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

Posted on

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.

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.

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

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

Daniel Ma Math

Daniel Ma Mathematics

Actuarial exam

\copyright 2018 – Dan Ma

Advertisements

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

Posted on Updated on

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%.

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

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

Daniel Ma Math

Daniel Ma Mathematics

Actuarial exam

\copyright 2018 – Dan Ma

Practice Problem Set 9 – Expected Insurance Payment – Additional Problems

Posted on

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.

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

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

Daniel Ma actuarial

Dan Ma actuarial

\copyright 2017 – Dan Ma

Practice Problem Set 8 – Expected Insurance Payment – Additional Problems

Posted on Updated on

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.

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.

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

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

Posted on Updated on

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.

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.

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

\text{ }

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

Mathematical models for insurance payments – part 3 – other modifications

Posted on Updated on

This post is a continuation of the discussion on models of insurance payments initiated in two previous posts. Part 1 focuses on the models of insurance payments in which the insurance policy imposes a policy limit. Part 2 continues the discussion by introducing models in which the insurance policy imposes an ordinary deductible. In each of these two previous posts, the insurance coverage has only one coverage modification. A more interesting and more realistic scenario would be insurance coverage that contains a combination of several coverage modifications. This post is to examine the effects on the insurance payments as a result of having some or all of these coverage modifications – policy limit, ordinary deductible, franchise deductible and inflation. Additional topics: expected payment per loss versus expected payment per payment and loss elimination ratio.

\text{ }

Ordinary Deductible and Policy Limit

The previous two posts discuss the expectations E[X \wedge u] and E[(X-d)_+]. The first is the expected insurance payment when the coverage has a policy limit u. The second is the expected insurance payment when the coverage has an ordinary deductible d. They are the expected values of the following two variables.

    \displaystyle  X \wedge u=\left\{ \begin{array}{ll}                     \displaystyle  X &\ X \le u \\           \text{ } & \text{ } \\           \displaystyle  u &\ X > u           \end{array} \right.

    \displaystyle  (X-d)_+=\left\{ \begin{array}{ll}                     \displaystyle  0 &\ X \le d \\           \text{ } & \text{ } \\           \displaystyle  X-d &\ X > d           \end{array} \right.

It is easy to verify that X=(X-d)_+ + X \wedge d. Buying a coverage with an ordinary deductible d and another coverage with policy limit d equals full coverage. Thus we have the following relation.

(1) \ \ \ \ E[(X-d)_+]=E[X]-E[X \wedge d]

The limited expectation E[X \wedge d] has expressions in closed form in some cases or has expressions in terms of familiar functions (e.g. gamma function) in other cases. Thus the expectation E[(X-d)_+] can be computed by knowing the original expected value E[X] and the limited expectation E[X \wedge d].

Inflation

Suppose that losses (or claims) in the next period are expected to increase uniformly by 100r%. For example, r=0.10 means 10%. What would be the effect of inflation on the expectations E[X \wedge u] and E[(X-d)_+]?

First, the effect on the limited expectation E[X \wedge u]. As usual, X is the random loss and u is the policy limit. With inflation rate r, the loss variable for the next period would be (1+r)X. One approach is to derive the distribution for the inflated loss variable and use the new distribution to calculate E[(1+r)X \wedge u]. Another approach is to express it in terms of the limited expectation of the pre-inflated loss X. The following is the expectation E[(1+r)X \wedge u], assuming that there is no change in the policy limit.

\displaystyle (2) \ \ \ \ E[(1+r) X \wedge u]=(1+r) \ E \biggl[X \wedge \frac{u}{1+r} \biggr]

Relation (2) relates the limited expectation of the inflated variable to the limited expectation of the pre-inflated loss X. It says that the limited expectation of the inflated variable (1+r) X is obtained by inflating the limited expectation of the pre-inflated loss but at a smaller policy limit \frac{u}{1+r}.

The following is the expectation E[(1+r) (X-d)_+]=E[(1+r) X]-E[(1+r) X \wedge d].

