Further Topics in Actuarial Mathematics: Premium Reserves. Matthew Mikola


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1 Further Topics in Actuarial Mathematics: Premium Reserves Matthew Mikola April 26, 2007
2 Contents 1 Introduction Expected Loss An Overview of the Project Net Premium Reserves Introduction The Net Premium Reserve for Various Life Insurance Policies Whole Life Insurance Policy Term Insurance Endowment Examples Other Forms of the Net Premium Reserve Equation Recursive Formulae Savings and Risk Premium Retrospective Reserves The Expected Value and Variance of Loss Variables Prospective Loss The Insurer s Net Cash Loss Allocation of the Risk Expense Loadings ExpenseLoaded Premium Reserves Uses of Reserves Introduction Alterations and Conversions Withprofits policies The Net Premium Method The Bonus Reserve Method Asset Share Method Applications of Reserves Beyond Life Insurance Insurance Actuaries Uses of Reserves The Balance Sheet The Income Statement Loss Reserves Adequacy Testing Reinsurance Conclusion 36 i
3 A Notation i A.1 Lifetime Models i A.2 Life Insurance ii A.3 Endowments iii A.4 Life Annuities iii B Useful Equations iv B.1 Proofs of Equations used in Section iv B.2 Proofs of Equations used in Section v B.3 Proofs of Equations used in Section v C Life Tables vi ii
4 Chapter 1 Introduction Actuarial Mathematics is a very interesting area of mathematics for me, as it has such a large impact in the real world. The concepts that have been developed are used daily by insurance companies in the calculations of their insurance policies and also when a monetary value is required for any choice that involves risk. Insurance mathematics relies heavily on probability theory, as the lifetime of a person or the occurrence of an event is unpredictable. Insurers calculate the expected probability that these events occur, so they can price their insurance policies correspondingly and also so no large unexpected losses occur. In this project, I am going to explore the topic of net premium reserves with regards to life insurance policies. This project assumes that the reader has understood the concepts covered in the 2nd year mathematics course Actuarial Mathematics II, Appendix A contains explanations and formulae for most of the theory covered for the reader s benefit. I am only going to consider discrete life insurance policies and the corresponding reserve formulae. Variants of the formulae for the continual policies can be used in most cases, this simply involves using the appropriate continuous net single premium formulae and lifetime probability distributions covered in the 2nd year course. Premium reserves, also known as benefit reserves, are of great interest to insurers as it enables them to work out their liability, this is how much money they should put aside, reserve, each year to cover the benefit payment to the insured in case of their death. The insurers take into account the premiums already paid by the insured and the investment return they can receive on these payments and the insured s expected future lifetime. Premium reserves look at the obligations of the insurer and the insured, at a time later than policy issue, as either party may still have financial duties that need settling at that time. In the next section I will look at one of the most useful concepts for calculating premium reserves, the expected loss of the insurer, which was covered in the course Actuarial Mathematics. This has particular significance for the formulation of net premium reserves as the expected loss to the insurer is used in its definition. 1
5 1.1 Expected Loss In this section I have used my notes from the 2nd year mathematics course Actuarial Mathematics II which, were based on concepts covered in [Gerber]. In this section I am going to recap the relevant actuarial concepts covered in the 2nd year mathematics course Actuarial Mathematics II. One of the important ideas developed near the end of the course was calculating the expected loss to the insurer. For an insurance policy we define the prospective loss to the insurer, L, to be the difference between the present value of future premium payments (paid by the insured) and the present value of future benefit payments (paid by the insurer). The prospective loss is a really important concept for the derivation of the formulae for the net premium reserve, as we shall see later. Premiums can be paid in a variety of different ways to suit the preferences of the insured, three of these ways are: 1. One single premium. 2. Periodic premiums of a constant amount, also known as level premiums. 3. Periodic premiums of varying amounts. In the case of the periodic premiums the frequency and the duration of the payments needs to be specified as well as the premium amount, commonly the premiums are paid annually and while the insured is still alive. A premium is called a net premium if it satisfies the following equivalence principle: E(L) = 0 (1.1.1) If the policy is financed by a single premium, then the premium used is the net single premium, the formulae for the different policies can be found in Appendix A.2. As the net single premium is used to calculate the benefit payment as well, then the premium payment and the benefit payment are equal in value so they satisfy (1.1.1). If periodic premiums of constant amount are used to finance the policy, then we use (1.1.1) to calculate the net premiums. We cannot use (1.1.1) to calculate the net premiums for periodic premiums of varying amounts though. (1.1.1) is called an equivalence relation, this is because the present value of the benefit payments exactly equals the present value of the premium payments, so the expected loss to the insurer is zero. In practice this is rarely true because the insurance company would not make any money, but we use this assumption in our calculations for simplicity. The prospective loss is very important in the formulation of premium reserves as we shall see in the next chapter. If we now consider the more realistic case where the insurer wants to make sure a profit is made, we can change (1.1.1) so that the difference between the premium payments and the benefit payments is a positive value. If the insurer wants to make a profit of θ then in (1.1.1) we would want the loss to the insurer to be θ: E(L) = θ The insurer could then use this variation, instead of (1.1.1), in their calculation of the net premiums. We shall look later at another way the insurer can factor profit and the costs of settingup and running an insurance policy, by using expense loadings in the premiums. 2
