Practice Problem Set 12 – (a,b,1) class

The practice problems in this post are to reinforce the concepts of (a,b,0) class and (a,b,1) class discussed this blog post and this blog post in a companion blog. These two posts have a great deal of technical details, especially the one on (a,b,1) class. The exposition in this blog post should be more accessible.

Notation: $p_k=P(N=k)$ for $k=0,1,2,\cdots$ whenever $N$ is the counting distribution that is one of the (a,b,0) distributions – Poisson, binomial and negative binomial distribution. The notation $P_k^T$ is the probability that a zero-truncated distribution taking on the value $k$. Likewise $P_k^M$ is the probability that a zero-modified distribution taking on the value $k$.

 Practice Problem 12-A Consider a Poisson distribution with mean $\lambda=1.2$. Evaluate the probabilities $P_k$ where $k=0,1,2,3,4,5$. Consider the corresponding zero-truncated Poisson distribution. Evaluate the probabilities $P_k^T$ where $k=1,2,3,4,5$. Consider the corresponding zero-modified Poisson distribution with $P_0^M=0.4$. Evaluate the probabilities $P_k^M$ where $k=1,2,3,4,5$.
 Practice Problem 12-B This problem is a continuation of Problem 12-A. The following is the probability generating function (pgf) of the Poisson distribution in Problem 12-A. …………….$\displaystyle P(z)=e^{1.2 \ (z-1)}$ Determine the mean, variance and the pgf of the zero-truncated Poisson distribution in Problem 12-A. Determine the mean, variance and the pgf of the zero-modified Poisson distribution in Problem 12-A.
 Practice Problem 12-C Consider a negative binomial distribution with $a=2/3$ and $b=4/3$. Evaluate $P_0$, $P_1$, $P_2$ and $P_3$. Evaluate the mean and variance of the corresponding zero-truncated negative binomial distribution. Evaluate the mean and variance of the corresponding zero-modified negative binomial distribution with $P_0^M=0.1$.
 Practice Problem 12-D The following is the probability generating function (pgf) of the negative binomial distribution in Problem 12-C. …………….$\displaystyle P(z)=[1-2 (z-1)]^{-3}$ Evaluate $P'(z)$, $P''(z)$ and $P^{(3)}(z)$, the first, second and third derivative of $P(z)$, respectively. Evaluate $P'(0)$, $\frac{P''(0)}{2!}$ and $\frac{P^{(3)}(0)}{3!}$. Compare these with $P_1$, $P_2$ and $P_3$ found in Problem 12-C.
 Practice Problem 12-E This is a continuation of Problem 12-C and Problem 12-D. Using the pgf $P(z)$ in Problem 12-D, find the pgf of the corresponding zero-truncated negative binomial distribution. Call this pgf $g(z)$. Evaluate $g'(z)$, $g''(z)$ and $g^{(3)}(z)$, the first, second and third derivative of $g(z)$, respectively. Obtain $P_1^T$, $P_2^T$ and $P_3^T$ by evaluating $g'(0)$, $\frac{g''(0)}{2!}$ and $\frac{g^{(3)}(0)}{3!}$.
 Practice Problem 12-F This problem is similar to Problem 12-E. Using the pgf $g(z)$ in Problem 12-E, find the pgf of the corresponding zero-modified negative binomial distribution. Call this pgf $h(z)$. Evaluate $h'(z)$, $h''(z)$ and $h^{(3)}(z)$, the first, second and third derivative of $h(z)$, respectively. Obtain $P_1^M$, $P_2^M$ and $P_3^M$ by evaluating $h'(0)$, $\frac{h''(0)}{2!}$ and $\frac{h^{(3)}(0)}{3!}$.
 Practice Problem 12-G Suppose that the following three probabilities are from a zero-truncated (a,b,0) distribution. $P_3^T=0.147692308$ $P_4^T=0.132923077$ $P_5^T=0.111655385$ Determine the recursion parameters $a$ and $b$ of the corresponding (a,b,0) distribution. What is the (a,b,0) distribution? Evaluate the mean and variance of this (a,b,0) distribution.
 Practice Problem 12-H Consider a zero-modified distribution. The following three probabilities are from this zero-modified distribution. $P_2^M=0.08669868$ $P_3^M=0.162560025$ $P_4^M=0.205740032$ Determine the recursion parameters $a$ and $b$ of the corresponding (a,b,0) distribution. What is the (a,b,0) distribution? Determine $P_1^T$, the probability that the corresponding zero-truncated distribution taking on the value of 1. Determine $P_1^M$, the probability that the zero-modified distribution taking on the value of 1. Determine $P_0^M$.
 Practice Problem 12-I For a distribution from the (a,b,0) class, you are given that $a=0.5$ and $b=1.5$, $P_5^T=7/60$ for the corresponding zero-truncated distribution, $P_7^M=0.05$ for the corresponding zero-modified distribution for some value of $P_0^M$. Determine $P_0^M$.
 Practice Problem 12-J Generate an extended truncated negative binomial (ETNB) distribution with $r=-0.5$ and $\theta=2$. Note that this is to start with a negative binomial distribution with parameters $r=-0.5$ and $\theta=2$ and then derive its zero-truncated distribution. The parameters $r=-0.5$ and $\theta=2$ will not give a distribution but over look this point and go through the process of creating a zero-truncated distribution. In particular, determine the following. Determine $P_k^T$ for $k=1,2,3,4$. Determine the mean and variance of the ETNB distribution.

