Chapter 7 LIFE TABLES AND POPULATION PROBLEMS

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Presentation transcript:

Chapter 7 LIFE TABLES AND POPULATION PROBLEMS Introduction Life Tables The Stationary Population Expectation of Life

7.1 Introduction How do we find probabilities? Data obtained from practice Data required to find probabilities of surviving to certain ages (or, equivalently, of dying before certain ages) are contained in life tables

7.2 Life Tables lx dx lx – number of lives survived to age x 1000 qx 1,000,000 1580 1.58 1 998,420 680 . 68 2 997,740 485 .49 3 997,255 435 .44 lx – number of lives survived to age x Thus l0 = 1,000,000 is a starting population Survival function S(x) = lx / l0 is the probability of surviving to age x dx = lx – lx+1 – number of lives who died between (x, x+1) qx = dx /lx – probability that x – year-old will not survive to age x + 1

Examples (p. 129 – p. 130) lx dx Find Age lx dx 1000 qx 1,000,000 1580 1.58 1 998,420 680 68 2 997,740 485 .49 3 997,255 435 .44 Find the probability that a newborn will live to age 3 the probability that a newborn will die between age 1 and age 3

Find an expression for each of the following: the probability that an 18-year-old lives to age 65 the probability that a 25-year-old dies between ages 40 and 45 the probability that a 25-year-old does not die between ages 40 and 45 the probability that a 30-year-old dies before age 60 There are four persons, now aged 40, 50, 60 and 70. Find an expression for the probability that both the 40-year-old and the 50-year-old will die within the five-year period starting ten years from now, but neither the 60-year-old nor the 70-year-old will die during that five-year period

Note: If we are already given by probabilities qx and starting population l0 we can construct the whole life table step-by-step since dx = qx lx and lx+1 = lx - dx Example Given the following probabilities of deaths q0 = .40, q1 = .20, q2 = .30, q3 = .70, q4 = 1 and starting with l0 = 100 construct a life table

More notations… qx = dx / lx – probability that x – year-old will not survive to age x + 1 px = 1- qx – probability that x – year-old will survive to age x + 1 Note: qx = (lx – lx+1) / lx and px = lx+1 / lx nqx – probability that x – year-old will not survive to age x + n npx = 1- nqx – probability that x – year-old will survive to age x + n Note: nqx = (lx – lx+n) / lx and n px = lx+n / lx Formulas: lx – lx+n = dx + dx+1 + …+ dx+n n+mpx = mpx ∙ npx+m

What is mPx when m is not integer? tPx where 0 < t <1 Assuming that deaths are distributed uniformly during any given year we can use linear interpolation to find tpx :

Examples (p. 132 – p. 133) 30% of those who die between ages 25 and 75 die before age 50. The probability of a person aged 25 dying before age 50 is 20%. Find 25P50 Using the following life table and assuming a uniform distribution of deaths over each year, find: 4/3P1 The probability that a newborn will survive the first year but die in the first two months thereafter Age lx dx 1000 qx 1,000,000 1580 1.58 1 998,420 680 .68 2 997,740 485 .49 3 997,255 435 .44

7.4 The Stationary Population Assume that in every given year (or, more precisely, in any given 12-months period) the number of births and deaths is the same and is equal to l0 Then after a period of time the total population will remain stationary and the age distribution will remain constant px and qx are defined as before lx denotes the number of people who reach their xth birthday during any given year dx = qx lx represent the number of people who die before reaching age x+1 Also, dx represent the number of people who die during any given year between ages x and x + 1 Similarly lx – lx + n represent the number of people who die during any given year between ages x and x + n

Number of people aged x Let Lx denote the number of people aged x (last birthday) at any given moment Note: Lx ≠ lx Assuming uniform distribution of deaths we obtain: More precisely:

Number of people aged x and over Let Tx denote the number of people aged x and over at any given moment Then Assuming uniform distribution we obtain: More precisely:

Example (p. 139) An organization has a constant total membership. Each year 500 new members join at exact age 20. 20% leave after 10 years, 10% of those remaining leave after 20 years, and the rest retire at age 65. Express each of the following in terms of life table functions: The number who leave at age 40 each year The size of the membership The number of retired people alive at any given time The number of members who die each year

7.5 Expectation of Life What is the average future life time ex of a person aged x now? The answer is given by expected value (or mathematical expectation) and is called the curtate expectation:

Complete expectation: (4) (2) (1) (3) Complete expectation: (4) (2)&(3)&(4) imply: Approximation: (since )

Remarks Thus Tx can be interpreted as the total number of years of future life of those who form group lx Note: this interpretation makes sense in any population (stationary or not)

Average age at death Average age at death of a person currently aged x is given by Letting x = 0 we obtain the average age at death for all deaths among l0 individuals

Examples (p. 142 – 143) If tp35 = (.98)t for all t, find e35 and eo35 without approximation. Compare the value for eo35 with its approximate value Interpret verbally the expression Tx-Tx+n – nlx+n Fin the average age at death of those who die between age x and age x+n