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Introduction to survival models

Overview A survival model is a probabilistic model of a random variable that represents the time until the occurrence of an unpredictable event. For example, we may wish to study the life expectancy of a newborn baby, or the future working lifetime of a machine until it fails. In both cases, we study how long the subject may be expected to survive.

The theory that we will develop throughout this course can be applied in a wide range of situations, in which the concept of “survival” may not be immediately obvious, for example:

• the time until a claim is made on an automobile insurance policy

• the time until a patient in a coma recovers from the coma, given that he recovers

• the time until a worker leaves employment.

The focus of our study—the time until the specified event—is known as a waiting time or a random time-to-event variable. Probabilities associated with these models play a central role in actuarial calculations such as pricing insurance contracts.

Introduction to survival models Chapter 1

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1.1 The role of a survival model in a contingent payment model

Let’s start by considering the most basic contingent payment model, in which a specified amount is paid if and only if a particular event occurs.

Suppose that an amount P is to be paid in n years if a random event E occurs. Otherwise, if the complementary event occurs, then nothing is to be paid.

At an effective annual rate of interest i , the random present value of the payment is:

⋅=

′

if occurs0 if occurs

nP v EZE

where −= + 1(1 )v i is the one-year present value discount factor.

The random present value of the payment, Z , is a discrete random variable. Its expected value is known as the actuarial present value of the payment, which incorporates the amount of the payment, the discount factor associated with the timing of the payment, and the probability of the payment being made:

[ ] ( ) ( )

( )

′= ⋅ ⋅ + ⋅

= ⋅ ⋅amount discount probability

Pr 0 Pr

Pr

n

n

E Z P v E E

P v E

Throughout Chapters 2 through 8 of this course, we will be concerned with the distribution of random present value of payment variables. We will frequently calculate the mean, the variance, a percentile, or the probability of some event regarding Z such as [ ]>Pr( )Z E Z .

In Chapters 2 and 3 we will introduce contingent payment models that arise in the context of life insurance and life annuities. For these models, we will need to compute probabilities of events that are expressed in terms of the random future lifetime after age x , when an insurance contract has just been issued. For example, in order to calculate the appropriate life insurance premium for a 30 year old policyholder, we need to calculate the probability that the policyholder will die before age 31, or 32, or 33, and so on.

In this first chapter, our aim is to familiarize you with the theory of survival models, the notation employed in these models, and standard terminology.

There are three principal variables, all of which are measured in years:

• the random lifetime (ie time until death) of a newborn life is denoted X

• the random future lifetime at age x , given that a newborn has survived to age x , is denoted ( )T x , ie = − >( )T x X x X x

• the curtate future lifetime at age x , given that a newborn has survived to age x , is the complete number of years of future lifetime at age x and is denoted ( )K x , ie

( ) ( )= K x T x (greatest integer)

The variables X and T are assumed to be continuous random variables, whereas K is obviously discrete. For example, suppose a newborn life eventually dies at age 74.72. Then 74.72X = ,

( )30 74.72 30 44.72T = − = , and ( ) ( )= = =30 [ 30 ] [44.72] 44K T .

Notice that T is a function of X , and K is a function of T . So, the distributions of these three variables are closely related. These relationships will be developed over the rest of this chapter.

Chapter 1 Introduction to survival models

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1.2 The life table – a discrete survival model

We begin by studying the life table, a discrete survival model commonly used in insurance applications. This model gives us the opportunity to gain an intuitive understanding of some of the most fundamental concepts before we study continuous time models.

We start by defining xl as the number of lives expected to survive to age x from a group of 0l newborn lives. A life table displays in a table format the values of xl at ages x equal to 0,1, 2, ,ω… , where ω is the first whole number age at which there are no remaining lives in the group. If we are modeling human mortality, we may choose a value of ω of around 120 years.

We’ll start by taking a deterministic view of future mortality. By this, we mean that the table tells us exactly how many of the 0l lives will be surviving at ages 1, 2, and so on.

Here is a portion of a hypothetical life table:

x 0 1 2 3 4 5 6 7 8 9

xl 1,000 991 985 982 979 976 972 968 964 959

xd 9 6 3 3 3 4 4 4 5 6

The row labeled xd represents the number of lives among 0l newborn lives that die in the age range +[ , 1)x x . It is computed as:

1x x xd l l += −

For example, since =2 985l lives survive to age 2, and =3 982l lives survive to age 3, then exactly = − =2 2 3 3d l l lives must die between age 2 and age 3.

A number of probabilities can be computed from the entries in such a table. Let’s begin by introducing the standard notation for the most significant types of probabilities that you will see in the actuarial models throughout Chapters 2 – 8.

The probability that a life currently age x will survive n years is denoted n xp . With our deterministic interpretation of the life table, along with the point of view that the probability of an event is the relative frequency with which it occurs, we have:

+=

x nn x

x

lp

l

It is a standard convention to omit the n subscript when 1n = , so the probability that a life currently age x will survive 1 year is:

+= 1xx

x

lpl

For example, the probability that a life age 5 survives for 2 years to age 7 is:

= =72 5

5

968976

lpl


4

The probability that a life currently age x will die within n years is denoted n xq , and we have:

+ +−= − = − =1 1 x n x x n

n x n xx x

l l lq pl l

Intuitively, this is the probability that a life is one of the +−( )x x nl l lives to die between age x and age +x n , out of the xl lives age x .

For example, the probability that a life age 1 dies within 3 years is:

− −= = =1 4

3 11

991 979 12991 991

l lql

Again, we omit the n subscript when 1n = , so the probability that a life currently age x will die within 1 year is:

+−= =1 or x x x

x xx x

l l dq ql l

Finally, the probability that a life currently age x will survive for m years and then die within the following n years is denoted xm nq , and we have:

+ + +−=

x m x m nxm n

x

l lq

l

Intuitively, xm nq is the probability that a life age x survives for m years, multiplied by the

probability that a life age +x m dies within n years:

+ + + + + + ++

+

− −= × = × =x m x m x m n x m x m n

x m x n x mm nx x m x

l l l l lq p ql l l

Again, we omit the n subscript when 1n = , so the probability that a life currently age x will survive for m years and then die within 1 year is:

+ + + +−= =

1 or x m x m x mx xm m

x x

l l dq ql l

For example, the probability that a life age 4 survives for 3 years and then dies within the following 2 years is:

− −

= = =7 943 2

4

968 959 9979 979

l lq

l

Example 1.1

Compute the following probabilities from the life table above:

(a) 5 0p

(b) 05 q

(c) 14 2 q

(d) 1p

(e) 2q


5

Solution

(a) = =55 0

0

9761,000

lpl

(b) = =505

0

41,000

dql

(c) − −= = =5 7

14 21

976 968 8991 991

l lql

(d) = =21

1

985991

lpl

(e) = =22

2

3985

dql

♦♦

We can also take a stochastic (ie random) view of future mortality. Under a stochastic approach, the number of survivors at age x is a random variable ( )L x , and the life table function xl represents the expected number of survivors, ie:

( )xl E L x=

The random variable ( )L x follows a binomial distribution. Each life is viewed as an independent Bernoulli trial. The number of trials is 0n l= . We define “success” as survival to age x , with probability = 0xp p .

For example, let’s consider the distribution of ( )5L , the random number of survivors at age 5 in this life table from the 1,000 newborn lives.

The random variable ( )5L follows a binomial distribution with = =0 1,000n l trials and a probability of success of 5 0 976 /1,000 0.976p p= = = .

The mean and variance of ( )5L are:

( )

( )( )

= = × = × = =

= = × × = × × =

0 5 0 5

5 0 5 0

5 1,000 0.976 976 ( )

var 5 1,000 0.976 0.024 23.424

E L np l p l

L npq n p q

Finally, we can deduce little from the life table about the continuous random lifetime X , but we can identify the distribution of the curtate lifetime of a newborn, (0)K .

The probability =Pr( (0) )K k is the probability that a newborn life dies in the age range +[ , 1)k k , which we have already defined as 0k q . Hence:

( )= = =00

Pr (0) kk

dK k ql

For example, the probability that a newborn life has a curtate lifetime of 3 years is:

( )= = = = =303

0

3Pr (0) 3 0.0031,000

dK ql


6

1.3 The theory of continuous survival models

In this section we will study five different mathematical functions that can all be employed to specify the distribution of X , the random lifetime (ie age at death) of a newborn life:

• the cumulative distribution function of X

• the probability density function of X

• the survival function

• the life table function

• the force of mortality.

We will focus on the relations between these functions as well as their meaning.

The pdf and cdf of the random lifetime The random lifetime (ie age at death) of a newborn life, X , is assumed to be a continuous random variable. We will review the basic properties of continuous random variables and explain their interpretation in the context of the random lifetime.

