Choosing suitable members of the family

266 B.L. Jones, J.A. Mereu Insurance: Mathematics and Economics 27 2000 261–276

4. Choosing suitable members of the family

In order to effectively use the family described in Section 2 in actuarial calculations, it is necessary to choose appropriate α x values for each x at which q x is needed. Suppose we have a life table with mortality rates at n consecutive ages, say from y to y + n − 1. We must therefore specify the corresponding n values of α x . As discussed in Section 2, it is possible to require that the force of mortality be continuous. If we do so, then the n − 1 equations, 1 − p α x x α x p α x x = 1 − p α x+1 x+1 α x+1 , x = y, y + 1, . . . , y + n − 2, 8 must be satisfied. Thus, we only need to specify α y , and the remaining α x values are determined by the continuity requirement. The choice of α y is important. To illustrate this, consider the ultimate mortality rates from the Canadian Institute of Actuaries CIA 1986–1992 male aggregate mortality table age nearest birthday for ages 15–39 see Table 1. The force of mortality obtained assuming UDD α x = 1 for all x is shown in Fig. 2. Notice that the force is very flat between integer ages, and the discontinuities are large. Column 3 of Table 1 shows the α x values obtained assuming that α 15 = 1 and using the continuity requirement to obtain the remaining values. These were found by repeatedly solving the equation in 8 numerically for α x+1 . The resulting force of mortality is shown in Fig. 3. Such a jagged function is certainly unacceptable. One way to reduce this problem is to choose the value of α y that minimizes the sum of squared differences between the left and the right derivative of the force of mortality at ages y + 1, y + 2, . . . , y + n − 1. That is, we Table 1 Mortality rates along with α x values that produce continuous forces of mortality x 1000q x α x α 15 = 1 α x improved 15 0.52 1.000 493.992 16 0.65 712.771 287.404 17 0.77 −182.722 193.990 18 0.87 452.507 105.540 19 0.94 −262.511 66.096 20 0.98 341.156 20.700 21 0.99 −318.278 −0.158 22 0.98 299.697 −20.556 23 0.97 −322.933 −0.545 24 0.96 303.561 −21.015 25 0.96 −303.571 20.454 26 0.98 341.761 20.454 27 1.01 −274.811 39.836 28 1.03 309.531 −1.699 29 1.05 −268.761 38.347 30 1.09 331.367 31.612 31 1.13 −259.165 33.311 32 1.19 338.369 55.332 33 1.24 −262.189 12.263 34 1.26 284.946 12.263 35 1.26 −284.995 −12.518 36 1.25 273.726 −0.211 37 1.24 −288.216 −11.440 38 1.26 310.705 35.821 39 1.31 −242.788 24.845 B.L. Jones, J.A. Mereu Insurance: Mathematics and Economics 27 2000 261–276 267 Fig. 2. Force of mortality assuming UDD. minimize y+n−2 X x=y ∂ ∂t µ x+t t =1 − ∂ ∂t µ x+1+t t =0 2 = y+n−2 X x=y [1 − p α x x ] 2 α x p 2α x x − [1 − p α x+1 x+1 ] 2 α x+1 2 with respect to α y , α y+1 , . . . , α y+n−1 , subject to the constraint equations given by 8, which reduce this to a function of α y only. The idea here is that it is desirable to have small discontinuities in the derivative of the force of mortality. Note that when the force of mortality is steep, a given discontinuity in the derivative will have a smaller effect on the smoothness of the force than when the force is flatter. Furthermore, the force of mortality tends to increase approximately exponentially over most ages, so that the logarithm of the force is approximately linear. For these reasons it may make sense to minimize y+n−2 X x=y ∂ ∂t log µ x+t t =1 − ∂ ∂t log µ x+1+t t =0 2 = y+n−2 X x=y p α x x + p α x+1 x+1 − 2 2 . The α x values for the above example obtained using the latter approach are given in the fourth column of Table 1, and the force of mortality is shown in Fig. 4. This force of mortality function is reasonably smooth. It is important to keep in mind that no “graduation” of the mortality rates is being done; they are fixed. We are merely attempting to choose FAAs that yield a well-behaved force across ages. Fig. 3. Continuous force of mortality with α 15 = 1. 268 B.L. Jones, J.A. Mereu Insurance: Mathematics and Economics 27 2000 261–276 Fig. 4. Improved continuous force of mortality. In choosing suitable α x values for an entire life table, it may be difficult to achieve a smooth force over the entire range of ages without sacrificing continuity at one or more ages. For the CIA 1986–1992 male aggregate table, we obtained α x values for x = 0, 1, . . . , 105. In doing so, we allowed the force to be discontinuous at ages 15 and 40. However, the discontinuities are small, especially at age 15. The resulting force of mortality is shown in Fig. 5. An alternative approach to choosing suitable FAAs is to fix the force of mortality at the beginning of each age interval based on the mortality rates for this and the preceding interval. For example, Jordan 1975, p. 18 gives the Fig. 5. Improved force of mortality for entire age range 1. B.L. Jones, J.A. Mereu Insurance: Mathematics and Economics 27 2000 261–276 269 Fig. 6. Improved force of mortality for entire age range 2. following approximation to the force at age x µ x ≈ ℓ x−1 − ℓ x+1 2ℓ x = q x−1 p x−1 + q x 2 . This formula is exact if ℓ z is a polynomial of degree 2 in z. By fixing the force at the beginning of the age interval, the value of α x is determined by the equation µ x = 1 − p α x x α x , which follows from 2. Note that the force will be discontinuous at the integer ages. However, the discontinuities tend to be small. The value of α cannot be determined using this approach because we do not have a q −1 . One can, however, obtain a value for α by assuming continuity of the force at age 1. The force of mortality obtained using this approach is shown in Fig. 6. While some roughness remains in both Figs. 5 and 6, these forces are a great improvement over those obtained with a constant α x e.g. UDD, constant force, or Balducci.

5. Some quantities arising in life contingencies and demography