Chapter 5
Net Premium Reserves
5.1 Net Premium Reserves
Consider an insurance benefit purchased by (x) with a sequence of annual premiums. Then under the Equivalence Principle, at issue,
E(p . v . of premiums)=E(p . v . of insurance benefits)
and L=(p . v . of insurance benefits)−(p . v . of premiums) satisfies E(L)=0 at issue.
Define tL=prospective loss at timet
¿(p . v .at x+t of future insurance benefits)−(p . v . at x+t of future premiums)
Then E
(
tL)
=net premium reserve¿E¿
−E(p . v . at x+t of future premiums)
In particular, for a fully continuous whole life insurance policy to (x), define
U=timeuntil deathof x+t
Then U kpx+tμx+t+k and tL=vU−P
(
Ax)
aU|E
(
tL)
=V(t¿Ax)¿¿E
(
vU)
−P(
Ax)
E(a¿¿U|
)¿¿Ax+t−P
(
Ax)
ax+tVar
(
tL)
=Var[
vU−P(
Ax) (
1−v U)
/δ
]
¿Var¿
5.2 Fully Continuous Net Premiums Reserves
1. Whole life insurance, premiums payable for lifeV(t¿Ax)=Ax+t−P
(
Ax)
ax+t¿2. h payment, whole life insurance
V
t h
(
Ax)
=Ax+t−P(
Ax)
ax+t:h−t|for t<h h¿Ax+tfor t ≥ h
3. n-year term insurance, premiums payable for n years
V(t¿Ax1:n|)=A1x+t:n−t|−P
(
Ax1:n|)
ax+t:n−t|for t<n¿ ¿0for t ≥ n4. h payment, n-year term insurance
V
t h
(
Ax1:n|)
=A1x+t:n−t|−hP(
A1x:n|)
ax+t:h−t|for t<h ¿A1x+t:n−t|for h ≤ t<n¿0for t ≥ n
5. n payment, n-year endowment insurance
V(t¿Ax:n|)=Ax+t:n−t|−P
(
Ax:n|)
ax+t:n−t|for t<n¿ ¿1for t=n¿0for t>n
6. h payment, n-year endowment insurance
V
t h
(
Ax:n|)
=Ax+t:n−t|−hP(
Ax:n|)
ax+t:h−t|for t<h¿Ax
+t:n−t|for h ≤ t<n ¿1for t=n
7. n payment, n-year deferred annuity
V(t¿n|ax)=ax+nvn−tn−tpx+t−P
(
n|ax)
ax+t:n−t|for t<n¿ ¿ax+tfor t ≥ n8. h payment, n year deferred annuity
V
t h
(
n|ax)
=ax+nvn−tn−tpx+t−hP(
n|ax)
ax+t:h−t|for t<h ¿ax+nvn−tn−tpx+tfor h≤ t<n¿ax+tfor t ≥ n
Example 5.2.1 You are given:
1000V(t¿Ax)=100¿ (ii) 1000P
(
Ax)
=10.5ax+t (iii) δ=0.03Calculate ax+t Solution
V(t¿Ax)=Ax+t−P
(
Ax)
ax+t¿ ¿(
1−δ ax+t)
−P(
Ax)
ax+t ¿1−[
δ+P(
Ax)
]
ax+ti.e., ax+t=¿ ¿
¿(1−0.1)/(0.03+0.0105)
5.3 Reserve Relationships
1. Whole lifeV(t¿Ax)=Ax+t−P
(
Ax)
ax+t¿¿
[
(
Ax+t/ax+t)
−P(
Ax)
]
ax+t ¿[
P(
Ax+t)
−P(
Ax)
]
ax+tAlternatively, V(t¿Ax)=Ax+t
[
1−P(
Ax) (
ax+t/Ax+t)
]
¿ ¿Ax+t[
1−(
P(
Ax)
/P(
Ax+t)
)
]
2. n-year term
V(t¿Ax1:n|)=A1x+t:n−t|−P
(
A1x:n|)
ax+t:n−t|¿¿
[
P(
A1x+t:n−t|)
−P(
A1x:n|)
]
ax+t:n−t|¿A1x+t:n−t|
[
1−P(
A1x:n|)
/P(
A1x+t:n−t|)
]
3. n-year endowment
V(t¿Ax:n|)=Ax+t:n−t|– P
(
Ax:n|)
ax+t:n−t|¿ ¿[
P(
Ax+t:n−t|)
– P(
Ax:n|)
]
ax+t:n−t|¿Ax+t:n−t|
[
1−P(
Ax:n|)
/P(
Ax+t:n−t|)
]
Example 5.3.1 Given u=
[
P(
Ax:t|)
−tP(
Ax)
]
/P(
Ax1:t|)
, Ax=(1/2)Ax+t, δ=¿ force of interest,express V(t¿Ax)¿ in terms of u∧δ.
