Survival models. F x (t) = Pr[T x t].

Transcription

1 2 Survival models 2.1 Summary In this chapter we represent the future lifetime of an individual as a random variable, and show how probabilities of death or survival can be calculated under this framework. We then define an important quantity known as the force of mortality, introduce some actuarial notation, and discuss some properties of the distribution of future lifetime. We introduce the curtate future lifetime random variable. This is a function of the future lifetime random variable which represents the number of complete years of future life. We explain why this function is useful and derive its probability function. 2.2 The future lifetime random variable In Chapter 1 we saw that many insurance policies provide a benefit on the death of the policyholder. When an insurance company issues such a policy, the policyholder s date of death is unknown, so the insurer does not know exactly when the death benefit will be payable. In order to estimate the time at which a death benefit is payable, the insurer needs a model of human mortality, from which probabilities of death at particular ages can be calculated, and this is the topic of this chapter. We start with some notation. Let (x) denote a life aged x, where x 0. The death of (x) can occur at any age greater than x, and we model the future lifetime of (x) by a continuous random variable which we denote by T x. This means that x + T x represents the age-at-death random variable for (x). Let F x be the distribution function of T x, so that F x (t) = Pr[T x t]. Then F x (t) represents the probability that (x) does not survive beyond age x + t, and we refer to F x as the lifetime distribution from age x. In many life 17

2 18 Survival models insurance problems we are interested in the probability of survival rather than death, and so we define S x as S x (t) = 1 F x (t) = Pr[T x > t]. Thus, S x (t) represents the probability that (x) survives for at least t years, and S x is known as the survival function. Given our interpretation of the collection of random variables {T x } x 0 as the future lifetimes of individuals, we need a connection between any pair of them. To see this, consider T 0 and T x for a particular individual who is now aged x. The random variable T 0 represented the future lifetime at birth for this individual, so that, at birth, the individual s age at death would have been represented by T 0. This individual could have died before reaching age x the probability of this was Pr[T 0 < x] but has survived. Now that the individual has survived to age x, so that T 0 > x, his or her future lifetime is represented by T x and the age at death is now x + T x. If the individual dies within t years from now, then T x t and T 0 x + t. Loosely speaking, we require the events [T x t] and [T 0 x + t] to be equivalent, given that the individual survives to age x. We achieve this by making the following assumption for all x 0 and for all t > 0 Pr[T x t] =Pr[T 0 x + t T 0 > x]. (2.1) This is an important relationship. Now, recall from probability theory that for two events A and B Pr[A B] = Pr[A and B], Pr[B] so, interpreting [T 0 x + t] as event A, and [T 0 > x] as event B, we can rearrange the right-hand side of (2.1) to give that is, Also, using S x (t) = 1 F x (t), Pr[T x t] = Pr[x < T 0 x + t], Pr[T 0 > x] F x (t) = F 0(x + t) F 0 (x). (2.2) S 0 (x) S x (t) = S 0(x + t), (2.3) S 0 (x)

3 which can be written as 2.2 The future lifetime random variable 19 S 0 (x + t) = S 0 (x) S x (t). (2.4) This is a very important result. It shows that we can interpret the probability of survival from age x to age x + t as the product of (1) the probability of survival to age x from birth, and (2) the probability, having survived to age x, of further surviving to age x + t. Note that S x (t) can be thought of as the probability that (0) survives to at least age x + t given that (0) survives to age x, so this result can be derived from the standard probability relationship Pr[A and B] =Pr[A B] Pr[B] where the events here are A =[T 0 > x + t] and B =[T 0 > x], so that Pr[A B] =Pr[T 0 > x + t T 0 > x], which we know from (2.1) is equal to Pr[T x > t]. Similarly, any survival probability for (x), for, say, t + u years can be split into the probability of surviving the first t years, and then, given survival to age x + t, subsequently surviving another u years. That is, S x (t + u) = S 0(x + t + u) S 0 (x) S x (t + u) = S 0(x + t) S 0 (x + t + u) S 0 (x) S 0 (x + t) S x (t + u) = S x (t)s x+t (u). (2.5) We have already seen that if we know survival probabilities from birth, then, using formula (2.4), we also know survival probabilities for our individual from any future age x. Formula (2.5) takes this a stage further. It shows that if we know survival probabilities from any age x ( 0), then we also know survival probabilities from any future age x + t ( x). Any survival function for a lifetime distribution must satisfy the following conditions to be valid. Condition 1. S x (0) = 1; that is, the probability that a life currently aged x survives 0 years is 1. Condition 2. lim t S x (t) = 0; that is, all lives eventually die.

4 20 Survival models Condition 3. The survival function must be a non-increasing function of t; it cannot be more likely that (x) survives, say 10.5 years than 10 years, because in order to survive 10.5 years, (x) must first survive 10 years. These conditions are both necessary and sufficient, so that any function S x which satisfies these three conditions as a function of t ( 0), for a fixed x ( 0), defines a lifetime distribution from age x, and, using formula (2.5), for all ages greater than x. For all the distributions used in this book, we make three additional assumptions: Assumption 1. S x (t) is differentiable for all t > 0. Note that together with Condition 3 above, this means that d dt S x(t) 0 for all t > 0. Assumption 2. lim t ts x (t) = 0. Assumption 3. lim t t 2 S x (t) = 0. These last two assumptions ensure that the mean and variance of the distribution of T x exist. These are not particularly restrictive constraints we do not need to worry about distributions with infinite mean or variance in the context of individuals future lifetimes. These three extra assumptions are valid for all distributions that are feasible for human lifetime modelling. Example 2.1 Let Calculate the probability that F 0 (t) = 1 (1 t/120) 1/6 for 0 t 120. (a) a newborn life survives beyond age 30, (b) a life aged 30 dies before age 50, and (c) a life aged 40 survives beyond age 65. Solution 2.1 (a) The required probability is S 0 (30) = 1 F 0 (30) = (1 30/120) 1/6 = (b) From formula (2.2), the required probability is F 30 (20) = F 0(50) F 0 (30) 1 F 0 (30) (c) From formula (2.3), the required probability is = S 40 (25) = S 0(65) S 0 (40) =

