Multiple Decrement Models

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1 Multiple Decrement Models Lecture: Weeks 7-8 Lecture: Weeks 7-8 (Math 3631) Multiple Decrement Models Spring Valdez 1 / 26

Multiple decrement models Lecture summary Multiple decrement model - epressed in terms of multiple state model Multiple Decrement Tables (MDT) several causes of decrement probabilities of decrement

2 Multiple decrement models Lecture summary Multiple decrement model - epressed in terms of multiple state model Multiple Decrement Tables (MDT) several causes of decrement probabilities of decrement forces of decrement The Associated Single Decrement Tables (ASDT) Uniform distribution of decrements in the multiple decrement contet in the associated single decrement contet Chapter 8 (DHW), Sections Lecture: Weeks 7-8 (Math 3631) Multiple Decrement Models Spring Valdez 2 / 26

3 Multiple decrement models eamples Eamples of multiple decrement models Multiple decrement models are etensions of standard mortality models whereby there is simultaneous operation of several causes of decrement. A life fails because of one of these decrements. Eamples include: life insurance contract is terminated because of death/survival or withdrawal (lapse). an insurance contract provides coverage for disability and death, which are considered distinct claims. life insurance contract pays a different benefit for different causes of death (e.g. accidental death benefits are doubled). pension plan provides benefit for death, disability, employment termination and retirement. Lecture: Weeks 7-8 (Math 3631) Multiple Decrement Models Spring Valdez 3 / 26

4 Multiple decrement models notation Introducing notation age no. of lives heart disease accidents other causes l (τ) d (1) d (2) d (3) 50 4, 832, 555 5, 168 1, 157 4, , 821, 937 5, 363 1, 206 5, , 810, 206 5, 618 1, 443 5, , 797, 185 5, 939 1, 679 6, , 782, 727 6, 277 2, 152 7, 631 Conventional notation: l (τ) represents the surviving population present at eact age. represents the number of lives eiting from the population between ages and + 1 due to decrement j. It is also conventional to denote the total number of eits by all modes between ages and + 1 by d (τ) i.e. m d (τ) = d (j) d (j) where m is the total number of possible decrements, and therefore, = l (τ) d (τ) l (τ) +1. Lecture: Weeks 7-8 (Math 3631) Multiple Decrement Models Spring Valdez 4 / 26 j=1

Multiple decrement models probabilities Probabilities of decrement The probability that a life () will leave the group within one year as a result of decrement j: q (j) = d (j) /l (τ).

5 Multiple decrement models probabilities Probabilities of decrement The probability that a life () will leave the group within one year as a result of decrement j: q (j) = d (j) /l (τ). The probability that () will leave the group (regardless of decrement): q (τ) = d (τ) /l (τ) = m j=1 d (j) /l (τ) = m j=1 q (j). The probability that () will remain in the group for at least one year: p (τ) = 1 q (τ) = l (τ) +1 /l(τ) = (l (τ) d (τ) )/l (τ). Lecture: Weeks 7-8 (Math 3631) Multiple Decrement Models Spring Valdez 5 / 26

Multiple decrement models probabilities - continued We also have the probability of remaining in the group after n years p (τ) n and the complement = l (τ) +n /l(τ) = p (τ) p (τ) +1 p(τ) +n 1.

6 Multiple decrement models probabilities - continued We also have the probability of remaining in the group after n years p (τ) n and the complement = l (τ) +n /l(τ) = p (τ) p (τ) +1 p(τ) +n 1. q (τ) n = 1 p (τ) n. The number of failures due to decrement j over the interval (, + n] is n 1 nd (j) = d (j) +t. t=0 These relationships should be straightforward to follow: nd (j) = l (τ) nq (j) nd (τ) = l (τ) q (τ) n Lecture: Weeks 7-8 (Math 3631) Multiple Decrement Models Spring Valdez 6 / 26

Multiple decrement models MDT Illustration of Multiple Decrement Table Epand Multiple Decrement Table (MDT) into: l (τ) d (1) d (2) d (3) q (1) q (2) q (3) q (τ) p (τ) 50 4, 832, 555 5, 168 1, 157 4,

