# Actuarial Science from the Air

```Actuarial Science from the Air
An Overview of Life Contingencies,
Including
Life Annuities and Life Insurance
Introduction
From the ground, individual streams may
seem isolated; but from the air they are
visible as interconnected features of a
larger watershed. Similarly, in the study of
Actuarial Science, a number of apparently
dissimilar insurance products actually
have a strong theoretical relationship. The
unifying principles that provide this broad
view are now explained and applied.
Properties of Insurance Products
• A policy can be in effect for a certain term or for
the whole life of the insured
• Benefits can begin immediately or be deferred
• Life insurance pays upon death
• Life annuities provide a series of payments
contingent upon survival
While Life insurance and Life annuities differ, the
same three unifying principles apply when
determining their costs.
Unifying Principle #1:
Invested money grows over time.
Simplifying Assumption #1: All
investments grow at an annual
compound interest rate of 5%.
Example 1 (Present Value):
To accumulate \$1000 in 10 years, invest
\$613.91 because
\$613.91(1.05)10 = \$1000
The amount \$613.91 is called the present
value of \$1000 to be paid in 10 years.
Example 2: To receive \$1000 at the end of each year for the next 10 years,
invest \$7,721.73, the sum of the 10 present values:
Present Values for a Sequence of Ten \$1000 Payments
Year
Present Values
1
\$952.38
2
\$907.03
3
\$863.84
4
\$822.70
5
\$783.53
6
\$746.22
7
\$710.68
8
\$676.84
9
\$644.61
10
\$613.91
SUM = \$7,721.73
Accumulated Values
\$952.38 (1.05)1 =
\$1,000
2
\$907.03 (1.05) =
\$1,000
3
\$863.84 (1.05) =
\$1,000
4
\$822.70 (1.05) =
\$1,000
5
\$783.53 (1.05) =
\$1,000
6
\$746.22 (1.05) =
\$1,000
7
\$710.68 (1.05) =
\$1,000
\$676.84 (1.05)8 =
\$1,000
9
\$644.61 (1.05) =
\$1,000
10
\$613.91 (1.05) =
\$1,000
SUM = \$10,000
Present Value Formula
The present value P of an amount X to be
paid after n years (at 5% interest) satisfies
the following equivalent equations
P(1.05)n = X
P = X/(1.05)n
Example 3: Other Present Values
Year PresentValues PresentValues PresentValues
of X
of \$1
of \$1000
0.95238X
\$0.95238
\$952.38
1
0.90703X
\$0.90703
\$907.03
2
0.86384X
\$0.86384
\$863.84
3
0.82270X
\$0.82270
\$822.70
4
0.78353X
\$0.78353
\$783.53
5
0.74622X
\$0.74622
\$746.22
6
0.71068X
\$0.71068
\$710.68
7
0.67684X
\$0.67684
\$676.84
8
0.64461X
\$0.64461
\$644.61
9
0.61391X
\$0.61391
\$613.91
10
Unifying Principle #2:
The cost of providing insurance benefits
depends upon when the insured dies.
For example, the longer a person lives,
• the smaller the present value of his life
insurance payment (decreasing the cost)
• the greater the number of his life annuity
payments (increasing the cost)
Life Table
In determining what to charge for insurance,
a company needs to know the likelihood or
probability of death at various times. This
information may be given in a life table.
Simplifying Assumption #2
Uniform Distribution of Deaths: Of the people
alive at age 75, 5% of them will die in each of
the next 10 years.
For example, if 10,000 are alive at age 75, then
• 500 (= 5% of 10,000) will die each year
• 2000 (= 4(500)) will die in the next 4 years
• 8000 (= 10,000 – 2000) will survive the next 4
years
• 0.80 = 8000/10000 = probability of surviving the
next 4 years
Year
n
1
2
3
4
5
6
7
8
9
10
Simplified Life Table
In Year n
End of Year n
Probability Number of Number Probability of
of death
deaths
alive
survival
0.05
500
9500
0.95
0.05
500
9000
0.90
0.05
500
8500
0.85
0.05
500
8000
0.80
0.05
500
7500
0.75
0.05
500
7000
0.70
0.05
500
6500
0.65
0.05
500
6000
0.60
0.05
500
5500
0.55
0.05
500
5000
0.50
Cost of a Future Benefit
Problem 1: Suppose each of two 75-year-old men wishes
to receive a benefit of \$1000 if he is still living after 10
years. How much should each pay now to receive that
future benefit?
•
•
•
•
•
The probability of surviving 10 years is 0.50.
On average, only one of the two will survive 10 years.
On average, only one benefit of \$1000 will be paid.
The present value of this benefit is \$613.91.
Thus, the men split this cost:
Cost for each = \$613.91 / 2 = \$306.96
Second Solution to Problem 1:
A 75-year-old man has a 50% chance of
surviving 10 years. Therefore, the cost of
paying him \$1000 if he is alive after 10
years is 50% of the present value of the
payment:
Cost = [\$1000/(1.05)10] 0.50
= [\$613.91] 0.50 = \$306.96
Unifying Principle #3
Suppose a payment A will be made with
probability p after n years. Then
• The cost varies directly as p
• More precisely, the cost equals the
present value of the payment times p:
Cost = [A/(1.05)n] p
(This cost is known as the actuarial present
value of the payment A.)
