Chapter 4
Net Premiums
4.1 Net Premiums
Insurance and annuity benefits are more commonly purchased by a series of periodic premium payments than by a single premium payment (n.s.p.). Since premium payments are generally made only while the insured is living, these premiums form a life annuity.
4.1.1 Equivalence Principle
E
[
present value of benefits]
=E[
present value of net premiums]
i.e., n . s . p . for benefits=n . s . p . for net premiums
Define L = loss-at-issue random variable
= (p.v. of benefits) – (p.v. of net premiums)
The Equivalence Principle states that E(L)=0.
4.1.2 Continuous Premiums, Insurance Payable at the Moment
of Death
General:
Z=p . v . of insurance benefit
Y=p . v . of continuous life annuity following the premium payment pattern
P=continuous annual premium
Then L=Z – P Y∧¿
E(L)=0⟺E(Z)=P E(Y)
i .e .:P=E(Z)
1. Whole life insurance, level premiums payable for life.
Z=vT
Y=aT|
Continuous annual premium=P
(
Ax)
=E(Z)
E(Y)=
Ax
ax
L=vt−P aT|
E(L)=0
To obtain Var(L)
write L=vT−P
[
1−vT
δ
]
=vT
[
1+Pδ
]
− P δThen Var(L)=
[
1+Pδ
]
2
Var
(
vT)
=[
1+Pδ
]
2
[
2Ax−(
Ax)
2
]
But P=Ax
ax
=1−δ ax
ax
i .e . ,1+P
δ=1+
1−δ ax
δ ax
= 1
δ ax
i .e . ,Var(L)=
[
A 2x−
(
Ax)
2
]
(
δ ax)
2
2. n-year term insurance, level premiums payable for n years.
Continous annual premium P
(
A1x:n|)
solves .P ax:n|=A1x:n|
i .e . , P
(
Ax1:n|)
=Ax:n| 1L=vt−P at|whenT=t ≤ n
¿−P an|when T=t>n
3. n-year endowment insurance, level premiums payable for n years. Continuous annual premium P
(
Ax:n|)
satisfiesP
(
Ax:n|)
=Ax:n|ax:n|
L=vt−P at|whenT=t ≤ n
¿vt−P an|whenT=t>n
4. h payment, n-year endowment insurance. Continuous annual premium hP
(
Ax:n|)
satisfiesP ax:h|=Ax:n|
i .e . , Ph
(
Ax:n|)
=Ax:n|ax:h|
L=vt−P at|whenT=t ≤ h
¿vt−P ah|whenh h<T=t ≤ n
¿vn−P ah|when h T=t>n
5. h payment, whole life insurance.
P
h
(
Ax)
=Ax ax:h|
L=vt−P at|whenT=t ≤ h
¿vt−P ah|whenh T=t>h
6. h payment, n-year term insurance.
P
h
(
Ax:n| 1)
=Ax:n| 1L=vt−P at|whenT=t ≤ h
¿vt−P ah|whenh h<T=t ≤ n
¿−P ah|when h T=t>n
Example 4.1.1L1 is the loss-at-issue random variable for a fully continuous whole life insurance of 1 on the life of (x) with a net level annual premium determined by the equivalence principal.
You are given:
(i) ax=5.0 (ii) δ=0.08 (iii) Var
(
L1)
=0.5625L2 is the loss-at-issue random variable for this insurance with a premium which is 4/3 times the net level annual premium.
Calculate the sum of the expected value of L2 and the standard deviation of L2.
Solution
L1=v
T
−P aT|=vT−P
[
1−vT
δ
]
¿vT
(
1+Pδ
)
− P δBut P=Ax/ax=
(
1−δ ax)
/axi.e., P/δ=1/(δ ax)−1 and 1+P/δ=1/(δ ax)
i.e., Var
(
L1)
=0.5625=[
1/(
δ ax)
]
2[
2Ax−(
Ax)
2]
¿
[
1/0.08(5)]
2[
2Ax−
(
Ax)
2
]
i.e., 2Ax−
(
Ax)
2=0.5625(0.16)=0.09L2=v
T
−4/3P aT|=L1−1/3P aT|
¿−1/3Ax
¿−1/3
(
1−δ ax)
¿−1/3
(
1−5(0.08))
=−0.2L2=v
T
−4/3P
[
1−vT
δ
]
¿vT
[
1+(4/3P)/δ]
−4/3P/δ¿vT
[
1+4/3((
1/δ ax)
−1)]
−4/3P/δ¿vT
[
4/3(
1/δ ax)
−1/3¿]
−4/3P/δi.e., Var
(
L2)
=[
2Ax−(
Ax)
2
]
[
4/3(
1/0.08(5))
−1/3¿]
2=0.09(9) =0.81i.e., SD
(
L2)
=0.9∧E(
L2)
+SD(
L2)
=−0.2+0.9=0.7Example 4.1.2 L is the loss-at-issue random variable for a fully continuous whole life insurance of 1 with premiums based on the equivalence principle.
