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: [email protected]
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
