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LIFE CONGINGENCY MODELS I Contents 1 Survival Models LIFE CONGINGENCY MODELS I VLADISLAV KARGIN Contents 1 Survival Models . 3 1.1 Survival function . 3 1.2 Expectation and survival function . 4 1.3 Actuarial notation . 8 1.4 Force of mortality . 13 1.5 Expectation of Life . 15 1.6 Future curtate lifetime. 18 1.7 Common analytical survival models . 20 2 Using Life Tables . 25 2.1 Life Tables. 25 2.2 Continuous Calculations Using Life Tables . 30 2.3 Select and ultimate tables . 35 3 Life Insurance . 38 3.1 Introduction . 38 3.2 Whole life insurance . 40 3.3 Term Life Insurance. 48 3.4 Deferred Life Insurance . 51 3.5 Pure Endowment Life Insurance . 55 3.6 Endowment Life Insurance. 56 3.7 Deferred Term Life Insurance . 58 3.8 Non-level payments paid at the end of the year . 59 3.9 Life insurance paid m times a year . 61 3.10 Level benefit insurance in the continuous case . 62 3.11 Calculating APV from life tibles. 64 3.12 Insurance for joint life and last survivor. 67 1 2 VLADISLAV KARGIN 4 Annuities . 72 4.1 Introduction . 72 4.2 Whole life annuities. 74 4.3 Annuities - Woolhouse approximation . 81 4.4 Annuities and UDD assumption . 84 5 Benefit Premiums . 86 5.1 Funding a liability. 86 5.2 Fully discrete benefit premiums: Pricing by equiva- lence principle . 87 5.3 Compensation for risk. 91 5.4 Premiums paid m times a year . 95 5.5 Benefit Premiums - Adjusting for Expenses . 96 VLADISLAV KARGIN 3 ~ SECTION 1 Survival Models The primary responsibility of the life insurance actuary is to maintain the solvency and profitability of the insurer. Consider, for example, a whole life insurance contract issued to a life aged 50. This contract will pay a fixed sum at the death of the insured individual. This individual will pay monthly premiums, which will be invested by the insurer to earn interest; the accumulated premiums must be sufficient to pay the benefit, on average. To ensure this goal, the actuary needs to model the survival probabil- ities of the policyholder, likely investment returns and likely expenses. The actuary may also take into consideration the probability that the policyholder decides to terminate the contract early. In addition, the actuary must determine how much money the insurer should hold to ensure that future liabilities will be covered with ade- quately high probability. To achieve these goals we need to have a mathematical model of a life, a human life or a work-life of an equipment. section 1.1 Survival fun ion Let a positive random variable X represent the duration of a life. We call X the age-at-death or age-at-failure. Definition 1.1. The survival function of a positive random variable X is SX (t) = P (X > t) = 1 − FX (t); where FX (t) is the cumulative distribution function of X: Intuitively, this is the probability to survive more than t units of time. From the properties of the cumulative distribution function, it follows that the survival function is non-decreasing, S (1) = 0; and S (0) = 1: 4 1. SURVIVAL MODELS 1 Example1.2. Determine which of the following functions is a survival function of a nonnegative r.v.: (i) 2 S(x) = ; x + 2 (ii) 1 + 2 S(x) = x+2 ; 2 (iii) S(x) = (1 − x)e−x; (iv) S(x) = (1 + x)e−x; (v) 8 < − x2 ≤ ≤ 1 10;000 for 0 x 100; s(x) = : 0 for 100 < x: 2 Exercise1.3. Find the density function for the following survival func- tions: − (1) S(x) =8 (1 + x)e x, for x ≥ 0. < − x2 ≤ ≤ 1 10;000 for 0 x 100; (2) S(x) = : 0 for 100 < x: 2 ≥ (3) S(x) = x+2 , for x 0. section 1.2 Expe ation and survival fun ion Recall the usual definition of the expectation of a random variable. Definition 1.4. The expectation of a function g(X) of a random vari- able X is X Z E[g(X)] = g(x)PfX = xg + g(x)fX (x) dx; x where f(x) is the density of X. 1Biespiel = Example in German 2Ubung = Exercise in German VLADISLAV KARGIN 5 This definition is general in the sense that it applicable for random variables that have both discrete and continuous components. Often, to find expectations, instead of the density we will use the sur- vival