You are given:
(i) E
(
v2T)
=0.34 (ii) E
(
vT)
=0.40Calculate Var(L)
Solution
L=vT−P ax
¿vT1/(δ ax)−P/δ
Var(L)=
(
1/δ ax)
2[
2Ax−(
Ax)
2]
A
2
x=E
(
v2T
)
=0.34 ;
Ax=E
(
vT)
=0.40i.e., Var(L)=(1/0.6)2
[
0.34−0.16]
¿0.18/0.36
¿0.5
4.1.3 Continuous Premiums, Life Annuities Payable
Continuously
1. n payment, n year pure premium Continuous annual premium P
(
Ax1:n|)
satisfies P ax:n|=A1x:n|i .e . , P
(
Ax1:n|)
=Ax:n| 1ax:n|
L=−P at|when T=t<n
¿vn−P an|when T=t ≥ n
2. h payment, n-year endowment insurance.
P
h
(
Ax:n|1
)
=Ax:n| 1ax:h|
L=−P at|when T=t<h
¿−P ah|when h h ≤T=t<n
¿vn−P ah|when h T ≥ n
3. n payment, n year deferred whole life insurance.
P
(
n|ax)
= n|axax:n|
L=−P at|when T=t<n
4.1.4 Relationships between Net Premiums, Insurances and
Annuities
1. Whole life insurance, premiums payable for life
P
(
Ax)
=Ax ax
=1−δ ax
ax
= δ Ax 1−Ax
2. n year endowment insurance, premiums payable for n years
P
(
Ax:n|)
=Ax:n|ax:n|=
1−δ ax:n|
ax:n| =
δ Ax:n|
1−Ax:n|
3. n year term insurance, premiums payable for n years
P
(
Ax1:n|)
=Ax:n| 1ax:n|=1−δ
ax:n|−¿E
x n
ax:n| =
δ Ax1:n|
1−Ax:n|¿
Example 4.1.3 Simplify 1−
[
P(
A30 : 15|)
−15P(
A30)
]
(
a30 : 15|)
/[
v15(1−15q30)]
SolutionP
(
A30 : 15|)
=A30 : 15|a30 : 15|
P
15
(
A30)
=A30 a30 : 15|
i.e.,
[
P(
A30: 15|)
−15P(
A30)
]
(
a30: 15|)
=A30 : 15|– A30¿A30 : 151 |+15E30−¿
¿15E30−15E30A45
v15
(
1−15q30)
=v15(
15p30)
=15E30i.e., expression ¿1−
[
15E30−15E30A45]
/15E30Example 4.1.4 A fully continuous whole life insurance of 1 is issued to (x). You are given the following:
L=vT−P
(
Ax)
aT| under the assumption that the force of mortality μ45 equals μ for all x and the force of interest equals δ.Calculate Var(L)
Solution
L=vT−P
(
Ax)
aT|=vT−P(
1−vT
)
δ
¿vT
[
1+Pδ
]
− P δVar(L)=
[
1+Pδ
]
2
[
2Ax−(
Ax)
2
]
Ax=
∫
0∞
vttpxμx+tdt
px
t =exp
{
−∫
x x+tμydy
}
=e−μti.e., Ax=μ
∫
0
∞
vte−μtdt
=μ
∫
0∞
e−δ t
e−μtdt
¿μ
∫
0
∞
e−(δ+μ)t
dt
¿−
[
μμ+δ e
−(δ+μ)t
]
|
∞0
¿
[
μ μ+δ]
A
2
x=
∫
0
∞
v2ttpxμx+tdt=μ
∫
0∞
e−(2δ+μ)t
¿
[
μ μ+2δ]
1+P
δ=1+ Ax
δ ax
