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.

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

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
• $\displaystyle E[N_T]=\frac{-1}{1-3^{0.5}}=1.366025404$
• $\displaystyle E[N_T^2]=\frac{-2}{1-3^{0.5}}$
• $\displaystyle Var[N_T]=E[N_T^2]-E[N_T]^2=0.8660254038$

Dan Ma practice problems

Daniel Ma practice problems

Dan Ma actuarial science

Daniel Ma actuarial science

Dan Ma Math

Dan Ma Mathematics

Daniel Ma Math

Daniel Ma Mathematics

Actuarial exam

$\copyright$ 2019 – Dan Ma

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.

actuarial practice problems

Daniel Ma actuarial

Dan Ma actuarial

Dan Ma actuarial science

Daniel Ma actuarial science

Daniel Ma Math

Daniel Ma Mathematics

Actuarial exam

$\copyright$ 2019 – Dan Ma

Practice Problem Set 11 – (a,b,0) class

The practice problems in this post focus on counting distributions that belong to the (a,b,0) class, reinforcing the concepts discussed in this blog post in a companion blog.

The (a,b,1) class is a generalization of (a,b,0) class. It is discussed here. A practice problem set on the (a,b,1) class is found here.

Notation: $p_k=P(X=k)$ for $k=0,1,2,\cdots$ where $X$ is the counting distribution being focused on.

 Practice Problem 11-A Suppose that claim frequency $X$ follows a negative binomial distribution with parameters $r$ and $\theta$. The following is the probability function. $\displaystyle P(X=k)=\binom{r+k-1}{k} \ \biggl(\frac{1}{1+\theta} \biggr)^r \biggl(\frac{\theta}{1+\theta} \biggr)^k \ \ \ \ \ \ k=0,1,2,\cdots$ Evaluate the negative binomial distribution in two ways. Evaluate $P(X=k)$ for $k=0,1,2,3,4$ with $r=\frac{11}{6}$ and $\theta=1$. Convert $r=\frac{11}{6}$ and $\theta=1$ into the parameters $a$ and $b$. Evaluate the (a,b,0) distribution for $k=1,2,3,4$.
 Practice Problem 11-B Suppose that $X$ follows a distribution in the (a,b,0) class. You are given that $p_2=0.185351532$ $p_3=0.105032535$ $p_4=0.055142081$ Evaluate the probability that $X$ is at least 1.
 Practice Problem 11-C The following information is given about a distribution from the (a,b,0) class. $p_3=0.160670519$ $p_4=0.072301734$ $p_5=0.026028624$ What is the form of the distribution? Evaluate $p_1$.
 Practice Problem 11-D For a distribution from the (a,b,0) class, the following information is given. $p_1=0.214663$ $p_2=0.053666$ $p_3=0.012522$ Determine the variance of this distribution.
 Practice Problem 11-E For a distribution from the (a,b,0) class, you are given that $a=-1/3$ and $b=2$. Find the value of $p_0$.
 Practice Problem 11-F You are given that the distribution for the claim count $X$ satisfies the following recursive relation: $\displaystyle p_k=\frac{2 p_{k-1}}{k} \ \ \ \ \ \ \ \ \ k=1,2,3,\cdots$ Determine $P[X=2]$.
 Practice Problem 11-G Suppose that the random variable $X$ is from the (a,b,0) class. You are given that $a=-1/4$ and $b=7/4$. Calculate the probability that $X$ is at least 3.
 Practice Problem 11-H For a distribution from the (a,b,0) class, you are given that $p_2=0.20736$ $p_3=0.13824$ $p_4=0.082944$ Determine $p_1$.
 Practice Problem 11-I For a distribution from the (a,b,0) class, you are given that $a=1/6$ and $b=1/2$. Evaluate its mean.
 Practice Problem 11-J The random variable $X$ follows a distribution from the (a,b,0) class. You are given that $a=0.6$ and $b=-0.3$. Evaluate $E(X^2)$.
 Practice Problem 11-K The random variable $X$ follows a distribution from the (a,b,0) class. Suppose that $E(X)=3$ and $Var(X)=12$. Determine $p_2$.
 Practice Problem 11-L Given that a discrete distribution is a member of the (a,b,0) class. Which of the following statement(s) are true? If $a>0$ and $b>0$, then the variance of the distribution is greater than the mean. If $a>0$ and $b=0$, then the variance of the distribution is less than the mean. If $a<0$ and $b>0$, then the variance of the distribution is greater than the mean. A. ……….. 1 only B. ……….. 2 only C. ……….. 3 only D. ……….. 1 and 2 only E. ……….. 1 and 3 only
 Practice Problem 11-M Given that a discrete distribution is a member of the (a,b,0) class, determine the variance of the distribution if $a=1/6$ and $b=1/4$.
 Practice Problem 11-N For a distribution in the (a,b,0) class, $p_2=0.2048$ and $p_3=0.0512$. Furthermore, the mean of the distribution is 1. Determine $p_1$.