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12-A
• Poisson
• $P_0=e^{-1.2}$
• $P_1=1.2 \ e^{-1.2}$
• $P_2=0.72 \ e^{-1.2}$
• $P_3=0.288 \ e^{-1.2}$
• $P_4=0.0864 \ e^{-1.2}$
• $P_5=0.020736 \ e^{-1.2}$
• Zero-Truncated Poisson
• $\displaystyle P_1^T=\frac{1.2 \ e^{-1.2}}{1-e^{-1.2}}=0.517215313$
• $\displaystyle P_2^T=\frac{0.72 \ e^{-1.2}}{1-e^{-1.2}}=0.310329288$
• $\displaystyle P_3^T=\frac{0.288 \ e^{-1.2}}{1-e^{-1.2}}=0.124131675$
• $\displaystyle P_4^T=\frac{0.0864 \ e^{-1.2}}{1-e^{-1.2}}=0.037239503$
• $\displaystyle P_5^T=\frac{0.020736 \ e^{-1.2}}{1-e^{-1.2}}=0.008937481$
• Zero-Modified Poisson
• $P_0^M=0.4$
• $\displaystyle P_1^M=0.6 \ \frac{1.2 \ e^{-1.2}}{1-e^{-1.2}}=0.310329188$
• $\displaystyle P_2^M=0.6 \ \frac{0.72 \ e^{-1.2}}{1-e^{-1.2}}=0.186197513$
• $\displaystyle P_3^M=0.6 \ \frac{0.288 \ e^{-1.2}}{1-e^{-1.2}}=0.074479005$
• $\displaystyle P_4^M=0.6 \ \frac{0.0864 \ e^{-1.2}}{1-e^{-1.2}}=0.022343702$
• $\displaystyle P_5^M=0.6 \ \frac{0.020736 \ e^{-1.2}}{1-e^{-1.2}}=0.005362488$
12-B
• Zero-Truncated Poisson
• mean = $\displaystyle E[N_T]=\frac{1.2}{1-e^{-1.2}}=1.717215313$
• second moment = $\displaystyle =E[N_T^2]=\frac{2.64}{1-e^{-1.2}}=3.777873688$
• variance = $\displaystyle Var[N_T]=E[N_T^2]-E[N_T]^2=0.829045258$
• pgf = $\displaystyle P^T(z)=\frac{1}{1-e^{-1.2}} \ [e^{1.2 (z-1)} - e^{-1.2}]$
• Zero-Modified Poisson
• mean = $\displaystyle E[N_M]=\frac{0.72}{1-e^{-1.2}}=1.030329188$
• second moment = $\displaystyle =E[N_M^2]=\frac{1.584}{1-e^{-1.2}}=2.266724213$
• variance = $\displaystyle Var[N_M]=E[N_M^2]-E[N_M]^2=1.205145978$
• pgf = $\displaystyle P^M(z)=0.4+\frac{0.6}{1-e^{-1.2}} \ [e^{1.2 (z-1)} - e^{-1.2}]$
12-C
• Negative binomial
• $\displaystyle P_0=\frac{1}{27}$
• $\displaystyle P_1=\frac{2}{27}$
• $\displaystyle P_2=\frac{8}{81}$
• $\displaystyle P_3=\frac{80}{729}$
• Zero-truncated negative binomial
• $\displaystyle E[N_T]=\frac{81}{13}=6.23$
• $\displaystyle E[N_T^2]=\frac{729}{13}$
• $\displaystyle Var[N_T]=\frac{2916}{169}=17.2544$
• Zero-modified negative binomial
• $\displaystyle E[N_M]=\frac{72.9}{13}=5.607692308$
• $\displaystyle E[N_M^2]=\frac{656.1}{13}$
• $\displaystyle Var[N_M]=\frac{3214.89}{169}=19.02301775$
12-D
• $\displaystyle P'(z)=6 \ [1-2 (z-1)]^{-4}$