=1+ Ax 1−Ax
= 1
1−Ax
= 1
[
1− μμ+δ
]
¿μ+δ δ
i.e., Var(L)=
[
(μ+δ)/δ]
2[
μ/(μ+2δ)−(
μ/(μ+δ))
2]
¿(μ+δ)
2
δ2
[
μ(μ+δ)2−μ2(μ+2δ)]
[
(μ+2δ) (μ+δ)2]
¿
[
μ(
μ2
+2μδ+δ2
)
−μ2(μ+2δ)]
δ2
[
μ+2δ]
¿ μ δ
2
δ2
[
μ+2δ]
¿ μ
[
μ+2δ]
4.2 Fully Discrete Premiums
Recall T = future years lived by (x)K = future complete years lived by (x)
P
[
K=k]
=P[
k<T ≤ k+1]
, k=0,1,2, … … … … .¿k|qx
1. Whole life insurance, benefits payable at the end of the year of death, level annual premiums payable for life
Annual Premium Px satisfies Pxa¨x=Ax. i.e., Px=Ax/ ¨ax
When K=k, L=vK+1−Pxa¨k+1|
Px Px Px ……….Px
--- --- --- -- x x+1 x+2 ……… x+k x+k+1
L=vK+1−Pxa¨k+1|=vK+1−Px
[
(
1−vK+1)
/d]
¿vK+1
[
1+Pxd
]
− Pxd
E(L)=Ax−Pxa¨x=0
Var(L)=¿ ¿
But 1+¿
Var(L)=
[
2Ax−(
Ax)
2]
/(
da¨x)
2
¿
[
2Ax−(
Ax)
2]
/[
1−(
Ax)
2]
2. h payment, whole life insurance
Annual premium hPx satisfies hPxa¨x:h|=Ax
i.e., hPx=Ax/ ¨ax:h|
L=vK+1−hPxa¨k+1|when K=k<h
¿vK+1−hPxa¨h|when K=k ≥ h
3. n-year term insurance, level premiums payable for n years Annual premium P1x:n| satisfies P1x:n|a¨x:n|=A1x:n|
i.e., P1x:n|=A1x:n|/ ¨ax:n|
¿−
(
P1x:n|)
a¨h|when K=k ≥ n4. h payment, n year term insurance
P1x:n|=h¿Ax:n| 1
/ ¨ax:h|¿
L=vK+1−
(
hP1x:n|)
a¨k+1|when K=k<h¿vK+1−
(
hP1x:n|)
a¨h|when h≤ K=k<n¿−
(
hP1x:n|)
a¨h|when K=k ≥ n5. n payment, n year pure endowment
P1x:n|=A1x:n|/ ¨ax:n|
L=−
(
P1x:n|)
a¨k+1|when K=k<n¿vn−
(
P1x:n|)
a¨n|when K=k ≥ n6. h payment, n year pure endowment
P1x:n|=h¿A1x:n|/ ¨ax:h|¿
L=−
(
hP1x:n|)
a¨k+1|when K=k<h¿−
(
hP1x:n|)
a¨h|whenh ≤ K=k<n¿vn−
(
hP1x:n|)
a¨h h|when K=k ≥n7. n payment, n year endowment insurance
Px:n|=Ax:n|/ ¨ax:n|
¿vn−
(
Px:n|)
a¨n|when K=k ≥ n8. h payment, n year endowment insurance
Px:n|=h¿Ax:n|/ ¨ax:h|¿
L=vK+1−
(
hPx:n|)
a¨k+1|when K=k<h¿vK+1−
(
hPx:n|)
a¨h|when h≤ K=k<n¿vn−
(
hPx:n|)
a¨h h|when K=k ≥nNote
When the insurance benefit is any amount other than $1, the resultant annual premium is multiplied by this amount.
e.g., whole life insurance of $1000, level annual premiums payable for life.
