Topics in Mathematics II - Actuarial Mathematics

Solutions to Exercises on Hand-Out 3

Frank Coolen (CM206 - [e-mail]), February 2008

3-1. This is De Moivre’s lifetime model, so T (30) ∼ U [0, 70], so ¯

A

30

=

1

70

R

70

0

v

t

dt =

1
7

(1 − e

−7

) =

0.1427, using v = e

−δ

, so the net single premium of this contract with death benefits of £50,000 is

£7,136.34. The approximation A

x

≈ (1+i)

−1/2

¯

A

x

, with i = e

δ

−1, gives £50, 000×A

x

≈ £6, 788.30,

the approximation A

x

≈

δ

i

¯

A

x

gives £50, 000×A

x

≈ £6, 785.47. However, following the justifications

of these approximations given in the lectures, it will be clear that for this model, this second
approximation is actually exact, as T (x) − K(x) ∼ U [0, 1) for the De Moivre model. Hence, the
first approximation is pretty good as well.

3-2. ¯

A

x

=

R

100

0

e

−0.1t

t

5000

dt =

1

50

(1 − 11e

−10

) = 0.01999 (via integration by parts).

3-3. Let AP be the Actual Payment in 10 years time, then P (AP = 1, 000) =

10

p

60

and P (AP =

500) = 1 −

10

p

60

, where

10

p

60

=

l

70

l

60

= 0.80803.

This gives E(AP ) = 904.01.

Let Z be the

present value of this payment, then clearly Z = (1.06)

−10

AP , so P (Z = 558.39) =

10

p

60

and

P (Z = 279.20) = 1 −

10

p

60

, with E(Z) = 504.79 (of course, E(Z) is just the discounted present

value of E(AP )).

3-4. (a) A

20

= 0.09419, A

50

= 0.30304, A

80

= 0.70860.

(b) A

1

20:10

= 0.00921, A

1

20:30

= 0.02893, A

1

20:60

= 0.07875; of course, these values increase towards

A

20

for n → ∞. For A

1

20:10

= 0.00921, the probability of a payment is pretty low (see lifetables),

but if such a death benefit is to be paid, its present value is relatively high due to discounting over
only a few years.
(c) A

1

40:10

= 0.029179, A

1

60:10

= 0.14427, A

1

80:10

= 0.57129; compare this latter one with A

80

(part

(a)), clearly relatively large due to the small probability that this person survives age 90, and the
discounting effect.
(d)

10|

A

20

= 0.084972,

30|

A

20

= 0.065256,

10|

A

40

= 0.17881,

10|

A

80

= 0.13731; the first of these is

small because payment is likely, but probably a long time away (so much effect of discounting), the
latter is also small, but this is because payment is not very likely. Of course, these calculations
illustrate that A

x

= A

1

x:n

+

n|

A

x

.

3-5. Let the net single premium be denoted by Π. The present value random variable Z of this
whole life insurance contract is Z = v

K+1

[10, 000 + Π] if K = 0, 1, . . . , 19 and Z = v

K+1

20, 000 if

K ≥ 20. This leads to

Π = E(Z) = [10, 000 + Π]

19

X

k=0

v

k+1

k

p

x

q

x+k

+ 20, 000

∞

X

k=20

v

k+1

k

p

x

q

x+k

= [10, 000 + Π]

∞

X

k=0

v

k+1

k

p

x

q

x+k

+ [10, 000 − Π]

∞

X

k=20

v

k+1

k

p

x

q

x+k

= [10, 000 + Π]

M

x

D

x

+ [10, 000 − Π]

M

x+20

D

x

,

and solving this gives

Π = 10, 000



M

x

+ M

x+20

D

x

− M

x

+ M

x+20



.

(Note: this exercise shows how such whole life insurance contracts, with variable payments, can
often be written as the sum of contracts presented in the lectures, where the deferred life insurances
(the second term above) play a useful role.)

3-6. ¯

A

x

=

R

80

0

e

−δt 1

80

dt =

1−e

−80δ

80δ

.

3-7.

m|

¯

A

x

=

R

∞

m

v

t

g

x

(t)dt =

R

∞

m

e

−δt

µe

−µt

dt =

µ

δ+µ

e

−(δ+µ)m

, of course again independent of x as this

is the exponential model for the future lifetime. For µ = 0.04 and δ = 0.1, this gives

1|

¯

A

x

= 0.2484,

5|

¯

A

x

= 0.1419 and

10|

¯

A

x

= 0.0705. According to this lifetime model, with E(T (x)) = 25 and

10

p

x

= 0.6703, there is a relatively large probability to die early, when compared to probabilities

inferred from life tables, for younger people. This shows here in the pretty large differences between
these three net single premiums.

3-8. ¯

A

1

40:25

=

R

25

0

v

t 1

60

dt =

1
3

(1 − e

−1.25

) = 0.2378.