• $\displaystyle P''(z)=48 \ [1-2 (z-1)]^{-5}$
• $\displaystyle P^{(3)}(z)=480 \ [1-2 (z-1)]^{-6}$
• $\displaystyle P'(0)=\frac{2}{27}$
• $\displaystyle \frac{P''(0)}{2!}=\frac{8}{81}$
• $\displaystyle \frac{P^{(3)}(0)}{3!}=\frac{80}{729}$
21-E
• $\displaystyle g(z)=P^T(z)=\frac{27}{26}\biggl([1-2 (z-1)]^{-3}-\frac{1}{27} \biggr)$
• $\displaystyle g'(z)=\frac{27}{26} \ P'(z)$
• $\displaystyle g''(z)=\frac{27}{26} \ P''(z)$
• $\displaystyle g^{(3)}(z)=\frac{27}{26} \ P^{(3)}(z)$
• $\displaystyle g'(0)=\frac{1}{13}=0.076923077$
• $\displaystyle \frac{g''(0)}{2!}=\frac{4}{39}=0.102564103$
• $\displaystyle \frac{g^{(3)}(0)}{3!}=\frac{40}{351}=0.113960114$
12-F
• $\displaystyle h(z)=P^M(z)=0.1+\frac{24.3}{26}\biggl([1-2 (z-1)]^{-3}-\frac{1}{27} \biggr)$
• $\displaystyle h'(z)=\frac{24.3}{26} \ P'(z)$
• $\displaystyle h''(z)=\frac{24.3}{26} \ P''(z)$
• $\displaystyle h^{(3)}(z)=\frac{24.3}{26} \ P^{(3)}(z)$
• $\displaystyle h'(0)=\frac{0.9}{13}=0.069230769$
• $\displaystyle \frac{h''(0)}{2!}=\frac{3.6}{39}=0.092307692$
• $\displaystyle \frac{h^{(3)}(0)}{3!}=\frac{36}{351}=0.102564103$
12-G
• $\displaystyle a=\frac{3}{5}$ and $\displaystyle b=\frac{6}{5}$
• Negative binomial distribution with $r=3$ and $\displaystyle \theta=\frac{3}{2}$
• mean = $\displaystyle \frac{9}{2}=4.5$ and variance = $\displaystyle \frac{45}{4}=11.25$
12-H
• $\displaystyle a=-0.5625$ and $\displaystyle b=7.3125$
• Binomial distribution with $n=12$ and $\displaystyle p=0.36$
• $\displaystyle P_1^T=\frac{6.75 \ (0.64)^{12}}{1-0.64^{12}}=0.032027218$
• $\displaystyle P_1^M=\frac{P_2^M}{a+\frac{b}{2}}=0.028023158$
• $P_0^M=0.125$
12-I
• $P_0^M=0.2$
12-J
• ETNB Probabilities
• $\displaystyle P_1^T=\frac{1}{1-3^{0.5}} \biggl(\frac{-1}{3} \biggr) \ 3^{0.5}=0.7886751346$
• $\displaystyle P_2^T=\frac{1}{1-3^{0.5}} \biggl(\frac{-1}{18} \biggr) \ 3^{0.5}=0.1314458558$
• $\displaystyle P_3^T=\frac{1}{1-3^{0.5}} \biggl(\frac{-1}{54} \biggr) \ 3^{0.5}=0.0438152853$
• $\displaystyle P_4^T=\frac{1}{1-3^{0.5}} \biggl(\frac{-5}{648} \biggr) \ 3^{0.5}=0.0182563689$
• ETNB Mean and Variance
• $\displaystyle E[N_T]=\frac{-1}{1-3^{0.5}}=1.366025404$
• $\displaystyle E[N_T^2]=\frac{-2}{1-3^{0.5}}$
• $\displaystyle Var[N_T]=E[N_T^2]-E[N_T]^2=0.8660254038$