5 2.3 The force of mortality 21 We remark that in the above example, S 0 (120) = 0, which means that under this model, survival beyond age 120 is not possible. In this case we refer to 120 as the limiting age of the model. In general, if there is a limiting age, we use the Greek letter ω to denote it. In models where there is no limiting age, it is often practical to introduce a limiting age in calculations, as we will see later in this chapter. 2.3 The force of mortality The force of mortality is an important and fundamental concept in modelling future lifetime. We denote the force of mortality at age x by µ x and define it as 1 µ x = lim dx 0 + dx Pr[T 0 x + dx T 0 > x]. (2.6) From equation (2.1) we see that an equivalent way of defining µ x is 1 µ x = lim dx 0 + dx Pr[T x dx], which can be written in tems of the survival function S x as 1 µ x = lim dx 0 + dx (1 S x(dx)). (2.7) Note that the force of mortality depends, numerically, on the unit of time; if we are measuring time in years, then µ x is measured per year. The force of mortality is best understood by noting that for very small dx, formula (2.6) gives the approximation µ x dx Pr[T 0 x + dx T 0 > x]. (2.8) Thus, for very small dx, we can interpret µ x dx as the probability that a life who has attained age x dies before attaining age x + dx. For example, suppose we have a male aged exactly 50 and that the force of mortality at age 50 is per year. A small value of dx might be a single day, or years. Then the approximate probability that (50) dies on his birthday is = We can relate the force of mortality to the survival function from birth, S 0.As S x (dx) = S 0(x + dx) S 0 (x),

6 22 Survival models formula (2.7) gives Thus, µ x = 1 S 0 (x) = 1 S 0 (x) S 0 (x) S 0 (x + dx) lim dx 0 + dx ( d ) dx S 0(x). µ x = 1 d S 0 (x) dx S 0(x). (2.9) From standard results in probability theory, we know that the probability density function for the random variable T x, which we denote f x, is related to the distribution function F x and the survival function S x by So, it follows from equation (2.9) that f x (t) = d dt F x(t) = d dt S x(t). µ x = f 0(x) S 0 (x). We can also relate the force of mortality function at any age x + t, t > 0, to the lifetime distribution of T x. Assume x is fixed and t is variable. Then d(x + t) = dt and so Hence 1 d µ x+t = S 0 (x + t) d(x + t) S 0(x + t) 1 d = S 0 (x + t) dt S 0(x + t) 1 d = S 0 (x + t) dt (S 0(x)S x (t)) = S 0(x) d S 0 (x + t) dt S x(t) = 1 d S x (t) dt S x(t). µ x+t = f x(t) S x (t). (2.10)

7 2.3 The force of mortality 23 This relationship gives a way of finding µ x+t given S x (t). We can also use equation (2.9) to develop a formula for S x (t) in terms of the force of mortality function. We use the fact that for a function h whose derivative exists, so from equation (2.9) we have d 1 log h(x) = dx h(x) d dx h(x), µ x = d dx log S 0(x), and integrating this identity over (0, y) yields y 0 µ x dx = (log S 0 (y) log S 0 (0)). As log S 0 (0) = log Pr[T 0 > 0] =log 1 = 0, we obtain { y } S 0 (y) = exp µ x dx, from which it follows that S x (t) = S 0(x + t) S 0 (x) 0 { x+t } { t } = exp µ r dr = exp µ x+s ds. (2.11) x 0 This means that if we know µ x for all x 0, then we can calculate all the survival probabilities S x (t), for any x and t. In other words, the force of mortality function fully describes the lifetime distribution, just as the function S 0 does. In fact, it is often more convenient to describe the lifetime distribution using the force of mortality function than the survival function. Example 2.2 As in Example 2.1, let F 0 (x) = 1 (1 x/120) 1/6 for 0 x 120. Derive an expression for µ x. Solution 2.2 As S 0 (x) = (1 x/120) 1/6, it follows that d ( ) dx S 0(x) = 1 6 (1 x/120) 5/ , and so µ x = 1 d S 0 (x) dx S 0(x) = (1 1 x/120) 1 = 720 6x.

8 24 Survival models As an alternative, we could use the relationship µ x = d dx log S 0(x) = d ( ) 1 1 log(1 x/120) = dx 6 720(1 x/120) 1 = 720 6x. Example 2.3 Let µ x = Bc x, x > 0, where B and c are constants such that 0 < B < 1 and c > 1. This model is called Gompertz law of mortality. Derive an expression for S x (t). Solution 2.3 From equation (2.11), Writing c r as exp{r log c}, giving x+t x { x+t } S x (t) = exp Bc r dr. x x+t Bc r dr = B exp{r log c}dr x = B x+t exp{r log c} log c = B ( c x+t c x), log c { } B S x (t) = exp log c cx (c t 1). The force of mortality under Gompertz law increases exponentially with age. At first sight this seems reasonable, but as we will see in the next chapter, the force of mortality for most populations is not an increasing function of age over the entire age range. Nevertheless, the Gompertz model does provide a fairly good fit to mortality data over some age ranges, particularly from middle age to early old age. Example 2.4 Calculate the survival function and probability density function for T x using Gompertz law of mortality, with B = and c = 1.07, for x = 20, x = 50 and x = 80. Plot the results and comment on the features of the graphs. x

