Chapter 19 Solutions
1. A call option in Black-Scholes formula needs following five variables:
(1) Current price of the firm’s common stock (S);
(2) Exercise price (or strike price) of the option (X);
(3) Term to maturity in years (T);
(4) Variance of the stock’s price (  2 ); and
(5) Risk-free rate of interest (r).
and the accumulative normal distribution functions are also used in Black-Scholes call option as
follows:
C  SN (d1 )  XR T N (d 2 ) ,
Where C = Price of the call option, d1 
S
)
Xr t  1 t , d  d   t , N(d)
2
1
2
 t
log(
is
the
value of the cumulative standard normal distribution.
On the other hand, the binomial option pricing model can be written as
C  S[
nT
 k !(nT  k )! p ' (1  p ')
nT !
k
nT  k
]
k m
nT
X
nT !
[
p k (1  p) nT  k ]
r
(1  )Tn k  m k !(nT  k )!
n
X
 SB(nT , p ', m) 
B(nT , p, m)
r Tn
(1  )
n


and C=0 if m>T,
where n is the number of periods per year of term to maturity (T), m = minimum number of
upward movements in stock price that is necessary for the option to terminate “in the money,”
p
Rd
uR
and 1  p 
, X = option exercise price (or strike price), R = 1 + r = 1 + riskud
ud
free rate of return, u = 1 + percentage of price increase, d = 1 + percentage of price decrease,
nT
u
p '    p , and B(nT , p, m)   nT Ck p k (1  p) n  k .
R
k m
By a form of the central limit theorem, when n   , the cumulative binomial density function can be
approximated by the cumulative normal density function as
B1 (nT , p, m)  N  Z1 , Z1 
B2 (nT , p, m)  N  Z 2 , Z 2 
where
Z1 
m  nTp
,
nTp(1  p)
Z1 
nT  nTp
nTp(1  p)
Z2 
m  nTp
,
nTp(1  p)
Z 2 
nT  nTp
nTp(1  p)
It can be shown that
lim
n 
Then Equation can be rewritten as
1
 e  rT
r nT
(1  )
n
and lim Z1  lim Z 2  
n 
n 
C  SN ( Z1 )  Xe rT N ( Z 2 )
Use the definition of m and property of lognormal distribution, it can be shown that Z1=d1 and Z2=d2
C  SN (Z1 )  Xe rT N (Z 2 )  SN (d1 )  XR T N (d 2 )
2. The stock price follows a log-normal distribution under Black-Scholes model framework.
That is,
Sj
S j 1
 exp[ K j ] is a random variable with a log-normal distribution, where S j is
the stock price in jth period, K j is the rate of return in jth period and is a random variable with
normal distribution. Let Kt have the expected value  k and variance  k2 for each j. Then
K1  K 2  ...  K t is a normal random variable with expected value t k and variance t k2 .
Thus, we can define the expected value (mean) of
E(
St
 exp[ K1  K 2  ...  K t ] as
S
St
t 2
)  exp[ t k  k ] .
S
2
(19.52)
Under the assumption of a risk-neutral investor, the expected return E (
St
) is assumed to be
S
exp( rt ) (where r is the riskless rate of interest). Therefore, the relationship between the expected
stock return and risk-free rate under the risk-neutral assumption is
k  r 
 k2
2
.
3. The stock price follows a log-normal distribution under Black-Scholes model framework.
4.
a. P(X > 1.65) = 0.0495
b. P(X < -2.38) = 0.0087
c. P(X > -1.37) = 0.9147
d. P(1.72>X > -2.43) = 0.9498
5. As the assumption in question 4, find z
a. P(X < z) = 0.99 , by the inverse function of standard normal cumulative distribution,
z= 2.33
b. P(X > z) = 0.05, P(X < z)=1- P(X > z)=1-0.05=0.95
by the inverse function of standard normal cumulative distribution, z= 1.64
c. P (z <X<0)=0.0130, P (z <X<0)=P (X<0)- P (X<z)=0.5- P (X<z) =0.0130
P (X<z)=0.5-0.013=0.487, by the inverse function of standard normal cumulative
distribution, z= -0.0326
d. P (-z <X<z)=0.5408, P (-z <X<z)=1- P(X > z)- P(X < -z)=1-2 P(X < -z),
Therefore, P(X < -z)= [1- P (-z <X<z)]/2= (1-0.5408)/2=0.2296
by the inverse function of standard normal cumulative distribution, -z = -0.74016,
z=0.74016.
6. Y ~N(5, 2) , then standard normal z=(Y-5)/2
a. P(5<Y<9)= P((5-5)/2 < (Y-5)/2 < (9-5)/2 )= P(0<Z<2) =0.4772
b. P(-1<Y<3)= P((-1-5)/2 < (Y-5)/2 < (3-5)/2 )= P(-3<Z<-1)=0.15
c. P(Y>6)=P( (Y-5)/2 > (6-5)/2 )=P(Z>0.5)=0.3085
d. P(Y <10) = P( (Y-5)/2 < (10-5)/2) =P(Z<2.5) =0.9938
7. Let x= the amount withdrawn, xp= the amount of cash prepared
P(x> xp)=0.001
Therefore, the standard normal value (xp-5) /1=3.08 by using the inverse function of standard
normal cumulative distribution. xp=$8.08 (millions).
8.
𝑋−20.05
a. 𝑃(𝑋 > 20) = 𝑃 (
0.05
>
20−20.05
0.05
) = 𝑃(𝑍 > −1) = 1 − 𝑃(𝑍 < −1) = 1 −
0.1587 = 0.8413
b. 𝑃(𝑋 < 20) = 1 − 𝑃(𝑋 > 20) = 0.1587
If we define y as the number of bottles that contain less than 20 ounces of milk, then
y can be approximated by a normal distribution.
Y~N( np, np(1-p))=N(400(0.1587), 400 (0.1587) (1-0.1587))=N(63.48, 7.3082)
𝑃(𝑌 < 12) = 𝑃 (
𝑌 − 63.48 12 − 63.48
<
) = 𝑃(𝑍 < −7.044) = 0%
7.308
7.308
9. By Equation (19.61), call option value is as follows
𝐶 = 𝑆𝑁(𝑑1) − 𝑋𝑒 −𝑟𝑡 𝑁(𝑑2), 𝑑1 =
𝑑1 =
ln(
105
1
)+(0.05+ (0.45)2 )0.8
110
2
√0.8(0.45)
𝑆
1
ln (𝑋) + (𝑟 + 2 𝜎 2 ) 𝑡
√𝑡𝜎
, 𝑑2 = 𝑑1 − √𝑡𝜎
= 0.1850, 𝑑2 = 0.1850 − √0.8(0.45) = −0.2175
𝐶 = 105𝑁(0.1850) − 110𝑒 −0.05(0.8) 𝑁(−0.2175) = 16.46
10. When X=90, use Equation (19.61), call option value is as follows
𝐶 = 𝑆𝑁(𝑑1) − 𝑋𝑒
𝑑1 =
ln(
−𝑟𝑡
105
1
)+(0.05+ (0.45)2 )0.8
90
2
√0.8(0.45)
𝑁(𝑑2), 𝑑1 =
𝑆
1
ln (𝑋) + (𝑟 + 2 𝜎 2 ) 𝑡
√𝑡𝜎
, 𝑑2 = 𝑑1 − √𝑡𝜎
= 0.6836, 𝑑2 = 0.1850 − √0.8(0.45) = 0.2811
𝐶 = 105𝑁(0.6836) − 90𝑒 −0.05(0.8) 𝑁(0.2811) = 26.25