Chapter 1

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Chapter 1 Survival Distributions and Life Tables

1.1 Introduction to Probability

1.1.1 Random Experiment

Any process with several possible outcomes, none of which is known in advance with certainty, e.g.: tossing a coin; winning a Lotto number; death claims during a year incurred by an insurance company.

1.1.2 Sample Space

Set of outcomes of a random experiment.

Example 1.1.1 Tossing a coin S = {H, T}

Winning Lotto number S =

{

(

44₆

)

possible outcomes

}

Death claims in a year S = {0, 1, 2 …N}

where N = total number of insured Life expectancy of a newborn S = [0, ω]

where ω = limiting age at death

1.1.3 Event

Any subset of a sample space

Example 1.1.2 S = Death claims in a year

A = {female deaths}

B = {deaths with face amounts > $250,000}

The probability of an event A is denoted by P(A) is some value in [0, 1] indicating the

likelihood of the occurrence of event A.

P(A) = 1 ⟹ A is certain to occur P(A) = 0 ⟹ A is certain not to occur

Example 1.1.3 Toss a coin 3 times: A = {≤3 heads occur}, B = {4 heads occur}. Then

⟹P(A) = 1 and P(B) = 0.

Facts:

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3. If A and B are disjoint events, then P(A∪B)=P(A)+P(B)

1.1.4 Finite Sample Spaces

Let S = { s1, s2, … sn}, P

{

si

}

=pi , i = 1, 2, …n. Then

∑

i=1 n

p_i=P{S}=1∧if A⊆S , P(A)=

∑

siϵA

P{s_i}

Example 1.1.4 Roll a fair die: S = {1, 2, 3, 4, 5, 6}, then P{1}=P{2}=…=P{6}=1/6 A = {even numbers occurs} = {2, 4, 6}

P(A)=P{2}+P{4}+P{6}=1/2

1.1.5 Continuous Sample Spaces

The probability of an event is defined by a probability density function (p.d.f.) f(t) such that 0≤ f(t)≤1 , t ϵ S and

_∫

t∈S

f(t)dt=1 . Then

P(A)=

∫

t∈A

f(t)dt

Example 1.1.5 Randomly generate a number in [0, 1];

f(t) = 1, t ϵ [0, 1], A = {number≤0.25} ⟹ P(A)=

∫

0 0.25

f(t)dt=0.25

1.1.6 Independent Events

Two events A, B are independent if P(A ∩ B)=P(AB)=P(A)P(B)

Example 1.1.6 Toss a fair coin twice: S = {HH, HT, TH, TT} and P{HH} = 1 4. Using independence:

P{HH}=P

{

heads∈1st_toss∧_heads∈2nd_toss

}

=

(

1 2

)(

1

2

)

=

1 4

1.1.7 Conditional Probability

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P(A|B)=P(AB) P(B)

Example 1.1.7 Roll a fair die: A = {even number occurs}, B = {number > 2 occurs} Directly, P(A|B)=P¿

By formula, P

(

A|B

)

=P(AB) /P(B)=P{4,6}/P{3,4,5,6}=

2 6 4 6

=1

2

Similarly, P(B|A)=P(AB)/P(A)= 2 6 3 6

=2 3

Facts:

1. P

(

B|B

)

=1

2. If A , B are independent P A|B¿=P(AB) P(B) =

P(A)P(B)

P(B) =P(A) i.e. the occurrence of B has no influence on the occurrence of A. Similarly, P

{

B|A

}

=P{B}.

1.1.8 Random Variable

A random variable X is a real valued function defined on a sample space S, i.e. for every outcome s ∈ S, X assigns a real number X(s).

Example 1.1.8 Toss a coin twice: S ={HH, HT, TH, TT} Let X = number of heads

obtained. Then X{HH} = 2, X{HT} = X{TH} = 1, X{TT} = 0,. i.e. X = {0, 1, 2}. If the coin is fair,

P{X=0}=P{TT}=1 4

P{X=1}=P{HT}+P{TH}=1 2

P{X=2}=P{HH}=1 4

Continuous random variable is a real valued continuous function defined on a sample

space S.

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1.1.9 Probability Distribution of a Random Variable

Given a probability distribution on a sample space S, this uniquely determines a probability distribution for any random variable X defined on S, since

P{X ≤t}=P

{

s:X(s)≤t

}

.

1.1.10 Discrete Distributions

X takes on a finite set of values and the probability distribution f(x) = P{X=x}

Example 1.1.10 Roll a fair die, X = outcome. Then

f(x)=P{X=x}=

{

1/6for x=1,2, … ,6

0otherwise

Fact: f(x)≥0∧

∑

x

f(x)=¿P{S}=1¿_.

