Approaches to representing mortality (cont.)

Mathematical models (cont.)

Makeham’s Law

In 1860, Makeham proposed a small modification to Gompertz’ law, noting that although the Gompertz function represents adequately the progression of mortality in adult ages, it does not accurately reflect the age pattern of mortality at younger ages. The modification involves adding a constant term to the force of mortality:

μx = A + BCx

A is generally small (usually in the range 0.001 to 0.003), and has the effect of increasing the force of mortality substantially at younger ages (when x is small and thus when the term BCx is relatively insignificant. At older ages, this term swamps the value of A entirely, meaning that at older ages the Makeham curve and Gompertz curve are virtually indistinguishable. Applying the same algebraic logic to the Makeham curve allows us to derive an expression for n px in terms of the constants A, B and C and age (x):

n px = exp ( - n

0
A + BC x+t dt ) = exp ( - nA -
B
ln(C)
.C x (C n - 1) )

This curve represents a significant improvement on the Gompertz function at younger ages, but still does not capture the pattern of mortality at young ages well: we know, for example, that the mortality of very young children is (relatively) very high in the first year of life, before declining dramatically.

Note that if x=0, then the function

n p0 = exp ( - n

0
A + BC t dt ) = exp ( - nA -
B
ln(C)
. (C n - 1) )

represents the probability of survival from birth to age n. This survival is frequently represented by l(n), the parenthesis implying survival from birth to age n in a life table with a radix of 1.

Interaction Question. If mortality in a given population follows a Makeham function with parameters A = 0.005; B = 0.001 and C = 1.085, calculate the values of a) 5q0 and b) l50 to five decimal places in a life table with a radix of 1. Enter your values in the input box in the format n.nnnnn and n.nnnnn.

The correct answer is: 0.03069 and 0.38212

5 q0 = 1 - 5 p0 = 1 - exp ( - 5A -
-B
ln(C)
.(C 5 - 1) ) = 1 - exp ( - 0.025 -
0.001
ln(1.085)
(1.0855 - 1) )
= 0.03069
l50 = l(50) = 50 p0 = exp ( - 50A -
-B
ln(C)
.(C 50 - 1) ) = exp ( - 0.25 -
0.001
ln(1.085)
(1.08550 - 1) )
= 0.38212

Please attempt the answer.

Well done, 0.03069 and 0.38212 is correct.

Sorry, that is not correct. Please try again.