9 2.3 The force of mortality 25 Solution 2.4 For x = 20, the force of mortality is µ 20+t = Bc 20+t and the survival function is { } B S 20 (t) = exp log c c20 (c t 1). The probability density function is found from (2.10): µ 20+t = f { } 20(t) B S 20 (t) f 20(t) = µ 20+t S 20 (t) = Bc 20+t exp log c c20 (c t 1). Figure 2.1 shows the survival functions for ages 20, 50 and 80, and Figure 2.2 shows the corresponding probability density functions. These figures illustrate some general points about lifetime distributions. First, we see an effective limiting age, even though, in principle there is no age to which the survival probability is exactly zero. Looking at Figure 2.1,we see that although S x (t) >0 for all combinations of x and t, survival beyond age 120 is very unlikely. Second, we note that the survival functions are ordered according to age, with the probability of survival for any given value of t being highest for age 20 and lowest for age 80. For survival functions that give a more realistic representation of human mortality, this ordering can be violated, but it usually holds at ages of interest to insurers. An example of the violation of this ordering is that S 0 (1) may be smaller than S x (1) for x 1, as a result of perinatal mortality. Looking at Figure 2.2, we see that the densities for ages 20 and 50 have similar shapes, but the density for age 80 has a quite different shape. For ages 20 and 50, the densities have their respective maximums at (approximately) Survival probability Time, t Figure 2.1 S x (t) for x = 20 (bold), 50 (solid) and 80 (dotted).

10 26 Survival models Time, t Figure 2.2 f x (t) for x = 20 (bold), 50 (solid) and 80 (dotted). t = 60 and t = 30, indicating that death is most likely to occur around age 80. The decreasing form of the density for age 80 also indicates that death is more likely to occur at age 80 than at any other age for a life now aged 80. A further point to note about these density functions is that although each density function is defined on (0, ), the spread of values of f x (t) is much greater for x = 20 than for x = 50, which, as we will see in Table 2.1, results in a greater variance of future lifetime for x = 20 than for x = Actuarial notation The notation used in the previous sections, S x (t), F x (t) and f x (t), is standard in statistics. Actuarial science has developed its own notation, International Actuarial Notation, that encapsulates the probabilities and functions of greatest interest and usefulness to actuaries. The force of mortality notation, µ x, comes from International Actuarial Notation. We summarize the relevant actuarial notation in this section, and rewrite the important results developed so far in this chapter in terms of actuarial functions. The actuarial notation for survival and mortality probabilities is tp x = Pr[T x > t] =S x (t), (2.12) tq x = Pr[T x t] =1 S x (t) = F x (t), (2.13) u t q x = Pr[u < T x u + t] =S x (u) S x (u + t). (2.14)

11 2.4 Actuarial notation 27 So t p x is the probability that (x) survives to at least age x + t, t q x is the probability that (x) dies before age x + t, u t q x is the probability that (x) survives u years, and then dies in the subsequent t years, that is, between ages x + u and x + u + t. This is called a deferred mortality probability, because it is the probability that death occurs in some interval following a deferred period. We may drop the subscript t if its value is 1, so that p x represents the probability that (x) survives to at least age x + 1. Similarly, q x is the probability that (x) dies before age x + 1. In actuarial terminology q x is called the mortality rate at age x. The relationships below follow immediately from the definitions above and the previous results in this chapter: Similarly, tp x + t q x = 1, u t q x = u p x u+t p x, t+up x = t p xu p x+t from (2.5), (2.15) µ x = 1 xp 0 d dx x p 0 from (2.9). (2.16) µ x+t = 1 tp x d dt t p x d dt t p x = t p x µ x+t, (2.17) µ x+t = f x(t) S x (t) f x(t) = t p x µ x+t from (2.10), (2.18) { t } tp x = exp µ x+s ds from (2.11). (2.19) 0 As F x is a distribution function and f x is its density function, it follows that F x (t) = t which can be written in actuarial notation as tq x = t 0 0 f x (s)ds, sp x µ x+s ds. (2.20) This is an important formula, which can be interpreted as follows. Consider time s, where 0 s < t. The probability that (x) is alive at time s is s p x,

12 28 Survival models Time 0 s s+ds t Age x Event Probability x+s x+s+ds (x) survives s years (x) dies µ sp x x+s ds Figure 2.3 Time-line diagram for t q x x+t and the probability that (x) dies between ages x + s and x + s + ds, having survived to age x + s, is (loosely) µ x+s ds, provided that ds is very small. Thus sp x µ x+s ds can be interpreted as the probability that (x) dies between ages x + s and x + s + ds. Now, we can sum over all the possible death intervals s to s + ds which requires integrating because these are infinitesimal intervals to obtain the probability of death before age x + t. We can illustrate this event sequence using the time-line diagram shown in Figure 2.3. This type of interpretation is important as it can be applied to more complicated situations, and we will employ the time-line again in later chapters. In the special case when t = 1, formula (2.20) becomes q x = 1 0 sp x µ x+s ds. When q x is small, it follows that p x is close to 1, and hence s p x is close to 1 for 0 s < 1. Thus q x 1 0 µ x+s ds µ x+1/2, where the second relationship follows by the mid-point rule for numerical integration. Example 2.5 As in Example 2.1, let F 0 (x) = 1 (1 x/120) 1/6 for 0 x 120. Calculate both q x and µ x+1/2 for x = 20 and for x = 110, and comment on these values.

