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by: Mary Veum

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# Actuarial Mathematics II MATH 3631

Mary Veum
UCONN
GPA 3.97

Emiliano Valdez

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COURSE
PROF.
Emiliano Valdez
TYPE
Class Notes
PAGES
50
WORDS
KARMA
25 ?

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This 50 page Class Notes was uploaded by Mary Veum on Thursday September 17, 2015. The Class Notes belongs to MATH 3631 at University of Connecticut taught by Emiliano Valdez in Fall. Since its upload, it has received 34 views. For similar materials see /class/205813/math-3631-university-of-connecticut in Mathematics (M) at University of Connecticut.

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Date Created: 09/17/15
Contingent Contract Reserves Lecture Week 1 l l Chapter summary 0 Insurance reserVes a what are they a how do we calculate them 9 why are they important a Net premium reserves or benefit reserves a prospective calculation of reserves a retrospective calculation of reserves in I 39 between F and 0 Chapters 8 Cunningham et al l l Insurance reserves 0 Money set aside to be able to cover insurer39s future financial obligations as promised through the insurance contract a reserves show up as a liability item in the balance sheet 3 increases in reserves are an expense item in the income statement 0 Reserve calculations may vary because of a purpose of reserve valuation statutory solvency GAAP realistic shareholdersinvestors mergersacquisitions a assumptions and basis mortality interest may be prescribed a Actuary is responsible for preparing an Actuarial Opinion and Memorandum that the company39s assets are sufficient to back reserves a Reserves are more often called provisions in Europe l l Why hold reserves 0 For several life insurance contracts a the expected cost of paying the benefits generally increases over the contract term but a the periodic premiums used to fund these benefits are level 0 The portion of the premiums not required to pay expected cost in the early years are therefore set aside or provisioned to fund the expected shortfall in the later years of the contract 0 Reserves also help reduce cost of benefits as they also earn interest while being set aside 0 Although reserves are usually held on a per contract basis it is still the overall responsibility of the actuary to ensure that in the aggregate the company39s assets are enough to back these reserves l l The insurer s prospective loss 9 At any future time 25 define the insuer39s prospective loss to be tL PVFBtiPVFPt o For most types of policies it is generally true that for t 2 O tL 2 0 Le PVFBt Z PVFPt o For our purposes we shall define the expected value of this prospective loss to be the reserves at time t tV E L E PVFBt 7 E PVFPt and is to be the smallest amount for which the insurer is required to hold to be able to cover future obligations 0 Note that EtL is actually conditional on the survival of the life at time 25 Because otherwise there is no need to hold reserves when policy has been paid out 0 Reserves are indeed released from the balance sheet when policy is paid out Il l I Fully discrete reserves whole life insurance 9 Consider the case of a discrete whole life insurance where the premium Pm is paid beginning of each year and benefit 1 is paid at the end of year of death 0 Define by Jm the curtate future lifetime of aged m k 7 7 Kmik 1 0erle JriJi foerltk 0 he insurer39s prospective loss at time k or at age z k is P P J1 J1 w w klliv iPmaJJrliv lt1jgt7i for J 07 127 l l continued 0 The net premium or benefit reserve at time k is ka E kL E vjll Pm imo Awk 7 Pm lm a This is called the prospective formula the difference between the APV of F8 and the APV of FP39s 0 One can also show that the variance of kL is VarkL Var UJ1lt1 7 i d 1 gt2 lesz Awk l l Other special formulas 0 Note that it can be shown that other special formulas for the net premium reserves for the discrete whole life hold 1 Maida rlt37dgtam 1i 1 am Pxd Pure Vi 477 k Pxkd P kd 1 Axk 7 Axk Ax kl 17 17A 17A l l Others 0 Others with more obvious interpretations o ngE 17 P75 d wrk net premium reserves equals sum insured minus expected PV of future premiums and unused interest Pm remaining face amount of 1 7 PgEchurki P a kV lt17 77 A k net premium reserves Is used to finance k a kVE Piurk 7 P1 wrk net premium reserves is the shortfall of the premiums if WLl were to be bought at age I 16 l I Retrospective formula 0 Consider the reserves at time k of the whole life insurance ka Azk Pm 5izk kEm P fix iizkkEmgt 7 Am A1kkEzgt 7 m m kEm ii A1 7 a 7 my 7 7 i Pm kEm gt i Pm 8mm kkm o This is referred to