Actuarial Notation: Annuities

8/11/2019 Actuarial Notation: Annuities

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In order to prove these results note that

a n|i = v +v2 + +vn =

v(1 vn)

1 v =

1 vn

i

and

a n|i= 1 +v+v2

+ +vn1

=1 vn

1 v =1 vn

dFurthermore

a n|i =

n0

etdt=1 en

=

1 vn

The accumulation in value based formulae can be derived by simply multiplying the above

expressions by a factor (1 +i)n.

Deferred level annuity based notation:

m|a n|= vm

a n|i

0 m

1

m+ 1

1

1

m+n=P V

Worked Examples

Example 1:

Calculate the present value as at 1 Jan 1990 of a series of payments of $1,000 that are payableon 31 Dec of each year starting from 31 Dec 1990 to 31 Dec 1995 inclusive. Assume a com-

pound rate of interest charge of 10% effective per annum.

Solution:

(i) Chosen time period unit: 1 year.

(ii) Effective rate of interest for that choice of time period unit: i= 0.10 v = (1.10)1.

(iii) Total number of payment periods: n= 6.

(iv) Total payment rate per period: R= 1000

(v) Focal date: 1 Jan 1990

Because these payments are being made in arrears, we have the result that

P V= 1000 a 6|0.10 = 1000

1 (1.10)6

0.10

= 4355.26

years

0 131/12/90

231/12/95

6

6@R1000

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Example 2:

An investor deposits R2000 and then withdraws a fixed amount Xat the end of every year

starting one year after the deposit has been made. If, immediately after the 11th annual draw-

ing, the investor has R400 left in the account, what fixed amount Xmust be withdrawn if

interest is being calculated at a compound rate of 8% effective per annum?

Solution:

(i) Time period unit: 1 year.

(ii) Effective rate of interest: i= 0.08.


(iv) Total payment rate per period: X.

Using the date of his original deposit as a focal date, the following equation of value

2000 = 400(1 +i)11 +X a 11|i=0.08

implies that

X= 256.12

years

0 1 2 11

400

2000

11 @ X

Note: a 11|8% =1 (1.08)11

0.08 = 7.13896.

Example 3:

An investment of R200 is due to be made at the beginning of each year for 10 years. If interest

is calculated at a compound rate of 6% effective per annum, how much will the investment be

worth at the end of 10 years?

Solution:

(i) Time period unit: 1 year.

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(ii) Effective rate of interest: i= 0.06.


(iv) Total payment rate in each period: R= 200.

Because these payments are being made in advance, the accumulated value after 10 years is

given by

AV = 200 s 10|0.06= 200(1.06)10 1

0.06/1.06 = 2794.33

years

0 1 2 9 10

10@200

Example 4:

Find the present value of an annuity contract that is payable continuously for a term of 10

years at a rate of $15 per annum. Assume an effective rate of interest of 5% per annum.

Solution:

STEP 1: Time period unit: 1 year

STEP 2: Effective rate of interest for that period: i= 0.05 = ln(1.05) = 0.04879

STEP 3: Total number of payment periods n= 10

STEP 4: Rate of payment per period: 15

P V = 15 a 10|= 151 (1.05)10

0.04879 = 118.69

Example 5:

Assuming an interest rate of 12% per annum that is convertible on a monthly basis, calculate

(i) the present value of an immediate annuity that is payable on a monthly basis in arrears

with payments of R1000 per annumfor the first 6 years, R400 per annumfor the next 4

years and a lump sum payment of R2000 at the end of the 10 years.

(ii) Now calculate the amount of the level annuity that is payable on a continuous basis for

10 years such that it has the same present value as the payment that has been made in

(i).

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(iii) Calculate the accumulated value after 7 years for the payments that have been used in

(i).

(iv) Calculate the accumulated value after 7 years for the payments that have been used in

(ii).

Solution(i):

STEP 1: Time period unit: 1 month

STEP 2: Effective rate of interest for that period: i= 0.12/12 = 0.01 v = 0.9901.

STEP 3: Rate of payment per chosen time period interval: Note that R1000 p.a. implies that

R = 1000/12 per month for the first 6 years, and R400 p.a. implies that R = 400/12 per

month for the next 4 years. Because the payments are being made in arrears

P V = 100012

a 72|+ 40012

v72 a 48|+ 2000v120

= 100012 51.504 + 400

12 0.48850 37.9740 + 2000 0.30299

= 4, 262.53 + 618.34 + 605.99 = R5,487

Solution(ii):

STEP 1: Time period unit: 1 year

STEP 2: Effective rate of interest for that period: 1 +i = (1 + 0.12/12)12 i = 0.126825

= ln(1 +i) = 0.1194 p.a. Now

X a 10|0.126825 = 5, 847 X= 939.947p.a.

Solution(iii):

STEP 1: Time period unit: 1 month

STEP 2: Effective rate of interest for that period: i= 0.12/12 = 0.01 v = 0.9901

The accumulated value of 7 years (i.e. 84 monthly periods) for the payments that are being

made in (i) is given by1

AV = (1 +i)84100012

a 72|+ 40012

v72 a 12|

= 2.3607100012 51.1504 + 400

12 0.4880 11.2551

= R10,255

1Note that as solution can also be given by the expression

AV= (1 +i)121000

12 s

72|+400

12 s

12|

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0.2 Annuities Payable pthly

Consider the following cashflow diagrams.