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The (a,b,0) and (a,b,1) classes

This post is on two classes of discrete distributions called the (a,b,0) class and (a,b,1) class. This post is a follow-up on two previous posts – summarizing the two posts and giving more examples. The (a,b,0) class is discussed in details in this post in a companion blog. The (a,b,1) class is discussed in details in this post in a companion blog.

Practice problems for the (a,b,0) class is found here. The next post is a practice problem set on the (a,b,1) class.

The (a,b,0) Class

A counting distribution is a discrete probability distribution that takes on the non-negative integers (0, 1, 2, …). Counting distributions are useful when we want to model occurrences of a certain random events. The three commonly used counting distributions would be the Poisson distribution, the binomial distribution and the negative binomial distribution. All three counting distributions can be generated recursively. For these three distributions, the ratio of any two consecutive probabilities multiplied by integers can be expressed as a linear quantity.

To make the last point in the preceding paragraph clear, let’s set some notations. For any integer $k=0,1,2,\cdots$, let $P_k$ be the probability that the counting distribution in question takes on the value $k$. For example, if we are considering the counting random variable $N$, then $P_k=P[N=k]$. Let’s look at the situation where the ratio of any two consecutive values of $P_k$ can be expressed as an expression for some constants $a$ and $b$.

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

Multiplying (1) by $k$ gives the following.

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

Note that the right-hand side of (1a) is a linear expression of $k$. This provides a way to fit observations to (a,b,0) distributions.

Any counting distribution that satisfies the recursive relation (1) is said to be a member of the (a,b,0) class of distributions. Note that the recursion starts at $k=1$. Does that mean $P_0$ can be any probability value we assign? The value of $P_0$ is fixed because all the $P_k$ must sum to 1.

The three counting distribution mentioned above – Poisson, binomial and negative binomial – are all members of the (a,b,0) class. In fact the (a,b,0) class essentially has three distributions. In other words, any member of (a,b,0) class must be one of the three distributions – Poisson, binomial and negative binomial.

An (a,b,0) distribution has its usual parameters, e.g. Poisson has a parameter $\lambda$, which is its mean. So we need to way to translate the usual parameters to and from the parameters $a$ and $b$. This is shown in the table below.

Table 1

Distribution Usual Parameters Probability at Zero Parameter a Parameter b
Poisson $\lambda$ $e^{-\lambda}$ 0 $\lambda$
Binomial $n$ and $p$ $(1-p)^n$ $\displaystyle -\frac{p}{1-p}$ $\displaystyle (n+1) \ \frac{p}{1-p}$
Negative binomial $r$ and $p$ $p^r$ $1-p$ $(r-1) \ (1-p)$
Negative binomial $r$ and $\theta$ $\displaystyle \biggl(\frac{1}{1+\theta} \biggr)^r$ $\displaystyle \frac{\theta}{1+\theta}$ $\displaystyle (r-1) \ \frac{\theta}{1+\theta}$
Geometric $p$ $p$ $1-p$ 0
Geometric $\theta$ $\displaystyle \frac{1}{1+\theta}$ $\displaystyle \frac{\theta}{1+\theta}$ 0

Table 1 provides the mapping to translate between the usual parameters and the recursive parameters $a$ and $b$.

