Fully Discrete Benefit Premiums, Infinite Divisibility Property for Some Probability Distributions, and Assorted Exam-Style Questions

The negative binomial distribution and the Poisson distribution both exhibit the infinite divisibility property.

Let X follow a negative binomial distribution with parameters r and β. When we consider only a part of the distribution of X, say over the range (a, b), then the variable X distributed over (a, b) also follows a negative binomial distribution with parameters r and Pr(a < X < b)*β.

Let T follow a Poisson distribution with parameter λ. When we consider only a part of the distribution of T, say over the range (a, b), then the variable T distributed over (a, b) also follows a Poisson distribution with parameter Pr(a < X < b)*λ.

The following relationships hold between actuarial present values of whole and term life insurance policies, respectively, where for one of the policies (Ā) benefits are paid at the time of death and for the other (A) benefits are paid at the end of the year of death. In order for these relationships to hold, deaths for fractional ages must be uniformly distributed. Here, i is the annual effective rate of interest and δis the annual force of interest.

Ā_x = (i/δ)A_x

Ā¹_x:n¬ = (i/δ)A¹_x:n¬

For fully discrete benefit premiums where benefits on the associated life insurance policy are always paid at the end of the year of death, the following formulas hold, where "P" denotes the fully discrete benefit premium and the subscripts and superscripts of P correspond with the subscripts and superscripts of the associated life insurance policy.

P_x = A_x/ä_x

P¹_x:n¬ = A¹_x:n¬/ä_x:n¬

P_x:n¬ = A_x:n¬/ä_x:n¬

When using life tables, the following identities will also prove useful.

_nE_x = _mE_x*_n-mE_x+m, where m is any number less than n.

A¹_x:n¬ = A_x - _nE_x*A_x+n

ä_x:n¬ = ä_x - _nE_x*ä_x+n

Some of the problems in this section were designed to be similar to problems from past versions of the Casualty Actuarial Society's Exam 3L and the Society of Actuaries' Exam MLC. They use original exam questions as their inspiration - and the specific inspiration for each problem is cited so as to give students a chance to see the original. All of the original problems are publicly available, and students are encouraged to refer to them. But all of the values, names, conditions, and calculations in the problems here are the original work of Mr. Stolyarov.

Sources: Bowers, Gerber, et. al. Actuarial Mathematics. 1997. Second Edition. Society of Actuaries: Itasca, Illinois. pp. 120, 181-183.

Broverman, Sam. Actuarial Exam Solutions - CAS Exam 3 - Fall 2006.

Original Problems and Solutions from The Actuary's Free Study Guide

Problem S3L52-1.

Similar to Question 31 from the Casualty Actuarial Society's Fall 2006 Exam 3.

The frequency of a firm's annual losses due to defective paper clips follows a negative binomial distribution with r = 5 and β = 8. The severity of each loss follows a Pareto distribution with θ = 1550 and α = 3. The firm is insured for these losses under a policy with a deductible of $1000. What is the variance of the number of payments the insurance company will have to make because of this firm's losses due to defective paper clips?

Solution S3L52-1. The insurance company will only have to make a payment for a loss that is greater than $1000. Since we are given the severity distribution of each loss, we can find the probability that any random loss will be greater that $1000. This probability is

s(1000) = θ^α/(1000 + θ)^α = (1550/2550)³ = about 0.2245817973.

Since the negative binomial distribution is infinitely divisible, the frequency of losses that the insurance company will have to pay for follows a negative binomial distribution with r = 5 and

β = 8*0.2245817973 = 1.796654379.

We want to find the variance of this distribution, which we can do using the formula

Var[M] = rβ(1+ β) = 5*1.796654379*2.796654379 = about 25.12310668.

Problem S3L52-2. Similar to Question 32 from the Casualty Actuarial Society's Fall 2006 Exam 3. The number of leaf-eating ants in a garden on any given day follows a Poisson distribution with λ = 5. The number of leaves eaten by each ant follows a normal distribution with F(3400) = 0.8. Find the probability that at least one ant has eaten more than 3400 leaves today.

Solution S3L52-2. The probability (under the normal distribution - which is our severity distribution here) that any given any has eaten more than 3400 leaves is s(3400) = 1 - F(3400) = 1 - 0.8 = 0.2. Thus, to find the probability that at least one ant has eaten more than 3400 leaves today, we only consider that part of the Poisson frequency distribution which is relevant to ants that eat more than 3400 leaves. Since the Poisson distribution is infinitely divisible, the mean of the distribution we wish to consider is λ = 5*0.2 = 1. We want to find the complement of the probability that no ant has eaten more than 3400 leaves. The latter probability is equal to e^-λ = e^-1,

so our desired probability is 1 - e^-1 = about 0.6321205588.

Problem S3L52-3. Similar to Question 35 from the Casualty Actuarial Society's Fall 2006 Exam 3. The annual force of interest is 0.07. A spaceship has the following probabilities of failure q_x during each age x:

Age x......q_x

0.........0.005

1.........0.0044

2.........0.0123

3.........0.0334

4.........0.0241

5.........0.064

6.........0.031

Spaceship failures are permanent once they occur and are uniformly distributed for fractional ages.

There are two kinds of warranties available of the spaceship. A normal 4-year warranty will pay the full purchase price of the spaceship if it fails within the first four years of its life. An extended 3-year warranty for the spaceship can also be purchased. It will only take effect if the spaceship survives to four years since the purchase time, and it will pay 0.6 of the spaceship's purchase price if the spaceship fails between prior to reaching age 7. What is the actuarial present value of such an extended warranty on a new spaceship, as a fraction of the spaceship's purchase price.