6 1.2 An Overview of the Project In chapter 2, I shall derive the formula for the net premium reserve for various life insurance policies. I will then proceed to look at some of the interesting properties of reserves and why they are useful to insurers. Near the end of the chapter I look at expected value and variance of the loss to the insurer, which is very interesting as it gives an indication of the risk associated with the policy. I also consider expense loadings, which are extra amounts added to the premium payments, to cover various expenses of the insurer. This part gives an insight into how, in the real world, insurance companies have to factor their overheads and profit margins into their policies. In chapter 3, I will look at some of the uses of reserves for life insurance policies. I examine how the reserves are used when the insured may want to make alterations to his existing policy and I also look at how reserves can be used to determine the level of extra bonuses, added to the benefit payment, in withprofit policies. In chapter 4, I look at why insurance is so important to society as a whole and how specialists known as actuaries work in insurance companies to ensure financial stability. I also look at some of the uses of reserves outside of life insurance. In writing this chapter I looked at papers written by researchers, who are looking at ways to apply reserves and various actuarial methods to real life situations and problems. They look at ways of using different types of reserves to try to estimate the amount of money an insurance company sets aside to cover their overall losses and expenses. Other forms of reserves are used by the Financial Services Authority (F.S.A.), who regulate all providers of financial services to make sure they are making reasonable levels of profit and have enough reserves to cover their policies and actions. Finally I look at when reserves are used by insurers to see if large risky policies are worth reinsuring, this is when an insurer looks to another insurance company to cover all or part of the risk of one of their policies or group of policies. In this project I have used information from the sources listed in the bibliography. At the beginning of each section I have noted where I have directly used equations or text from a source and what is my own work. All examples demonstrated in this project are my own work and have been calculated by myself. 3
7 Chapter 2 Net Premium Reserves 2.1 Introduction As we found in section 1.1, the expected loss to the insurer at the time of policy issue is zero (1.1.1), this is because the expected value of future premium payments equals the expected value of future benefit payments. In this chapter we look at the expected loss at a later time than policy issue. We therefore define a random variable t L as the difference at time t between the present value of future premium payments and the present value of future benefit payments. According to [Gerber, p.59] the net premium reserve at time t is defined as the conditional expectation of t L, given that T > t. This is the conditional expectation of the difference between the present value of future benefit payments and the present value of future premium payments at time t, given that the insured survives to t. It is denoted by the symbol t V. The standard convention for reserves is to calculate them just before the payment of the premium due at time t. The net premium reserve can be seen as the present value of the liability for the insurer at time t, i.e. the expected extra amount of money the insurer needs to set aside each year, to cover the benefit payment taking into account the premium payments already paid by the insured. More specifically, they are reserves that are calculated without the allowance for expenses and where the reserve basis and the premium basis agree. The reserve basis is the mortality, interest and expense assumptions used to calculate t V and the premium basis is a set of similar assumptions applied to the premium payments. The net premium reserve is often positive to give the insured some incentive to stay in the scheme, this also means that the insured will not have any additional payments to pay to the insurer, if they leave the scheme. Typically the premiums that the insured pays each year in the early years of the policy, are more than the insurer needs in order to cover the benefit payment, but this margin decreases as time progresses. At some point later in time, the future premiums will not be sufficient to cover the remaining benefit payment, therefore the excess collected in the early years, which is invested to gain interest, will be used to cover this deficit; this is another reason why net premium reserves are often positive. Negative reserves can occur on policies where the benefit payment decreases each year or the premium payments increase each year. Some of the uses of reserves are for when the insured wants to leave the insurance policy before a claim is made and so would want some compensation amount, this is known as the surrender value of the policy. Also the insured, if they were part of their employer s pension scheme, may change jobs and so may want to transfer their existing pension to their new employer s pension scheme. Most modern insurance policies are quite flexible and allow the insured to make changes to their policy (within reasonable boundaries, so the insurer still makes a profit), knowing the reserve makes this possible. Reserves can also be used to determine the bonus rates of withprofits policies and for the distribution of profits to shareholders. Withprofits contracts are life insurance 4