Example 1
Let $a=-1/3$ and $b=5/3$. Let the initial probability be $P_0=81/256$. Generate the first 4 probabilities according to the recursion formula (1)

$\displaystyle P_0=\frac{81}{256}$

$\displaystyle P_1=\biggl(-\frac{1}{3}+\frac{5}{3} \biggr) \ \frac{81}{256}=\frac{108}{256}$

$\displaystyle P_2=\biggl(-\frac{1}{3}+\frac{5}{3} \cdot \frac{1}{2} \biggr) \ \frac{108}{256}=\frac{54}{256}$

$\displaystyle P_3=\biggl(-\frac{1}{3}+\frac{5}{3} \cdot \frac{1}{3} \biggr) \ \frac{54}{256}=\frac{12}{256}$

$\displaystyle P_4=\biggl(-\frac{1}{3}+\frac{5}{3} \cdot \frac{1}{4} \biggr) \ \frac{12}{256}=\frac{1}{256}$

Note that the sum of $P_0$ to $P_4$ is 1. So this has to be a binomial distribution and not Poisson or negative binomial. The binomial parameters are $n=4$ and $p=1/4$. According to Table 1, this translate to $a=-1/3$ and $b=5/3$. The initial probability is $P_0=(1-p)^4$.

Example 2
This example generates several probabilities recursively for the negative binomial distribution with $r=2$ and $\theta=4$. According to Table 1, this translates to $a=4/5$ and $b=4/5$. The following shows the probabilities up to $P_6$.

$\displaystyle P_0=\biggl(\frac{1}{5}\biggr)^2=\frac{1}{25}=0.04$

$\displaystyle P_1=\biggl(\frac{4}{5}+\frac{4}{5} \biggr) \ \frac{1}{25}=\frac{8}{125}=0.064$

$\displaystyle P_2=\biggl(\frac{4}{5}+\frac{4}{5} \cdot \frac{1}{2} \biggr) \ \frac{8}{125}=\frac{48}{625}=0.0768$

$\displaystyle P_3=\biggl(\frac{4}{5}+\frac{4}{5} \cdot \frac{1}{3} \biggr) \ \frac{48}{625}=\frac{256}{3125}=0.08192$

$\displaystyle P_4=\biggl(\frac{4}{5}+\frac{4}{5} \cdot \frac{1}{4} \biggr) \ \frac{256}{3125}=\frac{256}{3125}=0.08192$

$\displaystyle P_5=\biggl(\frac{4}{5}+\frac{4}{5} \cdot \frac{1}{5} \biggr) \ \frac{256}{3125}=\frac{6144}{78125}=0.0786432$

$\displaystyle P_6=\biggl(\frac{4}{5}+\frac{4}{5} \cdot \frac{1}{6} \biggr) \ \frac{6144}{78125}=\frac{28672}{390625}=0.07340032$

The above probabilities can also be computed using the probability function given below.

$\displaystyle P_k=(1+k) \ \frac{1}{25} \ \biggl(\frac{4}{5} \biggr)^k \ \ \ \ \ \ \ \ k=0,1,2,3,\cdots$

For the (a,b,0) class, it is not just about calculating probabilities recursively. The parameters $a$ and $b$ also give information about other distributional quantities such as moments and variance. For a more detailed discussion of the (a,b,0) class, refer to this post in a companion blog.

The (a,b,1) Class

If the (a,b,0) class is just another name for the three distributions of Poisson, binomial and negative binomial, what is the point of (a,b,0) class? Why not just work with these three distributions individually? Sure, generating the probabilities recursively is a useful concept. The probability functions of the three distributions already give us a clear and precise way to calculate probabilities. The notion of (a,b,0) class leads to the notion of (a,b,1) class, which gives a great deal more flexibility in the modeling counting distributions. It is possible that the (a,b,0) distributions do not adequately describe a random counting phenomenon being observed. For example, the sample data may indicate that the probability at zero may be larger than is indicated by the distributions in the (a,b,0) class. One alternative is to assign a larger value for $P_0$ and recursively generate the subsequent probabilities $P_k$ for $k=2,3,\cdots$. This recursive relation is the defining characteristics of the (a,b,1) class.

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

Note that the recursion begins at $k=2$. Can the values for $P_0$ and $P_1$ be arbitrary? The initial probability $P_0$ is an assumed value. The probability $P_1$ is the value such that the sum $P_1+P_2+P_3+\cdots$ is $1-P_0$.

The (a,b,1) class gives more flexibility in modeling. For example, the initial probability is $P_0=0.04$ in the negative binomial distribution in Example 2. If this $P_0$ is felt to be too small, then a larger value for $P_0$ can be assigned and then let the remaining probabilities be generated by recursion. We demonstrate how this is done using the same (a,b,0) distribution in Example 2.

