Download Markov process

Market risk modelling - The Black
Scholes Equation
By: A V Vedpuriswar
Draws heavily from John C Hull, “ Options, Futures and Other Derivatives”
and Neil A Chriss, “ Black Scholes and Beyond-Option Pricing Models”
October 31, 2010
Modelling asset price movements
To measure the market risk of an asset portfolio, we should
be able to model the price of the underlying.
The most celebrated modeling has been done for stocks.
But this work can be extended to other asset classes too.
Before we look at the modeling techniques, we need to gain a
basic understanding of stochastic processes.
1
Discrete and Continuous time Stochastic processes
When the value of a variable changes over time in an
uncertain way, the variable follows a stochastic process.
 In a discrete time stochastic process, the value of the
variable changes only at certain fixed points in time.
In case of a continuous time stochastic process, the changes
can take place at any time.
Stochastic processes may involve discrete or continuous
variables.
The continuous variable continuous time stochastic process is
usually used for describing stock price movements.
2
Markov Process
 In a Markov process, the past cannot predict the future.
 Stock prices are usually assumed to follow a Markov process.
 All past data have been discounted by the current stock price.
 Suppose we have a coin tossing game.
 For every head we gain $1 and for every tail, we lose $1.
 Then expected value of the gains after i tosses will be zero.
 For every toss, the expected value of the gains is zero.
3
Markov Process ( Contd)
Let Si denote the total amount of money we have actually won
up to and including the ith toss.
Then the expected value of Si is zero.
 On the other hand, let us say we have already had 4 tosses
and S4 is the total amount of money we have actually won.
 The expected value of the fifth toss is zero.
Thus the expected value after five tosses is nothing but S4.
That is no change is expected in the variable.
The expected value in future equals the current value.
This leads to the idea of Wiener process.
4
Wiener Process
A Markov process with mean change = 0 and variance = 1 per
year is called a Wiener process.
 A variable z follows a Wiener process if the change Δz during
time Δt is given by Δz = ε  Δt,
 ε is a standard normal random variable(mean = 0, std devn = 1)
 The values of Δz for any two different short intervals of time, Δt are
independent.
 Mean of Δz = 0
 Variance of Δz = Δt or Standard deviation =  Δt
5
Illustration
To illustrate, say a variable follows a Wiener process and has
an initial value of 30.
Where will it be at the end of one year?
The expected change in the value of the variable is 0.
The variable will be normally distributed with a mean of 30
and a standard deviation of 1.0.
Where will it be at the end of 4 years?
 At the end of 5 years, the mean will remain 30 but the
standard deviation will be 4 = 2.
6
Generalized Wiener Process
Here the mean does not remain constant.
Instead, it “drifts” at a constant rate.
This is unlike the basic Wiener process which has a drift rate
of 0 and variance of 1.
The generalized Wiener process can be written as:

dx
=
a dt + bdz
or
dx
=
a dt + bεΔt
7
Generalized Wiener Process (Continued)
Suppose the value of a variable is currently 40.
 The drift rate is + 10 per year while the variance is 900 per year.
This means the expected change in the variable is 10 in a year.
 If we consider a period of 1 year the variable will be normally
distributed, with
– mean of 40+10 = 50
– std deviation of 30.
If we consider a period of 6 months, the variable will be normally
distributed with
– mean of 40+5 = 45
– std devn of 30.5 = 21.21.
8
Problem
Variables X1 and X2 follow generalized Wiener
processes, with drift rates µ1 and µ2 and variances σ21
and σ22. What process does X1 + X2 follow if:
(a) the changes in X1 and X2 in any short interval of time
are uncorrelated?
(b) there is correlation ρ between the changes in X1 and X2
in any short time interval?
Ref : John C Hull, Options, Futures and Other Derivatives
9
Solution
(a) Suppose that X1 and X2 equal a1 and a2 initially. After a time period of
length T, X 1 has the probability distribution

Ø (a1 + µ1T, σ1√T)

and X2 has a probability distribution

Ø (a2 + µ2T, σ2√T)
For independent normally distributed variables, mean and variance can
be added.
So .X1 + X2 has the probability distribution

 a1  1T  a2  2T ,  T   T
=

2
1
2
2

 a1  a2  ( 1   2 )T , ( 12T   22T )

S0 X1 + X2 follows a generalized wiener process with drift rate µ1+ µ2 and
variance rate , σ12 + σ22
10
(b) In this case the change in the value of X1 + X2 in a short interval of time
Δt has the probability distribution:


