Chapter 8 LIFE ANNUITIES Basic Concepts Commutation Functions Annuities Payable m thly Varying Life...

Chapter 8

LIFE ANNUITIES

• Basic Concepts

• Commutation Functions

• Annuities Payable mthly

• Varying Life Annuities

• Annual Premiums and Premium Reserves

8.1 Basic Concepts

• We know how to compute present value of contingent payments

• Life tables are sources of probabilities of surviving

• We can use data from life tables to compute present values of payments which are contingent on either survival or death

Example (pure endowment), p. 155

• Yuanlin is 38 years old. If he reaches age 65, he will receive a single payment of 50,000. If i = .12, find an expression for the value of this payment to Yuanlin today. Use the following entries in the life table: l38 = 8327, l65 = 5411

Pure Endowment

• Pure endowment: 1 is paid t years from now to an individual currently aged x if the individual survives

• Probability of surviving is t px

• Therefore the present value of this payment is the net single premium for the pure endowment, which is:

t Ex = (t px ) (1 + t) – t = v t t px

Example (life annuity), p. 156

• Aretha is 27 years old. Beginning one year from today, she will receive 10,000 annually for as long as she is alive. Find an expression for the present value of this series of payments assuming i = .09

• Find numerical value of this expression ifpx = .95 for each x

Life annuity

Series of payments of 1 unitas long as individual is alive

x x + 1 x + 2 x + n

1 1 1

…..

present value(net single premium)

of annuity ax

age

probability px 2px npx

…..

xtt

txn

nxxxx pvpvpvpvvpa

1

33

22

Temporary life annuity

Series of n payments of 1 unit(contingent on survival)

x x + 1 x + 2 x + n

1 1 1

…..

present value

ax:n|

age


xt

n

t

txn

nxxxnx

pvpvpvpvvpa

1

33

22

|:

last payment

n - years deferred life annuity

Series of payments of 1 unit as long as individual is alivein which the first payment is at x + n + 1

x x + 1 x + 2 x + n +1

1

…

present value

n|ax

age

probability n+1px

11

33

22

11|

nsxs

s

txtn

tn

xtntn

xnn

xnn

xnxn

pvpv

pvpvpvpva n

first payment

x + n + 2

1

n+2px

x + n …

Note: |:|

nxxxn aaa

Life annuities-due

x x + 1 x + 2 x + n

1 1 1

…..

äx

px 2px npx

…

x x + 1 x + 2 x + n

1 1

…..

äx:n|

px 2px

x x + 1 x + 2 x + n +1

1

…

n|äx

n+1px

x + n + 2

1

n+2px

x + n …

1

1

xtt

txx pvaa

1

11

xt

n

t

tnxnx

pvaa

1

1|1:|:

11

xnxn aa || 11

x + n-1

1

n-1px

npx

Note

|1:|:1

nxnxaa

but

|:|:)1(

nxnxaia

|:1|:1 nxxnxavpa

8.2 Commutation Functions

Recall: present value of a pure endowment of 1 to be paid n years hence to a life currently aged x

Denote Dx = vxlx

Then nEx = Dx+n / Dx

xx

nxnx

x

nxnxn

nxn lv

lv

l

lvpvE

Life annuity and commutation functions

11 t

xtxtt

tx Epva Since nEx = Dx+n / Dx

we have

3211

1xxx

t xx

txx DDD

DD

Da

Define commutation

function Nx as follows:

00 ttx

tx

ttxx lvDN

Then:x

xx D

Na 1

Identities for other types of life annuities

x

nxxn

t x

txnx D

NN

D

Da 11

1|:

x

nxxn D

Na 1|

x

nxxnx D

NNa

|:

temporary life annuity

n-years delayed l. a.

temporary l. a.-due

Examples (p. 162 – p. 164)

• (life annuities and commutation functions) Marvin, aged 38, purchases a life annuity of 1000 per year. From tables, we learn that N38 = 5600 and N39 = 5350. Find the net single premium Marvin should pay for this annuity

– if the first 1000 payment occurs in one year

– if the first 1000 payment occurs now

• Stay verbally the meaning of (N35 – N55) / D20

• (unknown rate of interest) Given Nx = 5000, Nx+1=4900,

Nx+2 = 4810 and qx = .005, find i

Select group

• Select group of population is a group with the probability of survival different from the probability given in the standard life tables