Let’s begin with the cumulative distribution function (cdf):

= ≤( ) Pr( )XF x X x

The cdf ( )XF x represents the probability that a newborn life will die at or before age x .

( )XF x is continuous and non-decreasing with ( )0 0XF = and ( ) 1XF ω = where ω is the first age at which death is certain to have occurred for a newborn life.

The probability density function (pdf) is:

′=( ) ( )X Xf x F x wherever the derivative exists

The pdf ( )Xf x is non-negative and continuous on the interval [0 , )ω .

Recall that a value of ( )Xf x is not a probability in itself. The probability that a newborn life dies between ages a and b is:

( )≤ ≤ = = −∫Pr ( ) ( ) ( )b

Xaa X b f x dx F b F a

You should note that since X is assumed to be a continuous random variable, all of the intervals ,a b , [ , )a b , ( , ]a b , and ( ),a b have the same probability, ( ) ( )F b F a− .

Hence:

( )0

1Xf x dxω

=∫

and:

( ) ( ) ( )0

Prx

X XF x X x f u du= ≤ = ∫

Finally, the probability that a newborn life dies in the interval + ∆[ , ]x x x can be estimated as:

( )≤ ≤ + ∆ ≈ ⋅ ∆Pr ( )Xx X x x f x x


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Example 1.2

Suppose that the lifetime X of a newborn life is uniformly distributed on the interval 0 ,100 .

(a) Identify the probability density function.

(b) Identify the cumulative distribution function.

(c) Calculate the probability of death occurring between ages 60 and 80.

Solution

(a) The pdf for a uniform distribution is constant and equal to the reciprocal length of the interval:

( ) = = ≤ ≤1 0.01 for 0 100

100Xf x x

(b) The cdf is:

( ) ( )= = = ≤ ≤∫ ∫0 00.01 0.01 for 0 100

x xX XF x f u du du x x

(c) The probability of death between ages 60 and 80 is:

( ) ( ) ( )≤ ≤ = − = − =Pr 60 80 80 60 0.01(80 60) 0.20X XX F F

Note that the uniform distribution is not particularly well suited as a model of human mortality, but it is useful as a simple context to illustrate the theory. This mortality model is commonly known as de Moivre’s law. It was actually the first mortality model to be used in insurance practice. ♦♦

Example 1.3

Suppose that the lifetime X of a newborn life is exponentially distributed with mean 75 years.

(a) Identify the probability density function.

(b) Identify the cumulative distribution function.

(c) Calculate the probability of death between ages 60 and 80.

Solution

The pdf for an exponential distribution with mean θ is ( ) θθ

−= >/1 for 0xXf x e x . Hence:

(a) The pdf is:

( ) −= >/751 for 075

xXf x e x

(b) The cdf is:

( ) ( )− − −= = − = −∫ /75 /75 /750 0

1 175

xx u u xXF x e du e e

(c) The probability of death between ages 60 and 80 is:

( ) ( ) ( ) − −≤ ≤ = − = − =60/75 80 /75Pr 60 80 80 60 0.10518X XX F F e e ♦♦


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The survival function

In actuarial mathematics it is common to describe a survival model by giving the survival function rather that the density function or distribution function. The survival function is denoted ( )Xs x and is defined as:

( ) ( )PrXs x X x= >

The survival function gives the probability that a newborn life dies after age x . This is the same as saying that the newborn survives to age x , or is alive at age x .

From the preceding discussion of the lifetime variable X , we can deduce the following properties of the survival function.

Key properties of the survival function

1. ( )Xs x is continuous and non-increasing with ( )0 1Xs = and ( ) 0Xs ω =

2. ( ) ( )1X Xs x F x= −

3. ( ) ( ) ( ) ( )Prb

X X Xaa X b f x dx s a s b≤ ≤ = = −∫

4. ( ) ( )X Xf x s x′= −

Example 1.4

Suppose that the lifetime X of a newborn is exponentially distributed with mean 75 years.

(a) Identify the survival function ( )Xs x .

(b) Calculate the probability that a newborn is still alive at age 100.

(c) Calculate the probability that a newborn dies between ages 60 and 75.

Solution

(a) In Example 1.3 we saw that the cdf is ( ) /751 xXF x e−= − . So, we have:

( ) ( ) −= − = >/751 for 0xX Xs x F x e x

(b) The probability a newborn is still alive at age 100 is:

( ) 100/75100 0.26360Xs e−= =

(c) The probability a newborn dies between ages 60 and 75 is:

( ) − −− = − =60/75 75/7560 (75) 0.08145X Xs s e e ♦♦


9

The life table function

In contrast with the discussion of the discrete case in Section 1.2, here we will define the life table function xl for all ages between 0 and w .

As before, let ( )L x denote the random number of survivors at any age x from a group of 0l

newborn lives. The random variable ( )L x follows a binomial distribution with 0n l= trials. We

define “success” as survival to age x , with probability ( ) ( )Pr Xp X x s x= > = .

The life table function xl is defined as the expected number of survivors at age x . Hence:

( ) ( )0x Xl E L x np l s x= = =

Example 1.5

Suppose that the lifetime X of a newborn is uniformly distributed on 0 ,100 .

(a) Identify the survival function ( )Xs x .

(b) Identify the life table function xl if 0 100l = .

Solution

(a) The survival function is given by:

( ) ( ) ( )

( )

= > = =

= − = − ≤ ≤

∫ ∫100 100

Pr 0.01

0.01 100 1 0.01 for 0 100

X Xx xs x X x f u du du

x x x

(b) The life table function is:

= = − = − ≤ ≤0 ( ) 100(1 0.01 ) 100 for 0 100x Xl l s x x x x ♦♦

Let’s summarize some of the key properties of the life table function, xl .

Key properties of the life table function

1. xl is the expected number of survivors at age x from a group of 0l newborn lives

2. ( )0x Xl l s x= is continuous and non-increasing with ω =0l

3. ( )0

xX

ls xl

=

Note that the value of 0l (sometimes called the radix of the life table) is not important to the survival model, since (by property 3) the survival function is independent of this quantity. So, we can choose the value of 0l for convenience. The survival function will be identical whether we choose =0 100l or =0 1,000,000l .


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Example 1.6

Suppose that the life table function is ( )210,000 100 for 0 100xl x x= − ≤ ≤ . Identify the cumulative distribution function and the probability density function for the associated lifetime variable X .

Solution

It is elementary to compute the survival function from the life table by property 3:

( ) ( ) ( )− −= = = ≤ ≤

×

2 2

2 20

10,000 100 100 for 0 100

10,000 100 100x

Xx xls x x

l

We can then calculate ( )XF x using the relationship ( ) ( )1X XF x s x= − :

( ) ( )−= − ≤ ≤

2

2100

1 for 0 100100X

xF x x

Finally, we can calculate ( )Xf x using the relationship ( ) ( )X Xf x F x′= :

( ) ( ) ( ) ( )′ − − − ′= = − = = ≤ ≤

2

2 2100 2 100 1001 for 0 100

5,000100 100X X

x x xf x F x x ♦♦

Example 1.7

Suppose that there are 1,000 newborn lives whose lifetime follows the survival model given in Example 1.6. Determine the interval that lies within two standard deviations either side of the mean for ( )10L , the random number of survivors at age 10.

Solution

( )10L follows a binomial distribution with:

= =0 1,000n l

( ) ( )−= = =

2

2100 10

10 0.81100Xp s

So we have:

( )( )( )

( )σ

= = × = = = × × − =

= =10

10 1,000 0.81 810

var 10 1,000 0.81 (1 0.81) 153.90

153.90 12.406L

E L np

L npq

Hence the required interval is:

− × + × = 810 2 12.406, 810 2 12.406 785.19, 834.81 ♦♦


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The force of mortality

We can also specify a survival model in terms of the force of mortality.

The force of the mortality is denoted ( )xµ . It is an instantaneous measure of mortality at age x , and it can be defined in several equivalent ways:

( ) ( )( )

( )( )

( )( )( )lnX X xX

X X x

f x s x lx s xs x s x l

µ′ ′′= = − = − = −

These equalities can be verified using simple calculus.

For example, using the information in Example 1.6, we have:

( ) ( )( )

( )( )

2

2 22 100 /100 2 for 0 100

100100 /100X

X

xs xx x

s x xxµ

− −′= − = − = ≤ <

−−

Or, using the information in Example 1.4, we have:

( ) ( )( )( ) 1ln for 075 75Xxx s x xµ

′ ′= − = − − = ≥

Let’s now see how to calculate the survival function from the force of mortality.