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

11-A
• $a=1/2$ and $b=5/12$
• $\displaystyle p_0=\biggl(\frac{1}{2} \biggr)^{11/6}$
• $\displaystyle p_1=\biggl(\frac{11}{12} \biggr) \biggl(\frac{1}{2} \biggr)^{11/6}$
• $\displaystyle p_2=\biggl(\frac{17}{24} \biggr) \biggl(\frac{11}{12} \biggr) \biggl(\frac{1}{2} \biggr)^{11/6}$
• $\displaystyle p_3=\biggl(\frac{23}{36} \biggr) \biggl(\frac{17}{24} \biggr) \biggl(\frac{11}{12} \biggr) \biggl(\frac{1}{2} \biggr)^{11/6}$
• $\displaystyle p_4=\biggl(\frac{29}{48} \biggr) \biggl(\frac{23}{36} \biggr) \biggl(\frac{17}{24} \biggr) \biggl(\frac{11}{12} \biggr) \biggl(\frac{1}{2} \biggr)^{11/6}$
11-B
• $1-(0.6)^{2.25}=0.683159775$
11-C
• Poisson distribution with mean 1.8
• $1.8 e^{-1.8}=0.2975$
11-D
• 0.46875
11-E
• $\displaystyle p_0=\frac{243}{1024}=0.237305$
11-F
• $2 e^{-2}=0.270671$
11-G
• 0.09888
11-H
• 0.2592
11-I
• 0.8
11-J
• 2.4375
11-K
• $\displaystyle p_2=\frac{5.625}{256}=0.02197$
11-L
• A. 1 only
11-M
• 0.6
11-N
• 0.4096

Daniel Ma Math

Daniel Ma Mathematics

Actuarial exam

$\copyright$ 2018 – Dan Ma

Practice Problem Set 6 – Negative Binomial Distribution

This post has exercises on negative binomial distributions, reinforcing concepts discussed in
this previous post. There are several versions of the negative binomial distribution. The exercises are to reinforce the thought process on how to use the versions of negative binomial distribution as well as other distributional quantities.

 Practice Problem 6A The annual claim frequency for an insured from a large population of insured individuals is modeled by the following probability function. $\displaystyle P(X=x)=\binom{1.8+x}{x} \ \biggl(\frac{5}{6}\biggr)^{2.8} \ \biggl(\frac{1}{6}\biggr)^x \ \ \ \ \ \ x=0,1,2,3,\cdots$ Determine the following: The percentage of the population of insureds that are expected to have exactly 2 claims during a year. The mean annual claim frequency of a randomly selected insured. The variance of the number of claims in a year for a randomly selected insured.
 Practice Problem 6B The number of claims in a year for an insured from a large group of insureds is modeled by the following model. $\displaystyle P(X=x \lvert \lambda)=\frac{e^{-\lambda} \lambda^x}{x!} \ \ \ \ \ x=0,1,2,3,\cdots$ The parameter $\lambda$ varies from insured to insured. However, it is known that $\lambda$ is modeled by the following density function. $\displaystyle g(\lambda)=62.5 \ \lambda^2 \ e^{-5 \lambda} \ \ \ \ \ \ \lambda>0$ Given that a randomly selected insured has at least one claim, determine the probability that the insured has more than one claim.
 Practice Problem 6C Suppose that the number of accidents per year per driver in a large group of insured drivers follows a Poisson distribution with mean $\lambda$. The parameter $\lambda$ follows a gamma distribution with mean 0.6 and variance 0.24. Determine the probability that a randomly selected driver from this group will have no more than 2 accidents next year.
 Practice Problem 6D Suppose that the random variable $X$ follows a negative binomial distribution such that $P(X=0)=0.2397410$ $P(X=1)=0.1038878$ $P(X=2)=0.0398236$ Determine the mean and variance of $X$.
 Practice Problem 6E Suppose that the random variable $X$ follows a negative binomial distribution with mean 0.36 and variance 1.44. Determine $P(X=3)$.
 Practice Problem 6F A large group of insured drivers is divided into two classes – “good” drivers and “bad”drivers. Seventy five percent of the drivers are considered “good” drivers and the remaining 25% are considered “bad”drivers. The number of claims in a year for a “good” driver is modeled by a negative binomial distribution with mean 0.5 and variance 0.625. On the other hand, the number of claims in a year for a “bad” driver is modeled by a negative binomial distribution with mean 2 and variance 4. For a randomly selected driver from this large group, determine the probability that the driver will have 3 claims in the next year.
 Practice Problem 6G The number of losses in a year for one insurance policy is the random variable $X$ where $X=0,1,2,\cdots$. The random variable $X$ is modeled by a geometric distribution with mean 0.4 and variance 0.56. What is the probability that the total number of losses in a year for three randomly selected insurance policies is 2 or 3?
 Practice Problem 6H The random variable $X$ follows a negative binomial distribution. The following gives further information. $E(X)=3$ $\displaystyle P(X=0)=\frac{4}{25}$ $\displaystyle P(X=1)=\frac{24}{125}$ Determine $P(X=2)$ and $P(X=3)$.
 Practice Problem 6I Coin 1 is an unbiased coin, i.e. when tossing the coin, the probability of getting a head is 0.5. Coin 2 is a biased coin such that when tossing the coin, the probability of getting a head is 0.6. One of the coins is chosen at random. Then the chosen coin is tossed repeatedly until a head is obtained. Suppose that the first head is observed in the fifth toss. Determine the probability that the chosen coin is Coin 2.
 Practice Problem 6J In a production process, the probability of manufacturing a defective rear view mirror for a car is 0.075. Assume that the quality status of any rear view mirror produced in this process is independent of the status of any other rear view mirror. A quality control inspector is to examine rear view mirrors one at a time to obtain three defective mirrors. Determine the probability that the third defective mirror is the 10th mirror examined.

