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1/27 Chapter 2. Survival models. Manual for SOA Exam MLC. Chapter 2. Survival models. Section 2.7 Common Analytical Survival Models c 2009. Miguel A. Arcones. All rights reserved. Extract from: ”Arcones’ Manual for SOA Exam MLC. Fall 2009 Edition”, available at http://www.actexmadriver.com/ c 2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.

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Page 1: MLC Section 2-7

1/27

Chapter 2. Survival models.

Manual for SOA Exam MLC.Chapter 2. Survival models.

Section 2.7 Common Analytical Survival Models

c©2009. Miguel A. Arcones. All rights reserved.

Extract from:”Arcones’ Manual for SOA Exam MLC. Fall 2009 Edition”,

available at http://www.actexmadriver.com/

c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.

Page 2: MLC Section 2-7

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Chapter 2. Survival models. Section 2.7 Common Analytical Survival Models.

Common Analytical Survival Models

Sometimes it is of interest to assume that the survival functionfollows a parametric model, i.e. it is of the form S(x , θ), where θis an unknown parameter. There are several reasons to make thisassumption:1. Data supports this assumption. Actuaries realized that themodels observed in real life follow this assumption.2. Computations are simpler using a parametric model.3. There are valid scientific reasons to justify the use of aparticular parametric model.

c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.

Page 3: MLC Section 2-7

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Chapter 2. Survival models. Section 2.7 Common Analytical Survival Models.

The most common approach is not to use parametric models dueto the following reasons:1. Modern computers allow to handle the computations neededusing the collected data.2. It is difficult to justify that a parametric model applies.3. Knowing that a particular model applies we can get moreaccurate estimates. But, this increase in accuracy is not much. If aparametric model does not apply, using the parametric approach wecan get much worse estimates than the nonparametric estimates.

c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.

Page 4: MLC Section 2-7

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Chapter 2. Survival models. Section 2.7 Common Analytical Survival Models.

De Moivre model.

De Moivre’s law (1729) assumes that deaths happen uniformlyover the interval of deaths, i.e. the density of the age–at–failure isfX (x) = 1

ω , for 0 ≤ x ≤ ω. Therefore,

SX (x) =ω − x

ω, for 0 ≤ x < ω,

FX (x) =x

ω, for 0 ≤ x < ω,

µ(x) =1

ω − x, for 0 ≤ x < ω,

tpx =s(x + t)

s(x)=

ω − x − t

ω − x, for 0 ≤ t ≤ ω − x ,

tqx =t

ω − x, for 0 ≤ t ≤ ω − x .

c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.

Page 5: MLC Section 2-7

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Chapter 2. Survival models. Section 2.7 Common Analytical Survival Models.

Example 1

Find the median of an age–at–death subject to de Moivre’s law ifthe probability that a life aged 20 years survives 40 years is 1

3 .

Solution: We know that 40p20 = 23 . For the uniform distribution,

tpx = ω−x−tω−x . Hence, 1

3 = ω−20−40ω−20 , ω − 20 = 3ω − 180 and

ω = 80. Let m be the median of the age–at–death. Then,

1

2= SX (m) =

ω −m

ω=

80−m

80

and m = 40.

c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.

Page 6: MLC Section 2-7

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Chapter 2. Survival models. Section 2.7 Common Analytical Survival Models.

Example 1

Find the median of an age–at–death subject to de Moivre’s law ifthe probability that a life aged 20 years survives 40 years is 1

3 .

Solution: We know that 40p20 = 23 . For the uniform distribution,

tpx = ω−x−tω−x . Hence, 1

3 = ω−20−40ω−20 , ω − 20 = 3ω − 180 and

ω = 80. Let m be the median of the age–at–death. Then,

1

2= SX (m) =

ω −m

ω=

80−m

80

and m = 40.

c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.

Page 7: MLC Section 2-7

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Chapter 2. Survival models. Section 2.7 Common Analytical Survival Models.