Computing Costs by
Unifying Principle #3
We now compute the costs of
• a 10-year term life annuity that pays \$1000 at the end of
each year of survival for the next 10 years
• 10-year term life insurance that pays \$1000 at the end of
the year of death if death occurs within 10 years
These are alike in that both offer a possible payment at the
end of each of the next ten years. Thus, by Unifying
Principle #3 the cost of each is the sum of 10 actuarial
present values.
Costs of 10-year Term Life Annuity and 10-year Term Life Insurance
Each with Potential Payments of \$1000
Year n
Present value
Life Annuity
Life Insurance
of \$1000 to be Probability Cost of
Probability Cost of
paid in n years of payment payment of payment payment
1
\$952.38
0.95 \$904.76
0.05 \$47.62
2
\$907.03
0.90 \$816.33
0.05 \$45.35
3
\$863.84
0.85 \$734.26
0.05 \$43.19
4
\$822.70
0.80 \$658.16
0.05 \$41.14
5
\$783.53
0.75 \$587.64
0.05 \$39.18
6
\$746.22
0.70 \$522.35
0.05 \$37.31
7
\$710.68
0.65 \$461.94
0.05 \$35.53
8
\$676.84
0.60 \$406.10
0.05 \$33.84
9
\$644.61
0.55 \$354.53
0.05 \$32.23
10
\$613.91
0.50 \$306.96
0.05 \$30.70
TOTALS: \$7,722
COST = \$5,753
COST = \$386
5-year Term Life Annuities
(undeferred) 5-year term pays at the end of each
year of survival for years 1-5. The probability of
payment equals 0 for years 6-10.
5-year deferred, 5-year term pays at the end of
each year of survival for years 6-10. The
probability of payment equals 0 for years 1-5.
The following table applies Unifying Principle #3 to
compute the two costs:
n
1
2
3
4
5
6
7
8
9
10
Cost of 5-year Term Life Annuities
Present value
Undeferred
5-year Deferred
of \$1000 to be Probability Cost of
Probability Cost of
paid in n years of payment payment
of payment payment
\$952.38
0.95
\$904.76
0
\$907.03
0.90
\$816.33
0
\$863.84
0.85
\$734.26
0
\$822.70
0.80
\$658.16
0
\$783.53
0.75
\$587.64
0
\$746.22
0
0.70 \$522.35
\$710.68
0
0.65 \$461.94
\$676.84
0
0.60 \$406.10
\$644.61
0
0.55 \$354.53
\$613.91
0
0.50 \$306.96
COST = \$3701
COST = \$2052
TOTALS:
Final Problem
Problem 2: What yearly premium X, paid at
the beginning of each year for 10 years,
would cover the cost of 10 year term life
insurance for a person aged 75?
Simplifying Assumption #3
Equivalence Principle: The cost to the insured
equals the cost to the insurance company.
That is, the following are equal:
• Actuarial present value of the 10 premium
payments of X
• Actuarial present value of the benefit payment
We now compute these expected present values
and use their equality to solve for X:
Yearly Premium X for 10-year Term Life Insurance Paying \$1000
Life Insurance
n Pres value
Prob of
Cost of
Pres value Prob of
Cost of
of \$1000 pd payment payment of X paid payment premium
in n years
in n years
0
X
1.00
X
1
\$952.38
0.05 \$47.62 0.95238X
0.95 0.90476X
2
\$907.03
0.05 \$45.35 0.90703X
0.90 0.81633X
3
\$863.84
0.05 \$43.19 0.86384X
0.85 0.73426X
4
\$822.70
0.05 \$41.14 0.82270X
0.80 0.65816X
5
\$783.53
0.05 \$39.18 0.78353X
0.75 0.58764X
6
\$746.22
0.05 \$37.31 0.74622X
0.70 0.52235X
7
\$710.68
0.05 \$35.53 0.71068X
0.65 0.46194X
8
\$676.84
0.05 \$33.84 0.67684X
0.60 0.40610X
9
\$644.61
0.05 \$32.23 0.64461X
0.55 0.35453X
10
\$613.91
0.05 \$30.70
 Equal Totals 
6.44607X
Yearly Premium = X = \$386.09/6.44607 = \$59.90
Homework Problem 1
How would each of the following affect the
cost of life annuities and life insurance?
• Change the annual interest rate from 5%
to 3%
• Change the number of deaths in the life
table from 500 to 600 per year
Homework Problem 2: Complete the following more
realistic life table:
Year
n
1
2
3
4
5
6
7
8
9
10
Life Table for 10,000 Alive at Age 75
In Year n
End of Year n
Probability Number of Number Probability of
of death
deaths
alive
survival
451
465
479
494
509
523
534
542
543
537
Homework Problem 3
Assuming 6% annual compound interest,
use the life table from Problem 2 to
compute the costs of a 10-year term life
annuity and 10-year term life insurance.
Conclusion
In a brief presentation, we have
• introduced important actuarial science
concepts
• seen a unified discussion of life insurance
and life annuities
```

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