( ) ( )( )( )

( ) ( )( ) ( )( ) ( )( )0 0

ln

ln ln ln 0

X

xxX X X

x s x

y dy s y s x s

µ

µ

′= −

⇒ = − = − +∫

But since ( ) ( )= =0 1 and ln 1 0Xs , we have:

( ) ( )( )

( ) ( )

µ

µ

= −

⇒ = −

∫

∫

0

0

ln

exp

xX

xX

y dy s x

s x y dy

Example 1.8

Suppose that the force of mortality for a survival model is given by the formula:

( ) 0.9 for 0 9090

x xx

µ = ≤ <−

Calculate the survival function.

Solution

The survival function is calculated as:

( ) ( )

( )

0 0

0

0.9

0.9exp exp90

90exp 0.9 ln 90 exp 0.9 ln90

90 for 0 9090

x xX

x

s x y dy dyy

xy

x x

µ = − = − −

− = − =

− = ≤ <

∫ ∫

♦♦


12

Since we have already studied simple relationships between the survival function, the life table function, and the pdf and cdf of the lifetime function, we can easily calculate any of these functions from the force of mortality.

For example, using the information in Example 1.8, we can calculate the life table function (with 0 1,000l = ) as:

− = = ≤ <

0.9

090( ) 1,000 for 0 90

90x Xxl l s x x

It is clear from the definition that the force of mortality is not a probability, so how should it be interpreted? In order to understand the meaning of ( )xµ , it is useful to rewrite the defining formula in the form:

( ) ( ) ( )X Xf x s x xµ=

Now, recall that:

( ) ( )∆ ≈ ≤ ≤ + ∆PrXf x x x X x x

Rewriting the probability term using a conditional probability, we have:

( ) ( )( ) ( )

( )

∆ ≈ ≤ ≤ + ∆

= ≤ + ∆ ≥ ≥

= ≤ + ∆ ≥

Pr

Pr Pr

Pr ( )

X

X

f x x x X x x

X x x X x X x

X x x X x s x

Substituting ( ) ( ) ( )X Xf x s x xµ= , we have:

( ) ( ) ( )

( ) ( )

Pr ( )

Pr

X Xs x x x X x x X x s x

x x X x x X x

µ

µ

∆ ≈ ≤ + ∆ ≥

⇒ ∆ ≈ ≤ + ∆ ≥

So, ( )µ ∆x x is approximately equal to the conditional probability that a newborn that has

survived to age x subsequently dies during the next ∆x years. For example, ( )20µ multiplied by 1/365x∆ = (a day), is approximately equal to the conditional probability that a newborn that has survived to exact age 20 will then die during the next day.

Example 1.9


( ) 0.9 for 0 9090

x xx

µ = ≤ <−

Calculate the approximate probability that a life age 40 dies within the next week.

Solution

Setting ∆ =7 /365x , the required probability is:

( )µ ∆ = × =−

0.9 740 0.0003590 40 365

x ♦♦


13

Let’s summarize the main properties of the force of mortality.

Key properties of the force of mortality

1. ( ) ( )( )

( )( )

( )( )( )lnX X xX

X X x

f x s x lx s xs x s x l

µ′ ′′= = − = − = −

2. ( ) ( )0

expx

Xs x y dyµ = − ∫

3. ( ) ( )Pr |x x X x x X xµ ⋅ ∆ ≈ ≤ + ∆ ≥

4. ( )xµ is non-negative and piece-wise continuous where defined

5. ( )0

y dyω

µ = ∞∫ in order that ( ) 0Xs ω =

Standard probabilities in a continuous survival model

Let’s now reconsider the ideas we met in Section 1.2 (in the context of a discrete life table) in the form of a continuous algebraic function defined for all x in 0 ,ω .

First, the probability that a life currently age x will survive t years is:

( )

( )( )

( ) ( )+ + > += = = = > + >

>

PrPr |

PrXx t

t xx X

s x t X x tlp X x t X xl s x X x

The probability that a life currently age x will die within the next t years is:

( ) ( )

( ) ( )+ > − > +−= = = ≤ + >

>

Pr PrPr |

Prx x t

t xx

X x X x tl lq X x t X x

l X x

Finally, the probability that a life currently age x will survive s years but die within the following t years is:

( )+ + +−= = + < ≤ + + >Pr |x s x s t

xs tx

l lq x s X x s t X x

l

These three functions are defined for all ages x in 0 ,ω and for 0 t xω≤ ≤ − .

It should be emphasized now that all of these probabilities are conditional, ie we are given that a newborn has survived to age x.

The symbol ( )x is commonly used to denote a newborn life that has survived to age x . So, 5 xp is the probability that ( )x will still be alive in 5 years’ time, at age + 5x , and 5 xq is the probability that ( )x will die within the next 5 years.

As in Section 1.2, the general convention is to drop the subscript t from the symbol when 1t = . So, for example, 3| xq is the probability that ( )x will die between ages 3x + and 4x + .


14

Example 1.10


( ) 0.9 for 0 9090

x xx

µ = ≤ <−

Compute the following probabilities:

(a) 2.5 20p

(b) 202.5q

(c) 202.5|q

Solution

Note that this is the force of mortality in Example 1.8. The life table function is:

− = = ≤ <

0.9

0 090( ) for 0 90

90x Xxl l s x l x

Recall that we can choose any convenient value of 0l without changing the distribution of X . So

let’s simplify our computations by choosing = 0.90 90l , which gives:

= = − ≤ <0.90 ( ) (90 ) for 0 90x Xl l s x x x

We can now calculate the required probabilities as follows.

(a) ( )( )

−= = =

−

0.922.5

2.5 20 0.920

90 22.50.96780

90 20

lpl

(b) −

= = − =20 22.52.5 20 2.5 20

201 0.03220

l lq p

l

(c) − −

= = =0.9 0.9

22.5 23.52.5 20 0.9

20

67.5 66.5| 0.01291

70l l

ql

♦♦

The challenge in dealing with these standard probabilities is that there are so many relationships that involve them. The key relations are listed below without proof. Most of the proofs rely on simple probability theory – you may like to attempt them to improve your understanding.

Key relations concerning standard probabilities

1. + = 1t x t xp q

2. + += ⋅s t x s x t x sp p p

3. |s t x s x t x s s x s t x s t x s xq p q p p q q+ + += ⋅ = − = −

4. 1 1 when is an integern x x x x np p p p n+ + −= ⋅ ⋅ ⋅

5. 0 1 1| | | when is an integern x x x n xq q q q n−= + + +


15

For example, if the probability that ( )20 survives for 10 years is 0.97, and if the probability that

( )30 survives for 10 years is 0.95, then the probability that ( )20 is still alive at age 40 is:

20 20 10 20 10 30 0.97 0.95 0.92150p p p= ⋅ = × =

Or, the probability that ( )20 dies between ages 30 and 40 is:

( )10 10 20 10 20 10 30| 0.97 1 0.95 0.04850q p q= ⋅ = − =

On the other hand, if = = =0 1 20.99 , 0.98 , and 0.97p p p , then the probability that a newborn dies within three years is:

3 0 3 0 0 1 21 1 1 0.99 0.98 0.97 0.05891q p p p p= − = − ⋅ ⋅ = − × × =

Or the probability that a newborn dies during the second year of life is:

( )1 0 0 1| 0.99 1 0.98 0.01980q p q= ⋅ = × − =

Relations 4 and 5 are useful in constructing a discrete life table for human lives. A statistical study conducted over a time span of several years could be used to produce estimates of the mortality rates 0 1 2, ,q q q , … and so on. Values of xl at whole number ages can then be produced as follows:

( )( ) ( )0 0 0 0 0 1 1 0 0 1 10

1 1 1nn n n n n

lp l l p l p p p l q q ql − −= ⇒ = ⋅ = ⋅ ⋅ ⋅ ⋅ = − − −

1.4 The continuous future lifetime after age x

Let the continuous random variable X again denote the random lifetime of a newborn. Now suppose that we are given that a newborn has survived to age x , that is, X x> . The future time lived after age x is X x− .

The conditional distribution of the time lived after age x , given survival to age x, is:

( ) |T x X x X x= − >

The continuous random variable ( )T x is a survival model defined on the interval ω −[0 , ]x . As such, it can also be specified in the same ways that we specified the survival model for a newborn life. It should be clear that the distribution of ( )T x is closely related to the distribution of X .

The quickest way to see the relation between the distributions of ( )T x and X is to calculate the

survival function for ( )T x , ( ) ( ) ( ) ( )( )PrTT xs t s t T x t= = > . In fact, we have already computed

this survival function in terms of the distribution of X , since the event ( )T x t> is equivalent to saying that ( )x is alive at age +x t . The probability of this event is simply t xp . So, we have:

( ) ( )( ) ( )( )

+ += > = = = = 0( ) Pr since ( )Xx t

t x x XT xx X

s x tls t T x t p l l s xl s x

Note: When there is no ambiguity, we will write T for ( )T x . However, subscripts are often

important, for example to distinguish ( )20Xs , the probability that a newborn survives to age 20,

from ( ) ( )10 20Ts , the probability that ( )10 survives to age 30.


