Topics in Mathematics II - Actuarial Mathematics
Solutions to Exercises on Hand-Out 4
Frank Coolen (CM206 - Frank.Coolen@durham.ac.uk), March 2008
4-1.
n
p
x
d¨
a
n
+
n−1
X
k=0
(1 − v
k+1
)
k
p
x
q
x+k
= d(
n
p
x
¨
a
n
+
n−1
X
k=0
1 − v
k+1
d
k
p
x
q
x+k
) =
d(¨
a
n
P (K ≥ n) +
n−1
X
k=0
¨
a
k+1
P (K = k)) = dE(¨
a
min(K+1,n)
) = d¨
a
x:n
= 1 − A
x:n
4-2. Use Y = (1 − Z)/d, with Z the net single premium of a corresponding whole life insurance. We have
Var(Z) = E(Z
2
) − (E(Z))
2
, with E(Z) = A
x
and (see lectures) E(Z
2
) = E[e
−2δ(K+1)
], which is the net
single premium of a similar whole life insurance, but calculated at twice the original force of interest (2δ).
Hence, Var(Y ) = d
−2
Var(Z) = d
−2
(E[e
−2δ(K+1)
] − A
2
x
), where A
x
= 1 − d¨
a
x
= 1 −
1
25
× 10 = 0.6. This
A
x
was evaluated with force of interest δ, and d =
i
1+i
=
1
25
and v =
1
1+i
=
24
25
= 0.96. So, evaluating such
a net single premium with force of interest 2δ leads to corresponding discount factor v
2
= (0.96)
2
, and
E[e
−2δ(K+1)
] = 1 − (1 − (0.96)
2
) × 6 = 0.5296. So Var(Y ) = (0.04)
−2
[0.5296 − (0.6)
2
] = 106.
The non-negative present value random variable Y has expected value 10, and standard deviation 10.3.
This seems to suggest a relatively high risk to the insurer, if we simply interpret the standard deviation
as representing this risk. However, we should be careful as standard deviation / variance does not clearly
report skewness of the probability distribution: they are no measure of skewness, but in this case we can
infer the following. As Y is a non-negative random quantity, with its standard deviation actually greater
than its mean value, the distribution will be skewed to the right, that is it will have a long tail to large
values of Y , which then occur with small probabilities. Hence, there will be considerable probability (more
than 0.5) that Y is less than 10. For the insurer, for this contract, low values of Y are ‘good’, as they relate
to relatively few annual payments to be made (hence correspond indeed to small values of K), so any real
risks to the insurer relate to (very) large values of Y - clearly situations where the insured survives a very
long time. This makes clear that variance is not always the most appropriate risk measure, as risks are
often logically restricted to a single tail of the distribution of Y . So, in such situations, it might be better
to measure risk to an insurer via, for example, the probability that Y exceeds a particularly large value.
4-3. ¨
a
x
=
∞
X
k=0
v
k
k
p
x
= 1+
∞
X
l=0
v
l+1
(
l+1
p
x
) = 1+vp
x
∞
X
l=0
v
l
l
p
x+1
= 1+vp
x
¨
a
x+1
, so p
x
=
¨
a
x
−1
v¨
a
x+1
, which leads to
p
73
=
7.73−1
(1.03)
−1
×7.43
= 0.93296. Similarly,
3
q
72
= 1−(p
72
p
73
p
74
) = 1−(0.94072×0.93296×0.92628) = 0.18704.
4-4. (a) Use ¨
a
x
=
N
x
D
x
and a
x
= ¨
a
x
− 1, leading to ¨
a
20
= 19.022, a
20
= 18.022, ¨
a
21
= 18.943, ¨
a
80
= 6.119,
a
80
= 5.119, ¨
a
81
= 5.845; the differences between these net single premiums for a young (20) and an old
(80) person are obviously related to future life expectancy, hence the number of annual payments to be
made under the contract. The difference between the nsp for such a person, and the nsp for someone
one year older, is larger for (80) than for (20), representing the fact that (80) has a substantially higher
probability to die within the next year than (20) has.