 ( 1  2 )t ,  12   22  2 1 2 )t

If µ1, µ2, σ1, σ2 and ρ are all constant, the change in a longer period of
time T is

 ( 1   2 )T ,  12   22  2  1 2 )T

The variable, X1 + X2, therefore follows a generalized Wiener process
with drift rate µ1+ µ2 and variance rate σ12 + σ22 + 2ρ σ1 σ2
11
Consider a variable S that follows the process: dS = µ dt + σ dz.
For the first four years, µ = 2 and σ = 4; for the next four years µ =
3 and σ =5.
If the initial value of the variable is 5, what is the probability
distribution of the value of the variable at the end of year 8?
The change in S during the first three years has the probability
distribution Ø(2x4, 4x√4) = Ø( 8, 8)
The change during the next three years has the probability
distribution Ø(4x3, 5x√4) = Ø (12, 10)
The change during the six years is the sum of a variable with
probability distribution Ø (8, 8) and a variable with probability
distribution Ø (12, 10).
 The probability distribution of the change is therefore:
 = Ø(8 +12, √(64 + 100) ) = Ø(20, 12.81 )
Since the initial value of the variable is 5, the probability distribution
at the end of the 8th year is Ø (25, 12.81)
12
Ito Process
The generalised Wiener process needs to be modified to make
it useful for modelling stock prices.
An Ito process is nothing but a generalized Wiener process.
Each of the parameters, a, b, in dx = a dt + bdz is a function of
the value of both the underlying variable x and time, t.
Earlier a was a function only of t.
 The expected drift rate and variance rate of an Ito process are
both liable to change over time.

Δ x = a (x, t) Δt + b (x, t) εΔt
13
Brownian Motion
In the coin tossing experiment, the expected winnings after
any number of tosses is just the amount we already hold.
This is called the Martingale property.
The quadratic variation of a random walk is defined by
[(S1 – S0)2 + (S2 – S1)2 + … + (Si – Si-1)2]
For each toss, the outcome is + $1 or - $1. So each of the
terms in the bracket will be (1)2 or (-1)2 i.e., exactly equal to 1.
Since there are i terms within the square bracket, the
quadratic variation is nothing but i.
14
Brownian Motion ( Contd)
Let us now advance the discussion by bringing in the time
element.
Suppose we have n tosses in the allowed time, t.
We define the game in such a way that each time we toss the
coin, we may gain or lose an amount of (t/n) .
Now each term in the small bracket is [(t/)n]2
or
t/n.
Since there are n tosses, the quadratic variation is (t/n) (n) = t.
Thus the expected value of the pay off is zero and that of the
variance is t.
The limiting process as time steps go to zero is called
Brownian motion.
15
Brownian Motion
Wilmott has summarized the properties of Brownian motion:
 The increment scales with the square root of the time step.
 The paths are continuous.
 The distribution follows the Markov property.
 The quadratic variation is t.
 Over finite time increments, ti-1 to ti, x(ti) – x (ti-1) is normally
distributed with mean zero and variance (ti – ti-1).
16
Geometric Brownian Motion
The most widely used model of stock price behaviour is given by
the equation:
ds/s =
μdt + dz
 is the volatility of the stock price
μ is the expected return.
This model is called Geometric Brownian motion.
The first term on the right is the expected return and the second is
the stochastic component.
The return on a stock price between now and a short period of time,
Δt later is normally distributed with mean μΔt and std devn,  Δt.
Over a long time, the return will be normally distributed with mean,
(μ - σ2/2) ( T) and standard deviation,  ΔT
17
Illustration
Suppose a stock has a volatility of 20% per annum and
provides an expected return of 15% per annum with
continuous compounding. If the time interval = 1 week =
.0192 years and the initial stock price is 50, what will be the
process for a short period of time?
 The process for the stock price can be written as:

ds/s =
.15dt + .20dz
or
Δs/s =
.15 Δt + .20 Δz
or
Δs/s =
.15 Δt + .20 ε Δt
.
Δs =