16

Example 1.11

Suppose that the life table function is ( )= − ≤ ≤210,000 100 for 0 100xl x x .

(a) Compute the survival function for newborn lives.

(b) Compute the survival function for lives currently aged 20.

Solution

(a) The survival function for newborn lives, ( )Xs x , is:

( ) ( )− − = = = = ≤ ≤ −

2 2

0 20

10,000 100 100for 0 100

10010,000( 100 0)x

X xx xls x p x

l

(b) The survival function for lives currently aged 20, (20)( )Ts t , is:

( ) ( ) + − + − = = = = ≤ ≤ −

2220

2020 220

10,000(100 (20 )) 80 for 0 808010,000(100 20)

ttT

tl ts t p tl

♦♦

When we deal with the future lifetime after age x , we’ll frequently see expressions of the type +x tl and µ +( )x t . In these expressions the value of x is fixed, and the value of t is allowed to

vary so that we can view the expressions as being functions of t .

For example, let’s see how to relate the pdf and cdf for the distributions of X and T .

( ) ( )( ) ( )

( )( )

( ) ( )( )

Pr Pr |

PrPr 1

T

X X

X

F t T x t X x t X x

x X x t F x t F xX x F x

= ≤ = − ≤ >

< ≤ + + −= =

> −

( ) ( )( ) ( ) ( )( )

( )( )

( )( )

+ − + +′= = = = − − 1 1X X X X

T TX X X

F x t F x f x t f x tdf t F tdt F x F x s x

In the figure below, we have areas ( ) ( )X XA F x t F x= + − and ( )+ = XA B s x .

x x+t ω

Area Area = A = B

fX(x)

Notice that we have:

( ) ( ) ( )( )

+ −= =

+X X

TX

F x t F x AF ts x A B

If we view the age x as being fixed, then as the value of t increases, the value of A increases, the value of B decreases, while the sum +A B remains constant.


17

Furthermore, if we examine the relation:

( ) ( )( )

ω+

= ≤ ≤ − for 0XT

X

f x tf t t x

s x

in light of the figure above, we can see how the graph of the pdf of T is related to the graph of the pdf of X .

If we take the portion of the graph of ( )Xf x to the right of x , divide by the total area under the

remainder of the graph, ( )XA B s x+ = , and then relabel the horizontal axis as t and the vertical

axis as ( )Tf t , we have a graph of the pdf of T :

0 t ω-x

Area = Area = A / sX(x) B / sX(x)

fT(t)

One final point worth noting is the similarity between the pdf’s for the distributions of X and T .

For X we have:

( ) ( )( )

( ) ( ) ( ) ( )µ µ µ= ⇒ = = 0X

X X xX

f xx f x s x x p x

s x

and for T we have:

( ) ( )( )

( ) ( ) ( )µ µ

µ++ + += = = = +0 0

0 0

X x t x t xT t x

X x x

f x t p x t p p x tf t p x t

s x p p

Example 1.12

Suppose that the life table function is ( )210,000 100 for 0 100xl x x= − ≤ ≤ .

(a) Compute the distribution function for the future lifetime of a life aged 20.

(b) Compute the density function for the future lifetime of a life aged 20.

Solution

(a) In Example 1.11 we computed:

( ) ( ) − = = ≤ ≤

2

20 2080 for 0 80

80t Ttp s t t

Hence, the distribution function for ( )20T is:

( ) ( ) ( ) ( ) − = − = − ≤ ≤

2

20 20801 1 for 0 80

80T TtF t s t t


18

(b) We have two options for calculating the density function.

The first option is to differentiate the distribution function:

( ) ( ) ( ) ( )( ) ( )′ −−′ = = − =

−= ≤ ≤

2

20 20 280801 2

80 8080 for 0 803, 200

T Tttf t F t

t t

The other option is to use the relation:

( ) ( )µ= +T t xf t p x t

It is consistent with the formula for xl that:

( ) 2100

xx

µ =−

Hence, we have:

( ) ( )2

(20) 2080 2 8020

80 80 3, 200T tt tf t p t

tµ − − = + = ⋅ = −

♦♦

Key results concerning the relation of the distributions of X and T(x)

1. ( ) ( )( ) ( )( )

Pr Xx tT t x

x X

s x tls t T x t pl s x+ +

= > = = =

2. ( ) ( ) ( ) ( )( )

1 X XT t x T

X

F x t F xF t q s t

s x+ −

= = − =

3. ( ) ( )( ) ( )X

T t xX

f x tf t p x t

s xµ

+= = +

1.5 The curtate future lifetime after age x

In addition to computing the distribution of the continuous future lifetime ( )T x , we may also wish to derive the distribution of the curtate future lifetime after age x .

The curtate lifetime is a discrete random variable that is defined by:

( ) ( )= the integer part (or greatest integer) of ( )K x T x ie T x

Since it is a function of ( )T x , it is simple to calculate the probability function of ( )K x from what

we know about ( )T x . The possible values of ( )K x are the numbers ω − −0,1,2, , 1x .

For example, if 70x = and 90ω = , then the possible values of ( )70K are the twenty whole

numbers 0 through 19. If the life ( )70 eventually dies at age 85.8, then the continuous future

lifetime is =(70) 15.8T and the curtate lifetime is [ ](70) 15.8 15K = = .


19

The key observation is that if ( ) =K x k , then we must have:

≤ < +( ) 1k T x k

This leads to the following formula for the probability function:

( ) ( )( )Pr ( ) Pr 1

| for 0 ,1 ,2, , 1x kk x

x

K x k k T x k

dq k xl

ω+

= = ≤ < +

= = = − −

Example 1.13

Suppose that the life table function is given by the formula:

= − ≤ ≤100 for 0 100xl x x

Compute the probability function for ( )75K .

Solution

The probability function for ( )75K is:

( )( )

( ) ( )( )

+ + ++ −= = =

− − − − − −= =

−

75 75 175

75 75Pr 75

100 75 100 75 1 1100 75 25

k kk l ldK kl l

k k

So ( )75K has 25 possible values ( 0,1, 2, , 24 ) that are equally likely to occur. ♦♦

Example 1.14


0.0151,000 for 0xxl e x−= ≤ < ∞

Compute the probability function for ( )75K .

Solution

The probability function for ( )75K is:

( )( )

( ) ( )( )

75 75 175

75 75

0.015 75 10.015 750.015 0.015

0.015 75

Pr 75

1

k kk

kkk

l ldK kl l

e e e ee

+ + ++

− + +− +− −

− ×

−= = =

−= = −

Note that this is a geometric distribution. ♦♦

It is also useful to develop formulas for the cumulative distribution function and survival function of the curtate future lifetime.


20

Recall that for any random variable = ≤( ) Pr( )XF x X x , hence:

( ) ( )( ) ( )( ) ( )( ) ( )( )11 2

1

( ) Pr Pr 0 Pr 1 Pr

for 0 ,1 , , 1

K x

x x kx x x x k

x x x x x

k x

F k K x k K x K x K x k

l ld d d dl l l l l

q k xω

+ ++ + +

+

= ≤ = = + = + + =

−= + + + + =

= = − −

The survival function of the curtate future lifetime is then easily derived as:

( ) ( ) ( )( ) ( ) ( ) ω+ += > = − = − = = − −1 1Pr 1 1 for 0 ,1 ,... , 1k x k xK x K xs k K x k F k q p k x

Example 1.15


= − ≤ ≤100 for 0 100xl x x

Compute the survival function for ( )75K .

Solution

The survival function for ( )75K is:

( ) ( ) + ++

− + + −= = = = =

−75 1

1 757575

100 (75 1) 24 for 0 ,1 , , 24100 75 25

kkK

kl ks k p kl

♦♦

Example 1.16


0.0151,000 for 0xxl e x−= ≤ < ∞

Compute the survival function for ( )75K .

Solution

The survival function for ( )75K is:

( ) ( ) ( )

− + +− ++ +

+ −= = = = =0.015(75 1)

0.015 175 11 7575 0.015(75)

75for 0 ,1 ,2 ,

kkk

kKl es k p e k

l e ♦♦

Let’s conclude this section with a summary of the key relations concerning the curtate future lifetime, ( )K x .