C CC C

0 1

x

2x

. . . . . .

. . . . . . (m1)x

mx

S1 :

An amount C paid in arrearsof each successive time interval of length 1/x.

C CC

0 1

x

2x

. . . . . .

. . . . . . (m1)x

mx

S2 :

C

An amount C paid in advanceof each successive time interval of length 1/x.

In either case: xC = constant rate of payment per b.t.u. Therefore we can calculate thepresent value and accumulation in value of these payments as explained in the previous sec-

tion. In this case PV(S1) =C am| i(x)/x

and PV(S2) =Cam| i(x)/x

.

Similarly we have AV(S1 = C sm| i(x)/x

and AV(S2 = Csm| i(x)/x

4. Some Examples(Constant )

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(a)

0

13

23 1

43

53 2

C = 100x = 3

100 100 100 100 100100

P V= 100a6|i(3)3

= 300a (3)2 | i

(b)

0 1 2 3 4 5 6 7 8

C = 50

x= 12

PV = 50 a4|2i(1/2) = 25a (1/2)

8 | i

50 50 50 50

(c)

0 25

45 1

65

85 2

125

145 3

165

PV = 20a9| 25 i

(5/2) = 50a(2 12)18/5| i

9 @ 2 0

C = 20

x= 52

(d)

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0 132 2 3 4

92 5 6 7 15

2 8 x= 2

3

C = 60

6060 60 60 60

P V= 60a5|32

i(3/2)= 40a (2/3)

15/2| i

10. Non-integer values ofn

Let p be a positive integer. Until now the symbol a n|(p)

has been defined only when n is a

positive integer. For certain non-integral values ofn the symbola n|(p)

has an intuitively obvious

interpretation. For example, it is not clear what meaning, if any, may be given to a23.25|, but

the symbol a23.25|(4) ought to represent the present value of an immediate annuity of 1 per an-

num payable quarterly in arrear for 23.25 years (i.e. a total of 93 quarterly payments, each of

amount 0.25). On the other hand, a23.25|(2)

has no obvious meaning.

Suppose that n is an integer multiple of 1/p, say n = r/p, where r is an integer. In this

case we define a n|(p)

to be the value at time 0 of a series ofr payments, each of amount 1/p, at

times 1/p, 2/p, 3/p, . . . , r/p= n. Ifi= 0, then clearly a n|(p)

=n. Ifi = 0, then

a n|(p)

= 1

p (v1/p +v2/p +v3/p + +vr/p)

= 1

pv1/p

1 vr/p

1 v1/p

= 1

p

1 vr/p

(1 +i)1/p 1

Thus

a n|(p)

=

1 vn

i(p) if i = 0

n if i= 0

()

The standard formula fora n|(p)

therefore applies for non-integer values ofn whenn is an integer

multiple of 1/p.

Note that, as seen previously, by working in terms of a new time unit equal to 1/p times

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the original time unit and with the equivalent effective interest rate of i(p)/p per new time

unit, we see that

a n|(p)

at ratei is equal to1

pa np| at rate i

(p)/p

The definition ofa n|(p)

given by equation () is mathematically meaningful however for all non-

negative values ofn. It is therefore convenient to adopt equation () as a definition ofa n|(p)

for

all n.

This is only a mathematical definition. It is not easily translated into the present value of

a series of payments.

Consider the following payment stream:

0 1

p

2p

. . . . . .

. . . . . . 1

. . . . . .

. . . . . . n-1

. . .

n T = n + f

sf|(p)

nppayments of

1

p

(n -1+1

p)

Here we haveT =n + f, wheren is an integer multiple of 1p

and 0< f < 1p

, we can then write

a T|(p)

=a n|(p)

+s f|(p)

vT (ora n|(p)

+a f|(p)

vn)

Algebraically we can prove this in the following way:

a n|(p)

+s f|(p)

vn+f

= 1 vn

i(p) +

(1 +i)f 1

i(p) vn+f

= 1 vn +vn vn+f

i(p)

= 1 vn+f

i(p)

= a n+f|(p)

=a T|(p)

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We can extend the above results to find formulae for a n|(p)

, s n|(p)

and s n|(p)

for all non-negative n.

Ifi= 0, we define for all non-negative n

a n|(p)

= (1 +i)1/p a n|(p)

=(1 v

n

)d(p)

s n|(p)

= (1 +i)n a n|(p)

=(1 +i)n 1

i(p)

s n|(p)

= (1 +i)n a n|(p)

=(1 +i)n 1

d(p)

Ifi= 0, each of these last three functions is defined to equal n.

Examples to be done in class

1. Prove that 1

an|i=

1

sn|i+i.

2. A savings plan provides that in return for n annual premiums of RX(payable annually

in advance), an investor will receive m annual payments of RY , the first such payment

being made one year after payment of the last premium.

Show that the equation of value for the transaction can be expressed in either of the

following forms:

(a) Y a n+m| (X+Y)a n|= 0 or

(b) (X+ Y)s m| Xs n+m|= 0

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Actuarial Notation: Annuities

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