Px¿
=1000Ax/ ¨ax=1000Px
L¿
=1000vK+1−Px¿a¨k+1|
¿1000
(
vK+1−Pxa¨k+1|)
¿1000L
Var(L¿)=(1000)2Var(L)
9. n payment , n year deferred whole life annuity due
P( ¨n|ax)= ¨n|ax/ ¨ax:n|
L=−P( ¨n|ax) ¨ak+1|when K=k<n
¿vna¨k+1−n|−
(
P( ¨n|ax))
a¨n|when K=k ≥ n10.h payment, n year deferred whole life annuity due
L=−
(
hP( ¨n|ax))
a¨k+1|when K=k<h¿−
(
hP( ¨n|ax))
a¨h|when h≤ K=k<n¿vna¨k+1−n|−
(
hP( ¨n|ax))
a¨h h|when K=k ≥ nExample 4.2.1 Given Px:n|=0.042, 20P35=0.0299, A55=0.6099, find the value of P35 : 201 | Solution
P35 : 201 |
=A35 : 201 |/ ¨a35: 20|=
[
A35 : 20|−20E35]
/ ¨a35 : 20|¿P35 : 20|−¿
But E35A55=A35−A35 : 20| 1 20
i.e., 20E35/ ¨a35 : 20|= 1
A55
[
(
A35/ ¨a35 : 20|)
−(
A35 : 20| 1/ ¨a35: 20|
)
]
= 1A55
[
20P35−P35: 20| 1]
Substituting into (1),
P35 : 201 |
=0.042−
(
10.6099
)
[
0.0299−P35: 20| 1]
i.e., P35 : 201 |
(0.6099)=0.042(0.6099)−
[
0.0299−P35 : 201 |]
i.e., P35 : 201 |=
[
0.0299−0.042(0.6099)]
/[
1−0.6099]
¿0.011
Example 4.2.2 A fully discrete whole life insurance of 1 with a level annual premium is issued to (x). You are given:
(i) L is the loss-at-issue random variable if the premium is based on the equivalence Principle.
(ii)Var(L)=0.75
(iii) L¿ is the loss-at-issue random variable if the premium is determined such that
E(L¿)=−0.5
Calculate Var(L¿ )
L=vK+1−Pxa¨k+1|; L ¿
=vK+1−Px¿a¨k+1|
E(L¿)=Ax−Px¿a¨x=−0.5
i.e., Px
¿
=(Ax+0.5)/ ¨ax=Px+(0.5/ ¨ax)
Var(L¿
)=
[
1+(
Px ¿d
)
]
2
[
Ax2
−
(
Ax)
2]
¿
[
1+(
Pxd
)
+(
0.5
da¨x
)
]
2
[
Ax2
−
(
Ax)
2
]
¿
[
(
1 da¨x)
+
(
0.5da¨x
)
]
2
[
2Ax−
(
Ax)
2]
¿2.25
(
1 da¨x)
2
[
Ax2
−
(
Ax)
2
]
¿2.25Var(L)
¿2.25(0.75)=1.6875
Example 4.2.3 You are given:
(i) Deaths are uniformly distributed over each year of age. (ii)i=0.04∧δ=0.0392
(iii) nEx=0.6
(iv) Ax:n|=0.804
Calculate 1000P(Ax:n|)
Solution
1000P
(
Ax:n|)
=1000 Ax:n| ¨ax:n|
¨
ax:n|=1−Ax:n|
d
¿
(
δ i)
Ax:n|1 +nEx
¿
(
δi
)
[
Ax:n|−nEx]
+nEx¿
(
0.03920.04
)
[
0.804−0.6]
+0.6¿0.8
d= i
(1+i)= 0.04 1.04
i.e., a¨x:n|=(1−0.08)(1.04)/0.04=5.2
i.e., 1000P
(
Ax:n|)
=1000(0.804)/5.2¿154.6
≈155
Example 4.2.4 For a special fully discrete whole life insurance of 1000 issued on the life of (75), increasing premiums πk are payable at time k, for k=0,1,2,……
You are given
(i)πk=π0(1+i)k
(ii) Mortality follows de Moivre’s law with ω=105
(iii) i=0.05; a30|=15.4at i=0.05
(iv) Premiums are calculated in accordance with the equivalence principle.