1.1.11 Continuous Distributions

X is a continuous random variable and there exists a function f, s.t. 0≤ f(t)≤1, for all t,

∫

−∞

∞

f(x)dx=1and P{X ≤t}=

∫

−∞

t

f(x)dx,

then f(x) is the probability density function of x.

Example 1.1.11 Generate a random variable from [0, 1], X = outcome. Then

f(x)=

{

1for0≤ x ≤1

0otherwise

Example 1.1.12 X = standard normal random variable. Then

f(x)= 1

√

2πe

−x2

2 _{,−∞ ≤ x ≤ ∞}

1.1.12 Distribution Function of a Random Variable

F(x)=P{X ≤ x}=

{

∑

t ≤ x

f(t)if x is discrete

∫

−∞

x

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By the fundamental Theorem of Calculus: f(x)=∂ F(x) ∂ x .

1.1.13 Expected Value of a Random Variable

E(X)=Expected value of X=

{

∑

x

x P{X=x}if x is discrete

∫

−∞

∞

t f(t)dt if x is continuous

Interpretation: The expected value is the weighted average of the values of X where the weights are based on the probability of occurrence of these values.

Example 1.1.13 Consider an insurance policy which pays out $100,000 if an individual

dies during the year. Let X = payout. SupposeP{individual dies}=0.01. Then

{

P[X=$100,000]=0.01

P[X=0]=0.99

E(X)=0.01(100,000)+0.99(0)=1,000.

1.1.14 Variance of a Random Variable

Discrete Case:

Var(X)=variance of X=

∑

x

(

x−E(X)

)

2P{X=x}

¿

₍

∑

x

x2P{X=x}

₎

−

₍

E(X)

)

2

¿E

(

X2

)

−

(

E(X)

)

2

Continuous Case:

Var(X)=

∫

−∞

¿98,010,000+990,000=99,000,000

Alternatively,

Var(X)=E

(

X2

)

−

(

E(X)

)

2=(100,000)20.01+(0)20.99−10002

¿100,000,000−1,000,000=99,000,000

Standard deviation of X SD(X)=

√

Var(X)

In the previous example, SD(X) =

√

99,000,000≈9,950

1.1.15 Survival Distribution of a Random Variable

Define the random variable X = age at death of a newborn F(x) = distribution function of X

= P{X ≤ x}

¿P{newborn diesby age x}

S(x) = 1 – F(x)

= survival function of X

¿P{X>x}

¿P{newborn dies after age x}

¿P{newborn survives¿age x}

Note

1) S(0)=1

2) 0≤ S(x)≤1, x ≥0

3) S(x)>S(y), y>x

i.e.S(x) is a strictly decreasing function of x.

4) lim_{x→ ω}S(x)=0, where ω=limitingage at death

Any function possessing the above four properties qualifies for a survival function.

Example 1.1.15

S(x)=e−x_{, x ≥}₀

S(x)=1−x/100,0≤ x ≤100

Any probability function of interest can be expressed in terms of survival functions.

Example 1.1.16 P{newborn dies between30∧40}=P{30<X<40}

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¿S(30)−S(40)

S(30)

______________________________________________________________ 0 30 40

S(40)

P{X ≤30}=1−S(30)etc .

Example 1.1.17 S(x)= 1

10

√

100−x ,0≤ x ≤100

Find the probability that a life aged 0 will die between age 19 and age 36.

S(19)

______________________________________________________________ 0 19 36

S(36)

Probability = S(19)−S(36)

¿ 1

10

[

√

81−

√

64

]

¿ 1

10

1.2 Future Life time of Individual Age x

T(x) = future lifetime of an individual age (x)

= time until death for a person age (x)

1.2.1 Distribution function of T(x)

P[T(x)≤ t]=_tq_x

________________ tqx ___________________

__________________________________________________________________ x x+t

tpx=1−tqx

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¿P[T(x)>t]

1.2.2 Relationship Between T(x) and x

X = time until death of a newborn T(x) = time until death for a person age x

1.X=T(0)

2.tqx=P

{

(x)dies within t years

}

¿P

{

newborn dies between x∧x+t|newborn survived¿age x

}

¿P{newborn dies between x∧x+t}

P{newborn survived¿age x}

¿S(x)−S(x+t)

s(x)

3. tpx=1−tqx

¿S(x+t)

S(x)

¿P

{

newborn survives¿x+t|newborn survived¿age x

}

1.2.3 Other Probability Functions of T(x)

1qx=qx

¿P

_{

(x)dies within a year

}

¿S(x)−S(x+1)

S(x)

