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) of a single individual. However, in order to deal with products involving multiple lives, we need more comple models. These future lifetimes may be dependent or independent. The force of transition from state i to state j based on the joint life statuses of () and (y) is denoted by µ ij y or µ ij :y, with the latter being preferred when we have numerical values for and/or y. 2
Joint Life Model () alive (y) alive () alive (y) dead 0 1 () dead (y) alive 2 () dead (y) dead 3 The model above could be used in cases where cash flows depend on the status of two lives () and (y). If the model allows a transition from state 0 directly to state 3, it s known as a common shock model. 3
Traditional Multiple Life Status Actuarial Notation The subscript y denotes a status which fails upon the first of the future lifetimes of () and (y) to fail. Likewise, the status y fails upon the failure of the second failure of () and (y). We will use these statuses in the same manner we used the single life status and the term status n This notation follows the same convention we used earlier for term products and guaranteed annuities. Then T y is the random variable describing the time until the first death of () and (y), and T y is the RV giving the time until the second death of () and (y). Important Relationship T + T y = T y + T y 4
Multiple Life Probability Definitions and Relationships µ y = force of mortality for the joint status y = µ 01 y + µ 02 y + µ 03 y tp y = P[the status y does not fail within t years] = P[() and (y) are both alive in t years] = t p 00 y tq y = P[the status y fails within t years] = 1 t p y = P[() and (y) are not both alive in t years] = t py 01 + t py 02 + t py 03 5
Multiple Life Probability Definitions and Relationships tp y = P[the status y does not fail within t years] = P[() or (y) or both are still alive in t years] = t py 00 + t py 01 + t py 02 tq y = P[the status y fails within t years] = 1 t p y = P[() and (y) are both dead in t years] = t p 03 y We have the following relationship for () and (y): tp + t p y = t p y + t p y Further, we define n E y = v n np y and n E y = v n np y 6
More Multiple Life Probability Definitions and Relationships We also use the numbers 1, 2,... above the individual life statuses to indicate a specific order of status failure: t q 1 y = P[() dies before (y) and within t years] = t 0 sp 00 y µ 02 +s:y+s ds t q 2 y = P[() dies after (y) and within t years] = t 0 sp 01 y µ 13 +s ds Note that we can let t be in either of these symbols. 7
Insurances and Annuities Based on Multiple Lives We can derive some relationships among the various multiple life insurances and annuities: A y + A y = A + A y a y + a y = a + a y a y = 1 A y δ a y = a y a y a y = 1 A y δ a y + a y = a y a y 8
Two Independent Lives In the special case where we re dealing with two future lifetimes that are independent, many of our epressions become much simpler. In this event, all of the forces of transition only depend on a single life, so in our joint life model, we would have: µ 01 y = µ 23 y = µ y µ 02 y = µ 13 = µ µ 03 y = 0 As a result, our transition probabilities become functions of the individual survival probabilities: tp 00 y = t p t p y t p 01 y = t p t q y t p 00 y = t p t p y t p 01 y = t p t q y And for the case of independent lives, in the traditional actuarial notation, we have: µ y = µ + µ y and tp y = t p t p y 9
Two Independent Lives Eample Suppose that () and (y) have independent future lifetimes with µ = 0.02, µ y = 0.03, and δ = 0.05. (a) Calculate a y. a y = 0 e δt tp 02 y dt = 0 e δt tq t p y dt = 2.50 (b) Calculate the probability that () dies before (y). q 1y = 0 tpy 00 µ 02 +t:y+t dt = tp t p y µ +t dt = 0.4 0 10
Multiple Decrement Models Another common application of multi-state model theory is in situations where a members of a population may leave due to one of several different causes. These so-called multiple decrement models can applied to various situations. These models typically have one active / alive state, where each individual begins. An individual may leave the population by transitioning to any of the eit states. We usually only model a single transition out of the active state; we either ignore or disallow transitions among the eit states. 11
Multiple Decrement Model Active Eit 1 0 1... Eit n Eit 2 2 n 12
Multiple Decrement Eample We issue a fully continuous whole life policy to (). The death benefit is $100,000, but doubles if () dies as a result of an accident. No benefit or surrender value is payable if the policyholder lapses his policy. (a) Draw an appopriate state diagram for modeling this type of policy. (b) Write an epression for the probability that this policy ever pays a benefit. (c) Write an epression for the EPV of this insurance. 13
Multiple Decrement Tables and Traditional Actuarial Notation The probabilities for multiple decrement models are traditionally epressed in multiple decrement tables, which are analogous to the mortality tables used in our alive-dead model. We again start with some cohort of l 0 people at age 0. The number of people remaining in the population at age is given by l. The number of people leaving at age due to decrement i is denoted by d (i) ; the total number leaving during this year is denoted by d (τ), where d (τ) = i d (i) Then l +1 = l d (1) d (2) d (n) = l d (τ) 14
