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.
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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:
Pv⋅ n if E occurs Z = 0if E′ occurs
where vi=+(1 )−1 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:
EZ[]=⋅⋅ Pvn Pr() E +⋅ 0 Pr() E′ =⋅Pvn ⋅Pr E () amount discount probability
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(ZEZ> [ ] ) .
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 Tx(), ie Tx()=− 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 Kx(), ie Kx()= Tx () (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 X =74.72 , T ()30=−= 74.72 30 44.72 , and KT()30=== [( 30) ] [44.72] 44 .
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.
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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 lx as the number of lives expected to survive to age x from a group of l0 newborn lives. A life table displays in a table format the values of lx 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 l0 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
lx 1,000 991 985 982 979 976 972 968 964 959
dx 9 6 3 3 3 4 4 4 5 6
The row labeled dx represents the number of lives among l0 newborn lives that die in the age range [,xx+ 1). It is computed as:
dllxxx=−+1
For example, since l2 = 985 lives survive to age 2, and l3 = 982 lives survive to age 3, then exactly dll223=−=3 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 nxp . 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:
lxn+ nxp = lx It is a standard convention to omit the n subscript when n = 1 , so the probability that a life currently age x will survive 1 year is:
lx+1 px = lx For example, the probability that a life age 5 survives for 2 years to age 7 is:
l7 968 25p == l5 976
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The probability that a life currently age x will die within n years is denoted nxq , and we have: l xn+ llxxn− + nxqp=−11 n x =− = llxx
Intuitively, this is the probability that a life is one of the ()llxxn− + lives to die between age x and age xn+ , out of the lx lives age x . For example, the probability that a life age 1 dies within 3 years is:
ll14− 991− 979 12 31q == = l1 991 991 Again, we omit the n subscript when n = 1 , so the probability that a life currently age x will die within 1 year is:
llxx− +1 d x qqxx== or llxx Finally, the probability that a life currently age x will survive for m years and then die within the following n years is denoted mnqx , and we have:
llxm+++− xmn mnqx = lx
Intuitively, mnqx is the probability that a life age x survives for m years, multiplied by the probability that a life age xm+ dies within n years:
lllxm+++++++ xm− xmn ll xm− xmn mnqpqxmxnxm=×+ = × = llxxmx+ l Again, we omit the n subscript when n = 1 , so the probability that a life currently age x will survive for m years and then die within 1 year is: ll− xm+++ xm1 dxm+ mmqqxx== or llxx For example, the probability that a life age 4 survives for 3 years and then dies within the following 2 years is:
ll79− 968− 959 9 32q4 == = l4 979 979
Example 1.1
Compute the following probabilities from the life table above:
(a) 50p
(b) 5 q0
(c) 42q1
(d) p1
(e) q2
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Solution
l5 976 (a) 50p == l0 1,000
d5 4 (b) 5 q0 == l0 1,000
ll57− 976− 968 8 (c) 42q1 == = l1 991 991
l2 985 (d) p1 == l1 991
d2 3 (e) q2 == ♦♦ l2 985
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 Lx( ) , and the life table function lx represents the expected number of survivors, ie:
lELxx =()
The random variable Lx() follows a binomial distribution. Each life is viewed as an
independent Bernoulli trial. The number of trials is nl= 0 . We define “success” as survival to age x , with probability p = x p0 . For example, let’s consider the distribution of L(5) , the random number of survivors at age 5 in this life table from the 1,000 newborn lives.
The random variable L()5 follows a binomial distribution with nl= 0 = 1,000 trials and a
probability of success of pp==50 976 /1,000 = 0.976 . The mean and variance of L()5 are:
EL==×=()5 np l050 p 1,000 × 0.976 = 976 ( = l 5 )
var()Lnpqnpq() 5==××=50 50 1,000 × 0.976 × 0.024 = 23.424 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, K(0) .
The probability Pr(Kk (0)= ) is the probability that a newborn life dies in the age range [,kk+ 1),
which we have already defined as k q0 . Hence:
dk Pr()Kk (0)==k q0 = l0 For example, the probability that a newborn life has a curtate lifetime of 3 years is:
d3 3 Pr()Kq (0)== 33 0 = = = 0.003 l0 1,000
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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):
FxX ()=≤ Pr( Xx )
The cdf FxX () represents the probability that a newborn life will die at or before age x .
FxX () is continuous and non-decreasing with FX (00) = and FX (ω ) = 1 where ω is the first age at which death is certain to have occurred for a newborn life.
The probability density function (pdf) is:
fxXX()= Fx′ () wherever the derivative exists
The pdf fX ()x is non-negative and continuous on the interval [0,ω ).
Recall that a value of fX ()x is not a probability in itself. The probability that a newborn life dies between ages a and b is:
b Pr()aXb≤≤= fxdxFbFaX () = () − () ∫ a
You should note that since X is assumed to be a continuous random variable, all of the intervals ab, , [,)ab, (,]ab, and ()ab, have the same probability, Fb( ) − Fa( ) . Hence:
ω fxdxX () = 1 ∫ 0
and:
x FxXX()=≤=Pr() Xx f() udu ∫0
Finally, the probability that a newborn life dies in the interval [,xx+ ∆ x ] can be estimated as:
Pr()xXx≤≤+∆≈ x fX (xx ) ⋅∆
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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 fx()==0.01 for 0 ≤≤ x 100 X 100 (b) The cdf is:
xx FxXX()===≤≤ f () udu0.01 du 0.01 x for 0 x 100 ∫∫00
(c) The probability of death between ages 60 and 80 is:
Pr() 60≤≤XFF 80 =XX() 80 −( 60) = 0.01(80 − 60) = 0.20 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 1 The pdf for an exponential distribution with mean θ is fx()= e−x /θ for x> 0 . Hence: X θ (a) The pdf is: 1 fx()=> e−x /75 for x 0 X 75 (b) The cdf is:
x x 1 −−uu/75 /75 − x /75 FxX ()==−=− e due()1 e ∫0 75 0
(c) The probability of death between ages 60 and 80 is:
−−60/75 80/75 Pr() 60≤≤XFFe 80 =XX() 80 − () 60 = − e = 0.10518 ♦♦
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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 sxX () and is defined as:
sxX ()=>Pr() Xx 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. sxX () is continuous and non-increasing with sX (01)= and sX (ω ) =0
2. sxXX()=−1 Fx ()
b 3. Pr()aXb≤≤= fxdxsaXXX() = () − sb () ∫ a
4. fXX()xsx=− ′ ()
Example 1.4
Suppose that the lifetime X of a newborn is exponentially distributed with mean 75 years.