Under the De Moivre’s law, T (x) has a uniform distribution on theinterval [0, ω − x ]. Hence,

Theorem 1For the De Moivre’s law,

◦ex =

ω − x

2and Var(T (x)) =

(ω − x)2

12.

c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.

Page 8: MLC Section 2-7

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Chapter 2. Survival models. Section 2.7 Common Analytical Survival Models.

Example 2

Suppose that the survival of a cohort follows the De Moivre’s law.Suppose that the expected age–at–death of a new born is 70 years.Find the expected future lifetime of a 50–year old.

Solution: Since 70 = e0 = ω2 , ω = 140. The expected future

lifetime of a 50–year old is

◦e50 =

140− 50

2= 45.

c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.

Page 9: MLC Section 2-7

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Chapter 2. Survival models. Section 2.7 Common Analytical Survival Models.

Example 2

Suppose that the survival of a cohort follows the De Moivre’s law.Suppose that the expected age–at–death of a new born is 70 years.Find the expected future lifetime of a 50–year old.

Solution: Since 70 = e0 = ω2 , ω = 140. The expected future

lifetime of a 50–year old is

◦e50 =

140− 50

2= 45.

c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.

Page 10: MLC Section 2-7

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Chapter 2. Survival models. Section 2.7 Common Analytical Survival Models.

Theorem 2For the de Moivre’s law, for 0 ≤ x ≤ ω, x, ω integers,

ex =ω − x − 1

2.

Proof: We have that

ex =ω−x∑k=1

ω − x − k

ω − x=

ω−x∑k=1

1−ω−x∑k=1

k

ω − x

=(ω − x)− ω − x)((ω − x) + 1)

2(ω − x)= (ω − x)− ω − x) + 1

2

=ω − x − 1

2.

c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.

Page 11: MLC Section 2-7

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Chapter 2. Survival models. Section 2.7 Common Analytical Survival Models.

Theorem 2For the de Moivre’s law, for 0 ≤ x ≤ ω, x, ω integers,

ex =ω − x − 1

2.

Proof: We have that

ex =ω−x∑k=1

ω − x − k

ω − x=

ω−x∑k=1

1−ω−x∑k=1

k

ω − x

=(ω − x)− ω − x)((ω − x) + 1)

2(ω − x)= (ω − x)− ω − x) + 1

2

=ω − x − 1

2.

c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.

Page 12: MLC Section 2-7

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Chapter 2. Survival models. Section 2.7 Common Analytical Survival Models.

Theorem 3For the de Moivre’s law with terminal age ω, where ω is a positiveinteger, and each for 0 ≤ x ≤ ω, where x is an integer,

P{K (x) = k} =1

ω − x, k = 0, 1, 2, . . . , ω − x .

and

P{Kx = k} = pk−11 (1− p1), k = 1, 2, . . . , ω − x − 1.

Proof:

P{K (x) = k} = P{k < Tx ≤ k + 1} =

∫ k+1

k

1

ω − xdt =

1

ω − x

c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.

Page 13: MLC Section 2-7

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Chapter 2. Survival models. Section 2.7 Common Analytical Survival Models.

Exponential model.

An exponential model assumes that the age–at–death has anexponential distribution, i.e.

SX (x) = e−µx , x ≥ 0

where µ > 0. In this case,

FX (x) = 1− e−µx , for 0 ≤ x ,

fX (x) = µe−µx , for 0 ≤ x ,

µ(x) = µ, for 0 ≤ x ,

tpx =s(x + t)

s(x)= e−µt = pt

x .

For an exponential model, the force of mortality is constant. Theexponential model is also called the constant force model.

c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.

Page 14: MLC Section 2-7

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Chapter 2. Survival models. Section 2.7 Common Analytical Survival Models.

Example 3

Suppose that:(i) the force of mortality is constant.(ii) the probability that a 30–year–old will survive to age 40 is 0.95.Calculate:(i) the probability that a 40–year–old will survive to age 50.(ii) the probability that a 30–year–old will survive to age 50.(iii) the probability that a 30–year–old will die between ages 40 and50.