________________ q_x ___________________

__________________________________________________________________ x x+1

1px=px

¿1−q_x

¿s(x+1)

s(x)

t|uqx=P

{

(x)dies between x+t∧x+t+u

}

¿S(x+t)−S(x+t+u)

(9)

¿tpx−t+upx=tpx uqx+t

________ t|uqx ___________

__________________________________________________________________ x x+t x+t+u

t|1qx=t|qx=P[(x)diesbetween x+t∧x+t+1]

¿S(x+t)−S(x+t+1)

S(x)

¿tpx−t+1px=tpxqx+t

1.2.4 Curtate Future Lifetime of x

K(x) = curtate future lifetime of (x) P[K(x)=k]=P[k<T(x)≤ k+1]

¿P

{

(x)dies between x+k∧x+k+1

}

¿k|qx

Note

∑

k=0 ∞

P[K(x)=k]=

∑

k=0 ∞

k|qx=1

where 0|qx=qx.

i.ek|qxisthe probability distributionfor the discrete random variable K(x).

If ω=limiting age at death, then

_∑

k=0 ω−x−1

P

{

K(x)=k

}

=

∑

k=0 ω−x−1

k|qx=1

Example 1.2.1 S(x)= 1

10

√

100−x ,0≤ x ≤100

a) Find the probability that a life age 19 will die before age 36.

b) Find the probability that a life age 19 dies between ages 36 and 75.

Answer:

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________ 17q19 __________

__________________________________________________________________

0 19 36

Probability=₁₇q₁₉=S(19)−S(36) S(19) =

1

10

[

√

81−

√

64

]

1

10

√

81

=1 9

b¿ ________ ₁₇_|₃₉q19 __________ __________________________________________________________________

0 19 36 75

Probability=₁₇_|₃₉q₁₉=S(36)−S(75) S(19)

¿

1

10

[√

64−

√

25

]

1

10

√

81

¿1

3

Alternatively, Probability=₁₇_|₃₉q19=17p19.39q36

¿ S(36)

S(19).

S(36)−S(75) S(36) =

1 3

1.2.5 Force of Mortality

μ_x = Force of Mortality at age x

¿

−∂ ∂ x S(x)

S(x)

¿

−∂

∂ x

[

1−F(x)

]

1−F(x)

¿ f(x)

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Where F(x) = distribution function of X, and X = age of a newborn, f(x) = probability density function of X.

Interpretation: Since ∂

∂ x g(x)=∆ x →lim0

g(x+∆ x)−g(x)

∆ x ,

μ_x=lim ∆ x→0

S(x)−S(x+∆ x)

∆ x S(x)

¿ lim

∆ x→0

P{x dies between x∧x+∆ x} ∆ x

¿instataneous rate of deathof x .

1.2.6 Relationship to Other Mortality Functions

Since ∂

∂ x log

[

S(x)

S(x+n)

]

=¿log⁡(npx)¿

_⟹ npx=e

−∫

x x+n

μydy

¿particular:

∫

0 x

μ_ydy=−log_xp₀=−log⁡[S(x)]

⟹S(x)=e

−∫

0

x μydy

T(x) = time until death for a person age x

tqx = distribution function of T(x) = P

{

T(x)≤ t

}

=G(t)

∂

∂ ttqx = g(t)

= probability density function of T(x)

¿ ∂

∂ t

S(x)−S(x+t)

S(x)

¿−S '

(x+t)

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¿− S'₍_x

+t)

S(x+t)∗S(x+t)

S(x)

¿tpxμx+t

i.e. tpxμx+t= p.d.f. of T(x). Then

∫

0 ∞

tpxμx+tdt=1

∫

n n+m

tpxμx+tdt=n|mqx

∫

0 r

tpxμx+tdt=rqx

∫

r ∞

tpxμx+tdt=rpx

In particular, for a newborn,

_∫

0 ∞

S(t)μ_tdt=

∫

0 ∞

[

−∂∂ t S(t)

]

[

−B

logC(C

x₋₁₎

]

Check

S(x)≥0, S(0)=1

lim

x→ ∞S(x)=

exp(B)

logC ≠0, since0≤C ≤1 Hence μx=BC

x_is_not_{a force of mortality function.}

2.¿

∫

0 x

B(y+1)−0.5dy=B

√

y+1

0.5

|

0x=2B

[

√

x+1−1

]

,

i.e.S(x)=exp⁡(−2B

[

√

x+1−1

]

) Check

S(x)≥0, S(0)=1

lim

x→ ∞S(x)=0,

S(x) is strictly decreasing.

Hence μx=B(x+1)−0.5 is a legitimate force of mortality function.