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

6A
• 8.8696%
• 0.56
• 0.672
6B $\displaystyle \frac{171}{546}$
6C 0.9548
6D mean = 0.65, variance = 0.975
6E 0.016963696
6F 0.04661
6G $\displaystyle \frac{31000}{117649}$
6H
• $\displaystyle P(X=2)=\frac{108}{625}$
• $\displaystyle P(X=3)=\frac{432}{3125}$
6I 0.329543
6J 0.008799914

Daniel Ma Math

Daniel Ma Mathematics

Actuarial exam

$\copyright$ 2017 – Dan Ma

Several versions of negative binomial distribution

This post shows how to work with negative binomial distribution from an actuarial modeling perspective. The negative binomial distribution is introduced as a Poisson-gamma mixture. Then other versions of the negative binomial distribution follow. Specific attention is paid to the thought processes that facilitate calculation involving negative binomial distribution.

Negative Binomial Distribution as a Poisson-Gamma Mixture

Here’s the setting for the Poisson-gamma mixture. Suppose that $X \lvert \Lambda$ has a Poisson distribution with mean $\Lambda$ and that $\Lambda$ is a random variable that varies according to a gamma distribution with parameters $\alpha$ (shape parameter) and $\rho$ (rate parameter). Then the following is the unconditional probability function of $X$.

$\displaystyle (1) \ \ \ \ P[X=k]=\frac{\Gamma(\alpha+k)}{k! \ \Gamma(\alpha)} \ \biggl( \frac{\rho}{1+\rho} \biggr)^\alpha \ \biggl(\frac{1}{1+\rho} \biggr)^k \ \ \ \ k=0,1,2,3,\cdots$

The distribution described in (1) is one parametrization of the negative binomial distribution (derived here). It has two parameters $\alpha$ and $\rho$ (coming from the gamma mixing weights). The following is another parametrization.

$\displaystyle (2) \ \ \ \ P[X=k]=\frac{\Gamma(\alpha+k)}{k! \ \Gamma(\alpha)} \ \biggl( \frac{1}{1+\theta} \biggr)^\alpha \ \biggl(\frac{\theta}{1+\theta} \biggr)^k \ \ \ \ k=0,1,2,3,\cdots$

The distribution described in (2) is obtained when the gamma mixing weight $\Lambda$ has a shape parameter $\alpha$ and a scale parameter $\theta$. Since the gamma scale parameter and rate parameter is related by $\rho=1/ \theta$, (2) can be derived from (1) by setting $\rho=1/ \theta$.

Both (1) and (2) contain the ratio $\frac{\Gamma(\alpha+k)}{k! \ \Gamma(\alpha)}$ that is expressed using the gamma function. The next task is to simplify the ratio using a general notion of binomial coefficient.

The Poisson-gamma mixture is discussed in this blog post in a companion blog called Topics in Actuarial Modeling.

General Binomial Coefficient

The familiar binomial coefficient is the following:

$(3) \ \ \ \ \displaystyle \binom{n}{j}=\frac{n!}{j! (n-j)!}$

where the top number $n$ is a positive integer and the bottom number $j$ is a non-negative integer such that $n \ge j$. Other notations for binomial coefficient are $C(n,j)$, $_nC_j$ and $C_{n,j}$. The right hand side of the above expression can be simplified by canceling out $(n-j)!$.