Key relations concerning the curtate future lifetime

1. ( ) ( ) ( )Pr | for 0 ,1 ,... , 1x kk xK x

x

df k K k q k xl

ω+= = = = = − −

2. ( ) ( ) 11 for 0 ,1 ,... , 1x k

k xK xx

ls k p k x

lω+ +

+= = = − −

3. ( ) ( ) 1 11 for 0 ,1 ,... , 1k x k xK xF k q p k xω+ += = − = − −


21

1.6 Important life table functions

Now that we have developed the basic properties of X , ( )T x , and ( )K x , it is time to study additional features of these distributions, such as the life expectancy for a newborn.

The life table functions xL and xT

The functions xL and xT are useful devices in the calculation of life expectancy. They are defined in terms of the life table function, xl , as follows:

ω

ω

+

+ −

=

= = + + +

∫

∫

1

1 1

xx yx

x y x xx

L l dy

T l dy L L L

These functions have interpretations in terms of the aggregate future lifetime of a group of lives that die exactly as scheduled in the life table (ie we take a deterministic view of the life table).

Consider a brief time interval + ∆[ , ]y y y that is part of the interval ω[ , ]x . At the start of this brief period there are yl survivors. We can estimate the total people-years lived by the survivors

during this brief period by yl y∆ . This approximation ignores the possibility that anyone dies in the short time available.

If we now sum these people-years lived over a set of disjoint sub-intervals of length y∆

comprising the age interval ω[ , ]x , we will have a Riemann sum for the integral yxl dy

ω∫ .

This Riemann sum can be interpreted as an approximation to the total number of people-years lived after age x by the survivors to age x .

Taking a limit as y∆ goes to zero, then we have the integral:

x yxT l dy

ω= ∫

which can be interpreted as the total people-years lived after age x by the survivors to age x . Beware of confusing xT with ( )T x , the random future lifetime of a single life age x.

We can break the interval ω[ , ]x into subintervals of length one year, ie +[ , 1]x x , ω ω+ + −[ 1, 2], , [ 1, ]x x . The function xL is calculated over just one of these one-year periods:

+

= ∫1x

x yxL l dy

which is the number of people-years lived by the survivors to age x during the next year.

Example 1.17

Compute the function xT for the following life table functions:

(a) 1,000 10 for 0 100xl x x= − ≤ ≤

(b) 0.0151,000 for 0xxl e x−= ≤ < ∞


22

Solution

(a) We have:

( ) ( ) ( )

1002 2100 21,000 10 1,000 101,000 10 5 100

20 20x xx

y xT y dy x

− − = − = − = = −

∫

(b) We have:

0.015 0.015

0.015 1,000 1,0001,000

0.015 0.015

y xy

x xx

e eT e dy

∞− −∞ − = = − =

∫ ♦♦

Complete life expectancy

Now let’s develop some formulas to calculate life expectancies. Before we do that, though, it’s useful to develop an alternative formula for expected value calculations. The work we do now will make the subsequent derivations simpler.

Let’s assume that X is a continuous, positive-valued random variable whose mean and variance both exist. The usual formula for calculating the expected value is:

[ ] ( )0 XE X x f x dx∞

= ∫

We can develop an alternative formula using integration by parts:

( ) ( )

[ ] ( ) ( )( ) ( )

( ) ( ) ( )

00 0

0 0

, ,

lim 0

X X

X X X

X X Xx

u x dv f x dx du dx v s x

E X x f x dx x s x s x dx

x s x s x dx s x dx

∞ ∞∞

∞ ∞

→∞

= = ⇒ = = −

= = − +

= − + + =

∫ ∫

∫ ∫

In the above calculation, the limit is zero. Apply L’Hopital’s rule writing ( )Xx s x as ( ) 1/Xs x x− .

Then use the fact that ( )2lim 0x

x f x→∞

= if 2E X exists.

In other words, we can calculate [ ]E X as the area under the graph of the survival function.

If the density function is supported on the finite interval ω[0, ] , then we can replace ∞ in the above integrals by ω , since both of the functions ( )Xf x and ( )Xs x are zero for x ω> .

So we have:

( )ω

= ∫0[ ] XE X s x dx

Another type of expected value that occurs in both survival model theory and loss distribution theory is the limited expected value.

Let’s define X n∧ as the minimum of the random variable X and the number n :

{ }if

min ,if

X X nX n X n

n X n≤

∧ = = >


23

Since X n∧ is a function of X , we can compute its expected value as above with limits 0 and n:

( ) ( ) ( ) ( )

( )( ) ( ) ( ) ( )

0 0

0 0 0

n nX X X Xn

n nnX X X X

E X n x f x dx n f x dx x f x dx ns n

x s x s x dx ns n s x dx

∞∧ = + = +

= − + + =

∫ ∫ ∫

∫ ∫

Now we’re ready to calculate the two types of life expectancy: complete and curtate.

The complete expected future lifetime at age x is denoted by xe , and is defined as:

( )xe E T x=

Using results derived earlier in this section, we can develop several methods to calculate this expected value:

( ) ( ) ( )

( )

( )

ω

ω

ω ωω

−

−

−+− +

= =

=

= = = = = +

∫

∫

∫ ∫∫

0

0

00

(integration by parts)

substitute

xx T x

xT

xx t yx xx t x

x x x x

e E T x t f t dt

s t dt

l dt l dyl Tdt y x tl l l l

So, life expectancy can be computed in three distinct ways, as:

• the standard expected value of the continuous random variable ( )T x

• the area beneath the graph of the survival function

• the ratio of people-years lived after age x by the group to the number of people in the group at age x (an average number of years lived by a member of the group of survivors at age x ).

Example 1.18

Compute the complete life expectancy at age x for the following life table functions:

(a) = − ≤ ≤1,000 10 for 0 100xl x x

(b) −= ≤ < ∞0.0151,000 for 0xxl e x

Solution

Using the results from Example 1.17, we have:

(a) ( )( )21,000 10 20 100

1,000 10 2x

xx

x xTel x

− −= = =

−

(b) 0.015

0.0151,000 /0.015 1 66.67

0.0151,000

xx

x xx

T eel e

−

−= = = = ♦♦


24

The temporary complete life expectancy at age x is denoted :x ne , and is used to represent the

expected number of years lived by ( )x in the next n years.

The random number of years lived by ( )x in the next n years is:

( ) ( ){ } ( ) ( )( )

if (death within years)min ,

if (survives years)T x T x n n

T x n T x nn T x n n

≤∧ = = >

As a result, we have the following:

( )

( ) ( )

( )+

+++

= ∧

=

−= = = = = +

∫

∫ ∫∫

:

0

00

substitute

x n

nT x

n x nx t yn x x x nx t

x x x x

e E T x n

s t dt

l dt l dy T Tl dt y x tl l l l

Example 1.19

Compute the 10-year temporary complete life expectancy at age 50 for the following life table functions:

(a) = − ≤ ≤1,000 10 for 0 100xl x x

(b) −= ≤ < ∞0.0151,000 for 0xxl e x

Solution

Using the results from Example 1.17, we have:

(a) ( )−−

= = =2 2

50 6050:10

50

500 400 /209

500T T

el

(b) ( )− × − ×

− ×

−−= = =

0.015 50 0.015 6050 60

50:10 0.015 5050

1,000 1,000 /0.0159.280

1,000

e eT Te

l e ♦♦

Key formulas for complete life expectancy

1. ( ) ( ) ( ) ( ) ( ) ( )0 0

where x x x

x T t xT x T xx

Te E T x t f t dt s t dt s t pl

ω ω− −= = = = = ∫ ∫

2. ( ) ( ) ( ) ( ): 0 where

n x x nT t xT xx n

x

T Te E T x n s t dt s t p

l+−

= ∧ = = = ∫

3. :x n x x nx ne e p e += +

4. ( ) ( ) ( ) ( )2 20 0

2x x

TT xE T x t f t dt t s t dtω ω− − = =

∫ ∫


25

The proof of the third formula goes as follows:

( )+ + + + +

+

+

− + −= = = + ×

= +:

x x n x n x x nx x n x nx

x x x x x n

n x x nx n

T T T T TT l Tel l l l l

e p e

This result has an intuitive interpretation. The life ( )x expects to live :x ne years in the next n

years, and, given survival to age x n+ , expects to live an additional x ne + more years.

The proof of the fourth formula relies on integration by parts with 2u t= and ( ) ( )T xdv f t dt= .

Curtate life expectancy

The curtate expected future lifetime at age x is denoted by xe , and is defined in terms of the curtate future lifetime ( )K x :

( )= xe E K x

So, xe is the expected number of full years (the fractional part of the final year of life is not counted) lived by the life ( )x after age x .