3.¿

∫

0 x

k(y+1)ndy=k(y+1) n+1

n+1

|

x

0=

k

n+1

[

(x+1) n+1₋₁

_]

⟹S(x)=exp

(

−k

n+1

[

(x+1) n+1₋₁

]

)

Check

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lim

x→ ∞S(x)=0, S(x)is strictly decreasing

Hence μx=k(x+1)

n_{is a legitimate force of mortality function.}

Example 1.2.3 The mortality pattern of a certain population may be described as follows: Out of every 98 lives born together, one dies annually until there are no survivors.

a) Find S(x) for this population Answer:

Number of survivors out of 98 at the end of x years is 98 – x. i.e.

S(x)=98−x

98 ,0≤ x ≤98

b) What is the probability that a life aged 30 will survive to attain age 35?

Answer:

________ 5p30 _______ ______________________________________________

30 35 Probability = 5p30

¿S(35)

S(30)

¿98−35

98−30

¿63

68

c) What is the force of mortality at 48?

Answer:

μx= −S'(x)

S(x)

¿ 1

98−x

S(x)=98−x

98 ∧S

'

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⟹μ₄₈= 1 98−48

¿0.02

1.3 Life Tables

Notations:

l₀= initial group of newborns

L(x)= number of survivors to age x

Then L(x) Bin

_[

l0, S(x)

]

∧E

[

L(x)

]

=l0S(x)=lx Dx

¿l_x−l_x₊_n ¿ndx Since S(x)– S(x+n)=P[newborn dies between x∧x+n]

¿P[newborn survives¿x dies between x∧x+n]

¿P[newborn survives¿x]∗P[newborn dies between x∧x+n|survival¿x]

¿S(x)nqx

⟹E

_[

nDx

]

=l0

[

S(x)nqx

]

¿lxnqx=lx−lx+n

E

[

L(x+n)

]

=l0S(x)npx=lxnpx=lx+n

Notation:

1¿1Dx=Dx∧E

[

1Dx

]

=1dx=dx=lx−lx+1=lxqx 2¿l_x₊₁=l_xp_x

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μ_x= − ∂

∂ x S(x) S(x) =

− ∂

∂ xl0S(x) l0S(x)

= − ∂

∂ xlx l_x

l_x=l₀S(x)=l₀exp

[

−

∫

0 x

μ_ydy

]

=l₀

_∫

0 x

S(t)μ_tdt=

∫

x ∞

l_tμ_tdt

l_x₊_n=l_x_np_x=l_xexp

[

−

∫

x x+n

μ_ydy

]

=l_x

_∫

n ∞

p_x

n μx+tdt=

∫

n ∞

l_x₊_tμ_x₊_tdt

l_x_nq_x=l_x−l_x₊_n=l_x

_∫

0 n

p_x

t μx+tdt=

∫

0 n

A life table is a table of expected survivors and expected deaths at each age starting with an initial cohort of l₀ lives (called the radix of the table).

Example 1.3.1 Consider the following life table

Age x l_x d_x

0 100,000 501

1 99,499 504

2 98,995 506

3 98,489 509

4 97,980 512

5 97,468 514

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¿(509−512)

100,000

¿0.0102 1

c . P{1year old dies between3∧5}=s(2)−s(4) s(1) =

(

l2−l4

)

l₁

¿

(

d2−d3

)

l₁

¿(506−509)

99,499

¿0.0102 0

1.3.2 Moments of T(x)

Recall T(x) = time until death for a person age x, T is continuous with p.d.f. f(t) = npxμx+t

and d.f. G(t) = P

[

T ≤t

]

=tqx Integration by parts: Given functions f(x), g(x)

∫

0 t

f (x)

[

∂

∂ xg(x)

]

dx=f(x)g(x)

|

0t−

∫

0 t

g(x)

[

∂

∂ xf(x)

]

dx

For time untildeath r . v . T , for a given function z(T)s . t . z(T)

_[

1−G(t)

_]

→0as t → ∞.