$(4) \ \ \ \ \displaystyle \binom{n}{j}=\frac{n (n-1) (n-2) \cdots (n-(j-1))}{j!}$

The expression in (4) is obtained by canceling out $(n-j)!$ in (3). Note that $n$ does not have to be an integer for the calculation in (4) to work. The bottom number $j$ has to be a non-negative number since $j!$ is involved. However, $n$ can be any positive real number as long as $n>j-1$.

Thus the expression in (4) gives a new meaning to the binomial coefficient where $n$ is a positive real number and $j$ is a non-negative integer such that $n>j-1$.

$\displaystyle (5) \ \ \ \ \binom{n}{j}=\left\{ \begin{array}{ll} \displaystyle \frac{n (n-1) (n-2) \cdots (n-(j-1))}{j!} &\ n>j-1, j=1,2,3,\cdots \\ \text{ } & \text{ } \\ \displaystyle 1 &\ j=0 \\ \text{ } & \text{ } \\ \displaystyle \text{undefined} &\ \text{otherwise} \end{array} \right.$

For example, $\binom{2.3}{1}=2.3$ and $\binom{5.1}{3}=(5.1 \times 4.1 \times 3.1) / 3!=10.8035$. The thought process is that the numerator is obtained by subtracting 1 $j-1$ times from $n$. If $j=0$, this thought process would not work. For convenience, $\binom{n}{0}=1$ when $n$ is a positive real number.

We now use the binomial coefficient defined in (5) to simplify the ratio $\frac{\Gamma(\alpha+k)}{k! \ \Gamma(\alpha)}$ where $\alpha$ is a positive real number and $k$ is a non-negative integer. We use a key fact about gamma function: $\Gamma(1+w)=w \Gamma(w)$. Then for any integer $k \ge 1$, we have the following derivation.

\displaystyle \begin{aligned} \Gamma(\alpha+k)&=\Gamma(1+\alpha+k-1)=(\alpha+k-1) \ \Gamma(\alpha+k-1) \\&=(\alpha+k-1) \ (\alpha+k-2) \ \Gamma(\alpha+k-2) \\&\ \ \ \vdots \\&=(\alpha+k-1) \ (\alpha+k-2) \cdots (\alpha+1) \ \alpha \ \Gamma(\alpha) \end{aligned}

$\displaystyle \frac{\Gamma(\alpha+k)}{k! \ \Gamma(\alpha)}=\frac{(\alpha+k-1) \ (\alpha+k-2) \cdots (\alpha+1) \ \alpha}{k!}$

The right hand side of the above expression is precisely the binomial coefficient $\binom{\alpha+k-1}{k}$ when $k \ge 1$. Thus we have the following relation.

$\displaystyle (6) \ \ \ \ \frac{\Gamma(\alpha+k)}{k! \ \Gamma(\alpha)}=\frac{(\alpha+k-1) \ (\alpha+k-2) \cdots (\alpha+1) \ \alpha}{k!}=\binom{\alpha+k-1}{k}$

where $k$ is an integer with $k \ge 1$.

Negative Binomial Distribution

With relation (6), the two versions of Poisson-gamma mixture stated in (1) and (2) are restated as follows:

$\displaystyle (7) \ \ \ \ P[X=k]=\binom{\alpha+k-1}{k} \ \biggl( \frac{\rho}{1+\rho} \biggr)^\alpha \ \biggl(\frac{1}{1+\rho} \biggr)^k \ \ \ \ k=0,1,2,3,\cdots$

$\displaystyle (8) \ \ \ \ P[X=k]=\binom{\alpha+k-1}{k} \ \biggl( \frac{1}{1+\theta} \biggr)^\alpha \ \biggl(\frac{\theta}{1+\theta} \biggr)^k \ \ \ \ k=0,1,2,3,\cdots$

The above two parametrizations of negative binomial distribution are used if information about the Poisson-gamma mixture is known. In (7), the gamma distribution in the Poisson-gamma mixture has shape parameter $\alpha$ and rate parameter $\rho$. In (8), the gamma distribution has shape parameter $\alpha$ and scale parameter $\theta$. The following is a standalone version of the binomial distribution.

$\displaystyle (9) \ \ \ \ P[X=k]=\binom{\alpha+k-1}{k} \ p^\alpha \ (1-p)^k \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ k=0,1,2,3,\cdots$

In (9), the negative binomial distribution has two parameters $\alpha>0$ and $p$ where $0. In this parmetrization, the parameter $p$ is simply a real number between 0 and 1. It can be viewed as a probability. In fact, this is the case when the parameter $\alpha$ is an integer. Version (9) can be restated as follows when $\alpha$ is an integer.