13 Solution 2.5 We have 2.5 Mean and standard deviation of T x 29 p x = S 0(x + 1) S 0 (x) = ( ) 1 1/6 1, 120 x giving q 20 = and q 110 = , and from the solution to Example 2.2, µ 20 1 = and µ = We see that µ x+1/2 2 is a good approximation to q x when the mortality rate is small, but is not such a good approximation, at least in absolute terms, when the mortality rate is not close to Mean and standard deviation of T x Next, we consider the expected future lifetime of (x),e[t x ], denoted in actuarial notation by e x. We also call this the complete expectation of life. In order to evaluate e x, we note from formulae (2.17) and (2.18) that f x (t) = t p x µ x+t = d dt t p x. (2.21) From the definition of an expected value, we have e x = = 0 0 tf x (t)dt t t p x µ x+t dt. We can now use (2.21) to evaluate this integral using integration by parts as ( ) d e x = t 0 dt t p x dt ( ) = t t p x 0 0 t p x dt. In Section 2.2 we stated the assumption that lim t t t p x = 0, which gives e x = tp x dt. (2.22) 0

14 30 Survival models Similarly, for E[Tx 2 ], we have E[Tx 2 ]= t 2 tp x µ x+t dt = = = t 2 ( d dt t p x ) dt ( t 2 tp x ) tp x 2tdt t t p x dt. (2.23) So we have integral expressions for E[T x ] and E[Tx 2 ]. For some lifetime distributions we are able to integrate directly. In other cases we have to use numerical integration techniques to evaluate the integrals in (2.22) and (2.23). The variance of T x can then be calculated as [ ] ( V [T x ] = E Tx 2 ex ) 2. Example 2.6 As in Example 2.1, let F 0 (x) = 1 (1 x/120) 1/6 for 0 x 120. Calculate e x and V[T x ] for (a) x = 30 and (b) x = 80. Solution 2.6 As S 0 (x) = (1 x/120) 1/6, we have tp x = S 0(x + t) S 0 (x) = ( ) t 1/ x Now recall that this formula is valid for 0 t 120 x, since under this model survival beyond age 120 is impossible. Technically, we have tp x = { ( 1 t 120 x ) 1/6 for x + t 120, 0 for x + t > 120. So the upper limit of integration in equation (2.22)is120 x, and e x = 120 x 0 ( ) t 1/6 1 dt. 120 x

15 2.5 Mean and standard deviation of T x 31 We make the substitution y = 1 t/(120 x), so that t = (120 x)(1 y), giving e x = (120 x) 1 0 = 6 7 (120 x). y 1/6 dy Then e 30 = and e 80 = Under this model the expectation of life at any age x is 6/7 of the time to age 120. For the variance we require E[Tx 2 ]. Using equation (2.23) we have [ ] E Tx 2 = 2 = x x 0 t t p x dt Again, we substitute y = 1 t/(120 x) giving [ E Tx 2 t ( ) t 1/6 1 dt. 120 x ] 1 = 2(120 x) 2 (y 1/6 y 7/6 ) dy = 2(120 x) 2 ( ). Then V[T x ]=E[T 2 x ] ( ex ) 2 = (120 x) 2 ( 2(6/7 6/13) (6/7) 2) = (120 x) 2 ( ) = ((120 x)( )) 2. So V[T 30 ]= and V[T 80 ]= Since we know under this model that all lives will die before age 120, it makes sense that the uncertainty in the future lifetime should be greater for younger lives than for older lives. A feature of the model used in Example 2.6 is that we can obtain formulae for quantities of interest such as e x, but for many models this is not possible. For example, when we model mortality using Gompertz law, there is no explicit formula for e x and we must use numerical integration to calculate moments of T x. In Appendix B we describe in detail how to do this.

16 32 Survival models Table 2.1. Values of e x,sd[t x ] and expected age at death for the Gompertz model with B = and c = x e x SD[T x ] x + e x Table 2.1 shows values of e x and the standard deviation of T x (denoted SD[T x ]) for a range of values of x using Gompertz law, µ x = Bc x, where B = and c = For this survival model, 130 p 0 = ,so that using 130 as the maximum attainable age in our numerical integration is accurate enough for practical purposes. We see that e x is a decreasing function of x, as it was in Example 2.6. In that example e x was a linear function of x, but we see that this is not true in Table Curtate future lifetime K x and e x In many insurance applications we are interested not only in the future lifetime of an individual, but also in what is known as the individual s curtate future lifetime. The curtate future lifetime random variable is defined as the integer part of future lifetime, and is denoted by K x for a life aged x.ifwelet denote the floor function, we have K x = T x. We can think of the curtate future lifetime as the number of whole years lived in the future by an individual. As an illustration of the importance of curtate future lifetime, consider the situation where a life aged x at time 0 is entitled to payments of 1 at times 1, 2, 3,...provided that (x) is alive at these times. Then

17 2.6 Curtate future lifetime 33 the number of payments made equals the number of complete years lived after time 0 by (x). This is the curtate future lifetime. We can find the probability function of K x by noting that for k = 0, 1, 2,..., K x = k if and only if (x) dies between the ages of x + k and x + k + 1. Thus for k = 0, 1, 2,... Pr[K x = k] =Pr[k T x < k + 1] = k q x = k p x k+1 p x = k p x k p x p x+k = k p x q x+k. The expected value of K x is denoted by e x, so that e x = E[K x ], and is referred to as the curtate expectation of life (even though it represents the expected curtate lifetime). So E[K x ]=e x = k Pr[K x = k] = k=0 k ( k p x k+1 p x ) k=0 = ( 1 p x 2 p x ) + 2( 2 p x 3 p x ) + 3( 3 p x 4 p x ) + = kp x. (2.24) k=1 Note that the lower limit of summation is k = 1. Similarly, E[Kx 2 ]= k 2 ( k p x k+1 p x ) k=0 = ( 1 p x 2 p x ) + 4( 2 p x 3 p x ) + 9( 3 p x 4 p x ) + 16( 4 p x 5 p x ) + = 2 k k p x kp x = 2 k=1 k=1 k k p x e x. k=1