=
50 (.15 x .0192 + .20 ε .0192)
.144 + 1.3856 ε
18
Problem
Stock A and stock B both follow geometric Brownian motion.
Changes in any short interval of time are uncorrelated with each
other. Does the value of a portfolio consisting of one of stock A and
one of stock B follow geometric Brownian motion?
Let SA, SB, µA , µB and σA, σB be stock price, expected return and
volatility for stocks A, B respectively..
Then ΔSA = µA SA Δt + σASAεA√Δt and
σBSBεB√Δt
ΔSB = µB SB Δt +
ΔSA + ΔSB = (µA SA + µBSB) Δt + (σASAεA + σBSBεB)√Δt
This cannot be written as
ΔSA + ΔSB = µ (SA + SB) Δt + σ(SA + SB) ε√Δt
Hence the value of the portfolio does not follow GBM.
19
Ref : John C Hull, Options, Futures and Other Derivatives
Understanding Geometric Brownian Motion
To get a good intuitive understanding of Geometric Brownian
motion, we draw on the work of Neil A Chriss.
Consider a heavy particle suspended in a medium of light
particles.
These particles move around and crash into the heavy article.
 Each collision slightly displaces the heavy particle.
The direction and magnitude of this displacement is random.
It is independent of other collisions.
Each collision is an independent, identically distributed (i.i.d)
random event.
20
Geometric Brownian Motion (Contd)
The stock price is equivalent to the heavy article.
 Trades are equivalent to the light particles.
 We can expect stock prices to change in proportion to their size.
 As the returns we expect do not change with the stock prices.
Thus we would expect 20% return on Reliance shares whether
they are trading at Rs. 50 or Rs. 500.
 So the expected price change will depend on the current price
of the stock.
So we write: Δs =
S (µdt + dz).
Because we “scale” by S, it is called Geometric Brownian
Motion.
21
The Short run
In the short run, the return of the stock price is normally
distributed.
The mean of the distribution is µ∆t.
The std devn is Δt.
µ is the instantaneous expected return.
 is the instantaneous standard deviation.
22
The long run
In the long term, things are different.
 Let S be the stock price at time, t.
Let µ be the instantaneous mean.
Let  be the instantaneous standard deviation.
The return on S between now (time t) and future time, T is
normally distributed with a mean of (µ-2/2) (T-t) and std
devn of T-t.
 Why do we write (µ-2/2) and not µ?
What is the intuitive explanation?
23
Geometric Brownian Motion
We need to first understand that volatility tends to depress the
returns below what the short term returns suggest.
Expected returns reduce because volatility jumps do not cancel
themselves.
A 5% jump multiplies the current stock price by 1.05; A 5% fall
multiplies the amount by .95.
If a 5% jump is followed by a 5% fall or vice versa, the stock price
will reach 0.9975, not 1!
 In general, if a positive return x is followed by a negative return
x, the price will reach (1+x) (1-x) = 1- x2
How do we estimate the value of x?
Consider a random variable x. We can calculate the variance of
x as follows:
24
2 = E [x2] – {E[x]}2 = E [x2] (assuming E[x] = 0, ie., ups and
downs in x cancel out)
Thus the expected value of x2 is the variance.
Amount by which the returns are depressed when a positive
movement of x is followed by an equal negative movement is x2.
 For two moves, the depression is x2.
So we could say that the average depression per move is x2/2.
But the expected value of x2 is 2 .
So we can write 2 /2 as the expected value of the amount by
which the returns fall from the mean.
That is why we write (µ-2/2) and not µ.
25
A simple example to explain (µ-2/2)
Suppose the returns on a stock in 5 years are 10%, 15%,
20%, -15% and 25%.
Then the average return ( arithmetic mean) is 11%.
Expected returns = 1.115 = 1.685.
Actual returns = 1.1X1.15X1.2X0.85X1.25 =1.6129
Or Geometric mean = 1.1003
The geometric mean is less than the arithmetic mean.
26
Geometric Brownian Motion
 Can we make some prediction about the kind of distribution followed by the stock
price under the assumption of a Geometric Brownian Motion?
 Let us begin with the assumption that the stock returns are normally distributed.
 Annualised return from t0 to T =
1
S
ln T
T  t 0 S t0
 ST = future price St0 = current price, T-t0 is expressed in years.
 Annualised return
=
1
1
lnST 
lnSt0
T  t0
T  t0
 Let random variable X
=
1
1
lnST 
lnS t0
T  t0
T  t0
 Let us define a new random variable

X+
1
lnSt0
T  t0
 The second term of the expression is a constant. So the basic characteristics of
the distribution are not affected. Only the mean
changes.
1
lnSt0
T  t0
 Also
X+

or (T-t0) X + ln St0
=
=
1
lnST
T  t0
ln ST
27
Geometric Brownian Motion
 The mean return on S from time t to time T is (T-t) (r-2/2), while the std devn is
T-t
 The return on S from time t to T = ln ST /St
 The random variable
is normally distributed with mean = 0 and std devn = 1.
 Suppose a call option on the stock with strike price, K is in the money at expiration.
 We want to estimate the probability of the stock price exceeding the strike price.

