Practice Problems

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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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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Practice Problem Set 10 – value at risk and tail value at risk

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In actuarial applications, an important focus is on developing loss distributions for insurance products. It is also critical to employ risk measures to evaluate the exposure to risk. This post provides practice problems on two risk measures that are useful from an actuarial perspective. They are: value-at-risk (VaR) and tail-value-at-risk (TVaR).

Practice problems in this post are to reinforce the concepts of VaR and TVaR discussed in this blog post in a companion blog.

Most of the practice problems refer to parametric distributions highlighted in a catalog for continuous parametric models.

Practice Problem 10-A
Losses follow a paralogistic distribution with shape parameter \alpha=2 and scale parameter \theta=1500.

Determine the VaR at the security level 99%.

Practice Problem 10-B
Annual aggregate losses for an insurer follow an exponential distribution with mean 5,000. Evaluate VaR and TVaR for the aggregate losses at the 99% security level.

Practice Problem 10-C

For a certain line of business for an insurer, the annual losses follow a lognormal distribution with parameters \mu=5.5 and \sigma=1.2.

Evaluate the value-at-risk and the tail-value-at-risk at the 95% security level.

Practice Problem 10-D

Annual losses follow a normal distribution with mean 1000 and variance 250,000. Compute the tail-value-risk at the 95% security level.

Practice Problem 10-E

An insurance company models its liability insurance business using a Pareto distribution with shape parameter \alpha=1.5 and scale parameter \theta=5000.

Evaluate the value-at-risk and the tail-value-at-risk at the 99.5% security level.

Practice Problem 10-F

Losses follow an inverse exponential distribution with parameter \theta=2000. Calculate the value-at-risk at the 99% security level.

Practice Problem 10-G
Losses follow a mixture of two exponential distributions with equal weights where one exponential distribution has mean 10 and the other has mean 20. Evaluate the value-at-risk and the tail-value-at-risk at the 95% security level.

Practice Problem 10-H
Losses follow a mixture of two Pareto distributions with equal weights where one Pareto distribution has shape parameter \alpha=1.2 and scale parameter \theta=5000 and the other has shape parameter \alpha=2.4 and scale parameter \theta=5000. Evaluate the value-at-risk and the tail-value-at-risk at the 99% security level.

Practice Problem 10-I

Losses follow a Weibull distribution with parameters \tau=2 and \theta=1000. Determine the value-at-risk at the security level 99.5%.

Practice Problem 10-J

Losses follow an inverse Pareto distribution with parameters \tau=2.5 and \theta=5000. Determine the value-at-risk at the security level 99%.

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Problem Answer
10-A
  • VaR = 4500
10-B
  • VaR = 23025.85093
10-C
  • VaR = 1761.639168
  • TVaR = 2248.088854
10-D
  • VaR = 1882.5
  • TVaR = 2031.108111
10-E
  • VaR = 165997.5947
  • TVaR = 507992.784
10-F
  • VaR = 198998.3249
10-G
  • VaR = 47.80473823
  • TVaR = 66.96553606
10-H
  • VaR = 127375.8029
  • TVaR = 257568.7795
10-I
  • VaR = 2301.807413
10-J
  • VaR = 1241241.206

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Practice Problem Set 9 – Expected Insurance Payment – Additional Problems

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This practice problem set is to reinforce the 3-part discussion on insurance payment models (Part 1, Part 2 and Part 3). The practice problems in this post are additional practice problems in addition to Practice Problem Set 7 and Practice Problem Set 8.

Practice Problem 9A

Losses follow a distribution that is a mixture of two equally weighted Pareto distributions, one with parameters \alpha=2 and \theta=2000 and the other with with parameters \alpha=2 and \theta=4000. An insurance coverage for these losses has an ordinary deductible of 1000.

Calculate the expected payment per loss.

Practice Problem 9B
Losses, prior to any deductible being applied, follow an exponential distribution with mean 17.5. An insurance coverage has a deductible of 8. Inflation of 15% impacts all claims uniformly from the current year to next year.