Before we continue with Example 2, we comment that there are two subclasses in the (a,b,1) class. The subclasses are distinguished by whether $P_0=0$ or $P_0>0$. The (a,b,1) distributions are called zero-truncated distributions in the first case and are called zero-modified distributions in the second case.

Because there are three related distributions, we need to establish notations to keep track of the different distributions. We use the notations established in this post. The notation $P_k$ refers to the probabilities for an (a,b,0) distribution. From this (a,b,0) distribution, we can derive a zero-truncated distribution whose probabilities are notated by $P_k^T$. From this zero-truncated distribution, we can derive a zero-modified distribution whose probabilities are denoted by $P_k^M$. For example, for the negative binomial distribution in Example 2, we derive a zero-truncated negative binomial distribution (Example 3) and from it we derive a zero-modified negative binomial distribution (Example 4).

Example 3
In Example 3, we calculated the (a,b,0) probabilities $P_k$ up to $k=6$. We now calculate the probabilities $P_k^T$ for the corresponding zero-truncated negative binomial distribution. For a zero-truncated distribution, the value of zero is not recorded. So $P_k^T$ is simply $P_k$ divided by $1-P_0$.

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

The sum of $P_k^T$, $k=1,2,3,\cdots$, must be 1 since $P_0,P_1,P_3,\cdots$ is a probability distribution. The (a,b,0) $P_0$ is $1/25$. Then $1-P_0=24/25$, which means $1/(1-P_0)=25/24$. The following shows the zero-truncated probabilities.

$\displaystyle P_1^T=\frac{8}{125} \cdot \frac{25}{24}=\frac{8}{120}$

$\displaystyle P_2^T=\frac{8}{125} \cdot \frac{25}{24}=\frac{48}{600}$

$\displaystyle P_3^T=\frac{256}{3125} \cdot \frac{25}{24}=\frac{256}{3000}$

$\displaystyle P_4^T=\frac{256}{3125} \cdot \frac{25}{24}=\frac{256}{3000}$

$\displaystyle P_5^T=\frac{6144}{78125} \cdot \frac{25}{24}=\frac{6144}{75000}$

$\displaystyle P_6^T=\frac{28672}{390625} \cdot \frac{25}{24}=\frac{28672}{375000}$

The above are the first 6 probabilities of the zero-truncated negative binomial distribution with $a=4/5$ and $b=4/5$ or with the usual parameters $r=2$ and $\theta=4$. The above $P_k^T$ can also be calculated recursively by using (2). Just calculate $P_1^T$ and the rest of the probabilities can be generated using the recursion relation (2).

Example 4
From the zero-truncated negative binomial distribution in Example 3, we generate a zero-modified negative binomial distribution. If the original $P_0=0.04$ is considered too small,e.g. not big enough to account for the probability of zero claims, then we can assign a larger value to the zero probability. Let’s say 0.10 is more appropriate. So we set $P_0^M=0.10$. Then the rest of the $P_k^M$ must sum to $1-P_0^M$, or 0.9 in this example. The following shows how the zero-modified probabilities are related to the zero-truncated probabilities.

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

The following gives the probabilities for the zero-modified negative binomial distribution.

$\displaystyle P_0^M=0.1$ (assumed value)

$\displaystyle P_1^M=0.9 \cdot \frac{8}{120}=\frac{7.2}{120}$

$\displaystyle P_2^M=0.9 \cdot \frac{48}{600}=\frac{43.2}{600}$

$\displaystyle P_3^M=0.9 \cdot \frac{256}{3000}=\frac{230.4}{3000}$

$\displaystyle P_4^M=0.9 \cdot \frac{256}{3000}=\frac{230.4}{3000}$

$\displaystyle P_5^M=0.9 \cdot \frac{6144}{75000}=\frac{5529.6}{75000}$

$\displaystyle P_6^M=0.9 \cdot \frac{28672}{375000}=\frac{25804.8}{375000}$

The same probabilities can also be obtained by using the original (a,b,0) probabilities $P_k$ directly as follows:

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

ETNB Distribution

Examples 2, 3 and 4 show, starting with with an (a,b,0) distribution, how to derive a zero-truncated distribution and from it a zero-modified distribution. In these examples, we start with a negative binomial distribution and the derived distributions are zero-truncated negative binomial distribution and zero-modified negative binomial distribution. If the starting distribution is a Poisson distribution, then the same process would produce a zero-truncated Poisson distribution and a zero-modified Poisson distribution (with a particular assumed value of $P_0^M$).

