# The (a,b,1) class

This post is a continuation of the preceding post on (a,b,0) class. This post introduces the class of discrete discrete distributions called the (a,b,1) class.

The discussion in this post has a great deal of technical details. A concise summary of the (a,b,0) class and (a,b,1) class is found here.

The (a,b,1) Class

A counting distribution is a discrete probability distribution that takes on the non-negative integers. Let $N$ be a random variable that is a counting distribution. For each integer $k=0,1,2,\cdots$, let $P_k=P[N=k]$ if $N$ is the counting distribution being considered. Recall that a counting distribution is a member of the (a,b,0) class of distributions if the following recursive relation holds for some constants $a$ and $b$.

(1)……….$\displaystyle \frac{P_k}{P_{k-1}}=a + \frac{b}{k} \ \ \ \ \ \ \ \ \ \ \ \ \ k=1,2,3,\cdots$

For a member of the (a,b,0) class, the initial probability $P_0$ is fixed (the sum of all the $P_k$ must sum to 1). Thus a member of the (a,b,0) class has two parameters, namely $a$ and $b$ in the recursive relation (1). A counting distribution is a member of the (a,b,1) class of distributions if the following recursive relation holds for some constants $a$ and $b$.

(2)……….$\displaystyle \frac{P_k}{P_{k-1}}=a + \frac{b}{k} \ \ \ \ \ \ \ \ \ \ \ \ \ k=2,3,4 \cdots$

The recursion in the (a,b,1) class begins at $k=2$. That means that the initial probability $P_0$ must be an assumed value. The probability $P_1$ is then the value such that the sum $P_1+P_2+\cdots$ is $1-P_0$. Thus a member of the (a,b,1) class has three parameters: $a$, $b$ and $P_0$.

There are two subclasses in the (a,b,1) class of distributions. They are determined by whether $P_0=0$ or $P_0>0$ The (a,b,1) distributions in the first category are called the zero-truncated distributions. The (a,b,1) distributions in the second category are called zero-modified distributions.

This is how we will proceed. Using a given distribution in the (a,b,0) class as a starting point, we show how to derive the zero-truncated distribution. From a given zero-truncated distribution, we show how to derive the zero-modified distribution.

The name of the (a,b,1) distribution has the same (a,b,0) name with either zero-truncated or zero-modified as the prefix. For example, if the starting point is the negative binomial distribution in the (a,b,0) class, then the derived distributions in the (a.b.1) class are the zero-truncated negative binomial distribution and the zero-modified negative binomial distribution.

There are only three distributions in the (a,b,0) class – Poisson, binomial and negative binomial. Then the (a,b,1) class contains the zero-truncated and zero-modified versions of these three distributions. However, the (a,b,1) class contained distributions that are not modifications of the (a,b,0) distributions. We discuss three such additional distributions – extended truncated negative binomial (ETNB) distribution, logarithmic distribution and Sibuya distribution. These three distributions that are not derived from an (a,b,0) distribution are discussed in a separate section below.

We present three examples demonstrating how using a base (a,b,0) negative binomial distribution to derive a zero-truncated negative binomial distribution (Example 1) and a zero-modified negative binomial distribution (Example 2). We also give an example for ETNB distribution (Example 3).

Notations

To facilitate discussion, let’s fix some notations. To clearly denote the distributions, notations without superscripts and subscripts refer to the (a,b,0) distributions. Notations with the superscript T (or subscript T) refer to the zero-truncated distributions in the (a,b,1) class. Likewise notations with the superscript M (or subscript M) refer to the zero-modified distributions in the (a,b,1) class.

For example, the following are the probability function (pf) and the probability generating function (pgf) of a distribution from the (a,b,0) class.

(3)……….\displaystyle \begin{aligned}&P_k=P[N=k] \\&P(z)=\sum \limits_{k=0}^\infty P_k z^k \end{aligned}

The following shows the notations for the pf and pgf for a zero-truncated distribution from the (a,b,1) class.