8 policies where the insured has the right to share in the profits of the company, usually in the form of bonuses added to the benefit payment. 2.2 The Net Premium Reserve for Various Life Insurance Policies In this section equations (2.2.3) and (2.2.4) are based on equations found in [Bowers, p.206], but the derivation of (2.2.4) has been extended by me to provide clearer explanation. Using the definition of the net premium reserve given in the Introduction I am now going to find the net premium reserve for various life insurance policies Whole Life Insurance Policy The present value of a whole life insurance policy for a life aged x, which covers the insured until death, with a benefit payment of 1 unit payable at the end of year of death, is given by: Z = v K+1 K 0 (2.2.1) Where K(x) is the curtate future lifetime of the insured and v is the discount factor, with v = 1 1+i where i is the Annual Equivalent Rate of interest (see the end of A.1 for further details). A whole life insurance is financed by level annual premiums P x, payable at the beginning of each year whilst the insured is still alive. An equivalent way of describing this mathematically is setting up a whole life annuity with benefit payment P x, this is because the payments are made at the start of every year whilst the insured is still alive. The present value of this is given by: Y = P x ä K+1 K 0 (2.2.2) The prospective loss to the insurer, as described in section 1.1, is the difference between the present value of future premium payments and future present value of the benefit payments. For the whole life insurance policy we just appropriately combine the two present value formulae (2.2.1) and (2.2.2) to get: L = v K+1 P x ä K+1 K 0 If we now consider the prospective loss k years later than policy issue, this is given by the symbol k L and is defined in exactly the same way as before but we have just adjusted which point in time we are looking at. If the insured s curtate future lifetime is K(x) years, then if we look k years later than policy issue, we only need to consider the remaining K k years of the insured s life, so the index K in (2.2.1) and (2.2.2) changes to K k. For example if the insured was 50 when the policy was issued and we expect them to die at 80 (in reality this is a random event as we don t know when someone is going to die, but for simplicity assume we can for this case), then K = 30 and if we now look at the person k = 10 years later when they are 60, we now only need to consider the remaining K k = = 20 years. Therefore the present value of the prospective loss for a whole life insurance is: kl = v K+1 k P x ä K+1 k K(x) k (2.2.3) For the formulation of the net premium reserve we need the expected value of (2.2.1) and (2.2.2) and these are shown by : E(Z) = A x = v k+1 kp x q x+k k=0 5
9 and E(Y ) = ä x = ä k+1 kp x q x+k. k=0 as shown by (A.2.3) and (A.4.2) respectively. The net premium reserve; the conditional expectation of the prospective loss, for a whole life insurance policy, is given by: kv = E[ k L K(x) > k] = E[v K+1 k K(x) > k] P x E[ä K+1 k K(x) > k] = v j+1 Pr[v K+1 k K(x) > k] P x ä j+1 Pr[ä K+1 k K(x) > k] = j=0 v j+1 jp x+k q x+j+k P x j=0 j=0 j=0 ä j+1 jp x+k q x+j+k = A x+k P x ä x+k (2.2.4) Term Insurance Similarly the net premium reserve at the end of year k of a term insurance is denoted by k V 1 x:n. A term life insurance pays 1 unit benefit payment at the end of year of death, if the insured dies within n years and is financed by premiums payable at the beginning of each year the insured is alive, up to the start of year n. Using the expected values for the term life insurance and the n year temporary life annuity due, which can be found in (A.2.4) and (A.4.4), its net premium reserve is given by: Endowment kv 1 x:n = A 1 x+k:n k P 1 x:n ä x+k:n k k = 0, 1,..., n 1 (2.2.5) The net premium reserve for an n year endowment is denoted by k V x:n. An n year endowment pays 1 unit benefit payment at the end of the year of death if the insured dies within n years, or if the insured survives n years the benefit payment is at the end of the n th year. Using the expected values for the n year endowment and the n year temporary life annuity due can be found in (A.3.3) and (A.4.4), it is given by: Examples kv x:n = A x+k:n k P x:n ä x+k:n k k = 0, 1,..., n 1 (2.2.6) I will now look at a numerical illustration of the reserves for a term insurance and a n year endowment. I will assume a sum insured of 100 units, the initial age of the insured as x = 50 years old and a duration of n = 10 years. We will use the life tables found in appendix C and i = 5%. As a first step we work out the net annual premiums for each policy. As the premiums are constant we can use the equivalence relation. This is calculated in a very similar way to the method used in the whole life insurance policy example above. This process is shown below: 6
10 E(L) = 0 P 1 = 100 A 1 x:n P 1 x:n ä x:n ä 50:10 50:10 = 100 A 1 50:10 = 100(M 50 M 60 ) N 50 N 60 = Similarly: P 50:10 = Using (2.2.5), (A.2.5) and (A.4.5): kv 1 50:10 = 100(M 50+k M 60 ) P 1 50:10 50+k N 60 ) D 50+k k = 0, 1,..., 9 (2.2.7) and using (2.2.6), (A.3.4) and (A.4.5): kv 50:10 = 100(M 50+k M 60 + D 60 ) P 50:10 (N 50+k N 60 ) D 50+k k = 0, 1,..., 9 (2.2.8) Using these equations we can produce the following table which will enable us to directly compare the net premium reserves for the term insurance with the endowment. Table 2.1: The net premium reserves for a term insurance and an endowment k 100(A 1 50+k:10 k 50+k:10 k kv 1 100(A 50:10 50+k:10 k ) kv 50: Things to notice from these results are: The net premium reserve for the term insurance and the endowment at k = 0, i.e policy issue, is 0 as expected. This follows directly from the net premium equivalence principle that E(L) = 0 at policy issue. The net single premium of the term insurance decreases (1st column), this is because the probability of the insured surviving to the end of the term increases (see Table 2.2), and this out weighs the decreasing time for interest to accumulate on the premium payment. 7
