{1-2p-\xi p{q}^m}{1-2p}\right)\}$ and $w\left(z\right)=\frac{h\left(z\right)-1}{h\left(z\right)+1}.$ This implies that $w\left(z\right)=\frac{\frac{1-2p}{\xi p{q}^m}\sum _{j=n+1}^{\mathrm{\infty }}\frac{a_j}{j}{z}^{j-1}}{2+2\sum _{j=2}^n\frac{a_j}{j}{z}^{j-1}+\frac{1-2p}{\xi p{q}^m}\sum _{j=n+1}^{\mathrm{\infty }}\frac{a_j}{j}{z}^{j-1}}.$ Applying the fact that $\left|w\right(z\left)\right|\le 1$, we have $\left|\frac{\frac{1-2p}{\xi p{q}^m}\sum _{j=n+1}^{\mathrm{\infty }}\frac{a_j}{j}{z}^{j-1}}{2+2\sum _{j=2}^n\frac{a_j}{j}{z}^{j-1}+\frac{1-2p}{\xi p{q}^m}\sum _{j=n+1}^{\mathrm{\infty }}\frac{a_j}{j}{z}^{j-1}}\right|<1,$ which reduces to $\frac{1-2p}{\xi p{q}^m}\sum _{j=n+1}^{\mathrm{\infty }}\left|\frac{a_j}{j}\right|+\sum _{j=2}^n\left|\frac{a_j}{j}\right|\le 1.$ Since the left-hand side of the above expression is bounded above by$\frac{1-2p}{\xi p{q}^m}$ $\sum _{j=2}^{\mathrm{\infty }}\left|\frac{a_j}{j}\right|\le 1,$ by virtue of Lemma 2.2(3). This completes the proof.

$\left(2\right)$ Consider the function $\begin{array}{rl}h\left(z\right)& =\frac{1}{\frac{1-2p}{1-2p+\xi p{q}^m}}\{\frac{Af(z,m;p{)}_n}{Af(z,m;p)}-(1-\frac{1-2p}{1-2p+\xi p{q}^m})\}\\ & =\frac{1+\sum _{j=2}^n\frac{a_j}{j}{z}^{j-1}-\frac{(1-2p)}{\xi p{q}^m}\sum _{j=n+1}^{\mathrm{\infty }}\frac{a_j}{j}{z}^{j-1}}{1+\sum _{j=2}^{\mathrm{\infty }}\frac{a_j}{j}{z}^{j-1}}\\ & =\frac{1+w\left(z\right)}{1-w\left(z\right)},\left(\mathrm{s}\mathrm{a}\mathrm{y}\right).\end{array}$ This implies that $w\left(z\right)=\frac{-(\frac{(1-2p)}{\xi p{q}^m}+1)\sum _{j=n+1}^{\mathrm{\infty }}\frac{a_j}{j}{z}^{j-1}}{2+2\sum _{j=2}^n\frac{a_j}{j}{z}^{j-1}-(\frac{(1-2p)}{\xi p{q}^m}-1)\sum _{j=n+1}^{\mathrm{\infty }}\frac{a_j}{j}{z}^{j-1}}.$ Applying the fact that $\left|w\right(z\left)\right|\le 1$, we have $\left|\frac{-(\frac{(1-2p)}{p{q}^m}+1)\sum _{j=n+1}^{\mathrm{\infty }}\frac{a_j}{j}{z}^{j-1}}{2+2\sum _{j=2}^n\frac{a_j}{j}{z}^{j-1}-(\frac{(1-2p)}{p{q}^m}-\xi )\sum _{j=n+1}^{\mathrm{\infty }}\frac{a_j}{j}{z}^{j-1}}\right|\le 1,$ which reduces to $(\frac{(1-2p)}{\xi p{q}^m}+1)\sum _{j=n+1}^{\mathrm{\infty }}\left|\frac{a_j}{j}\right|\le 2-2\sum _{j=2}^n\left|\frac{a_j}{j}\right|-(\frac{(1-2p)}{\xi p{q}^m}-1)\sum _{j=n+1}^{\mathrm{\infty }}\left|\frac{a_j}{j}\right|.$ This implies that $\frac{(1-2p)}{\xi p{q}^m}\sum _{j=n+1}^{\mathrm{\infty }}\left|\frac{a_j}{j}\right|+\sum _{j=2}^n\left|\frac{a_j}{j}\right|\le 1.$ Since the left-hand side of the above expression is bounded above by$\frac{1-2p}{\xi p{q}^m}$ $\sum _{j=2}^{\mathrm{\infty }}\left|\frac{a_j}{j}\right|\le 1,$ by virtue of Lemma 2.2(3).