\displaystyle \begin{aligned} (3) \ \ \ \ E[((1+r) X-d)_+]&=E[(1+r) X]-E[(1+r) X \wedge d] \\&=(1+r) E[X]-(1+r) \ E \biggl[X \wedge \frac{d}{1+r} \biggr] \\&=(1+r) \biggl( E[X]- \ E \biggl[X \wedge \frac{d}{1+r} \biggr] \biggr) \\&=(1+r) E \biggl[ \biggl(X-\frac{d}{1+r} \biggr)_+ \biggr] \end{aligned}

Similarly, (3) expresses the expected payment on the inflated loss in terms of the expected payment of the pre-inflated loss. It says that when the loss is inflated, the expected payment per loss is obtained by inflating the expected payment on the pre-inflated loss but at a smaller deductible.

Insurance Payment Per Loss versus Per Payment

The previous post (Part 2) shows how to evaluate the average amount paid to the insured when the coverage has an ordinary deductible. The average payment discussed in Part 2 is the average per loss (over all losses). As a simple illustration, let’s say the amounts of losses in a given period for an insured are 7, 4, 33 and 17 subject to an ordinary deductible of 5. Then the insurance payments are: 2, 0, 28 and 12. The average payment per loss would be (2+0+28+12)/4 = 10.5. If we only count the losses that require a payment, the average is (2+28+12)/3 = 14. Thus the average payment per payment is greater than the average payment per loss since only the losses exceeding the deductible are counted in the average payment per payment. In the calculation discussed here, the average payment per payment is obtained by dividing the average payment per loss by the probability that the loss exceeds the deductible. Note that 10.5/0.75 = 14.

Suppose that the random variable X is the size of the loss (if a loss occurs). Under an insurance policy with an ordinary deductible d, the first d dollars of a loss is responsible by the insured and the amount of the loss in excess of d is paid by the insurer. Under such an arrangement, a certain number of losses are not paid by the insurer, precisely those losses that are less than or equal to d. If we only count the losses that are paid by the insurer, the payment amount is the conditional random variable X-d \lvert X>d. The expected value of this conditional random variable is denoted by e_X(d) or e(d) if the loss X is understood.

Given that P(X>d)>0, the variable X-d \lvert X>d is called the excess loss variable. Its expected value e_X(d) is called the mean excess loss function. Other names are mean residual life and complete expectation of life (when the context is that of a mortality study).

For the discussion in this post and other posts in the same series, we use Y_P to denote X-d \lvert X>d. The P stands for payment so that its expected value would be average insurance payment per payment, i.e. the expected amount paid given that the loss exceeds the deductible. When the random variable X is the age at death, e_X(x) would be the expected remaining time until death given that the life has survived to age x.

The expected value e_X(d) is thus the expected payment per payment (or expected cost per payment) under an ordinary deductible. In contrast, the expected value E[(X-d)_+] is the expected payment per loss (or expected cost per loss), which is discussed in the previous post. The two expected values are related. The calculation of one will give the other. The following compares the two calculation. Let f_X(x) and F_X(x) be the PDF and CDF of X, respectively.

\displaystyle (4) \ \ \ \ E[(X-d)_+]=\int_d^\infty (x-d) \ f_X(x) \ dx

\text{ }

\displaystyle (5) \ \ \ \ e_X(d)=\int_d^\infty (x-d) \ \frac{f_X(x)}{S_X(d)} \ dx=\frac{E[(X-d)_+]}{S_X(d)}

Note that e_X(d) is calculated using a conditional density function. As a result, e_X(d) can be obtained by dividing the expected payment per loss divided by the probability that there is a payment. Thus the expected payment per payment is the expected payment per loss divided by the probability that there is a payment. This is described in the following relation.

\displaystyle (6) \ \ \ \ e_X(d)=\frac{E[(X-d)_+]}{S_X(d)}=\frac{E[X]-E[X \wedge d]}{S_X(d)}

\text{ }

\text{ }

Distributions of Insurance Payment Variables

Relation (1) and Relation (4) are used for the calculation of expected insurance payment when the coverage has an ordinary deductible. For deriving other information about insurance payment in the presence of an ordinary deductible, it is helpful to know the distributions of the insurance payment (per loss and per payment).