(a) Identify the survival function sxX (). (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
−x /75 (a) In Example 1.3 we saw that the cdf is FxX ()=−1 e . So, we have:
−x /75 sxXX()=−1for Fxe () = x >0 (b) The probability a newborn is still alive at age 100 is:
−100/75 seX ()100== 0.26360 (c) The probability a newborn dies between ages 60 and 75 is:
−−60/75 75/75 sseXX()60−= (75) − e = 0.08145 ♦♦
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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 lx for all ages between 0 and w .
As before, let Lx() denote the random number of survivors at any age x from a group of l0
newborn lives. The random variable Lx( ) follows a binomial distribution with nl= 0 trials. We
define “success” as survival to age x , with probability p =>=Pr(Xx) sX ( x) .
The life table function lx is defined as the expected number of survivors at age x . Hence:
lELxnplsxxX=() = =0 ( )
Example 1.5
Suppose that the lifetime X of a newborn is uniformly distributed on 0,100 .
(a) Identify the survival function sxX ().
(b) Identify the life table function lx if l0 =100 .
Solution
(a) The survival function is given by:
100 100 sXX() x=>=Pr() X x∫∫ f() u du = 0.01 du xx =−=−≤≤0.01() 100xx 1 0.01 for 0 x 100
(b) The life table function is:
llsxxX==−=−≤≤0 ( ) 100(1 0.01 x ) 100 x for 0 x 100 ♦♦
Let’s summarize some of the key properties of the life table function, lx .
Key properties of the life table function
1. lx is the expected number of survivors at age x from a group of l0 newborn lives
2. llsxxX= 0 () is continuous and non-increasing with lω =0
lx 3. sxX ()= l0
Note that the value of l0 (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 l0 for convenience. The survival function will be identical whether we choose l0 = 100 or l0 = 1,000,000 .
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Example 1.6
2 Suppose that the life table function is lxxx =−≤≤10,000() 100 for 0 100 . 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:
22 l 10,000() 100−−xx() 100 sx==x = for 0 ≤≤ x 100 X () 22 l0 10,000× 100 100
We can then calculate FxX () using the relationship FxXX( )=−1 sx( ) :
()100 − x 2 FxX ()=−1 for 0 ≤≤ x 100 1002
Finally, we can calculate fxX () using the relationship fXX(xFx) = ′ ( ) :
2 ′ ()100−−xx 2() 100 100 − x fxFx()==−′ () 1 = = for 0 ≤≤ x 100 ♦♦ XX22 100 100 5,000
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 L()10 , the random number of survivors at age 10. Solution
L()10 follows a binomial distribution with:
nl==0 1,000
2 ()100− 10 ps==X ()10 = 0.81 1002 So we have:
EL==×=()10 np 1,000 0.81 810 var()Lnpq() 10== 1,000 ××−= 0.81 (1 0.81) 153.90
σL()10 ==153.90 12.406
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 µ (x) . It is an instantaneous measure of mortality at age x , and it can be defined in several equivalent ways:
fxXX() sx′ () ′ lx′ µ ()xsx==−=−=−()ln ()X () sxXX() sx() l x These equalities can be verified using simple calculus. For example, using the information in Example 1.6, we have:
2 sx′ () −−2() 100x /100 2 µ ()xx=−X =− = for 0 ≤ < 100 2 2 sxX () ()100− x /100 100 − x
Or, using the information in Example 1.4, we have:
′ x ′ 1 µ ()xsx=−()ln()X () = − − = for x ≥ 0 75 75 Let’s now see how to calculate the survival function from the force of mortality. ′ µ ()xsx=−()ln()X () x x ⇒=−=−+µ ()ydyln() sXXX () y ln() s() x ln() s () 0 ∫0 0
But since sX ()01== and ln10 () , we have:
x µ ()ydy=−ln () sX () x ∫0
x ⇒=−sxX () expµ () ydy ∫ 0
Example 1.8
Suppose that the force of mortality for a survival model is given by the formula: 0.9 µ ()xx=≤ Solution The survival function is calculated as: xx0.9 sxX ()=−exp∫∫µ () ydy =− exp dy 0090 − y x 90 − x =−=exp0.9ln90()y exp0.9ln 0 90 0.9 90 − x =≤< for 0x 90 ♦♦ 90 11 Introduction to survival models Chapter 1 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 l0 =1,000 ) as: 0.9 90 − x llsxxX==0 ( ) 1,000 for 0 ≤< x 90 90 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 µ (x) , it is useful to rewrite the defining formula in the form: fXX()xsxx= ()µ () Now, recall that: fX ()xx∆≈Pr() xXx ≤ ≤ +∆ x Rewriting the probability term using a conditional probability, we have: fxxX ()∆≈Pr() xXx ≤ ≤ +∆ x =≤+∆≥≥Pr()Xx xXx Pr() Xx =≤+∆≥Pr()Xx xXxsxX ( ) Substituting fXX()xsxx= ()µ (), we have: sxXX()µ () xx∆≈Pr() Xx ≤ +∆ xXxsx ≥ ( ) ⇒∆≈≤+∆≥µ ()xxPr() Xx xXx So, µ ()xx∆ 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, µ ()20 multiplied by ∆=x 1/365 (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 Suppose that the force of mortality for a survival model is given by the formula: 0.9 µ ()xx=≤ Solution Setting ∆=x 7/365, the required probability is: 0.9 7 µ ()40∆=x × = 0.00035 ♦♦ 90− 40 365 12 Chapter 1 Introduction to survival models Let’s summarize the main properties of the force of mortality. Key properties of the force of mortality fxXX() sx′ () ′ lx′ 1. µ ()xsx==−=−=−()ln ()X () sxXX() sx() l x x 2. sxX ()=−expµ () ydy ∫0 3. µ ()xx⋅∆ ≈Pr() XxxXx ≤ +∆ | ≥ 4. µ ()x is non-negative and piece-wise continuous where defined ω 5. µ ()ydy=∞ in order that s (ω ) = 0 ∫ 0 X 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: lxt+ sxtX ()+>+Pr( Xxt) txp == = =Pr()XxtXx >+> | lsxXxxX() Pr()> The probability that a life currently age x will die within the next t years is: llxxt− + Pr()Xx>− Pr( Xxt >+) txqXxtXx== =≤+>Pr() | lXxx Pr()> Finally, the probability that a life currently age x will survive s years but die within the following t years is: llxs+++− xst stqxsXxstXxx ==+<≤++>Pr() | lx These three functions are defined for all ages x in 0,ω and for 0 ≤ tx≤−ω . 