(4)……….\displaystyle \begin{aligned}&P_k^T=P[N_T=k] \\&P^T(z)=\sum \limits_{k=1}^\infty P_k^T z^k \end{aligned}

The following shows the notations for the pf and pgf for a zero-modified distribution from the (a,b,1) class.

(5)……….\displaystyle \begin{aligned}&P_k^M=P[N_M=k] \\&P^M(z)=\sum \limits_{k=0}^\infty P_k^M z^k \end{aligned}

Whenever it is convenient to do so, $N$ is a random variable from (a,b,0) while $N_T$ and $N_M$ are to denote random variables for the zero-truncated distribution and zero-modified distribution, respectively.

Zero-Truncated Distributions

The focus in this section is on the zero-truncated distributions that originate from the (a,b,0) class. The three distributions indicated above (ETNB, logarithmic and Shibuya) are discussed in a separate section below.

Suppose we start with a distribution from the (a,b,0) class, with the notations $P_k$, $P(z)$ and $N$ as indicated above. We show how to derive the corresponding zero-truncated distribution in the (a,b,1) class. For the zero-truncated distribution, there are two ways to compute probabilities. One is the recursion relation:

(6)……….$\displaystyle \frac{P_k^T}{P_{k-1}^T}=a + \frac{b}{k} \ \ \ \ \ \ \ \ \ \ \ \ \ k=2,3,4 \cdots$

The recursion relation (6) is identical to the one in (2). The recursion begins at $k=2$. The zero-truncated probabilities can also be derived from the (a,b,0) probabilities as follows:

(7)……….$\displaystyle P_k^T=\frac{1}{1-P_0} P_k \ \ \ \ \ \ \ \ \ \ \ \ \ k=1,2,3,4 \cdots$

The probabilities $P_k^T$ in (7) can be regarded as conditional probabilities – the probability that $N=k$ given that $N>0$. From a procedural standpoint, the probabilities $P_k^T$ are the (a,b,0) probabilities $P_k$ multiplied by $1/(1-P_0)$ to make the probabilities sum to 1. With the probabilities established, the probability generating function (pgf) and mean and moments of the zero-truncated distribution can also be expressed in terms of the corresponding quantities of the (a,b,0) distribution.

(8)……….$\displaystyle P^T(z)=\sum \limits_{k=1}^\infty P_k^T \ z^k =\frac{1}{1-P_0} \ [P(z)-P_0]$

(9)……….$\displaystyle E[N_T]=\frac{1}{1-P_0} \ E[N]$

(10)……..$\displaystyle E[N_T^2]=\frac{1}{1-P_0} \ E[N^2]$

(11)……..\displaystyle \begin{aligned} Var[N_T]&=\frac{1}{1-P_0} \ E[N^2]-\biggl( \frac{1}{1-P_0} \ E[N]\biggr)^2 \\&=\frac{1}{1-P_0} \ Var[N]+\biggl(1-\frac{1}{1-P_0} \biggr) \ \frac{1}{1-P_0} \ E[N]^2 \end{aligned}

The goal of the above items is to inform on the zero-truncated distribution based on information from the (a,b,0) distribution. They can also be derived based on definitions using the probability function (7). The following shows the factorial means of the zero-truncated distribution.

(12)……..$\displaystyle \mu_{(1)}=E[N_T]=\frac{a+b}{(1-a) (1-P_0)}$
……..