3-9. ¯

A

1

30:10

=

R

10

0

v

t 1

70

dt =

1

70 ln v

(v

10

− 1) = 0.0921, as v =

1

1.1

. When compared to the net single

premium in Ex. 3-8, where the same model is used but for an older person, and a longer term
contract, the net single premium ¯

A

1

30:10

is smaller both due to the fact that this younger person

has a lower force of mortality, and the term of the contract is substantially shorter. Of course, this
model is also not very realistic as a description for human lifetime, in particular for younger people
probabilities to die in the next year are normally too high in De Moivre’s model with a reasonable
maximum age parameter w.

3-10. Π = ΠA

1

x:20

+ 10, 000v

20

20

p

x

= Π



M

x

−M

x+20

D

x



+ 10, 000

D

x+20

D

x

, which leads to

Π = 10, 000



D

x+20

D

x

+M

x+20

−M

x



.

3-11. The present value random variable for this contract, Z, takes on value 0 for K = 0, 1, . . . , 19,
and value 1.05

−20

10, 000 = 3, 768.89 for K ≥ 20, with probabilities P (Z = 3, 768.89) =

l

70

l

50

= 0.739

and P (Z = 0) = 0.261. Hence, the net single premium of this contract is 10, 000A

1

50:20

= E(Z) =

2, 785.82, which can also be derived using A

1

50:20

=

D

70

D

50

= 0.27858.

3-12. (a) A

1

20:10

= 0.6065, A

1

20:60

= 0.02179, A

1

50:10

= 0.5616, A

1

80:10

= 0.1660; these reflect the

very high probabilities that such payment actually takes place, in case of 10-year pure endowments,
for (20) and also for (50), but this probability is pretty small for (80). Also, for (20) it is far less
likely that the payment would take place for the 60-year pure endowment, and in case this payment
takes place, it is far more discounted of course. (Note: to appreciate these values, it is useful to
consider that 1.05

−10

= 0.6139 and 1.05

−60

= 0.0535.)

(b) A

20:10

= 0.6157, A

20:60

= 0.1005, A

50:10

= 0.6254, A

80:10

= 0.7373; for (20) and (50), the

payment is very likely to be at the end of the period for the 10-year endowment, whereas for (80)
it is far more likely to be a death benefit, hence paid before the end of this period, so less effect of
discounting. For the 60-year endowment for (20), the nsp is far smaller due to discounting, as the
payment is likely to be many years away.
(c) A

25:15

= 0.4857, A

25:30

= 0.2483, A

25:45

= 0.1481, A

25:60

= 0.1178; these show that payment

is expected to take place later with increasing n, so the differences are due to discounting over
longer periods of time (for larger n). Of course, the limitting value, for n → ∞, is A

25

= 0.1148, so

A

25:60

is already very close to this limitting value, which is logical as any payments which would be

different for the 60-year endowment and the whole life insurance, for (25), would take place after
he has reached age 85, so with pretty small probability, and affected by large discounting.

3-13. We have A

x:n

= u, A

1
x:n

= y and A

x+n

= z, and we know the relations (check lecture notes

and/or confirm) A

x

= A

1

x:n

+ v

n

n

p

x

A

x+n

and A

x:n

= A

1
x:n

+ v

n

n

p

x

. This straightforwardly leads to

A

x

= (1 − z)y + uz.

3-14. (a) ¯

A

50:10

=

R

10

0

v

t 1

50

dt + v

10

10

p

50

=

1

50

h

v

t

ln v

i

10

0

+ v

10

(1 −

10
50

) =

2
5

(1 + e

−0.5

) = 0.6426, using

v = e

−δ

. So the net single premium for this contract, with payment £50,000, is £32,130.61.

(b) (Note: payment of death benefit here at the end of the year of death.) A

50:10

= 0.6254 (see Ex.

3-12 (b)), so the net single premium is £31,271.71.
(c) The force of interest in (b) is δ = ln 1.05 = 0.04879, which would lead to slightly higher net
single premium than δ = 0.05 as used in (a). The payment of death benefit at the end of the year of
death, in (b), would lead to a slightly lower net single premium than for payment at the moment of
death, as in (a). The main difference here, however, is caused of course by the use of life tables, in
(b), instead of the De Moivre model with w = 100 in (a), as this latter model tends to give higher
probabilities to die ‘young’ (so in the next 10 years here) than life tables.

3-15. ¯

A

x:n

=

i

δ

A

1
x:n

+ A

1

x:n

=

i

δ

h

M

x

−M

x+n

D

x

i

+

D

x+n

D

x

, which gives, for x = 35 and n = 30, and using

our commutation columns, ¯

A

35:30

= 0.266235, so the net single premium of this contract, with

payment of £10,000, is £2,662.35.

3-16. Let Z

A

= Z

A1

+ Z

A2

, with Z

A1

the present value of an n-year term insurance issued to A,

and Z

A2

the present value of an n-year pure endowment issued to A. Because of the independence

of the lives of A, B and C, and their identical ages, we have (assuming also that we can use the
same future lifetime models or life tables for them, which is reasonable from the perspective of the
insurer, if he does not have further information about these people) that Var(Z

A1

) = Var(Z

B

) and

Var(Z

A2

) = Var(Z

C

). Clearly, only one of the payments for the contracts related to Z

A1

and Z

A2

will actually take place, hence Z

A1

× Z

A2

= 0 with probability 1, and therefore E(Z

A1

× Z

A2

) = 0.