18 34 Survival models As with the complete expectation of life, there are a few lifetime distributions that allow E[K x ] and E[Kx 2 ] to be calculated analytically. For more realistic models, such as Gompertz, we can calculate the values easily using Excel or other suitable software. Although in principle we have to evaluate an infinite sum, at some age the survival probability will be sufficiently small that we can treat it as an effective limiting age The complete and curtate expected future lifetimes, e x and e x As the curtate future lifetime is the integer part of future lifetime, it is natural to ask if there is a simple relationship between e x and e x. We can obtain an approximate relationship by writing e x = 0 t p x dt = j=0 j+1 j tp x dt. If we approximate each integral using the trapezium rule for numerical integration (see Appendix B), we obtain and hence e x j+1 j j=0 tp x dt 1 2 ( jp x + j+1 p x ), 1 2 ( jp x + j+1 p x ) = jp x. Thus, we have an approximation that is frequently applied in practice, namely j=1 e x e x (2.25) In Chapter 5 we will meet a refined version of this approximation. Table 2.2 shows values of e x and e x for a range of values of x when the survival model is Gompertz law with B = and c = Values of e x were calculated by applying formula (2.24) with a finite upper limit of summation of 130 x, and values of e x are as in Table 2.1. This table illustrates that formula (2.25)is a very good approximation in this particular case for younger ages, but is less accurate at very old ages. This observation is true for most realistic survival models.

19 2.7 Notes and further reading 35 Table 2.2. Values of e x and e x for Gompertz law with B = and c = x e x ex Notes and further reading Although laws of mortality such as Gompertz law are appealing due to their simplicity, they rarely represent mortality over the whole span of human ages. A simple extension of Gompertz law is Makeham s law (Makeham, 1860), which models the force of mortality as µ x = A + Bc x. (2.26) This is very similar to Gompertz law, but adds a fixed term that is not age related, that allows better for accidental deaths. The extra term tends to improve the fit of the model to mortality data at younger ages. In recent times, the Gompertz Makeham approach has been generalized further to give the GM(r, s) (Gompertz Makeham) formula, µ x = h 1 r (x) + exp{h2 s (x)}, where h 1 r and h2 s are polynomials in x of degree r and s respectively. Adiscussion of this formula can be found in Forfar et al. (1988). Both Gompertz law and Makeham s law are special cases of the GM formula. In Section 2.3, we noted the importance of the force of mortality. A further significant point is that when mortality data are analysed, the force of mortality

20 36 Survival models is a natural quantity to estimate, whereas the lifetime distribution is not. The analysis of mortality data is a huge topic and is beyond the scope of this book. An excellent summary article on this topic is Macdonald (1996). For more general distributions, the quantity f 0 (x)/s 0 (x), which actuaries call the force of mortality at age x, is known as the hazard rate in survival analysis and the failure rate in reliability theory. 2.8 Exercises Exercise 2.1 Let F 0 (t) = 1 (1 t/105) 1/5 for 0 t 105. Calculate (a) the probability that a newborn life dies before age 60, (b) the probability that a life aged 30 survives to at least age 70, (c) the probability that a life aged 20 dies between ages 90 and 100, (d) the force of mortality at age 50, (e) the median future lifetime at age 50, (f) the complete expectation of life at age 50, (g) the curtate expectation of life at age 50. Exercise 2.2 The function G(x) = x x has been proposed as the survival function S 0 (x) for a mortality model. (a) What is the implied limiting age ω? (b) Verify that the function G satisfies the criteria for a survival function. (c) Calculate 20 p 0. (d) Determine the survival function for a life aged 20. (e) Calculate the probability that a life aged 20 will die between ages 30 and 40. (f) Calculate the force of mortality at age 50. Exercise 2.3 Calculate the probability that a life aged 0 will die between ages 19 and 36, given the survival function S 0 (x) = x, 10 0 x 100 (= ω). Exercise 2.4 Let S 0 (x) = exp { ( Ax Bx2 + where A, B, C and D are all positive. C log D Dx C )} log D

21 (a) Show that the function S 0 is a survival function. (b) Derive a formula for S x (t). (c) Derive a formula for µ x. (d) Now suppose that 2.8 Exercises 37 A = , B = , C = , D = (i) Calculate t p 30 for t = 1, 5, 10, 20, 50, 90. (ii) Calculate t q 40 for t = 1, 10, 20. (iii) Calculate t 10 q 30 for t = 1, 10, 20. (iv) Calculate e x for x = 70, 71, 72, 73, 74, 75. (v) Calculate e x for x = 70, 71, 72, 73, 74, 75, using numerical integration. Exercise 2.5 Let F 0 (t) = 1 e λt, where λ>0. (a) Show that S x (t) = e λt. (b) Show that µ x = λ. (c) Show that e x = (e λ 1) 1. (d) What conclusions do you draw about using this lifetime distribution to model human mortality? Exercise 2.6 Given that p x = 0.99, p x+1 = 0.985, 3 p x+1 = 0.95 and q x+3 = 0.02, calculate (a) p x+3, (b) 2 p x, (c) 2 p x+1, (d) 3 p x, (e) 1 2 q x. Exercise 2.7 Given that F 0 (x) = x for x 0, find expressions for, simplifying as far as possible, (a) S 0 (x), (b) f 0 (x), (c) S x (t), and calculate: (d) p 20, and (e) 10 5 q 30. Exercise 2.8 Given that S 0 (x) = e x2 for x 0,

22 38 Survival models find expressions for, simplifying as far as possible, (a) f 0 (x), and (b) µ x. Exercise 2.9 Show that d dx t p x = t p x (µ x µ x+t ). Exercise 2.10 Suppose that Gompertz law applies with µ 30 = and µ 50 = Calculate 10 p 40. Exercise 2.11 A survival model follows Makeham s law, so that µ x = A + Bc x for x 0. (a) Show that under Makeham s law tp x = s t g cx (c t 1), (2.27) where s = e A and g = exp{ B/ log c}. (b) Suppose you are given the values of 10 p 50, 10 p 60 and 10 p 70. Show that c = ( ) log( 10 p 70 ) log( 10 p 60 ) 0.1. log( 10 p 60 ) log( 10 p 50 ) Exercise 2.12 (a) Construct a table of p x for Makeham s law with parameters A = , B = and c = 1.075, for integer x from age 0 to age 130, using Excel or other appropriate computer software. You should set the parameters so that they can be easily changed, and you should keep the table, as many exercises and examples in future chapters will use it. (b) Use the table to determine the age last birthday at which a life currently aged 70 is most likely to die. (c) Use the table to calculate e 70. (d) Using a numerical approach, calculate e 70. Exercise 2.13 A life insurer assumes that the force of mortality of smokers at all ages is twice the force of mortality of non-smokers. (a) Show that, if * represents smokers mortality, and the unstarred function represents non-smokers mortality, then tp x = ( tp x ) 2.