Solution
u=
[
Ax:t|/ax+t−Ax/ax+t
]
[
A1x:t|/ax+t]
=[
Ax:t|– Ax]
Ax1:t| =[
A1x:t|−nExAx+t]
i.e., Ax+t=1−u Ax=(1−u)/2
V(t¿Ax)=Ax+t−P
(
Ax)
ax+t¿ Ax+t=1−uax+t=
(
1−Ax+t)
δ =
u δ
P
(
Ax)
=Ax ax=δ Ax
1−Ax=
δ
[
(1−u)/2]
[
(
1−(1−u))
/2]
=δ(1−u) (1+u)
V(t¿Ax)=(1−u)−
[
δ(1−u) (1+u)]
u δ ¿
¿
[
(
1−u2
)
−
(
u−u2)
]
(1+u)¿(1−u) (1+u)
5.4 Retrospective Formulas
Recall from Theory of interest,
an| payable now = $1 per year payable continuously for n years
= sn| payable at the end of n years
an|=
∫
0
n
vtdt , sn|=
∫
0
n
(1+i)tdt and vnsn|=an|
Similarly, in Actuarial Math,
ax:n|payable now=$1per year payable continuously for n yearswhile(x)is living
¿ax:n|/vnnpx=sx:n|payable at the end n years if(x)is living
Alternatively,
$1 payable at the end of n years if (x) is living = nEx payable now. and
$1 payable now = 1/nEx payable at the end of n years if (x) is living
Ex
n =actuarial present value of $1payable at the end of n years if(x)is living
1/nEx=actuarial accumulated valueof $1payable at the end of n yearsif(x)isliving
1. Whole life insurance
V(t¿Ax)=Ax+t−P
(
Ax)
ax+t¿¿ Ax−Ax:t|
1
Ex t
−P
(
Ax)
[
ax−ax:t| Ext
]
since Ax=Ax:t|
1
+tExAx+t and ax=ax:t|+tExax+t
i.e., V(t¿Ax)=
(
1/tEx)
[
Ax−P(
Ax)
ax+P(
Ax)
ax:t|−Ax:t|1
]
¿¿P
(
Ax)
ax:t| Ext
−Ax:t|
1
Ex
t
¿retrospective reserve at x+t
In words,
V(t¿Ax)=Ax+t−P
(
Ax)
ax+t¿¿prospective reserve at(x+t)
¿(actuarial p . v . at x+t of future benefits)−(actuarial p . v . at x+t of future premiums)
¿P
(
Ax)
ax:t| Ex t−Ax:t|
1
Ex t
¿(actuarial accumulated value at x+t of past premiums)−(acturialaccumulated value at x+t of past benefits)
2. n-year term insurance
V(t¿Ax1:n|)=A1x+t:n−t|−P
(
A1x:n|)
ax+t:n−t|¿ ¿prospective reserve at(x+t)¿ Ax:n|
1
−Ax1:t| Ex t
−P
(
Ax1:n|)
[
ax:n|– ax:t| Ext
]
¿P
(
Ax1:n|)
ax:t| Ext
−Ax:t|
1
Ex
t
¿retrospective reserve at x+t
3. n-year endowment insurance
V(t¿Ax:n|)=Ax+t:n−t|– P
(
Ax:n|)
ax+t:n−t|¿ ¿prospective reserve at(x+t)¿ Ax:n|−Ax:t|
1
Ex t
−P
(
Ax:n|)
[
ax:n|– ax:t| Ext
]
¿P
(
Ax:n|)
ax:t| Ext
−Ax:t|
1
Ex
t
¿retrospective reserve at x+t
4. n pay, n-year deferred life annuity
For t<n
¿n|ax Ex t
−P
(
n|ax)
[
ax:n|– ax:t| Ext
]
¿P
(
n|ax)
ax:t| Ext
¿retrospective reserve at x+t
For t ≥ n
V(t¿n|ax)=ax+t¿
¿prospective reserve at(x+t)
But ax+t+
(
n|ax:n−t|/tEx)
=P(
n|ax)
(
ax:n|/tEx)
=n|ax/tEx V(t¿n|ax)=ax+t¿¿P
(
n|ax)
[
ax:n| Ext
]
−n|ax:n−t| Ex t
¿retrospective reserve at x+t
Example 5.4.1 Which of the following are true?
A. V(t¿Ax)=1−
[
ax+t/ax]
¿B. V(t¿Ax)=
(
Ax−Ax+t)
/(
1−Ax)
¿C. V(t¿Ax)=
[
ax:t|/tEx]
[
P(
Ax)
−P(
A1x:t|)
]
¿Solution
A. V(t¿Ax)=Ax+t−P
(
Ax)
ax+t¿¿(1−δ ax+t)
[
(1−δ ax)/ax]
ax+t¿
(
ax−δ ax+tax+δ axax+t)
/ax ¿(
ax−ax+t)
/ax¿1−
[
ax+t/ax]
B. From (A), V(t¿Ax)=
(
ax−ax+t)
/ax¿¿
[
(
1−Ax)
/δ−(
1−Ax+t)
/δ]
/[
(
1−Ax)
/δ]
¿(
Ax+t−Ax)
/(
1−Ax)
i.e., B is true.