(b) Using ¨
a
x:n
=
N
x
−N
x+n
D
x
, we get ¨
a
20:30
= 15.870, ¨
a
20:31
= 16.086, ¨
a
21:30
= 15.856, ¨
a
80:10
= 5.517,
¨
a
80:11
= 5.683, ¨
a
81:10
= 5.346; the very small difference between ¨
a
20:30
and ¨
a
21:30
is based on the fact that
(21) is only very little less likely to survive 30 years than (20) is. Of course, these nsp’s also show the facts
mentioned in part (a), although the different lengths of the contracts are very influential as well.
(c) We use
n|
¨
a
x
=
N
x+n
D
x
, so
30|
¨
a
20
= 3.152,
30|
¨
a
21
= 3.086,
10|
¨
a
20
= 10.952,
10|
¨
a
80
= 0.603,
11|
¨
a
80
= 0.437;
of course, corresponding nsp’s in (b) and (c) sum up to nsp’s in (a). The effect of ageing is again clear on
these nsp’s, in particular actual payment taking place is far more likely for (20), even with such payments
deferred for 30 years, than for (80) with 10-year deferred payments. The large difference between the last
two nsp’s is caused by the large probability for (80) not to survive a further 11 years, which is even large
when compared to the probability for (80) not to survive a further 10 years.
4-5. We use ¨
a
x
=
1−A
x
d
and the relation A
x
=
δ
i
¯
A
x
, which is only exactly true for the De Moivre model
(for other models we can use it as an approximation, as discussed in the lectures). The De Moivre model
with maximum possible age w = 100 has as probability density function for the future lifetime of (x)
(of course with x < 100): g
x
(t) =
1
100−x
for t ∈ [0, 100 − x] (and zero else). So, ¯
A
x
=
R
∞
0
v
t
g
x
(t)dt =
1
100−x
R
100−x
0
v
t
dt =
1
100−x
h
v
t
ln v
i
100−x
t=0
=
1
100−x
v
(100−x)
− 1
. With d =
i
1+i
, this leads to ¨
a
20
= 15.856 and
¨
a
80
= 7.915. For (20), this net single premium is lower for this De Moivre model than when based on our
life tables (see Ex. 4-4), which is caused by the uniformity assumption for the moment of death over the
possible residual lifetime, which is particularly pessimistic for a young person. For (80), the De Moivre
model based nsp is greater than the corresponding nsp based on our life tables, which is a consequence of
the far greater probability for (80) to survive a given number of years under the De Moivre assumption
than when based on our life table, for example our life table has
10
p
80
=
l
90
l
80
= 0.270, whereas this 10-year
survival probability for (80) is equal to 0.5 for this De Moivre model.
4-6. This is a deferred annuity-due, as payments start on his birthday (be careful to check when precisely
the first payment is). The net single premium is
20|
¨
a
40
=
N
60
D
40
= 3.995 for unit payments, hence a net
single premium of £100, 000 provides annual payments of
100,000
20|
¨
a
40
= 25, 033.29. Clearly, such a contract
may appear to be relatively cheap, which in this case is a consequence of the interest compounded over 20
years before the first payment is taken from this fund (as (40) is pretty likely to survive till 60, and even
to have a reasonable number of annual payments thereafter).
4-7. Let Y be the present value random variable of this contract, then
Y =
a
K
=
1−v
K
i
for 0 ≤ K < n
a
n
=
1−v
n
i
for K ≥ n
So, trivially (since v =
1
1+i
), Y =
1−Z
1
−Z
2
i
. Writing this contract in this manner, as the sum of an n-year
pure endowment, with unit payment (with present value Z
2
), and an n-year term insurance, with payment
of (1 + i) at the end of the year of death (with present value Z
1
), leads immediately to
a
x:n
= E(Y ) =
1 − E(Z
1
) − E(Z
2
)
i
=
1 − (1 + i)A
1
x:n
− A
1
x:n
i
.
