Month: November 2018

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

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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?

  1. If a>0 and b>0, then the variance of the distribution is greater than the mean.
  2. If a>0 and b=0, then the variance of the distribution is less than the mean.
  3. 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.

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Problem Answer
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

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