THE BLACK-SCHOLES-MERTON
MODEL
指導老師：王詩韻老師
學生：曾雅琪(69936017)，藍婉綺(69936011)
Contents
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Lognormal property of stock prices
The distribution of the rate of return
The expected return
Volatility
Concept underlying the Black-Scholes-Merton differential
equation
Derivation of the Black-Scholes-Merton differential equation
Risk-neutral valuation
Black-Scholes pricing formulas
Cumulative normal distribution function
Implied volatilities
Dividends
Assume
Define
S
：in a short period of time(Δt)
S
S
：normal distribution
S
μ：expected return on stock per year
σ：volatility of the stock price per year
s
 t  z
s
Ｗiener process

ΔS
~ μΔt ，  Δt
S


ΔS
~ μΔt ，  Δt
S
G  ln S
ItÔ’s Process
2
G
1

dx a, ( xG
, t )dt 1b,(xG, t )dz
0
t
2
S S S
S t
 ST
ln ST  ln S 0  ln 
 S0

ItÔ’s Lemma
 G
G 1  2G 2 2 
 G 

dG  
S 


S
dt

S dz

2


S

t
2

S

S







2 
 ~    
T ,  T 
2 






2 
T ,  T 
ln ST ~  ln S 0    
2 






2 
T ,  T 
ln ST ~  ln S 0    
2 



Example 13.1
A stock with an initial price of $40
An expected return of 16% per annum
A volatility of 20% per annum
Ask：the probability distribution of
the stock price in 6 months？
ST  40
Thus,
 0.16
there is a 95%
probability
 0 .2
that the stock
Tprice
 0.5in 6 months will lie
between $32.55~$56.56
  0.05


0.2 2
ln ST ~  ln 40  (0.16 
)  0.5,0.2 0.5 
2


ln ST ~  3.759,0.141
 3.759  1.96  0.141  ln ST  3.759  1.96  0.141
 e3.7591.960.141  ST  e3.7591.960.141
 32.55  ST  56.56
Lognormal distribution
A variable that has s lognormal distribution can take
any value between zero and infinity.
E ( ST )  ?
Var ( ST )  ?

X  ln( V ), X ~  (m, s )
1
 f (X ) 
e
2 s
 h(V ) 
1
2 sV
n
V
 h(V )dV
( X m)2

2s2
e
0
 n  1  E (V )
 n  2  E (V 2 )
 Var (V )
[ln(V )  m )]2

2s2
E ( ST )  S 0 e
T
2 2 T
0
Var ( ST )  S e
(e
 2T
 1)
The nth
moment
of V
X  ln( V ), X ~  (m, s )
f (X ) 
h(V ) 
1
e
2 s
1
2 sV
( X m)

2s2

e


2
0
[ln(V )  m )]
2s2
nx
e
e
x
2 se

2




V  ex





1 x
x
de

ln
e
x
0 e x



[ x  m ]2

2s2
nx

e
e
2 s
1
e
2 s
1
e
2 s
de x
[ x  m ]2
2s2
dx
[ x  m ]2  2 s 2 nx

2s2
( x  m  ns 2 ) 2

2s2
n2s 2 
nm
2
2

1
e
e

2
2
2
2
 ( x  m)  2s nx   x  2mx  m  2s snx
e
dx
2 mns2  n 2 s 4
2s2
( x  m  ns 2 ) 2

2s2
dx
dx
 ( x  m  ns )   x  2mx  nm  2s nx  2mns  n 2 s 4
  V h(V )dV e
2 2
2


2
2
n2s2 2
nm
2

n
V
 h(V )dV e
nm
n2s2
2
0
ln ST ~ [ln S0  (  
n  1  E (V )  e
2
2
m
n  2  E (V )  e
2
)T ,  T ]  m  ln S0  (  
2
2
)T , s   T
s2
2
2m2 s 2
 Var (V )  E (V )  [ E (V )]  e
2
2
E (V )  E (ST )  e
Var (V )  Var (ST )  e
ln S0  (  
2
2[ln S0  (  
2
)T 
2
2
2m2 s 2
 2T
2
e
s2
2( m )
2
 S0e
)T ] 2T
T 
 2T
[e
e
2m s 2
 2T  2T
2

