Insurance Companies and Pension Plans
3.3 Mortality Tables
Mortality tables are the key to valuing life insurance contracts. Table 3.1 shows an extract from the mortality rates estimated by the U.S. Department of Social Security from statistics on death rates as a function of age collected in 2013. To understand the table, consider the row corresponding to age 31. The second column shows that the probability of a man who has just reached age 31 dying within the next year is 0.001505 (or 0.1505%). The third column shows that the probability of a man
surviving to age 31 is 0.97376 (or 97.376%). The fourth column shows that a man age 31 has a remaining life expectancy of 46.89 years. This means than on average he will
live to age 77.89. The remaining three columns show similar statistics for a woman.
The probability of a 31‐year‐old woman dying within one year is 0.000705
(0.0705%), the probability of a woman surviving to age 31 is 0.98569 (98.569%), and the remaining life expectancy for a 31‐year‐old woman is 51.04 years.
BUSINESS SNAPSHOT 3.1 Equitable Life
Equitable Life was a British life insurance company founded in 1762 that at its peak had 1.5 million policyholders. Starting in the 1950s, Equitable Life sold annuity products where it guaranteed that the interest rate used to calculate the size of the annuity payments would be above a certain level. (This is known as a Guaranteed Annuity Option, GAO.) The guaranteed interest rate was gradually increased in response to competitive pressures and increasing interest rates.
Toward the end of 1993, interest rates started to fall. Also, life expectancies were rising so that the insurance companies had to make increasingly high provisions for future payouts on contracts. Equitable Life did not take action. Instead, it grew by selling new products. In 2000, it was forced to close its doors to new business. A report issued by Ann Abraham in July 2008 was highly critical of regulators and urged compensation for policyholders.
An interesting aside to this is that regulators did at one point urge insurance companies that offered GAOs to hedge their exposures to an interest rate decline.
As a result, many insurance companies scrambled to enter into contracts with banks that paid off if long‐term interest rates declined. The banks in turn hedged their risk by buying instruments such as bonds that increased in price when rates fell. This was done on such a massive scale that the extra demand for bonds caused long‐term interest rates in the UK to decline sharply (increasing losses for insurance companies on the unhedged part of their exposures). This shows that when large numbers of different companies have similar exposures, problems are created if they all decide to hedge at the same time. There are not likely to be enough investors willing to take on their risks without market prices changing.
The full table shows that the probability of death during the following year is a decreasing function of age for the first 10 years of life and then starts to increase.
Mortality statistics for women are a little more favorable than for men. If a man is lucky enough to reach age 90, the probability of death in the next year is about 16.7%.
The full table shows this probability is about 35.4% at age 100 and 57.6% at age 110.
For women, the corresponding probabilities are 13.2%, 30.5%, and 54.6%, respectively.
Table 3.1 Mortality Table
Male Female
Probability Probability
Age of Death Survival Life of Death Survival Life (Years) within 1
Year Probability Expectancy within 1
Year Probability Expectancy
0 0.006519 1.00000 76.28 0.005377 1.00000 81.05
1 0.000462 0.99348 75.78 0.000379 0.99462 80.49
2 0.000291 0.99302 74.82 0.000221 0.99425 79.52
3 0.000209 0.99273 73.84 0.000162 0.99403 78.54
… … … … … … …
30 0.001467 0.97519 47.82 0.000664 0.98635 52.01
31 0.001505 0.97376 46.89 0.000705 0.98569 51.04
32 0.001541 0.97230 45.96 0.000748 0.98500 50.08
33 0.001573 0.97080 45.03 0.000794 0.98426 49.11
… … … … … … …
40 0.002092 0.95908 38.53 0.001287 0.97753 42.43
41 0.002240 0.95708 37.61 0.001393 0.97627 41.48
42 0.002418 0.95493 36.70 0.001517 0.97491 40.54
43 0.002629 0.95262 35.78 0.001662 0.97343 39.60
… … … … … … …
Male Female
Probability Probability
Age of Death Survival Life of Death Survival Life (Years) within 1