Let Y_L=(X-d)_+ (payment per loss) and let Y_P=X-d \lvert X>d (payment per payment). We now discuss the distribution for Y_P. First, the following gives the PDF, CDF and the survival function of Y_L.

Payment Per Loss
PDF \displaystyle  f_{Y_L}(y)=\left\{ \begin{array}{ll}                     \displaystyle  F_X(d) &\ \ \ \ y=0 \\           \text{ } & \text{ } \\           \displaystyle  f_X(y+d) &\ \ \ \ y > 0           \end{array} \right.
CDF \displaystyle  F_{Y_L}(y)=\left\{ \begin{array}{ll}                     \displaystyle  0 &\ \ \ \ y<0 \\           \text{ } & \text{ } \\           \displaystyle  F_X(y+d) &\ \ \ \ y \ge 0                      \end{array} \right.
Survival Function \displaystyle  S_{Y_L}(y)=\left\{ \begin{array}{ll}                     \displaystyle  1 &\ \ \ \  y<0 \\           \text{ } & \text{ } \\           \displaystyle  S_X(y+d) &\ \ \ \ y \ge 0                      \end{array} \right.

Note that the above functions have a point mass at y=0 to account for the losses that are not paid. For Y_P=X-d \lvert X>d, there is no point mass at y=0. We only need to consider y>0. Normalizing the function f_X(y+d) would give the PDF of Y_P=X-d \lvert X>d. Thus the following gives the PDF, the survival function and the CDF of Y_P=X-d \lvert X>d.

Payment Per Payment
PDF \displaystyle f_{Y_P}(y)=\frac{f_X(y+d)}{S_X(d)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y > 0
CDF \displaystyle F_{Y_P}(y)=\frac{F_X(y+d)-F_X(d)}{S_X(d)} \ \ \ \ \ \ \ \ \ \ \ \ \ y > 0
Survival Function \displaystyle S_{Y_P}(y)=\frac{S_X(y+d)}{S_X(d)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y > 0

The following calculates the two averages using the respective PDFs, as a result deriving the same relationship between the two expected values.

\displaystyle (7) \ \ \ \ E(Y_L)=E[(X-d)_+]=\int_0^\infty y \ f_X(y+d) \ dy

\text{ }

\displaystyle (8) \ \ \ \ E(Y_P)=e_X(d)=\int_0^\infty y \ \frac{f_X(y+d)}{S_X(d)} \ dy

Relation (1) and Relation (7) are identical. The former is calculated using the distribution of the original loss X and the latter is calculated using the distribution of the payment per loss variable. Similarly, compare Relation (5) and Relation (8). The former is computed from the distribution of the unmodified loss X and the latter is computed using the distribution of the payment per payment variable.

Note that Relation (7) and Relation (8) also lead to Relation (6), which relates the expected payment E[(X-d)_+] and the expected payment e_X(d).

With the PDFs and CDFs for the payment per loss and the payment per payment developed, other distributional quantities can be derived, e.g. hazard rate, variance, skewness and kurtosis.

Loss Elimination Ratio

When the insurance coverage has an ordinary deductible, a natural question is (from the insurance company’s perspective), what is the impact of the deductible on the payment that is made to the insured? More specifically, on average what is the reduction of the payment? The loss elimination ratio is the proportion of the expected loss that is not paid to the insured by the insurer. For example, if the expected loss before application of the deductible is 45 (of which 36 is expected to be paid by the insurer), the loss elimination ratio is 9/45 = 0.20 (20%). In this example, the insurer has reduced its obligation by 20%. More formally, the loss elimination ratio (LER) is defined as:

    \displaystyle LER=\frac{E(X)-E[(X-d)_+]}{E(X)}=\frac{E[X \wedge d]}{E(X)}

LER is the ratio of the expected reduction in payment as a result of imposing the ordinary deductible to the expected payment without the deductible. Though it is possible to define LER as the ratio of the reduction in expected payment as a result of a coverage modification to the expected payment without the modification for a modification other than an ordinary deductible, we do not attempt to further generalize this concept.