Determine the percentage change in the expected claim cost per loss from the current year to next year.

Practice Problem 9C
Losses follow a distribution that has the following density function.

    \displaystyle  f(x)=\left\{ \begin{array}{ll}                     \displaystyle  0.15 &\ \ \ \ 0<x < 2 \\           \text{ } & \text{ } \\           \displaystyle  0.10 &\ \ \ \ 2 \le x < 5 \\           \text{ } & \text{ } \\           \displaystyle  0.08 &\ \ \ \ 5 \le x < 10           \end{array} \right.

An insurance policy is purchased to cover these losses. The policy has a deductible of 3.

Calculate the expected insurance payment per payment.

Practice Problem 9D
Losses follow a distribution with the following density function.

    \displaystyle f(x)=\frac{1250}{(x+1250)^2} \ \ \ \ \ \ \ \ \ x>0

An insurance coverage pays losses up to a maximum of 100,000. Determine the average payment per loss.

Practice Problem 9E

You are given the following information.

  • Losses, prior to any application of a deductible, follow a Pareto distribution with \alpha=3 and \theta=5000.
  • An insurance coverage is purchased to cover these losses.
  • If the size of a loss is between 5,000 and 15,000, the coverage pays for the loss in excess of 5,000. Otherwise, the coverage pays nothing.

Determine the average insurance payment per loss.

Practice Problem 9F

You are given the following information.

  • An insurance coverage is purchased to cover a certain type of liability losses.
  • The coverage has a deductible of 1000.
  • Other than the deductible, there are no other coverage modifications.
  • If the deductible is an ordinary deductible, the expected insurance payment per loss is 1,215.
  • If the deductible is a franchise deductible, the expected insurance payment per loss is 1,820.

Determine the proportion of the losses that exceed 1,000.

Practice Problem 9G

Losses follow a uniform distribution on the interval (0, 1000). The insurance coverage has a deductible of 250.

Determine the variance of the insurance payment per loss.

Practice Problem 9H

Losses follow an exponential distribution with mean 500. An insurance coverage that is designed to cover these losses has a deductible of 1,000.

Determine the coefficient of variation of the insurance payment per loss.

Practice Problem 9I
Losses are modeled by an exponential distribution with mean 3,000. An insurance policy covers these losses according to the following provisions.

  • The insured pays 100% of the loss up to 1,000.
  • For the loss amount between 1,000 and 10,000, the insurance pays 80%.
  • The loss amount above 10,000 is paid by the insured until the insured has paid 10,000 in total.
  • For the remaining part of the loss, the insurance pays 90%.

Determine the expected insurance payment per loss.

Practice Problem 9J

You are given the following information.

  • The underlying loss distribution for a block of insurance policies is a Pareto distribution with \alpha=2 and \theta=5000.
  • In the next calendar year, all claims in this block of policies are expected to be impacted uniformly by an inflation rate of 25%.
  • In the next calendar year, the insurance company plans to purchase an excess-of-loss reinsurance policy that caps the insurer’s loss at 10,000 per claim.

Determine the insurance company’s expected claim cost per claim after the effective date of the reinsurance policy.

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Problem Answer
9A
  • \displaystyle \frac{6800}{3}=2266.67
9B
  • \displaystyle 1.15 e^{1.2/20.125}=1.22
  • About 22% increase
9C
  • \displaystyle \frac{10}{3}=3.33
9D
  • 5493.061443
9E
  • 312.5
9F
  • 0.605
9G
  • 61523.4375
9H
  • \sqrt{2 e^2-1}=3.77887956
9I
  • 1642.795495
9J
  • \displaystyle \frac{50000}{13}=3846.15

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Practice Problem Set 8 – Expected Insurance Payment – Additional Problems

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This practice problem set is to reinforce the 3-part discussion on insurance payment models (Part 1, Part 2 and Part 3). The practice problems in this post are additional practice problems in addition to Practice Problem Set 7. Another problem set on expected insurance payment: Practice Problem Set 9.

Practice Problem 8A

Losses follow a uniform distribution on the interval (0,50000).