11 The net single premium of the n year temporary life annuity due decreases (2nd column), this is because fewer benefit payments have to be paid and therefore there is less time for interest to accumulate, this out weighs the probability of the insured surviving to the end of the term, which increases. The net single premium of the endowment increases (4th column), this is because the decreasing time for interest to accumulate on the premium payment out weighs the decreasing probability of the insured dying before the end of the term. The net premium reserve of the term insurance is very small and does not change by much (3rd column). It grows to start with since the premiums slightly exceed that of a corresponding 1 year term insurance. Near the end of the period the net premium reserve drops again as the insurer does not pay the benefit payment if the insured survives and the probability of that occurring increases. The net premium reserve of the endowment increases over time (5th column), this reflects the increasing net single premium of the endowment and this is due to the decreasing time for the interest to accumulate. Note that the net premium reserve of at the end of the 9th year, plus the last premium payment of , plus interest of 5% on both, is sufficient to cover the benefit payment of 100 at the end of the 10 years = = Table 2.2: The probability of a life aged 50 surviving to 60 given they have survived k years. k l x 10 k P 50+k Other Forms of the Net Premium Reserve Equation In this section equations (2.3.1) and (2.3.2) can be found in [Gerber, p.63,64], but the derivations of these have been extended by me for clearer explanation. So far I have only used the prospective method to write formulae for the net premium reserve, stating that the net premium reserve is defined as the difference between the present value of future benefit payments and of future premium payments. We can use this method to develop other formulae for the net premium reserve, in terms of the premiums paid by the insured. These other formulae are of interest because they give us different interpretations of the net premium reserve and therefore give us greater insight into their use to actuaries. This is useful because the actuary can now consider the reserve in terms of the amount paid by the insured each 8
12 year and therefore if they wanted to reserve only a certain proportion of the premium, then they can adjust other factors accordingly to take this into account. Also by having an alternative way to look at the concept of the net premium reserve I hope it will aid the reader in their understanding. If we consider the net premium reserve of a whole life insurance policy as defined in section 2.2.1: kv = A x+k P x ä x+k Using results derived in [Gerber, p.64], the premium difference formula for k V can be found by using A x+k = 1 d ä x+k (as shown in (B.1.3)) and 1 d ä x+k = P x+k ä x+k (from (B.1.5)), giving: kv = A x+k P x ä x+k = (1 d ä x+k ) P x ä x+k = 1 (P x + d)ä x+k = (P x+k P x )ä x+k (2.3.1) As we can see from (2.3.1), if a whole life insurance, of 1 unit, was bought k years later than the original, now costing P x+k ä x+k, the net premium reserve is the expected present value of the deficit of the premiums. Another way to look at this is to think of P x+k as the amount that should be charged, for a policy with the same benefits if issued at time k to a person aged x + k, in order that the premium equals the benefit payment after k years. Then the quantity P x+k P x is the difference between what should be charged and what is actually charged. This means that the reserve at time k is the yearly deficit between these premium amounts. Also using results derived in [Gerber, p.64], the paid up insurance formula for k V can be found by using ä x+k = (1 A x+k )/d (as shown in (B.1.3)) and d = (P x+k (1 A x+k ))/A x+k (as shown in (B.1.6)), giving: kv = A x+k P x ä x+k = A x+k P x ( 1 A x+k ) d = A x+k P x d (1 A x+k) = A x+k A x+kp x (1 A x+k ) P x+k (1 A x+k ) = A x+k A x+kp x P x+k = (1 P x P x+k )A x+k (2.3.2) (2.3.2) shows the net premium reserve as the present value of a portion of the remaining future benefit payments, the portion which is not funded by future premium payments. P x+k is the premium required if the future benefit payments were to be funded from only the future premium payments, but P x is the benefit payment actually paid. Therefore P x /P x+k is the portion of future benefit payments actually funded by future premium payments. Alternatively, as mentioned before in the early years of the policy the premium payments exceed what should be paid for the benefit payment, but this changes later in the policy, so the amount (1 Px P x+k ) must be the that portion of future benefits that has already been provided for by the excess past premiums. 9
13 Example If we now calculate the net premium reserve for our example in section using the premium difference formula and the paid up insurance formula. We can calculate the values of P x+k from the equivalence relation: P 1 = A 1 x+k:n k (2.3.3) x+k:n k ä x+k:n k for the term insurance and P x+k:n k = A x+k:n k ä x+k:n k for the endowment. The results are shown below (see Table 2.1 for the unchanged values of 100(A 1 ), 50+k:10 k ä 50+k:10 k and 100(A 50+k:10 k )): Table 2.3: The net premium reserve for the term life insurance and endowment calculated via the premium difference and paid up formulae k P 1 50+k:10 k kv 1 P 50:10 50+k:10 k kv 50: Unsurprisingly we see that the values for the net premium reserve are the same as they were when we performed the calculation the first time. 2.4 Recursive Formulae In this section I have used the theory covered in [Gerber, p.61], but the derivations of all the equations have been extended by me for greater clarity. In this section I am going to develop a relation between k V and k+h V which is the net premium reserve h years after k. If we consider the net premium reserve for a whole life insurance, with varying benefit payment c j, being the amount insured in the jth year after policy issued, financed by varying premiums Π 0, Π 1,..., Π k, being the premium due at time k. Then the net premium reserve at the end of year k is: If we rewrite ä j+1 kv = c k+j+1 v j+1 jp x+k q x+k+j Π k+j ä j+1 jp x+k q x+k+j (2.4.1) j=0 as: j=0 ä j+1 = 1 + v + + v j = 10 v j I {J=j} k=0