E

_[

z(T)

_]

=

∫

0 ∞

z(t)f(t)dt

¿

∫

0 ∞

−z(t) ∂

∂t

[

1−G(t)

]

dt

¿−z(t)

[

1−G(t)

]

|

∞

0+

∫

0 ∞

[

∂

∂t z(t)

]

[

1−G(t)

]

dt

¿z(0)+

∫

0 ∞

[

∂

∂t z(t)

]

t pt xdt

Var(T)=E

(

T2

)

−

[

E(T)

]

2=2

_∫

0 ∞

t pt xdt−

[

e˙x

]

2

Example 1.3.2 Given the survival function

S(x)=e−0.05x_{, x ≥}₀

1.¿Calculate5|10q30

Answer:

________ 5|10q30 __________

__________________________________________________________________

30 35 45

q30 5|10 =

S(35)−S(45)

S(30) =

e−1.75_−e−2.25 e−1.5 =e

−0.25_−e−0.75

2.¿Calculate F(30)

Answer:

F(30)=P{X ≤30}=1−s(30)=1−e−1.5

3.¿Calculatee˙30

Answer:

˙ e_x=

∫

0 ∞

p_x t dt=

∫

0 ∞

[

S(x+t)

S(x)

]

dt

˙ e₃₀=

∫

0 ∞

e−0.05_dt =−

(

e

−0.05

0.05

)

(19)

¿ 1

0.05

¿20

4.¿Calculate Var[T(x)]

Answer:

E

[

T2

]

=2

_∫

0 ∞

t e−0.05t

dt=2

(

[

−t e

−0.05t

0.05

]

|

∞0+

∫

₀ ∞

e−0.05t

0.05 dt

)

¿ 2

0.05

∫

₀

∞ e−0.05t

dt

¿2

(

1

0.05

)

2

¿800

2=800−400=400

i .e .:Var

[

T(30)

]

=400

1.3.3 Median Future Lifetime of x

Median of T(x)is m(x)such that P

[

T>m

]

=1/2

i .e .:S(x+m) S(x) =1/2

Example 1.3.3 For S(x)=e−0.05x

, m(x)satifies

e−0.05(x+m)

e−0.05x = 1

2⟺e

−0.05m

=1 2

⟺−0.05m=log

(

1

2

)

=−log 2

(20)

Recall the discrete curtate future lifetime of (x) random variable K(x) where P

{

K(x)=k

=k+1px

Theorem 1.3.1

For any given function z(K)

E

_[

z(K)

_]

=

∑

t=0 w−x−1

z(t)_t_|q_x=z(0)+

∑

t=0 w−x−2

Δ z(t)_t₊₁p_x, where Δ z(t)=z(t+1)−z(t)

Proof:

E

[

z(K)

]

=z(0)qx+z(1)1|qx+z(2)2|qx+…+z(w−x−1)w−x−1|qx

w−x−1|qx

¿z(0)

_[

qx+1|qx+2|qx+…+w−x−1|qx

]

+Δ z(0)

[

1|qx+2|qx+…+w−x−1|qx

]

+Δ z(1)

[

2|qx+3|qx+…+w−x−1|qx

]

+…+Δ z(w−x−2)w−x−1|qx

¿z(0)+Δ z(0)px+Δ z(1)2px+…+Δ z(w−x−2)w−x−1px

¿z(0)+

∑

t=0 w−x−2

Δ z(t)_t₊₁p_x

In particular, forz(K) = K, z(0)=0, ∆ z(t)=1

E

[

K

]

=

∑

t=0 w−x−2

t q_t_| _x=

∑

t=0 w−x−2

p_x t+1 =ex

e_x=thecurtate expectation of life at age x .

For z(K)=K2, z(0)=0, ∆ z(t)=1

E

[

K2

_]

=

∑

t=0 w−x−2

t2 _q x t| =

∑

t=0 w−x−2

(2t+1)_t₊₁p_x

Var(K)=E

[

K2

]

−

_[

E(K)

_]

2=

∑

t=0 w−x−2

(2t+1)_t₊₁p_x+

₍

e_x

₎

2

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d₉₀=6, d₉₁=d₉₂=3, d₉₃=d₉₄=d₉₅=d₉₆=2, d₉₇=1

Calculate ex=E(K)∧Var(K)

Answer:

x l_x d_x _t₊₁_px _(2t+1)_t₊₁_px t

90 21 6 15

21

15 21

0

91 15 3 12

21

36 21

1

92 12 3 9

21

45 21

2

93 9 2 7

21

49 21

3

94 7 2 5

21

45 21

4

95 5 2 3

21

33 21

5

96 3 2 1

21

13 21

6

97 1 1 0

21

0 21

7

98 0 0 −−¿ −−¿ 8

Total52 21

236 21

E(K)=e_x=

∑

t=0 6

p₉₀ t+1 =

52 21=2.48

Var(K)=E

[

K2

]

−

[

E(K)

]

2=

∑

t=0 6

(2t+1)_t₊₁p_x−(2.48)2

¿236

21 −(2.48)

−∂

∂ t lx+t

]

/lx+t¿

¿l_x₊₁+t

₍

−l_x₊_t

₎

|

1

0+

∫

0 1

l_x₊_td t ,

¿

∫

0 1

l_x₊_tdt

i.e. L_x=

∫

0 1

l_x₊_tdt.