    An insurance policy has an 80% coinsurance and an ordinary deductible of 10,000. The coinsurance is applied after the deductible so that a positive payment is made on the loss amount above 10,000.

Determine the expected payment per loss.

Practice Problem 8B
Losses follow a uniform distribution on the interval (0,50000).

    An insurance policy has an 80% coinsurance and an ordinary deductible of 10,000. The coinsurance is applied after the deductible so that a positive payment is made on the loss amount above 10,000. In addition to the deductible and coinsurance, the coverage has a policy limit of 24,000 (i.e. the maximum covered loss is 40,000).

Determine the expected payment per loss.

Practice Problem 8C
Losses in the current year follow a uniform distribution on the interval (0,50000). Further suppose that inflation of 25% impacts all losses uniformly from the current year to the next year. Losses in the next year are paid according to the following provisions:

  • Coverage has an ordinary deductible of 10,000.
  • Coverage has an 80% coinsurance.
  • The coinsurance is applied after the deductible.
  • The coverage has a policy limit of 24,000.

Determine the expected payment per loss.

Practice Problem 8D
Liability claim sizes follow a Pareto distribution with shape parameter \alpha=1.2 and scale parameter \theta=10000. Suppose that the insurance coverage has a franchise deductible of 20,000 per loss. Given that a loss exceeds the deductible, determine the expected insurance payment.

Practice Problem 8E
Losses in the current year follow a Pareto distribution with parameters \alpha=3 and \theta=5000. Inflation of 10% is expected to impact these losses in the next year. The coverage for next year’s losses has an ordinary deductible of 1,000.

Determine the expected amount per loss in the next year that will be paid by the insurance coverage.

Practice Problem 8F

Losses in the current year follow a Pareto distribution with parameters \alpha=3 and \theta=5000. Inflation of 10% is expected to impact these losses in the next year. The coverage for next year’s losses has a franchise deductible of 1,000.

Determine the expected amount per loss in the next year that will be paid by the insurance coverage.

Practice Problem 8G

Losses follow a distribution that is a mixture of two equally weighted exponential distributions, one with mean 6 and the other with mean 12. An insurance coverage for these losses has an ordinary deductible of 2. Calculate the expected payment per loss.

Practice Problem 8H

Losses follow a distribution that is a mixture of two equally weighted exponential distributions, one with mean 6 and the other with mean 12. An insurance coverage for these losses has a franchise deductible of 2. Calculate the expected payment per loss.

Practice Problem 8I
You are given the following information.

  • Losses follow a distribution with the following cumulative distribution function.
    • \displaystyle F(x)=1-\frac{1}{3} e^{-2x}-\frac{1}{3} e^{-x}-\frac{1}{3} e^{-x/2} \ \ \ \ x>0
  • For each loss, the insurance coverage pays 80% of the portion of the loss that exceeds a deductible of 1.

Determine the average payment per loss.

Practice Problem 8J

You are given the following information.

  • Losses follow a lognormal distribution with \mu=3 and \sigma=1.2.
  • An insurance coverage has a deductible of 10.

Determine the percentage change in the expected claim cost per loss when losses are uniformly impacted by a 20% inflation.

All normal probabilities are obtained by using the normal distribution table found here.

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Problem Answer
8A
  • 12,800
8B
  • 12,000
8C
  • 14,400
8D
  • 170,000
8E
  • 1968.9349
8F
  • 2574.761038
8G
  • \displaystyle 3 e^{-1/3}+6 e^{-1/6}=7.2285
8H
  • \displaystyle 4 e^{-1/3}+7 e^{-1/6}=8.7915
8I
  • \displaystyle 0.8 \biggl(\frac{1}{6} e^{-2}+\frac{1}{3} e^{-1}+\frac{2}{3} e^{-1/2} \biggr)=0.43963
8J
  • Claim Cost before inflation: 32.52697933.
  • Claim Cost after inflation: 40.51721002.
  • 24.56% change.