\displaystyle \begin{aligned} (10) \ \ \ \ P[X=k]&=\binom{\alpha+k-1}{k} \ p^\alpha \ (1-p)^k \\&=\frac{(\alpha+k-1)!}{k! \ (\alpha-1)!} \ p^\alpha \ (1-p)^k \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ k=0,1,2,3,\cdots \end{aligned}

In version (10), the parameters are $\alpha$ (a positive integer) and a real number $p$ with $0. Since $\alpha$ is an integer, the usual binomial coefficient appears in the probability function.

Version (10) has a natural interpretation. A Bernoulli trial is an random experiment that results in two distinct outcome – success or failure. Suppose that the probability of success is $p$ in each trial. Perform a series of independent Bernoulli trials until exactly $\alpha$ successes occur where $\alpha$ is a fixed positive integer. Let the random variable $X$ be the number of failures before the occurrence of the $\alpha$th success. Then (10) is the probability function for the random variable $X$.

A special case of (10). When the parameter $\alpha$ is 1, the negative binomial distribution has a special name.

$\displaystyle (11) \ \ \ \ P[X=k]=p \ (1-p)^k \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ k=0,1,2,3,\cdots$

The distribution in (11) is said to be a geometric distribution with parameter $p$. The random variable $X$ defined by (11) can be interpreted as the number of failures before the occurrence of the first success when performing a series of independent Bernoulli trials. Another important property of the geometric distribution is that it is the only discrete distribution with the memoryless property. As a result, the survival function of the geometric distribution is $P[X>k]=(1-p)^{k+1}$ where $k=0,1,2,\cdots$.

The probability functions of various versions of the negative binomial distribution have been developed in (1), (2), (7), (8), (9), (10) and (11). Other distributional quantities can be derived from the Poisson-gamma mixture. We derive the mean and variance of the negative binomial distribution.

Suppose that the negative binomial distribution is that of version (8). The conditional random variable $X \lvert \Lambda$ has a Poisson distribution with mean $\Lambda$ and the random variable $\Lambda$ has a gamma distribution with a shape parameter $\alpha$ and a scale parameter $\theta$. Note that $E(\Lambda)=\alpha \theta$ and $Var(\Lambda)=\alpha \theta^2$. Furthermore, we have the conditional mean and conditional variance

The following derives the mean and variance of $X$.

$E(X)=E[E(X \lvert \Lambda)]=E[\Lambda]=\alpha \theta$

\displaystyle \begin{aligned} Var(X)&=E[Var(X \lvert \Lambda)]+Var[E(X \lvert \Lambda)] \\&=E[\Lambda]+Var[\Lambda] \\&=\alpha \theta+\alpha \theta^2 \\&=\alpha \theta (1+\theta) \end{aligned}

The above mean and variance are for parametrization in (8). To obtain the mean and variance for the other parametrizations, make the necessary translation. For example, to get (7), plug $\theta=\frac{1}{\rho}$ into the above mean and variance. For (9), let $p=\frac{1}{1+\theta}$. Then solve for $\theta$ and plug that into the above mean and variance. Version (10) should have the same formulas as for (9). To get (11), set $\alpha=1$. The following table lists the negative binomial mean and variance.

Version Mean Variance
(7) $\displaystyle E(X)=\frac{\alpha}{\rho}$ $\displaystyle Var(X)=\frac{\alpha}{\rho} \ \biggl(1+\frac{1}{\rho} \biggr)$
(8) $E(X)=\alpha \ \theta$ $Var(X)=\alpha \ \theta \ (1+\theta)$
(9) and (10) $\displaystyle E(X)=\frac{\alpha (1-p)}{p}$ $\displaystyle Var(X)=\frac{\alpha (1-p)}{p^2}$
(11) $\displaystyle E(X)=\frac{1-p}{p}$ $\displaystyle Var(X)=\frac{1-p}{p^2}$

The table shows that the variance of the negative binomial distribution is greater than its mean (regardless of the version). This stands in contrast with the Poisson distribution whose mean and the variance are equal. Thus the negative binomial distribution would be a suitable model in situations where the variability of the empirical data is greater than the sample mean.

Modeling Claim Count

The negative binomial distribution is a discrete probability distribution that takes on the non-negative integers $0,1,2,3,\cdots$. Thus it can be used as a counting distribution, i.e. a model for the number of events of interest that occur at random. For example, the $X$ described above can be a good model for the frequency of loss, i,e, the random variable of the number of losses, either arising from a portfolio of insureds or from a particular insured in a given period of time.

The Poisson-gamma model has a great deal of flexibility. Consider a large population of individual insureds. The number of losses (or claims) in a year for each insured has a Poisson distribution with mean $\Lambda$. From insured to insured, there is uncertainty in the mean annual claim frequency $\Lambda$. However, the random variable $\Lambda$ varies according to a gamma distribution. As a result, the annual number of claims for an “average” insured or a randomly selected insured from the population will follow a negative binomial distribution.