Policy with a Limit and a Deductible

In part 1, expected insurance payment under a policy limit is developed. In part 2, expected insurance payment under a policy with an ordinary deductible is developed. We now combine both provisions in the same insurance policy. First, let’s define two terms. A policy limit is the maximum amount that will be paid by a policy. For example, if the policy limit is 10,000, the policy will paid at most 10,000 per loss. If the actual loss is 15,000, then the policy will paid 10,000 and the insured will have to be responsible for the remaining 5,000. On the other hand, maximum covered loss is the level above which no loss will be paid. For example, Suppose that the policy covers up to 10,000 per loss subject to a deductible of 1,000. If the actual loss is 20,000, then the maximum covered loss is 10,000 with the policy limit being 9,000. This is because the policy only covers the first 10,000 with the first 1,000 paid by the insured.

Suppose that an insurance policy has an ordinary deductible d, a maximum covered loss u with d<u and no other coverage modifications. Any loss below d is not paid by the insurer. For any loss exceeding d, the insurer pays the loss amount in excess of d up to the maximum covered loss u, with the policy limit being u-d. The following describes the payment rule more explicitly.

    \displaystyle  (X \wedge u)-(X \wedge d)=\left\{ \begin{array}{ll}                     \displaystyle  0 &\ \ \ \ X \le d \\           \text{ } & \text{ } \\           \displaystyle  X-d &\ \ \ \ d < X \le u \\           \text{ } & \text{ } \\           \displaystyle  u-d &\ \ \ \ X > u           \end{array} \right.

Under such a policy, the maximum amount paid (policy limit) is u-d, which is the maximum covered loss minus the deductible. The policy limit is reached when X > u. When d < X \le u, the payment X \wedge u is X and the payment X \wedge d is d. Then the expected payment per loss under such a policy is:

\displaystyle  (9) \ \ \ \ E[X \wedge u]-E[X \wedge d]

There is not special notation for the expected payment. In words, it is the limited expected value at u (the maximum covered loss) minus the limited expected value at the ordinary deductible d. The higher moments can be derived by evaluating E(Y^k) using the PDF of the loss X where Y=(X \wedge u)-(X \wedge d).

The payment Y=(X \wedge u)-(X \wedge d) is on a per loss basis. It is also possible to consider payment or payment by removing the point mass at zero. The expected payment per payment is obtained by dividing (9) by the probability of a positive payment.

\text{ }

Franchise Deductible

An alternative to the ordinary deductible is the franchise deductible. It works like an ordinary deductible except that when the loss exceeds the deductible, the policy pays the loss in full. The following gives the payment rule.

    \displaystyle  Y=\left\{ \begin{array}{ll}                     \displaystyle  0 &\ \ \ \ X \le d \\           \text{ } & \text{ } \\           \displaystyle  X &\ \ \ \ X > d           \end{array} \right.

Note that when the loss exceeds the deductible (X > d), the policy with a franchise deductible pays more than a policy with the same ordinary deductible. By how much? By the amount d. Thus the expected payment per loss under a franchise deductible is E[(X-d)_+]+d \ S_X(d). The addition of d \cdot S_X(d) reflects the additional benefit when X > d. Instead of deriving calculation specific to franchise deductible, we can derive the payments under franchise deductible by adding the additional benefit appropriately.

\displaystyle  (10) \ \ \ \ E[(X-d)_+]+d \ S_X(d)

\text{ }

Benefit Tables

The previous two posts and this post discuss coverage with two types of deductible (ordinary and franchise) as well as coverage that may include a limit. In order to keep the expected payments straight, the following table organizes the different combinations of coverage options discussed up to this point.

