Actuarial Mathematics II
Solutions to Exercises on Hand-Out 1
Frank Coolen (CM206 - Frank.Coolen@durham.ac.uk), January 2008
1-1. Let F
0
= C be the initial capital invested in the savings account(s). Under option (1), the balance at the
end of n years is F
n,1
= (1 + i)
n
C; under option (2), the total balance in both accounts at the end of n years is
F
n,2
=
c
2
[(1 + i + )
n
+ (1 + i − )
n
]. We first prove the hint: using (a + b)
n
=
P
n
j=0
n
j
a
j
b
n−j
,
(1 + α)
n
+ (1 − α)
n
=
n
X
j=0
n
j
1
j
α
n−j
+
n
X
j=0
n
j
1
j
(−α)
n−j
= 2 +
n−1
X
j=0
n
j
α
n−j
[1 + (−1)
n−j
].
The sum term is 0 for n = 1, and positive for all n ≥ 2, so the hint is true. Now consider the original problem.
Clearly, F
n,2
≥ F
n,1
if and only if (1 + i + )
n
+ (1 + i − )
n
≥ 2(1 + i)
n
, which can be rewritten as
(1 +
1 + i
)
n
+ (1 −
1 + i
)
n
≥ 2,
and this is true by using the hint with α =
1+i
. Hence, option (2) is at least as good as option (1), and it is actually
better for all n ≥ 2.
1-2. (a) F
3
= 5, 000(1 + 0.04)
3
= 5, 624.32.
(b) We need the minimal integer n for which (1.04)
n
≥ 2, so for which n ≥
ln 2
ln 1.04
= 17.67, so n = 18 years.
(c) We would need i such that (1 + i)
y
≥ 2, hence such that i ≥ 2
1/y
− 1. For example, for y = 10 we would need
i ≥ 0.0718, and for y = 12 we would need i ≥ 0.0595. (Note: if y does not have to be an integer number of years,
then i may have to be a bit larger as the payment of the final year will not be made; for simplicity let us just assume
that y is an integer.)
(d) Let F
(m)
3
be the balance in the account at the end of 2007, with 4% nominal interest rate, and with interest
compounded m times per year, then F
(m)
3
= 5, 000 1 +
0.04
m
3m
, so F
(2)
3
= 5, 630.81, F
(12)
3
= 5, 636.36, F
(52)
3
=
5, 637.22 and F
(365)
3
= 5, 637.44. With continuous compounding, this balance becomes F
(∞)
3
= 5, 000e
0.12
= 5, 637.48.
(e) From part (c), we know that we need i = 2
1/y
− 1 (with regard to the comment made in part (c), of course this
little problem if y is not an integer does not occur with continuous compounding). For the force of interest we have
δ = ln(1 + i), so we need δ = ln(2
1/y
) =
0.693
y
. For example, for y = 10 we would need δ = 0.0693, and for y = 12 we
would need δ = 0.0578 (or δ greater then these values, of course).
1-3. (a) For i = 0.03, we get F
5
= (1.03)
5
1, 000 +
P
4
k=1
(1.03)
5−k
500 = 3, 313.84. For i = 0.05 this balance is
3, 539.10, and for i = 0.06 it is 3, 656.77. We see that the total interest earned over this period with i = 0.06, is more
than twice the total interest earned if i = 0.03.
(b)F
5
= (1.03)
2
(1.05)
3
1, 000 + [(1.03)(1.05)
3
+ 1.05
3
+ 1.05
2
+ 1.05]500 = 3, 479.36. Of course, this is between the
values corresponding to i = 0.03 and i = 0.05 for all years, as calculated in part (a).
(c) The end balance in this case is 3, 451.25.
1-4. 15, 000 = (1 + i)
8
2, 000 +
P
8
k=1
(1 + i)
8−k
c leads directly to
c =
15, 000 − (1 + i)
8
2, 000
P
8
k=1
(1 + i)
8−k
.
For i = 0.04 this gives c = 1, 330.86, and for i = 0.08 this gives c = 1, 062.19.
1-5. In this problem, i
(m)
= 0.04 and m = 12. Be careful here that the payments are made at the start of each
month, so e.g. there will still be one month of interest paid on the 12th (final) payment. Hence, the total sum at the
end of the year is
P
11
k=0
(1 +
0.04
12
)
12−k
200 = 2, 452.64. (Check that, if payments were made at the end of the month,
the sum would go from k = 1 to 12.)
1-6. (a) The present value of this immediate annuity with unit payments is a
10|
=
P
10
j=1
v
j
=
1−v
10
i
, where v =
1
1+i
.