7 Multiple decrement models MDT Illustration of Multiple Decrement Table Epand Multiple Decrement Table (MDT) into: l (τ) d (1) d (2) d (3) q (1) q (2) q (3) q (τ) p (τ) 50 4, 832, 555 5, 168 1, 157 4, , 821, 937 5, 363 1, 206 5, , 810, 206 5, 618 1, 443 5, , 797, 185 5, 939 1, 679 6, , 782, 727 6, 277 2, 152 7, Lecture: Weeks 7-8 (Math 3631) Multiple Decrement Models Spring Valdez 7 / 26

Multiple decrement models illustrative problems Illustrative problems Using the previously given multiple decrement table, compute and interpret the

9 Multiple decrement models illustrative problems Illustrative problems Using the previously given multiple decrement table, compute and interpret the following: 1 d (3) p (τ) q (1) q (2) Lecture: Weeks 7-8 (Math 3631) Multiple Decrement Models Spring Valdez 8 / 26

The continuous case force of decrement Total force of decrement The total force of decrement at age is defined as µ (τ) = lim 1 h 0 h q(τ) h = 1 l (τ) d d l(τ) = d log l(τ) d Therefore, analogous to

13 The continuous case force of decrement Total force of decrement The total force of decrement at age is defined as µ (τ) = lim 1 h 0 h q(τ) h = 1 l (τ) d d l(τ) = d log l(τ) d Therefore, analogous to the single decrement table, we have p (τ) t ( t ) = ep µ (τ) +s ds 0 and or, more generally 1 q (τ) = sp (τ) µ (τ) +s ds 0 t q (τ) t = sp (τ) µ (τ) +s ds. 0 Lecture: Weeks 7-8 (Math 3631) Multiple Decrement Models Spring Valdez 9 / 26

14 The continuous case force of single decrement Force of a single decrement The force of decrement due to decrement j is defined as: µ (j) = 1 d l (τ) d l(j). Notice that the denominator is NOT l (j) As a consequence, we see that µ (τ) = m j=1 µ (j). but is rather l (τ). The total force of decrement is (indeed) the sum of all the other partial forces of decrement. We can also show that 1 q (j) = sp (τ) µ (j) +s ds. 0 Lecture: Weeks 7-8 (Math 3631) Multiple Decrement Models Spring Valdez 10 / 26

The continuous case illustrative eercise Illustrative eercise Suppose that in a triple-decrement model, you are given constant forces of decrement, for a person now age, as follows: µ (1) +t = b, for

17 The continuous case illustrative eercise Illustrative eercise Suppose that in a triple-decrement model, you are given constant forces of decrement, for a person now age, as follows: µ (1) +t = b, for t 0, µ (2) +t = b, for t 0, µ (3) +t = 2b, for t 0. You are also given that the probability () will eit the group within 3 years due to decrement 1 is Compute the length of time a person now age is epected to remain in the triple decrement table. Answer (to be discussed in lecture): 83 1/3 years. Lecture: Weeks 7-8 (Math 3631) Multiple Decrement Models Spring Valdez 11 / 26

20 The ASDT The associated single-decrement table (ASDT) For each of the causes of decrement in an MDT, a single-decrement table can be defined showing the operation of that decrement independent of the others. called the associated single-decrement table (ASDT) Each table represents a group of lives reduced continuously by only one decrement. For eample, a group subject only to death, but not to other decrements such as withdrawal. The associated probabilities in the ASDT are called absolute rates of decrements. For eample, the absolute rate of decrement due to decrement j over the interval (, + t] is q (j) t. One should be able to eplain intuitively why the following always hold true: q (j) t q (j) t. Lecture: Weeks 7-8 (Math 3631) Multiple Decrement Models Spring Valdez 12 / 26

22 Link between the MDT and the ASDT Link between the MDT and the ASDT If given the absolute rates of decrements, say q (1), q (2),..., q (m), how do we derive the probabilities of decrements q (1), q (2),..., q (m) the MDT? And vice versa. The fundamental link: µ (j) Therefore, it follows that = µ (j) for all j = 1, 2,..., m. in Furthermore, we note that p (τ) t = p (1) t p (2) t p (m) t. q (j) t = = t 0 t 0 p (τ) s p (τ) s µ (j) +s ds µ (j) +s ds = t 0 p (τ) s p (j) s sp (j) µ (j) +s ds. Lecture: Weeks 7-8 (Math 3631) Multiple Decrement Models Spring Valdez 13 / 26

Link between the MDT and the ASDT In the multiple decrement contet We assume the following UDD assumption: q (j) t This leads us to the following result: p (j) t Proof to be done in class.