C. Retrospectively, V(t¿Ax)=
[
P(
Ax)
ax:t|/tEx]
−[
Ax:t|1
/tEx
]
¿¿
[
ax:t|/tEx][
P(
Ax)
−Ax1:t|/ax:t|]
¿
[
ax:t|/tEx]
[
P(
Ax)
−P(
A1x:t|)
]
i.e., C is true.
Example 5.4.2 Consider a special fully continuous 30 year endowment to (35) which pays $1000 upon death the first 20 years and $2000 at 65 if (35) survives.
You are given the following:
(i) The net level premium for this special endowment policy is $30. (ii) The n.s.p. for 10 year pure endowment of $1 to (35) is $0.5. (iii) The n.s.p. for 10 year continuous annuity of $1 to (35) is $7.
(iv) The n.s.p. for 10 year continuous term insurance of $1 to (35) is $0.1.
Calculate the 10th year net premium reserve for this policy.
Solution
Retrospectively, 10V=
[
P a35: 10|/10E35]
−[
A35 :10|1
/10E35
]
¿
[
30(7)/0.5]
−[
1000(0.1)/0.5]
¿420−200
¿$220
Premiums are payable at the beginning of the year and insurance benefits are paid at the end of the year of death. Then
L
k =prospective loss at time k
¿(p . v .at x+k of future insurance benefits)−(p . v . at x+k of future premiums)
Define J=¿ curtate future lifetime of x+k.
Then J=0,1,2, … … and P
[
J=j]
=j|qx+k¿jpx+kqx+k+j
Suppose $1 = insurance benefit paid and P = annual premium paid.
Then when J=j
L
k =vj+1−Pa¨j+1|
--- --- --- --- x x+1 x+2 x+k+j x+k+j+1
i.e., kL=v
J+1
−Pa¨J+1|
E
(
kL)
=net premium reserve at end of k years¿E¿
1. Whole life insurance, premiums payable for life
Vx
k =Ax+k−Pxa¨x+k
L
k =vJ+1−Pa¨J+1|=vJ+1−Px
[
(
1−vJ+1)
/d]
¿vJ+1
[
1+(
Px/d)
]
+(
Px/d)
Var
(
kL)
=[
1+Px/d]
2[
2Ax+k−(
Ax+k)
2]
Vx
k =
Ax−Ax:k|
1
Ex k
−Px
[
a¨x− ¨ax:k| Exk
]
¿Pxa¨x:k| Ex k
−Ax:k|
1
Ex k
2. h-payment, whole life insurance
Vx k h
=Ax+k−hPxha¨x+k:h−k|for k<h h ¿Ax+kfor k ≥ h
Retrospectively,
Vx
k h
=hPxa¨x:k| Ex k
−Ax:k|
1
Ex k
¿hPxs¨x:k|−kPxfor k<h h
¿hPxa¨x:h| Ex k
−Ax:k|
1
Ex k
for k ≥ h
3. n-year term insurance, premiums payable for n years
V
k 1x:n|=A1x+k:n−k|−P1x:n|a¨x+k:n−k| Retrospectively,
V
k x:n|
1
=Px:n|
1
¨ ax:k| Ex k
−Ax:k|
1
Ex k
4. n-year term insurance, premiums payable for h years.