Using commutation columns, we have (see lecture notes) A
1
x:n
=
M
x
−M
x+n
D
x
and A
1
x:n
=
D
x+n
D
x
, which lead
to a
x:n
=
D
x
−D
x+n
−(1+i)[M
x
−M
x+n
]
iD
x
. Using our provided commutation colums, we get a
20:30
= 15.086,
a
20:10
= 7.677 and a
80:10
= 4.683. It is pretty likely that (20) will survive the terms of these contracts,
but for (80) it is far less likely that the maximum possible number (ten) of payments will indeed be paid,
hence the smaller nsp for the last contract than for the 10-year contract to (20).
4-8. Under this contract, unit payments are made at the end of the year if the annuitant is alive, so the
present value of this contract is Y = 0 if K = 0, 1, . . . , m, and Y =
P
K
l=m+1
v
l
if K ≥ m + 1. We can
calculate E(Y ), the net single premium of this contract, via
m|
a
x
=
m+1
¨
a
x
=
P
∞
l=m+1
v
l
l
p
x
=
N
x+m+1
D
x
.
This gives
30|
a
20
= 2.936,
30|
a
21
= 2.872,
10|
a
20
= 10.345,
11|
a
20
= 9.768,
10|
a
80
= 0.437 and
11|
a
80
= 0.308.
When compared to Ex. 4-4(c), we see that indeed
10|
a
80
=
11
¨
a
80
. It is clear that the nsp’s of 30-year
deferred contracts to young people are low due to the discounting, as it is likely that payments will take
place but they are far in the future, whereas the last two nsp’s are low as it is unlikely that payments will
actually take place. The difference between the nsp’s for the last two contracts is relatively large (also
when compared to the corresponding difference for such contracts to (20)), which is due to the pretty high
probability that (80) would die before age 91, if he survives to 90.
4-9. Use the following equalities (see lecture notes): d = 1 − v, 1 = d¨
a
x:n
+ A
x:n
and ¨
a
x:n
= 1 + a
x:n−1
,
leading to
v¨
a
x:n
− a
x:n−1
= v¨
a
x:n
− ¨
a
x:n
+ 1 = ¨
a
x:n
(v − 1) + d¨
a
x:n
+ A
x:n
= A
x:n
.
A
x:n
is the nsp of an n-year endowment, with unit payment at the end of year n, or at the end of the
year of death if this occurs earlier. The right-hand side of the given equality is based on expressing this
contract as the difference between two other contracts: from the perspective of a person, aged x, buying
this contract, +v¨
a
x:n
is the nsp of receiving v units at the start of year 1, . . . , n, if alive, and −a
x:n−1
is the
nsp of paying 1 unit at the end of year 1, . . . , n − 1 if alive. Hence, the interpretation is that, at the start of
year 1, this person gets v, which is worth 1 unit at the end of year 1. At that moment, however, he must
pay this 1 unit back. This is repeated for every year 2, . . . , n − 1, as long as this person survives. If he dies
before the end of year n, he does not pay back the unit at the end of the year of death. And, similarly, if
he survives year n, he also keeps that unit. Hence, this is clearly the same as the n-year endowment.
4-10. (a) Payment b
k
at the end of year k, for k = 0, 1, 2, . . ., if alive (so b
0
is paid now), has present value
v
k
b
k
. Hence, the present value for any year k payment, using the indicator function to distinguish between
actual payments (if alive) and payments 0 (if death), is v
k
b
k
I
k
. Hence, the present value random variable
is Y =
P
∞
k=0
v
k
b
k
I
k
, and as E(I
k
) =
k
p
x
, its corresponding net single premium is E(Y ) =
P
∞
k=0
v
k
b
k
(
k
p
x
).