Thus in a Poisson-gamma model, the claim frequency for an individual in the population follows a Poisson distribution with unknown gamma mean. The weighted average of these conditional Poisson claim frequencies is a negative binomial distribution. Thus the average claim frequency over all individuals has a negative binomial distribution.

The table in the preceding section shows that the variance of the negative binomial distribution is greater than the mean. This is in contrast to the fact that the variance and the mean of a Poisson distribution are equal. Thus the unconditional claim frequency $X$ is more dispersed than its conditional distributions. The increased variance of the negative binomial distribution reflects the uncertainty in the parameter of the Poisson mean across the population of insureds. The uncertainty in the parameter variable $\Lambda$ has the effect of increasing the unconditional variance of the mixture distribution of $X$. Recall that the variance of a mixture distribution has two components, the weighted average of the conditional variances and the variance of the conditional means. The second component represents the additional variance introduced by the uncertainty in the parameter $\Lambda$.

We present two examples. More examples to come at the end of the post.

Example 1
For a given insured driver in a large portfolio of insured drivers, the number of collision claims in a year has a Poisson distribution with mean $\Lambda$. The Poisson mean $\Lambda$ follows a gamma distribution with mean 4 and variance 80. For a randomly selected insured driver from this portfolio,

• what is the probability of having exactly 2 collision claims in the next year?
• what is the probability of having at most one collision claim in the next year?

The number of collision claims in a year is a Poisson-gamma mixture and thus is a negative binomial distribution. From the given gamma mean and variance, we can determine the parameters of the gamma distribution. In this example, we use the parametrization of (8). Expressing the gamma mean and variance in terms of the shape and scale parameters, we have $\alpha \theta=4$ and $\alpha \theta^2=80$. These two equations give $\alpha=0.2$ and $\theta=20$. The probabilities are calculated based on (8).

$\displaystyle P[X=0]=\biggl( \frac{1}{21} \biggr)^{0.2}=0.5439$

$\displaystyle P[X=1]=\binom{0.2}{1} \ \biggl( \frac{1}{21} \biggr)^{0.2} \ \biggl( \frac{20}{21} \biggr)=0.2 \ \biggl( \frac{1}{21} \biggr)^{0.2} \ \biggl( \frac{20}{21} \biggr)=0.1036$

$\displaystyle P[X=2]=\binom{1.2}{2} \ \biggl( \frac{1}{21} \biggr)^{0.2} \ \biggl( \frac{20}{21} \biggr)^2=\frac{1.2 (0.2)}{2!} \ \biggl( \frac{1}{21} \biggr)^{0.2} \ \biggl( \frac{20}{21} \biggr)^2=0.0592$

The answer for the first question is $P[X=2]=0.0592$. The answer for the second question is $P[X \le 1]=P[X=0]+P[X=1]=0.6475$. Thus there is a closed to 65% chance that an insured driver has at most one claim in a year.

Example 2
For an automobile insurance company, the distribution of the annual number of claims for a policyholder chosen at random is modeled by a negative binomial distribution that is a Poisson-gamma mixture. The gamma distribution in the mixture has a shape parameters of $\alpha=1$ and scale parameter $\theta=3$. What is the probability that a randomly selected policyholder has more than two claims in a year?

Since the gamma shape parameter is 1, the unconditional number of claims in a year is a geometric distribution with parameter $p=1/4$. The following is the desired probability.

$\displaystyle P[X>2]=\biggl( \frac{1}{4} \biggr) \ \biggl( \frac{3}{4} \biggr)^3=\frac{27}{256}=0.1055$

A Recursive Formula

The probability functions described in (1), (2), (7), (8), (9), (10) and (11) describe clearly how the negative binomial probabilities are calculated based on the two given parameters. The probabilities can also be calculated recursively. Let $P_k=P[X=k]$ where $k=0,1,2,\cdots$. We introduce a recursive formula that allows us to compute the value $P_k$ if $P_{k-1}$ is known. The following is the form of the recursive formula.

$\displaystyle (12) \ \ \ \ \frac{P_k}{P_{k-1}}=a+\frac{b}{k} \ \ \ \ \ \ \ \ \ \ k=1,2,3,\cdots$

In (12), the numbers $a$ and $b$ are constants. Note that the formula (12) calculates probabilities $P_k$ for all $k \ge 1$. It turns out that the initial probability $P_0$ is determined by the constants $a$ and $b$. Thus the constants $a$ and $b$ completely determines the probability distribution represented by $P_k$. Any discrete probability distribution that satisfies this recursive relation is said to be a member of the (a,b,0) class of distributions.