A = Ordinary Deductible Only
B = Franchise Deductible Only
C = Ordinary Deductible + Limit
D = Franchise Deductible + Limit

Expected Cost Per Loss Expected Cost Per Payment
A \displaystyle E[(X-d)_+]=E(X)-E(X \wedge d) \displaystyle \frac{E(X)-E(X \wedge d)}{S_X(d)}=e_X(d)
B \displaystyle E(X)-E(X \wedge d)+d \ S_X(d) \displaystyle \frac{E(X)-E(X \wedge d)}{S_X(d)}+d=e_X(d)+d
C \displaystyle E(X \wedge u)-E(X \wedge d ) \displaystyle \frac{E(X \wedge u)-E(X \wedge d)}{S_X(d)}
D \displaystyle E(X \wedge u)-E(X \wedge d )+d \ S_X(d) \displaystyle \frac{E(X \wedge u)-E(X \wedge d)}{S_X(d)}+d

Row A shows the expected payment in a coverage with an ordinary deductible for both per loss and per payment. The expected per loss payment in Row A is the subject of the previous post and the expected payment per payment is discussed earlier in this post. Row B is for expected payments in a coverage with a franchise deductible. Note that the coverage with a franchise deductible pays more than a coverage with an identical ordinary deductible. Thus Row B is Row A plus the added benefit, which is the deductible d.

Row C is for the coverage of an ordinary deductible with a policy limit. The per loss expected payment is \displaystyle E(X \wedge u)-E(X \wedge d ). The per payment expected value is a conditional one, conditional on the loss greater than the deductible. Thus the per payment expected value is the per loss expected value divided by S_X(d).

Note that Row D is Row C plus the amount of d, the additional benefit as a result of having a franchise deductible instead of an ordinary deductible.

Rather than memorizing these formulas, focus on the general structure of the table. For example, understand the payments involving ordinary deductible (Row A and Row C). Then the payments for franchise deductible are obtained by adding an appropriate additional benefit.

The following table shows the expected payments under the influence of claim cost inflation. The maximum covered loss u and the deductible d are identical to the above benefit table. The losses are subject to a (100r)% inflation in the next exposure period. Note that u^* and d^* are the modified maximum covered loss and deductible that will make the formulas work.

E = Ordinary Deductible Only
F = Franchise Deductible Only
G = Ordinary Deductible + Limit
H = Franchise Deductible + Limit

\displaystyle u^*=\frac{u}{1+r}

\displaystyle d^*=\frac{d}{1+r}

Expected Cost Per Loss (with Inflation) Expected Cost Per Payment (with Inflation)
E \displaystyle (1+r) \biggl \{ E(X)-E [X \wedge d^* ] \bigg \} \displaystyle (1+r) \biggl \{ \frac{\displaystyle  E(X)-E [X \wedge d^* ]}{\displaystyle S_{X} (d^*)} \biggr \}
F \displaystyle (1+r) \biggl \{ E(X)-E [X \wedge d^* ]+d^* \ S_{X}(d^* ) \bigg \} \displaystyle (1+r) \biggl \{ \frac{\displaystyle  E(X)-E [X \wedge d^* ]}{\displaystyle S_{X} (d^* )}+d^* \biggr \}
G \displaystyle (1+r) \biggl \{ E [X \wedge u^* ]-E [X \wedge d^* ] \bigg \} \displaystyle (1+r) \biggl \{ \frac{\displaystyle  E [X \wedge u^* ]-E [X \wedge d^*]}{\displaystyle S_{X} (d^* )} \biggr \}
H \displaystyle (1+r) \biggl \{ E [X \wedge u^* ]-E [X \wedge d^* ]+d^* \ S_{X}   (d^* ) \bigg \} \displaystyle (1+r) \biggl \{ \frac{\displaystyle  E[X \wedge u^* ]-E [X \wedge d^* ]}{\displaystyle S_{X} (d^* )}+d^* \biggr \}

There is really no need to mathematically derive the formulas for the second insurance benefit table (Rows E through H). Recall the effect of inflation on the expected payments E[X \wedge u] and E[ (X-d)_+] (see Relation (2) and Relation (3)). Then apply the inflation effect on the first insurance benefit table. Note that the second table is obtained by inflating the first table by 1+r but with smaller upper limit u^* and deductible d^*. Also note the relation between Row E and Row F (same relation between Row A and Row B) and the relation between Row G and Row H (same relation between Row C and Row D).