23 2.8 Exercises 39 (b) Calculate the difference between the life expectancy of smokers and nonsmokers aged 50, assuming that non-smokers mortality follows Gompertz law, with B = and c = (c) Calculate the variance of the future lifetime for a non-smoker aged 50 and for a smoker aged 50 under Gompertz law. Hint: You will need to use numerical integration for parts (b) and (c). Exercise 2.14 (a) Show that (b) Show that (c) Explain (in words) why e x e x e x e x. e x e x (d) Is e x always a non-increasing function of x? Exercise 2.15 (a) Show that o e x = 1 S 0 (x) x S 0 (t)dt, where S 0 (t) = 1 F 0 (t), and hence, or otherwise, prove that d o o e x = µ x ex 1. dx Hint: d { x } g(t)dt = g(x). What about d { a dx a dx x (b) Deduce that } g(t)dt? x + o e x is an increasing function of x, and explain this result intuitively. 2.1 (a) (b) (c) Answers to selected exercises

24 40 Survival models (d) (e) (f) (g) (a) 90 (c) (d) 1 3x/308 x 2 / (e) (f) (d) (i) , , , , , (ii) , , (iii) , , (iv) , , , , , (v) , , , , , (a) 0.98 (b) (c) (d) (e) (d) (e) (b) 73 (c) (d) (b) (c) (non-smokers), (smokers)

25 3 Life tables and selection 3.1 Summary In this chapter we define a life table. For a life table tabulated at integer ages only, we show, using fractional age assumptions, how to calculate survival probabilities for all ages and durations. We discuss some features of national life tables from Australia, England & Wales and the United States. We then consider life tables appropriate to individuals who have purchased particular types of life insurance policy and discuss why the survival probabilities differ from those in the corresponding national life table. We consider the effect of selection of lives for insurance policies, for example through medical underwriting. We define a select survival model and we derive some formulae for such a model. 3.2 Life tables Given a survival model, with survival probabilities t p x, we can construct the life table for the model from some initial age x 0 to a maximum age ω. We define a function {l x } for x 0 x ω as follows. Let l x0 be an arbitrary positive number (called the radix of the table) and, for 0 t ω x 0, define l x0 +t = l x0 tp x0. From this definition we see that for x 0 x x + t ω, l x+t = l x0 x+t x 0 p x0 = l x0 x x 0 p x0 tp x = l xt p x, 41

26 42 Life tables and selection so that tp x = l x+t /l x. (3.1) For any x x 0, we can interpret l x+t as the expected number of survivors to age x + t out of l x independent individuals aged x. This interpretation is more natural if l x is an integer, and follows because the number of survivors to age x + t is a random variable with a binomial distribution with parameters l x and tp x. That is, suppose we have l x independent lives aged x, and each life has a probability t p x of surviving to age x + t. Then the number of survivors to age x + t is a binomial random variable, L t, say, with parameters l x and t p x. The expected value of the number of survivors is then E[L t ]=l xt p x = l x+t. We always use the table in the form l y /l x which is why the radix of the table is arbitrary it would make no difference to the survival model if all the l x values were multiplied by 100, for example. From (3.1) we can use the l x function to calculate survival probabilities. We can also calculate mortality probabilities. For example, q 30 = 1 l 31 l 30 = l 30 l 31 l 30 (3.2) and q 40 = 15 p q 55 = l 55 l 40 ( 1 l ) 85 = l 55 l 85. (3.3) l 55 l 40 In principle, a life table is defined for all x from the initial age, x 0, to the limiting age, ω. In practice, it is very common for a life table to be presented, and in some cases even defined, at integer ages only. In this form, the life table is a useful way of summarizing a lifetime distribution since, with a single column of numbers, it allows us to calculate probabilities of surviving or dying over integer numbers of years starting from an integer age. It is usual for a life table, tabulated at integer ages, to show the values of d x, where d x = l x l x+1, (3.4) in addition to l x, as these are used to compute q x. From (3.4) we have ( d x = l x 1 l ) x+1 = l x (1 p x ) = l x q x. l x

27 3.2 Life tables 43 Table 3.1. Extract from a life table. x l x d x We can also arrive at this relationship if we interpret d x as the expected number of deaths in the year of age x to x + 1 out of l x lives aged exactly x, so that, using the binomial distribution again d x = l x q x. (3.5) Example 3.1 Table 3.1 gives an extract from a life table. Calculate (a) l 40, (b) 10 p 30, (c) q 35, (d) 5 q 30, and (e) the probability that a life currently aged exactly 30 dies between ages 35 and 36. Solution 3.1 (a) From equation (3.4), (b) From equation (3.1), (c) From equation (3.5), l 40 = l 39 d 39 = p 30 = l 40 l 30 = = q 35 = d 35 l 35 = =

28 44 Life tables and selection (d) Following equation (3.2), 5q 30 = l 30 l 35 l 30 = (e) This probability is 5 q 30. Following equation (3.3), 5 q 30 = l 35 l 36 l 30 = d 35 l 30 = Fractional age assumptions A life table {l x } x x0 provides exactly the same information as the corresponding survival distribution, S x0. However, a life table tabulated at integer ages only does not contain all the information in the corresponding survival model, since values of l x at integer ages x are not sufficient to be able to calculate probabilities involving non-integer ages, such as 0.75 p Given values of l x at integer ages only, we need an additional assumption or some further information to calculate probabilities for non-integer ages or durations. Specifically, we need to make some assumption about the probability distribution for the future lifetime random variable between integer ages. We use the term fractional age assumption to describe such an assumption. It may be specified in terms of the force of mortality function or the survival or mortality probabilities. In this section we assume that a life table is specified at integer ages only and we describe the two most useful fractional age assumptions Uniform distribution of deaths The uniform distribution of deaths (UDD) assumption is the most common fractional age assumption. It can be formulated in two different, but equivalent, ways as follows. UDD1 For integer x, and for 0 s < 1, assume that sq x = sq x. (3.6) UDD2 Recall from Chapter 2 that K x is the integer part of T x, and define a new random variable R x such that T x = K x + R x.