as the retrospective formula Amk 7 Pm i H U A 9 H k mi 9 H I l l I Interpretation of the retrospective formula bi Pm 5 I Pm m of the retrospective formula refers to the actuarial accumulated value of the premiums paid during the first k years 9 The first term o The second term A1 k 7 1 k m 7 kEm is called the accumulated cost of insurance and it represents the net single premium for the first k years39 insurance accumulated to the end of k years l I Endowment insurance 0 The insurer39s prospective loss at time k lt n or at age z k is 7 UJ1 7 7 J17 kLiv Pmm JH 7 d for J O7 17 27 Loss is zero for k 2 n o The prospective net premium or benefit reserve at time k is k Vm Amwm PM Wam39 o Variance of kL is P 2 VarkL 1 2Awkm 7 ltAmk3mgt 2 l l Other special formulas 0 Premium difference formula P k V mm Hem mm Pmkml 0 Paid up insurance formula P V lt17 W gtA k mm zkml Pmknk o Retrospective formula k Vz i m igmm kkz a Try to derive these formulas similar approach to the whole life insurance 7 j Other types of insurances Insurance Plan Notation Prospective Formula A 1 7 P1 1 gt k lt 7L wyear term kvwlm wk nrk a it km gt k 2 71 A P a k lth hrpay whole Ilfe fenw k h 0 wk ft Awkgt k 2 h klthltn hSkltn kn Auk k thmawkry we W gt hrpay wyr endowment wam 1 7P la 1ch m Merv 3gt wyr pure endowment k V Ea OH wee V E y a a o f a o w V l l Fully continuous reserves whole life 9 Consider now the case of a continuous whole life insurance with an annual premium rate of PAm o The insurer39s prospective loss at time t or at age x l t tL UT 7 1521me Um 1 7 15 6 o The prospective net premium or benefit reserve at time t is tTAm E L AEH 7 PQTE 711 o The variance of tL is 13ng 2 VartL 1 gt FAN 7 Amt2 l l Other special formulas 0 Some continuous analogues of the discrete case EWAE 17 a VU PULHE 7 Pl a t 1 Am 5 o my a t l l Retrospective formula for fully continuous 9 Consider the reserves at time t of the fully continuous WLI tVAm Amt PAm zt 7 7 P Azt PAm zt 15pm iw WtEw 7 PM gist 715px g 7 z 7 m 7 z t tEm tEm m o The first term is the actuarial accumulated value of the premiums paid during the first 25 years The second term is the accumulated cost of insurance I l l l Other types of insurances o For other types of insurances similar development of prospective and retrospective formulas can be made a For these plans and other special types the fundamental principles always hold Keep in mind that when developing the reserves the prospective formula of APVFB APVFP and equivalently the retrospective formula of AAVPP AAVPB always hold a AAVPP refers to the actuarial accumulated value of past premiums while AAVPB refers to the actuarial accumulated value of past cost of insurance l l A general form of life insurance 0 Consider a general fully discrete insurance issued to for which a death benefit is payable at the end of the policy year of death in premiums are payable each year at the beginning of the year a death benefit in the jth policy year is bpj 12M a benefit net premium payment in the jth policy year is F1 0 Net premiums satisfy the equation APVFPO APVFBO 0 0 k k 1 2w km wa m k0 k0 l I Prospective loss and reserves 0 The insurer39s prospective loss at time h from issue is given by Kan hL bK lqutl h 7 Zyrjw h for Km h h 1 jh o The net premium reserve at h is hV EihLiKz 2 hi 00 0 7 H1 739 bhalv jiqzh Z My jpmh 3970 j0 APVFBh iAPVFPh l l Recursive calculations 0 The reserve in the next period h l 1 can be shown to be V 7 h V M 1 i 7 bh1qzh h 1 7 39 l pmh o Intuitively we have a accumulate previous reserves plus premium with interest a deduct death benefits to be paid at the end of the year and a divide the reserves by the proportion of survivors l l Net amount at risk 0 The difference bh17 h V is called the net amount at risk 0 The recursive formula can alternatively be written as h V l 7W 1 l h1 V t 5M1 h V qzh where the term bh 7h V qmh is called the expected net amount at risk l l Illustrative examples 0 Some illustrative examples will be done in lecture 0 Time permitting we39ll try to do 9 Examples 812 813 814 and 815 M ultiple Decrement Models Lecture Weeks 4 5 l l Chapter summary 0 Multiple decrement tables a several causes of decrement in probabilities of decrement a forces of decrement o The Associated Single Decrement Tables 0 Statistical treatment of multiple decrement theory 0 Uniform distribution of decrements o in the multiple decrement context a in the associated single decrement context 0 Chapters 10 Cunningham et al l l Examples of multiple decrement models 9 Multiple decrement models are extensions of standard mortality models whereby there is simultaneous operation of several causes of decrement o A life fails because of one of these decrements 