Stat 475

Life Contingencies I

Chapter 2: Survival models

The future lifetime random variable — Notation

We are interested in analyzing and describing the futurelifetime of an individual.

We use (x) to denote a life age x for x ≥ 0.

Then the random variable describing the future lifetime, ortime until death, for (x) is denoted by Tx for Tx ≥ 0.

We also have notation for the density, distribution andsurvival functions of Tx :

Density: fx(t) = ddtFx(t)

Distribution: Fx(t) = Pr [Tx ≤ t]

Survival: Sx(t) = Pr [Tx > t] = 1− Fx(t)

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The future lifetime random variable — Relationships

We now consider the set of random variables {Tx}x≥0 and therelationships between them:

Important Assumption

Pr [Tx ≤ t] = Pr [T0 ≤ x + t|T0 > x ]

Important Consequence

S0(x + t) = S0(x) · Sx(t)

This idea can be generalized:

Important General Relationship

Sx(t + u) = Sx(t) · Sx+t(u)

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Survival function conditions and assumptions

We will require any survival function corresponding to a futurelifetime random variable to satisfy the following conditions:

Necessary and Sufficient Conditions

1 Sx(0) = 1

2 limt→∞

Sx(t) = 0

3 Sx(t) must be a non-increasing function of t

In addition, we will make the following assumptions for the survivalfunctions used in this course:

Assumptions

1 Sx(t) is differentiable for all t > 0.

2 limt→∞

t · Sx(t) = 0

3 limt→∞

t2 · Sx(t) = 0

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Future lifetime random variable example

Assume that the survival function for a newborn is given by

S0(t) = 8(t + 2)−3

1 Verify that this is a valid survival function.

2 Find the density function associated with the future lifetimerandom variable T0, that is, find f0(t).

3 Find the probability that a newborn dies between the ages of1 and 2.

4 Find the survival function corresponding to the randomvariable T10.

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Force of Mortality

We will define the force of mortality as

µx = limdx→0+

Pr [T0 ≤ x + dx |T0 > x ]

dx

Then for small values of dx , we can use the approximation

µxdx ≈ Pr [T0 ≤ x + dx |T0 > x ]

We can relate the force of mortality to the survival function:

µx =− d

dx S0(x)

S0(x)and Sx(t) = exp

{−∫ t

0µx+s ds

}

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Some commonly used forms for the force of mortality

It is sometimes convenient to describe the future lifetime randomvariable by explicitly modeling the force of mortality. Some of theparametric forms that have been used are:

Gompertz: µx = Bcx , 0 < B < 1, c > 0

Makeham: µx = A + Bcx , 0 < B < 1, c > 0

de Moivre1: µx =1

ω − x, 0 ≤ x < ω

Exponential1: µx = k, 0 ≤ x ≤ ∞

If there is a limiting age, it is typically denoted by ω.

1When these models are used to model human mortality, it is primarily formathematical convenience rather than realism.

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Force of mortality example

Again, assume that the survival function for a newborn is given by

S0(x) = 8(x + 2)−3

1 Find the force of mortality, µx .

2 Consider the shape of µx as a function of x .

Does it seem to be a realistic survival model for describinghuman lifetimes?

What would you expect the force of mortality corresponding tohumans to look like?

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Actuarial Notation

Thus far we have described the properties of the future lifetimerandom variable using statistical notation. Actuaries have alsodeveloped their own (different) notation to describe many of thesame concepts:

tpx = Pr [Tx > t] = Sx(t)

tqx = Pr [Tx ≤ t] = Fx(t)

u|tqx = Pr [u < Tx ≤ u + t] = Fx(u + t)− Fx(u)

In any of the above expressions, the subscript t may be omittedwhen its value is 1.

The density of the random variable Tx can be written asfX (t) = tpx µx+t .

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Some Basic Relationships

We can express some relationships in this actuarial notation:

tpx = 1− tqx

u|tqx = upx − u+tpx

u|tqx = upx tqx+u

t+upx = tpx upx+t

tpx = exp

{−∫ t

0µx+s ds

}

tqx =

∫ t

0spx µx+s ds

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Mean of the Future Lifetime Random Variable

One quantity that is often of interest is the mean of the futurelifetime random variable, E [Tx ], which is called the completeexpectation of life.