We have:

( ) ( )( )

( )

ω

ω ωω

ωω

ω

− −

=

− − − −+ + + −+

= =

+ + + −+ + + + −

++

= = =

+ + + + − −= ⋅ = =

+ + + += = + + +

∑

∑ ∑

1

0

1 11 2 3 1

0 0

1 2 3 11 1

the survivors at age die at age or later

Pr

2 3 1|

(note:

x

xk

x xx x xx k

k xx xk k

x x xx k x k x k

x x kx k

e E K x k K x k

d d d x dk dk q

l l

l l l ll d d d

l

ω− −= + + + +2 3 1

)

x x x x xp p p p

The temporary curtate life expectancy at age x is denoted :x n

e , and is defined as:

( )= ∧ :x ne E K x n

It is the expected number of full years (the fractional part of the final year of life is not counted) lived by the life ( )x in the next n years. In a similar fashion, we can show that:

( ) 2: x x n xx ne E K x n p p p= ∧ = + + +

There is a recursive relation:

+

+ +

= +

= + = + =

:

1 1:1 (the special case with 1)

x n x x nx n

x x x x x xx

e e p e

e e p e p p e n


26

Example 1.20


100 for 0 100xl x x= − ≤ ≤

(a) Compute the curtate life expectancy for a newborn.

(b) Compute the 10-year temporary curtate life expectancy at age 50.

Solution

(a) The curtate life expectancy for a newborn is:

0 0 2 0 99 0

1 2 99

0

99 98 1 4950 49.5100 100

e p p p

l l ll

= + + +

+ + + + + += = = =

(b) The 10-year temporary curtate life expectancy at age 50 is:

50 2 50 10 5050:10

51 52 60

50

49 48 40 445 8.950 50

e p p p

l l ll

= + + +

+ + + + + += = = = ♦♦

Key formulas for curtate life expectancy

1. ωω

+ + + −− −

+ + + += + + + + = 1 2 3 1

2 3 1x x x

x x x x x xx

l l l le p p p p

l

2. 1 22:

x x x nx x n xx n

x

l l le p p p

l+ + ++ + +

= + + + =

3. :x n x x nx n

e e p e += +

4. ( )2 1 2 33 5x x x

x

l l lE K x

l+ + ++ + + =

The proof of the second moment formula above could be established using summation by parts, a technique that will not be covered since it is not often useful in contingent payment model theory. It can also be proven with some fancy algebraic manipulation starting with the formula for a second moment:

( )1

2 2

01

2

01

2 1

0

Pr

|

x

kx

k xk

xx k x k

xk

E K k K k

k q

l lk

l

ω

ω

ω

− −

=− −

=− −

+ + +

=

= =

=

−= ×

∑

∑

∑


27

The central mortality rate

The -yearn central mortality rate denoted by n xm computes a weighted average of the force of mortality over the range from age x to age x n+ .

It can be calculated in several ways:

( )

( )

µ

+ + + + −

+= = = = =

− − + + +

∫

∫0

1 1:0

//

nt x

n x n x x n x n xn x n

x x n x x x n x x x nx nt x

p x t dt q d l d dmT T l T T L L Lep dt

The symbol n xd is defined as:

+= −n x x x nd l l

ie it is the expected number of deaths in the age range from x to x n+ .

When 1n = we simply write xm instead of 1 xm . In this case, we have:

= xx

x

dmL

We now have three annual measures of mortality at age x : ( )µ , , andx xx q m .

The force of mortality ( )xµ is not a probability, but ( )x xµ ∆ is approximately the probability that a life currently aged x will die in the next instant x∆ .

The probability that a life currently aged x will die within the next year is xq :

= xx

x

dql

where xl is the total number of survivors aged x at the start of the year.

The only difference between xq and xm is the denominator. The quantity xL in the definition of

xm , is often called an exposure, since it represents the total number of people-years lived by the

xl lives age x over the next year (when they are exposed to death).

Example 1.21


( )0.5100xl x= −

Compute 50q , ( )50µ and 50m .

Solution

For 50m we will need the value of 50:1e :

( ) ( )11.5

1 1 1 0.5505050:1 0 0 050

0

1 0.021 0.02 0.99498

1.5 0.02t

ttle p dt dt t dt

l+

− = = = − = − = ×

∫ ∫ ∫


28

Now we have the necessary information to complete the three calculations:

( )( )

( ) ( )( )

0.551

50 0.550

0.550

0.550

5050

50:1

491 1 0.01005

50

100 50 /2 0.550 0.01000100 50100 50

0.01005 0.010100.99498

lql

ll

qm

e

µ−

= − = − =

− −′= − = − = =

−−

= = = ♦♦

The function ( )a x

The function ( )a x is defined as the average part of the final year of life lived by someone who dies between ages x and 1x + . From this verbal description we can see that it is the conditional expected value:

( ) ( )| 1E T x T x ≤

since the event ( ) 1T x ≤ indicates that the life ( )x dies within a year.

Rather than calculating ( )a x by integrating a conditional density, let’s look at an intuitive way.

Consider the xl lives aged x . Let’s split them into the xd lives that die in the next year, and the

1xl + lives that survive the next year. Now consider :1xe , the average time lived in the next year.

• For the lives that die in the next year, the average time lived is ( )a x , by definition.

• For the survivors to age 1x + , the average time lived in the next year is 1.

The number :1xe is a weighted average of time lived in the next year by these two groups, so:

( ) ( ) ( )

( )

+ +

+

+

⋅ + ⋅ ⋅ + ⋅= = = +

+

− −⇒ = =

1 1:1

1

1:1

1 1x x x xx xx

x x x

x x xx

x x

d a x l d a x le q a x p

d l l

e p L la x

q d

For example, if 100xl x= − , we have:

( ) ( )+ + + −= = − = − = − −∫ ∫

2 21 1 1100 100 100 0.5

2x x

x yx x

x xL l dy y dy x

and:

+ − − − − −−= = =1 100 0.5 (100 1)( ) 0.5

1x x

x

x xL la xd

So, in this case the average time of death is midway through the final year of life.


29

1.7 Extending a life table to a continuous survival model

In Sections 1.3 through 1.6 we have developed the theory of continuous survival models, beginning with the distribution of the lifetime of a newborn. Now let’s consider the practical application of this theory.

Suppose that we have a discrete life table that lists the number of survivors xl at whole number ages ω0,1, 2, , , where 0lω = .

How can we compute life table functions such as 2.5 3p , ( )2.5µ , or 25e , which rely on a continuous model?

To calculate these functions, we’ll need to start by extending xl to a continuous model defined for all ages x between 0 and ω . In order to do this, we’ll have to make an assumption about the pattern in which deaths occur between the integer ages.

A common assumption is that the xd deaths that occur between age x and age + 1x are spread uniformly over the year. This is known as the uniform distribution of deaths assumption, or the UDD assumption for short.

The implication of the UDD assumption is that we define the life table function at fractional ages by interpolating linearly between the values at integer ages.

So, if x is an integer age and ≤ ≤0 1t , then we have:

( )

( )+ +

+

= − +

= − − = −

1

1

1 (linear interpolation)x t x x

x x x x x

l t l t l

l t l l l t d

The resulting piecewise linear graph of xl is continuous and non-increasing (usual properties of the life table function). The same must be true of the survival function, ( )Xs x .

lx

0 1 2 3 4 x

•

• •• •

The UDD assumption has several important consequences.

For integer x , and ≤ ≤0 1t , we have:

( ) ( )

1

1

1

x xx tt x x

x x

t x t x x

x x xx t x

x t x x x x x

l t dlp t ql l

q p t q

l t d ql dx tl l t d l t d t q

µ

+

+

+

−= = = − ⋅

= − = ⋅

′−′+ = − = − = =

− − −


30

We can now perform some of the calculations that would have been impossible from discrete data without the UDD assumption. For example, we have:

( )µ−

= = = =− −

5 5 25.5 22.5 3

3 3 2 2 2

0.5 and 2.5

0.5 1 0.5l d ql dp

l l l d q

Example 1.22

Suppose that = = =90 91 920.4, 0.7, and 1.0q q q . Calculate 90e under the UDD assumption.

Solution

A standard method of calculation is to compute the area beneath the graph of the survival function. This can be accomplished with the help of calculus (Method 1 below), or it can be done geometrically (Method 2).

Method 1.

We have:

90 90

91 91

92 92

1 1 0.4 for 0 11 1 0.7 for 0 11 1 for 0 1

t

t

t

p t q t tp t q t tp t q t t

= − ⋅ = − ≤ ≤= − ⋅ = − ≤ ≤= − ⋅ = − ≤ ≤

Hence:

( ) ( ) ( )

( ) ( ) ( )( )

= + = + +

= + + =

= − + − + −

= − + − + − =

∫ ∫ ∫

∫ ∫ ∫

90 90 91 90 91 9290:1 90:1 91:1

1 1 190 90 91 91 92 3 900 0 0

1 1 1

0 0 0

( )

(note that 0)

1 0.4 0.6 1 0.7 0.3 1

1 0.2 0.6 1 0.35 0.3 1 0.5 1.28

t t t

e e p e e p e p e

p dt p p dt p p dt p

t dt t dt t dt

Method 2.