(b) Clearly, for j ≤ k we have that I
j
I
k
= 1 if T (x) ≥ k, and I
j
I
k
= 0 else, so E(I
j
I
k
) =
k
p
x
, and
Cov(I
j
, I
k
) = E(I
j
I
k
) − E(I
j
)E(I
k
) =
k
p
x
− (
j
p
x
)(
k
p
x
) = (
k
p
x
)(
j
q
x
).
(c)
Var
∞
X
k=0
v
k
b
k
I
k
!
=
∞
X
k=0
Var(v
k
b
k
I
k
) + 2
∞
X
k=0
X
j<k
Cov(v
j
b
j
I
j
, v
k
b
k
I
k
) =
∞
X
k=0
v
2k
b
2
k
Var(I
k
) + 2
∞
X
k=0
X
j<k
v
j+k
b
j
b
k
Cov(I
j
, I
k
) =
∞
X
k=0
v
2k
b
2
k
(
k
p
x
)(
k
q
x
) + 2
∞
X
k=0
X
j<k
v
j+k
b
j
b
k
(
k
p
x
)(
j
q
x
),
where we used Var(X) = Cov(X, X).
4-11. For (x) in this fund, the expected value of the present value random variable, for a unit payment, is
simply ¨
a
x
=
N
x
D
x
, so ¨
a
65
= 10.594, ¨
a
75
= 7.562 and ¨
a
85
= 4.804. Hence, the total expected present value of
these 60 obligations is 10, 000 × [30 × ¨
a
65
+ 20 × ¨
a
75
+ 10 × ¨
a
85
] = 5, 171, 000.
4-12. This is the same contract, and payment arrangement, as considered as an example in the lectures,
so the net annual premium is
Π = C
A
1
40:10
¨
a
40:10
.
However, instead of basing it on our life tables, the future lifetime probabilities are now based on the De
Moivre model with w = 100. Hence, for (40) the probability to die in any of the possible 60 future years
is 1/60. This leads to
A
1
40:10
=
9
X
k=0
v
k+1
1
60
=
1
60
a
10
=
1
60
×
1 − v
10
i
= 0.1287,
where a
10
is the net single premium of a 10-year immediate annuity with unit payments. To calculate the
denominator of Π, we use ¨
a
40:10
= (1 − A
40:10
)/d, where we calculate the net single premium of the 10-year
endowment via A
40:10
= A
1
40:10
+ A
1
40:10
. For this pure endowment term, we have A
1
40:10
= v
10
10
p
40
=
(1.05)
−10 5
6
= 0.5116, so ¨
a
40:10
= 7.5537. This leads to net annual premium Π = 0.017038007 × C. For
example, for C = 100, 000, the net annual premium based on this De Moivre model and AER 5% is
£1,703.80. For the same setting, but the future lifetime inferences for (40) based on our provided life
tables (lectures), we had a net annual premium of only £364.90 per year. This large difference is due to
the unrealistically high probability of death of this person within the next 10 years, when based on the
De Moivre model: for the life table, we have
10
p
40
= l
50
/l
40
= 0.961, whereas this De Moivre model gives
10
p
40
= 0.833.
4-13. The results below are similarly derived as the example in the lectures for the term insurance (related
to Ex. 4-12). Of course, they relate to the results for the n-year term insurance via n → ∞.
(a) L = v
K+1
− P
x
¨
a
K+1
, for all K = 0, 1, 2, . . ..
(b) E(L) =
P
∞
k=0
[v
k+1
− P
x
¨
a
k+1
]
k
p
x
q
x+k
= A
x
− P
x
¨
a
x
, so the net premium requirement E(L) = 0 leads
to P
x
=
A
x
¨
a
x
.
(c) It is easy to see that all the premiums that are actually paid under this contract, together are the
difference between a perpetuity starting at time 0, and a perpetuity starting at time K + 1, so for the
corresponding net single premiums we have ¨
a
K+1
= ¨
a
∞
− v
K+1
¨
a
∞
=
1
d
− v
K+1 1
d
.
