# 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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