2.2. Partial sums of negative binomial distribution:

One of the most important discrete distributions is Negative Binomial distribution. This distribution is also known as Pascal distribution. It was formulated by Montmort in (1714), see [32]. There are two parameter of this distribution p and m, where, p is number of success and $m\ge 0.$ There are numerous applications of this distribution in the literature. It is used to analyse count data at large rang. The axioms of this distribution are given by,

1. Negative Binomial distribution has two out comes success and failure.

2. Successive trials are independent.

3. The experiment is repeated until to obtain the fix number of trial.

4. One of the most important properties is that its variance $(mp+m{p}^2)$ exceeds its mean $\left(mp\right),$ sometime this property is referred to over dispersion.

The probability mass function for a random variable X of negative binomial, having m trials and j successes is $F(X=j)=(\genfrac{}{}{0ex}{}{m-1}{j-1})p^j{q}^{m-j},m,j=1,2,3,\dots .$ In this section, it comprises the findings of partial sums of Negative Binomial Distribution. The series representation of analytic functions associated with the negative binomial distribution function is given by, (9) $g(z,m;p)=z+\sum _{j=2}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{m-1}{j-2})p^{j-1}{q}^{m-j+1}{z}^j,z\in E.$(9) The partial sum of the above series is defined as (10) $g(z,m;p{)}_n=z+\sum _{j=2}^n(\genfrac{}{}{0ex}{}{m-1}{j-2})p^{j-1}{q}^{m-j+1}{z}^j.$(10) Consider the Alexander transform of the function $g(z,m;p)$ as defined by (11) $Ag(z,m;p)=z+\sum _{j=2}^{\mathrm{\infty }}\frac{1}{j}(\genfrac{}{}{0ex}{}{m-1}{j-2})p^{j-1}{q}^{m-j+1}{z}^j.$(11) Now, compute the lower bounds of certain ratios of these partial sums like $\mathrm{\Re }\left\{\frac{g(z,m;p)}{g(z,m;p{)}_n}\right\},$ $\mathrm{\Re }\left\{\frac{g(z,m;p{)}_n}{g(z,m;p)}\right\},$ $\mathrm{\Re }\left\{\frac{g^{\mathrm{\prime }}(z,m;p)}{g^{\mathrm{\prime }}(z,m;p{)}_n}\right\},$ $\mathrm{\Re }\left\{\frac{g^{\mathrm{\prime }}(z,m;p{)}_n}{g^{\mathrm{\prime }}(z,m;p)}\right\},$ $\mathrm{\Re }\left\{\frac{Ag(z,m;p)}{Ag(z,m;p{)}_n}\right\},$ $\mathrm{\Re }\left\{\frac{Ag(z,m;p{)}_n}{Ag(z,m;p)}\right\}.$ The next result is the preliminary one, which will be used in the proof of our main results.

Lemma 2.6

Let $n>0,$ $0\le p\le 1$. Then, the function $g(z,m;p):E\to \mathbb{C}$ defined by (9(9) $g(z,m;p)=z+\sum _{j=2}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{m-1}{j-2})p^{j-1}{q}^{m-j+1}{z}^j,z\in E.$(9) ) satisfies the following inequalities. (12) $\begin{array}{rl}& \left(1\right)\left|g\right(z,m;p\left)\right|\le 1+\varphi \frac{p{q}^m}{(1-2p)},\end{array}$(12) (13) $\begin{array}{rl}& \left(2\right)\left|g^{\mathrm{\prime }}\right(z,m;p\left)\right|\le 1+\varphi \frac{p{q}^m(2q-p)}{{(q-p)}^2},\end{array}$(13) (14) $\begin{array}{rl}& \left(3\right)\left|Ag\right(z,m;p\left)\right|\le 1-\varphi \frac{q^{m+1}}{p}[\ln (1-\frac{p}{q})+\frac{p}{q}],\end{array}$(14) where $\varphi =\frac{(m-1)!}{2^{m-3}}.$

Proof.