E [Tx ] =◦ex =

∫ ∞0

t fx(t) dt

=

∫ ∞0

t tpx µx+t dt

=

∫ ∞0

tpx dt

We can also find the term expectation of life.

E [min(Tx , n)] =◦ex :n =

∫ n

0ttpxµx+tdt +

∫ ∞n

ntpxµx+tdt

=

∫ n

0tpxdt

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Variance of the Future Lifetime Random Variable

In order to calculate the variance of Tx , we need its secondmoment:

E[T 2x

]=

∫ ∞0

t2 fx(t) dt

=

∫ ∞0

t2 tpx µx+t dt

= 2

∫ ∞0

t tpx dt

Then we can calculate the variance in the usual way:

V [Tx ] = E[T 2x

]− (E [Tx ])2

= E[T 2x

]−(◦ex)2

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Summary of notation for the future lifetime RV

We can summarize the statistical and actuarial notation for theproperties of Tx covered thus far:

Concept Statistical Notation Actuarial Notation

Expected Value E [Tx ]◦ex

Distribution Function Fx(t) tqx

Survival Function Sx(t) tpx

Force of Mortality λ(x) or λx µx

Density fx(t) tpx µx+t

Deferred Mortality Prob. Pr [u < Tx ≤ u + t] u|tqx

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Curtate Future Lifetime Random Variable

We are often interested in the integral number of years lived in thefuture by an individual. This discrete random variable is called thecurtate future lifetime and is denoted by Kx .

We can consider the pmf (probability mass function) of Kx :

Pr [Kx = k] = Pr [k ≤ Tx < k + 1]

= k |qx= kpx qx+k

We also have the relation Kx = bTxc.

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Moments of the Curtate Future Lifetime Random Variable

We can find the mean (called the curtate expectation of life anddenoted by ex) and variance of Kx :

E [Kx ] = ex =∞∑k=1

kpx

E[K 2x

]= 2

∞∑k=1

k kpx − ex

V [Kx ] = E[K 2x

]− (ex)2

Using the trapezoidal rule, we can also find an approximaterelation between the means of Tx and Kx :

◦ex ≈ ex +

1

2

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Exercise 2.2

S0(x) =18000− 110x − x2

18000

(a) What is the implied limiting age? [90]

(b) Is it a valid survival function?

(c) Calculate 20p0. [0.8556]

(d) S20(t)[15400−150t−t2

15400

](e) 10|10q20 [0.1169]

(f) µ50 [0.021]

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Exercise 2.1/2.6

F0(t) =

{1−

(1− t

120

)1/6for 0 ≤ t ≤ 120

1 for t > 120

(a) 30p0 [0.9532]

(b) 20q30 [0.041]

(c) 25p40 [0.9395]

(d) Derive µx[

1720−6x

](e) Calculate

◦ex and Var [Tx ] for (30)

[77.143, 21.3962

](f) Calculate

◦ex and Var [Tx ] for (80)

[34.286, 9.5092

]

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Exercise 2.5

F0(t) = 1− e−λt

(a) S0(t) e−λt

(b) µx λ

(c) ex1

eλ−1(d) Is this reasonable for human mortality?

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Exercise 2.6

px = 0.99 px+1 = 0.985 3px+1 = 0.95 qx+3 = 0.02

(a) px+3 0.98

(b) 2px 0.97515

(c) 2px+1 0.96939

(d) 3px 0.95969

(e) 1|2qx 0.03031

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SOA #32

S0(t) =

1 for 0 ≤ t < 1

1− et

100 for 1 ≤ t < 4.5

0 for 4.5 ≤ t

Calculate µ4. [1.20]

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SOA #200

The graph of a piecewise linear survival function, S0(t), consists of3 line segments with endpoints (0, 1), (25, 0.50), (75, 0.40), (100,0).

Calculate 20|55q15

55q35. [0.6856]

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