There are members of the (a,b,1) class that do not originate from a member of the (a,b,0) class. Three such distributions are discussed in this post on the (a,b,1) class. We give an example discussing one of them.

Example 5
This example demonstrates how to work with the extended truncated negative binomial distribution (ETNB). The usual negative binomial distribution has two parameters $r$ and $\theta$ in one version ($r$ and $p$ in another version). Both parameters are positive real numbers. To define an ETNB distribution, we relax the $r$ parameter to include the possibility of $-1 in addition to $r>0$. Of course if $r>0$, then we just have the usual negative binomial distribution. So we focus on the new situation of $-1.

Let’s say $r=-0.2$ and $\theta=1$. We take these two parameters and generate the “negative binomial” probabilities, from which we generate the zero-truncated probabilities $P_k^T$ as shown in Example 3. Now the parameters $r=-0.2$ and $\theta=1$ do not belong to a legitimate negative binomial distribution. In fact the resulting $P_k$ are negative values. So this “negative binomial” distribution is just a device to get things going.

According to Table 1, $r=-0.2$ and $\theta=1$ translate to $a=1/2$ and $b=-3/5$. We generate the “negative binomial” probabilities using the recursive relation (1). Don’t be alarmed that the probabilities are negative.

$\displaystyle P_0=\biggl(\frac{1}{2}\biggr)^{-0.2}=2^{0.2}=1.148698355$

$\displaystyle P_1=\biggl(\frac{1}{2}-\frac{3}{5} \biggr) \ 2^{0.2}=\frac{-1}{10} \ 2^{0.2}$

$\displaystyle P_2=\biggl(\frac{1}{2}-\frac{3}{5} \ \frac{1}{2} \biggr) \ \frac{-1}{10} \ 2^{0.2}=\frac{-2}{100} \ 2^{0.2}$

$\displaystyle P_3=\biggl(\frac{1}{2}-\frac{3}{5} \ \frac{1}{3} \biggr) \ \frac{-2}{100} \ 2^{0.2}=\frac{-6}{1000} \ 2^{0.2}$

$\displaystyle P_4=\biggl(\frac{1}{2}-\frac{3}{5} \ \frac{1}{4} \biggr) \ \frac{-6}{1000} \ 2^{0.2}=\frac{-21}{10000} \ 2^{0.2}$

$\displaystyle P_5=\biggl(\frac{1}{2}-\frac{3}{5} \ \frac{1}{5} \biggr) \ \frac{-21}{10000} \ 2^{0.2}=\frac{-399}{500000} \ 2^{0.2}$

The initial $P_0$ is greater than 1 and the other so called probabilities are negative. But they are just a device to get the ETNB probabilities. Using the formula stated in (3) gives the following zero-truncated ETNB probabilities.

$\displaystyle P_1^T=\frac{1}{1-2^{0.2}} \ \biggl( \frac{-1}{10} \ 2^{0.2}\biggr)=0.7725023959$

$\displaystyle P_2^T=\frac{1}{1-2^{0.2}} \ \biggl( \frac{-2}{100} \ 2^{0.2}\biggr)=0.1545004792$

$\displaystyle P_3^T=\frac{1}{1-2^{0.2}} \ \biggl( \frac{-6}{1000} \ 2^{0.2}\biggr)=0.0463501438$

$\displaystyle P_4^T=\frac{1}{1-2^{0.2}} \ \biggl( \frac{-21}{10000} \ 2^{0.2}\biggr)=0.0162225503$

$\displaystyle P_5^T=\frac{1}{1-2^{0.2}} \ \biggl( \frac{-399}{500000} \ 2^{0.2}\biggr)=0.0061645691$

The above gives the first 5 probabilities of the zero-truncated ETNB distribution with parameters $a=1/2$ and $b=-3/5$. It is an (a,b,1) distribution that does not originate from any (legitimate) (a,b,0) distribution.

Practice Problems

The next post is a practice problem set on the (a,b,1) class.

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