$\left(1\right)$ From (9(9) $g(z,m;p)=z+\sum _{j=2}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{m-1}{j-2})p^{j-1}{q}^{m-j+1}{z}^j,z\in E.$(9) ), one can have $\begin{array}{rl}\left|g\right(z,m;p\left)\right|& \le 1+\sum _{j=2}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{m-1}{j-2})p^{j-1}{q}^{m-j+1}\\ & =1+\frac{(m-1)!\left(q^{m+1}\right)}{2^{m-3}p}\sum _{j=2}^{\mathrm{\infty }}{\left(\frac{p}{q}\right)}^j\\ & =1+\frac{(m-1)!\left(q^{m+1}\right)}{2^{m-3}p}\frac{p^2}{q(1-2p)}\\ & =1+\varphi \frac{p{q}^m}{(1-2p)},\end{array}$ where $\varphi =\frac{(m-1)!}{2^{m-3}}.$

$\left(2\right)$ Also from (9(9) $g(z,m;p)=z+\sum _{j=2}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{m-1}{j-2})p^{j-1}{q}^{m-j+1}{z}^j,z\in E.$(9) ), one can have $g^{\mathrm{\prime }}(z,m;p)=1+\sum _{j=2}^{\mathrm{\infty }}j(\genfrac{}{}{0ex}{}{m-1}{j-2})p^{j-1}{q}^{m-j+1}{z}^{j-1}.$ This implies that $\begin{array}{rl}\left|g^{\mathrm{\prime }}\right(z,m;p\left)\right|& \le 1+\sum _{j=2}^{\mathrm{\infty }}j(\genfrac{}{}{0ex}{}{m-1}{j-2})p^{j-1}{q}^{m-j+1}\\ & =1+\frac{(m-1)!}{2^{m-3}}\frac{q^{m+1}}{p}\sum _{j=2}^{\mathrm{\infty }}j{\left(\frac{p}{q}\right)}^j\\ & =1-\frac{(m-1)!}{2^{m-3}}\frac{q^{m+1}}{p}\frac{p^2(p-2q)}{q{(q-p)}^2}\\ & =1+\varphi \frac{p{q}^m(2q-p)}{{(q-p)}^2}.\end{array}$ $\left(3\right)$ From (11(11) $Ag(z,m;p)=z+\sum _{j=2}^{\mathrm{\infty }}\frac{1}{j}(\genfrac{}{}{0ex}{}{m-1}{j-2})p^{j-1}{q}^{m-j+1}{z}^j.$(11) ), one can have $\begin{array}{rl}\left|Ag\right(z,m;p\left)\right|& \le 1+\sum _{j=2}^{\mathrm{\infty }}\frac{1}{j}(\genfrac{}{}{0ex}{}{m-1}{j-2})p^{j-1}{q}^{m-j+1}\\ & =1+\frac{(m-1)!}{2^{m-3}}\frac{q^{m+1}}{p}\sum _{j=2}^{\mathrm{\infty }}\frac{1}{j}{\left(\frac{p}{q}\right)}^j\end{array}$ Now consider, $\sum _{j=2}^{\mathrm{\infty }}\frac{1}{j}{\left(\frac{p}{q}\right)}^j=\frac{1}{2}{\left(\frac{p}{q}\right)}^2+\frac{1}{3}{\left(\frac{p}{q}\right)}^3+\dots$ As we know that, $-\ln (1-x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\dots$ Using this, one may have the following. $\begin{array}{rl}-\ln (1-\frac{p}{q})-\frac{p}{q}& =\frac{{\left(\frac{p}{q}\right)}^2}{2}+\frac{{\left(\frac{p}{q}\right)}^3}{3}+\dots \\ & =\sum _{j=2}^{\mathrm{\infty }}\frac{1}{j}{\left(\frac{p}{q}\right)}^j.\end{array}$ Thus, $\left|Ag\right(z,m;p\left)\right|\le 1-\varphi \frac{q^{m+1}}{p}[\ln (1-\frac{p}{q})+\frac{p}{q}].\text{■}$

By using the above lemma and the similar technique as followed in the proof of Lemma 2.2, one can easily establish the following inequalities.