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 px is the probability that ()x will still be alive in 5 years’ time, at age x + 5 , and 5 qx 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 t = 1 . So, for example, 3|qx is the probability that ()x will die between ages x + 3 and x + 4 . 13 Introduction to survival models Chapter 1 Example 1.10 Suppose that the force of mortality for a survival model is given by the formula: 0.9 µ ()xx=≤ (a) 2.5p 20 (b) 2.5q20 (c) 2.5|q20 Solution Note that this is the force of mortality in Example 1.8. The life table function is: 0.9 90 − x llsxlxX==00( ) for 0 ≤< x 90 90 Recall that we can choose any convenient value of l0 without changing the distribution of X . So 0.9 let’s simplify our computations by choosing l0 =90 , which gives: 0.9 llsxxX==−0 ( ) (90 x ) for 0 ≤< x 90 We can now calculate the required probabilities as follows. 0.9 l ()90− 22.5 (a) p ==22.5 =0.96780 2.5 20 0.9 l20 ()90− 20 ll20− 22.5 (b) 2.5qp 20==−=1 2.5 20 0.03220 l20 ll−−67.50.9 66.5 0.9 (c) |q ==22.5 23.5 = 0.01291 ♦♦ 2.5 20 0.9 l20 70 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. txpq+= tx 1 2. st++p x=⋅ spp x t xs 3. st| qpq x=⋅ s x t x+++ s =− s p x s t p x = s t q x − s q x 4. nxppp=⋅ x x++−11 ⋅⋅ p xn when n is an integer 5. nxxxnxqqq=+++01| | − 1 | q when n is an integer 14 Chapter 1 Introduction to survival models 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: 20ppp 20=⋅=×= 10 20 10 30 0.97 0.95 0.92150 Or, the probability that ()20 dies between ages 30 and 40 is: 10| 10qpq 20=⋅= 10 20 10 30 0.97( 1 −= 0.95) 0.04850 On the other hand, if pp01==0.99 , 0.98 , and p 2 =0.97 , then the probability that a newborn dies within three years is: 30qpppp=−1 30 = 1 − 0 ⋅ 1 ⋅ 2 =− 1 0.99 × 0.98 × 0.97 = 0.05891 Or the probability that a newborn dies during the second year of life is: 10|qpq=⋅= 01 0.99 ×−() 1 0.98 = 0.01980 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 qqq012,,, … and so on. Values of lx at whole number ages can then be produced as follows: ln nnnnpllplppplqqq00000110011=⇒=⋅=⋅⋅⋅⋅=−− − ()()()11 1 − n− l0 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, Xx> . The future time lived after age x is Xx− . The conditional distribution of the time lived after age x , given survival to age x, is: Tx()=− X x| X > x The continuous random variable Tx() is a survival model defined on the interval [0,ω − 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 Tx() is closely related to the distribution of X . The quickest way to see the relation between the distributions of Tx() and X is to calculate the survival function for Tx(), ststTxtTx()( ) = T ( ) =>Pr( ( ) ) . In fact, we have already computed this survival function in terms of the distribution of X , since the event Tx()> t is equivalent to saying that ()x is alive at age xt+ . The probability of this event is simply txp . So, we have: lxt+ sxtX ( + ) stTx()()=>===Pr() Txtp ( ) tx since llsx x =0 X ( ) lsxxX() Note: When there is no ambiguity, we will write T for Tx( ) . However, subscripts are often important, for example to distinguish sX (20) , the probability that a newborn survives to age 20, from sT()10 ()20 , the probability that (10) survives to age 30. 15 Introduction to survival models Chapter 1 Example 1.11 2 Suppose that the life table function is lxxx =−≤≤10,000() 100 for 0 100 . (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, sxX (), is: 2 2 l 10,000() 100 − x 100 − x sx=== px =for 0 ≤≤ x 100 Xx() 0 2 l0 10,000( 100− 0) 100 (b) The survival function for lives currently aged 20, stT(20)(), is: 2 l 10,000(100−+ (20t ))2 80 − t stp==20+t = =for 0 ≤≤ t 80 ♦♦ T()20 () t 20 2 l20 10,000(100− 20) 80 When we deal with the future lifetime after age x , we’ll frequently see expressions of the type lxt+ and µ()xt+ . 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 . FtT ()=≤=−≤>Pr() Tx ( ) t Pr() XxtXx | Pr()xXxt<≤+ FxtFx()+− () ==XX Pr()Xx>− 1 FxX () ′ d FxtFxXXX()+− () f() xt + f X() xt + ftTT()==() Ft () = = dt11−− FxXXX() Fx() sx() In the figure below, we have areas A=+−FxtFxXX( ) ( ) and A+=BsxX ( ) . fX(x) Area Area = A = B x x+t ω Notice that we have: FxtFxXX()+− () A FtT ()== sxX () AB+ 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. 