This gives Cov(Z

A1

, Z

A2

) = E(Z

A1

× Z

A2

) − E(Z

A1

)E(Z

A2

) < 0, and hence Var(Z

A

) = Var(Z

A1

) +

Var(Z

A2

) + 2Cov(Z

A1

, Z

A2

) < Var(Z

B

) + Var(Z

C

).

3-17. Let Z

i

be the present value of this contract to person i, for i = 1, . . . , 100. Then, for

the exponential lifetime model, E(Z

i

) =

µ

µ+δ

= 0.4 and E(Z

2

i

) =

µ

µ+2δ

0.25 (see lecture notes),

so Var(Z

i

) = 0.25 − 0.16 = 0.09 and SD(Z

i

) = 0.3. As these lives are independent, we have,

for Z =

P

100
i=1

Z

i

, E(Z) = 40 and Var(Z) = 9, so by the Central Limit Theorem, we have that

Z is approximately Normally distributed with these expected value and variance. Let h be the
minimum initial amount required in this fund to ensure sufficient funds for all 100 death benefit
payments, assuming 1-unit payments, with approximate probability 0.95. Then we want, approx-
imately, P (Z ≤ h) = 0.95, so

h−40

3

= 1.645, giving h = 44.935. As the payments are £100,000,

a good approximation for the required initial amount, to meet this criterion, is £4,493,500. The
difference between this amount, and the expected total present value of all these 100 payments,
which is £4,000,000, is called the ‘risk loading’ (for this criterion). In this exercise, this risk loading
is 12.34% of the net single premiums, which may perhaps be surprisingly large. Of course, in this
example, there is more than one would expect in real life, due to the exponential distribution being
used for the future lifetimes, giving high probabilities that people die young. Also, because of this
distribution, it is actually not relevant what the current ages are of the people involved (this is only
true for the exponential distribution).

3-18. At the inception of the plan, each of these 100 people pay 1, 000A

30

= 1, 000

M

30

D

30

= 140.1077,

giving a total fund of £14,010.77. Let us first calculate the expected size of the fund after 5 years
as determined at the time of the inception of the plan. In general terms, the contribution to the net
single premium A

x

, in case of such a fund for one person and with unit payments, of death benefits

paid after year r of the contract, is

∞

X

k=r

v

k+1

k

p

x

q

x+k

=

∞

X

l=0

v

r+l+1

(

r

p

x

)(

l

p

x+r

)q

x+r+l

= v

r

r

p

x

A

x+r

.

So, the expected size of the fund after r years is this contribution to the net single premium, with the
appropriate interest added, so (1 + i)

r

v

r

r

p

x

A

x+r

=

r

p

x

A

x+r

. In this problem, the expected total size

of the fund after 5 years, as expected at the inception of the plan, is therefore 1, 000×100×

5

p

30

A

35

=

100, 000

l

35

M

35

l

30

D

35

= 16, 947.00.

Let us now consider the actual development of this fund. Let F

k

be the balance at the end of year

k. Then

F

1

= 1.05 × 14, 010.77 = 14, 711.31

F

2

= 1.055F

1

− 1, 000 = 14, 520.43

F

3

= 1.055F

2

= 15, 319.05

F

4

= 1.06F

3

= 16.238.20

F

5

= 1.06F

4

− 1, 000 = 16, 212.49

So, the fund actually holds £734.51 less at the end of year 5 as was expected at the start of the fund.
This is due to two effects, a positive effect on the balance of the fund as the interest has been higher
than expected (or at least than the assumed AER used to calculate each person’s contribution to
the fund), and a negative effect on the fund’s balance due to more deaths in these first 5 years than
expected: as

5

p

30

= 0.9915, the probability of one or more deaths amongst 100 individuals in this

5-year period was 0.5742 (Binomial(100,0.9915) distribution) and the expected number of deaths in
this period was 100 × 0.0085 = 0.85, so 2 deaths is indeed more than expected.

3-19. Let m < n, then

A

x:n

=

n−1

X

k=0

v

k+1

k

p

x

q

x+k

+ v

n

n

p

x

=

m−1

X

k=0

v

k+1

k

p

x

q

x+k

+

n−1

X

k=m

v

k+1

k

p

x

q

x+k

+ v

n

n

p

x

= A

1
x:m

+ v

m

m

p

x

"

n−m−1

X

l=0

v

l+1

l

p

x+m

q

x+m+l

+ v

n−m

n+m

p

x+m

#

= A

1
x:m

+ v

m

m

p

x

A

x+m:n−m

Clearly, this shows that an n-year endowment for (x) is identical to an m-year term insurance,
paying death benefit if death occurs during the first m < n years of the contract, followed by an
(n − m)-year endowment starting when this person has reached age (x + m), under the condition
that this person will indeed survive till this age, and with the present value of the latter contract
discounted by a factor v

m

of course.