Lemma 2.7

Let $n>0,$ $0\le p\le 1$. Then, the function $g(z,m;p):E\to \mathbb{C}$ defined by (9(9) $g(z,m;p)=z+\sum _{j=2}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{m-1}{j-2})p^{j-1}{q}^{m-j+1}{z}^j,z\in E.$(9) ) satisfies the following inequalities. $\begin{array}{rl}& \left(1\right)\frac{(1-2p)}{\varphi p{q}^m}\sum _{j=2}^{\mathrm{\infty }}\left|b_j\right|\le 1\\ & \left(2\right)\frac{{(q-p)}^2}{-\varphi p{q}^m(p-2q)}\sum _{j=2}^{\mathrm{\infty }}j\left|b_j\right|\le 1\\ & \left(3\right)\frac{p}{\varphi q^{m+1}}\frac{1}{[\ln (1-\frac{p}{q})+\frac{p}{q}]}\sum _{j=2}^{\mathrm{\infty }}\frac{1}{j}\left|b_j\right|\le 1\end{array}$

Theorem 2.8

Let $n>0,$ $0\le p\le 1$. Then, $\begin{array}{rl}& \left(1\right)\mathrm{\Re }\left\{\frac{g(z,m;p)}{g(z,m;p{)}_n}\right\}\ge 1-\frac{p\varphi q^m}{1-2p},\\ & \left(2\right)\mathrm{\Re }\left\{\frac{g(z,m;p{)}_n}{g(z,m;p)}\right\}\ge \frac{1-2p}{1-2p+\varphi p{q}^m},\end{array}$ where $\varphi =\frac{(m-1)!}{2^{m-3}}.$

Proof follows similarly as done in the proof of Theorem 2.3.

Theorem 2.9

Let $n>0,$ $0\le p\le 1$. Then, $\begin{array}{rl}& \left(1\right)\mathrm{\Re }\left\{\frac{g^{\mathrm{\prime }}(z,m;p)}{g^{\mathrm{\prime }}(z,m;p{)}_n}\right\}\ge 1+\frac{\varphi q^{m}p(p-2q)}{{(q-p)}^2},\\ & \left(2\right)\mathrm{\Re }\left\{\frac{g^{\mathrm{\prime }}(z,m;p{)}_n}{g^{\mathrm{\prime }}(z,m;p)}\right\}\ge \frac{{(q-p)}^2}{{(q-p)}^2-\varphi p{q}^m(p-2q)},\end{array}$ where $\varphi =\frac{(m-1)!}{2^{m-3}}.$

Proof follows similarly as done in the proof of Theorem 2.4.

Theorem 2.10

Let $n>0,$ $0\le p\le 1$. Then, $\begin{array}{rl}& \left(1\right)\mathrm{\Re }\left\{\frac{Ag(z,m;p)}{Ag(z,m;p{)}_n}\right\}\ge \varphi \frac{q^{m+1}}{p}(\ln (1-\frac{p}{q})+\frac{p}{q})+1,\\ & \left(2\right)\mathrm{\Re }\left\{\frac{Ag(z,m;p{)}_n}{Ag(z,m;p)}\right\}\ge \frac{1}{1+\varphi \frac{q^{m+2}}{p}[-\ln (1-\frac{p}{q})-\frac{p}{q}]},\end{array}$ where $Ag(z,m;p)$ is the Alexander transform of $g(z,m;p)$ as defined by (11(11) $Ag(z,m;p)=z+\sum _{j=2}^{\mathrm{\infty }}\frac{1}{j}(\genfrac{}{}{0ex}{}{m-1}{j-2})p^{j-1}{q}^{m-j+1}{z}^j.$(11) ).

Proof.