ST ≥ K

ST/St ≥ K/St

ln (ST/St) ≥ ln (K/St)
ST
2
ln
 (T  t )( r 
)
St
2
 T t
≥
K
2
ln  (T  t )( r  )
St
2
 T t
28
Geometric Brownian Motion

St
2
ln
 (T  t )( r  )
ST
2
 T t
≤
 of both sides and noting that
St
2
ln  (T  t )( r 
)
K
2
 T t
ln
S
K
 ln t
ST
K
(Taking the negative
ln
S
ST
 ln t
St
ST
 The probability of the stock price exceeding the strike price can be written
as:
(lnSt / K  (T  t )( r   2 / 2)
]
 P (ST ≥K) = N [
( T  t )
 We will see that this is N(d2 ) in the Black Scholes formula.
 N(d2 ) is nothing but the probabiliy of the call option being in the money at
the time of expiration.
29
Geometric Brownian Motion : Concluding notes
 This expression reminds us of the Black Scholes formula!
 Indeed, GBM is central to Black Scholes pricing.
 GBM assumes stock returns are normally distributed.
 But empirical data reveals that large movements in stock price are more
likely than a normally distributed stock price model suggests.
 The likelihood of returns near the mean and of large returns is greater than
that predicted by GBM while other returns tend to be less likely.
 There is also evidence that monthly and quarterly volatilities are higher
than annual volatility.
 Daily volatilities are lower than annual volatilities.
 So stock returns do not scale as they are supposed to.
30
Problem
A stock price follows geometric Brownian motion with an
expected return of 16% and a volatility of 35%. The current
price is $38.
(a) What is the probability that a European call option on the
stock with an exercise price of $40 and a maturity date in 6
months will be exercised?
(b) What is the probability that a European put option on the
stock with the same exercise price and maturity will be
exercised?
Ref : John C Hull, Options, Futures and Other Derivatives
31
Solution
(a) The required probability is the probability of the stock price being above
$40 in six months’ time. Suppose that the stock price in six months is
ST.
lnST ~ Ø (ln38 + (0.16 – 0.352/2) 0.5, 0.35√0.5
i.e., lnST ~ Ø (3.687, 0.247)
Since ln40 = 3.689, the required probability is
 3.689  3.687 
1 N
  1  N (0.008)
0.247


From normal distribution tables N(0.008) = 0.5032 so that the required
probability is 0.4968.
(b) In this case the required probability is the probability of the stock price
being less than $40 in six months’ time. It is
1 – 0.4968 = 0.5032
32
Problem
 An exchange rate is currently 0.8000. The annualised volatility of the exchange
rate is quoted as 12% and interest rates in the two countries are the same. Using
the log normal assumption, estimate the probability that the exchange rate in 3
months will be
 (a) less than 0.7000,
 (b) between 0.7000 and 0.7500,
 (c) between 0.7500 and 0.8000,
 (d) between 0.8000 and 0.8500,
 (e) between 0.8500 and 0.9000, and
 (f) greater than 0.9000.
Based on the volatility smile usually observed in the
market for exchange rates, which of these estimates would you expect to be too
low and which would you expect to be too high?
Ref : John C Hull, Options, Futures and Other Derivatives
33
Solution
An exchange rate behaves like a stock that provides a dividend yield
equal to the foreign risk-free rate. Whereas the growth rate in a nondividend-paying stock in a risk-neutral world is r, the growth rate in the
exchange rate in a risk-neutral world is r – rf.
 In this case the foreign risk-free rate equals the domestic risk-free rate
(r=rf).
 The expected growth rate in the exchange rate is therefore zero. If ST is
the exchange rate at time T its probability distribution is given by :

lnST~Ø(lnS0 – σ2T/2, σ√T)
where S0 is the exchange rate at time zero and σ is the volatility of the
exchange rate. In this case S0 = 0.8000 and σ= 0.12 and T = 0.25 so that
lnST~Ø(ln0.8 – 0.122 X 0.25/2, 0.12√0.25)
or
lnST ~ Ø (-0.2249, 0.06)
34
Cont..
(a) In 0.70 = -03567. The probability that ST < 0.70 is the same as the
probability that lnST < -3567. It is

  0.3567  0.2249 
N

0.06


= N (-2.1955)
This is 1.41%
b) In 0.75 = -0.2877. The probability that ST < 0.75 is the same as the
probability that lnST < -0.2877. It is
  0.2877  0.2249 
N