Sample Double Decrement Table Here s an ecerpt of a table with two decrements: l d (1) d (2) 60 1000 11 10 61 979 12 10 62 957 13 10 63 934 14 10 64 910 15 10.. We can calculate that of the original 1000 members of this population, d (2) 2 = 10 of them leave due to decrement 2 in the third year, and d (τ) 3 = d (1) 3 + d (2) 3 = 24 of them leave in the fourth year for any reason... 15
Multiple Decrement Tables More Notation We denote the probability of leaving in a given year as a result of decrement i by q (i) = d (i) l and the probability of leaving due to any decrement as q (τ) = i q (i) = d (τ) l We denote the probability of surviving all decrements in a given year by p (τ) = 1 q (τ) = l +1 l 16
Multiple Decrements in Continuous Models In a continuous model, the force of failure for decrement i at age is denoted by µ (i) and the total force of failure is µ (τ) = µ (1) + µ (2) + + µ (n) The probability of failing within t years due to decrement i is: t tq (i) = 0 sp (τ) µ (i) +s ds and the probability of failing within t due to any decrement is: t tq (τ) = 0 sp (τ) µ (τ) +s ds (Note that these aren t actually new formulas; they re just epressed in different notation.) 17
Dependent and Independent Probabilities for Multiple Decrement Models In our multiple decrement table above, the d (i) the calculated q (i) the table; hence, q (i) of decrement. values (and hence values) depended on the other decrements in is sometimes called a dependent probability If we calculate the probability of someone age failing due to decrement i in the absence of other decrements, the resulting probability is called the independent probability of decrement, the probability of decrement in the associated single decrement table, or sometimes the absolute rate of decrement and is denoted by q (i). Note that t q (i) t q (i) for any, t, and i. 18
Formulas for Independent Probabilities in Multiple Decrement Models The probability of still being in the population in t years in the associated single decrement model for decrement i is denoted tp (i). tp (i) tp (i) = 1 t q (i) [ t ] = ep µ (i) +s ds 0 tp (τ) = tp (τ) n tp (i) i=1 [ t ] = ep µ (τ) +s ds 0 19
Fractional Age Assumptions in Multiple Decrement Models If we have a discrete model (i.e., a multiple decrement table), we ll often need to make some assumptions regarding how the decrements act within the year: We need this in order to get the independent failure probabilities from the dependent ones (and vice-versa). We also need this additional information to do calculations for non-integral lengths of time. 20
Fractional Age Assumptions Eample (Decrement 1 is retirement, decrement 2 is death) l d (1) d (2) 60 1000 11 10 61 979 12 10 62 957 13 10 63 934 14 10 64 910 15 10.. Eample: For the double decrement table above, calculate and interpret q (1) 62 and q (2) 62 assuming: (a) All retirements occur at the beginning of the year. (b) All retirements occur at the end of the year. (c) All retirements occur uniformly throughout the year. (Assume in all cases that deaths happen throughout the year.) 21..
Uniform Distribution of Decrements in the Multiple-Decrement Table One option is to assume that each decrement in the multiple decrement table is distributed uniformly over the year in the presence of other decrements, i.e., t q (j) is a linear function of t. Under this UDD assumption in the multiple decrement contet: tq (j) = t q (j) From this assumption, we can derive the following relationships: tq (τ) = t q (τ) tp (τ) ( tp (j) = = 1 t q (τ) 1 t q (τ) ) q (j) µ (j) +t = /q (τ) q (j) 1 t q (τ) 22
Uniform Distribution in the Associated Single-Decrement Tables Another option is to assume that each decrement is distributed uniformly in its associated single decrement table, i.e., in the absence of other decrements, so that t q (j) is a linear function of t. Under this UDD assumption in the single decrement contet: tq (j) = t q (j) From this assumption, we can derive various relationships. For eample, for a double decrement table, we have: ( q (1) = q (1) 1 1 ) ( 2 q (2) and q (2) = q (2) 1 1 ) 2 q (1) Similar equations can be derived for triple decrement tables. 23
Constant Force of Decrement Assumption We could instead assume that the forces of decrement are constant within the year. This assumption results in the following formulas: [ tp (j) = p (j) ] t [ tp (τ) = p (τ) ] t tq (j) ( = q(j) [ q (τ) 1 [ p (j) = p (τ) ] q (j) p (τ) /q (τ) ] t ) 24
Multiple Decrement Eamples I This table is from a policy which epires on first of death (1), surrender (2), or illness (3). Calculate l d (1) d (2) d (3) 40 100000 51 4784 44 41 95121 52 4526 47 42 90496 53 4268 50 43 86125 54 4010 53 44 82008 55 3753 56 45 78144 56 3496 59 46 74533 57 3239 62 47 71175 57 2983 65 48 68070 58 2729 67 49 65216 58 2476 69 50 62613 58 2226 70 (a) 3 p (τ) 45, q(1) 40, and 5q (3) 41. (b) the probability that a policy issued to (45) generates a claim for death or critical illness before age 47. (c) the probability that a policy issued to (40) is surrendered between ages 45 and 47. 25
Multiple Decrement Eamples II This table is from a policy which epires on first of death (1), surrender (2), or illness (3). Calculate l d (1) d (2) d (3) 40 100000 51 4784 44 41 95121 52 4526 47 42 90496 53 4268 50 43 86125 54 4010 53 44 82008 55 3753 56 45 78144 56 3496 59 46 74533 57 3239 62 47 71175 57 2983 65 48 68070 58 2729 67 49 65216 58 2476 69 50 62613 58 2226 70 Calculate 0.2 q (j) 50 for j = 1, 2, 3 assuming, between integer ages, (a) UDD in all decrements (b) constant transition forces in all decrements 26