• Such groups can have higher than average probability of survival (e.g. due to excellent health) or, conversely, higher mortality rate (e.g. due to dangerous working conditions)

Notations• Suppose that a person aged x is

in the first year of being in the select group

• Then p[x] denotes the probability of survival for 1 year and q[x] = 1 – p[x] denotes the probability of dying during 1 year for such a person

• If the person stays within this group for subsequent years, the corresponding probabilities of survival for 1 more year are denoted by p[x]+1, p[x]+2, and so on

• Similar notations are used for life annuities:a[x] denotes the net single premium for a life annuity of 1 (with the first payment in one year) to a person aged x in his first year as a member of the select group

• A life table which involves a select group is called a select-and-ultimate table

Examples (p. 165 – p. 166)

• (select group) Margaret, aged 65, purchases a life annuity which will provide annual payments of 1000 commencing at age 66. For the next year only, Margaret’s probability of survival is higher than that predicted by the life tables and, in fact, is equal to p65 + .05, where p65 is taken from the standard life table. Based on that standard life table, we have the values D65 = 300, D66 = 260 and N67 = 1450. If i = .09, find the net single premium for this annuity

• (select-and-ultimate table) A select-and-ultimate table has a select period of two years. Select probabilities are related to ultimate probabilities by the relationships p[x] = (11/10) px and p[x]+1 = (21/20) px+1. An ultimate table shows D60 = 1900, D61 = 1500, and ä 60:20| = 11, when i = .08. Find the select temporary life annuity ä[60]:20|

• The following values are based on a unisex life table: N38 = 5600, N39 = 5350, N40 = 5105, N41 = 4865,N42 = 4625.It is assumed that this table needs to be set forward one year for males and set back two years for females. If Michael and Brenda are both age 40, find the net single premium that each should pay for a life annuity of 1000 per year, if the first payment occurs immediately.

8.3 Annuities Payable mthly

• Payments every mth part of the year

• Problem: commutation functions reflect annual probabilities of survival

• First, we obtain an approximate formula for present value

• Assume for a moment that the values Dy are also given for non-integer values of y

Usual life annuity

x x + 1 x + 2 x + n

1 1 1

…..

ax

age …..

Annuity payable every 1/m part of the year

x x + 1/m

x + 2/m

x + (m-1)/m

1/m 1/m

…..

a(m)x

age x + 1…..

1/m 1/m


x x + 1/m

x + 2/m

x + (m-1)/m

1/m 1/m

…..

a(m)x

age x + 1…..

1/m 1/m

x

mjix

x

mx

x

mx

xx

mjixmjix

xx

mxmx

xx

mxmx

x

mjixmji

x

mxm

x

mxm

xmjimji

xmm

xmmm

x

D

D

D

D

D

D

m

lv

lv

lv

lv

lv

lv

m

l

lv

l

lv

l

lv

m

pvpvpvm

a

//2/1

//

/2/2

/1/1

///2/2/1/1

//

/2/2

/1/1)(

1

1

1

1


x x + 1/m

x + 2/m

x + (m-1)/m

1/m 1/m

…..

a(m)x

age x + 1…..