14 Where I {J=j} is an indicator function: Then: So: j=0 Now using (2.4.2) in (2.4.1) we get: I {J=j} = { 1 if J = j 0 elsewhere E(I {J=j} ) = P r(j = j) = j p x ä j+1 jp x+k q x+k+j = ä x = E(ä j+1 ) = v j jp x (2.4.2) kv = c k+j+1 v j+1 jp x+k q x+k+j Π k+j v j jp x+k (2.4.3) j=0 If we then use (A.1.5) with s = h, t = j h and x = x + k and rearrange we have: j=0 j=0 jp x+k = h p x+k j h p x+k+h (2.4.4) Then substitute (2.4.4) in all except the first h terms of (2.4.3) then we have: kv = h 1 h 1 c k+j+1 v j+1 jp x+k q x+k+j Π k+j v j jp x+k j=0 + j=0 c k+j+1 v j+1 hp x+k j h p x+k+h q x+k+j j=h Π k+j v j hp x+k j h p x+k+h j=h (2.4.5) If we then use j = j h as a summation index in the second and third lines of (2.4.5) then this part becomes: c k+j +h+1 v j +h+1 h p x+k j p x+k+h q x+k+j +h Π k+j +h v j +h h p x+k j p x+k+h (2.4.6) j =0 There is a common term of v h hp x+k in (2.4.6), if this is taken out as a factor we are left with: j =0 j =0 c k+j +h+1v j +1 j p x+k+h q x+k+j +h Π k+j +h v j j p x+k+h Which is the net single premium of k+h V. If we then combine (2.4.6) and the first line of (2.4.5) and rearrange, we get: h 1 kv + Π k+j v j jp x+k = j=0 j =0 h 1 c k+j+1 v j+1 jp x+k q x+k+j + h p x+k v h k+hv (2.4.7) j=0 If we look at (2.4.7) we can see that if the insured survived to the end of year k, then the net premium reserve plus the expected present value of the premium payments for the next h years is 11
15 just sufficient to cover a life insurance for those years, plus a pure endowment of k+h V, at the end of year k + h. A recursive equation for the net premium reserve can be achieved by letting h = 1: kv + Π k = v[c k+1 q x+k + k+1 V p x+k ] (2.4.8) We can therefore calculate the net premium reserve recursively in 2 ways: 1. We can calculate 1 V, 2 V,... successively by starting with the initial value 0 V = 0, by rearranging (2.4.8) to make k+1 V the subject. 2. Or if the duration of the insurance is of finite duration n, then we may calculate n 1 V, n 2 V,... from a known value of n V, by letting k + 1 = n The recursive formula is really useful for actuaries as it enables the insured to have greater flexibility with their insurance policies. By being able to work out the net premium reserve each year from the previous value can allow the insured to change the policy and use the current value of the net premium reserve to begin a new policy. An example may be when the insured wants to convert the insurance policy to a policy which has no further premium payments, know as a paidup insurance policy. The actuary can then use the value of the last net premium reserve to calculate the net single premium for the paid up insurance. There is a type of insurance known as universal life or flexible life, which offers the maximum degree of flexibility to the insured. The insured can change any two of the following parameters: Π k, the next premium to be paid, c k+1, the benefit paid in case of death in the next year, k+1 V, the net premium reserve for next year, this can be seen as the target value of the insured s saving s in the next year (see the savings premium in Section 2.5 for further details on the consideration of the net premium reserve as a saving). We calculate the value of k+1 V from the recursive formula (2.4.8), so we can see how useful it is for actuaries when they are helping the insured modify their insurance policy. There are usually some restrictions applied as to how much these values can be changed, so as not to leave the insurer out of pocket, but yet still allow flexibility. 2.5 Savings and Risk Premium All the equations from this section are from [Gerber, p.61,62] If we examine (2.4.8) more closely we can see that the net premium reserve at time k plus the premium payment equals the expected present value of the funds needed by the insurer at the end of that year, c k+1 if the insured dies, or k+1 V otherwise. As p x+k = 1 q x+k then (2.4.8) can be rewritten as: kv + Π k = v[ k+1 V + (c k+1 k+1 V ) q x+k ] (2.5.1) (2.5.1) can be interpreted as; k+1 V is needed whatever happens to the insured and an additional amount of c k+1 k+1 V if the insured dies. According to [Gerber, p.61], c k+1 k+1 V is known as the net amount at risk. (2.5.1) can be rearranged to decompose the premium into two components, Π k = Π s k + Πr k, where Π s k = k+1 V v k V (2.5.2) 12
16 and Π r k = (c k+1 k+1 V )v q x+k (2.5.3) Π s k is known as the savings premium and Πr k is known as the risk premium. We can see in (2.5.2) that the savings premium is the part of the premium used to increase the net premium reserve and cover the change in liability for the insurer. It is known as the savings premium as it is the extra amount the insurer has to save each year to be able to cover any future benefit payment. This part can also be seen as a savings account, with the premium accumulating interest only, which can be used by the insurer in later years to cover the deficit between the premium payments and the benefit payment. Some financial advisers suggest that people should not buy whole life or endowment policies, but instead should purchase term life insurance. That way the premiums are less and the insured can choose where to invest this extra money and so has greater control of a portion of the savings premium and can make money from this. This led to the insurance companies introducing the universal life policy, mentioned at the end of section 2.4, in an attempt to offer the