16 Chapter 1 Introduction to survival models Furthermore, if we examine the relation: fxtX ()+ ftT ()=≤≤− for 0 tω x sxX () 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 fX (x) to the right of x , divide by the total area under the remainder of the graph, A+=BsX ( x) , and then relabel the horizontal axis as t and the vertical axis as fT ()t , we have a graph of the pdf of T : fT(t) Area = Area = A / sX(x) B / sX(x) 0 t ω-x One final point worth noting is the similarity between the pdf’s for the distributions of X and T . For X we have: fxX () µµµ()xfxsxxpx=⇒=XX() () () = x0 () sxX () and for T we have: fxtX ()+ xt+ pxtppxt00µµ( ++) x t x ( ) ftTtx()== = =+ pµ () xt sxXx() p00 x p Example 1.12 2 Suppose that the life table function is lxxx =−≤≤10,000() 100 for 0 100 . (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 80 − t t ps20 ==() tfor 0 ≤≤ t 80 T()20 80 Hence, the distribution function for T (20) is: 2 80 − t Ft()=−11for080 st() = − ≤≤ t TT()20() 20 80 17 Introduction to survival models Chapter 1 (b) We have two options for calculating the density function. The first option is to differentiate the distribution function: 2 ′ ′ 80 − t ()80 − t ftFt()==−=() 12 TT()20()() 20 2 80 80 80 − t =≤≤ for 0t 80 3,200 The other option is to use the relation: fTtx()t =+p µ ()xt It is consistent with the formula for lx that: 2 µ ()x = 100 − x Hence, we have: 2 80− tt 2 80 − ftpTt(20)()=+=⋅= 20 µ ()20 t ♦♦ 80 80− t 3,200 Key results concerning the relation of the distributions of X and T(x) lxt+ sxtX ( + ) 1. stTtx()=>===Pr() Txt ( ) p lsxxX() FxtFxXX()+− ( ) 2. FtTtxT()==−= q1 st () sxX () fxtX ()+ 3. ftTtx()==+ pµ () xt sxX () 1.5 The curtate future lifetime after age x In addition to computing the distribution of the continuous future lifetime Tx( ) , 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: Kx()= Tx () ie the integer part (or greatest integer) of Tx ( ) Since it is a function of Tx(), it is simple to calculate the probability function of Kx() from what we know about Tx(). The possible values of Kx( ) are the numbers 0,1,2, ,ω − x − 1. For example, if x =70 and ω = 90 , then the possible values of K (70) are the twenty whole numbers 0 through 19. If the life ()70 eventually dies at age 85.8, then the continuous future lifetime is T(70)= 15.8 and the curtate lifetime is K(70)= [ 15.8] = 15 . 18 Chapter 1 Introduction to survival models The key observation is that if Kx( ) = k, then we must have: kTxk≤<+() 1 This leads to the following formula for the probability function: Pr()Kx ( )== k Pr() k ≤ T( x) <+ k 1 dxk+ ==kx|qkx for = 0,1,2,,1 ω −− lx Example 1.13 Suppose that the life table function is given by the formula: lxx =−100 for 0 ≤≤ x 100 Compute the probability function for K (75) . Solution The probability function for K ()75 is: d ll− Pr()Kk() 75 ==75+k =75+++kk 75 1 ll75 75 ()()100−−− 75kk 100 −−− 75 1 1 == ()100− 75 25 So K ()75 has 25 possible values ( 0,1,2, ,24 ) that are equally likely to occur. ♦♦ Example 1.14 Suppose that the life table function is given by the formula: −0.015x lex =≤<∞1,000 for 0 x Compute the probability function for K (75) . Solution The probability function for K ()75 is: d ll− Pr()Kk() 75 ==75+k =75+++kk 75 1 ll75 75 −+0.015() 75 k −++0.015() 75k 1 ee− ==−ee−−0.015k ()1 0.015 e−×0.015 75 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. 19 Introduction to survival models Chapter 1 Recall that for any random variable FxX ()= Pr( Xx≤ ), hence: FkKx()( )=≤==+=++= Pr() Kxk() Pr() Kx() 0 Pr( Kx( ) 1) Pr( Kxk( ) ) dd d d ll− =+xx++12 + x ++ xk + =xxk++1 llxx l x l x l x ==−−kx+1qk for 0 ,1 , ,ω x 1 The survival function of the curtate future lifetime is then easily derived as: skKx()()=>=−=−=Pr() Kxk () 1 Fk Kx()( ) 1kxkx++11 q p for k =−− 0 ,1,...,ω x 1 Example 1.15 Suppose that the life table function is given by the formula: lxx =−100 for 0 ≤≤ x 100 Compute the survival function for K ()75 . Solution The survival function for K ()75 is: l75++k 1 100−++ (75k 1) 24 − k skK()75 ()==k+175 p = =for k = 0 ,1, ,24 ♦♦ l75 100− 75 25 Example 1.16 Suppose that the life table function is given by the formula: −0.015x lex =≤<∞1,000 for 0 x Compute the survival function for K ()75 . Solution The survival function for K ()75 is: −++0.015(75k 1) l75++k 1 e −+0.015()k 1 sk()==+ p = = efor k = 0 ,1 ,2 , K()75 k 175 l −0.015(75) 75 e ♦♦ Let’s conclude this section with a summary of the key relations concerning the curtate future lifetime, Kx(). Key relations concerning the curtate future lifetime dxk+ 1. fkKx()()====Pr() Kkqkx | for k = 0 ,1,...,ω −− x 1 lx lxk++1 2. skKx()()==kx+1 pfor k = 0 ,1,...,ω −− x 1 lx 3. FkKx()()==−kx++11 q1 kx p for k = 0 ,1,...,ω −− x 1 20 Chapter 1 Introduction to survival models 1.6 Important life table functions Now that we have developed the basic properties of X , Tx( ) , and Kx(), it is time to study additional features of these distributions, such as the life expectancy for a newborn. The life table functions Lx and Tx The functions Lx and Tx are useful devices in the calculation of life expectancy. They are defined in terms of the life table function, lx , as follows: x+1 Lldyxy= ∫x ω TldyLLxy==+++ xx+−11 Lω ∫x 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 yy+ ∆ ] that is part of the interval [,x ω ]. At the start of this brief period there are ly survivors. We can estimate the total