29 3.3 Fractional age assumptions 45 The UDD2 assumption is that, for integer x, R x U(0, 1), and R x is independent of K x. The equivalence of these two assumptions is demonstrated as follows. First, assume that UDD1 is true. Then for integer x, and for 0 s < 1, Pr[R x s] = = = = = s = s = s. Pr[R x s and K x = k] k=0 Pr[k T x k + s] k=0 kp xs q x+k k=0 kp x s (q x+k ) k=0 kp x q x+k k=0 Pr[K x = k] k=0 using UDD1 This proves that R x U(0, 1). To prove the independence of R x and K x, note that Pr[R x s and K x = k] =Pr[k T x k + s] = k p xs q x+k = s k p x q x+k = Pr[R x s] Pr[K x = k] since R x U(0, 1). This proves that UDD1 implies UDD2. To prove the reverse implication, assume that UDD2 is true. Then for integer x, and for 0 s < 1, sq x = Pr[T x s] = Pr[K x = 0 and R x s] = Pr[R x s] Pr[K x = 0]

30 46 Life tables and selection as K x and R x are assumed independent. Thus, sq x = sq x. (3.7) Formulation UDD2 explains why this assumption is called the Uniform Distribution of Deaths, but in practical applications of this assumption, formulation UDD1 is the more useful of the two. An immediate consequence is that for 0 s < 1. This follows because and substituting sq x for s q x gives Hence l x+s = l x sd x (3.8) sq x = 1 l x+s l x s d x l x = l x l x+s l x. l x+s = l x sd x for 0 s 1. Thus, we assume that l x+s is a linearly decreasing function of s. Differentiating equation (3.6) with respect to s, we obtain d ds s q x = q x, 0 s 1 and we know that the left-hand side is the probability density function for T x at s, because we are differentiating the distribution function. The probability density function for T x at s is s p x µ x+s so that under UDD q x = s p x µ x+s (3.9) for 0 s < 1. The left-hand side does not depend on s, which means that the density function is a constant for 0 s < 1, which also follows from the uniform distribution assumption for R x. Since q x is constant with respect to x, and s p x is a decreasing function of s, we can see that µ x+s is an increasing function of s, which is appropriate for ages of interest to insurers. However, if we apply the approximation over successive ages, we obtain a discontinuous function for the force of mortality,

31 3.3 Fractional age assumptions 47 with discontinuities occurring at integer ages, as illustrated in Example 3.4. Although this is undesirable, it is not a serious drawback. Example 3.2 Given that p 40 = , calculate 0.4 q 40.2 under the assumption of a uniform distribution of deaths. Solution 3.2 We note that the fundamental result in equation (3.7), that for fractional of a year s, s q x = sq x, requires x to be an integer. We can manipulate the required probability 0.4 q 40.2 to involve only probabilities from integer ages as follows 0.4q 40.2 = p 40.2 = 1 l 40.6 l 40.2 = p 40 = q p q 40 = Example 3.3 Use the life table in Example 3.1 above, with the UDD assumption, to calculate (a) 1.7 q 33 and (b) 1.7 q Solution 3.3 (a) We note first that 1.7q 33 = p 33 = 1 (p 33 )( 0.7 p 34 ). We can calculate p 33 directly from the life table as l 34 /l 33 = and 0.7p 34 = q 34 = under UDD, so that 1.7 q 33 = (b) To calculate 1.7 q 33.5 using UDD, we express this as 1.7q 33.5 = p 33.5 = 1 l 35.2 l 33.5 = 1 l d 35 l d 33 = Example 3.4 Under the assumption of a uniform distribution of deaths, calculate lim µ t 1 40+t using p 40 = , and calculate lim µ t t using p 41 =

32 48 Life tables and selection Solution 3.4 From formula (3.9), we have µ x+t = q x / t p x. Setting x = 40 yields while setting x = 41 yields lim t 1 µ 40+t = q 40 /p 40 = , lim t 0 + µ 41+t = q 41 = Example 3.5 Given that q 70 = and q 71 = , calculate 0.7 q 70.6 assuming a uniform distribution of deaths. Solution 3.5 As deaths are assumed to be uniformly distributed between ages 70 to 71 and ages 71 to 72, we write 0.7q 70.6 = 0.4 q (1 0.4 q 70.6 ) 0.3 q 71. Following the same arguments as in Solution 3.3, we obtain 0.4q 70.6 = 1 1 q q 70 = , and as 0.3 q 71 = 0.3q 71 = , we obtain 0.7 q 70.6 = Constant force of mortality A second fractional age assumption is that the force of mortality is constant between integer ages. Thus, for integer x and 0 s < 1, we assume that µ x+s does not depend on s, and we denote it µ x. We can obtain the value of µ x by using the fact that { } p x = exp 1 0 µ x+s ds Hence the assumption that µ x+s = µ x for 0 s < 1 gives p x = e µ x or µ x = log p x. Further, under the assumption of a constant force of mortality, for 0 s < 1 we obtain { s } sp x = exp µ x du = e µ x s = (p x ) s. Similarly, for t, s > 0 and t + s < 1, { sp x+t = exp 0 s 0 µ x du } = (p x ) s..