V
k h
x:n|
1
=A1x+t:n−t|−hPx1:n|a¨x+k:n−k|for k<h h
¿A1x+t:n−k|for k ≥ h
V
k h
x:n|
1
= Px:n|
1
h a¨x:k| Ex k
−Ax:k|
1
Ex k
for k<h h
¿ Px:n|
1
h a¨x:h|
Ex k
−Ax:k|
1
Ex k
for k ≥ h
5. n payment, n-year endowment insurance
V
k x:n|=Ax+k:n−k|−Px:n|a¨x+k:n−k|
Retrospectively,
V
k x:n|=
Px:n|a¨x:k| Ex
k
−Ax:k|
1
Ex
k
6. h payment, n-year endowment insurance
V
k h
x:n|=Ax+t:n−t|−hPx:n|a¨x+k:n−k|for k<h
¿Ax
+k:n−k|for k ≥ h
Retrospectively,
V
k h
x:n|= P
h x:n|a¨x:k| Ex k
−Ax:k|
1
Ex k
for k<h
¿hPx:n|a¨x:h| Ex k
−Ax:k|
1
Ex k
for k ≥ h
7. n payment, n-year deferred annuity
V(k¿n|a¨x)= ¨ax+nn−kEx+k−P
(
n|a¨x)
a¨x+k:n−k|for k<n¿ ¿a¨x+nfor k ≥ nV(k¿n|a¨x)=
P
(
n|a¨x)
a¨x:k| Ex kfor k<n¿
¿P
(
n|a¨x)
a¨x:n| Ex k−n|a¨x:k−n| Ex k
for k ≥ n
Example 5.5.1 You are given:
(i) Px=0.01212 (ii)20Px=0.01508 (iii) P1x: 10|
=0.06942 (iv) 10Vx=0.11430
Calculate 10Vx
20 Solution Retrospectively, Vx 10 20
=20Pxa¨x: 10| Ex
10
−Ax: 10|
1
Ex
10
Given 10Vx=
[
Pxa¨x: 10|/10Ex]
−[
Ax:10|1
/10Ex
]
and Px: 10|1
=10Ex/ ¨ax: 10| then
Vx
10 =
Px
P1x: 10|− Ax1: 10|
Ex
10
Ax1: 10|
Ex
10
= Px
P1x: 10|−10Vx
¿
[
0.012120.06942
]
−0.011430¿0.06029
Thus,
Vx
10 20
=20Pxa¨x: 10| Ex
10
−Ax: 10|
1
Ex
10
¿ 20Px
Px1: 10|−0.06029
¿
[
0.01508¿0.15694
Example 5.5.2 Find the value of P1x:n|
if nVx=0.08, Px=0.024 and Px:n|
1
=0.2
Solution
Retrospectively,
Vx
n =
Pxa¨x:n|
Ex
n
−Ax:n|
1
Ex
n
Since P1x:n|=nEx/ ¨ax:n|, it follows that
Ax1:n| Ex
n
=
[
0.0240.20
]
−0.08=0.04Want P1x:n|
=¿ A1x:n|/ ¨ax:n|. But A1x:n|=0.04nEx and a¨x:n|=nEx/Px:n|
1
=nEx/0.2. i.e.,
P1x:n|
=0.04nEx Ex
n /0.2
=0.008
Example 5.5.3 A four year term life insurance of 1000 is issued on a fully discrete basis to (82). The insured survives to the end of second policy year. You are given
(i) i=0 (ii) net annual premium = 120 (iii) q84=0.12 (iv) q85=0.13
Calculate Var
(
2L)
Solution
120 120
--- --- --- --- --- 82 83 84 85 86
1000 1000
L
2 =(p . v . at84of future benefits)−(p . v . at84of future premiums)
Var
(
2L)
=E(
2L2)
−[
E(
2L)
]
2Case 1: (84) dies before 85
L
L
2 2
=(880)2
Probability = 0.12
Case 2: (84) dies between 85 and 86
L
2 =1000−2(120)=760
L
2 2
=(760)2
Probability = (0.88)(0.13) = 0.1144
Case 1: (84) dies after 86
L
2 =0−2(120)=−240
L
2 2=(240)2
Probability = 1-0.12- 0.1144 = 0.7656
E
(
2L)
=880(0.12)+760(0.1144)−240(0.7656)=8.8 E(
2L2
)
=(880)2(0.12)+(760)2(0.1144)+(240)2(0.7656)=203,104i.e., Var
(
2L)
=E(
2L2)
−[
E(
2L)
]
2=203,104−(8.8)2
¿203,027
5.5.1 Relationship between reserves and pricing
Consider the example of the whole life insurance of $1000 issued to (x) at i=0.25 and
mortality rates qx=0.5,qx+1=0.9, qx+2=1.
We have shown the net annual premium is 1000px=1000(0.49832)=$498.32
The net premium reserves at each duration are:
V
0 =0
V
1 =1000
[
(0.9) (0.8)+(0.1) (1)(0.64)]
−498.32[
1+(0.1)(0.8)]
=245.81V
2 =100(1) (0.8)−498.32=301.68
V
If 100 such policies are sold, we have the following fund progression:
t lx+t Beg. Fund Interest Benefits End. Fund Share/supervisor
0 100 49,832.40 12,458.10 50,000 12,290.50 245.81
1 50 37,206.70 9,301.68 45,000 1,508.38 301.68
2 5 4,000.00 1,000.00 5,000 0 0
i.e., share per survivor = (end of the year fund)÷(survivors at end of year)
= net premium reserve
In other words, if net premiums are determined in accordance with the Equivalence Principle, and if actual mortality and interest experienced are the same as that used in pricing, then the total reserve at the end of each year (which represents the insurer’s expected net future obligations) is precisely the end of year fund that is being built up.