This gives L =
v
K+1
− P
x
1
d
− v
K+1 1
d
, so L = 1 +
P
x
d
v
K+1
−
P
x
d
.
(d) Part (c) leads immediately to Var(L) =
1 +
P
x
d
2
Var(v
K+1
). If this contract was paid by a net
single premium, then the one-off payment was of course A
x
, but any such constant premium would not
affect the variance of the loss (say ˜
L) to the insurer: let ˜
L = v
K+1
− A
x
, then Var( ˜
L) = Var(v
K+1
), so
Var(L) = 1 +
P
x
d
2
Var( ˜
L). So, if risk to the insurer is measured via the variance of the loss, the insurer
has greater risk with net annual premiums than for a net single premium. This is, of course, logical, as the
any random element in the total amount the insured actually ends up paying (due to possibly interrupted
payments in case of death) are an extra risk factor to the insurer. In particular, in case of early death, the
loss to the insurer can be large in case of annual premium payments.
4-14. L = c
K+1
v
K+1
−
P
K
k=0
Π
k
v
k
for K = 0, 1, 2, . . ., and E(L) = 0 gives
∞
X
k=0
c
k+1
v
k+1
k
p
x
q
x+k
=
∞
X
l=0
l
X
k=0
Π
k
v
k
l
p
x
q
x+l
.
By changing the order of summation (formally, there is a convergence issue that needs to be checked for
such an infinite sum, but we can assume this is no problem here as all terms in the sum decrease rapidly
to zero for increasing k and l), the right-hand side is equal to
∞
X
k=0
∞
X
l=k
Π
k
v
k
l
p
x
q
x+l
=
∞
X
k=0
Π
k
v
k
∞
X
l=k
l
p
x
q
x+l
!
=
∞
X
k=0
Π
k
v
k
k
p
x
.
Hence, the net annual premiums must satisfy
∞
X
k=0
c
k+1
v
k+1
k
p
x
q
x+k
=
∞
X
k=0
Π
k
v
k
k
p
x
.
4-15. The loss for policy A is
L
A
= 4v
K+1
− 0.18¨
a
K+1
= 4v
K+1
− 0.18
1 − v
K+1
d
=
−0.18
d
+
4 +
0.18
d
v
K+1
= −2.25 + 6.25v
K+1
,
for K = 0, 1, 2, . . .. So 3.25 = Var(L
A
) = (6.25)
2
Var(v
K+1
).
Similarly, L
B
= 6v
K+1
− 0.22
1−v
K+1
d
= 2.75 + 8.75v
K+1
, so
Var(L
B
) = (8.75)
2
Var(v
K+1
) = (8.75)
2
×
3.25
6.25
2
= 6.37.
4-16. The hint follows directly from the relation ¨
a
x
=
1−A
x
d
(lectures) and P
x
=
A
x
¨
a
x
(Ex. 4.13). The
present value of death benefits, if death occurs in year j, is 1.06
−j
× 1.06
j
× 1, 000 = 1, 000 for all years of
the policy. Let Π be the net annual premium for this contract, then the insurer’s loss random variable is
L = 1, 000 − Π¨
a
K+1
for K ≥ 0. So E(L) = 0 implies
Π =
1, 000
P
∞
k=0
¨
a
k+1 k
p
x
q
x+k
=
1, 000
¨
a
x
= 1, 000(d + P
x
) = 1, 000
0.06
1.06
+ 1, 000P
x
= 66.604.
4-17. (a) The random loss to the insurer is
L =
−Π¨
a
K+1
if K = 0, 1, . . . , m − 1
v
K+1
− Π¨
a
K+1
if K = m, m + 1, . . .
