# 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 Expense-Loaded Premium Reserves Uses of Reserves Introduction Alterations and Conversions With-profits 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

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 with-profit 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

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

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 re-write 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

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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