(13)……..\displaystyle \begin{aligned} \mu_{(j)}&=E \{ N_T \ [N_T-1] \ [N_T-2] \cdots [N_T-(j-1)] \}\\&\text{ } \\&=\frac{(a j+b) \ \mu_{(j-1)}}{1-a} \end{aligned}

The first factorial mean is identical to the mean of $N_T$. The $P_0$ in $\mu_{(1)}$ is the value of zero probability for the corresponding member in the (a,b,0) class. The higher factorial moments are derived recursively as in the (a,b,0) case. The raw moments $E[N_T^k]$ can be derived using the factorial moments. The variance, as derived from the factorial moments, is:

(14)……..$\displaystyle Var[N_T]=\frac{(a+b) \ [1-(a+b+1) \ P_0]}{[(1-a) \ (1-P_0) ]^2}$

Example 1
It is helpful to go through an example. First, we set up an (a,b,0) distribution – an negative binomial distribution with parameters $r=2$ and $\theta=3$. The (a,b,0) parameters are $a=3/4$ and $b=3/4$. The following gives the pf and the recursive relation for this negative binomial distribution, as well as the mean, variance and pgf.

……….$\displaystyle P_k=(1+k) \ \frac{1}{16} \ \biggl(\frac{3}{4} \biggr)^k \ \ \ \ \ \ \ \ k=0,1,2,3,\cdots$

……….$\displaystyle P_k=\biggl(\frac{3}{4}+\frac{3}{4} \ \frac{1}{k} \biggr) \ P_{k-1} \ \ \ \ \ \ \ \ k=1,2,3,\cdots$

……….$\displaystyle E[N]=6$

……….$\displaystyle Var[N]=24$

……….$\displaystyle P(z)=[1-3 \ (z-1)]^{-2}$

The following table shows the first 5 probabilities for the zero-truncated negative binomial distribution.

Table – Zero-Truncated Negative Binomial

$\bold k$ (a,b,0) $\bold P_{\bold k}$ Zero-Truncated $\bold P_{\bold k}^{\bold T}$
0 $\displaystyle \frac{1}{16}$
1 $\displaystyle \frac{3}{32}$ $\displaystyle \frac{3}{30}$
2 $\displaystyle \frac{27}{256}$ $\displaystyle \frac{27}{240}$
3 $\displaystyle \frac{27}{256}$ $\displaystyle \frac{27}{240}$
4 $\displaystyle \frac{405}{4096}$ $\displaystyle \frac{405}{3840}$
5 $\displaystyle \frac{729}{8192}$ $\displaystyle \frac{729}{7680}$

The probabilities $P_k$ are generated by either the (a,b,0) pf or the recursive relation. The probabilities $P_k^T$ are generated by the recursive relation (7) or by the recursive relation (6). The following lists the mean, variance and pgf of the zero-truncated negative binomial example.

……….$\displaystyle E[N_T]=\frac{16}{15} \ 6=\frac{32}{5}=6.4$

……….$\displaystyle Var[N_T]=\frac{576}{25}=23.04$

……….$\displaystyle P^T(z)=\frac{16}{15} \biggl( [1-3 \ (z-1)]^{-2}-\frac{1}{16} \biggr)$

Zero-Modified Distributions

The goal of this section is to derive a zero-modified distribution from a zero-truncated distribution, either derived from an (a,b,0) distribution as discussed in the preceding section, or a truncated distribution not originated from (a,b,0) class.

We now take a zero-truncated distribution as a given and derive the probabilities $P_k^M$ and other distributional quantities. As in the case of zero-truncated distribution, one way to generate probabilities is through the recursion process:

(15)……..$\displaystyle \frac{P_k^M}{P_{k-1}^M}=a + \frac{b}{k} \ \ \ \ \ \ \ \ \ \ \ \ \ k=2,3,4 \cdots$

The probability $P_0^M>0$ is an assumed value. The probability $P_1^M$ is the value that ensures that all the probabilities sum to 1. As indicated the recursion begins at $k=2$. Another way to calculate probabilities is through the zero-truncated distribution:

(16)……..$\displaystyle P_k^M=(1-P_0^M) \ P_k^T \ \ \ \ \ \ \ \ \ \ \ \ \ k=1,2,3,4 \cdots$

Of course, if the zero-truncated distribution is based on a distribution from the (a,b,0) class, we can express the zero-modified probabilities as, after plugging (7) into (16):

(17)……..$\displaystyle P_k^M=\frac{1-P_0^M}{1-P_0} \ P_k \ \ \ \ \ \ \ \ \ \ \ \ \ k=1,2,3,4 \cdots$