The expected value of this loss is
E(L)
=
m−1
X
k=0
(−Π¨
a
k+1
)
k
p
x
q
x+k
+
∞
X
k=m
(v
k+1
− Π¨
a
k+1
)
k
p
x
q
x+k
=
∞
X
k=m
v
k+1
k
p
x
q
x+k
− Π
∞
X
k=0
¨
a
k+1 k
p
x
q
x+k
=
m|
A
x
− Π¨
a
x
So E(L) = 0 leads to
Π =
m|
A
x
¨
a
x
(b) The two net single premiums on the right-hand side of the result for Π in part (a) were presented in
the lectures, so we immediately get Π =
M
x+m
/D
x
N
x
/D
x
=
M
x+m
N
x
, and based on our life tables and corresponding
commutation columns, we get the following net annual premiums
x
m
Π
20
10
0.004467
20
20
0.003990
70
10
0.028358
70
20
0.005495
For (20), these annual premiums are small due to the discounting effect, it is likely that a death benefit
payment will actually be made under these contracts, but still many years away. For (70), an actual
payment taking place is still pretty likely for the 10-year deferred life insurance, in which case the payment
would, however, not be very far away, but for the 20-year deferred life insurance payment becomes unlikely,
hence the far greater difference in these net annual premiums for (70) than for (20).
(c) We can express the loss to the insurer as L = L
1
+ L
2
, with
L
1
=
0
if K = 0, 1, . . . , m − 1
v
K+1
if K = m, m + 1, . . .
and L
2
=
−Π¨
a
K+1
if K = 0, 1, . . . , s − 1
−Π¨
a
s
if K = s, s + 1, . . .
So
E(L) =
∞
X
k=m
v
k+1
k
p
x
q
x+k
− Π
s−1
X
k=0
¨
a
k+1 k
p
x
q
x+k
+ ¨
a
s s
p
x
!
=
m|
A
x
− Π¨
a
x:s
So, setting E(L) = 0 gives
Π =
m|
A
x
¨
a
x:s
=
M
x+m
N
x
− N
x+s
.
This leads to the following net annual premiums (with m = 10 in all cases)
x
s
Π
20
5
0.01873
20
15
0.00785
70
5
0.07936
70
10
0.03119
Clearly, if one pays at most 5 premiums, these are a lot higher than if one pays up to 15 premiums, or until
death (part (b)). In particular, if at most 5 such annual premiums, then the whole premium is certainly
paid before a possible death benefit is paid for a 10-year deferred life insurance, hence this has small risk
for an insurer (no matter how risk is measured). These premiums are again far greater for (70) than for
(20), as the possible death benefit payment would likely occur far sooner than for (20).
4-18. (a) The random loss to the insurer is
L =
v
K+1
− P
x:n
¨
a
K+1
if K = 0, 1, . . . , n − 1
v
n
− P
x:n
¨
a
n
if K ≥ n
Net premiums must satisfy.
0 = E(L) =
n−1
X
k=0
[v
k+1
− P
x:n
¨
a
k+1
]
k
p
x
q
x+k
+ (v
n
− P
x:n
¨
a
n
)
n
p
x
With A
x:n
=
P
n−1
k=0
v
k+1
k
p
x
q
x+k
+ v
n
n
p
x
and ¨
a
x:n
=
P
n−1
k=0
¨
a
k+1 k
p
x
q
x+k
+ ¨
a
n n
p
x
, this leads to
P
x:n
=
A
x:n
¨
a
x:n
.
(b) Using the commutation functions for the endowment and life annuity due, as presented in the lectures,
we immediately get P
x:n
=
M
x
−M
x+n
+D
x+n
N
x
−N
x+n
.
(c) 50, 000 × P
40:25
= 50, 000 ×
M
40
−M
65
+D
65
N
40
−N
65
= 50, 000 × 0.02329543 = 1, 164.77.
(d) The general equation in Ex. 4-14, that must be satisfied by net premiums Π
k
paid at the start of year
k, for a contract paying c
k+1
at the end of the year k + 1 of death, is
∞
X
k=0
c
k+1
v
k+1
k
p
x
q
x+k
=
∞
X
k=0
Π
k
v
k
k
p
x
.