23 Link between the MDT and the ASDT In the multiple decrement contet We assume the following UDD assumption: q (j) t This leads us to the following result: p (j) t Proof to be done in class. = t q (j), for 0 t 1. = (1 t q (τ) ) q(j) /q (τ). This result allows us to compute the absolute rates of decrements q (j) given the probabilities of decrements in the multiple decrement model. In particular, when t = 1, we have q (j) = 1 (1 q (τ) ) q(j) /q (τ). Lecture: Weeks 7-8 (Math 3631) Multiple Decrement Models Spring Valdez 14 / 26

Link between the MDT and the ASDT Illustrative eample In a double decrement table where cause d is death and cause w is withdrawal, you are given: both deaths and withdrawals are each uniformly

26 Link between the MDT and the ASDT Illustrative eample In a double decrement table where cause d is death and cause w is withdrawal, you are given: both deaths and withdrawals are each uniformly distributed over each year of age in the double decrement table. l (τ) = 1000 q (w) = 0.48 d (d) Calculate q (d) = 0.35d (w) and q (w). Note: One way to check your results make sense is to ensure the inequality q (j) q (j) is satisfied. Lecture: Weeks 7-8 (Math 3631) Multiple Decrement Models Spring Valdez 15 / 26

Link between the MDT and the ASDT In the associated single decrement contet We assume the following UDD assumption: This implies: q (j) t = t q (j), for 0 t 1.

28 Link between the MDT and the ASDT In the associated single decrement contet We assume the following UDD assumption: This implies: q (j) t = t q (j), for 0 t 1. p (j) t µ (j) +t = t p (j) µ(j) +t = q (j). Using the previous link, one can derive q (j) t = = t 0 t 0 = q (j) sp (τ) µ (j) +s ds i j t 0 p (i) s i j sp (j) µ (j) +s ds (1 s q (i) )ds. Use this integration to derive the probabilities of decrement given the absolute rates of decrements. Lecture: Weeks 7-8 (Math 3631) Multiple Decrement Models Spring Valdez 16 / 26

29 Link between the MDT and the ASDT The case of two decrements When we have m = 2, we can derive q (1) t = q (1) = q (1) t 0 ( 1 s q (2) ( t 1 2 t2 q (2) ), ) ds and similarly, ( q (2) t = q (2) t 1 ) 2 t2 q (1). Check the case when t = 1. As an eercise, etend the derivation to the case of a triple decrement case. Lecture: Weeks 7-8 (Math 3631) Multiple Decrement Models Spring Valdez 17 / 26

Link between the MDT and the ASDT Illustrative eample 1 In a triple decrement table where each of the decrement in their associated single decrement tables satisfy the uniform distribution of

31 Link between the MDT and the ASDT Illustrative eample 1 In a triple decrement table where each of the decrement in their associated single decrement tables satisfy the uniform distribution of decrement assumption, you are given: q (1) = 0.03 and q (2) = 0.06 l (τ) Calculate d (3). = 1, 000, 000 and l (τ) +1 = 902, 682 Lecture: Weeks 7-8 (Math 3631) Multiple Decrement Models Spring Valdez 18 / 26

Link between the MDT and the ASDT Illustrative eample 2 In a triple decrement table, you are given that decrement (1) is death, decrement (2) is disability, and decrement (3) is withdrawal.

33 Link between the MDT and the ASDT Illustrative eample 2 In a triple decrement table, you are given that decrement (1) is death, decrement (2) is disability, and decrement (3) is withdrawal. In addition, you have: q (1) 60 = 0.01, q (2) 60 = 0.05 and q (3) 60 = Withdrawals occur only at the end of the year. Mortality and disability are uniformly distributed over each year of age in the associated single decrement tables. Calculate q (3) 60. Lecture: Weeks 7-8 (Math 3631) Multiple Decrement Models Spring Valdez 19 / 26

Actuarial present value calculations illustrative eample 1 Illustrative eample 1 An insurance policy issued to (50) will pay $40, 000 upon death if death is accidental and occurs within 25 years.