Further distributional quantities can now be derived:

(18)……..$\displaystyle P^M(z)=P_0^M \cdot 1+(1-P_0^M) \ P^T(z)$

(19)……..$\displaystyle E[N_M]=(1-P_0^M) \ E[N_T]$

(20)……..$\displaystyle Var[N_M]=(1-P_0^M) \ Var[N_T]+P_0^M \ (1-P_0^M) \ E[N_T]^2$

The result (18) is the pgf of the zero-modified distribution based on the pgf of the given zero-truncated distribution. In words, (19) says that the mean of the modified distribution is $1-P_0^M$ times the mean of the given zero-truncated distribution. In words, (20) says that the variance of the zero-modified distribution is $1-P_0^M$ times the variance of the given zero-truncated distribution plus $P_0^M (1-P_0^M)$ times the square of the mean of the truncated distribution.

If the given zero-truncated distribution is actually obtained from a member of the (a,b,0) class, then the above three results can be expressed in terms of (a,b,0) information, after plugging the corresponding information for $N_T$ into (18), (19) and (20).

(21)……..$\displaystyle P^M(z)=\biggl(1-\frac{1-P_0^M}{1-P_0} \biggr) \cdot 1+\frac{1-P_0^M}{1-P_0} \ P(z)$

(22)……..$\displaystyle E[N_M]=\frac{1-P_0^M}{1-P_0} \ E[N]$

(23)……..$\displaystyle Var[N_M]=\frac{1-P_0^M}{1-P_0} \ Var[N]+ \biggl(1-\frac{1-P_0^M}{1-P_0} \biggr) \ \frac{1-P_0^M}{1-P_0} \ E[N]^2$

Example 2
Consider the zero-truncated negative binomial distribution considered in Example 1. We now generate information on the corresponding zero-modified negative binomial distribution with the assumed value of $P_0^M=0.2$. The following table gives several probabilities.

Table – Zero-Modified Negative Binomial

$\bold k$ (a,b,0) $\bold P_{\bold k}$ Zero-Truncated $\bold P_{\bold k}^{\bold T}$ Zero-Modified $\bold P_{\bold k}^{\bold M}$
0 $\displaystyle \frac{1}{16}$ 0.2
1 $\displaystyle \frac{3}{32}$ $\displaystyle \frac{3}{30}$ $\displaystyle \frac{2.4}{30}$
2 $\displaystyle \frac{27}{256}$ $\displaystyle \frac{27}{240}$ $\displaystyle \frac{21.6}{240}$
3 $\displaystyle \frac{27}{256}$ $\displaystyle \frac{27}{240}$ $\displaystyle \frac{21.6}{240}$
4 $\displaystyle \frac{405}{4096}$ $\displaystyle \frac{405}{3840}$ $\displaystyle \frac{326.4}{3840}$
5 $\displaystyle \frac{729}{8192}$ $\displaystyle \frac{729}{7680}$ $\displaystyle \frac{583.2}{7680}$

The zero-modified probabilities $P_k^M$ are calculated according to (16). With the assumed value $P_0^M=0.2$, $1-P_0^M=0.8$. We simply multiply each zero-truncated probability by 0.8. Using (18), (19) and (20), we obtain the mean, variance and pgf of the zero-modified negative binomial example.

……….$\displaystyle E[N_M]=0.8 \ E[N_T]=0.8 (6.4)=5.12$

……….$\displaystyle Var[N_M]=24.9856$

……….$\displaystyle P^M(z)=0.2+0.8 \biggl[ \frac{16}{15} \biggl( [1-3 \ (z-1)]^{-2}-\frac{1}{16} \biggr) \biggr]$

Additional Zero-Truncated Distributions

As indicated earlier, the (a,b,1) class contains distributions other than the ones derived from the three (a,b,0) distributions. These distributions also have the zero-truncated versions as well as the zero-modified versions. We discuss the truncated versions. They are: the extended truncated negative binomial (ETNB) distribution, the logarithmic distribution and the Sibuya distribution. The extended truncated negative binomial (ETNB) distribution is resulted from relaxing the r parameter of the negative binomial distribution. The logarithmic distribution and Sibuya distribution are derived from the ETNB distribution. The modified versions of these three distributions can then be obtained by going through the process outlined in the preceding section.