2
[e  1]
s2
 S0e T
2 2 T
 1]  S0 e
 2T
(e
 1)
E ( ST )  S 0 e
T
2 2 T
0
Var ( ST )  S e
(e
Example 13.2
A stock where the current price is $20
An expected return of 20% per annum
A volatility of 40% per annum
Ask：the expected stock price, and the
variance of the stock price, in 1year？
 2T
 1)
S 0  20
  0. 2
  0.4
T 1
E ( ST )  20e 0.21  24.43
Var ( ST )  20 e
2
20.21
(e
0.4 2 1
 1)  103.54
The distribution of the rate of return
Information
Define
The probability distribution of
the continuously compounded
rate of return earned on a stock
between times 0 and T.
ST
 e xT
S0
whenT  
ST  S 0 e xT
S
ln T  xT
S0
x
 s tan dard
1 ST
x  ln
T S0
1 ST
ln
T S0
ST

~  [(   )T ,  T ]
S0
2
2
ln
X：the continuously compounded
rate of return per annum realized
between times 0 and T.
x ~  ( 
2 
2
,
T
)

T
0
x ~  ( 
2 
2
,
T
)
Example 13.3
A stock with an expected return 17%
per annum
A volatility of 20% per annum
Ask：the average rate of
return(continuously compounded)
realized over 3 years？
  0.17
Thus, we can be 95%
 0.the
2 average
confidentthat
return realized
T  3 over 3 years
will be between
  0.05
-7.6%~37.6%

0.2 2 0.2 
x ~  (0.17 
),

2
3


 x ~  0.15,0.1155
 0.15  1.96  0.1155  x  0.15  1.96  0.1155
 7.6%  ST  37.6%
The expected return


ΔS
~ μΔt ，  Δt
S
Expected return=E(R)= μ
2 
,
)
ST
 x ~  ( 
2
ln( )  (   )T  E[ln( 2ST )] Tln S0  (   )T
S0
2
2
2
2
T
E ( SExpected
return=E(x)=
ln E ( ST )  ln S0  
T
T )  S0e

2
E ( ST )  S 0 e T  ln[ E ( ST )]  ln S 0  T
Assume
ln[ E ( ST )]  E[ln( ST )]
In fact
ln[ E ( ST )]  E[ln( ST )]
ln[ E ( ST )]  ln( S 0 )  T
E[ln(
ST
E[ln( )]  T
S0
 E ( x)   Geometric mean
 E ( R)  
Arithmetic mean
ST
)]  T
S0
( E ( x)   
2
2
)
Arithmetic mean>geometric mean
 E ( x)   
2
2
Example 13.4
(0.1772) 2
Initial investment of a mutual fundisE ($100
x)  0.14 
 12.43%
2
The returns per annum report over the last five years：
15%, 20%, 30%, -20%, 25%
Arithmetic mean
Geometric mean
E ( R)
15%  20%  30%  20%  25%

5
 14%
FV5  100  (1.14)  192.54
5
E ( x)
(1  15%)(1  20%)(1  30%)

1
(1  20%)(1  25%)
5
 12.4%
FV5  100 1.15 1.2 1.3
 0.8 1.25  179.4
Volatility
a.
b.
Estimating volatility from historical
data
Trading days vs. calendar days
σ：a measure of our uncertainty about
the returns provided by the stock
x ~  ( 
2 
2
,
T
Between 15%~60%
The standard deviation of the return
)
The standard deviation of the
percentage change in the stock price

ΔS
~ μΔt ，  Δt
S

With the square root of how far ahead
we are looking
The standard deviation of the stock price
in 4 weeks is approximately twice the
standard deviation in 1 week.
Estimating volatility from historical data
Define
n+1：number of observations
Si：stock price at end of ith interval, with i=0,1,2……,n
τ：length of time interval in years
ui  ln(
s
Si
)
Si 1
1
n
2
(
u

u
)

i
n  1 i 1
1
1
n
n
2
2

u

(
u
)
 i n(n  1) i1 i
n  1 i 1
S
ln  T
 S0



2 
 ~    
T ,  T 
2 



 standard deviation (ui )   

Var (ui ),

s




s




Standard error( ) 


2n
s
1
n
2
(
u

u
)
 i
n  1 i 1
2
1
n
2

(ui  2ui u  u )

i 1
n 1
1
n
n
2

[i 1 ui  2i 1 ui (
n 1
i 1 ui
n
n
)  i 1
n

(
1
2
1
n
n
n
2
2

[i 1 ui  (i 1 ui )  (i 1 ui ) 2 ]
n 1
n
n

1
1
n
n
2
[i 1 ui  (i 1 ui ) 2 ]
n 1
n

1
1
n
n
2
2
u

(
u
)
 i n(n  1) i 1 i
n  1 i 1
n
u
i 1 i
n
)2 ]
Trading days vs. calendar days
Research
1. The variance of stock price returns between the close of trading
‧ Volatility
is muchon
higher
when
on one day and the
close of trading
the next
day when there are
the exchange
is open for trading
no intervening nontrading
days.
than when it is closed.
‧Practitioners
ignore
days between the close of
2. The variance
of the tend
stocktoprice
returns
the exchange
tradingwhen
on Friday
and closeisofclosed.
trading on Monday.
Reasonably expect
The first is a variance over a
1-day period. We might
reasonably expect the second
variance to be three times as
great as the first variance.
In fact
The second variance to be,
respectively, 22%, 19%,
and 10.7% higher than the
first variance.
Volatility
per annum

Volatility per
trading day

Number of trading
days per annum
Number of trading days until option matutity
T
252
The number of trading days in a years is
usually assumed to be 252 for stocks.