5.6 Annual Premiums, Continuous Benefits
1. a. Whole life insurance, premiums payable for lifeV(t¿Ax)=Ax+t−P
(
Ax)
a¨x+t¿ Under UDD, Ax+t=(i/δ)Ax+tP
(
Ax)
=Ax/ ¨ax=(i/δ)Ax/ ¨ax=(i/δ)Pxand
V(t¿Ax)=(i/δ)Ax+t−(i/δ)Pxa¨x+t¿
b. h-payment, whole life insurance
V
t h
(
Ax)
=Ax+t−hP(
Ax)
a¨x+t:h−t|for t<h h ¿Ax+tfor t ≥ hUnder UDD,
V
t h
(
Ax)
=(i/δ)tVx h2. a. n-year term insurance
V(t¿Ax1:n|)=A1x+t:n−t|−P
(
Ax1:n|)
a¨x+t:n−t|¿Under UDD,
V(t¿Ax1:n|)=(i/δ)tVx1:n|¿
b. h payment, n-year term insurance
V
t h
(
Ax1:n|)
=A1x+t:n−t|−hP(
A1x:n|)
a¨x+t:h−t|for t<h ¿A1x+t:n−t|for t ≥ h
Under UDD,
V
t h
(
Ax1:n|)
=(i/δ)htVx1:n|3. a. n-year endowment insurance
V(t¿Ax:n|)=Ax+t:n−t|−P
(
Ax:n|)
a¨x+t:n−t|¿Under UDD,
Ax+t:n−t|=A1x+t:n−t|+A1x+t:n−t| ¿(i/δ)A1x+t:n−t|+A1x+t:n−t|
P
(
Ax:n|)
=Ax:n| ¨ ax:n|=[
(i/δ)A1x:n|+Ax1:n|]
¨¿(i/δ)P1x:n|+P1x:n|
V(t¿Ax:n|)=(i/δ)Ax+t:n−t|
1
+A1x+t:n−t|−¿ ¿
[
(i/δ)P1x:n|+Px1:n|
]
a¨x+t:n−t| ¿(i/δ)tV1x:n|+tV1x:n|b. h payment, n-year endowment insurance
V
t h
(
Ax:n|)
=Ax+t:n−t|−hP(
Ax:n|)
a¨x+t:h−t|for t<h¿Ax
+t:n−t|for t ≥ h
Under UDD,
V
t h
(
Ax:n|)
=(i/δ)tVh x:n|
1
+htVx:n|
1
Example 5.6.1 You are given:
(i) Ax=0.25 (ii) Ax+20=0.40 (iii) Ax: 20|=0.55
(iv) i/δ=1.01 (v) d=0.03 (vi) P
(
Ax:n|)
=0.02Assume deaths are uniformly distributed over each year of age.
Calculate V(20¿Ax:n|)¿
Solution
Retrospectively,
V(20¿Ax:n|)=P
(
Ax:n|)
a¨x: 20| Ex20
−Ax:20|
1
Ex
20
¿
¨
ax: 20|=
(
1−Ax: 20|)
/d=(1−0.55)/0.03=15Ax: 20|
1
=(i/δ)A1x: 20|=1.01A1x: 20|
Ax=Ax: 20|
1
+20ExAx+20
Ax: 20|=Ax1: 20|+20Ex
i.e., 0.55=A1x: 20|+20Ex… … …(2)
(2)-(1) 0.3=20Ex(0.6)
i.e., 20Ex=0.5, Ax: 20|
1
=0.05
Substituting,
V(20¿Ax:n|)=0.02 15
0.5−
1.01(0.05)
0.5 ¿
¿0.6−0.101
¿0.499
5.6.1 Mthly Payment of Premiums
Reserves can be calculated when premiums are paid in a mode other than annual.
e.g., Vx
(12)
t =¿ reserve at the end of t years for a whole life insurance to (x) with premiums
payable monthly.
Prospectively, tVx(m)=Ax+t−Px (m)
¨ ax+t
(m)
where P(xm)=Ax/ ¨a(xm); a¨x
(m)
=(1−Ax(m))/d(m)=
[
1−(i/i(m))Ax]
/d(m) under UDDRelationship to annual reserves
Vx
t =Ax+t−Pxa¨x+t
i.e., tVx(m)−tVx=Pxa¨x+t−Px (m)
¨ ax+t
(m)
¿P(xm)
(
a¨x (m)¨
ax
)
a¨x+t−Px(m) ¨ ax+t
(m)
¿P(xm)
[
a¨x+ta¨x (m)¨
ax − ¨ax+t
(m)
¨ ax
(m)
=
(
1−Ax (m))
d(m) =
(
Ax+da¨x−Ax(m)
)
d(m)
¿ d
d(m)a¨x− Ax d(m)
[
i/i(m) −1
]
Thus a¨x+t
(
a¨(xm)/ ¨ax)
=(
a¨x+td/d(m)
)
−(
Px/d(m))
a¨x+t[
i/i(m)−1]
and¨ ax(m+)t
=
(
a¨x+td/d(m)
)
−(
Ax+t/d(m))[
i/i(m)−1]
. Hence ¨ax+ta¨x
(m)
¨
ax − ¨ax(m+)t= 1
d(m)
[
i/i (m)−1
][
Ax+t−Pxa¨x+t]
¿tVx
[
i−i (m)i(m)d(m)
]
¿tVxβ(m)
i.e., Vx
(m)
t −tVx=Px
(m) Vx
t β
(m)
Relationships between mthly and annual reserves for other forms of insurances can
similarly be obtained.
