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

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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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Problem Answer
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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4 thoughts on “Practice Problem Set 12 – (a,b,1) class

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

    […] 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. […]

    The (a,b,0) class | Topics in Actuarial Modeling said:
    January 12, 2019 at 5:43 pm

    […] Practice problems on (a,b,1) class […]

    The (a,b,1) class | Topics in Actuarial Modeling said:
    January 12, 2019 at 5:45 pm

    […] Practice problems on (a,b,1) class […]

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