Solution S3L52-3. Let R be the spaceship's purchase price. The extended warranty only takes effect if the spaceship survives to age 4. The probability of this happening is

(1 - q₀) (1 - q₁) (1 - q₂) (1 - q₃) = (1 - 0.005)(1 - 0.0044)(1 - 0.0123)(1 - 0.0334) = 0.9457575419.

This probability will need to be discounted for four years to take into account the time value of money:
0.9457575419*e^-0.07*4 = 0.7147881735.

Moreover, the extended warranty only pays 0.6R, so our value above will need to be multiplied by this factor: 0.7147881735*0.6R = 0.4288729041R.

Now we need to consider under which circumstances payment will be made.
We can calculate A¹_4:3¬, the actuarial present value (at time 4) of the extended warranty (analogous to a three-year term life insurance policy for a life aged 4), if payments are only made at the end of the year of failure.

A¹_4:3¬ = e^-0.07q₄+ e^-0.07*2p₄q₅ + e^-0.07*3p₄p₅q₆

A¹_4:3¬ = e^-0.07*0.0241 + e^-0.07*2*0.9759*0.064+ e^-0.07*3*0.9759*0.936*0.031 = 0.0997218026.

Now we can use the formula Ā¹_x:n¬ = (i/δ)A¹_x:n¬ to find Ā¹_4:3¬, the actuarial present value of the extended warranty in question at time 4.

We find i = e^δ - 1 = e^0.07 - 1 = 0.0725081813.

Thus, Ā¹_4:3¬ = (0.0725081813/0.07)0.0997218026 = 0.1032949505.

Our desired answer is therefore 0.4288729041R* Ā¹_4:3¬ = 0.4288729041R*0.1032949505 =

0.0443004054R. Expressed as a fraction of R, our answer is thus 0.0443004054.

Problem S3L52-4.

This question uses the Illustrative Life Table, which can be found here.

Fully discrete benefit premiums of amount P are paid each year to insure a 40-year-old robot whose mortality follows the Illustrative Life Table at i = 6%. The insurance policy is will pay $60000 at the end of the year of the robot's failure, until the robot is 80 years old. Find P.

Solution S3L52-4. We use the formula P¹_x:n¬ = A¹_x:n¬/ä_x:n¬.

We need to find A¹_40:40¬. We do so using the formula A¹_x:n¬ = A_x - _nE_x*A_x+n. Thus,

A¹_40:40¬ = A₄₀ - ₄₀E₄₀*A₈₀ = A₄₀ - ₂₀E₄₀*₂₀E₆₀*A₈₀ = 0.16132 - 0.27414*0.14906*0.66575=

A¹_40:40¬ = 0.1341152524.

Now we need to find ä_40:40¬. We do so using the formula ä_x:n¬ = ä_x - _nE_x*ä_x+n. Thus,

ä_40:40¬ = ä₄₀ - ₄₀E₄₀*ä₈₀ = ä₄₀ - ₂₀E₄₀*₂₀E₆₀*ä₈₀ = 14.8166 - 0.27414*0.14906*5.9050 =

ä_40:40¬ = 14.57530216.

We want to find P = 60000*P¹_40:40¬ = 60000*A¹_40:40¬/ ä_40:40¬ = 60000*0.1341152524/14.57530216 = P = $552.0925092 = about $552.09.

Problem S3L52-5. Similar to Question 36 from the Casualty Actuarial Society's Fall 2006 Exam 3. Fully discrete benefit premiums are paid each year to insure a 40-year-old robot whose mortality follows the Illustrative Life Table at i = 6%. The insurance policy is will pay $60000 at the end of the year of the robot's failure, until the robot is 80 years old. For the first 20 years of the policy, premiums of amount Q are paid. For the last 20 years of the policy, premiums of amount 1.5Q are paid. Find Q.

Solution S3L52-5. We use the formula P¹_x:n¬ = A¹_x:n¬/ä_x:n¬.

We need to find A¹_40:40¬. We do so using the formula A¹_x:n¬ = A_x - _nE_x*A_x+n. Thus,

A¹_40:40¬ = A₄₀ - ₄₀E₄₀*A₈₀ = A₄₀ - ₂₀E₄₀*₂₀E₆₀*A₈₀ = 0.16132 - 0.27414*0.14906*0.66575=

A¹_40:40¬ = 0.1341152524.

Here, ä_40:40¬ is not calculated on the basis of uniform payments each period. Rather, the payments from time 20 to time 40 are 1.5 times the payments from time 0 to time 20.

Thus, ä_40:40¬ = ä_40:20¬ + 1.5*₂₀E₄₀*ä_60:20¬.

Why was 1.5*ä_60:20¬ also multiplied by ₂₀E₄₀? Because we need to discount 1.5*ä_60:20¬ by 20 years and only consider it in the event that the robot survives to age 60.

We know that _nE_x = vⁿ*_np_x, so including a factor of ₂₀E₄₀ takes both of these considerations into account.

Now we find ä_40:20¬ = ä₄₀ - ₂₀E₄₀*ä₆₀ = 14.8166 - 0.27414*11.1454 = 11.7612

Now we find ä_60:20¬ = ä₆₀ - ₂₀E₆₀*ä₈₀ = 11.1454 - 0.14906*5.9050 = 10.2652

Thus, ä_40:40¬ = ä_40:20¬ + 1.5*₂₀E₄₀*ä_60:20¬ = 11.7612 + 1.5*0.27414*10.2652 = 15.98235289.

Thus, Q = 60000A¹_40:40¬/ä_40:40¬ = 60000*0.1341152524/15.98235289 = Q = $503.4875152 = about $503.49.

See other sections of The Actuary's Free Study Guide for Exam 3L.