Fully Discrete Benefit Premiums, Infinite Divisibility Property For Some Probability Distributions, And Assorted Exam-style Questions
This section of sample problems and solutions is a part of The Actuary’s Free Study Guide for Exam 3L, authored by Mr. Stolyarov. This is Section 52 of the Study Guide. Look an index of all sections by following the link in this paragraph.
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 kill 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.
Āx = (i/δ)Ax
Ā1x:n¬ = (i/δ)A1x:n¬
For fully discrete back premiums where benefits on the associated life insurance policy are always paid at the demolish 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.
Px = Ax/äx
P1x:n¬ = A1x:n¬/äx:n¬
Px:n¬ = Ax:n¬/äx:n¬
When using life tables, the following identities will also prove useful.
nEx = mEx*n-mEx+m, where m is any number less than n.
A1x:n¬ = Ax – nEx*Ax+n
äx:n¬ = äx – nEx*äx+n
Some of the problems in this portion 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 expend 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 new. 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 Contemplate Guide
Problem S3L52-1.
Similar to Quiz 31 from the Casualty Actuarial Society’s Fall 2006 Exam 3.
The frequency of a firm’s annual losses due to disagreeable 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 invent 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) = θα/(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. Rep 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 collect 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 qx during each age x:
Age x……qx
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 stout 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 select time, and it will pay 0.6 of the spaceship’s purchase trace 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 bewitch price. The extended warranty only takes accomplish 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 A14:3¬, the actuarial show 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.
A14:3¬ = e-0.07q4+ e-0.07*2p4q5 + e-0.07*3p4p5q6
A14:3¬ = 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 Ā1x:n¬ = (i/δ)A1x:n¬ to gather Ā14:3¬, the actuarial present value of the extended warranty in question at time 4.
We find i = eδ – 1 = e0.07 – 1 = 0.0725081813.
Thus, Ā14:3¬ = (0.0725081813/0.07)0.0997218026 = 0.1032949505.
Our desired answer is therefore 0.4288729041R* Ā14:3¬ = 0.4288729041R*0.1032949505 =
0.0443004054R. Expressed as a fraction of R, our acknowledge is thus 0.0443004054.
Problem S3L52-4.
This query uses the Illustrative Life Table, which can be found here.
Fully discrete abet 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 destroy of the year of the robot’s failure, until the robot is 80 years faded. Find P.
Solution S3L52-4. We exercise the formula P1x:n¬ = A1x:n¬/äx:n¬.
We need to find A140:40¬. We do so using the formula A1x:n¬ = Ax – nEx*Ax+n. Thus,
A140:40¬ = A40 – 40E40*A80 = A40 – 20E40*20E60*A80 = 0.16132 – 0.27414*0.14906*0.66575=
A140:40¬ = 0.1341152524.
Now we need to find ä40:40¬. We do so using the formula äx:n¬ = äx – nEx*äx+n. Thus,
ä40:40¬ = ä40 – 40E40*ä80 = ä40 – 20E40*20E60*ä80 = 14.8166 – 0.27414*0.14906*5.9050 =
ä40:40¬ = 14.57530216.
We want to find P = 60000*P140:40¬ = 60000*A140:40¬/ ä40:40¬ = 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 aid 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 P1x:n¬ = A1x:n¬/äx:n¬.
We need to secure A140:40¬. We do so using the formula A1x:n¬ = Ax – nEx*Ax+n. Thus,
A140:40¬ = A40 – 40E40*A80 = A40 – 20E40*20E60*A80 = 0.16132 – 0.27414*0.14906*0.66575=
A140:40¬ = 0.1341152524.
Here, ä40:40¬ 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, ä40:40¬ = ä40:20¬ + 1.5*20E40*ä60:20¬.
Why was 1.5*ä60:20¬ also multiplied by 20E40? Because we need to discount 1.5*ä60:20¬ by 20 years and only consider it in the event that the robot survives to age 60.
We know that nEx = vn*npx, so including a factor of 20E40 takes both of these considerations into epic.
Now we find ä40:20¬ = ä40 – 20E40*ä60 = 14.8166 – 0.27414*11.1454 = 11.7612
Now we find ä60:20¬ = ä60 – 20E60*ä80 = 11.1454 – 0.14906*5.9050 = 10.2652
Thus, ä40:40¬ = ä40:20¬ + 1.5*20E40*ä60:20¬ = 11.7612 + 1.5*0.27414*10.2652 = 15.98235289.
Thus, Q = 60000A140:40¬/ä40:40¬ = 60000*0.1341152524/15.98235289 = Q = $503.4875152 = about $503.49.
See other sections of The Actuary’s Free Study Guide for Exam 3L.
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