University of Economics, Faculty of Informatics

Dolnozemská cesta 1, 852 35 Bratislava

Slovak Republic

Financial Mathematics in Derivative Securities and Risk Reduction

Financial Mathematics

Ass. Prof. Ľudovít Pinda, CSc.

Department of Mathematics,

Tel.:++421 2 67295 813, ++421 2 67295 711

Fax:++421 2 62412195

e-mail: pinda@dec.euba.sk

 

Sylabus of the lecture

 

        Simple and compound interest.

        Comparison simple interest with compound interest.

        Nominal interest rates.

        Accumulation factor, force of interest.

        Stoodleys formula for the force of interest.

        The basis compound interest functions.

        Annuities-certain and annuities-due, present values and accumulations.

        Continuously payable annuities.

        Discounted cash flow, net present values.

Simple and compound interest

K0 – the amount in t = 0,

Kn – the amount in t =n,

i – the interest rate p. a.,

n – the time of duration less then one year,

d – the time of duration measured in days,

K0 n i – the interest of amount,

K0 n i

Kn

K0

n t

36511 00

diKniKK n

n – the time of duration is greater then one year,

,

Fig. 1

nn iKK 10

110 niK – the interest of amount,

110 niK

Kn

1 t

K0

Fig. 2

Comparison simple interest with compound interest

Let K0 =1

From the Binomic theorem

nn iinn

inn

in

i

32

!3

1

!2

1

!111

From then ni 0 2

!2

111 i

nnnii n

oOi

,

if 10 n then

0!2

12i

nn and inin11 ,

if 1n then niin11 ,

if 1n then

0!2

12i

nn and inin11 .

.

.

- the rate of interest for the period (the effective rate of interest for the period) ti 1, tt

1111010 niiiKKn .

Kn

n0 = 1 n

K0

Fig. 3

Nominal interest rates

h - the term of lenght the time unit of transaction,

ti h - the nominal rate of interest per unit time,

tih h - the effective rate of interest for the period of length h beginning at time t.

Thus K0 invested at time t for a term h, htt , is

httAKtihKK h ,:1 001 .

For K0 = 1 we have

httAtih h ,1 and h

httAtih

1, . (1)

Let 210 ttt and cosider an investment of 1 at time t0. Then principle of consistency is

211020 ,,, ttAttAttA .

Denote by

t - the force of interest per unit time at time t,

h

httAtit

hh

h

1,limlim

00

.

T h e o r e m :

I f t a n d ttA ,0 a r e c o n t i n u o u s f u n c t i o n s o f t f o r 0tt , a n d t h e p r i n c i p l e o f c o n s i s t e n c y

h o l d s , t h e n f o r 210 ttt

2

1

exp, 21

t

t

dttttA .

P r o o f :

S u p p o s e t h a t t 1 a n d t 2 a r e g i v e n w i t h 210 ttt . F o r 0tt l e t ttAtf ,0 a n d f i s

c o n t i n u o u s . F o r 0tt w e h a v e

h

httAtit

hh

h

1,limlim

00

h

ttAhttA

ttAttAh

ttAhttAttAhh

,,lim

,

1

,

,,,lim 00

000

00

0

tftfh

tfhtf

tf h

´

0

1lim

1

.

T h e r e f o r e

tfttf ´ .

S i n c e a n d f a r e c o n t i n u o u s f u n c t i o n s , s o t o o i s ´f a n d

ttftf .

B y i n t e g r a t i n g

t

t

dssctf0

exp , c – a n a r b i t r a r y c o n s t a n t .

F r o m c o n s i s t e n c y p r i n c i p l e

2

1

exp,

,,

1

2

10

2021

t

t

dsstf

tf

ttA

ttAttA .

From (1) is

h

dss

ti

ht

th

1exp

.

For the practice is very important the case where is

t , is constant.

Then

hdthttAht

t

expexp,

.

The relation between the effective interest rate and force of interest is

1 ei .

Let the force of interest per unit time t is 0.12 for t. Finde the nominal rate of interest p.a.

of term

a) sevent days,

b) one month,

c) six month.

Solution

From formula h

htih

112.0exp we recive for

a) 365

7h , 01.12tih %,

b) 12

1h , 06.12tih %,

c) 2

1h , 30.12tih %.

Example 1.

The present value of 1 in time t is

tt

tdsdsstv00

expexpexp , (2)

and discont factor for t = 1 is exp v .

O n 1 . J u l y 2 0 0 2 a c u s t o m e r d e p o s i t e d 5 0 0 0 0 w i t h t h e b a n k . O n 1 . J u l y 2 0 0 4 h i s d e p o s i t h a d

g r o w t o 5 9 1 0 2 . A s s u m m i n g t h a t t h e f o r c e o f i n t e r e s t p . a . w a s a l i n e a r f u n c t i o n o f t h e d u r i n g

p e r i o d . F i n d e t h e f o r c e o f i n t e r e s t t o t h e 1 . 7 . 2 0 0 3 .

D e n o t e b y

t

t

dssttAtf0

exp,0 .

10 f / 1 . 7 . 2 0 0 2 / , 182040.100050

102592 f / 1 . 7 . 2 0 0 4 / .

Example 2.