33 3.4 National life tables 49 Thus, under the constant force assumption, the probability of surviving for a period of s < 1 years from age x + t is independent of t provided that s + t < 1. The assumption of a constant force of mortality leads to a step function for the force of mortality over successive years of age. By its nature, the assumption produces a constant force of mortality over the year of age x to x + 1, whereas we would expect the force of mortality to increase for most ages. However, if the true force of mortality increases slowly over the year of age, the constant force of mortality assumption is reasonable. Example 3.6 Given that p 40 = , calculate 0.4 q 40.2 under the assumption of a constant force of mortality. Solution 3.6 We have 0.4 q 40.2 = p 40.2 = 1 (p 40 ) 0.4 = Example 3.7 Given that q 70 = and q 71 = , calculate 0.7 q 70.6 under the assumption of a constant force of mortality. Solution 3.7 As in Solution 3.5 we write 0.7q 70.6 = 0.4 q (1 0.4 q 70.6 ) 0.3 q 71, where 0.4 q 70.6 = 1 (p 70 ) 0.4 = and 0.3 q 71 = 1 (p 71 ) 0.3 = , giving 0.7 q 70.6 = Note that in Examples 3.2 and 3.5 and in Examples 3.6 and 3.7 we have used two different methods to solve the same problems, and the solutions agree to five decimal places. It is generally true that the assumptions of a uniform distribution of deaths and a constant force of mortality produce very similar solutions to problems. The reason for this can be seen from the following approximations. Under the constant force of mortality assumption q x = 1 e µ µ provided that µ is small, and for 0 < t < 1, tq x = 1 e µ t µ t. In other words, the approximation to t q x is t times the approximation to q x, which is what we obtain under the uniform distribution of deaths assumption. 3.4 National life tables Life tables based on the mortality experience of the whole population of a country are regularly produced for many countries in the world. Separate life

34 50 Life tables and selection Table 3.2. Values of q x 10 5 from some national life tables. Australian Life Tables English Life Table US Life Tables 2002 x Males Females Males Females Males Females tables are usually produced for males and for females and possibly for some other groups of individuals, for example on the basis of smoking habits. Table 3.2 shows values of q x 10 5, where q x is the probability of dying within one year, for selected ages x, separately for males and females, for the populations of Australia, England & Wales and the United States. These tables are constructed using records of deaths in a particular year, or a small number of consecutive years, and estimates of the population in the middle of that period. The relevant years are indicated in the column headings for each of the three life tables in Table 3.2. Data at the oldest ages are notoriously unreliable. For this reason, the United States Life Tables do not show values of q x for ages 100 and higher. For all three national life tables and for both males and females, the values of q x follow exactly the same pattern as a function of age, x. Figure 3.1 shows the US 2002 mortality rates for males and females; the graphs for England & Wales and for Australia are similar. (Note that we have plotted these on a logarithmic scale in order to highlight the main features. Also, although the information plotted consists of values of q x for x = 0, 1,..., 99, we have plotted a continuous line as this gives a clearer representation.) We note the following points from Table 3.2 and Figure 3.1. The value of q 0 is relatively high. Mortality rates immediately following birth, perinatal mortality, are high due to complications arising from the

35 3.4 National life tables Mortality rates Age Figure 3.1 US 2002 mortality rates, male (dotted) and female (solid). later stages of pregnancy and from the birth process itself. The value of q x does not reach this level again until about age 55. This can be seen from Figure 3.1. The rate of mortality is much lower after the first year, less than 10% of its level in the first year, and declines until around age 10. In Figure 3.1 we see that the pattern of male and female mortality in the late teenage years diverges significantly, with a steeper incline in male mortality. Not only is this feature of mortality for young adult males common for different populations around the world, it is also a feature of historical populations in countries such as the UK where mortality data has been collected for some time. It is sometimes called the accident hump, as many of the deaths causing the hump are accidental. Mortality rates increase from age 10, with the accident hump creating a relatively large increase between ages 10 and 20 for males, a more modest increase from ages 20 to 40, and then steady increases from age 40. For each age, all six values of q x are broadly comparable, with, for each country, the rate for a female almost always less than the rate for a male of the same age. The one exception is the Australian Life Table, where q 100 is slightly higher for a female than for a male. According to the Australian Government Actuary, Australian mortality data indicate that males are subject to lower mortality rates than females at very high ages, although there is some uncertainty as to where the cross-over occurs due to small amounts of data at very old ages.

36 52 Life tables and selection Mortality rates Age Figure 3.2 US 2002 male mortality rates (solid), with fitted Gompertz mortality rates (dotted). The Gompertz model introduced in Chapter 2 is relatively simple, in that it requires only two parameters and has a force of mortality with a simple functional form, µ x = Bc x. We stated in Chapter 2 that this model does not provide a good fit across all ages. We can see from Figure 3.1 that the model cannot fit the perinatal mortality, nor the accident hump. However, the mortality rates at later ages are rather better behaved, and the Gompertz model often proves useful over older age ranges. Figure 3.2 shows the older ages US 2002 Males mortality rate curve, along with a Gompertz curve fitted to the US 2002 Table mortality rates. The Gompertz curve provides a pretty close fit which is a particularly impressive feat, considering that Gompertz proposed the model in A final point about Table 3.2 is that we have compared three national life tables using values of the probability of dying within one year, q x, rather than the force of mortality, µ x. This is because values of µ x are not published for any ages for the US Life Tables. Also, values of µ x are not published for age 0 for the other two life tables there are technical difficulties in the estimation of µ x within a year in which the force of mortality is changing rapidly, as it does between ages 0 and Survival models for life insurance policyholders Suppose we have to choose a survival model appropriate for a man, currently aged 50 and living in the UK, who has just purchased a 10-year term insurance