We show that the negative binomial distribution is a member of the (a,b,0) class of distributions. First, assume that the negative binomial distribution conforms to the parametrization in (8) with parameters $\alpha$ and $\theta$. Then let $a$ and $b$ be defined as follows:

$\displaystyle a=\frac{\theta}{1+\theta}$

$\displaystyle b=\frac{(\alpha-1) \theta}{1+\theta}$.

Let the initial probability be $P_0=(1+\theta)^{-\alpha}$. We claim that the probabilities generated by the formula (12) are identical to the ones calculated from (8). To see this, let’s calculate a few probabilities using the formula.

$\displaystyle P_0=\biggl(\frac{1}{1+\theta} \biggr)^\alpha$

\displaystyle \begin{aligned} P_1&=(a+b) P_0 \\&=\biggl(\frac{\theta}{1+\theta}+ \frac{(\alpha-1) \theta}{1+\theta} \biggr) \ \biggl(\frac{1}{1+\theta} \biggr)^\alpha \\&=\alpha \ \biggl(\frac{1}{1+\theta} \biggr)^\alpha \ \frac{\theta}{1+\theta} \\&=\binom{\alpha}{1} \ \biggl(\frac{1}{1+\theta} \biggr)^\alpha \ \frac{\theta}{1+\theta}=P[X=1] \end{aligned}

\displaystyle \begin{aligned} P_2&=\biggl(a+\frac{b}{2} \biggr) P_1 \\&=\biggl(\frac{\theta}{1+\theta}+ \frac{(\alpha-1) \theta}{2(1+\theta)} \biggr) \ \alpha \ \biggl(\frac{1}{1+\theta} \biggr)^\alpha \ \frac{\theta}{1+\theta} \\&=\frac{(\alpha+1) \alpha}{2!} \ \biggl(\frac{1}{1+\theta} \biggr)^\alpha \ \biggl( \frac{\theta}{1+\theta} \biggr)^2 \\&=\binom{\alpha+1}{2} \ \biggl(\frac{1}{1+\theta} \biggr)^\alpha \ \biggl( \frac{\theta}{1+\theta} \biggr)^2=P[X=2] \end{aligned}

\displaystyle \begin{aligned} P_3&=\biggl(a+\frac{b}{3} \biggr) P_2 \\&=\biggl(\frac{\theta}{1+\theta}+ \frac{(\alpha-1) \theta}{3(1+\theta)} \biggr) \ \frac{(\alpha+1) \alpha}{2!} \ \biggl(\frac{1}{1+\theta} \biggr)^\alpha \ \biggl( \frac{\theta}{1+\theta} \biggr)^2 \\&=\frac{(\alpha+2) (\alpha+1) \alpha}{3!} \ \biggl(\frac{1}{1+\theta} \biggr)^\alpha \ \biggl( \frac{\theta}{1+\theta} \biggr)^3 \\&=\binom{\alpha+2}{3} \ \biggl(\frac{1}{1+\theta} \biggr)^\alpha \ \biggl( \frac{\theta}{1+\theta} \biggr)^3=P[X=3] \end{aligned}

The above derivation demonstrates that formula (12) generates the same probabilities as (8). By adjusting the constants $a$ and $b$, the recursive formula can also generate the probabilities in the other versions of the negative binomial distribution. For the negative binomial version (9) with parameters $\alpha$ and $p$, the $a$ and $b$ should be defined as follows:

$a=1-p$

$b=(\alpha-1) \ (1-p)$

With the initial probability $P_0=p^\alpha$, the recursive formula (12) will generate the same probabilities as those from version (9).

More Examples

Example 3
Suppose that an insured will produce $n$ claims during the next exposure period is

$\displaystyle \frac{e^{-\lambda} \ \lambda^n}{n!}$

where $n=0,1,2,3,\cdots$. Furthermore, the parameter $\lambda$ varies according to a distribution with the following density function:

$\displaystyle g(\lambda)=\frac{9.261}{2} \ \lambda^2 \ e^{-2.1 \lambda} \ \ \ \ \ \ \lambda>0$

What is the probability that a randomly selected insured will produce more than 2 claims during the next exposure period?

Note that the claim frequency for an individual insured has a Poisson distribution with mean $\lambda$. The given density function for the parameter $\lambda$ is that of a gamma distribution with $\alpha=3$ and rate parameter $\rho=2.1$. Thus the number of claims in an exposure period for a randomly selected (or “average” insured) will have a negative binomial distribution. In this case the parametrization (7) is the most useful one to use.