1.3.5 Central-Death-Rate at age x

m_x=dx L_x=

∫

0 1

∫

0 1

lx+tdt

0 ∞

t p_t _xμ_x₊_tdt=l_x

_∫

0 ∞

p_x t dt=

∫

0 ∞

l_x₊_tdt

1.3.7 Average Years Lived After x Amongst Deaths in [x, x+1]

Total deathsbetween

[

x , x+1

]

=d_x=

∫

0 1

Expected number of years lived between

[

x , x+1

]

=

∫

0 1

i.e. average number of years lived between

[

x , x+1

]

amongst d_xdeaths

a(x)=

∫

0 1

Example 1.3.5 Out of 100 lives age x, four die at age x + 1/4, two die at age x + 1/2 and four die at age x + 3/4.

L_x=total years lived between

[

x , x+1

]

¿90+4

(

1

4

)

+2

(

1 2

)

+4

(

3

4

)

=95

m_x=central death rate at age x=10 95

a(x)=average years lived between

[

x , x+1

]

amongst d_xdeaths∈

[

x , x+1

]

¿4¿ ¿ ¿1

2

1.4 Assumptions for Fractional Ages

Given a lifetable with values for qx, pxat integer values of x , we need¿estimate qx

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1.4.1 Uniform Distribution of Deaths

qx

t =(t)qxfor0≤ t ≤1

i .e . pt x=1−tqx=1−(t)qx

Interpretation: For0≤ t+y ≤1

l_x₊_y−l_x₊_y₊_t=number of deaths betweenages x+y∧x+y+1

¿lxypx−lxy+tpx

¿l_x

_[

1−(y)q_x

_]

−l_x[1−(y+t)q_x]

¿l_xt q_x ¿t d_x

i .e . number of deaths

any age interval t∈

[

x , x+1

]

¿t

(

number of deaths between

[

x , x+1

])

i .e . deaths areuniformly distributed [x , x+1]

Other Results

1.) lx+t=lx−(t)dx=lx−t

[

lx−lx+1

]

2.¿μ_x₊_t= − ∂

∂ tlx+t lx+t

= dx lx+t

=qx px t

= qx

(

1−(t)q_x

)

3.) tpxμx+t=qx

4.¿_yq_x₊_t=lx+t−lx+t+y l_x₊_t =

(y)dx l_x_tp_x =

(y)qx

1−(t)q_x

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μ_x₊_tis an increasing function of t:

lim t →1 μx+t

=μ_x₊₁= qx

(

1−qx

)

=qx

px

lim

t →1 μx+t+1=μx+1=qx+1

i .e . μ_x₊_tis discontinous at the end points between successive age intervals .

1.4.2 Balducci Assumption

qx+t

1−t =(1−t)qx

Interpretation: Based on the assumption that the function 1/l_x function is linear between x and x+1.

i .e . 1 l_x₊_t=

1 l_x+t

(

1 l_x₊₁−

1 l_x

)

ˇ :lx+1

l_x₊_t= l_x₊₁

l_x +t

(

1− l_x₊₁

l_x

)

i.e. 1−1−tqx+t=1−qx+(t)qx⟺1−tqx+t=(1−t)qx

Note

The Uniform Distribution of Deaths assumption assumes that the l_x_{function is linear} between x and x+1.

i .e .:l_x₊_t=l_x−t

_[

l_x−l_x₊₁

_]

Other Results

1) Using the relationships

qx=tqx+tpx.1−tqx+t

¿1−tpx+tpx.1−tqx+t i .e . pt x

(

1−1−tqx+t

)

=px

i .e . p_t _x= px

(

1−₁₋_tq_x₊_t

₎

= p_x

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2) tqx=1−tpx

¿1−(1−t)qx−px

1−(1−t)qx

=t qx 1−(1−t)qx

3) μ_x₊_t=−l'_x₊_t/l_x₊_t

Since 1 l_x₊_t=

1 l_x+t

(

1 l_x₊₁−

1

l_x

)

, differentiating implicitly we get

−1

(

lx+t

)

2. l'x+t= 1 l_x₊₁−

1 l_x

i .e . ,−l'x+t=

(

lx+t

)

2

(

l_x1₊₁− 1 l_x

)

μ_x₊_t=−l '

x+t l_x₊_t =lx+t

[

1 l_x₊₁−

1 l_x

]

But

lx+1

lx+t

=1−₁₋_tq_x₊₁=1−(1−t). q_x

and lx+t lx

=_tp_x=p_x/ [1−(1−t). qx]

i.e. μ_x₊_t=lx+t lx+1

−lx+t lx

¿ 1

1−(1−t)qx−

p_x

1−(1−t)qx

¿ qx

1−(1−t)qx

4) tpxμx+t=

[

p_x 1−(1−t)q_x

][

q_x

1−(1−t)q_x

]