Solution: (i) Since the force of mortality is constant,

10p40 = 10p30 = 0.95.(ii) 20p30 = 10p30 · 10p40 = (0.95)(0.95) = 0.9025.(iii) We can do either

10|10p30 = 10p30 − 20p30 = 0.95− 0.9025 = 0.0475.

or

10|10p30 = 10p30 · 10q40 = (0.95)(1− 0.95) = 0.0475.

c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.

Page 15: MLC Section 2-7

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Chapter 2. Survival models. Section 2.7 Common Analytical Survival Models.

Example 3

Suppose that:(i) the force of mortality is constant.(ii) the probability that a 30–year–old will survive to age 40 is 0.95.Calculate:(i) the probability that a 40–year–old will survive to age 50.(ii) the probability that a 30–year–old will survive to age 50.(iii) the probability that a 30–year–old will die between ages 40 and50.

Solution: (i) Since the force of mortality is constant,

10p40 = 10p30 = 0.95.

(ii) 20p30 = 10p30 · 10p40 = (0.95)(0.95) = 0.9025.(iii) We can do either

10|10p30 = 10p30 − 20p30 = 0.95− 0.9025 = 0.0475.

or

10|10p30 = 10p30 · 10q40 = (0.95)(1− 0.95) = 0.0475.

c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.

Page 16: MLC Section 2-7

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Chapter 2. Survival models. Section 2.7 Common Analytical Survival Models.

Example 3

Suppose that:(i) the force of mortality is constant.(ii) the probability that a 30–year–old will survive to age 40 is 0.95.Calculate:(i) the probability that a 40–year–old will survive to age 50.(ii) the probability that a 30–year–old will survive to age 50.(iii) the probability that a 30–year–old will die between ages 40 and50.

Solution: (i) Since the force of mortality is constant,

10p40 = 10p30 = 0.95.(ii) 20p30 = 10p30 · 10p40 = (0.95)(0.95) = 0.9025.

(iii) We can do either

10|10p30 = 10p30 − 20p30 = 0.95− 0.9025 = 0.0475.

or

10|10p30 = 10p30 · 10q40 = (0.95)(1− 0.95) = 0.0475.

c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.

Page 17: MLC Section 2-7

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Chapter 2. Survival models. Section 2.7 Common Analytical Survival Models.

Example 3

Suppose that:(i) the force of mortality is constant.(ii) the probability that a 30–year–old will survive to age 40 is 0.95.Calculate:(i) the probability that a 40–year–old will survive to age 50.(ii) the probability that a 30–year–old will survive to age 50.(iii) the probability that a 30–year–old will die between ages 40 and50.

Solution: (i) Since the force of mortality is constant,

10p40 = 10p30 = 0.95.(ii) 20p30 = 10p30 · 10p40 = (0.95)(0.95) = 0.9025.(iii) We can do either

10|10p30 = 10p30 − 20p30 = 0.95− 0.9025 = 0.0475.

or

10|10p30 = 10p30 · 10q40 = (0.95)(1− 0.95) = 0.0475.

c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.

Page 18: MLC Section 2-7

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Chapter 2. Survival models. Section 2.7 Common Analytical Survival Models.

Theorem 4Suppose that the lifetime random variable of a new born hasconstant mortality force µ. Then,

◦ex =

1

µand Var(X ) =

1

µ2.

Proof: Since tpx = e−µt , T (x) has an exponential distributionwith mean 1

µ .

c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.

Page 19: MLC Section 2-7

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Chapter 2. Survival models. Section 2.7 Common Analytical Survival Models.

Theorem 4Suppose that the lifetime random variable of a new born hasconstant mortality force µ. Then,

◦ex =

1

µand Var(X ) =

1

µ2.

Proof: Since tpx = e−µt , T (x) has an exponential distributionwith mean 1

µ .

c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.

Page 20: MLC Section 2-7

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Chapter 2. Survival models. Section 2.7 Common Analytical Survival Models.

Theorem 5Suppose that the lifetime random variable of a new born hasconstant mortality force µ. Then, the curtate future lifetime of (x)is

ex =1

eµ − 1.