function. Satz 1.5. a Let X be a non-negative random variable with survival function S that has R+ ! R a finite or countable number of discontinuities. Suppose that H : is a continuous function which is differentiable everywhere except for a finite number of points. Then, Z 1 E[H(X)] = H(0) + S(t)H0(t) dt: 0 aSatz = Theorem in German ⊆ R Proof. Given a set A , the indicator function of A is the function 8 < 1 t 2 A 2 if IA(t) = I(t A) = : 0 if t 62 A Since Z Z x 1 0 0 H(x) − H(0) = H (t) dt = I[0;x](t)H (t) dt; 0 0 therefore Z 1 [ ] 0 E[H(X)] − H(0) = E I[0;X](t) H (t) dt: [ ] 0 Note that E I[0;X](t) = PfX ≥ tg = S(t) + PfX = tg. So, Z 1 Z 1 E[H(X)] − H(0) = S(t)H0(t) dt + PfX = tgH0(t):dt 0 0 The second integral is zero because the function PfX = tg can be different from zero only at a finite or countable number of points. □ Corollary 1.6. Let X be a nonnegative r.v. with survival function S. Then, Z 1 E[X] = S(t) dt: 0 6 1. SURVIVAL MODELS Corollary 1.7. Let X be a nonnegative r.v. with survival function S. Then, Z 1 E[Xp] = S(t)ptp−1 dt; 0 if p > 0. 2 Question: E[X ] =? Corollary 1.8. X S a ≥ 0 Let be a non-negative r.v. with survivalZ function . Let . Then, a E[min(X; a)] = S(t) dt: 0 Corollary 1.9. Let X be a non-negativeR r.v. with survival function S. Let δ > 0. Then, E −δX − 1 −δt [e ] = 1 0 δe S(t) dt: −x ≥ Exercise 1.10. Let SX (x) = e (x + 1), x 0. E(X) = ? E(X ^ 10) = ? Theorem 1.5 can be written in a different way if X is a discrete random variable. Satz 1.11. Let X be a discrete random variable whose possible values are positive integers. R+ ! R Suppose that H : is a continuous function which is differentiable ev- f g erywhere except for a finite number of points. Assume that Pr X = 0 = 0. Then, X1 [ ] E[H(X)] = H(0) + PrfX ≥ kg H(k) − H(k − 1) : k=1 Proof: by Theorem 1.5 on page 5, Z 1 E[H(X)] = H(0) + S(t)H0(t) dt 0 X1 Z = H(0) + S(t)H0(t) dt 2 − k=1 t (k 1;k] VLADISLAV KARGIN 7 Next, we use the definition of S(t) and the fact that PrfX > tg = PrfX ≥ kg for every t 2 (k − 1; k]. Z X1 k 0 E[H(X)] = H(0) + PrfX ≥ kg H (t) dt − k=1 k 1 X1 = H(0) + PrfX ≥ kg(H(k) − H(k − 1)): k=1 □ By taking H(x) = x, we find X1 X1 ( E[X] = PrfX ≥ kg = S(k) + PrfX = kg) P k=1 k=1 1 f g f g Since k=1 Pr X = k = Pr X > 0 = S(0), we obtain a useful formula: X1 (1) E[X] = S(k): k=0 In a similar fashion we can get two analogous formulae: X1 X1 (2) E[X2] = PrfX ≥ kg(k2 − (k − 1)2) = PrfX ≥ kg(2k − 1); k=1 k=1 and Xn (3) E[min(X; n)] = PrfX ≥ kg; n ≥ 1: k=1 2 Exercise 1.12. Find E[X] and E[X ] if k 0 1 2 PrfX = kg 0.2 0.3 0.5 8 1. SURVIVAL MODELS section 1.3 A uarial notation Let (x) denote a life that survives to the age x. The life (x) is called a life-age-x or a life aged x. Definition 1.13. The future lifetime of (x) is denoted by Tx. By defi- nition, Tx = X − x. The quantity Tx is especially important for an insurer and is called time-until-death of the life-age-x. Note that Tx is defined only for a life which survives to age x. In particular Tx has the distribution of X − x conditioned on the event X > x. The survival function of Tx is denoted by tpx. It is the probability that a life aged x survives t more years. S(x + t) (4) tpx = PfX > x + tjX > xg = ; S(x) The cumulative distribution function of Tx is S(x + t) q := 1 − p = 1 − : t x t x S(x) The density of Tx is d S(x + t) f (x + t) f (t) = − = X : Tx dt S(x) S(x) S (t) = 90−t 0 < Exercise 1.14. Consider the survival function X 90 , for t < 90. Find the survival function and the probability density function of T30. Other notation: • For simplicity, we denote 1px and 1qx by px and qx, respectively. For example, px is the probability that a life aged x survives one year. The variable qx is often called the mortality rate. • Given x, s, t > 0; sjtqx represents the probability of a life just turning age x will die between ages x + s and x + s + t, i.e., VLADISLAV KARGIN 9 S(x + s) − S(x + s + t) j q = Pfs < T ≤ s + tg = : s t x x S(x) These quantities are called the deferred mortality probabilities.
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