\text{ }

Calculation

One approach of modeling insurance payments is to assume that the unmodified insurance losses are from a catalog of parametric distributions. For certain parametric distributions, the limited expectations E[X \wedge d] have convenient forms. Examples are: exponential distribution, Pareto distribution and lognormal distribution. Other distributions do not close form for E[X \wedge d] but the limited expectation can be expressed as a function that can be numerically calculated. One example is the gamma function.

The table in this link is an inventory of distributions that may be useful in modeling insurance losses. Included in the table are the limited expectations for various distributions. The following table shows the limited expectations for exponential, Pareto and lognormal.

Limited Expectation
Exponential E[X \wedge x]=\theta (1-e^{-x/\theta})
Pareto \displaystyle E[X \wedge x]=\frac{\theta}{\alpha-1} \biggl[1-\biggl(\frac{\theta}{x+\theta} \biggr)^{\alpha-1} \biggr], \alpha \ne 1
Lognormal \displaystyle E[(X \wedge x)^k]=\exp(k \mu+k^2 \sigma^2/2) \Phi \biggl(\frac{\log(x)-\mu-k \sigma^2}{\sigma} \biggr)+x^k \biggl[1-\Phi \biggl(\frac{\log(x)-\mu}{\sigma} \biggr) \biggr]

See the table in this link for the limited expectations of the distributed not shown in the above table. Once E[X \wedge d] is computed or estimated, the expected payment for various types of coverage provisions can be derived.

Examples

The first example demonstrates the four categories of expected payments in the table in the preceding section.

Example 1
Suppose that the loss distribution is described by the PDF \displaystyle f_X(x)=\frac{1}{5000} (100-x) with support 0<x<100. Walk through all the expected payments discussed in the above table (Rows A through D) assuming d=12 and u=60.

First, the basic calculation.

    \displaystyle E(X)=\int_0^{100} x \ \frac{1}{5000} (100-x) \ dx=\frac{100}{3}=33.3333

    \displaystyle F_X(x)=\frac{1}{5000} (100x- \frac{1}{2} x^2); \ \ \ 0<x<100

    F_X(12)=0.2256

    S_X(12)=1-0.2256=0.7744

    F_X(60)=0.84

    S_X(60)=1-0.84=0.16

    \displaystyle E(X \wedge 12)=\int_0^{12} x \ \frac{1}{5000} (100-x) \ dx +12 \cdot S_X(12)=\frac{6636}{625}

    \displaystyle E(X \wedge 60)=\int_0^{60} x \ \frac{1}{5000} (100-x) \ dx +12 \cdot S_X(60)=\frac{12864}{625}

Four categories of expected payments are derived and are shown in the following table.

A = Ordinary Deductible Only
B = Franchise Deductible Only
C = Ordinary Deductible + Limit
D = Franchise Deductible + Limit

Expected Cost Per Loss Expected Cost Per Payment
A \displaystyle \frac{100}{3}-\frac{6636}{625}=\frac{42592}{1875}=22.7157 \displaystyle e_X(d)=\frac{42592}{1875} \frac{1}{0.7744}=\frac{10648}{363}=29.3333
B \displaystyle \frac{42592}{1875}+12 \cdot 0.7744=\frac{60016}{1875}=32.0085 \displaystyle \frac{10648}{363}+12=\frac{15004}{363}=41.3333
C \displaystyle \frac{12864}{625}-\frac{6636}{625}=\frac{6228}{625}=9.9648 \displaystyle \frac{6228}{625} \frac{1}{0.7744}=\frac{6228}{484}=12.8678
D \displaystyle \frac{6228}{625}+12 \cdot 0.7744=\frac{12036}{625}=19.2576 \displaystyle \frac{6228}{484}+12=\frac{12036}{484}=24.8678

\text{ }

Example 2
Suppose that a coverage has an ordinary deductible and suppose that the CDF of the insurance payment per loss is given by:

    \displaystyle  G(y)=\left\{ \begin{array}{ll}                     \displaystyle  0 &\ \ \ \ y<0 \\           \text{ } & \text{ } \\           \displaystyle  1-e^{-\frac{y+20}{100}} &\ \ \ \ y \ge 0                      \end{array} \right.