people-years lived by the survivors during this brief period by ly ∆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 ω comprising the age interval [,x ω ], we will have a Riemann sum for the integral ldy . ∫x y 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: ω Tld= y xy∫x which can be interpreted as the total people-years lived after age x by the survivors to age x . Beware of confusing Tx with Tx( ) , the random future lifetime of a single life age x. We can break the interval [,x ω ] into subintervals of length one year, ie [,xx+ 1], [1,2],,[1,]xx++ ω −ω . The function Lx is calculated over just one of these one-year periods: x+1 Lld= y xy∫x which is the number of people-years lived by the survivors to age x during the next year. Example 1.17 Compute the function Tx for the following life table functions: (a) lxxx =−1,000 10 for 0 ≤≤ 100 −0.015x (b) lex =≤<∞1,000 for 0 x 21 Introduction to survival models Chapter 1 Solution (a) We have: 100 2 2 100 1,000− 10y − ()()1,000 10x 2 Tydyx =−=−1,000 10 = =− 5() 100 x ∫x 20 20 x (b) We have: ∞ −0.015y −0.015x ∞ −0.015y 1,000ee 1,000 Tedyx ==−=1,000 ♦♦ ∫x 0.015 0.015 x 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: ∞ EX[]= xfX () xdx ∫ 0 We can develop an alternative formula using integration by parts: u== x,, dv fXX() x dx ⇒==− du dx v s( x) ∞∞∞ EX[]==−+ xfxdxxsxXXX() () sxdx() ∫∫00()0 ∞∞ =−limxsXXX() x + 0 + s () x dx = s () x dx x→∞ ∫∫00 −1 In the above calculation, the limit is zero. Apply L’Hopital’s rule writing xsX ( x) as sxxX ()/ . Then use the fact that limxfx2 ()= 0 if EX 2 exists. x→∞ In other words, we can calculate EX[] as the area under the graph of the survival function. If the density function is supported on the finite interval [0,ω ] , then we can replace ∞ in the above integrals by ω , since both of the functions fX (x) and sxX ( ) are zero for x > ω . So we have: ω EX[]= sX () x dx ∫0 Another type of expected value that occurs in both survival model theory and loss distribution theory is the limited expected value. Let’s define Xn∧ as the minimum of the random variable X and the number n : XXnif ≤ Xn∧=min{} Xn , = nXnif > 22 Chapter 1 Introduction to survival models Since Xn∧ is a function of X , we can compute its expected value as above with limits 0 and n: nn∞ EXn∧= xfxdxnfxdxXX() + () = xfxdxnsn XX () + () ∫∫00n ∫ n nn =−xsXXXX() x + s() x dx + ns () n = s () x dx ()0 ∫∫00 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 ex , and is defined as: eETxx =() Using results derived earlier in this section, we can develop several methods to calculate this expected value: ω−x eETxx =() = tftdt() ∫ 0 Tx() ω−x = stdt() (integration by parts) ∫ 0 T ωω−x ldtxt+ ldy y ω −x lT∫∫0 x ==∫ xt+ dt == x () substitute y =+ x t 0 llxx ll xx So, life expectancy can be computed in three distinct ways, as: • the standard expected value of the continuous random variable Tx() • 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) lxxx =−1,000 10 for 0 ≤≤ 100 −0.015x (b) lex =≤<∞1,000 for 0 x Solution Using the results from Example 1.17, we have: ()1,000− 10x 2 20 Tx ( ) 100 − x (a) ex == = lxx 1,000− 10 2 −0.015x T 1,000e /0.015 1 (b) e ==x = =66.67 ♦♦ x −0.015x lx 1,000e 0.015 23 Introduction to survival models Chapter 1 The temporary complete life expectancy at age x is denoted exn: , 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: ≤Tx( ) if Tx( ) n (death within n years) Tx()∧= nmin{} Tx () , n = nTxnnif ()> (survives years) As a result, we have the following: = ∧ eETxnxn: () n = stdt() ∫0 Tx() nxn+ ldtxt+ ldy y n l ∫∫0 x TT− ==∫ xt+ dt = =xxn+ () substitute y =+ x t 0 llxx l x l x Example 1.19 Compute the 10-year temporary complete life expectancy at age 50 for the following life table functions: (a) lxxx =−1,000 10 for 0 ≤≤ 100 −0.015x (b) lex =≤<∞1,000 for 0 x Solution Using the results from Example 1.17, we have: 22 TT− ()500− 400 /20 ==50 60 = (a) e50:10 9 l50 500 −×0.015 50 −× 0.015 60 TT− ()1,000ee− 1,000 /0.015 (b) e ==50 60 =9.280 ♦♦ 50:10 −×0.015 50 l50 1,000e Key formulas for complete life expectancy ωω−−xxT = = = =x = 1. exTtx ETx() ∫∫ tfTx()() tdt s () tdt where s Tx()() t p 00lx n TTxxn− + 2. eETxn= ∧= stdt = where stp = xn: ()∫ Ttx () Tx()() 0 lx =+ 3. eexnxxnxn: pe+ 2 ωω−−xx2 4. ETx() == t f() tdt2 tsT () tdt ∫∫00Tx() 24 Chapter 1 Introduction to survival models The proof of the third formula goes as follows: TlTxxnxn()TTxxnxn−+++ T TTxxn− + ++ ex == = +× llxx lll xxxn+ =+ epexn: nxxn+ This result has an intuitive interpretation. The life (x) expects to live exn: years in the next n years, and, given survival to age xn+ , expects to live an additional exn+ more years. 