Calculate π0
Solution
∑
t=0 29
πkvt p
75
t =1000
∑
t=0 29vt+1 q 75
t|
π0
∑
t=0 29
p75
t =1000
∑
t=029
vt+1
q75
t|
De Moivre: tp75μ75+t=1/(ω¿−75)=1/(105−75)=1/30¿
q75
t| =
∫
0 1p75
k μ75+kdk=1/30
p75
t =1−tq75=1−(t/30)
∑
t=0 29
p75
t =
∑
t=0 29
1−
(
t30
)
=30−(
1 30)
[
29(30)
2
]
=15.5∑
t=0 29
vt+1 q 75
t| =
(
130
)
a30|=15.4/30i.e., π0=
[
1000(15.4)]
/[
15.5(30)]
¿33.1
4.3 Annual Premiums, Immediate Payment of
Benefits
1. Whole life insurance, premiums payable for life.
P
(
Ax)
=Ax/ ¨ax=(i/δ)Ax/ ¨ax=(i¿¿δ)Px¿ under UDD2. h payment, whole life insurance.
P(Ax)
h =Ax/ ¨ax:h|=(i¿ ¿δ)hPx¿ under UDD
3. n year term insurance, premiums payable for n years.
P
(
Ax1:n|4. h payment, n year term insurance.
P(A1x:n|)
h =Ax1:n|/ ¨ax:h|=(i¿¿δ)hP1x:n|¿ under UDD
5. n year endowment insurance, premiums payable for n years.
P
(
Ax:n|)
=(A1x:n|+A1x:n|)/ ¨ax:n|=(i/δ)P1x:n|+P1x:n| under UDD
6. h payment, n year endowment insurance.
P(Ax:n|)
h =(Ax:n| 1
+Ax1:n|)/ ¨ax:h|=(i¿¿δ)hP1x:n|+hPx1:n|¿ under UDD
Example 4.3.1 Simplify
[
P(
Ax:n|)
−nP(Ax)]
/Px:n| 1Solution
Expression =
[
(
Ax:n|/ ¨ax:n|)
−(
Ax/ ¨ax:n|)
]
/(
A1x:n|/ ¨ax:n|)
¿
[
(
A1x:n|+A1x:n|)
−(
Ax1:n|+A1x:n|. Ax+n)
]
/(
Ax1:n|/ ¨ax:n|)
¿1−Ax+n
4.3.1 Monthly Payment Premiums
Px
(m)
=¿ net level premium, payable in mthly installments of (1/m)P(xm) at the beginning of each mth of a year, for $1 of insurance payable at the end of the year death.
1. Whole life insurance, mthly premiums payable for life.
P(xm)
=Ax/ ¨ax
(m)
=(d(m)Ax) /(1−A(xm))=d(m)Ax/
[
1−(i/i(m))Ax]
under UDD2. Whole life insurance, mthly premiums payable for h years.
P(xm)
h =Ax/ ¨ax:h|
(m)
=Ax/
(
a¨x(m)
− Exa¨x+h
(m)
h
)
¿
(
d(m)Ax)
/[
(
1−Ax(m)
)
− Ex(
1−Ax+h h(m)
)
h
]
3. n year term insurance, mthly premiums payable for n years.
P1x:n|(m)
=Ax1:n|/ ¨a(xm:)h|=
(
d(m)A1x:n|)
/[
(
1−nEx)
−(i/i(m))A1x:n|]
under UDD 4. n year term insurance, mthly premiums payable for h years.P1x:n|(m)
h =Ax:n|
1
/ ¨a(xm:h)|=
(
d(m)A1x:n|)
/[
(
1−hEx)
−(i/i(m))A1x:h|]
under UDD 5. n year endowment, mthly premiums payable for n years.P(xm:n)|
=Ax:n|/ ¨a(xm:n)|=
(
d(m)Ax:n|)
/[
(
1−nEx)
−(i/i(m))Ax1:n|]
under UDD 6. n year endowment, mthly premiums payable for n years.P(xm:n)|
h =Ax:n|/ ¨ax:h|
(m)
=
(
d(m)Ax:n|)
/[
(
1−hEx)
−(i/i(m))Ax1:h|]
under UDDThe formulas are similar when the insurance is payable at the moment of death
e.g., Px
(m)
(
Ax)
=net level annual premium , payable∈m thly installments of (1/m)P(xm) at the
beginning of each mth of a year, for $1 of insurance payable at the moment of death.
Example 4.3.2 Find the net annual premium payable semi-annually for a whole life insurance of $100,000 to (30), based on the Illustrative Life Table at
i=6 %,i/i(2)=1.01,d(2)=0.057. Assume the Uniform distribution of Deaths within each age.
