week 3 actl3002 lecture slides

7/23/2019 Week 3 ACTL3002 Lecture Slides

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Life Insurance and Superannuation Models – Week 3: Premiums

ACTL 3002 Life Insurance and SuperannuationModels

Michael Sherris

School of Risk and Actuarial Studies, Australian School of BusinessUniversity of New South Wales

ARC Centre of Excellence in Population Ageing Research

Week 3:Premiums

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Plan

1 Overview

2 Net Premiums

3 Equivalence Principle

4 Insurances

5 Continuous insurance

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Overview

1 Overview

2 Net Premiums


4 Insurances


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Overview

Week 3: Premiums

Overview

Net premium:

Net (random) future lossPrinciple of equivalenceDiscrete premiumsWhole of life, limited premiums, term life, endowmentPremiums paid m times per yearContinuous premiums

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Overview

References

References

Dickson et al. Chapter 6 (6.1 to 6.7)

Will cover rest of Chapter 6 in Week 6

Chapters 6 and 7 covered over Weeks 3, 4, 5 and 6

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Lif I d S i M d l W k 3 P i

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Net Premiums

1 Overview

2 Net Premiums


4 Insurances


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Lif I d S ti M d l W k 3 P i

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Net Premiums

Net premiums

Net premiums

Values only the benefits providedNo allowance for allocated expenses, profit or contingencymargins

Principle of equivalence

Other premium principles (covered in other courses)

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Net Premiums

Net Random Future Loss

An insurance contract is an agreement between two partiesThe insurer agrees to pay for insurance benefits;In exchange for insurance premiums to be paid by the insured

Denote by PVFB0 the present value, at time of issue, of future benefits to be paid by the insurer.

Denote by PVFP0 the present value, at time of issue, of future premiums to be paid by the insured.

The insurer’s net random future loss is defined by

0L = L = PVFB 0 − PVFP 0

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Equivalence Principle

1 Overview

2 Net Premiums


4 Insurances


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The principle of equivalenceEPV of benefit outgo = EPV of net premium income

The net premium is the amount of premium required to meetthe expected cost of the insurance or annuity benefits under acontract, given mortality and interest rate assumptions.

The net premium is determined according to the principle of

equivalence by setting

E [L] = 0.

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For example, for a unit of benefit payment, let Z be the PVr.v. associated with the life insurance benefits and Y is the

PV r.v. associated with the life annuity premium payments,with π the premium payable annually, then

L = Z − πY

so that

π = E (Z )/E (Y )

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p


Different Possible Payment Assumptions

Premium payment Benefit payment

annually at the end of the year of deathat the end of the 1

m

th year of deathimmediately upon death

m-thly of the year at the end of the yearat the end of the 1

mth year of death

immediately upon death

continuously at the end of the yearat the end of the 1

mth year of death

immediately upon death

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p

Insurances

1 Overview

2 Net Premiums


4 Insurances


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Insurances

Fully Discrete Annual Premiums - Whole Life Insurance

(WLI)

Consider, for example, the case of a unit WLI with levelannual premium payment. Here the loss function is

L = v K + 1− πa

K +1 , for K = 0, 1, 2,...

By the principle of equivalence, we have

E (L) = E (v K + 1) − πE (aK +1 ) = 0

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Insurances

Discrete payments - WLI

Hence if we denote π by P x then

π = P x =

Ax

ax

The variance of the loss function

Var (L) =2Ax − (Ax )

2

(d ax )2 =

2Ax − (Ax )2

(1 − Ax )2

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Insurances

Example

Assume that the survival model is given by, 90 = 100,

91 = 72, 92 = 39, 93 = 0. And the annual interest ratei = 0.06. All of 90 = 100 people buy $1000 of fully discretewhole life, paying net level annual premiums 1000P 90 at thebeginning of each year. Find the annual net premium andillustrate the aggregate accounting in a table to show how the

insurer breaks even after three years.

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Insurances

Solution

Solution.

a90 =∞k =0

v k k p 90 = 1 × 1 + 1

1.06 ×

72

100 +

1

(1.06)2 ×

39

100 + 0

Note A90 = 1 − d a90, d = 1 − v = 1 − 11.06 = 0.05660377.

So A90 = 0.8853.P 90 = A90

a90= 0.436895 and 1000P 90 = 436.9.

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Insurances

WLI - discrete payments

Net premium

P x = Ax

ax

and sinceax =

1 − Ax

d

We have then

1ax

= P x + d

P x = dAx

1 − Ax

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Insurances

Exercise

Consider a fully discrete whole life insurance of 1, 000 issued

to (60), the annual benefit premium was calculated using thefollowing assumptions: i = 6%, q 60 = 0.01376,1000A60 = 369.33, and 1000A61 = 383.00.A particular insured is expected to experience a first-yearmortality rate 10 times the rate used to calculate the annual

benefit premium. The expected mortality rates for all otheryears are the ones originally used.

1 Calculate the expected loss at issue for this insured based onthe original benefit premium.

2 Why do you think there is a loss?

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Insurances

WLI with h Premium Payments - Limited premiums

The loss function in this case is

L = v K + 1 − πa

K +1 for K = 0, 1, · · · , h− 1

v K + 1 − πah for K = h, h + 1, · · ·

Applying the principle of equivalence, we have

π = hP x = Ax

ax :h

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Insurances

Exercise Consider a whole life insurance with annual premiumsand a 20-year premium paying term issued to a life aged 30, with

sum insured $200,000 payable at the end of the year of death,assuming AM92 Ultimate mortality and interest rate of 4%.