5.6.2 Apportionable Reserves
Reserves can be determined when premiums are paid on an apportionable basis. i.e., upon death, a death benefit is payable together with a refund of the ``unearned’’ premium.
We have shown that the annual premium payable mthly with such a refund feature is given
by:
P{xm}
(
Ax)
=Ax/ ¨ax{m}for a whole life insurance benefit¿
(
d(m)/δ)
P(
Ax)
Vx{m}
(
Ax)
t =t
t h
year reserve for a whole lifeinsuance payable mthly with a premium
refund feature
¿Ax+t−Px{ m}
¨ ax{+t
m}
¿Ax+t−
(
d (m)δ
)
Px(
Ax)
(
δ d(m)
)
ax+t¿Ax+t−Px
(
Ax)
ax+t ¿V(t¿Ax)¿¿ fully continuous reserve
In general, apportionable reserves for any premium paying mode are equal to fully continuous reserves.
In particular, for m=1,
P{x1}
(
Ax)
=apportionableannual premium¿
(
dδ
)
P(
Ax)
Px
(
Ax)
=premium without refund featureP{x1}
(
Ax)
−Px(
Ax)
=P(
Ax)
PR¿net level annual premium for therefund portion
It follows that
V
t {
1}
(Ax)=t t h
year reserve for an annual premium whole life insurance policy with
a refund of unearned premium
¿ ¿
¿refund feature]+[tt hyear reserve for refund of unearned premium feature
¿
¿ V(t¿Ax)¿
Example 5.6.2 Calculate 40V
(
A20)
PR given the following values:V(40¿A20)=0.5078¿ a20=25 a60=12.24 δ=0.03
Solution
V
40
(
A20)
PR
=40V{1}
(
A20)
−V(40¿A20)¿¿40V{ 1}
(
A20)
−0.5078V
40 {1}
(
A20)
=V(40¿A20)¿¿A60−P
(
A20)
a60A60=1−δ a60=1−0.03(12.24)=0.6328
P
(
A20)
=(
A20/a20)
=[
(1−δ a20)/a20]
=[
(1−0.03(25))/25]
=0.01i .e . , V40 { 1}
(
A20)
=0.6328−0.01(12.24)¿0.5104
i .e . , V40
(
A20)
PR
=0.5104−0.5078
¿0.0026
Example 5.6.3 For a fully continuous 20 year deferred life annuity of 1 issued to (35), you are given:
(i) Mortality follows de Moivre’s law with ω=75
(ii) i=0
Calculate the net premium reserve at the end of 10 years for this annuity.
Solution
Retrospectively,
V(10¿20|a35)=P
(
20|a35)
a35 : 10| E3510
¿
P
(
20|a35)
=20|a35/a35 : 20|Under de Moivre’s law,
px
t μx+t= 1
ω−x=
1 75−x
Since i=0, at|=t
a35 20| =v
20 p 35a55
20 =20p35a55
p35
20 =1−20q35=1−[20/(75−35)
]=
0.5a55=
∫
0 75−55
at|tp55μ55+tdt=
[
175−55
]
∫
020
t dt= t
2
40
|
20¿0=10
a35: 20|=
∫
0 20
at|tp55μ35+tdt+a20|20p35=
[
175−35
]
t22
|
20¿0+20[
1−20
(75−35)
]
¿5+10
¿15
P
(
20|a35)
=20|a35/a35 : 20|=10(0.5)/15=1/3a35: 10|
∫
0 10
t
[
1(75−35)
]
dt+10(
10p35)
= t280
|
10¿0+10
[
1−10 40
]
¿1.25+7.5
¿8.75
E35
10 =v 10
p35
10 =10p35=1−
10
V(10¿20|a35)=P
(
20|a35)
a35 : 10| E3510
¿
¿
(
13
)
8.75 0.75¿3.889
5.7 Recursive Formulas
For an insured age x, let π0, π1, π2, … … be a sequence of premiums where πt=¿ premium
paid at beginning of year t+1.
Let b1, b2, b3, … … be a sequence of death benefits where bt=¿ death benefit paid at end of
year t.
Assume premiums and death benefits are based on the Equivalence principle.
i.e., expected p.v. of premiums = expected p.v. of benefits
∑
t=0
∞
πtvt p x
t =
∑
t=0
∞
bt+1vt+1 q
x t|
V
t x=reserve at end of year t , t=1,2,3…..