Solution
Want 100,000P30(2)=100,000A30/ ¨a30(2)
¿100,000d(2)A30/
[
1−(
i/i(2))
A30]
under UDD¿
[
100(102.48)(0.057)]
/[
1−1.01(0.10248)]
from table¿$652
i.e., semi-annual premium is $326
Here, premiums are paid mthly, the death benefit is payable at the moment of death and a refund of premium is made based on the time of death and the time of the next premium payment.
Recall a¨x{m}=¿ n.s.p. for an mthly paying life annuity of (1/m) at the beginning of each mth of a
year with a refund of the ``unearned’’ premium at death.
P{xm}
(
Ax)
=¿ apportionable whole life premium paid mthly based on the same refund featureas a¨x{m}
i.e., P{xm}
(
Ax)
satisfies P{xm}(
Ax)
a¨{xm}=Axi.e., P{x m}
(
Ax)
=Ax/ ¨ax{ m}¿Ax/
[
(δ
/d(m))
ax]
¿
(
δ/d(m)) (
Ax/ax)
¿
(
d(m)/δ)
P(
Ax)
i.e., 1/m Px{m}
(
Ax)
=(
d(m)
/δ
)
P(
Ax)
is payable at the beginning of each mth of a year.In particular for m =1, P{x
1}
(
Ax)
= apportionable whole life premium payable at thebeginning of the year for an insurance benefit of $1 payable at the moment of death, together with a premium refund feature.
Since Px
(
Ax)
=¿ annual premium for a similar benefit without the refund feature, it followsthat P{x
1}
(
Ax)
−Px(
Ax)
=¿ net level annual premium for the refund-of-premium feature.P
(
Ax)
PR=(d/δ)
(
Ax/ax)
−(
Ax/ ¨ax)
¿
(
Ax/ax)
[
d/δ−(
ax/ ¨ax)
]
¿
(
Ax/ax)
[
(
da¨x– δ ax)
/δa¨x]
¿
(
Ax/ax)
[
(
(
1−Ax)
–(
1−Ax)
)
/δa¨x]
¿Ax
(
Ax−Ax)
/(
1−Ax)
a¨xNote
The formula for apportionable premiums for other types of insurance benefits can be similarly obtained.
e.g., P{m} (A1x:n|)
h =Ax:n| 1
/ ¨a{xm:h}|=
(
Ax1:n|/ax:h|)
d(m)/δ = apportionable premium for an h-pay mthlypaying n year continuous term insurance on (x)
Example 4.3.2 You are given
(i) Ax=0.2 (ii) i=0.05 (iii) i/i(∞)=1.0248
Assume deaths are uniformly distributed over each year of age. Calculate 10,000P{1}
(
Ax)
−10,000P(Ax)Solution
Expression = 10,000P
(
Ax)
PR=10,000
[
Ax(
Ax−Ax)
/(
1−Ax)
a¨x]
Ax=(i/δ)Ax=i/i(∞)Ax=0.2(1.0248)=0.20469
¨
ax=
(
1−Ax)
d =
(
1−Ax)
(1+i)i =
(1−0.2)(1.05)
0.05 =16.8
10,000P
(
Ax)
PR=10,000
[
0.20496(0.20496−0.2)/(1−0.20496)16.8]
¿0.76
4.3.3 Illustration of Net Level Premium Calculations
Example 4.3.4
A) Determine the net annual premium for a whole life insurance of $1000 issued to (x) using i=0.25 and the following mortality rates:
qx=0.5qx+1=0.9qx+2=1
Solution
Net annual premium = 1000Px where Px satisfies
Px
∑
t=0 2
vt p x
t =
∑
t=0 2
vt+1 q
v= 1 1+i=
1 1.25=0.8
Age q p tp t|q v
t vt
p
t v
t+1 q
t|
x 0.5 0.5 0.5 0.5 0.8 1 0.4
x+1 0.9 0.1 0.05 0.45 0.64 0.4 0.288
x+2 1 1 0 0.05 0.512 0.032 0.0256
1.432 0.7136
i.e., Px(1.432)=0.7136⟹Px=0.49832
i.e., net annual premium 1000Px=$498.32
B) Verify that E(L)=0 where L=¿ loss-at-issue random variable
L=vK+1−Pa¨k+1|
K 1000(vK+1
¿ a¨k+1| Pa¨k+1| L Prob.