Hence, for i = 0.02 this is 8.9826, for i = 0.04 we get 8.1109, and for i = 0.06 we get 7.3601. To get the present value
if the payments are £5,000 (which can be considered the unit of the payments), we just multiply these values with
5,000, giving £44,913, 40,555 and 36,801, respectively, for these AER’s.
(b) Clearly the answers here are just twice the present values in (a) (with some rounding): 89,826, 81,109 and 73,601,
respectively.
(c) We can use ¨
a
10
=
1−v
10
d
, with d = 1 + i, or, easier, the relation that follows immediately: ¨
a
10
= (1 + i)a
10
and
use the answers from (b). This leads to present values, for i = 0.02, 0.04 and 0.06: 91,623, 84,353, and 78,017.
(d) The present value is ¨
a
∞
=
1
d
=
1+i
i
for unit payments, hence for annual payments of £5,000 we get 255,000,
130,000 and 88,333 for these AERs, respectively.
(e) Simply twice the values in (d), so 510,000, 260,000 and 176,667.
(f ) a
∞
=
1
i
, so 500,000, 250,000 and 166,667. Of course, the difference between the present values in parts (e) and
(f) is just a single payment at ‘time 0’.
1-7. (a) An immediate annuity with 10 annual unit payments has present value a
10
=
1−v
10
i
(see 1-6(a)). So, with
present value £100,000, such an annuity gives 10 annual payments of £100,000/a
10
. So, for these AERs the annual
payments are 11,133, 12,329 and 13,587, respectively.
(b) As (a), a
20
=
1−v
20
i
leads to 6,116, 7,358 and 8,718.
(c) You can either do this ‘directly’, or using part (a) and the relation ¨
a
10
= (1 + i)a
10
, so the annual payments
here are those from part (a) multiplied by
1
1+i
: 10,915, 11.855 and 12,818, respectively.
(d) ¨
a
∞
=
1+i
i
, so the annual payments are £100,000/¨
a
∞
, which for these AERs are 1,961, 3,846 and 5,660 re-
spectively. Clearly, in comparison to corresponding annuities, these annual payments are much smaller but increase
relatively more as function of i, as the payments are based on the interest (see part (e)), whereas for annuities the
payments also include a substantial (growing) part of the invested sum.
(e) a
∞
=
1
i
, so the annual payments are £100,000/a
∞
, giving 2,000, 4,000 and 6,000 for these AERs. These annual
payments are exactly the annual interest on the invested sum. The main difference with (d) is due to the first
payment for the perpetuity-due, which is taken straight from the invested sum (as there is no interest yet at time 0).
1-8. This is easiest via the relation: final value = present value ×(1 + i)
n
. So, for the three AERs, we get:
(a) 54,749, 60,031 and 65,905. (b) 109,497, 120,061 and 131,808. (c) 111,688, 124,863 and 139,717.
1-9. (a) For the original annuity, a
10
=
1−
(
1
1.03
)
10
0.03
= 8.5302, so the annual payments (starting at end of year 1)
are £50,000/a
10
= 5, 861.53. Calculating the payments, and remaining sum in the annuity, at the end of each year
(do this for years 1 to 5), leads to a remaining amount, immediately after the 5th payment, of 26,844.07. But Stan
also invests the 5th payment again, which was 5, 861.53. Hence, with the penalty taken into account, Stan invests
31,705.60 in a new annuity, for 5 years. This gives 5 annual payments of 31,705.60/a
5
= 7, 121.97, where we used
a
5
=
1−
(
1
1.04
)
5
0.04
= 4.4518. This may therefore seem a good move. However, Stan has reinvested the 5th payment,
hence the direct comparison with the annual payments under the first contract is not completely fair. If Stan had
not invested the 5th payment again, then he would have invested 25,844.07 in the new annuity, for 5 years, giving 5
annual payments of 5,805.28, hence not such a good move.
(b) Following the same steps as in (a), the annual payments for the new scheme for years 6-10 are greater than those
under the original scheme, if the 5th payment is again reinvested, if
32,705.60−C
4.4518
> 5, 861.53, so if C < 6, 6.111.24.
If Stan did not reinvest the 5th payment, the reinvestment would be beneficial if
26,844.07−C
4.4518
> 5, 861.53, so if
C < 749, 58.