Begin by drawing the graph of the continuous and piecewise linear survival function.

t p90

0 1 2 3 t

The area of a trapezoid whose base is 1 unit long is equal to the height at the midpoint:

( ) ( ) ( )

= =

= + + = − × + − × + × −

=

∫3

90 900

0.5 90 1.5 90 2.5 90

sum of three trapezoidal areas

1 0.4 0.5 0.6 1 0.7 0.5 0.6 0.3 1 0.5

1.28

te p dt

p p p

♦♦


31

Example 1.23

Suppose that = = =90 91 920.4, 0.7, and 1.0q q q . Draw a graph of the pdf for ( )90T under the UDD assumption.

Solution

We should start by looking at the picture of the survival function in the Solution to Example 1.22.

In general, the pdf is equal to the negative derivative of the survival function. This derivative (the tangent slope) is undefined at t equal to 1 and 2 due to the sharp corners on the graph at these locations.

So, we just need to look at the survival function and note the slope on each of the three linear segments. The result is:

( ) ( ) ( )( )

( )

( )

( )( )

9090

90 90 1 91

1 0.4 for 0 1

1.02 0.42 for 1 2

0.54 0.18 for 2 3

note: for 1 2 we have 0.6 1 1 0.7 1.02 0.42

tT

t t

t t

f t p t t

t t

t p p p t t−

′− − < <′ ′= − = − − < <

′− − < <

≤ ≤ = = − − = −

Hence:

( ) ( )< <

= < < < <

90

0.40 for 0 10.42 for 1 20.18 for 2 3

T

tf t t

t

The graph is as shown below:

°ο ο

ο ο

f(t)

t0 1 2 3

ο

Note: A quicker way to calculate the pdf form is to note that:

( ) ( ) ( )| when k xT xf t q k T x= = .

For example, notice in the example above that:

= =

= =

= × × =

0 90 90

1 90 90 91

2 90 90 91 92

| 0.4

| 0.42

| 0.18

q q

q p q

q p p q ♦♦


32

Key formulas for the UDD assumption (if x is an integer age and ≤ ≤0 1t )

1. ( ) 11 (linear interpolation)x t x x x xl t l t l l t d+ += − + = −

2. 1 ,t x x t x xp t q q t q= − ⋅ = ⋅

3. ( )1

xx

x x x

qdx tl t d t q

µ + = =− −

4. ( ) ( ) | for 1k xT xf t q k t k= < < +

5. ( ) 0.5a x = for all integer ages x

6. 0.5x xe e= + for all integer ages x

Here is a brief explanation of the final two results in the summary. The UDD assumption gets its name from the fact that linear interpolation in effect spreads the deaths out uniformly and continuously over each year of age. For example, we assume that in the age interval

+ +[ 0.40, 0.65]x x there will be 25% of the deaths between ages x and 1x + .

With a little effort it can be shown that the fractional part of the final year of life is uniformly distributed on the interval from 0 to 1, regardless of the (integer) age at the start of that year. This should make it easy to see where the final two results come from.

A simple application of the final point provides an alternative solution to Example 1.22. It is easy to see from the given mortality rates that:

90 90 2 90 3 90

90 90

0.6 0.6 0.3 0.6 0.3 0 0.78

UDD 0.5 0.78 0.5 1.28

e p p p

e e

= + + = + × + × × =

⇒ = + = + =

1.8 Select and ultimate life tables

When an individual applies to buy a life insurance policy, there will typically be some sort of medical screening process, either by filling out a questionnaire or visiting a doctor for a physical examination. This protects the insurance company from people who know that they are in poor health and likely to result in a claim on the policy. This process (known as underwriting) acts as a filter. The healthy lives are accepted for the policy, and the unhealthy lives are either not accepted or are issued a policy at a higher premium rate.

As a result, the average mortality of individuals who have just passed the medical screening process will be lower than the mortality of the general population, which includes a mix of healthy and unhealthy lives. It’s important to note that the differential in mortality rates is greatest immediately after the underwriting process, ie as soon as the policy is sold. Within, say, 3 years of the medical screening process, the mortality rates of the policyholders will tend back towards those of the general population as some of the policyholders contract illnesses and their health deteriorates.

The underwriting process has an interesting implication. The mortality of a 40-year-old policyholder who has just bought a policy (and has therefore just passed the medical screening process) should be lower than the mortality of a 40-year-old policyholder who bought his policy several years ago.


33

The lower mortality rates that apply in the years following the “selection” process (ie underwriting) are called select mortality rates. The general rates that apply once the filtering effects of underwriting have worn off are called ultimate mortality rates. The period of time for which the differential exists (in this case 3 years) is called the select period.

Other events can give rise to mortality differentials too, eg giving up smoking. The average mortality of people immediately after giving up smoking is higher than that of the general population. After 10 years, the harmful effects of smoking may have reduced to such an extent that the ex-smokers’ average mortality returns to that of the general population. The select period here is 10 years.

In a general select and ultimate model, the mortality rate xq (integer age x ) for a life subject to the model xl is adjusted to a select mortality rate for several years (the select period) after some event takes place. At the end of the select period, the mortality rate reverts to the ultimate mortality rates.

Let’s now look at the notation with the help of a select and ultimate table.

[ ]x [ ]xl [ ] 1xl + [ ] 2xl + 3xl + 3x +

22 35,628 35,619 35,607 35,596 25

23 35,615 35,604 35,590 35,581 26

24 35,602 35,589 35,575 35,562 27

The select portion of the table refers to the columns headed [ ] [ ] 1 [ ] 2, , and x x xl l l+ + .

The subscript +[ ]x k indicates that a life “selected” at age x is currently age x k+ , ie the life was “selected” k years ago. In these models both x and k are assumed to be whole numbers.

The ultimate portion of the table refers to the column headed 3xl + . The select period is 3 years in length, so we start to use ultimate mortality rates 3 years after the life was selected.

What we have in effect is a distinct survival model for the curtate future lifetime after each integer age at which selection could take place.

For example, if a life is selected at age 23, then the curtate future lifetime for this individual at age 23 is based on the numbers [ ] [ ] [ ]+ + 26 2723 23 1 23 2, , , , ,l l l l l surviving to each whole number age.

In general, we follow the table across the row for the age of selection until we arrive at the ultimate column (at the end of the select period). At that point we continue down the ultimate column.

The age subscript in other life table functions conforms to the +[ ]x k notation. For example, for a life selected at age 23 we would have the following results according to the table given above.

The probability that a newly selected life aged 23 survives one year is:

[ ][ ]

[ ]

+= = =

23 123

23

35,604 0.9996935,615

lp

l

The probability that a newly selected life aged 23 dies within one year is:

[ ]= − = − =[23] 231 1 0.99969 0.00031q p


34

The probability that a life aged 24, selected at age 23 (one year ago) survives one year is:

[ ][ ]

[ ]

23 223 1

23 1

35, 590 0.9996135,604

lp

l+

++

= = =

The probability that a life aged 25, selected at age 23 (two years ago) survives one year is:

[ ][ ]

++

= = =2623 2

23 2

35, 581 0.9997535,590

lpl

The probability that a life aged 26, selected at age 23 (three years ago) survives one year is:

= = =2726

26

35, 562 0.9994735, 581

lpl

We can calculate the curtate life expectancy as:

[ ] [ ] [ ] [ ] [ ] [ ]( ) [ ] ( )= + = + + + + +3 26 2 3 3 26 2 2623 23 23 23 23 2323 :3e e p e p p p p p p

To compute a long-term survival probability such as [ ]25 23p , we must consider the split of 25-

year period between the 3-year select period and the following 22-year ultimate period:

[ ] [ ][ ] [ ]

= = × =26 48 4825 3 22 2623 23

2623 23

l l lp p pl l l

Example 1.24

Suppose we have a select and ultimate model based on the life table 100xl x= − at integer ages x, where the select period is 2 years and the select mortality rates are calculated according to the rule:

[ ] ( )( )1 0.05 2 x kx kq k q ++ = − − for 0 ,1k =

Determine the following values:

(a) [ ]25p

(b) [ ]2 25p

(c) [ ]3 25p

(d) [ ]25 :3e

Solution

This formula indicates that select mortality at the age of selection (ie when 0k = ) is 90% of ultimate mortality at the same age. On the other hand, one year after selection, the select rate is 95% of the ultimate rate at the same age. The effect of selection wears off after 2 years.