=

p_xq_x

[

1−(1−t)qx

]

2

Note

(27)

¿

(

1−

[

1−t

]

qx

)

qx

1−(1−t)qx

¿q_x

Recall under UDD assumption, tpxμx+t=qx

b¿Under B alducci , μ_x₊₁₋_t= qx

1−t . qx

¿μ_x₊_tunder UDD

5) Using the relationship

px+t

1−t =yqx+t+ypx+t.1−t−yqx+t+y

--- ---x ---x+t ---x+t+y ---x+1

We get 1−tqx=yqx+t+

(

1−yqx+t

)

+(1−t−y)qx

i .e . , qy x+t

[

1−(1−t−y)qx

]

=(1−t)qx−(1−t−y)qx

i .e . , q_y _x₊_t= yqx

1−(1−t−y)q_x

1.4.3 Constant Force of Mortality

μ_x₊_t=μ , for0≤ t ≤1

Other Results

1.¿_tp_x=exp

{

−

∫

x x+t

μ_ydy

}

=exp{−μt}=

₍

p_x

₎

t

2.¿_tq_x=1−_tp_x=1−exp{−μt}=1−

(

px

)

t

3.¿l_x₊_t=l_x_tp_x=l_xexp{−μt}=lx

(

px

)

t

4.¿_yq_x₊_t=1−_yp_x₊_t=1−exp

{

−

∫

₍

p_x

₎

t

Example 1.4.1 There are 100 persons now age 40 of whom 19 are expected to die

before age 41. Determine 0.5q40+0.375 assuming (a) UDD (b) Balducci and (c) Constant Force.

q40=0.19, p40=0.81, Want q0.5 40+0.375.

a¿Under UDD q_x₊_t=¿ ¿ ¿

i .e . q_0.5 ₄₀₊_0.375= (0.5)0.19

1−(0.375)0.19=0.102

Alternatively , q_0.5 ₄₀₊_0.375=l40+0.375−l40+0.875 l40+0.375

¿ (0.5)19

100−(0.375)19=

9.5

92.875=0.102

b¿Under Balducci q_x₊_t= (y)qx

1−(1−t−y)q_x

y

i .e . q_0.5 ₄₀₊_0.375= (0.5).19

1−(0.125)0.19=0.0097

c¿Under Constant Force q_x₊_t=1−_yp_x₊_t=1−_y¿

₍

p_x

₎

y¿

i .e . q0.5 40+0.375=1−0.810.5=0.1

Example 1.4.2 You are given

i. Deaths are uniformly distributed over each year of age. ii.

28

x l_x

35 100

36 99

37 96

(29)

Which of the following are true? I. 1|2q36=0.091

II. m₃₇=0.043 III. 0.33q38.5=0.021

Answer:

I . q₁_|₂ ₃₆=l37−l39 l36

=96−87

99 =0.091⟹I istrue

II . m₃₇=d37 L37

under UDD L_x=

∫

0 1

l_x₊_tdt=

_∫

0 1

(

lx−(t)dx

)

dt=lx−(1/2)dx

m₃₇=d37 L37

= 96−92

96−(1/2)4=0.043⟹II istrue

III ._0.33q_38.5=l38.5−l38.83 l38.5

= 0.33(5)

92−0.5(5), since lx+t=lx−(t)dx

¿0.018⟹I II is false

1.4.4 Analytical Laws of Mortality

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¿ω−x

ω

l_x=l₀S(x)=l₀ω−x

ω =l0−x l₀ ω

i .e . d_x=l_x−l_x₊₁=l0 ω

i .e .:de Moivr e's Law assumes a contant l0

ωdeaths each year .

Also, density function of T(x)=_tp_xμ_x₊_t

¿s(x+t)

s(x) μx+t

¿ω−x−t

ω−x . 1 ω−x−t

¿ 1

ω−x, which is constant

Gompertz

μx=B c x

, B>0

Then S(x)=exp

{

−

∫

0 x

B cydy

}

¿exp

{

−B c

y

logc

|

x0

}

¿exp

{

−B

logc

(

c

x₋₁

₎

}

¿exp

{

−m

(

cx−1

)

}

, where m= B

logc

S(0)=1∧sincelim

x→ ∞S(x)=0⇒c>1

Whenc=1, μ_x=B=constant forceof mortality assumption∧S(x)=exp⁡{−Bx}

Makeham

(31)

