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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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Life Insurance and Superannuation Models – Week 3: Premiums
Plan
1 Overview
2 Net Premiums
3 Equivalence Principle
4 Insurances
5 Continuous insurance
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Life Insurance and Superannuation Models – Week 3: Premiums
Overview
1 Overview
2 Net Premiums
3 Equivalence Principle
4 Insurances
5 Continuous insurance
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Life Insurance and Superannuation Models – Week 3: Premiums
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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Life Insurance and Superannuation Models – Week 3: Premiums
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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Life Insurance and Superannuation Models – Week 3: Premiums
Net Premiums
1 Overview
2 Net Premiums
3 Equivalence Principle
4 Insurances
5 Continuous insurance
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Lif I d S ti M d l W k 3 P i
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Life Insurance and Superannuation Models – Week 3: Premiums
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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Life Insurance and Superannuation Models Week 3: Premiums
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Life Insurance and Superannuation Models – Week 3: Premiums
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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Life Insurance and Superannuation Models Week 3: Premiums
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Life Insurance and Superannuation Models – Week 3: Premiums
Equivalence Principle
1 Overview
2 Net Premiums
3 Equivalence Principle
4 Insurances
5 Continuous insurance
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Life Insurance and Superannuation Models – Week 3: Premiums
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Life Insurance and Superannuation Models Week 3: Premiums
Equivalence Principle
Principle of equivalence
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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Life Insurance and Superannuation Models – Week 3: Premiums
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Life Insurance and Superannuation Models Week 3: Premiums
Equivalence Principle
Principle of equivalence
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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Life Insurance and Superannuation Models – Week 3: Premiums
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p
Equivalence Principle
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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Life Insurance and Superannuation Models – Week 3: Premiums
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p
Insurances
1 Overview
2 Net Premiums
3 Equivalence Principle
4 Insurances
5 Continuous insurance
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Life Insurance and Superannuation Models – Week 3: Premiums
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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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Life Insurance and Superannuation Models – Week 3: Premiums
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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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Life Insurance and Superannuation Models – Week 3: Premiums
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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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Life Insurance and Superannuation Models – Week 3: Premiums
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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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Life Insurance and Superannuation Models – Week 3: Premiums
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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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Life Insurance and Superannuation Models – Week 3: Premiums
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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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Life Insurance and Superannuation Models – Week 3: Premiums
I
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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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Life Insurance and Superannuation Models – Week 3: Premiums
I
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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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Life Insurance and Superannuation Models – Week 3: Premiums
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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Life Insurance and Superannuation Models – Week 3: Premiums
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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Life Insurance and Superannuation Models – Week 3: Premiums
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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Life Insurance and Superannuation Models – Week 3: Premiums
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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Life Insurance and Superannuation Models – Week 3: Premiums
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, · · · .
The corresponding premium formula is
P x :1n =
Ax :1n
ax :n
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Life Insurance and Superannuation Models – Week 3: Premiums
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
3 Equivalence Principle
4 Insurances
5 Continuous insurance
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Life Insurance and Superannuation Models – Week 3: PremiumsContinuous insurance
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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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Life Insurance and Superannuation Models – Week 3: PremiumsContinuous insurance
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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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Life Insurance and Superannuation Models – Week 3: PremiumsContinuous insurance
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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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Life Insurance and Superannuation Models – Week 3: PremiumsContinuous insurance
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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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Life Insurance and Superannuation Models – Week 3: PremiumsContinuous insurance
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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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Life Insurance and Superannuation Models – Week 3: PremiumsContinuous insurance
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Fully Continuous Contracts
n-yr term
L = v
T
− πaT , T ≤ n0 − πan T ≥ n
The corresponding premium formula is
P (A1x :n ) = A
1
x :nax :n
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Life Insurance and Superannuation Models – Week 3: PremiumsContinuous insurance
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Fully Continuous Contracts
h-pay, whole life
L = v T − πaT , T ≤ hv T − πa
h , T > h
The corresponding premium formula is
h P (Ax ) = Ax
ax :h
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Life Insurance and Superannuation Models – Week 3: PremiumsContinuous insurance
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Fully Continuous Contracts
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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Life Insurance and Superannuation Models – Week 3: PremiumsContinuous insurance
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Fully Continuous Contracts
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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Life Insurance and Superannuation Models – Week 3: PremiumsContinuous insurance
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Fully Continuous Contracts
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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