1/m 1/m

0 1/

1/1

1/1

2/)1(1/21/11

1/)1(/2/1

//2/1)(

11

1

1

i

m

jmjix

x

m

jmjx

m

jmjx

x

xmmxmxmx

xmmxmxmx

x

x

mjix

x

mx

x

mxmx

DmD

DDmD

DDDD

DDDD

mD

D

D

D

D

D

D

ma

Using linear interpolation for Dx+i+j/m

xx

xxxxxxxxx

xxxxxxxxx

i

m

jmjix

ixixix

ixixix

m

jmj

ixixix

m

jixixm

jix

m

jmjix

ixixmj

ixmjix

Dm

mN

DDDDDDm

mDDDm

mDDmD

mDDmD

mDDmD

D

mDDmD

m

mmDDmDDDmD

DDDD

DDDD

2

12

)1(2

)1(

2

)1(

2

)1(

2

)1(

2

)1(

1

2312121

2321211

0 1/

1

11

1

11

1/

1/

Using linear interpolation for Dx+i+j/m

xxi

m

jmjix D

mmND

2

11

0 1/

m

ma

m

m

D

ND

mmN

mDa x

x

xxx

x

mx 2

1

2

1

2

11 11

)(

m

maa x

mx 2

1)(

Continuous life annuity

2

1

2

1limlim )(

xx

m

mx

mx a

m

maaa

dtpvpvpvpvm

aa

xtt

xmjimji

xmm

xmm

m

mx

mx

1

lim

lim

0

//

/2/2

/1/1

)(

dtpva xtt

x 0

2

1 xx aa

Annuity payable m-thly, deferred

x x + n+1/m

x + n+2/m

x +n+ (m-1)/m

1/m 1/m

…..

n|a(m)x

age x + n+1…..

1/m 1/m

… x + n

a(m)x+n

m

m

D

Dan

m

ma

D

D

aD

Dapvan

x

nxxnx

x

nx

mnx

x

nxmnxxn

nmx

2

1

2

1

)()()(

m

m

D

Danan

x

nxx

mx 2

1)(

Annuity payable m-thly, temporary

a(m)x:n|

m

m

D

Da

m

m

D

Dana

m

m

D

Dan

m

maanaa

x

nxnx

x

nxxx

x

nxxx

mx

mx

mnx

2

11

2

11

2

1

2

1

|:

)()()(|:

m

m

D

Daa

x

nxnx

mnx 2

11|:

)(|:

x x + 1/m

x + 2/m

x +n+ (m-1)/m

1/m 1/m

…..age x + n

1/m 1/m

Examples

• Page 168, 8.10

8.4 Varying Life Annuities

• Arithmetic increasing annuities

• It is sufficient to look at the sequence 1,2,3,….

• Temporary decreasing annuities

Example

• Ernest, aged 50, purchases a life annuity, which pays 5,000 for 5 years, 3,000 for 5 subsequent years, and 8,000 each year after. If the first payment occurs in exactly 1 year, find the price in terms of commutation functions.

Arithmetic increasing annuity

x x + 1 x + 2 x + n

1 2 n

…..

(Ia)x

age


…..

0

1

0

21321

33

22

|

||

32

t x

tx

txt

xxxxtt

txt

t

txt

t

t

xnn

xxxx

D

Na

aaapvpvpv

pnvpvpvvpIa

0ttxx NS

x

xx D

SIa 1

Arithmetic increasing annuity, temporary

x x + 1 x + 2 x + n

1 2 n

…..

(Ia)x:n|

age


x

nxnxxnx D

nNSSIa 111

|:

x + n+1

Arithmetic decreasing annuity, temporary

x x + 1 x + 2 x + n

n n-1 1

…..

(Da)x:n|

x + n+1


x x + 1 x + 2 x + n

n n-1 1

…..

(Da)x:n|

x + n+1

x x + 1 x + 2 x + n

1 2 n

…..

(Ia)x:n|

x + n+1


x x + 1 x + 2 x + n

n n-1 1

…..

(Da)x:n|

|:|:|: )1( nxnxnx anDaIa

x + n+1

x x + 1 x + 2 x + n

1 2 n

…..

(Ia)x:n|

x + n+1

x x + 1 x + 2 x + n

n+1 n+1

…..

(n+1)ax:n|

n+1


x x + 1 x + 2 x + n

n n-1 1

…..

(Da)x:n|

age


|:|:|: )1( nxnxnx anDaIa

x + n+1

x

nxnxxnx D

nNSSIa 111

|:

x

nxxnx D

NNa 11

|:

x

nxxxnx D

SSnNDa

)( 221|:

Examples

• Georgina, aged 50, purchases a life annuity which will pay her 5000 in one year, 5500 in two years, continuing to increase by 500 per year thereafter. Find the price if S51 = 5000, N51 = 450, and D50 = 60

• Redo the previous example if the payments reach a maximum level of 8000, and then remain constant for life. Assume S58 = 2100

• Two annuities are of equal value to Jim, aged 25. The first is guaranteed and pays him 4000 per year for 10 years, with the first payment in 6 years. The second is a life annuity with the first payment of X in one year. Subsequent payments are annual, increasing by .0187 each year.If i = .09, and from the 7% -interest table, N26=930 and D25= 30, find X.