V
0 x=0
V
t x=(expected p . v . of future benefits)−(expected p . v . of future premiums)
¿
∑
k=0
∞
bt+k+1vk+1
qx
k| +t−
∑
k=0
∞
πt+kvkkpx+t
¿v bt+1qx+t+
∑
k=1
∞
bt+k+1vk+1 q
x+t
k| −
∑
k=1
∞
πt+kvk p x+t k −πt
¿v bt+1qx+t−πt+v px+t
∑
k=0
∞
bt+1+k+1vk+1
qx+t+1
k| −v px+t
∑
k=0∞
πt+1+kvkkpx+t+1
¿v bt+1qx+t−πt+v px+tt+1Vx
i.e.,
(
tVx+πt)
(1+i)=bt+1qx+t+px+tt+1Vx… …… .(1)Multiply both sides by lx+t and we get
lx+t
(
tVx+πt)
(1+i)=bt+1dx+t+(
lx+t+1)
t+1Vx… … .(2)Interpretation: For a group of lx+t policies, the tt hyear reserve, increased by one year’s
premium and interest, is sufficient provide benefit payments bt for each of the dx+t lives that
die during the year, plus the (t+1)st year reserve, t+1Vx, for each of the lx+t+1 lives that
survive.
Alternatively, equations (1) and (2) can be written as:
(
tVx+πt)
(1+i)=(
bt+1−t+1Vx)
qx+t+t+1Vx… …...(3)lx+t
(
tVx+πt)
(1+i)=dx+t(
bt+1−t+1Vx)
+(
lx+t+1)
t+1Vx…...(4)(
bt+1−t+1Vx)
=net amount at risk for(t+1)styear(
bt+1−t+1Vx)
qx+t=cost of insurance(1year term insurance)based upon the net amountat risk for the(t+1)styear
5.7.1 Income Statement Interpretation
In the (t+1)st year, we have the following ``Statement of Operations’’:
Premiums: πt
Investment Income:
(
tVx+πt)
iBenefits: bt+1qx+t
Cash flow: πt+
(
tVx+πt)
i−bt+1qx+tIncrease in reserves: px+tt+1Vx−tVx
Gain from Operations: πt+
(
tVx+πt)
i−bt+1qx+t−(
px+tt+1Vx−tVx)
¿
(
tVx+πt)
(1+i)−bt+1qx+t−px+tt+1Vx¿0
Terminology: For the (t+1)st year,
V
t +πt=¿ initial reserve
V
t+1 = terminal reserve
Example 5.7.1 For a fully discrete whole life insurance with level annual premiums on the life of (x), you are given:
(i) i=0.05
(ii) qx+h−1=0.004
(iii) The initial reserve for policy year h is 200
(iv) The net amount at risk for year h is 1,295
(v) a¨x=16.2
Calculate the terminal reserve for policy year h-1
Solution
200=h−1V+P
………. ………..… ……… ………
x x+h-2 x+h-1 x+h
We want h−1Vx
Also, F−hVx=1295, where F=¿ face amount of policy
(
h−1Vx+Px)
(1+i)=(
F−hVx)
qx+h−1+hVxi.e., 200(1.05)=1295(0.004)+hVx
i.e., hVx=210−5.18=204.82
F=1295+204.82=1499.82
Px=F
(
Ax ¨ax
)
=F(
1−da¨x ¨
ax
)
=1499.82(
1−(0.05/1.05)16.2
16.2
)
=21.16i.e., h−1Vx=200−2.16=178.84
Example 5.7.2 An insured under ordinary life policy for $1000, issued at age 20, is to be subject during the eleventh policy year to an extra mortality not covered by the terms of the policy. This extra risk may be expressed as an addition of 0.01 to the normal rate of mortality during the year. If the eleventh year terminal reserve is $81.54 and the tabular rate of mortality at age 30 is 0.008427, calculate on a net basis the theoretical reduction in the amount of insurance the company should require during the eleventh year if the policy is to be amended to include the extra risk.
Solution
(
10V20+P20)
(1+i)=11V20(
1−q30)
+1000q30… …… …(1)where q30=tabular mortality rate=0.008427
Let F¿
=reduced daethbenefit
q30¿
=mortality of impaired life=q30+0.01
Then
(
10V20+P20)
(1+i)=11V20(
1−q30¿)
+F¿q30¿ …… …(2)F¿
=11V20
(
q30¿
−q30
)
+1000q30q30¿
¿81.54(0.01)+1000(0.008427)
0.01+0.008427
¿$501.57
i.e., reduction in insurance = 1000-501.57
= $498.43
Example 5.7.3 Given that i=0.04,2320V15=0.585 and 2420V15=0.600, calculate p38
Solution
Since premium paying period is 20 years, P=0for t ≥23
i.e., 2320V15(1+i)=2420V15p38+q38
¿2420V15p38+(1−p¿¿38)¿
¿1−p38
(
1−2420V15)
p38=1−23V15
20
(1+i)
1−2420V15
¿1−5.85(1.04)
1−0.6 ¿0.979
5.7.2 Fackler Reserve Accumulation Formula
Vx t+1 =
(
tVx+πt)
(1+i)−bt+1qx+tpx+t
This is the Fackler reserve accumulation formula, starting off at 0V=0.