Year
Probability Expectancy within 1 Year
Probability Expectancy
50 0.005038 0.92940 29.58 0.003182 0.95829 33.16
51 0.005520 0.92472 28.73 0.003473 0.95524 32.27
52 0.006036 0.91961 27.89 0.003767 0.95193 31.38
53 0.006587 0.91406 27.05 0.004058 0.94834 30.49
… … … … … … …
60 0.011197 0.86112 21.48 0.006545 0.91526 24.46
61 0.012009 0.85147 20.72 0.007034 0.90927 23.62
62 0.012867 0.84125 19.97 0.007607 0.90287 22.78
63 0.013772 0.83042 19.22 0.008281 0.89600 21.95
… … … … … … …
70 0.023528 0.73461 14.24 0.015728 0.82864 16.43
71 0.025693 0.71732 13.57 0.017338 0.81561 15.68
72 0.028041 0.69889 12.92 0.019108 0.80147 14.95
73 0.030567 0.67930 12.27 0.021041 0.78616 14.23
… … … … … … …
80 0.059403 0.50629 8.20 0.043289 0.63880 9.64
Male Female
Probability Probability
Age of Death Survival Life of Death Survival Life (Years) within 1
Year
Probability Expectancy within 1 Year
Probability Expectancy
81 0.065873 0.47621 7.68 0.048356 0.61114 9.05
82 0.073082 0.44484 7.19 0.054041 0.58159 8.48
83 0.081070 0.41233 6.72 0.060384 0.55016 7.94
… … … … … … …
90 0.167291 0.17735 4.03 0.132206 0.29104 4.80
91 0.184520 0.14768 3.74 0.147092 0.25257 4.45
92 0.202954 0.12043 3.47 0.163154 0.21542 4.13
93 0.222555 0.09599 3.23 0.180371 0.18027 3.84
SOURCE: U.S. Department of Social Security, www.ssa.gov/OACT/STATS/table4c6.html.
All the numbers in the table can be calculated from the “Probability of Death within 1 Year” column. The probability that an individual will survive to age n + 1 is the probability that the individual will survive to age n multiplied by 1 minus the probability that an individual aged n will die within the next year. For example, the probability of surviving to age 61 can be calculated as 0.86112 × (1 − 0.011197).
Next consider life expectancy for an individual of a certain age. This can be determined from the probability of death during the first year, the second year, the third year, and so on. Consider a man age 90. The probability that the man will die within one year is, from the table, 0.167291. The probability that he will die in the second year (between ages 91 and 92) is the probability that he does not die in the first year multiplied by the probability that he does die in the second year. From the numbers in the second column of the table, this is
Similarly, the probability that he dies in the third year (between ages 92 and 93) is
Assuming that death occurs on average halfway through a year, the life expectancy of a man age 90 is
Example 3.1
Assume that interest rates for all maturities are 4% per annum (with semiannual compounding) and premiums are paid once a year at the beginning of the year.
What is an insurance company's break‐even premium for $100,000 of term life insurance for a man of average health age 90? If the term insurance lasts one year, the expected payout is 0.167291 × 100, 000 or $16,729. Assume that the payout occurs halfway through the year. (This is likely to be approximately true on average.) The premium is $16,729 discounted for six months. This is 16, 729/1.02 or $16,401.
Suppose next that the term insurance lasts two years. In this case, the present value of expected payout in the first year is $16,401 as before. The probability that the policyholder dies during the second year is (1 − 0.167291) × 0.184520 = 0.153651 so that there is also an expected payout of 0.153651 × 100, 000 or
$15,365 during the second year. Assuming this happens at time 18 months, the present value of the payout is 15, 365/(1.023) or $14,479. The total present value of payouts is 16, 401 + 14, 479 or $30,880.
Consider next the premium payments. The first premium is required at time zero, so we are certain that this will be paid. The probability of the second premium payment being made at the beginning of the second year is the probability that the man does not die during the first year. This is 1 − 0.167291 = 0.832709. When the premium is X dollars per year, the present value of the premium payments is
The break‐even annual premium is given by the value of X that equates the present value of the expected premium payments to the present value of the expected payout. This is the value of X that solves
or X = 17, 152. The break‐even premium payment is therefore $17,152.