Exam Style Questions on Life Table Functions for Independent Lives and Multiple Causes of Decrement

When there are multiple causes of decrement (i.e. some kind of failure or bad event happening to the entity in question), we can express life table functions pertaining to all causes of decrement with the superscript ^(τ). If assign numbers 1 through n to each of the causes of decrement, then we can express life table functions pertaining some cause of decrement I with the superscript ⁽ⁱ⁾, where i can be an integer from 1 through n.

If you are given each _mq_x⁽ⁱ⁾, then the following holds: _mq_x^(τ) = _i=1ⁿ∑ _mq_x⁽ⁱ⁾.

Moreover, if you are given each µ_x⁽ⁱ⁾, then the following holds:µ_x^(τ) = _i=1ⁿ∑µ_x⁽ⁱ⁾.

The following useful relationship holds between any _np_x and µ_x+t:

_np_x = exp[-₀ⁿ∫µ_x+tdt].

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.

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1997. Second Edition. Society of Actuaries: Itasca, Illinois. pp. 308-311.

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

Problem S3L46-1.

Similar to Question 14 from the Casualty Actuarial Society's Fall 2006 Exam 3.

Two independent lives are currently aged 70 and 90. Mortality follows DeMoivre's Law with ω = 150. (This is the future - when life expectancy has been increased dramatically due to technological advances.)

Let _30│10φ_70:90 be the probability that the first of the two deaths occurs between 30 and 40 years from now.

Let _30│10λ_70:90 be the probability that the last of the two deaths occurs between 30 and 40 years from now.

Find _30│10λ_70:90 - _30│10φ_70:90.

Solution S3L46-1.

DeMoivre's Law corresponds to the uniform distribution, with ω being the limited age.

A life aged 70 has probability 1/(150-70) = 1/80 of dying each year.

A life aged 90 has probability 1/(150-90) = 1/60 of dying each year.

We find _30│10φ_70:90:

For a first death to occur between 30 and 40 years from now, it follows that both lives must have survived during the next 30 years. The probability that this is the case is

(50/80)(30/60) = 0.3125

If both lives survive for 30 years, the probability of both surviving another 10 years is (40/50)(20/30) = 8/15.

The probability that at least one of the survivors to age 30 will perish between 30 and 40 years from now is 1 - 8/15 = 7/15.

Therefore, _30│10φ_70:90 = 0.3125(7/15) = 7/48

Now we find _30│10λ_70:90:

_30│10λ_70:90 is the complement of the sum of the following probabilities.

P(both lives die before 30 years from now) = (30/80)(30/60) = 0.1875

P(both lives die after 40 years from now) = (40/80)(20/60) = 0.166666667

P(no lives die before 30 years from now and only one life dies between 30 and 40 years from now) = 0.3125*(10/50)(20/30) + 0.3125*(40/50)(10/30) = 0.125

P(one life dies before 30 years from now and one life dies after 40 years from now) = (30/80)(20/60) + (40/80)(30/60) = 0.375

Thus, _30│10λ_70:90 = 1 - 0.1875 - 0.166666667 - 0.125 - 0.375 = 0.1458333333333 = 7/48.

Thus, _30│10λ_70:90 - _30│10φ_70:90 = 7/48 - 7/48 = 0.

Problem S3L46-2.

Similar to Question 15 from the Casualty Actuarial Society's Fall 2006 Exam 3.

Two government departments are operated by Bureaucrat X and Bureaucrat Y. Department A only operates so long as both bureaucrats are alive. Department B operates so long as at least one bureaucrat is alive. Bureaucrat mortality is independent for each bureaucrat and exhibits the property that _tq₀ = t²/400 for t < 20. Find the absolute value of the difference in the expected operational lifetimes of Departments A and B.

Solution S3L46-2.

We use the following formula from Section 10: ė_x = ₀^∞∫_tp_xdt. Here, the upper bound of our integral is 20, since no bureaucrat will survive past 20 years from now.

Since _tq₀ = t²/400, _tp₀ = 1 - t²/400.

For Department A, the relevant function to be integrated is (_tp₀)², since both bureaucrats must be alive for the department to function.
Thus, for Department A, E(A) = ₀²⁰∫(_tp₀)²dt = ₀²⁰∫(1 - t²/400)²dt = 10.6666666667 years.

We know from basic probability theory that (X or Y) = X + Y - (X and Y).

Thus, E(X or Y) = E(X) + E(Y) - E(X and Y).

Department B will keep operating if either X or Y are alive. Thus, the expected lifetime for this department will be (ė₀ for X) + (ė₀ for Y) - (ė₀ for X and Y).

We already found (ė₀ for X and Y) - the expected lifetime of department A - to be 10.6666666667 years.

We also know that (ė₀ for X) = (ė₀ for Y) = ₀²⁰∫(1 - t²/400)dt = 13.333333333 years.

Hence, E(B) = 2*13.333333333 - 10.6666666667 = 16 years.