20
u  0.09531
i 1 i
20
20.10
20.00
ln 1.00500
1
1
n
n
2
2
u

(
u
)


i
i
i 1
n  1 i 1
n(n  1)
0.00326 0.095312


 0.01216
20  1 20(20  1)
ln 0.99282


20.90
20.90
u  0.00326
i 1 i
s
20.75
20.90
2
s



ln 1.00000
0.01216
 0.193(year)
1
252


Standard error( )



2n

0.193
 0.031
2  20
Concepts
A riskless portfolio
stock
derivation
No arbitrage opportunities
Return(portfolio)=risk-free interest rate(r)
Example p.290
△c=0.4 △S
→1. a long position in 0.4 shares
2. a short position in 1 call option
△ c=o.5△S
→1. an extra 0.1 share be purchased
2. for each call option sold
Assumptions







The stock price follows the process developed in CH12 with
μ and σ constant
The short selling of securities with full use of proceeds if
permitted.
There are no transactions costs or taxes. All securities are
perfectly divisible.
There are no dividends during the life of the derivative.
There are no riskless arbitrage opportunities.
Security trading is continuous.
The risk-free rate of interest, r, is constant and the same for
all maturities.
Define
f：the price of the call option or other derivative contingent on S
→f must be come function of S and t.
π：the value of the portfolio
equation
S  St  Sz
 f
f 1  2 f 2 2 
 f


f   S  

S

t


S

z
2


S

t
2

S

S




portfolio
-1：derivative

f
：shares
S
process
 f
f 1  2 f 2 2 
 f


f   S  

S

t


S

z
2

t 2 S
 S 
 S

f
S
S  St  Sz
S
f
   f 
S
S
 f
 f
f 1  2 f 2 2 
f

St  Sz 
     S  

S

t


S

z

2

t 2 S
S

 S
 S
 f 
f
f
1 2 f 2 2
f
f
f
   
St  t 

S

t


S

z


S

t

Sz
2
S
t
2 S
S
S
S
 f 1  2 f 2 2 
     
 S t
2
 t 2 S

This equation does not involve △z
The portfolio must instantaneously
earn the same rate of return as other
short-term risk-free securities.
  rt
△π=rπ△t
f
S
S
 f 1  2 f 2 2 
    
 S t
2
 t 2 S

 f 
 f 1  2 f 2 2 
f 

   
 S t  r   f  S t
2
S 

 t 2 S

f 1  2 f 2 2
f
 

S


rf

rS
2
t 2 S
S
f f
1  2 f 2 2 (Black-Schiles-Merton
 rf 

rS 
 S
2
differential equation)
t S
2 S
f f
1 2 f 2 2
rf 

rS 
 S
2
t S
2 S
Example 13.5
f
  rKe  r (T t )
t
f
1
S
2 f
0
2
S
f  S  Ke r (T t )

rf   rKe
 r (T t )
P.108_equation(5.5)