38 Actuarial present value calculations illustrative eample 1 Illustrative eample 1 An insurance policy issued to (50) will pay $40, 000 upon death if death is accidental and occurs within 25 years. An additional benefit of $10, 000 will be paid regardless of the time or cause of death. The force of accidental death at all ages is The force of death for all other causes is 0.05 at all ages. You are given δ = 10%. Find the net single premium for this policy. [To be discussed in lecture.] Lecture: Weeks 7-8 (Math 3631) Multiple Decrement Models Spring Valdez 20 / 26

Actuarial present value calculations illustrative eample 2 Illustrative eample 2 An employer provides his employees aged 62 the following one-year term benefits, payable at the end of the year of

40 Actuarial present value calculations illustrative eample 2 Illustrative eample 2 An employer provides his employees aged 62 the following one-year term benefits, payable at the end of the year of decrement: $1 if decrement results from cause 1; $2 if decrement results from cause 2; and $6 if decrement results from cause 3. Only three possible decrements eist. In their associated single-decrement tables, all three decrements follow de Moivre s Law with ω = 65. In addition, assume UDD in each single decrement table. You are given i = 10%. Find the actuarial present value at age 62 of the benefits. Lecture: Weeks 7-8 (Math 3631) Multiple Decrement Models Spring Valdez 21 / 26

42 Asset shares Asset share calculations Asset shares refer to the projections of the assets epected to accumulate under a single policy (or a portfolio of policies). To illustrate, consider an insurance contract that pays: a benefit of b (d) k at the end of year k for deaths during the year, and a benefit of b (w) k at the end of year k for withdrawals of surrenders during the year. The policy receives an annual contract premium of G at the beginning of the year. It pays a percentage r k of the premium for epenses plus a fied amount of epense of e k. Epenses occur at the beginning of the year. Lecture: Weeks 7-8 (Math 3631) Multiple Decrement Models Spring Valdez 22 / 26

45 Asset shares - continued In addition,we have Interest rate is an effective annual rate of i. The probabilities of decrements are denoted by q (d) respectively, for deaths and withdrawals. +k 1 The probability of staying in force through year k is therefore p (τ) +k 1 = 1 q(d) +k 1 q(w) +k 1. and q(w) +k 1, Denote the asset share at the end of year k by AS k with an initial asset share at time 0 of AS 0 which may or may not be zero. For a new policy/contract, we may assume this is zero. Lecture: Weeks 7-8 (Math 3631) Multiple Decrement Models Spring Valdez 23 / 26

46 Asset shares recursive formula The recursion formula for asset shares Beginning with k = 1, we find [AS 0 + G(1 r 0 ) e 0 ](1 + i) = b (d) 1 q(d) + b (w) 1 q (w) + AS 1 p (τ), and we get AS 1 = [AS 0 + G(1 r 0 ) e 0 ](1 + i) b (d) 1 q(d) b (w) This is easy to generalize as follows: p (τ) AS k = [AS k 1 + G(1 r k 1 ) e k 1 ](1 + i) b (d) k q(d) p (τ) +k 1 1 q (w) +k 1 b(w) k q (w) +k 1 Do not memorize - use your intuition to develop the recursive formulas... Lecture: Weeks 7-8 (Math 3631) Multiple Decrement Models Spring Valdez 24 / 26

47 Asset shares recursive formula - continued Note that b (d) k is the amount of death benefit payable at the end of the year of death. We have sometimes used DB to refer to this amount. is the amount of withdrawal benefit payable at the end of the year of withdrawal. This amount is what we have called the cash surrender value and have used the symbol CV. b (w) k Lecture: Weeks 7-8 (Math 3631) Multiple Decrement Models Spring Valdez 25 / 26

Asset shares illustrative eample Illustrative eample For a portfolio of fully discrete whole life insurances of $1,000 on (30), you are given: the contract annual premium is $9.

48 Asset shares illustrative eample Illustrative eample For a portfolio of fully discrete whole life insurances of $1,000 on (30), you are given: the contract annual premium is $9.50; renewal epenses, payable at the start of the year, are 3% of premium plus a fied amount of $2.50; AS 20 = 145 is the asset share at the end of year 20; CV 21 = 100 is the cash value payable upon withdrawal at the end of year 21; interest rate is i = 7.5% and the applicable decrement table is given below: q (d) q (w) Calculate the asset share at the end of year 21. Lecture: Weeks 7-8 (Math 3631) Multiple Decrement Models Spring Valdez 26 / 26

Stat 476 Life Contingencies II. Multiple Life and Multiple Decrement Models

Stat 476 Life Contingencies II Multiple Life and Multiple Decrement Models Multiple Life Models To this point, all of the products we ve dealt with were based on the life status (alive / dead or multi-state)