\displaystyle \begin{aligned} P(X=k)&=\binom{k+2}{k} \ \biggl(\frac{2.1}{3.1} \biggr)^3 \ \biggl(\frac{1}{3.1} \biggr)^k \\&=\frac{(k+2) (k+1)}{2} \ \biggl(\frac{21}{31} \biggr)^3 \ \biggl(\frac{10}{31} \biggr)^k \ \ \ \ \ k=0,1,2,3,\cdots \end{aligned}

The following calculation gives the relevant probabilities to answer the question.

$\displaystyle P(X=0)=\biggl(\frac{21}{31} \biggr)^3$

$\displaystyle P(X=1)=3 \ \biggl(\frac{21}{31} \biggr)^3 \ \biggl(\frac{10}{31} \biggr)$

$\displaystyle P(X=2)=6 \ \biggl(\frac{21}{31} \biggr)^3 \ \biggl(\frac{10}{31} \biggr)^2$

Summing the three probabilities gives $P(X \le 2)=0.805792355$. Then $P(X>2)=0.1942$. There is a 19.42% chance that a randomly selected insured will have more than 2 claims in an exposure period.

Example 4
The number of claims in a year for each insured in a large portfolio has a Poisson distribution with mean $\lambda$. The parameter $\lambda$ follows a gamma distribution with mean 0.75 and variance 0.5625.

Determine the proportion of insureds that are expected to have less than 1 claim in a year.

Setting $\alpha \theta=0.75$ and $\alpha \theta^2=0.5625$ gives $\alpha=1$ and $\theta=0.75$. Thus the parameter $\lambda$ follows a gamma distribution with shape parameter $\alpha=1$ and scale parameter $\theta=0.75$. This is an exponential distribution with mean 0.75. The problems asks for the proportion of insured with $\lambda<1$. Thus the answer is $1-e^{-1/0.75}=0.7364$. Thus about 74% of the insured population are expected to have less than 1 claim in a year.

Example 5
Suppose that the number of claims in a year for an insured has a Poisson distribution with mean $\Lambda$. The random variable $\Lambda$ follows a gamma distribution with shape parameter $\alpha=2.5$ and scale parameter $\theta=1.2$.

One thousand insureds are randomly selected and are to be observed for a year. Determine the number of selected insureds expected to have exactly 3 claims by the end of the one-year observed period.

With this being a Poisson-gamma mixture, the number of claims in a year for a randomly selected insured has a negative binomial distribution. Using (8) and based on the gamma parameters given, the following is the probability function of negative binomial distribution.

$\displaystyle P(X=k)=\binom{k+1.5}{k} \ \biggl(\frac{1}{2.2} \biggr)^{2.5} \ \biggl(\frac{1.2}{2.2} \biggr)^k \ \ \ \ \ k=0,1,2,3,\cdots$

The following gives the calculation for $P(X=3)$.

\displaystyle \begin{aligned} P(X=3)&=\binom{4.5}{3} \ \biggl(\frac{1}{2.2} \biggr)^{2.5} \ \biggl(\frac{1.2}{2.2} \biggr)^3 \\&=\frac{4.5 (3.5) (2.5)}{3!} \ \biggl(\frac{1}{2.2} \biggr)^{2.5} \ \biggl(\frac{1.2}{2.2} \biggr)^3 \\&=6.5625 \ \biggl(\frac{1}{2.2} \biggr)^{2.5} \ \biggl(\frac{1.2}{2.2} \biggr)^3 \\&=0.148350259 \end{aligned}

With $1000 \times 0.148350259=148.35$, about 149 of the randomly selected insureds will have 3 claims in the observed period.

Example 6
Suppose that the annual claims frequency for an insured in a large portfolio of insureds has a distribution that is in the (a,b,0) class. Let $P_k$ be the probability that an insured has $k$ claims in a year.

Given that $P_1=0.3072$, $P_2=0.12288$ and $P_3=0.04096$, determine the probability that an insured has no claims in a one-year period.

Given $P_1$, $P_2$ and $P_3$, find $P_0$. Based on the recursive relation (12), we have the following two equations of $a$ and $b$.

$\displaystyle \frac{P_2}{P_1}=\frac{0.12288}{0.3072}=0.4=a+\frac{b}{2}$

$\displaystyle \frac{P_3}{P_2}=\frac{0.04096}{0.12288}=\frac{1}{3}=a+\frac{b}{3}$

Solving these two equations gives $a=0.2$ and $b=0.4$. Plugging $a$ and $b$ into the recursive relation gives the answer.

$\displaystyle \frac{P_1}{P_0}=\frac{0.3072}{P_0}=0.6$

$\displaystyle P_0=\frac{0.3072}{0.6}=0.512$.

Dan Ma actuarial

Daniel Ma actuarial

Dan Ma math

Dan Ma mathematics

$\text{ }$

$\text{ }$

$\text{ }$

$\copyright$ 2017 – Dan Ma

Revised Nov 2, 2018.