1-10. (a) This is an immediate perpetuity, which varies a little from those presented in the lectures as the annual
payment is not constant. However, it is easy to see that the present value of the payment at the end of year j ≥ 1 is
[v(1 + k)]
j
, hence the present value of this perpetuity is
∞
X
j=1
[v(1 + k)]
j
=
∞
X
j=1
1 + k
1 + i
j
=
1 + k
i − k
=
1 + k
0.04 − k
.
For this present value to be equal to 51, we need k = 0.02.
(b) If k ≥ AER, the present value of the perpetuity would not be finite. To get such increasing annual payments,
with a fixed AER, the annual interest cannot be entirely used for the annual payment, as the base sum must increase
as well to allow larger payments in later years.
1-11. (a) This exercise is a little more challenging than most exercises we have seen, but the principles are the
same (it generalizes the idea in 1-10). The year 1 grant is 15,000, which must be paid now (so this is also its present
value). At the start of year 2, the grant is 15, 000 × 1.03, which has present value 15, 000 × 1.03/1.08 due to the
investment of this sum by those clever people. Generally, for year j ≥ 1, the grant paid at the start of the year
will be 15, 000 × (1.03)
j−1
, which has present value 15, 000 × (1.03/1.08)
j−1
. Hence, the total present value of this
perpetuity, covering the Dagobert-studentship for ever (assuming constant inflation and investment returns) is
15, 000 ×
∞
X
j=1
1.03
1.08
j−1
= 15, 000 ×
1
1 −
1.03
1.08
= 324, 000.
(b) V
1
= 324, 000 and g
1
= 15, 000. For j ≥ 2, the recursive formulae are V
j
= 1.08×[V
j−1
−g
j−1
] and g
j
= 1.03×g
j−1
,
which of course gives g
j
= 15, 000 × (1.03)
j−1
as used in part (a), and hence V
j
= 1.08 × [V
j−1
− 15, 000(1.03)
j−2
].
For the first 10 years, these values are given in the table below.
j
g
j
V
j
1
15,000
324,000
2
15,450
333,720
3
15,914
343,732
4
16,391
354,044
5
16,883
364,665
6
17,389
375,605
7
17,911
386,873
8
18,448
398,479
9
19,002
410,434
10
19,572
422,747
1-12. (a) If the debt is 1 unit, then the required repayment in case you opt for 10 equal payments at the end of
years 1 to 10, with AER i, is
r =
i
1 −
1
1+i
10
.
For i = 0.05, we get r = 0.129505, so with the unit being £10,000, your annual payment is £1,295.05. And, for
i = 0.10, we get r = 0.162745, leading to annual payments of £1,627.45. With S
k
denoting the outstanding debt
(‘principal’) at the start of year k ≥ 0, the annual payments consist of interest, namely i × S
k−1
for k ≥ 1, and
reduction of principal, as given in the table below (with some rounding effects).
i = 0.05
i = 0.10
k
0.05 × S
k−1
S
k
0.1 × S
k−1
S
k
0
-
10,000
-
10,000
1
500.00
9,204.95
1,000.00
9,372.55
2
460.25
8,370.15
937.26
8,682.36
3
418.51
7,493.61
868.24
7,923.15
4
374.68
6,573.24
792.32
7,088.02
5
328.66
5,606.85
708.80
6,169.37
6
280.34
4,592.14
616.94
5,158.86
7
229.61
3,526.70
515.89
4,047.30
8
176.34
2,407.99
404.73
2,824.58
9
120.40
1,233.34
282.46
1,479.59
10
61.67
0
147.96
0
(b) For AER of 10%, the interest per year is of course £1,000, so exactly the annual payment. Hence, the final
payment required at the end of year 10 is £11,000, consisting of the principal, which has not been reduced at all,
and the interest for the final year. For AER of 5%, the detailed breakdown up to the end of year 9 is given in the
table below.
k
0.05 × S
k−1
S
k
0
-
10,000
1
500.00
9,500.00
2
475.00
8,975.00
3
448.75
8,423.75
4
421.19
7,844.94
5
392.25
7,237.19
6
361.86
6,599.05
7
329.95
5,929.00
8
296.45
5,225.45
9
261.27
4,486.72
Hence, at the end of year 10, a final payment is required of 1.05 × 4, 486.72 = 4, 711.06.
(c) If you take this opportunity to switch, then at the end of year 5 (so at the start of year 6), your total outstanding
debt is 6, 169.37 + C (see (a)), which must then be repaid in 5 years with AER 0.08. This requires 5 annual payments
of
"
0.08
1 −
1
1.08
5
#
× (6, 169.37 + C) = 1, 545.16 + 0.25046C,
hence you are better off if this annual payment is less than 1627.45, so if C < 328.56.
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