0 800 1 498.3 301.7 0.5
1 640 1.8 897.0 -257.0 0.45
2 512 2.44 1215.9 -703.9 0.05
E(L)=301.7(0.5)−257.0(0.45)−703.9(0.05)=0
C) If 100 such policies are sold, show that the premiums collected, with interest, will be sufficient to pay all death benefits, if actual mortality equals expected mortality assumed in pricing.
Duration BeginningFund Premiums Interest Benefits EndingFund
1 0 49,832 12,458 50,000 12,291
2 12,291 24,916 9,302 45,000 1,508
3 1,508 2,492 1,000 5,000 0
4.4 Commutation Functions
General formula:(net annual premium)×(expected p.v. of annuity based on premium paying pattern)
= (expected p.v. of insurance benefit)
Can express annuity and insurance pieces in commutation form and hence solve for the net annual premium in commutation symbols.
Example 4.4.1
a. Obtain a formula in commutation symbols for the net annual premium, payable for five years for a ten year deferred, 20 year temporary immediate annuity of 1 to (30)
Solution
E (p.v. of premium payments)
X: 30 31 32 33 34 35 ………….
D: P P P P P 0 ………….
N: P 0 0 0 0 -P ………….
i.e., E (p.v. of premium payments) = P(N30−N35)/D30
E (p.v. of deferred annuity)
X: 30 31…. 39 40 41 42…. 59 60 61…
D: 0 0 0 0 1 1…. 1 1 0….
N: 0 0 0 0 1 0…. 0 0 -1….
E (p.v. of deferred annuity) = (N41−N61)/D30
i.e., P(N30−N35)/D30=(N41−N61)/D30
b. Find the net annual premium if the annuity in (a) has the additional provision that the net premiums paid will be returned in the event of death within the 10 year period.
Solution
E (p.v. of return of premium benefit)
X: 30 31 32 33 34 35 ……… 38 39 40 41….
C: P 2P 3P 4P 5P 5P ……… 5P 5P 0 0…..
M: P P P P P 0 ……… 0 0 -5P 0…..
R: P 0 0 0 0 -P …….. 0 0 -5P 5P
i.e., E (p.v. of return of premium benefit) = P
(
R30−R35−5R40+5R41)
/D30¿P(R30−R35−5R40)/D30
i.e., P(N30−N35)/D30=
[
(N41−N61)/D30]
+[
P(R30−R35−5R40)/D30]
i.e., P=(N41−N61)/
[
(
N30−N35)
−(R30−R35−5R40)]
4.5 Accumulation Type Benefits
Arises when in the event of death within a prescribed period, we have the additional benefit of either
a. Return of net premiums without interest or
b. Return of net premiums with interest
Example 4.5.1 Find the net annual premium for a $5,000, 20 year term insurance on (x) which provides in case of death within the first 10 years, the additional benefit of net annual premiums paid.
a. without interest
b. with interest same as that used in determining premiums.
Solution
a. Net annual premium P solves
Pa¨x: 20|=5000A1x: 20|+P
∑
t=0 9
(t+1)vt+1t|qx
¿5000A1x: 20|+P(IA)1x: 10|
i.e., P=5000A1x: 20|/
[
a¨x: 20|−(IA)1x: 10|]
b. Now Pa¨x: 20|=5000A1x: 20|+P
∑
t=0
9 ¨
st+1|vt+1 q
x t|
¿5000A1x: 20|+P
∑
t=0 9
¨
at+1|t|qx
¿5000A1x: 20|+P
[
a¨x: 10|− ¨a10|10px]
i.e., P=5000A1x: 20|/
[
a¨x: 20|− ¨ax: 10|+ ¨a10|10px]
Example 4.5.2 Find the net annual for a special 20 year insurance on (x) which provides a death benefit of $5000 if death occurs after two years and the return of the net annual premium paid
a. without interest
b. with interest(same as that used in calculating premiums) if death occurs within the first two years.
Solution
i.e., P=5000
[
Ax1: 20|−A1x: 2|]
/(IA)x1: 2|b. Pa¨x: 20|=5000