0 Examples include a life insurance contract is terminated because of deathsurvival or withdrawal lapse a an insurance contract provides coverage for disability and death which are considered distinct claims a life insurance contract pays a different benefit for different causes of death eg accidental death benefits are doubled a pension plan provides benefit for death disability employment termination and retirement l l Introducing notation age number of employees deaths Withdrawals overseas transfers a 1 2 253 d5 20 100 000 452 5 517 2 569 21 91 462 433 4 780 2 431 22 83 818 414 4136 2 302 23 76 966 402 4 076 2 264 0 Conventional notation a 17 represents the surviving population present at exact age I a d represents the number of lives exiting from the population between ages I and 1 1 due to decrement j a It is also conventional to denote the total number of exits by all modes between ages I and 1 1 by d ie dy Edy j1 where m is the total number of possible decrements and therefore dv 7 7 Z 7 x 1 14 II l l Probabilities of decrement o The probability that a life will leave the group within one year as a result of decrement j MMW o The probability that will leave the group regardless of decrement wewwezuwezm j1 j1 a The probability that will remain in the group for at least one year mkmdnwewnwmn l I continued 0 We also have the probability of remaining in the group after 71 years pp alt69gt p ml manil and the complement ah 17 my 0 The number of failures due to decrementj over the interval m m n is n7 1 my 2 my t0 0 These relationships should be straightforward to follow my ah ml My w my l I Illustration of the MDT Expand MDT into z 51 d9 d9 d9 4150 4152 4155 415 175 100000 452 5517 2569 000452 005517 002569 008538 091462 91462 433 4 780 2431 000473 005226 002658 008358 091642 83 818 414 4 136 2 302 000494 004935 002746 008175 091825 23 76 966 402 4 076 2 264 000522 005296 002942 008760 091240 m 0 mm NH l I Illustrative problems Using the previously given multiple decrement table compute and interpret the following 0 zd 9 329 9 g 2 9 1l2q 0 l I Total force of decrement o The total force of decrement at age z is defined as 1 d d 59gt 7 10g 59gt 1 TA 7 77i W l hhqr ATMm 9 Therefore analogous to the single decrement table we have t m ms 0 and 1 q 0 ll19st or more generally t My 0 gol ngst l I Force of a one decrement o The force of decrement due to decrementj is defined as 1 d 739 iii 739 0 Notice that the denominator is NOT 17 but is rather Kg 0 As a consequence we see that My Z M j1 o The total force of decrement is indeed the sum of all the other partial forces of decrement 0 We can also show that 1 q A spyIXst l l Illustrative exercise Suppose that in a triple decrement model you are given constant forces of decrement for a person now age m as follows ulmllt b for t 2 07 Milt b for t 2 07 p52 2b fort 2 0 You are also given that the probability will exit the group within 3 years due to decrement 1 is 000884 Compute the length of time a person now age z is expected to remain in the triple decrement table Answer to be discussed in lecture 83 13 years l l The associated single decrement tables a 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 a called the associated singledecrement table ASDT 0 Each table represents a group of lives reduced continuously by only one decrement For example a group subject only to death but not to other decrements such as withdrawal 0 The associated probabilities in the ASDT are called absolute rates of decrements For example the absolute rate of decrement due to decrement j over the interval zm l t is tqmlj a One should be able to explain intuitively why the following always hold true I 1 161 2 161 I l l l Statistical treatment 0 It is often instructive to develop multiple decrement theory from a statistics point of view by defining the random variables a x T the future lifetime random variable and a x J the cause of decrement which is discrete with possible values J 17 2 Wm 0 Relationships between actuarial and statisical expressions 16711677 up 39 114 a fgfmmds 14 a 10 quot mews may Mi W 1 a 271 moi mt 405 552 0 MSW Fm 45 J39 Mac leTlt 31 Mmz l l l I continued 0 One can also similarly develop actuarial and statistical expressions in the case where the consider the discrete future lifetime Km together with the discrete random variable Jm indicating the decrement 0 See the development in Cunningham et al pp 342 343 l I Link between the MDT and the ASDT o If given the absolute rates of decrements say qm17qm2qlmm how do we derive the probabilities of decrements q lbq zl 7q m in the MDT And vice versa 0 The fundamental link lug Hg for all j 127 7m 0 Therefore it follows that an will x an x x ml 