8.5 Annual Premiums and Premium Reserves

• Paying for deferred life annuity with a series of payments instead of a single payment

• Premium reserve is an analog of outstanding principal• Premiums often include additional expenses and

administrative costs• In such cases, the total payment is called

gross premium• Loading = gross premium – net premium• General approach: actuarial present values of two

sequences of payments must be the same (equation of value)

Annual premiums P = tP(n|äx)

x x + t-1 x + n +1

1

age

txx

nx

x

txx

x

nx

tx

xn

NN

N

DNN

D

N

a

aP

|:

x + n + 2

1

…x + 1 x + nx +t ……

1

• t is the number of premium payments

• Present value of premiums is P äx:t|

• Present value of benefits is n|äx

• Therefore P äx:t| = n|äx

txx

nxxnt NN

NaP

)(

P PP

Example

• Arabella, aged 25, purchases a deferred life annuity of 500 per month, with the first benefit coming in exactly 20 years. She intends to pay for this annuity with a series of annual payments at the beginning of each year for the next 20 years. Find her net annual premium if D25 = 9000, D 45 = 5000, ä25 = 15 and ä45 = 11.5

Reserves

x x + t -1 x + n +1

1

age

ntD

NNPNaPa

ntD

Na

aV

tx

nxtxnxtntxtxtn

tx

txtx

xnnt

,)(

|

,)(

|:

x + n + 2

1

…x + 1 x + nx +n-1 ……

1

• Analog of outstanding principal immediately after premium t has been paid

• Assume that the number of premium payments is n

• ReserventV (n|äx) = PV of all future benefits – PV of all future premiums

P PP P

ntV (n|äx)

Loading and Gross premiums

• Arabella, aged 25, purchases a deferred life annuity of 500 per month, with the first benefit coming in exactly 20 years. She intends to pay for this annuity with a series of annual payments at the beginning of each year for the next 20 years. Assume that 50% of her first premium is required for initial underwriting expenses, and 10% of all subsequent premiums are needed for administration costs. In addition, 100 must be paid for issue expenses. Find Arabella’s annual gross premium, if D25 = 9000, D 45 = 5000, ä25 = 15, and ä45 = 11.5

Chapter 9

LIFE INSURANCE

• Basic Concepts

• Commutation Functions and Basic Identities

• Insurance Payable at The Moment of Death

• Varying Insurance

• Annual Premiums and Premium Reserves

9.1 Basic Concepts

• Benefits are paid upon the death of the insured

• Types of insurance

– Whole life policy

– Term insurance

– Deferred insurance

– Endowment insurance

Whole life policy

• Benefit (the face value) is paid to the beneficiary at the end of the year of death of inured person

• If the face value is 1 and insurance is sold to a person aged x, the premium is denoted by Ax

x x + 1 x + 2 x + t+1…..

Ax

age

probability px 2px tqx

x + t

1

Whole life policy

x x + 1 x + 2 x + t+1…..

Ax

age

probability px 2px qx+t

x + t

1

0

1

t

ttxxtx vqpA

Term insurance

• Benefit (the face value) is paid to the beneficiary at the end of the year of death of inured person, only if the death occurs within n years

• If the face value is 1 and insurance is sold to a person aged x, the premium is denoted by A1

x

1

0

11|:

n

t

ttxxtnx vqpA

Deferred insurance

• Does not come into force until age x+n

• If the face value is 1 and insurance is sold to a person aged x, the premium is denoted by A1

x:n|

xnnxx AAA |1|: 1

|:| nxxxn AAA

n-year endowment insurance• Benefit (the face value) is paid to the beneficiary at the end

of the year of death of inured person, if the death occurs within n years

• If the insured is still alive at the age x+n, the face value is paid at that time

• If the face value is 1 and insurance is sold to a person aged x, the premium is denoted by Ax:n|

xnnxnx EAA 1|:|: |:

1|: nxxnx AAA

Exercise:

Examples

• Rose is 38 years old. She wishes to purchase a life insurance policy which will pay her estate 50,000 at the end of the year of her death. If i=.12, find an expression for the actuarial present value of this benefit and compute it, assuming px = .94 for all x.

• Michael is 50 years old and purchases a whole life policy with face value 100,000. If lx= 1000(1-x/105) and i=.08, find the price of this policy.