1
 1rS  0 2 S 2
2
 rf  rS  rKe r (T t )
r
f  S  Ke r (T t )
Risk-neutral valuation
Application to Forward Contracts on a
Stock
Risk-neutral valuation
It is the single most important tool for the
t: time
analysis of derivatives.
S:the current stock
2
f
f 1 2 2  f
 rS
 σ S
 rf
2
t
S 2
S
price
 : stock price volatility
rf: the risk-free rate
of interest
They all are independent of risk preferences
If the expected return,u,involved in
the above eqution.
Independent of
risk preferences
Risk-neutral valuation
Assumption:
All investors are risk neutral
The expected return of all investment asset is
the risk-free rate of interest, r .
the expected return = r
Why?
Risk-neutral valuation
Reason:
1.
r=rf + s
The risk-neutral investors do not require a
premium to induce them to take risks.
2. Any cash flow
Discount by
present value
r
expected value
How to use risk-neutral valuation
A derivative provides a payoff at one particular time.
step1
Assume:
the expected
return from the
underlying asset
is the risk-free
interest rate, r.
Step2
Calculate the
expected
payoff from
the derivative
Step3
Discount the
expected
payoff at
the risk-free
interest rate.
Application to Forward Contracts
on a Stock
A long forward contract
Maturity: time T
DeliveryStep3
price: K
step1
Calculate:
the value at maturity
ST - K
ST : stock price at time T
Step2
Equation13.4
Calculate:
E(ST) = S0erT
the value at time 0
e-rT
f =
E(ST – K)
= e-rTE(ST) –Ke-rT
(13.17)
step3
E(ST) = S0erT
(13.18)
Substitute
equation(13.18)
into (13.17)
f = S0 – Ke-rt
European Option Pricing
c=European call Price
Formula
P=European put Price
S0=the stock price at time zero
 rT
c  S0N(d1)  Ke N(d2)
K=the strike price
r=risk-free rate
p  Ke-rT N(d2)  S0N(d1)
 =stock price volatility
2
ln S0/K   r  σ /2 T
T=time to maturity of the option
d1 
N(X)=
σ T
the cumulative probability
ln S0/K   r  σ 2 /2 T
d2 
 d1  σ T distribution function for a
σ T
standardized normal
distribution




Black-Scholes pricing formulas
f
f 1 2 2  2 f
 rS  σ S 2  rf
t
S 2
S
Deriving the Black-Scholes formulas,we can use
a. solve the differential equation(13.16)
b. use risk-neutral valuation
An European call option

Emax ST  K ,0
ST>K,
 the expected value of a
 rt
c  e Emax ST  K ,0
variable is equal to S(T)
ST<K,0
ce
 rt
S N(d1)e
0
rt
The probability that
the option will be
exercised in a riskneutral world

 KN(d2)
Derivation
Key result
If ln V~ N(m, w)
E [max(V − K, 0)] = E(V )N(d1) − KN(d2)
where
d1 
ln[E(V)/K]  W 2 / 2
W
d2 
ln[E(V)/K] - W 2 / 2
W
E : expected value
Derivation
Step1
Define g(V) as the probability density function of V .
It follows that

Emax V - K,0   (V - K)g(V)dV
(13A.2)
K
• ln V ~ N(m, w) m=ln[E(V)] - w2/2 (13A.3)
Derivation
Step2
Define a new variable
lnV  m
Q
w
(13A.4)
Q ~ N(0,1)
Denote the density function for Q, h(Q)
h(Q) 
1 Q 2 /2
e
2π
Derivation
Step3
Using equation(13A.4) to convert the expression on the
right-hand side of equation(13A.2)from an integral over V
to an integral over Q
Emax V  K,0  

(lnK m)/w
Emax V  K,0  

(lnK m)/w
e
(e
Qw  m
Qw  m
 K)h Q dQ 
hQdQ  K 

(lnK m)/w
h Q dQ 
(13A.5)
Derivation
step4
e
Qw  m
1 Q 2  2Qw 2m /2
e
2π
h Q  
1

 Q  w 2  2m w 2 /2
e
2π


e
e
Emax V - K,0  e
m  w2 / 2

m  w 2 /2
2π
m w 2 /2

(lnK m)/w
e Q  w  /2
2
h(Q  w)
h(Q - w)dQ  K 

(lnK m)/w
hQ d Q 
(13A.6)
Derivation
N(X): probability that a variable with a mean of zero and a
standard deviation of 1 is less than x.
Step5

 h (Q  w)dQ
ln K  m
= 1  NlnK  m/w  w
/w
=
N lnK  m/w  w
Substituting for m from equation (13A.3)
 ln EV /K   w 2 /2 
  Nd1
N
w


Emax V - K,0  e
m w 2 /2
N(d1)  KN(d2)
Properties of the Black-Scholes Formulas
When stock price becomes very large




d
lnSO/K  r  2 / 2 T
1
 T
d
lnSO/K  r  2 / 2 T
2
 T
N(-d1)=1-N(d1)
d1.d2 become very large
N(-d2)=1-N(d2)
N(d1)1
N(d2)1
N(-d1)0
N(-d2)0
c  S0N(d1)  Ke N(d2)
p  Ke rT N(d2)  S0N(d1)
So- Ke-rt
price p approaches zero
 rT
very similar to a forward
contract with delivery
price K.




d
lnSO/K  r  2 / 2 T
1
 T
d
lnSO/K  r  2 / 2 T
2
 T
When volatility approaches zero
When  0
the stock is riskless and the
 rT
rT
Call
price
=
max