Proof: We have that

ex =∞∑

k=1

kpx =∞∑

k=1

e−kµ =e−µ

1− e−µ=

1

eµ − 1.

c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.

Page 21: MLC Section 2-7

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Chapter 2. Survival models. Section 2.7 Common Analytical Survival Models.

Theorem 5Suppose that the lifetime random variable of a new born hasconstant mortality force µ. Then, the curtate future lifetime of (x)is

ex =1

eµ − 1.

Proof: We have that

ex =∞∑

k=1

kpx =∞∑

k=1

e−kµ =e−µ

1− e−µ=

1

eµ − 1.

c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.

Page 22: MLC Section 2-7

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Chapter 2. Survival models. Section 2.7 Common Analytical Survival Models.

Example 4

Suppose that:(i) the force of mortality is constant.(ii) the probability that a 30–year–old will survive to age 40 is 0.95.Calculate:(i) the future lifetime of a 40–year–old.(ii) the future curtate lifetime of a 40–year–old.

Solution: (i) We know that 10p30 = 0.95 = e−(10)µ. Hence,

µ = − ln(0.95)10 and

◦e40 = 1

µ = 10− ln(0.95) = 194.9572575.

(ii) Since e−µ = (0.95)0.1,e40 = 1

eµ−1 = 1(0.95)−0.1−1

= 194.4576849.

c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.

Page 23: MLC Section 2-7

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Chapter 2. Survival models. Section 2.7 Common Analytical Survival Models.

Example 4

Suppose that:(i) the force of mortality is constant.(ii) the probability that a 30–year–old will survive to age 40 is 0.95.Calculate:(i) the future lifetime of a 40–year–old.(ii) the future curtate lifetime of a 40–year–old.

Solution: (i) We know that 10p30 = 0.95 = e−(10)µ. Hence,

µ = − ln(0.95)10 and

◦e40 = 1

µ = 10− ln(0.95) = 194.9572575.

(ii) Since e−µ = (0.95)0.1,e40 = 1

eµ−1 = 1(0.95)−0.1−1

= 194.4576849.

c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.

Page 24: MLC Section 2-7

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Chapter 2. Survival models. Section 2.7 Common Analytical Survival Models.

Example 4

Suppose that:(i) the force of mortality is constant.(ii) the probability that a 30–year–old will survive to age 40 is 0.95.Calculate:(i) the future lifetime of a 40–year–old.(ii) the future curtate lifetime of a 40–year–old.

Solution: (i) We know that 10p30 = 0.95 = e−(10)µ. Hence,

µ = − ln(0.95)10 and

◦e40 = 1

µ = 10− ln(0.95) = 194.9572575.

(ii) Since e−µ = (0.95)0.1,e40 = 1

eµ−1 = 1(0.95)−0.1−1

= 194.4576849.

c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.

Page 25: MLC Section 2-7

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Chapter 2. Survival models. Section 2.7 Common Analytical Survival Models.

Gompertz model.

Gompertz’s model (1825) says that µx = Bcx , where B > 0 andc > 1. Hence, s(x) = e−m(cx−1) for x ≥ 0, where m = B

log c .

c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.

Page 26: MLC Section 2-7

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Chapter 2. Survival models. Section 2.7 Common Analytical Survival Models.

Makeham model.

Makehan (1860) introduced the model µx = A + Bcx , whereA ≥ −B, B > 0 and c > 1. Hence, s(x) = e−Ax−m(cx−1) forx ≥ 0, where m = B

log c . We also have that

fX (x) = s(x)µ(x) = (A + Bcx)e−Ax−m(cx−1), x ≥ 0.

c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.

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Chapter 2. Survival models. Section 2.7 Common Analytical Survival Models.

Weibull model.

Weibull (1939) introduced the model µ(x) = kxn, for x ≥ 0,where k > 0 and n > −1. Then,

s(x) = e−kxn+1

n+1 , x ≥ 0

fX (x) = s(x)µ(x) = kxne−kxn+1

n+1 , x ≥ 0.

c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam MLC.