Daniel Ma actuarial

Dan Ma actuarial

\copyright 2017 – Dan Ma

Practice Problem Set 7 – Expected Insurance Payment

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This practice problem set is to reinforce the 3-part discussion on insurance payment models (Part 1, Part 2 and Part 3). The practice problems in this post are basic problems on calculating average insurance payment (per loss or per payment).

Additional problem set: Practice Problem Set 8 and Practice Problem Set 9.

Practice Problem 7A

Losses follow a uniform distribution on the interval (0,50000).

  • An insurance policy has an ordinary deductible of 10,000. Determine the expected payment per loss.
  • Suppose that in addition to the deductible of 10,000, the coverage has a policy limit of 30,000 (i.e. the maximum covered loss is 40,000). Determine the expected payment per loss.
Practice Problem 7B
Losses for the current year follow a uniform distribution on the interval (0,50000). Further suppose that inflation of 25% impacts all losses uniformly from the current year to the next year.

  • Determine the expect insurance payment per loss for the next year assuming that the policy has a deductible of 10,000.
  • Determine the expect insurance payment per loss for the next year assuming that the policy has a deductible of 10,000 and a policy limit of 30,000 (maximum covered loss = 40,000).
Practice Problem 7C
Losses follow an exponential distribution with mean 5,000. An insurance policy covers losses subject to a franchise deductible of 2,000. Determine the expected insurance payment per loss.

Practice Problem 7D
Liability claim sizes follow a Pareto distribution with shape parameter \alpha=1.2 and scale parameter \theta=10000. Suppose that the insurance coverage pays claims subject to an ordinary deductible of 20,000 per loss. Given that a loss exceeds the deductible, determine the expected insurance payment.

Practice Problem 7E
Losses follow a lognormal distribution with \mu=7.5 and \sigma=1. For losses below 1,000, no payment is made. For losses exceeding 1,000, the amount in excess of the deductible is paid by the insurer. Determine the expected insurance payment per loss.

Practice Problem 7F

Losses in the current exposure period follow a lognormal distribution with \mu=7.5 and \sigma=1. Losses in the next exposure period are expected to experience 12% inflation over the current year. Determine the expected insurance payment per loss if the insurance contract has an ordinary deductible of 1,000.

Practice Problem 7G

Losses follow an exponential distribution with mean 2,500. An insurance contract will pay the amount of each claim in excess of a deductible of 750. Determine the standard deviation of the insurance payment for one claim such that a claim includes the possibility that the amount paid is zero.

Practice Problem 7H

Liability losses for auto insurance policies follow a Pareto distribution with \alpha=3 and \theta=5000. These insurance policies have an ordinary deductible of 1,250. Determine the expected payment made by these insurance policies per loss.

Practice Problem 7I
Liability losses for auto insurance policies follow a Pareto distribution with \alpha=3 and \theta=5000. These insurance policies make no payment for any loss below 1,250. For any loss greater than 1,250, the insurance policies pay the loss amount in excess of 1,250 up to a limit of 5,000. Determine the expected payment made by these insurance policies per loss.

Practice Problem 7J
Losses follow a lognormal distribution with \mu=7.5 and \sigma=1. For losses below 1,000, no payment is made. For losses exceeding 1,000, the amount in excess of the deductible is paid by the insurer. Determine the average insurance payment for all the losses that exceed 1,000.

All normal probabilities are obtained by using the normal distribution table found here.

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Problem Answer
7A
  • 16,000
  • 15,000
7B
  • 22,050
  • 18,000
7C
  • 4,692.24
7D
  • 150,000
7E
  • 0.9441 e^8-722.4 = 2091.92
7F
  • 1.071168 e^8-761.1 = 2432.01
7G
  • 2414.571397
7H
  • 1,600
7I
  • 1,106.17284
7J
  • \displaystyle \frac{0.9441 e^8-722.4}{0.7224}= 2895.80

Daniel Ma actuarial

Dan Ma actuarial

\copyright 2017 – Dan Ma

Practice Problem Set 6 – Negative Binomial Distribution

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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.

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Problem Answer
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

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Daniel Ma Mathematics

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\copyright 2017 – Dan Ma