2 The proof of the fourth formula relies on integration by parts with ut= and dv= fTx()( t) dt . Curtate life expectancy The curtate expected future lifetime at age x is denoted by ex , and is defined in terms of the curtate future lifetime Kx(): eEKxx =( ) So, ex 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−−x 1 eEKxx =() =∑ kPr() Kxk () = k =0 ωω−−xx11 −− kd ddd+++++++−−23 ()ω xd 1ω − =⋅=kq| xk+ =xxx123 1 ∑∑kx ll kk==00xx lll++++ l ==+++xxx+++123ω − 1 (note: ldxk++ xk d xk ++− 11 dω ) lx the survivors at age xk+ die at age xk+ or later =+ppxx23 + p x ++ ω−− xx 1 p The temporary curtate life expectancy at age x is denoted e , and is defined as: xn: = ∧ eEKxnxn: () 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: eEKxnpp=() ∧=+ + + p xn: xx2 nx There is a recursive relation: eexnxxn=+ pe+ xn: =+ =+ = eexxxxxxx:1 peppe++11 (the special case with n 1) 25 Introduction to survival models Chapter 1 Example 1.20 Suppose that the life table function is given by the formula: lxx =−100 for 0 ≤≤ x 100 (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: ep0020=+ p ++ 990 p ll+++ l 99 +++ 98 1 4950 ====12 99 49.5 l0 100 100 (b) The 10-year temporary curtate life expectancy at age 50 is: epp=+ ++ p 50:10 50 2 50 10 50 ll+++ l 49 +++ 48 40 445 ====51 52 60 8.9 ♦♦ l50 50 50 Key formulas for curtate life expectancy lllxxx+++123+ +++ lω − 1 1. epppxxxx=+23 + ++ ω−− xx 1 p = lx ll+ ++ l =+ ++= xx++12 xn + 2. epppxn: xx2 nx lx 3. ee=+ pe xnxxnxn: + lll+ 35++ 4. EKx()2 = xxx+++123 lx 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: ω−−x 1 22== EK ∑ kPr() K k k=0 ω −−x 1 2 = ∑ kqkx| k=0 ω −−x 1 ll− =×∑ k2 xk+++ xk1 k=0 lx 26 Chapter 1 Introduction to survival models The central mortality rate The n-year central mortality rate denoted by nxm computes a weighted average of the force of mortality over the range from age x to age xn+ . It can be calculated in several ways: n txpxtdtµ ()+ ∫0 q dl/ d d m =====nx nx x nx nx nx n TT−−+++/ l TT L L L pdt exn: ()xxnx+ xxnx+++− x11 xn ∫0 tx The symbol nxd is defined as: nxdll=− x xn+ ie it is the expected number of deaths in the age range from x to xn+ . When n = 1 we simply write mx instead of 1 mx . In this case, we have: dx mx = Lx We now have three annual measures of mortality at age x : µ (x) , qxx, and m . The force of mortality µ ()x is not a probability, but µ (xx)∆ 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 qx : dx qx = lx where lx is the total number of survivors aged x at the start of the year. The only difference between qx and mx is the denominator. The quantity Lx in the definition of mx , is often called an exposure, since it represents the total number of people-years lived by the lx lives age x over the next year (when they are exposed to death). Example 1.21 Suppose that the life table function is given by the formula: 0.5 lxx =−()100 Compute q50 , µ (50) and m50 . Solution For m50 we will need the value of e50:1 : 1 1.5 11l 10.5 ()10.02− t e== p dt50+t dt =−=−=1 0.02 t dt 0.99498 50:1 ∫∫t 50 ∫() 00l50 0 1.5× 0.02 0 27 Introduction to survival models Chapter 1 Now we have the necessary information to complete the three calculations: 0.5 l ()49 q =−151 =− 1 = 0.01005 50 0.5 l50 ()50 −0.5 l′ −−()100 50 /2 0.5 µ 50=−50 =− = = 0.01000 () 0.5 l50 ()100− 50 100− 50 q 0.01005 m ==50 =0.01010 ♦♦ 50 0.99498 e50:1 The function ax( ) The function ax() is defined as the average part of the final year of life lived by someone who dies between ages x and x + 1 . From this verbal description we can see that it is the conditional expected value: ETx≤()|1 Tx () since the event Tx()≤ 1 indicates that the life (x) dies within a year. Rather than calculating ax() by integrating a conditional density, let’s look at an intuitive way. Consider the lx lives aged x . Let’s split them into the dx lives that die in the next year, and the lx+1 lives that survive the next year. Now consider ex:1 , the average time lived in the next year. • For the lives that die in the next year, the average time lived is ax(), by definition. • For the survivors to age x + 1 , the average time lived in the next year is 1. The number ex:1 is a weighted average of time lived in the next year by these two groups, so: dax⋅+⋅() l11 dax ⋅+⋅( ) l ==xxxx++11 =+ eqaxpx:1 xx() dlxx+ +1 l x ep− x Ll− ⇒=ax() x:1 =xx+1 qdxx For example, if lxx =−100 , we have: 2 2 xx++11 ()xx+−1 Lldyydyxy==−=−=−−()100 100 100 x 0.5 ∫∫xx 2 and: Ll− 100−−xx 0.5 − (100 −− 1) ax()==xx+1 = 0.5 dx 1 So, in this case the average time of death is midway through the final year of life. 28 Chapter 1 Introduction to survival models 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 lx at whole number ages 0,1,2, ,ω , where lω =0 . How can we compute life table functions such as 2.5p 3 , µ (2.5) , or e25 , which rely on a continuous model? To calculate these functions, we’ll need to start by extending lx 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 dx deaths that occur between age x and age x + 1 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 01≤ t ≤ , then we have: ltltlxt++=−()1 x + x1 (linear interpolation) =−ltllxxxxx() −+1 =− ltd The resulting piecewise linear graph of lx is continuous and non-increasing (usual properties of the life table function). The same must be true of the survival function, sxX (). • lx • • • • 0 1 2 3 4 x The UDD assumption has several important consequences. For integer x , and 01≤≤t , we have: lxt+ ltdxx− txptq== =−⋅1 x llxx txqptq=−1 tx =⋅ x ′ ld′ ()ltd− q µ ()xt+=−xt+ =−xx = x = x lltdltdtqxtxx+ −−− xx1 x 29 Introduction to survival models Chapter 1 We can now perform some of the calculations that would have been impossible from discrete data without the UDD assumption. For example, we have: l5.5 ld55−0.5 d2 q 2 2.5p 3 == and µ () 2.5 = = ll33 ldq 22−−0.5 1 0.5 2 Example 1.22 Suppose that qq90==0.4, 91 0.7, and q 92 = 1.0 . Calculate e90 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: t ptqt90=−⋅110.4 90 = − for 01 ≤ t ≤ t ptqt91=−⋅110.7 91 = − for 01 ≤ t ≤ t ptqt92=−⋅1 92 = 1 − for 0 ≤ t ≤ 1 Hence: =+ =+ + ee9090:1 pee 90 91 90:1 pepe 90() 91:1 91 92 111 =+tttpdtp90 90 pdtp 91 + 91 pdt 92 (note that 3 p 90 = 0) ∫∫∫000 111 =−()1 0.4tdt + 0.6 () 1 − 0.7 tdt + 0.3 () 1 − tdt ∫∫∫000 =−()1 0.2 + 0.6() ( 1 − 0.35 )() + 0.3 1 − 0.5 = 1.28 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 epdt==sum of three trapezoidal areas 90∫0 t 90 =++=−×+−×+×−0.5ppp 90 1.5 90 2.5 90 ()()()1 0.4 0.5 0.6 1 0.7 0.5 0.6 0.3 1 0.5 =1.28 ♦♦ 30 Chapter 1 Introduction to survival models Example 1.23 Suppose that qq90==0.4, 91 0.7, and q 92 = 1.0 . Draw a graph of the pdf for T ()90 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: −−10.4tt′ for0 << 1 () ′ ′ ftT()90 ()=−()t p90 = −()1.02 − 0.42 t for 1 < t < 2 −−0.54 0.18tt′ for 2 << 3 () note: for 1≤≤tpppt 2 we havett90 = 90− 1 91 = 0.6() 1 −−() 1 0.7 = 1.02 − 0.42 t Hence: 0.40 for 0< t < 1 ftT()90 ()=<<0.42 for 1 t 2 0.18 for 2< t < 3 The graph is as shown below: f(t) οο ° ο οο 0 1 2 3 t Note: A quicker way to calculate the pdf form is to note that: fTx()()t ==kx|q when kTx ( ) . For example, notice in the example above that: 09090|0.4qq== 1909091|0.42qpq== 290909192|0.18qppq=××= ♦♦ 31 Introduction to survival models Chapter 1 Key formulas for the UDD assumption (if x is an integer age and 01≤ t ≤ ) 1. ltltlltdxt++=−()1 x + x1 =− x x (linear interpolation) 2. txp =−⋅1,tq x tx q =⋅ tq x d q 3. µ ()xt+=x = x ltdxx−−1 tq x 4. ftTx()()=<<+kx| q for ktk 1 5. ax()= 0.5 for all integer ages x 6. eexx=+0.5 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 [xx++ 0.40, 0.65] there will be 25% of the deaths between ages x and x + 1 . 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: ep90= 90 + 2 p 90 + 3 p 90 =0.6 +× 0.6 0.3 +××= 0.6 0.3 0 0.78 UDD⇒ee90 =+= 90 0.5 0.78 += 0.5 1.28 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. 32 Chapter 1 Introduction to survival models 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 qx (integer age x ) for a life subject to the model lx 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 l[]x l[]x +1 l[]x +2 lx+3 x + 3 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 ll[]xx,,and []+ 1 l []2 x+ . The subscript []xk+ indicates that a life “selected” at age x is currently age xk+ , 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 lx+3 . 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 ll[][]23,,,,, 23++ 1 l [] 23 2 ll26 27 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 []xk+ 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: l[]23+ 1 35,604 p[]23 ===0.99969 l[]23 35,615 The probability that a newly selected life aged 23 dies within one year is: qp[23] =−1[]23 =− 1 0.99969 = 0.00031 33 Introduction to survival models Chapter 1 The probability that a life aged 24, selected at age 23 (one year ago) survives one year is: l[]23+ 2 35, 590 p[]23+ 1 ===0.99961 l[]23+ 1 35,604 The probability that a life aged 25, selected at age 23 (two years ago) survives one year is: l26 35, 581 p[]23+ 2 == =0.99975 l[]23+ 2 35,590 The probability that a life aged 26, selected at age 23 (three years ago) survives one year is: l27 35, 562 p26 == =0.99947 l26 35, 581 We can calculate the curtate life expectancy as: =+ =++ + ++ ee[]23[]23 :3 3262 pep[] 23() [] 23 p [] 23 3 p [] 23 3 ppp[] 23 () 26226 To compute a long-term survival probability such as 25 p[]23 , we must consider the split of 25- year period between the 3-year select period and the following 22-year ultimate period: lll26 48 48 25ppp[]23==×= 3 [] 23 22 26 lll[]2326 [] 23 Example 1.24 Suppose we have a select and ultimate model based on the life table lxx =100 − at integer ages x, where the select period is 2 years and the select mortality rates are calculated according to the rule: qkq[]xk+ =−()10.052() − xk+ for k =0,1 Determine the following values: (a) p[]25 (b) 2 p[]25 (c) 3 p[]25 (d) e[]25 :3 Solution This formula indicates that select mortality at the age of selection (ie when k = 0 ) 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: llxx− +1 11001− x − qpqxxx== and =−= 1 lxx 100−− 100 x 34 Chapter 1 Introduction to survival models As a result, we have: 1 (a) pq=−1 =− 1 0.9 q25 =− 1 0.9 = 0.98800 []25 [] 25 75 1 (b) pq=−1 =− 1 0.95 q26 =− 1 0.95 = 0.98716 ⇒ 2 p = 0.97532 []25++ 1 [] 25 174 [] 25 l28 100− 28 (c) 32272pppp[]25==×=×= [] 25 [] 25 0.97532 0.96196 l27 100− 27 =+ + = + + = ♦♦ (d) eppp[]25 :3 []2523 [] 25 [] 25 0.98800 0.97532 0.96196 2.92527 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) l[]25+ 1 (b) l[]25 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: l27 73 (a) l[]25+ 1 == =73.94935 p[]25+ 1 0.98716 l[]25+ 1 73.94935 (b) l[]25 == =74.84752 p[]25 0.98800 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: µµ* ()xaxb=+ () where the constants are chosen so that µ *0(x) ≥ for all x >0 . Let sx* () be the survival function corresponding to the force µ * (x) . 