• Calculate the price of Rose’s and Michael’s policies if both policies are in force for a term of only 30 years.

• Calculate the price of Rose’s and Michael’s policies if both policies are to be 30 years endowment insurance.

9.2 Commutation Functions

• Recall:

0ttxx DN

xx

x lvD x

nxxn D

DE

Commutation Functions

• Recall:

• So we need:

1 xxx vdC

x

txttxxt D

Cvqp

1

0

1

t

ttxxtx vqpA

xx

txtxt

x

txt

tx

tx

x

txttxxt vl

vdv

l

dv

l

d

l

lvqp

1111

Whole life insurance

0ttxx CM

x

xx D

MA

000

1 1

ttx

xt x

tx

t

ttxxtx C

DD

CvqpA

Term insurance

1

0

11|:

n

t

ttxxtnx vqpA

x

nx

x

x

t nt x

tx

x

txn

t x

txnx D

M

D

M

D

C

D

C

D

CA

0

1

0

1|:

x

nxxnx D

MMA

1|:

n-year endowment insurance

xnnxnx EAA 1|:|:

x

nxnxx

x

nx

x

nxxnx D

DMM

D

D

D

MMA

|:

Note

• We can represent insurance premiums in terms of actuarial present values of annuities, e.g. Ax = 1 – d äx

• Hence they also can be found using “old” commutation functions

Examples

• Juan, aged 40, purchases an insurance policy paying50,000 if death occurs within the next 20 years, 100,000 if death occurs between ages 60 and 70, and 30,000 if death occurs after that. Find the net single premium for this policy in terms of commutation functions.

• Phyllis, aged 40, purchases a whole life policy of 50,000. If N40 = 5000, N41 = 4500, and i = .08, find the price.

9.3 Insurance Payableat the Moment of Death

• We consider scenario when the benefit is paid at the end of the year of death

• Alternatively, the benefit can be paid at the moment of death

Divide each year in m parts

mjix

mjixmjix

x

mjixmji

mx

mxmx

x

mxm

x

mxxm

mjxmxmjimji

mxmxmm

xmm

mx

l

ll

l

lv

l

ll

l

lv

l

llv

qpvqpvqv

A

/)1(

//)1(/)1(/

/1

/2/1/1/2/1/1

/)1(/1/)1(/

/1/1/1/2

/1/1

)(

0 1

'/),(

/

'/),(

/

'/)2,0(

/2'/)1,0(

/1

//)1(/

/2/1/2/1/1

//)1(/

/2/1/2/1/1

1

1

/1

/1/11

i

m

jmjisx

mji

x

mjisxmji

msxm

msxm

x

mjixmjixmji

mxmxmmxxm

x

x

mjixmjixmji

x

mxmxm

x

mxxm

lvml

lv

lvlv

ml

m

llv

m

llv

m

llv

ml

l

llv

l

llv

l

llv

Taking the limit as m→∞ we get:

dtpvdtl

l

l

lvdt

l

lv

lvml

AA

txxtt

tx

tx

x

txt

x

txt

i

m

jmjisx

mji

xm

mx

mx

0

'

0

'

0

0 1

'/),(

/)( 1limlim

• Whole life policy:

• Term policy:

Premium for insurance payable at the moment of death

dtpvA txxxt

0

t

dtpvnA txx

n

t |:0

t

Examples

• Find the net single premium for a 100,000 life insurance policy, payable at the moment of death, purchased by a person aged 30 if i = .06 and tp30 = (.98)t for all t

• Solve the previous example if it is 20 years endowment insurance, force of interest is .06 andlx = 105 – x, 0 ≤ x ≤ 105.

Remarks

• Using integration by parts, we can get

Āx = 1 – δ āx

• Approximate formula:

Āx ≈ (i/δ) Ax

• To obtain it, use linear interpolation in the following expression:

)()()(1

//)1(/

/2/1/2

/1/1

mjixmjixmji

mxmxm

mxxm

x

llvllvllvl

A

Chapter 8 LIFE ANNUITIES Basic Concepts Commutation Functions Annuities Payable m thly Varying Life...

Documents

Transcript of Chapter 8 LIFE ANNUITIES Basic Concepts Commutation Functions Annuities Payable m thly Varying Life...