(1) Consider the function $\begin{array}{rl}h\left(z\right)& =\frac{1}{1-\text{β}}\{\frac{Ag(z,m;p)}{Ag(z,m;p{)}_n}-\text{β}\}\\ & =\frac{1}{1-\text{β}}\{\frac{1+\sum _{j=2}^{\mathrm{\infty }}\frac{1}{j}{b}_j{z}^{j-1}}{1+\sum _{j=2}^n\frac{1}{j}{b}_j{z}^{j-1}}-\text{β}\},\end{array}$ where $\text{β}=1-\varphi \frac{q^{m+1}}{p}(-\ln (1-\frac{p}{q})-\frac{p}{q}).$ This implies that $h\left(z\right)=\frac{1+\sum _{j=2}^n\frac{1}{j}{b}_j{z}^{j-1}+\frac{1}{\varphi \frac{q^{m+1}}{p}(-\ln (1-\frac{p}{q})-\frac{p}{q})}\sum _{j=n+1}^{\mathrm{\infty }}\frac{1}{j}{b}_j{z}^{j-1}}{1+\sum _{j=2}^n\frac{1}{j}{b}_j{z}^{j-1}}.$ Now taking $w\left(z\right)=\frac{h\left(z\right)-1}{h\left(z\right)+1},$ we have, $w\left(z\right)=\frac{\frac{1}{\varphi \frac{q^{m+1}}{p}(-\ln (1-\frac{p}{q})-\frac{p}{q})}\sum _{j=n+1}^{\mathrm{\infty }}\frac{1}{j}{b}_j{z}^{j-1}}{2+2\sum _{j=2}^n\frac{1}{j}{b}_j{z}^{j-1}+\frac{1}{\varphi \frac{q^{m+1}}{p}(-\ln (1-\frac{p}{q})-\frac{p}{q})}\sum _{j=n+1}^{\mathrm{\infty }}\frac{1}{j}{b}_j{z}^{j-1}}.$ Using the fact that $\left|w\right(z\left)\right|\le 1,$ we get $\left|\frac{-\frac{1}{(\ln (1-\frac{p}{q})+\frac{p}{q})}\sum _{j=n+1}^{\mathrm{\infty }}\frac{1}{j}{b}_j{z}^{j-1}}{2+2\sum _{j=2}^n\frac{1}{j}{b}_j{z}^{j-1}-\frac{1}{\varphi \frac{q^{m+1}}{p}(\ln (1-\frac{p}{q})+\frac{p}{q})}\sum _{j=n+1}^{\mathrm{\infty }}\frac{1}{j}{b}_j{z}^{j-1}}\right|\le 1.$ This implies that $\frac{1}{\varphi \frac{q^{m+1}}{p}(\ln (1-\frac{p}{q})+\frac{p}{q})}\sum _{j=n+1}^{\mathrm{\infty }}\frac{1}{j}\left|b_j\right|+\sum _{j=2}^n\frac{1}{j}\left|b_j\right|\le 1.$ Since the left-hand side of the above expression is bounded above by $\frac{1}{\varphi \frac{q^{m+1}}{p}\left([\ln (1-\frac{p}{q})+\frac{p}{q}]\right)}\sum _{j=2}^{\mathrm{\infty }}\frac{1}{j}\left|b_j\right|\le 1,$ by virtue of Lemma 2.7(3). This completes the proof.