We want to find E(B) - E(A) = 16 - 10.6666666667 = 16/3 = 5.333333333 years.

Problem S3L46-3.

Similar to Question 16 from the Casualty Actuarial Society's Fall 2006 Exam 3.

Pink slugs aged x can perish from Bad Thing 1 or Bad Thing 2. The following 2-decrement table expresses the probabilities associated with perishing from each Bad Thing.

x........ q_x⁽¹⁾........q_x⁽²⁾

2.......0.09........0.04

3.......0.14........0.02

4.......0.01........0.04

5.......0.07........0.10

Use this 2-decrement table to find ₄q₂^(τ).

Solution S3L46-3.

We use the formula _mq_x^(τ) = _i=1ⁿ∑ _mq_x⁽ⁱ⁾ to find the following:

q₂^(τ) = 0.09 + 0.04 = 0.13

q₃^(τ) = 0.14 + 0.02 = 0.16

q₄^(τ) = 0.01 + 0.04 = 0.05

q₅^(τ) = 0.07 + 0.10 = 0.17

Now to find ₄q₂^(τ), we need to find 1 - ₄p₂^(τ) = 1 - (p₂^(τ))(p₃^(τ))(p₄^(τ))(p₅^(τ)) =

1 - (1 - 0.13)(1 - 0.16)(1 - 0.05)(1 - 0.17) = ₄q₂^(τ) = 0.4237642.

Problem S3L46-4.

Similar to Question 16 from the Casualty Actuarial Society's Fall 2006 Exam 3.

Pink slugs aged x can perish from Bad Thing 1 or Bad Thing 2. The following 2-decrement table expresses the probabilities associated with perishing from each Bad Thing.

x........ q_x⁽¹⁾........q_x⁽²⁾

2.......0.09........0.04

3.......0.14........0.02

4.......0.01........0.04

5.......0.07........0.10

Use this 2-decrement table to find _3│q₂⁽²⁾. Hint: You do need to take into account the values of q_x⁽¹⁾. Think about how this might be done.

Solution S3L46-4. _3│q₂⁽²⁾ is the probability that a pink slug will perish from Bad Thing 2 between 3 and 4 years from now. This implies that the pink slug needs to have survived both Bad Things until turning 5.

Hence, using some of the values calculated in Solution S3L46-3,

_3│q₂⁽²⁾ = (p₂^(τ))(p₃^(τ))(p₄^(τ))(q₅⁽²⁾) = (1 - 0.13)(1 - 0.16)(1 - 0.05)0.10 = _3│q₂⁽²⁾ = 0.069426.

Problem S3L46-5.

Similar to Question 17 from the Casualty Actuarial Society's Fall 2006 Exam 3.
Mortimer the Mortal can only die of shark attacks or falling elephants. His force of mortality due to shark attacks is µ_x+t⁽¹⁾ = 0.5t. His force of mortality due to falling elephants is µ_x+t⁽²⁾ = 0.2t. Mortimer takes out a 30-year term life insurance policy paying 30 Golden Hexagons (GH) if he dies of a shark attack and 20 GH if he dies of a falling elephant. Elephants have fallen on all the buildings containing financial markets, so the interest rate is 0. Find the actuarial present value of this policy to Mortimer.

Solution S3L46-5. We use the formula µ_x^(τ) = _i=1ⁿ∑µ_x⁽ⁱ⁾ to find µ_x+t^(τ) = 0.5t + 0.2t = 0.7t.

We use the formula _np_x = exp[-₀ⁿ∫µ_x+tdt].

Thus, _np_x^(τ) = exp[-₀ⁿ∫0.7t*dt] = exp[-0.35n²].

Now we use the formula Ā¹_x:n¬ = ₀ⁿ∫v^t*_tp_x*μ_x(t)dt.

We know that _tp_x^(τ) = exp[-0.35t²]. Moreover, v^t = 1 for all t, since the interest rate is 0.

Thus, Ā³⁰_x:n¬ ⁽¹⁾due to shark attacks is 30*₀ⁿ∫v^t*_tp_x^(τ)*μ_x+1⁽¹⁾dt =

30*₀ⁿ∫0.5t*exp[-0.35t²]dt = 30*(0.5/0.7)(-exp[-0.35t²])│₀³⁰ =

30(0.5/0.7)(1 - exp[-0.35*900]) = about 21.42857143 GH.

Ā²⁰_x:n¬⁽²⁾due to falling elephants is 20*₀ⁿ∫v^t*_tp_x^(τ)*μ_x+1⁽²⁾dt =

20*₀ⁿ∫0.2t*exp[-0.35t²]dt = 20*(0.2/0.7)(-exp[-0.35t²])│₀³⁰ =

20(0.2/0.7)(1 - exp[-0.35*900]) = about 5.714285714 GH.

Thus, the actuarial present value of this policy to Mortimer is 21.42857143 + 5.714285714 = about 27.14285714 GH.

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