Determine the expected value and the variance of the insurance cost per loss.

Note that there is a jump in the CDF at y=0. Thus y=0 is a point mass. The following is the PDF of the insurance payment.

    \displaystyle  g(y)=\left\{ \begin{array}{ll}                     \displaystyle  1-e^{-1/5} &\ \ \ \ y=0 \\           \text{ } & \text{ } \\           \displaystyle  \frac{1}{100} e^{-\frac{y+20}{100}}=e^{-1/5} \ \frac{1}{100} e^{-y/100} &\ \ \ \ y > 0                      \end{array} \right.

It is clear that g(y)=f(y+20) where f(y) is the PDF of the exponential distribution with mean 100. Thus g(y) is the PDF of the insurance payment per loss when the coverage has an ordinary deductible of 20 where the loss has exponential distribution with mean 100. We can use g(y) to calculate the mean and variance of Y. Using g(y), the mean and variance are:

    \displaystyle \begin{aligned} E(Y)&=\int_0^\infty y \ e^{-1/5} \ \frac{1}{100} e^{-y/100} \ dy \\&=e^{-1/5} \ \int_0^\infty y \ \frac{1}{100} e^{-y/100} \ dy \\&=e^{-1/5} \ (100)=81.873 \end{aligned}

    \displaystyle \begin{aligned} E(Y^2)&=\int_0^\infty y^2 \ e^{-1/5} \ \frac{1}{100} e^{-y/100} \ dy \\&=e^{-1/5} \ \int_0^\infty y^2 \ \frac{1}{100} e^{-y/100} \ dy \\&=e^{-1/5} \ 2(100)^2 \end{aligned}

    \displaystyle Var(Y)=e^{-1/5} \ 2(100)^2-\biggl( e^{-1/5} \ (100) \biggr)^2=9671.414601

Note that the calculation for E(Y) and E(Y^2) is done by multiplying e^{-1/5} by mean and second moment of the exponential distribution with mean 100.

\text{ }

Example 3
You are given the following:

  • Losses follow a Pareto distribution with parameters \alpha=3 and \theta=500.
  • The coverage has an ordinary deductible of 100.

Determine the PDF and CDF of the payment to the insured per payment. Determine the expected cost per payment.

Note that the setting of this example is identical to Example 3 in this previous post. Since this only concerns the insurance when the loss exceeds the deductible, the insurance payment is the conditional distribution Y=X-100 \lvert X>100 where X is the Pareto loss distribution.

The following gives the PDF and CDF and other information of the loss X.

    \displaystyle f_X(x)=\frac{3 \ 500^3}{(x+500)^4} \ \ \ \ \ \ \ \ \ \ \ \ \ \ x> 0

    \displaystyle F_X(x)=1-\biggl(\frac{500}{x+500} \biggr)^3 \ \ \ \ \ \ x> 0

    \displaystyle S_X(x)=\biggl(\frac{500}{x+500} \biggr)^3 \ \ \ \ \ \ \ \ \ \ \ \ x> 0

    \displaystyle S_X(100)=\biggl(\frac{500}{100+500} \biggr)^3=\biggl(\frac{500}{600} \biggr)^3=\frac{125}{216}

The following shows the PDF and CDF for the payment per payment variable Y.