ETNB
Recall that the (a,b,0) negative binomial distribution has two parameters $r$ and $\theta$. The following gives the parameters $a$ and $b$ used in the (a,b,0) recursion and the first two probabilities.

……….$\displaystyle a=\frac{\theta}{1+\theta} \ \ \ \ \ \ \ \ \ \ \ \ \ \ b=(r-1) \ \frac{\theta}{1+\theta} \ \ \ \ \ \ r>0, \ \theta >0$

……….$\displaystyle P_0=\biggl( \frac{1}{1+\theta} \biggr)^r \ \ \ \ \ \ P_1=r \ \biggl( \frac{1}{1+\theta} \biggr)^r \biggl( \frac{\theta}{1+\theta} \biggr)=\frac{r \ \theta}{(1+\theta)^{r+1}}$

The extended negative binomial distribution is resulted from extending the $r$ parameter so that $-1 is applicable in addition to the usual $r>0$. With the extension of $r$, the ETNB probabilities are generated according to the truncated probabilities of (7). In effect, we are pretending that we are starting from a base (a,b,0) negative binomial distribution even the $r$ parameter could be such that $-1. Thus the two parameters of the zero-truncated ETNB distribution are given by the following:

(24)……..$\displaystyle a=\frac{\theta}{1+\theta} \ \ \ \ \ \ b=(r-1) \ \frac{\theta}{1+\theta} \ \ \ \ \ \ -10$

What do we do with the ETNB parameters indicated in (24)? Using these $a$ and $b$, we can generate the “negative binomial” probabilities $P_k$ according to the recursive relation (1) with $P_0=[1/(1+\theta)]^r$. However, with $r$ being negative, these values of $P_k$ are not probabilities (in fact they are negative). However, the value of $1/(1-P_0)$ is also negative when $r$ is negative. Using (7), the zero-truncated probabilities $P_k^T$ are positive. Thus the “negative binomial” distribution using a negative $r$ is not really a distribution. It is just a device to define ETNB distribution.

Using the idea in the preceding paragraph, we can also come up with direct formula for the ETNB probabilities $P_k$. The following gives the first three probabilities.

……….$\displaystyle P_1^T=\frac{1}{1-P_0} \ P_1=\frac{1}{1-P_0} \ \frac{r \ \theta}{(1+\theta)^{r+1}}=\frac{r \theta}{(1+\theta)^{r+1}-(1+\theta) }$

……….\displaystyle \begin{aligned} P_2^T&=\frac{1}{1-P_0} \ P_2\\&=\frac{1}{1-P_0} \ \frac{r (r+1)}{2} \frac{1}{(1+\theta)^r} \ \biggl(\frac{\theta}{1+\theta} \biggr)^2 \\&=\frac{r (r+1)}{2} \ \frac{1}{(1+\theta)^r-1} \ \biggl(\frac{\theta}{1+\theta} \biggr)^2 \end{aligned}

……….\displaystyle \begin{aligned} P_3^T&=\frac{1}{1-P_0} \ P_3\\&=\frac{1}{1-P_0} \ \frac{r (r+1) (r+2)}{3!} \frac{1}{(1+\theta)^r} \ \biggl(\frac{\theta}{1+\theta} \biggr)^3 \\&=\frac{r (r+1) (r+2)}{3!} \ \frac{1}{(1+\theta)^r-1} \ \biggl(\frac{\theta}{1+\theta} \biggr)^3 \end{aligned}

Based on the pattern of the above three probabilities, the ENTB probability $P_k$, $k=1,2,3,\cdots$, is:

(25)……..$\displaystyle P_k=\frac{1}{1-P_0} \ P_k=\frac{r (r+1) \cdots (r+k-1)}{k!} \ \frac{1}{(1+\theta)^r-1} \ \biggl(\frac{\theta}{1+\theta} \biggr)^k$

All other distributional quantities such as pgf and means and higher moments can be derived based on the ETNB pf $P_k^T$ For example, the mean, variance and pgf are:

(26)……..$\displaystyle E[N_T]=\frac{1}{1-P_0} \ E[N]=\frac{1}{1-P_0} \ r \ \theta=\frac{r \ \theta}{1-(1+\theta)^{-r}}$

(27)……..\displaystyle \begin{aligned} Var[N_T]&=\frac{1}{1-P_0} \ Var[N]+\biggl(1-\frac{1}{1-P_0} \biggr) \ \frac{1}{1-P_0} \ E[N]^2 \\&=r \ \theta \ \frac{(1+\theta)-(1+\theta+ r \ \theta) \ (1+\theta)^{- r}}{[1-(1+\theta)^{-r}]^2} \end{aligned}

(28)……..\displaystyle \begin{aligned} P^T(z)&=\frac{1}{1-P_0} \ (P(z)-P_0) \\&=\frac{1}{1-(1+\theta)^{-r}} \ \biggl[ \biggl(1-\theta (z-1) \biggr)^{-r} -(1+\theta)^{-r}\biggr] \end{aligned}

Logarithmic Distribution
This is a truncated distribution that is derived from ETNB by letting $r \rightarrow 0$. The following shows the information that is needed for the recursive generation of probabilities.

(29)……..$\displaystyle a=\frac{\theta}{1+\theta} \ \ \ \ \ \ \ \ b=-\frac{\theta}{1+\theta}$

(30)……..$\displaystyle P_1^T=\frac{\theta}{(1+\theta) \ \ln(1+\theta)}$

The parameter $b$ is obtained by letting $r \rightarrow 0$ in the $b$ for ETNB. The logarithmic $P_1^T$ is from taking the limit of the ETNB $P_1^T$ as $r \rightarrow 0$ (using the L’Hopital’s rule). The rest of the pf $P_k$ for $k=2,3,\cdots$ can be generated from the recursive relation (6). Unlike a zero-truncated distribution that is derived from an (a,b,0) distribution, the distributional quantities of the logarithmic distribution cannot be derived from an (a,b,0) distribution. Thus in order to gain more information about the logarithmic, its pf must be used. The mean and variance for the logarithmic distribution are:

(31)……..$\displaystyle E[N_T]=\frac{\theta}{\ln(1+\theta) }$

(32)……..$\displaystyle Var[N_T]=\frac{\theta \biggl[1+\theta-\theta / \ln(1+\theta) \biggr]}{\ln(1+\theta)}$

Sibuya Distribution
This is a truncated distribution that is derived from ETNB by letting $\theta \rightarrow \infty$ and making $-1. The following shows the information that is needed for the recursive generation of probabilities.

(33)……..$\displaystyle a=1 \ \ \ \ \ \ \ \ b=r-1$

(34)……..$\displaystyle P_1^T=-r$

All of the three items are obtained by letting $\theta \rightarrow \infty$ in the corresponding items in ETNB. To see that $\displaystyle P_1^T=-r$, rewrite the ETNB $P_1^T$ as follows:

……….$\displaystyle P_1^T=\frac{r \theta}{(1+\theta)^{r+1}-(1+\theta)}=\frac{r \theta}{(1+\theta) \ [(1+\theta)^r-1]}=r \ \frac{\theta}{1+\theta} \ \frac{1}{(1+\theta)^r-1}$

As $\theta \rightarrow \infty$, the ratio $\theta / (1+\theta)$ goes to 1. As $\theta \rightarrow \infty$, $(1+\theta)^r$ goes to 0 because $r$ is negative. Thus the above $P_1^T$ goes to $-r$. With the $a$ and $b$ in (34) and the $P_1^T$ in (35), the rest of the Shibuya pf can be generated by the recursive relation in (6). Note that the mean does not exist for the Shibuya distribution. The following is the pgf of the Sibuya distribution.