policyholder more flexibility with regards to the levels of the risk and savings premium. In (2.5.3) we see that the risk premium is the part used to fund a oneyear term insurance, of amount c k+1 k+1 V, to cover the net amount at risk to the insurer, which is the benefit payment. This part of the policy has no reserves, since the risk premium is just sufficient to purchase appropriate coverage for the net amount at risk for that 1 year. If we multiply (2.5.2) by (1 + i) j k and just consider the first j years, then sum over k = 0, 1,..., j 1, we get: j 1 jv = (1 + i) j k Π s k k=0 We can see from this equation that the net premium reserve is the total value of the savings premiums since the beginning of the policy. Also we can rewrite (2.4.8) by using d = i/(1 + i), where d is the discount rate covered in the Actuarial Mathematics II course, which gives us v = 1 d, resulting in: Π k + d k+1 V = ( k+1 V k V ) + Π r k We can see that the premium plus the interest received on the net premium reserve, is used to change (increase or decrease) the net premium reserve and to cover the risk premium. Example The decomposition of the net annual premium into the savings premiums and the risk premiums for the same example of section is tabulated below: We can see from these results that: The savings premium for the term insurance decreases steadily and becomes negative from year 5 onwards, this is because the net premium reserve decreases from year 6 and once the net single premium in year 6 has been discounted it is less than the value for year 5. The risk premium for the term insurance increases steadily this is because the probability of the insured dying each year increases (See Table 2.5), which out weighs the increasing net premium reserve during the first five years. After that the net premium reserve decreases and the probability of the insured dying each year continues to increase, so an increasing risk premium is expected. The savings premium for the endowment increases slowly, this is because the net premium reserve increases at roughly the same rate. 13
17 Table 2.4: Decomposition into savings premium and risk premium Term insurance Endowment k Π s k Π r k Π s k Π r k The risk premium for the endowment decreases to zero, this is because the net premium reserve increases as does the probability of the insured dying each year. The risk premium in year 9 is zero because the net premium reserve in year 10 is 100 which is equal to the benefit payment in year 10. Table 2.5: The probability of the insured dying each year k l x q x+k Retrospective Reserves In this section the definition of retrospective reserves is from [Scott, p.102]. Equations (2.6.1) and (2.6.2) can be found in [Scott, p.103], but their derivations have been extended by me for further clarity. As mentioned in section 2.1 one of the uses of reserves is calculating the surrender value of the policy, in this section I will find formulae to achieve this. If the insured decides to surrender the contract before a claim is made, he will probably expect in the early of the years of the policy, to receive a surrender value related to the accumulation of the premiums he has paid, less expenses and the cost of the life insurance cover, since he hasn t received any benefit payment. Such a surrender value is related to the retrospective reserve of the contract, which is defined to be the total of the premiums paid, less expenses and the cost of the benefit payment, of a hypothetical large group of identical policies whose mortality exactly follows a set life table and then dividing the hypothetical funds amongst the survivors. 14
18 If we now consider the whole life insurance policy defined in section 2.2.1, the retrospective reserve is found by collecting together the funds of say l x identical policies until time k and then sharing the accumulated money out among the survivors. If we use standard notation used for life tables, there are d x deaths in the first policy year, d x+1 deaths in the second policy year and l x+k survivors at time k. The total accumulated premiums paid by the survivors is given by: l x P x (1 + i) k + l x+1 P x (1 + i) k l x+k 1 P x (1 + i) = l x P x (1 + i) k [ i (1+i) k 1 ] = l x P x (1 + i) k [1 + v + + v k 1 ] = l x P x (1 + i) k ä x:k The first line can be simply explained: all the l x people pay premium P x at policy issue (k = 0), which gains k years of interest, then at the start of the next year the surviving l x+1 people again pay premium P x, which can gain k 1 years of interest, until at the start of the k th year the surviving L x+k 1 people pay premium P x which only has that one year to accumulate interest. The total of the benefit payments, of 1 unit, that need to be paid out to the people that died each year is given by (we use the fact that d x+k = l x k p x q x+k which is proved in section B.2): d x (1 + i) k 1 + d x+1 (1 + i) k d x+k 1 = l x 0 p x q x+0 v(1 + i) k + l x 1 p x q x+1 v 2 (1 + i) k + + l x k 1 p x q x+k 1 v t (1 + i) k = l x (1 + i) k [v 0 p x q x+0 + v 2 1p x q x v k k 1p x q x+k 1 ] = l x (1 + i) k [ k 1 j=0 vj+1 jp x q x+j ] = l x (1 + i) k A 1 x:k The first line can be simply explained: at the end of the first year the d x people that died that year need to get the benefit payment of 1 unit, but they receive it at the end of year k, so it has k 1 years to gain interest. This carries on till the end of year k when the d x+k 1 people that died receive the benefit payment of 1 unit, but with no years of interest as this is the end of the final year. We now need to work out the difference between these quantities to work out the loss to the insurer, which can simply be written as: l x (1 + i) k [A 1 x:k P x ä x:k ] (2.6.1) As we said before we now share this total accumulated amount amongst the l x+k survivors at time k by dividing (2.6.1) by l x+k to find the retrospective reserve at time k, which is given by the v symbol k V R (we also use the fact that x l x v x+k l x+k = Dx D x+k ). kv x R = = = = l x (1 + i) k [A 1 l P x:k x ä x:k ] x+k l x 1 l x+k v k [A1 P x:k x ä x:k ] l x v x l x+k v x+k [A1 P x:k x ä x:k ] D x [A 1 D P x:k x ä x:k ] x+k (2.6.2) 15