For an n-year endowment, with unit payment (here, the unit of payment is £50,000), set c
j
= 1 for
j = 1, . . . , n, and c
j
= 0 for j ≥ n + 1, and set
Π
j
=
P
x:n
for j = 0, 1, . . . , n − 1
−1
for j = n
0
for j ≥ n + 1
With these values, the above equality becomes
n−1
X
k=0
v
k+1
k
p
x
q
x+k
=
n−1
X
k=0
P
x:n
v
k
k
p
x
− v
n
n
p
x
,
leading to
P
x:n
=
P
n−1
k=0
v
k+1
k
p
x
q
x+k
+ v
n
n
p
x
P
n−1
k=0
v
k
k
p
x
.
This is, of course, again P
x:n
= A
x:n
/¨
a
x:n
, but with the net single premium for the n-year life annuity due
in the form of the second expression presented in the lectures.
4-19.
1
¨
a
x
= P
x
+ d:
We use the word ‘equivalent’ to mean equal in terms of the expected value of the present value of contracts
(so equal net single premiums). For the left-hand side, note that one unit now is equivalent to a life annuity
of ¨
a
−1
x
payable at the beginning of each year while (x) survives. For the right-hand side, note that one unit
now is also equivalent to interest-in-advance of d at the beginning of each year while (x) survives with the
repayment of the unit at the end of the year of (x)’s death; that is, 1 =
¨
a
x
¨
a
x
= d¨
a
x
+ A
x
. The repayment of
the unit at the end of the year of death is, in turn, equivalent to a life annuity-due of P
x
while (x) survives.
Therefore, the unit now is equivalent to P
x
+ d at the beginning of each year during the lifetime of (x).
Then ¨
a
−1
x
= P
x
+ d, for each side of the equality, represents the annual payment of a life annuity produced
by a unit available now.
P
x
=
dA
x
1−A
x
:
Consider an insured (x) who borrows the single benefit premium A
x
for the purchase of a single-premium
unit whole life insurance. The insured agrees to pay interest-in-advance in the amount of dA
x
on the loan
at the beginning of each year during survival and to repay the A
x
from the unit death benefit at the end
of the year of death. In essence, the insured is paying an annual benefit premium of dA
x
for an insurance
of amount 1 − A
x
. Then for a full unit of insurance, the annual benefit premium must be dA
x
/(1 − A
x
).
4-20.
(a) Remember that
k|
q
x
=
k
p
x
q
x+k
and P
x
= A
x
/¨
a
x
.
The net single premiums are A
x
=
P
∞
k=0
v
k+1
k|
q
x
=
P
∞
k=0
1.06
−(k+1)
c0.96
k+1
= 0.4 and ¨
a
x
=
1−A
x
d
=
0.6
0.06/1.06
= 10.6, so P
x
= 0.0377.
(b) A
x
=
P
∞
k=0
1
1+i
k+1
(1 − r)r
k
=
1−r
1+i
P
∞
k=0
r
1+i
k
=
1−r
1+i−r
, which gives ¨
a
x
=
1+i
1+i−r
, and together
these lead to P
x
=
1−r
1+i
, which agrees with part (a) for r = 0.96.
4-21. In case (40) survives to age 60, this contract provides him with an annuity which, now, is a 20-year
deferred immediate life annuity, which now has net single premium (so expected present value) in terms of
unit payments, expressed in terms of the net single premium of a life annuity-due,
20|
a
40
=
20|
¨
a
40
−
20
p
40
v
20
,
with
20|
¨
a
40
=
20
p
40
v
20
¨
a
60
, so in terms of commutation columns,
20|
a
40
=
N
61
D
40
. The random loss to the insurer
is
L =
100, 000v
K+1
− Π¨
a
K+1
if K = 0, 1, . . . , 19
10, 000
P
K−1
j=20
v
j+1
− Π¨
a
20
if K ≥ 20
and
E(L)
=
100, 000
19
X
k=0
v
k+1
k
p
x
q
x+k
− Π
19
X
k=0
¨
a
k+1 k
p
x
q
x+k
+ 10, 000
∞
X
k=20
k−1
X
j=20
v
j+1
k
p
x
q
x+k
− Π¨
a
20 20
p
x
=
100, 000A
1
40:20
− Π¨
a
40:20
+ 10, 000
20|
a
40
.