(35)……..$\displaystyle P^T(z)=1-(1-z)^{-r}$

Once these three zero-truncated distributions are obtained, we can derive the zero-modified versions of these distributions in the process described earlier.

Example 3
We demonstrate how ETNB is calculated. Let $r=-\frac{1}{2}$ and $\theta=3$. Then the parameters for the “artificial” negative binomial distribution are:

……….$\displaystyle a=\frac{3}{4} \ \ \ \ \ \ \ \ b=\biggl(-\frac{1}{2}-1 \biggr) \frac{3}{4}=-\frac{9}{8}$

The $P_0$ for the artificial negative binomial distribution is $P_0=(1/4)^{-0.5}=2$, making $1/(1-P_0)=-1$. We generate the fake negative binomial probabilities recursively using the $a$ and $b$. Then we multiply by the $1/(1-P_0)=-1$ to get the zero-truncated probabilities according to (7).

Table – Zero-Truncated ETNB

$\bold k$ Artificial $\bold P_{\bold k}$ Zero-Truncated $\bold P_\bold k^{\bold T}$
0 $\displaystyle 2$
1 $\displaystyle -\frac{3}{4}$ $\displaystyle \frac{3}{4}$
2 $\displaystyle -\frac{9}{64}$ $\displaystyle \frac{9}{64}$
3 $\displaystyle -\frac{27}{512}$ $\displaystyle \frac{27}{512}$
4 $\displaystyle -\frac{405}{16384}$ $\displaystyle \frac{405}{16384}$
5 $\displaystyle -\frac{1701}{131072}$ $\displaystyle \frac{1701}{131072}$

The column labeled artificial $P_k$ is obviously not probabilities. It is generated recursively using $a=3/4$ and $b=-9/8$. Then multiply the column labeled artificial $P_k$ by $1/(1-P_0)=-1$ to obtain the ETNB probabilities, which can also be computed directly using (25).

Using (26) and (27), the ETNB mean and variance are $E[N_T]=3/2$ and $Var[N_T]=3/2$. With an assumed value of $P_0^M=0.1$, we generate the first 5 zero-modified ETNB probabilities in the following table.

Table – Zero-Modified ETNB

$\bold k$ Artificial $\bold P_{\bold k}$ Zero-Truncated $\bold P_\bold k^{\bold T}$ Zero-Modified $\bold P_\bold k^{\bold M}$
0 $\displaystyle 2$ 0.1
1 $\displaystyle -\frac{3}{4}$ $\displaystyle \frac{3}{4}$ $\displaystyle \frac{2.7}{4}$
2 $\displaystyle -\frac{9}{64}$ $\displaystyle \frac{9}{64}$ $\displaystyle \frac{8.1}{64}$
3 $\displaystyle -\frac{27}{512}$ $\displaystyle \frac{27}{512}$ $\displaystyle \frac{24.3}{512}$
4 $\displaystyle -\frac{405}{16384}$ $\displaystyle \frac{405}{16384}$ $\displaystyle \frac{364.5}{16384}$
5 $\displaystyle -\frac{1701}{131072}$ $\displaystyle \frac{1701}{131072}$ $\displaystyle \frac{1530.9}{131072}$

With the assumed value of $P_0^M=0.1$, the zero-modified probabilities are obtained by multiplying the zero-truncated probabilities by $1-P_0^M=0.9$. Using (19) and (20), the zero-modified ETNB mean and variance are: $E[N_M]=1.35$ and $Var[N_M]=1.5525$.

Practice Problems

The discussion in this post has a great deal of technical details. A concise summary of the (a,b,0) class and (a,b,1) class is found here.

Practice problems on (a,b,0) class

Practice problems on (a,b,1) class

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