0 Furthermore we note that lt39gt 0 1fqu 0 spy 1st quot t lt7 739 I 5131 Sp iii52 W doggy152 0 0 919m l l In the multiple decrement context 9 We assume the following UDD assumption my 254 for 0 g t g 1 o This leads us to the following result my 1tqggtq q 0 Proof to be done in class 0 This result allows us to compute the absolute rates of decrements q ljl given the probabilities of decrements in the multiple decrement model In particular when t 1 we have m 117 qgtgtq q l l Illustrative example In a double decrement table where cause dis death and cause w is withdrawal you are given a both deaths and withdrawals are each uniformly distributed over each year of age in the double decrement table 0 a 1000 o 61 048 o if 03m Calculate qul and q Note One way to check your results make sense is to ensure the inequality qmljl 2 q is satisfied gw l l In the associated single decrement context 9 We assume the following UDD assumption tqmlj 25 m for 0 g t g 1 o This implies I I I I lo17 th tpg l bglt 615 0 Using the previous link one can derive t A spill13st t Hsp mlllsp mlWmlilsds 0 1 t m Hu 7 sqglt1gtgtda 0 a39 0 Use this integration to derive the probabilities of decrement given the absolute rates of decrements tqg I l l I The case of two decrements 9 When we have m 2 we can derive t 1615 c151 17845218 0 61 6152 lt75 75M o In the case when t 1 we can check that we do get equations 1029a and 102 of Cunningham et a p 353 and similarly 0 As an exercise derive equation 1030 of Cunningham et a in the triple decrement case l l Illustrative example 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 0 61 003 and CA 006 o a 1000 000 and 621 902682 Calculate if l l Illustrative example 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 o c1301 001 c1502 005 and c1503 010 o Withdrawals occur only at the end of the year 0 Mortality and disability are uniformly distributed over each year of age in the associated single decrement tables Calculate q gl l l Illustrative example 1 9 An insurance policy issued to 50 will pay 407 000 upon death if death is accidental and occurs within 25 years a An additional benefit of 107 000 will be paid regardless of the time or cause of death 0 The force of accidental death at all ages is 001 o The force of death for all other causes is 005 at all ages a You are given 6 10 9 Find the net single premium for this policy 0 To be discussed in lecture l l Illustrative example 2 a An employer provides his employees aged 62 the following oneyear term benefits payable at the end of the year of decrement a 1 if decrement results from cause 1 a 2 if decrement results from cause 2 and a 6 if decrement results from cause 3 0 Only three possible decrements exist o In their associated singledecrement tables all three decrements follow de Moivre39s Law with w 65 0 You are given i 10 0 Find the actuarial present value at age 62 of the benefits l l Asset share calculations Asset shares refer to the projections of the assets expected to accumulate under a single policy or a portfolio of policies To illustrate consider an insurance contract that pays a a benefit of bid at the end of year k for deaths during the year and o a benefit of 135 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 7 of the premium for expenses plus a fixed amount of expense of ek Expenses occur at the beginning of the year l l continued In addition we have 0 Interest rate is an effective annual rate of i o The probabilities of decrements are denoted by q zkil and qgilkil respectively for deaths and withdrawals The probability of staying in force through year k is therefore d P kq 1 qllka qSllkel39 Denote the asset share at the end of year k by kAS with an initial asset share at time 0 of OAS which may or may not be zero For a new policycontract we may assume this is zero l l The recursion formula for asset shares Beginning with k 1 we find OAS an e n e 6112 WW b wgtqlwgt1As my and we get OAS an 7 r1 7 elilt1 gte 12 a 125 AS 1 739 py This is easy to generalize as follows HAS an e m 7 skim 2 e brag e bimq ils T pmk71 kAS Do not memorize use your intuition to develop the formulas l I Illustrative example For a portfolio of fully discrete whole life insurances of 1000 on 30 you are given 0 the contract annual premium is 950 0 renewal expenses payable at the start of the year are 3 of premium plus a fixed amount of 250 0 ZOAS 145 is the asset share at the end of year 20 O 21CV 100 is the cash value payable upon withdrawal at the end of year 21 0 interest rate is i 75 and the applicable decrement table is given below 1 w m qm qz 50 00062 00415 51 00065 00400 Calculate the asset share at the end of year 21

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