Solution.

t i s l i n e a r f u n c t i o n , t h e r e f o r e t

t

dsstf0

ln i s q u a d r a t i c f u n c t i o n .

D e n o t e b y tgtf ln . ( 3 )

F o r a n y q u a d r a t i c f u n c t i o n tg f o r hatha a n d f r o m L a g r a n g e t h e o r e m i s

h

haghagg

2

.

I f a t h e n

h

haghagag

2

.

F r o m ( 3 ) ttftf

tg 1, 11 g .

083621.00167242.02

10182040.1ln

2

10ln2ln

2

11 ff .

T h e f o r c e o f i n t e r e s t t o t h e t i m e 1 . 7 . 2 0 0 3 w a s 0 . 0 8 3 6 2 1 .

Consider the unit of the time one year. Let tt 9.006.0 . Calculate tv and present

value 100 with due the maturity 3.5 years.

t

s

ts

t

dsdsdsstv000

09.006.0exp9.006.0expexp

9.0ln

19.006.0exp

9.0ln

9.006.0exp

0

tts

.

Present value of 89.839.0ln

19.06.0exp1005.3100

5.3

v .

Example 3.

Solution.

Stoodley's formula for the force of interest

Using: The model a smoothly decreasing or smoothly increasing force of interest.

s ter

spt

1 ,

p , r , s – p a r a m e t e r s .

p = 0 . 0 7 6 9 6 1 , r = 0 . 5 , s = 0 . 1 2 1 8 9 0 ,

tt

121890.0exp5.01

121890.0076961.0

.

F i n d a f o r m u l a f o r tv i n t e n y e a r s ' t i m e .

F r o m ( 2 ) w e d e r i v e

tpr

rtsp

rtv

exp

1exp

1

1.

tttttv 08.13

122.1

3

2076961.0exp

3

2121890.0076961.0exp

3

2.

T h e n v a l u e o f 1 d u e i n t e n y e a r s ' t i m e i s

24566.008.13

122.1

3

2 1010 tv .

Situation:

Example 4

ster

spt

1

t = 0 the interest rate 0.11,

t = 4 the interest rate 0.10,

t = ∞ the interest rate 0.08.

From ie 1 and i 1ln is

104.011.1ln0 , 01

104.0

ser

sp ,

095.01.1ln4 , 41

095.0

ser

sp ,

077.008.1ln , 0077.0 p .

From this system

r = 3.578, s = 0.1236, p = 0.077.

,

The basis compound interest function

L e t t s o m e c o n s t a n t , t h e v a l u e a t t i m e s o f 1 d u e a t t i m e s + t i s

tdrdrrtvts

s

ts

s

expexpexp

tt dv 1 ,

expv a n d ed1 a n d ei 1 .

R e l a t i o n s h i p s b e t w e e n dvi ,,,

i v d

1exp exp exp1

i i1ln 11 i 11 ii

v vln 11v v1

d d 1ln 11 1 d d1

Tab. 1

A p p r o x i m a t e s h e d u l e s

2

2

1ii , 2iid ,

2

2

1 i , 2

2

1 d .

A n n u i t i e s – c e r t a i n a n d a n n u i t i e s – d u e , p r e s e n t v a l u e s a n d a c c u m u l a t i o n s

A n n u i t y - c e r t a i n

1 1 1 1

t t + 1 t + 2 t + 3 t + n

i

v

v

v

v

vvvvvva

nnnn

n

1

1

1

1

11

32 .

A n n u i t y – d u e

1 1 1 1 1

t t + 1 t + 2 t + 3 t + n

d

v

v

vvvva

nnn

n

1

1

11 12 ,

A c c u m u l a t e d a m o u n t h ns , ns

i

iais

n

nn

n

111

,

d

iais

n

nn

n

111

.

Continuously payable annuities

T h e v a l u e o f t i m e 0 o f a n a n n u i t y p a y a b l e c o n t i n u o u s l y b e t w e e n t i m e 0 a n d t i m e n , w h e r e t h e

r a t e o f p a y m e n t p e r u n i t t i m e i s c o n s t a n t a n d e q u a l t o 1 i s d e n o t e d b y

nn

n

vndtta

1exp1

exp0

.

I n c r e a s i n g a n n u i t y

1 2 3 n

t t + 1 t + 2 t + 3 t + n

i

vnavnvvvIa

nnn

n

32 32 ,

nn

nt

n

vnadtvtaI

0

.

Discounted cash flow, net present values

c t = ca sh in flo w a t tim e t – ca sh o u tflo w a t tim e t,

q (t) = ra te s o f in flo w a t tim e t – ra te s o f o u tflo w a t tim e t,

t

Ttt

t dtitqiciNPV0

11 ,

t

Ttt

t dtetqecNPV0

.

Example 5.

Consider the cash flow an initial outlay of 20 000, after one year a futher outlay 10 000, an inflow of 3 000

per annum payable continuously for ten years beginning in three years' time and final inflow of 6 000 in the

end of thirteen years' time. Express the net present value.

Solution.

13331

000600030001000020 vaaviNPV at the rate i.

For 2i % is

77303.0000691269.2461395.11000398039.0000100002002.0NPV

395.480 .

13