Notice first that the ultimate mortality and survival rates at age x are:

1 1 100 1 and 1100 100

x xx x x

x

l l xq p ql x x

+− − −= = = − =

− −


35

As a result, we have:

(a) [ ] [ ] = − = − = − =

2525 2511 1 0.9 1 0.9 0.98800

75p q q

(b) [ ] [ ] [ ]+ + = − = − = − = ⇒ =

26 225 1 25 1 2511 1 0.95 1 0.95 0.98716 0.97532

74p q q p

(c) [ ] [ ] [ ]28

3 2 27 225 25 2527

100 280.97532 0.96196100 27

lp p p pl

−= = × = × =

−

(d) [ ] [ ] [ ] [ ]2 325 25 2525 :3 0.98800 0.97532 0.96196 2.92527e p p p= + + = + + = ♦♦

Note: In this text it is a convention that the numbers displayed are often rounded, but the un-rounded form of the number is used in subsequent calculations.

Example 1.25

Using the same model as in Example 1.24, determine the following values:

(a) [ ]+25 1l

(b) [ ]25l

Solution

When the model is specified by select mortality rates in terms of ultimate mortality rates, to compute the select portion of the life table we have to work backwards from the ultimate portion of the table.

We’ll use the select survival rates calculated in Example 1.24. Hence:

(a) [ ][ ]

++

= = =2725 1

25 1

73 73.949350.98716

llp

(b) [ ][ ]

[ ]

+= = =

25 125

25

73.94935 74.847520.98800

ll

p

1.9 Mortality laws

Historically, various mathematical formulas have been used to model the force of mortality for human lives. In this section we’ll look at some of these formulas and compute the associated survival functions.

Before we get started, it’s useful to derive a formula for the effect on the survival function of a linear transformation on the force of mortality. Suppose that we have a force function:

( ) ( )* x a x bµ µ= +

where the constants are chosen so that ( )* 0xµ ≥ for all 0x > .

Let ( )*s x be the survival function corresponding to the force ( )* xµ .


36

Then the relation between ( )*s x and ( )Xs x is as follows:

( ) ( ) ( )( )

( ) ( ) ( )

( ) ( )

( )( )

0 0

0 0

0

* exp * exp

exp exp exp

exp exp

x x

x x

ax

abxX

s x y dy a y b dy

a y dy bx a y dy bx

y dy bx

e s x

µ µ

µ µ

µ

−

= − = − + = − − = − −

= − −

=

∫ ∫

∫ ∫

∫

For example, if female mortality is given by ( )F xµ and male mortality is given by ( )M xµ where:

( ) ( )0.01 1.005M Fx xµ µ= +

then the survival function for males is:

( ) ( )( )−=1.0050.01x

M Fs x e s x

Gompertz’s law of mortality Under Gompertz’s law, the force of mortality is determined by the formula:

( ) xx Bcµ = where 0B> , 1c > , 0x ≥

This formula has a geometrically increasing force of mortality to model the effect of aging.

The corresponding survival function is:

( )

( )( )

( )

( )( ) ( )

00

1exp exp exp

ln ln

exp 1 where ln

x xyx yG

x

B cBcs x Bc dyc c

Bm c mc

− = − = − = −

= − − =

∫

Makeham’s law of mortality Makeham’s law adds an extra constant to Gompertz’s law to reflect an additional constant level of risk at all ages (eg accidental death). Under Makeham’s law, the force of mortality is determined by the formula:

( ) xx A Bcµ = + where > >0 and 0A B

Since this force formula is a linear transformation of Gompertz’s formula, the Makeham survival function, ( )Ms x , is closely related to the Gompertz survival function, ( )Gs x :

( ) ( )AxM Gs x e s x−=

Weibull’s law of mortality Under Weibull’s law, the force of mortality is determined by the formula:

( ) nx kxµ = where 0 , 0 , and 0k n x> > ≥

The Weibull survival function is:

( )+ −

= +

1exp

1

nW

kxs xn


37

De Moivre’s law of mortality De Moivre’s law is based on a life table function that is linear:

xl xω= − for 0 x ω≤ ≤

This leads to a survival function of:

( ) ωω

= = − ≤ ≤0

1 for 0xX

l xs x xl

and a force of mortality of:

( )µ ωω

′−= = ≤ <

−1 for 0x

x

lx xl x

Since this law will be used extensively in examples illustrating theory in succeeding chapters, it is worth listing some more properties of this mortality law.

The pdf of the lifetime function is:

( ) ( ) ( ) ωµ ωω ω ω−

= = × = ≤ <−1 1 for 0X X

xf x s x x xx

and the pdf of the future lifetime function is:

( ) ( ) ( ) ( ) ( ) ωµ ωω ω ω− −

= + = × = ≤ < −− − − −

1 1 for 0T x T xx tf t s t x t t x

x x t x

It is important to note that these final two formulas indicate that:

• the lifetime of a newborn follows a uniform distribution on the interval ω[0, ]

• the future lifetime after age x follows a uniform distribution on the interval ω −[0, ]x .

Constant force of mortality Finally, it is worthwhile to mention the constant force model:

( )xµ µ= for all ages x

While not suitable as a model of human mortality, it is again used in later chapters to illustrate the theory being presented.

The survival function is:

( ) µ−= > for all 0xXs x e x

The pdf of the lifetime function is:

( ) ( ) ( ) µµ µ −= = >for all 0xX Xf x s x x e x

and the pdf of the future lifetime function is:

( ) ( ) ( ) ( ) ( )( )µ

µµµ µ µ

− +−

−= + = × = >for all 0x t

tT x T x x

ef t s t x t e te

The lifetime of a newborn and the future lifetime after age x follow an identical exponential distribution with mean 1/µ . This distribution will also be used extensively in loss model theory.


38

Chapter 1 Practice Questions

Question 1.1

Compute the following probabilities from the life table in Section 1.2:

2 0 2| 0 4| 2 3 4 5, , , ,p q q p q

Question 1.2

Suppose that 2 1| 0.015q = . Discuss the distribution of the random number of deaths between ages 3 and 4 for a group of 20 lives currently age 1.

Question 1.3

The life table below is a survival model for a group of newborns suffering from a certain heart impairment.

x 0 1 2 3 4

xl 100 46 19 6 0

Give the probability function for the curtate future lifetime of a member of this group of impaired lives.

Question 1.4

Suppose that ( ) ( )3/ 2x xµ = + for 0x > . Determine the pdf, cdf, and survival function for the

lifetime of a newborn. Check your work by verifying properties such as ( )0

1f x dx∞

=∫ ..

Question 1.5

Using the survival model in Question 4, determine the values of 1p and 2 1|q .

Question 1.6

Suppose that ( ) ( )0.5/ 100x xµ = − for 0 100x≤ < . Determine the pdf, cdf, and survival function for the lifetime of a newborn.

Question 1.7

Using the survival model in Question 6, determine the values of 20 40p and 20 20 40| q .


39

Question 1.8

Suppose that ( ) ( )1.1/ 100x xµ = + for 0x ≥ . Compute 20t p , the survival function for the future lifetime after age 20.

Question 1.9

For the survival model in Question 1.8, determine the probability that the curtate future lifetime at age 20 is less than 2.

Question 1.10

Assuming that ( ) 0.015xµ = for all 0x ≥ , determine the pdf for the continuous lifetime after age 20.

Question 1.11

For the survival model in Question 1.10, determine the probability function for the curtate lifetime after age 20. What is the probability that the curtate lifetime exceeds 1?

Question 1.12

For the life table function ( )0.951,000 100xl x= − for 0 100x≤ ≤ , develop a formula for xT .

Question 1.13

For the survival model in Question 1.12, determine 20:5e .

Question 1.14

Suppose that ( ) 0.015xµ = for all 0x ≥ . Determine :1xe and ( )a x .

Question 1.15

For the survival model in Question 1.14, determine 20q and 20m .

Question 1.16

The life table function ( )xl xµ is called the curve of deaths. It approximates the expected number of deaths in the year of age from x to 1x + . Calculate the curve of deaths for the life table given by ( )0.751,000 100xl x= − for 0 100x≤ ≤ .


40

Question 1.17

You are given: 30 31 320.01 , 0.02 , 0.03q q q= = = . Computing the following using the UDD assumption to calculate the number living at fractional ages:

( ) ( ) ( )1.4 30 3031.4 , , 1.4Tp fµ

Question 1.18

Assuming the same mortality rates as in Question 1.17, calculate 30:3

e . Then see if you can

reason how to adapt the UDD relation 0.5x xe e= + to calculate 30:3e .

Question 1.19

Suppose that 0.021,000 xxl e−= for whole number ages only. A select and ultimate model uses this

function for the ultimate portion and the relations below for the 2-year select period:

[ ] ( )( )1 0.1 2 x kx kq k q ++ = − − .

Determine [ ]3 20q .

Question 1.20

For the select and ultimate model in Question 1.19, determine [ ]20l .

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