19 Example I shall now find the retrospective reserves for the term insurance and the endowment policies we considered in section The equations for the respective reserves for the term insurance and the endowment policies are respectively given by: kv 1 x:n R = V x:n R = D x D x+k [A 1 x:k P 1 x:n ä x:k ] D x D x+k [A 1 x:k P x:n ä x:k ] The retrospective reserves for each year are shown below, included is the prospective reserves calculated earlier for comparison. We can see from these results that: Table 2.6: Retrospective Reserves k kvx:n 1 R kvx:n 1 V x:n R V x:n The retrospective reserve for the term insurance (1st column) matches quite closely to the prospective reserve (2nd column) for the first 3 years, which is what the retrospective reserve set out to accomplish. After that the retrospective reserve increases faster than the prospective reserve and the difference between the two widens. This shows that the retrospective reserve is not beneficial to the insurer later into the contract and so surrendering the contract would probably not be possible. A similar trend can be seen between the retrospective reserve (3rd column) and prospective reserve (4th column) for the endowment policy. 16
20 2.7 The Expected Value and Variance of Loss Variables In this section equation (2.7.1) is from [Bowers, p.230]. Equations (2.7.2), (2.7.3), (2.7.4), (2.7.5) are from [Bowers, p ], but the derivations of (2.7.3), (2.7.4) have been extended by me for greater clarity. Table 2.7 is from [Bowers, p.233]. Equations (2.7.6), (2.7.8), (2.7.9) and (2.7.10) are from [Bowers, p ], but (2.7.8), (2.7.9) and (2.7.10) have been extended by me for greater clarity. Table 2.8 is from [Bowers, p.242]. In this section I am going to find formulae for the expected value and variance of different loss variables Prospective Loss The first loss variable I am going to look at is the prospective loss, touched on in section 1.1. I shall consider a general discrete whole life insurance where: 1. The benefit payment is payable at the end of year of death. 2. The benefit payment in the jth year is c j, j = 1, 2, The premium payments are paid annually at the beginning of the year. 4. The premium payment in the jth year is Π j 1, j = 1, 2,.... Similar to before, the prospective loss in year k, k L, is the present value of future benefit payments less the present value of future premium payments. Expressed as a function of k this is: { 0 K = 0, 1,..., k 1 kl = c K+1 v K+1 k K j=k Π j v j k (2.7.1) K = k, k + 1,... The expected value of the prospective loss, E( k L K k) is by definition the net premium reserve k V. I will leave Var( k L K k) until later when we have seen the other loss variables The Insurer s Net Cash Loss The next loss variable I am going to look at is the insurer s net cash loss within each insurance year for the general discrete life insurance defined above. Table 2.7 is a time diagram that illustrates the annual cash income and expenses for the insurer. Table 2.7: The Insurer s Annual Cash Income and Expenses Year K 1 K K + 1 K + 2 K + 3 Expenses c K Income Π 0 Π 1 Π 2 Π K 1 Π K etc... Let C k be the present value, at the beginning of year k, of the net cash loss during the year (k, k + 1). If (k, k + 1) is after the year of death (i.e. K(x) < k) then, the insured has already died and so no premium payments are made and the benefit payment has already been paid, so C k = 0. If (k, k + 1) is the year of death (i.e. K(x) = k) then the premium payment, Π k, was paid at the start of the year and the benefit payment, c k+1, is paid by the insurer at the end of the year, so C k = c k+1 v Π k. If (k, k + 1) is before the year of death (i.e. K(x) > k) then, the insured is still alive and has to pay the premium Π k at the start of the year, so C k = Π k. The explicit function of K is shown below: 17
21 0 K = 0, 1,..., k 1 C k = c k+1 v Π k K = k Π k K = k + 1, k + 2,... For the conditional distribution of C k, given that K k, we see that: (2.7.2) C k = v c k+1 I Π k where I is again an indicator function but this time: { 1 with probability qx+k I = 0 with probability p x+k This is the case since C k only takes the value v c k+1 in year k when the insured dies and that occurs with probability q x+k. Therefore the expectation of C k is: E(C k K k) = v c k+1 q x+k Π k (2.7.3) Using that: then the variance is: Var(x) = E(x 2 ) [E(x)] 2 Var(C k K k) = E[(C k ) 2 K k] [E(C k K k)] 2 = E[(v c k+1 I Π k ) 2 ] (v c k+1 q x+k Π k ) 2 = E(v 2 c 2 k+1 I 2 2v Π k c k+1 I + Π 2 k) v 2 c 2 k+1 q 2 x+k + 2v Π k c k+1 q x+k Π 2 k = E(v 2 c 2 k+1 I) 2v Π k c k+1 q x+k + Π 2 k v 2 c 2 k+1 q 2 x+k + 2v Π k c k+1 q x+k Π 2 k = v 2 c 2 k+1 q x+k v 2 c 2 k+1 q 2 x+k = v 2 c 2 k+1 q x+k v 2 c 2 k+1 q x+k (1 p x+k ) = v 2 c 2 k+1 q x+k v 2 c 2 k+1 q x+k + v 2 c 2 k+1 q x+k p x+k = v 2 c 2 k+1 q x+k p x+k (2.7.4) By rearranging the terms in the definition of the prospective loss in (2.7.1), we find an equivalent one that states k L as the sum of the present value of the insurer s future net annual losses at year k. This gives us: kl = v j k C j (2.7.5) j=k We can verify (2.7.5) for K k by substituting in (2.7.2), this is shown explicitly below: kl = v j k C j j=k Obviously we get (2.7.1) as expected. K 1 = v K k (v c K+1 Π k ) v j k Π j = c K+1 v K+1 k j=k K Π j v j k j=k 18