    \displaystyle f_Y(y)=\frac{f_X(y+100)}{S_X(100)}=\frac{3 \ 500^3}{(y+100+500)^4} \ \biggl(\frac{600}{500} \biggr)^3=\frac{3 \cdot 600^3}{(y+600)^4} \ \ \ \ \ \ \ \ \ y> 0

    \displaystyle S_Y(y)=\frac{S_X(y+100)}{S_X(100)}=\biggl(\frac{500}{y+100+500} \biggr)^3 \ \biggl(\frac{600}{500} \biggr)^3=\biggl(\frac{600}{y+600} \biggr)^3 \ \ \ \ \ \ y> 0

    \displaystyle F_Y(y)=1-S_Y(y)=1-\biggl(\frac{600}{y+600} \biggr)^3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y> 0

Note that these PDF and CDF are for a Pareto distribution with \alpha=3 and \theta=600. Thus the expected payment per payment and the variance are:

    \displaystyle E(Y)=\frac{600}{3-1}=300

    \displaystyle E(Y^2)=\frac{600^2 \cdot 2}{(3-1) (3-2)}=600^2

    Var(Y)=600^2 - 300^2=270000

Example 4
Suppose that losses X follow a Pareto distribution with shape parameter \alpha and scale parameter \theta. Suppose that an insurance coverage pays each loss subject to an ordinary deductible d. Derive a formula for the expected payment per payment e_X(d).

Let Y=X-d \lvert X>d be the payment per payment. In Example 3, we see that the CDF and PDF of Y are also Pareto but with shape parameter \alpha (same as for X) and scale parameter \theta+d (the original scale parameter plus the deductible). Thus the following gives the expected payment per payment.

    \displaystyle e_X(d)=\frac{\theta+d}{\alpha-1}=\frac{\theta}{\alpha-1}+\frac{1}{\alpha-1} d

Note that e_X(d) in this case is an increasing linear function of the deductible d. The higher the deductible, the larger the expected payment. This is a clear sign that the Pareto distribution is a heavy tailed distribution.

Example 5
Suppose that losses follow a lognormal distribution with parameters \mu=5 and \sigma=0.6. An insurance coverage has an ordinary deductible of 100. Compute the expected payment per loss. Compute the expected payment per loss if the deductible is a franchise deductible.

The following calculates E(X) and E[X \wedge 100].

    E(X)=e^{5+0.6^2/2}=e^{5.18}=177.682811

    \displaystyle \begin{aligned} E[X \wedge 100]&=e^{5+0.6^2/2} \ \Phi \biggl(\frac{\log(100)-5-0.6^2}{0.6} \biggr)+100 \ \biggl[1-\Phi \bigg(\frac{\log(100)-5}{0.6} \biggr) \biggr] \\&=e^{5.18} \ \Phi (-1.26)+100 \ [1-\Phi (-0.66 )]\\&=e^{5.18} \ 0.1038+100(0.7454)=e^{5.18} \ 0.1038+74.54=92.98347578 \end{aligned}

As a result, the expected payment per loss is E(X)-E[X \wedge 100]=84.69933522=84.70. If the deductible of 100 is a franchise deductible, the expected payment per loss is obtained by adding an additional insurance payment, which is d \ S(d). The following is the added payment.

    \displaystyle \begin{aligned} \text{added payment}&=100 \ \biggl[1-\Phi \bigg(\frac{\log(100)-5}{0.6} \biggr) \biggr] \\&=100 \ \biggl[1-\Phi (-0.66 ) \biggr] \\&=100 (0.7454) \\&=74.54  \end{aligned}

Thus the expected payment per loss with a franchise deductible of 250 is 84.70 + 74.54 = 159.24.

Practice Problems

The following practice problem sets are to reinforce the concepts discussed here.

Practice Problem Set 7
Practice Problem Set 8
Practice Problem Set 9

\text{ }

\text{ }

\text{ }

models of insurance payment

ordinary deductible, franchise deductible

maximum covered loss

Daniel Ma Math

Daniel Ma Mathematics

Dan Ma Actuarial

\copyright 2017 – Dan Ma

Practice Problem Set 6 – Negative Binomial Distribution

Posted on

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{ }

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

Daniel Ma Math

Daniel Ma Mathematics

Actuarial exam

\copyright 2017 – Dan Ma