S
O

Ke
,0
price grows at rate r to S 0e .
put price=max Ke rT  S 0,0
 rT
Ke
S0 >
ln(S0/K)+rT >0
d1,d2 +
 rT
Ke
S0 <
ln(S0/K)+rT<0
d1,d2  
Cumulative normal distribution function
1  N ( x)(a1k  a2 k 2  a3k 3  a4 k 4  a5k 5 ) if x  0
N ( x)  
if x  0
1  N ( x)
1
k
,   0.2316419
1  x
a1  0.319381530 a2  0.356563782
a4  1.821255978
a5  1.330274429
1  x2 2
N ( x) 
e
2
a3  1.781477937
Example
Example 13.6
The stock price 6months from the
expiration of an option is $ 42
The exercise price of the option is $40
The risk-free interest rate is 10%
The volatility of 20% per annum




ln42/40 0.10.22 / 2 0.5
1
 0.7693
0.2 0.5
d
ln42/40 0.10.22 / 2 0.5
2
 0.6278
0.2 0.5
d
Ke-rt =40e-0.1*0.5 = 38.049
S 0  42
K  40
N(0.7693)=0.7791
r 
0. 1
N(-0.7693)=0.2209
 N(0.6278)=0.7349
 0 .2
N(0.-6278)=0.2651
T  0.5
c=
42N(0.7693)38.049N(0.6278)
p=
38.049 N(-0.6278)38.049N(-0.7693)
Implied volatilities

In the Black-Scholes pricing formulas
Weof
can’t
find
the volatility
the
stock price.
calculated by
option price observed in the market.
Functions:
a.Monitor the market’s opinion about the volatility of particular
stock.
b. From actively traded option on a certain asset,Traders use to
calculate the appropriate volatility for pricing a less actively traded
option on the same stock.
c  S0N(d1)  KerT N(d2)
European call option (No dividend)
C=1.875
So= 21
K= 20
r=10% (per annum)
T=0.25
Ask: σ =??Implied volatility
How to do??
Iterative search
σ=0.2
c=1.76
too low
σ=0.3
c=2.1
too high
σ=0.25
σ lies between 0.2 and 0.25.
In this example, the implied
volatilities is 0.235
too high
C is an
increasing
function of
σ
Dividends
Assumption:
a.The amount and timing of the dividends during the
life of an option can be predicted with certainty.
b.The date on which the dividend is paid should be
assumed to be the ex-dividend date.
c.On this date the stock price declines by the amount
of the dividend.
European options
Assumption
riskless component
Stock price
risky component
1. the present value of all the
1. S0 is equal to the risky
dividends during the life of the
component of the stock
option discounted from the
price.
ex-dividend to the present at the
2.  is the volatility of the
risk-free rate.
process followed by the
2.By the time the option matures,
risky component
the dividends will have been paid
and the this part will no longer exist.
Example 13.8
European call option on a stock
Ex-dividend dates in two months and
five months
The dividend is expected to be $0.5
The current share price is$40
The exercise price is $40
The risk-free interest rate is 9%
The volatility of 30% per annum
Time maturity is six months
ASK:European call option price??
S 0  40
D  0.5
K  40
  0.3
T  0.5
r  0.9
2/12
5/12
Calculate the present value of the dividends
0.5e-0.1667*0.09 + 0.5e-0.4167*0.09 = 0.9741
So minus the present value of the dividends
40 - 0.9741=39.0259
lnSO/K  r  2 / 2 T
d1 
Use Black-Scholes pricing formulas
d1=0.2017
N(d1)=0.58
d2=-0.0104
N(d2)=0.4959
d
 T


lnSO/K  r  2 / 2 T
2 
 T
C  S0N(d1)  Ke rT N(d2)
Call price:
39.0259 × 0.58 - 40e-0.5*0.09 ×0.4959 =3.67
American Option
Assumption
When there are dividends, it is optimal to exercise only at
a time immediately before the stock goes ex-dividend.
n ex-dividend dates are anticipated and they are at
times t1,t2 …tn, with(t1<t2< …<tn)
The dividends corresponding to these times
will be denoted by D1,D2 …,Dn,respectively.
American Option
c  S 0  D  Ke
 rT
(9.5)
Final ex-dividend date(tn)
The option is exercised
S(tn)-K
The option is not exercised
S(tn)-Dn
c  S tn  Dn  Ke r (T tn)
American Option
S tn  Dn  Ke r (T tn)  S (tn  K )
If
then
Dn  K (1  e  r (T tn) )
Very
It cannot be optimal
to exercise at time tn
large
T-tn is small
If
Dn  K (1  e  r (T tn) )
It can be optimal to exercise at time tn

Di  K 1  e  r (ti 1ti)

It is not optimal to exercise immediately prior to time ti