(2) Consider the function $\begin{array}{rl}h\left(z\right)& =\frac{1+\varphi \frac{q^{m+1}}{p}[-\ln (1-\frac{p}{q})-\frac{p}{q}]}{\varphi \frac{q^{m+1}}{p}[-\ln (1-\frac{p}{q})-\frac{p}{q}]}\{\frac{Ag(z,m;p{)}_n}{Ag(z,m;p)}-\frac{1}{1+\varphi \frac{q^{m+2}}{p}[-\ln (1-\frac{p}{q})-\frac{p}{q}]}\}\\ & =\frac{1+\varphi \frac{q^{m+1}}{p}[-\ln (1-\frac{p}{q})-\frac{p}{q}]}{\varphi \frac{q^{m+1}}{p}[-\ln (1-\frac{p}{q})-\frac{p}{q}]}\\ & \times \{\frac{1+\sum _{j=2}^n\frac{1}{j}{b}_j{z}^{j-1}}{1+\sum _{j=2}^{\mathrm{\infty }}\frac{1}{j}{b}_j{z}^{j-1}}-\frac{1}{1+\varphi \frac{q^{m+2}}{p}[-\ln (1-\frac{p}{q})-\frac{p}{q}]}\},\end{array}$ which reduces to $h\left(z\right)=\frac{1+\sum _{j=2}^n\frac{1}{j}{b}_j{z}^{j-1}+\frac{1}{\varphi \frac{q^{m+2}}{p}[\ln (1-\frac{p}{q})+\frac{p}{q}]}\sum _{j=n+1}^{\mathrm{\infty }}\frac{1}{j}{b}_j{z}^{j-1}}{1+\sum _{j=2}^{\mathrm{\infty }}\frac{1}{j}{b}_j{z}^{j-1}}.$ Now taking $w\left(z\right)=\frac{h\left(z\right)-1}{h\left(z\right)+1},$ we have $w\left(z\right)=\frac{-(+1+\frac{1}{\varphi \frac{q^{m+1}}{p}\left([\ln (1-\frac{p}{q})+\frac{p}{q}]\right)})\sum _{j=n+1}^{\mathrm{\infty }}\frac{1}{j}{b}_j{z}^{j-1}}{2+2\sum _{j=2}^{\mathrm{\infty }}\frac{1}{j}{b}_j{z}^{j-1}-(-1+\frac{1}{\varphi \frac{q^{m+1}}{p}\left([\ln (1-\frac{p}{q})+\frac{p}{q}]\right)})\sum _{j=n+1}^{\mathrm{\infty }}\frac{1}{j}{b}_j{z}^{j-1}}.$ Using the fact that $\left|w\right(z\left)\right|\le 1,$ we have $\left|\frac{-(+1+\frac{1}{\varphi \frac{q^{m+1}}{p}\left([\ln (1-\frac{p}{q})+\frac{p}{q}]\right)})\sum _{j=n+1}^{\mathrm{\infty }}\frac{1}{j}{b}_j{z}^{j-1}}{2+2\sum _{j=2}^{\mathrm{\infty }}\frac{1}{j}{b}_j{z}^{j-1}-(-1+\frac{1}{\varphi \frac{q^{m+1}}{p}\left([\ln (1-\frac{p}{q})+\frac{p}{q}]\right)})\sum _{j=n+1}^{\mathrm{\infty }}\frac{1}{j}{b}_j{z}^{j-1}}\right|\le 1.$ This implies that $\frac{1}{\varphi \frac{q^{m+1}}{p}\left([\ln (1-\frac{p}{q})+\frac{p}{q}]\right)}\sum _{j=n+1}^{\mathrm{\infty }}\frac{1}{j}\left|b_j\right|+\sum _{j=2}^{\mathrm{\infty }}\frac{1}{j}\left|b_j\right|\le 1.$ Since the left hand side of above expression is bounded above by $\frac{1}{\varphi \frac{q^{m+1}}{p}\left([\ln (1-\frac{p}{q})+\frac{p}{q}]\right)}\sum _{j=2}^{\mathrm{\infty }}\frac{1}{j}\left|b_j\right|\le 1,$ by virtue of Lemma 2.7(3). This completes the proof.

Remark 2.11

Inspired by the research going on and a wide range of applications, we study binomial and negative binomial distributions in complex variables. The partial sums have also been addressed by many researchers, the main area of concern for them was to discuss the extent of univalency, starlikeness and convexity of the partial sums but the problem of finding the lower bounds of ratios of partial sums to infinite series sums is the one of those geometrical aspects that has not been discussed or studied before for complex analogue of binomial and negative binomial distributions. This has been brought into discussion for the first time in this paper. A detailed survey on the importance and significance of partial sums can be seen in [33] which contains the developments on the said topic in a chronological order. More results on this topic can be accessed from [13–16, 26–31]. Moreover, many other geometrical properties of the complex analogues of these and earlier studied distributions can be discussed and studied. The method adopted in this paper to get the required results generalizes the analytic approach and extends the real part of the partial sum to the nth order.

3. Conclusion

This work has contributed to the theory of analytic functions by defining and studying analytic functions by using the well-known Binomial distribution function and negative Binomial distribution function. It established certain inequalities for the ratio of the series of partial sums to the infinite series sum, for the said analytic functions. This is a new dimension of the study of both analytic functions and probability distribution functions. This study will be useful in such investigations for other discrete distribution functions.

Disclosure statement

No potential conflict of interest was reported by the author(s).

Additional information

Funding

This research was supported by the researchers Supporting Project Number [RSP-2021/401], King Saud University, Riyadh, Saudi Arabia.