5.7.3 Payment of Reserve in additional to face Amount
We can write the formula for successive terminal reserves as:
Vx
t+1 −(1+i)tVx=πt(1+i)−
(
bt+1−t+1Vx)
qx+t¿πt(1+i)−Kx+t
where Kx+t=
(
bt+1−t+1Vx)
qx+t¿cost of insurance for the net amount at risk∈the(t+1)styear
Multiplying both sides by (1+i)n−t−1, we get
(1+i)n−t−1t+1Vx−(1+i)n−ttVx=πt(1+i)n−t−Kx+t(1+i)n−t−1
This is true for all t ≥0
∑
t=0
n−1
[
(1+i)n−t−1t+1Vx−(1+i)n−ttVx]
=∑
t=0
n−1
πt(1+i)n−t−
∑
t=0
n−1
Kx+t(1+i)n−t−1
Since 0Vx=0, this reduces to
V
n x=
∑
t=0n−1
πt(1+i)n−t−
∑
t=0
n−1
Kx+t(1+i)n−t−1
i.e., nt h year terminal reserve = accumulation of past premiums to the end of the nt h year
by interest only - accumulation of past cost of insurance
charges to the end of the nt h year by interest only
When the death benefit b 't+1 is the face amount bt+1 plus the reserve, we have
b 't+1=bt+1+t+1Vx and
Then
V
n x=
∑
t=0n−1
πt(1+i) n−t
−
∑
t=0
n−1
bt+1qx+t(1+i)
n−t−1
In the special case when πt=π and bt=b for all t,
V
n x=πs¨n|−b
∑
t=0
n−1
qx+t(1+i)n−t−1
Example 5.7.4 Consider a whole life fully discrete insurance of 1 to (x) with the provision that the reserve will be paid in additional to the face amount in the event of death during
the first n years. The extra premium for the additional benefit is payable for n years. Find
an expression for this extra premium.
Solution
Let P=premium payable∈the first n years.
Px=premium payable after n years
V
n x=nthterminal reserve for whole life insurance
¿Ps¨n|−
∑
t=0
n−1
qx+t(1+i)
n−t−1
i .e . , P= V
n x+
∑
t=0n−1
qx+t(1+i)n−t−1 ¨
sn|
Example 5.7.5 A special fully discrete two year endowment insurance with a maturity value of 1000 is issued to (x). The death benefit in each year is 1000 plus the net premium reserve at the end of that year. You are given
(i) i=0.1
(ii) qx+k=(0.1)(1.1) k
, k=0,1
Calculate the net level annual premium.
Solution
V
2 =1000π
[
(1+i) 2+(1+i)
]
−1000[
qx(1+i)+qx+1]
i..e, 1000=π
[
(1+i)2+(1+i)]
−1000[
0.1(1.1)+0.1(1.1)]
.Solving,
π=1000(1+0.22)
2.31 =$528.14
Example 5.7.6 A deferred life annuity is issued to (40) for an annual income of 1 commencing at age 60. Net annual premiums are paid during the deferred period. During the deferred period, a death benefit equal to the net premium reserve is payable at the end
of the year of death. Determine an expression for the net premium reserve at the end of 10th
year.
Determine an expression for the net premium reserve at the end of 10th year.
Solution
K40+t=q40+t
[
bt+1−t+1~V]
=q40+t[
t+1~V−t+1~V]
=0for t ≤0i.e., 10~V
(
20|a¨40)
=P¨s10|~V
10
(
20|a¨40)
=10V(
20|a¨40)
= ¨a60=Ps¨20|i.e., P= ¨a60/ ¨s20| and
~V
10
(
20|a¨40)
=¨ a60s¨10|
¨ s20|
Recall the formula connecting successive terminal reserves
(
tV+πt)
(1+i)=bt+1qx+t+px+tt+1VThis formula can be generalized to determine reserves at any duration between t and t+1,
premiums not necessarily payable at the beginning of the year and death benefits not necessarily payable at the end of the year.
Consider the case for evaluating reserves at t+s, premium πt payable at t+s1 and death
benefit payable at t+s2, 0≤ s1≤ s2≤1
-- --- --- ---
Then
V
t+s (1+i)
1−s
−πts1−spx+t+s(1+i)1−s1
=bt+1s2−sqx+t+s(1+i)1−s2
−t+1V1−spx+t+s
i.e., t+sV=bt+1s2−sqx+t+svs−s2
−t+1V1−spx+t+sv1−s−πts1−spx+t+svs1−s
One can use any of the mortality assumptions for fractional ages to evaluate this expression.
Special Cases:
A. Premiums payable at the beginning of the year, death benefits payable at the end of the
year
---
---V
t+s =bt+11−sqx+t+sv
1−s
−t+1V1−spx+t+sv
1−s
t V t πt t+1 V
t+1
πt+1
t V
t
t+1 V
t+1
t+s1 πt
t+s2 bt+1 t+s
V
t+s
t V
t
πt
t+s V
t+s
t+1 V
t+1