Setting E(L) = 0 gives
Π =
100, 000A
1
40:20
+ 10, 000
20|
a
40
¨
a
40:20
,
which can be calculated via commutation columns using
A
1
40:20
=
M
40
− M
60
D
40
,
20|
a
40
=
N
61
D
40
, ¨
a
40:20
=
N
40
− N
60
D
40
,
giving
Π =
100, 000(M
40
− M
60
) + 10, 000N
61
N
40
− N
60
= 3, 427.72.
4-22. (a) The annual loan payment P , for a loan of 1 unit, is
P =
1
a
4
=
4
X
j=1
1
1.06
j
−1
= 0.2885914.
For the life insurance considered, if death benefit is paid at the end of the year of death (K + 1), so within
the 4-year period of this contract, then this payment is equal to P ¨
a
4−K
if K < 4, and 0 else, so this death
benefit payment has present value
Z =
v
K+1
P ¨
a
4−K
if K = 0, 1, 2, 3
0
if K ≥ 4
=
P v
K+1
(1 − v
4−K
)/d
if K = 0, 1, 2, 3
0
if K ≥ 4
so the insurer’s random loss is
L = Z − G =
P (v
K+1
− v
5
)/d − G
if K = 0, 1, 2, 3
−G
if K ≥ 4
(b) The net premium requirement gives 0 = E(L) = E(Z) − G = P (A
1
25:4
− v
5
4
q
25
)/d − G, with
A
1
25:4
= A
25:4
− v
4
4
p
25
= 1 − d¨
a
25:4
− v
4
4
p
25
= 1 −
0.06
1.06
× 3.667
− 1.06
−4
0.995 = 0.00430076.
This leads to G = 0.00287791. For example, if the original loan was £10,000, to be repaid by 4 annual
payments and protected by such a 4-year life insurance, then the net single premium for this term insurance
is £28.78 for this person.
(c) We must calculate the net single premium of this term insurance when it is paid for in 4 annual
payments, precisely as the loan is paid for. (So you should not just use G from part (b)!)
Let G
∗
be the additional amount of borrowing to pay for the term insurance (so the net single premium
of this insurance, under this contract combining the loan of £10,000 and the loan to pay this net single
premium), and let L
∗
denote the corresponding random loss to the insurer (‘loan provider’). Note that
the loan repayments are at the end of each year of this contract, so differs from our usual assumption that
(premium) payments are at the start of the years. Then
L
∗
=
(
10,000+G
∗
a
4
(v
K+1
− v
5
)/d − G
∗
if K = 0, 1, 2, 3
−G
∗
if K ≥ 4
The net premium requirement gives
E(L
∗
) =
10, 000 + G
∗
a
4
d
(A
1
25:4
− v
5
4
q
25
) − G
∗
= 0,
and using results in parts (a) and (b) gives
(10, 000 + G
∗
) × 0.00287791 − G
∗
= 0
so
G
∗
=
28.7791
1 − 0.00287791
= 28.862.
Hence, the annual total payment for this loan, with insurance cover, is
10, 000 + 28.862
a
4
= 2, 894.24.
The price G
∗
is a little bit different to the required one-off price G of this insurance as calculated in part
(b), which reflects the risk to the insurer involved with the possibility that the death benefit may need to
be paid now before all payments have been received. Of course, the difference for such a small insurance
between G
∗
and G is almost neglectable, but it is important to see the steps involved in these computations,
and if the probability of the insured dying before all loan repayments have been made were not so small,
then of course this would cause larger differences between the expected present values of one-off and sum
of annual prices to pay for such cover insurance.
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