35 Introduction to survival models Chapter 1 Then the relation between sx* () and sxX ( ) is as follows: xx sx*()=− expµµ *() ydy =− exp () aybdy () + ∫∫00 xx =−expaydybxaydybxµµ() −=− exp () exp ( − ) ∫∫00 a x =−expµ ()ydy exp ( − bx ) ∫0 −bx a = esx()X () For example, if female mortality is given by µF (x) and male mortality is given by µM ()x where: µµMF()xx=+0.01 1.005 () then the survival function for males is: −0.01x 1.005 sxeMF()= () sx () Gompertz’s law of mortality Under Gompertz’s law, the force of mortality is determined by the formula: µ ()xBc= x where B>0 , c >1 , x ≥ 0 This formula has a geometrically increasing force of mortality to model the effect of aging. The corresponding survival function is: x x x Bcy Bc( − 1) =−y =− =− sxG ()exp∫ Bcdy exp exp 0 ln()cc ln () 0 B =−−expmcx 1 where m = ( ()) ln()c 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: µ ()xABc=+ x where AB>>0and 0 Since this force formula is a linear transformation of Gompertz’s formula, the Makeham survival function, sxM (), is closely related to the Gompertz survival function, sxG (): −Ax sxeMG()= sx () Weibull’s law of mortality Under Weibull’s law, the force of mortality is determined by the formula: µ ()xkx= n where kn>>0, 0,and x ≥ 0 The Weibull survival function is: −kxn+1 = sxW () exp n + 1 36 Chapter 1 Introduction to survival models De Moivre’s law of mortality De Moivre’s law is based on a life table function that is linear: lxx =−ω for 0 ≤≤x ω This leads to a survival function of: lx x sxX ()==−1 for 0 ≤≤ xω l0 ω and a force of mortality of: −l′ 1 µ ()xx==x for 0 ≤<ω lxx ω − 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: ω − x 11 fxsx()==×=≤< ()µ () xfor 0 xω XX ωω− x ω and the pdf of the future lifetime function is: ω − xt− 11 ftstxtTx()()=+=×=≤<− Tx()()µω ( ) for 0 t x ωω−−−−xxtx ω It is important to note that these final two formulas indicate that: • the lifetime of a newborn follows a uniform distribution on the interval [0,ω ] • the future lifetime after age x follows a uniform distribution on the interval [0,ω − x ]. Constant force of mortality Finally, it is worthwhile to mention the constant force model: µ ()x = 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: −µx sxeX ()=> for all x 0 The pdf of the lifetime function is: −µx fxsxXX()== ()µµ () x efor all x > 0 and the pdf of the future lifetime function is: −+µ(xt) e −µt ftstxtTx()()=+=×=> Tx()()µµµ ( ) efor all t 0 e−µx 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. 37 Introduction to survival models Chapter 1 Chapter 1 Practice Questions Question 1.1 Compute the following probabilities from the life table in Section 1.2: 20pq,, 2|0 4|23 qpq ,, 4 5 Question 1.2 Suppose that 21|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 lx 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 µ ()xx=+3/ ( 2 ) for x > 0 . Determine the pdf, cdf, and survival function for the ∞ lifetime of a newborn. Check your work by verifying properties such as fxdx() =1 .. ∫0 Question 1.5 Using the survival model in Question 4, determine the values of p1 and 21|q . Question 1.6 Suppose that µ ()xx=−0.5/( 100 ) for 0100≤ x < . 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 20p 40 and 20| 20q 40 . 38 Chapter 1 Introduction to survival models Question 1.8 Suppose that µ ()xx=+1.1/ ( 100 ) for x ≥ 0 . Compute t p20 , 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 µ ()x =0.015 for all x ≥ 0 , 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 0.95 For the life table function lxx =−1,000() 100 for 0100≤ x ≤ , develop a formula for Tx . Question 1.13 For the survival model in Question 1.12, determine e20:5 . Question 1.14 µ = ≥ Suppose that ()x 0.015 for all x 0 . Determine ex:1 and ax( ) . Question 1.15 For the survival model in Question 1.14, determine q20 and m20 . Question 1.16 The life table function lxx µ () is called the curve of deaths. It approximates the expected number of deaths in the year of age from x to x + 1 . Calculate the curve of deaths for the life table given 0.75 by lxx =−1,000() 100 for 0100≤≤x . 39 Introduction to survival models Chapter 1 Question 1.17 You are given: qqq30===0.01 , 31 0.02 , 32 0.03 . Computing the following using the UDD assumption to calculate the number living at fractional ages: µ ()31.4 ,1.4pf 30 ,T()30 () 1.4 Question 1.18 Assuming the same mortality rates as in Question 1.17, calculate e . Then see if you can 30:3 =+ reason how to adapt the UDD relation eexx0.5 to calculate e30:3 . Question 1.19 −0.02x Suppose that lex =1,000 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: qkq[]xk+ =−()10.12() − xk+ . Determine 3 q[]20 . Question 1.20 For the select and ultimate model in Question 1.19, determine l[]20 . 40