22 2.7.3 Allocation of the Risk The next loss variable I am going to look at explores the allocation of the risk within each year of the insurance policy. It is very similar to the net cash loss looked at previously, but now it also includes the change in liability for the insurer each year. The change in liability, as mentioned before, is basically the change in the net premium reserve over each year, this can be thought of as the extra money the insurer needs to set aside each year to cover the premium payment. I will now look at the risk to the insurer within each insurance year for the general discrete life insurance defined above, Table 2.8 is a time diagram that illustrates the annual cash income, expenses and change in liability for the insurer. Table 2.8: The Insurer s Annual Cash Income, Expenses and Change in Liability Year K K + 1 K + 2 Expenses c K+1 0 Income Π 0 Π 1 Π 2 Π K 0 0 etc... Liability 1V 2V (1 + i) 1 V KV (1 + i) K 1 V K V 0 Let Λ k be the present value, at the beginning of year k, of the net cash loss and change in liability during the year (k, k + 1). If (k, k + 1) is after the year of death (i.e. K(x) < k) then, the insured has already died and the benefit payment has already been paid, so the insurer has no liability and as before C k = 0 so Λ k = 0. If (k, k + 1) is the year of death (i.e. K(x) = k) then the insurer has the reserve from the beginning of the year, k V, as the premium payment was paid by the insured, but there is no reserve needed at the end of the year for next year as the benefit payment has already been made, so Liability = k V. C k = c k+1 v Π k for the reasons described previously. The change in liability has to be discounted back to the start of the year so Λ k = c k+1 v Π k v k V. If (k, k + 1) is before the year of death (i.e. K(x) > k) then, the insured is still alive and the insurer has the reserve k V at the beginning of the year, as the premium payment was paid by the insured. It is also paid at the beginning of next year as the insured will survive to at least that point so the insurer now has reserve k+1 V. The insurer s change in liability for the year (k, k + 1), with discounting, is therefore Liability= v k+1 V k V. C k = Π k for the reasons described previously. This gives us Λ k = v k+1 V k V Π k. The explicit function of K is shown below: 0 K = 0, 1,..., k 1 Λ k = (c k+1 v Π k ) + ( k V ) K = k (2.7.6) ( Π k ) + (v k+1 V k V ) K = k + 1, k + 2,... We can write (2.7.6) in terms of the risk and savings premiums from (2.5.3) and (2.5.2): 0 K = 0, 1,..., k 1 Λ k = Π r k + (c k+1 k+1 V )v K = k Π r k K = k + 1, k + 2,... The conditional distribution of Λ k, given that K k, we see that: Λ k = v c k+1 I + v k+1 V J Π k k V (2.7.7) where I is again an indicator function : { 1 with probability qx+k I = 0 with probability p x+k 19
23 As is J but this is: This is the case as Λ k takes: { 1 with probability px+k J = 0 with probability q x+k v c k+1 in year k when the insured dies and that occurs with probability q x+k v k+1 V if the insured survives year k and that occurs with probability p x+k. The expectation of Λ k is given by: E(Λ k K k) = v c k+1 q x+k + v k+1 V p x+k Π k k V = 0 from (2.4.8) (2.7.8) The variance of Λ k is given by: Var(Λ k K k) = E[(Λ k ) 2 K k] [E(Λ k K k)] 2 = E[(v c k+1 I + v k+1 V J Π k k V ) 2 ] (v c k+1 q x+k + v k+1 V p x+k Π k k V ) 2 = E[v 2 c 2 k+1 I + 2v 2 c k+1 k + 1V IJ 2v c k+1 Π k I 2v c k+1 k V I + v 2 k+1v 2 J 2v k+1 V Π k J 2v k+1 V k V J + Π 2 k + 2Π k k V + k V 2 ] [E(Λ k K k)] 2 = v 2 c 2 k+1 q x+k + 0 2v c k+1 Π k q x+k 2v c k+1 k V q x+k + v 2 k+1v 2 p x+k 2v k+1 V Π k p x+k 2v k+1 V k V p x+k + Π 2 k + 2Π k k V + k V 2 v 2 c 2 k+1 q 2 x+k 2v 2 c k+1 k+1 V p x+k q x+k + 2v c k+1 Π k q x+k + 2v c k+1 k V q x+k v 2 k+1v 2 p 2 x+k + 2v k+1 V Π k p x+k + 2v k+1 V k V p x+k Π 2 k 2Π k k V k V 2 = v 2 c 2 k+1 q x+k + v 2 k+1v 2 p x+k v 2 c 2 k+1 q 2 x+k 2v 2 c k+1 k+1 V p x+k q x+k v 2 k+1v 2 p 2 x+k = v 2 c 2 k+1 q x+k v 2 c 2 k+1 q x+k + v 2 c 2 k+1 q x+k p x+k + v 2 k+1v 2 p x+k v 2 k+1v 2 p x+k + v 2 k+1v 2 p x+k q x+k 2v 2 c k+1 k+1 V p x+k q x+k = v 2 c 2 k+1 p x+k q x+k + v 2 k+1v 2 p x+k q x+k 2v 2 c k+1 k+1 V p x+k q x+k = v 2 (c k+1 k+1 V ) 2 p x+k q x+k (2.7.9) We can now express the prospective loss variable k L in terms of Λ k using the definition of the Λ k s and (2.7.5): v j k Λ k = v j k [C k + v Liability(k, k + 1)] j=k j=k = k L + v j k+1 Liability(k, k + 1) j=k If we look at Table 2.8 we can see that for the sum of all the discounted change in liabilities, most of the terms cancel each other out and we are simply left with 0 k V for the year (k, k + 1) 20
24 with all the years after this obviously having no change in liability. rearranging the above formula we have the relationship: Using this deduction and kl = { 0 K < k j=k vj k Λ j + k V K k We can now find Var( k L) using the simple identity: Var(Λ k K j) = Var(Λ k K k) k j p x+j for j k this can be simply explained as k j p x+j is just the probability that the insured survives to the year k, given that he has survived j years. We also need: Var(aX + b) = a 2 Var(X) Using these identities, the arrangement of k L above and (2.7.9) we get: Var( k L K k) = Var[ v j k Λ j + k V K k] = = = = j=k Var[v j k Λ j + k V K k] j=k v 2(j k) Var[Λ j K k] j=k v 2(j k) j kp x+k Var[Λ k K k] j=k v 2(j k) j kp x+k [v 2 (c k+1 k+1 V ) 2 p x+k q x+k ] j=k (2.7.10) Example These quantities for the term insurance and endowment example from section are shown in table 2.9 below: We can see from these results that: The expected net cash loss for the insurer for the term insurance (1st column) is negative, showing that the insurer expects to make money each year (as a negative loss is a profit), but the amount increases each year as more needs to be set aside just in case the insured dies. The variance of the net cash loss is the same for both the term insurance (2nd column) and the endowment (6th column) as it only depends on the sum insured and the probabilities of surviving the period, which are the same for each. We can see that the variance increases over time showing that the risk of the insurer losing money each year increases. The variance of the allocation of the risk (3rd column) increases each year showing that the term insurance becomes more risky each year as there is more at stake should the insured die, as the insurer would lose a lot of the money he could have been able to keep, in order to pay the benefit payment. The variance of the prospective loss (4th column) increases each year again showing that the term insurance becomes more risky. 21
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