1 Write an expression for the loss function.

2 Calculate the net annual premium.

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Insurances

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Insurances

Term life insurance

n-yr term:

L =

v K + 1 − πa

K +1 for K = 0, 1, · · · , n− 10 − πan for K = n, n + 1, · · ·

The corresponding premium formula is

P 1x :n = A1x :n

ax :n|

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Insurances

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Insurances

Example

Calculate the annual premium for a 10-year term insurance fora 30-year old with a sum assured of $500,000, assumingAM92 Ultimate mortality and interest rate of 4% pa. Assumethat the death benefit is paid at the end of the year of death.

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Insurances

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Insurances

n-yr endowment

n-yr endowment loss RV

L =

v K + 1 − πa

K +1 for K = 0, 1, · · · , n − 1v n − πan for K = n, n + 1, · · ·

The premium formula is

P x :n = Ax :n

ax :n

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Insurances

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Insurances

Limited premium n-yr endowment

Limited premium h-pay, n-yr endowment (h < n)

L =

v K + 1 − πaK +1 for K = 0, 1, · · · , h − 1

v K + 1 − πah

for K = h, ...,n − 1v n − πa

h for K = n, n + 1, · · · .

Premium Formula is

hP x :n = Ax :n

ax :h

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Insurances

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n-yr pure endowment

n-yr pure endowment

L =

0 − πa

K +1 for K = 0, 1, · · · , n − 1

v n − πan for K = n, n + 1, · · · .


P x :1n =

Ax :1n

ax :n

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Insurances

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Exercise

Prove :P x :n = nP x + P x :

1n (1 − Ax +n)

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Life Insurance and Superannuation Models – Week 3: PremiumsInsurances

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n-yr deferred WL annuity with n-yr premium

n-yr deferred WL annuity with n-yr premium payment

L =

0 − πa

K +1 for K = 0, 1, · · · , n − 1

v naK +1−n − πan for K = n, n + 1, · · · .

Premium Formula is

P (n|ax ) = Ax :1n ax +n

ax :n

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Life Insurance and Superannuation Models – Week 3: PremiumsInsurances

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Premiums paid m times a year - SummaryInsurance Plan Benefit paid

WLI at the EOY of death P (m)x = Ax /a

(m)x

at the moment of death P (m)( Ax ) = Ax /a(m)x

n-year term I at the EOY of death P (m)1x :n

= A1x :n /a

(m)x :n

at the moment of death P (m)( A1x :n ) = A1

x :n /a(m)x :n

n-year at the EOY of death P (m)x :n = Ax :n /a(m)

x :n

endowment at the moment of death P (m)( Ax :n ) = Ax :n /a(m)x :n

h-pay, WHI at the EOY of death hP (m)x = Ax /a

(m)

x :h

at the moment of death hP (m)( Ax ) = Ax /a

(m)

x :h

h-pay, at the EOY of death hP (m)x :n = Ax :n /a

(m)

x :h

n-year at the moment of death hP (m)( Ax :n ) = Ax :n /a

(m)

x :hendowment

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Life Insurance and Superannuation Models – Week 3: PremiumsContinuous insurance

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1 Overview

2 Net Premiums


4 Insurances


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Fully continuous premiums - whole life insurance

Consider fully continuous level annual premiums for a unitwhole life insurance payable immediately upon death of (x ).

The loss function is expressed as

L = v T − πaT

By the principle of equivalence, and denoting the net premiumπ by P ( Ax ), we have

π = P ( Ax ) =Ax

ax

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Fully continuous premiums - whole life insurance

Variance of the insurer’s loss function:

Var (L) =

2 Ax − ( Ax )2

1 + (P (Ax )/δ )2

=2 Ax − ( Ax )

2

(δ ax )2 =

2 Ax − ( Ax )2

(1 − Ax )2

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Endowment Insurance Premiums

For an n-year endowment insurance, loss function is:

L =

v T − πa

T , T ≤ n

v n − πan T ≥ n

Net premium formula:

π = P ( Ax :n ) = Ax :n

ax :n

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Endowment Insurance Premiums

Variance of the insurer’s loss function:

Var (L) =

2Ax :n − (Ax :n )2

1 + (P (Ax :n )/δ )2

=2Ax :n − (Ax :n )2

(δ ax :n )2 =

2Ax :n − (Ax :n )2

(1 − Ax :n )2

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Premium Identities

Whole life insurance:

P ( Ax ) = 1

ax − δ =

δ Ax

1 − Ax

Endowment insurance:

P (Ax :n ) = 1

ax :n− δ =

δ Ax :n

1 − Ax :n

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Fully Continuous Contracts

n-yr term

L = v

T

− πaT , T ≤ n0 − πan T ≥ n


P (A1x :n ) = A

1

x :nax :n

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h-pay, whole life

L = v T − πaT , T ≤ hv T − πa

h , T > h


h P (Ax ) = Ax

ax :h

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h-pay, n-yr endowment

L =

v T

− πaT , T ≤ hv T − πa

h , h < T ≤ n

v n − πah , T > n

Premium Formula is

h P (Ax :n ) =

Ax :n

ax :h

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n-yr pure endowment

L =

0 − πaT , T ≤ nv n − πan , T > n

Premium Formula is

P (Ax :1n ) = Ax :

1n

ax :n

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n-yr deferred WL annuity

L =

0 − πaT , T ≤ n

v naT −n − πan , T > n

Premium Formula is

P (n|ax ) = Ax :1n ax +n

ax :n

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Overview and Summary

Main ideas

Net premiums

Principle of equivalencedifferent contracts - whole of life, term, limited premiumdifferent payment frequency

Next week

Reserves and Policy Values

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week 3 actl3002 lecture slides

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