Then S(x)=exp

{

−

∫

0 x

A+B cydy

}

¿exp

{

−Ay−B c y

logc

|

0x

}

¿exp

{

−Ax−B

(

c x

−1

)

logc

}

¿exp

{

−Ax−m

(

cx−1

)

}

, wherem= B logc

Since S(x)is strictly decreasing , S '(x)<0,

i .e . ,exp

{

−Ax−B

(

c x₋₁

₎

logc

(

−A−B c x

₎

}

<0⟺A>−B cxwhich holdif A>−B since c>1

Note:When A=0, then Makeham=Gompertz

Weibull

μx=k x n

Then S(x)=exp

{

−

∫

0 x

k yndy

}

=exp

{

−k x n+1

n+1

}

, k>0, n>0.

Example 1.4.3 You are given:

i. Mortality follows de Moivre’s Law ii. Var

[

T(15)

]

=675

Calculate e˙25.

Answer:

Under de Moivr e'_{s Law μ}

x=(ω−x)

−1_{, S}

(x)=ω−x ω

Then p_t _x=S(x+t) S(x) =

ω−x−t

ω−x , μx+t= 1

(32)

i .e . p . d . f . of T(x) U(0, ω−x)

E

[

T(x)

]

= ˙e_x=

∫

o ω−x

¿ 1

ω−x t2

2

|

ω−x 0

¿ω−x

2

Var

[

T(x)

]

=E

_[

(

T(x)

)

2

]

–

₍

e˙x

)

2

=

∫

o ω−x

t2_tp_xμ_x₊_tdt−

(

ω−x

2

)

2

¿ 1

ω−x t3

3

|

ω−x0 −

(ω−x)2

4

¿(ω−x)

2

3 –

(ω−x)2

4

¿(ω−x)

2

12

Given Var

[

T(15)

]

=675

⟹(ω−15) 2

12 =675⟺ω−15=

√

675(12)

⟺ω=105

Wante˙25= ω−25

2 =

105−25

2 =40

1.4.5 Select and ultimate Tables

A select mortality table recognizes the fact that a newly insured age x shows better (i.e. lower) mortality than an individual age x insured t years ago. This is because of the selection or underwriting process a newly insured has to undergo.

i.e., probability of dying in the coming year is a function of one’s attained age and the duration since the issue of the policy.

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q_[_x_]₊_t=P{life age x+t∧newly insured at age x dies∈the coming year}

Then

q_[_x_]≤ q_[_x₋₁_]₊₁≤ q_[_x₋₂_]₊₂≤…

i.e. for a given attained age x, the closer the duration since issue, the better the mortality.

The effect of selection usually wears out after a certain number of years, typically 15 for mortality. If the select period is for r durations, then

q_[_x_]₊_r=q_x₊_r=q_[_x₋₁_]₊_r₊₁=q_[_x₋₂_]₊_r₊₂=…

qx+r=ultimate motality at attained age x+r

¿mortality rate at attained age x+r amongst insured ’ s with policiesissued

at least r years ago.

For a select period of 3 years, a select and ultimate mortality table will look as follows: Issue

Age 1 DurationSelect

2 3

Ultimate 4+

x q_[_x_] q_[_x_]₊₁ q_[_x_]₊₂ q_x₊₃ x+1 q_[_x₊₁_] q_[_x₊₁_]₊₁ q_[_x₊₁_]₊₂ q_x₊₄ X+2 q_[_x₊₂_] q_[_x₊₂_]₊₁ q_[_x₊₂_]₊₂ q_x₊₅

Aggregate mortality at age x is the mortality rate at attained age x without regard to duration since issue, i.e., select and ultimate lives of the same attained age are grouped together.

Note

l_[_x_]₊_k=number of survivorsat attained age(x+k)amongst l_xlives insured at age x .

For a select period of r years,

l_[_x_]₊_r=l_x₊_r

(34)

Answer:

[30] 32 34 40

For a five year period, the l_x_{function is of the form}

l_[₃₀_], l_[₃₀_]₊₁,l_[₃₀_]₊₂,l_[₃₀_]₊₃,l_[₃₀_]₊₄, l₃₅, l₃₆, …

Then q₂_|₆ _[₃₀_]₊₂=l[30]+2+2−l[30]+2+8 l_[₃₀_]₊₂

¿l[30]+4−l40

l_[30]+2

Example 1.4.5 Express by a single symbol the probability that a life now age 50, insured three years ago, will die in his 59th_{year i.e., between ages 58 and 59 assuming a}

five year select period.

Answer:

[47] 50 58 59

Probability=₈_|q_[₄₇_]₊₃

¿d[47]+3+8

l_[₄₇_]₊₃

¿ d58

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