0.06


= N (-1.0456)
This is 14.79%.
The probability that the exchange rate is between 0.70 and 0.75 is
therefore 14.79 – 1.41 = 13.38%.
35
Cont..
(c) In 0.80 = -0.2231. The probability that ST < 0.80 is the same as the
probability that lnST < -0.2231. It is
  0.2231  0.2249  = N (0.0300)
N

0.06


This is 51.2%. The probability that exchange rate is between .75 and .80
is 51.20 – 14.79 = 36.41%.
(d) In 0.85 = -0.1625. The probability that ST < 0.85 is the same as the
probability that lnST < -0.1625. It is
  0.1625  0.2249 
N

0.06


= N (1.0404)
This is 85.09%.
The probability that the exchange rate is between 0.80 and 0.85 is
therefore 85.09 – 51.20 = 33.89%
36
Cont..
(e) In 0.90 = -0.1054. The probability that ST < 0.90 is the same as the
probability that lnST < -0.1054. It is
  0.1054  0.2249 
N

0
.
06


= N (1.9931)
This is 97.69%. The probability that the exchange rate is between 0.85
and 0.90 is therefore 97.69 – 85.09 = 12.60%.
(f) The probability that the exchange rate is greater than 0.90 is
100 – 97.69 = 2.31%.
37
Ito’s lemma
 Let us move closer to the Black Scholes formula.
 Black and Scholes formulated a partial differential equation which they
later solved, with the help of Merton by setting up boundary conditions.
 To understand the basis for their differential equation, we need to
appreciate Ito’s lemma.
 Consider G, a function of x.
 The change in G for a small change is x can be written as:
 ΔG =
G
x
dx
 We can understand this intuitively by stating that the change in G is
nothing but the rate of change with respect to x multiplied by the change in
x.
 If we want a more precise estimate, we can use the Taylor series:
 ΔG =
dG
x +
dx
1 d 2G
(x) 2 
2
2 dx
1 d 3G
3
(
x
)
 .....
3
6 dx
Ref : John C Hull, Options, Futures and Other Derivatives
38
Ito’s lemma
 Now suppose G is a function of two variables, x and t.
 We will have to work with partial derivatives.
 This means we must differentiate with respect to one variable at a time, keeping
the other variable constant. We could write:
 ΔG =
G
G
x 
t
dx
dt
 Again, if we want to get a more accurate estimate, we could use the Taylor series:
 ΔG =
G
G
1  2G
 2G
1  2G
2
x 
t 
(x) 
(x)( t ) 
(t ) 2 
2
2
x
t
2 x
xdt
2 t
 Suppose we have a variable x that follows the Ito process.

dx = a (x,t) dt+ b(x,t) dz
 or
Δx = a(x,t) Δt + b(x,t)Δt
 or
Δx = a Δt + b Δt
  follows a standard normal distribution, with mean = 0 and standard deviation = 1.
39
Ito’s lemma
 We can write (Δx)2 = b22 Δt + other terms where the power of Δt is higher.
 If we ignore these terms assuming they are too small, we can write:

Δx2 = b22 Δt
 All the other terms have Δt with power 2 or more. They can be ignored. But Δx2
itself is big enough and cannot be ignored.
 Let us now go back to G and write:
G
G
1 G
2

x


t

(

x
)

ΔG = x
t
2 x 2
 But (Δx)2 = b22 Δt as we just saw a little earlier.
 It can be shown (beyond the scope of this coverage) that the expected value of 2
Δt is Δt, as Δt becomes very small.
 Thus
(Δx)2 = b2Δt
40
Ito’s lemma and GBM
 But dx = a(x,t) dt + b(x,t) dz
 So we can rewrite:
 dG
=

=
G
G
1  2G 2
(adt  bdz ) 
dt 
b dt
2
x
t
2 x
G G 1  2G 2
G
(a


b
)
dt

b
dz
x t 2 x 2
x
 This is called Ito’s lemma.
 It is very much a type of generalised Weiner process.
 Let us say the stock price is lognormally distributed.
 If G=ln S and x= S we get :
 ds = µS dt + S dz; a = µS and b = S
 dG=
{ µS /S + 0 - ½[-1/S2] 2 S2 }dt + [S/S ] dz
 dG=
(µ - 2 /2) dt + dz
41
Problem
Suppose that a stock price S follows geometric Brownian motion
with expected return µ and volatility σ: dS = µSdt + σSdz. What is
the process followed by the variable Sn?
If G(S,t) = Sn then G/S = nSn-1, and 2G/S2 = n(n-1)Sn-2.
 Using Ito’s lemma:
dG = [µnG+ ½ n(n-1) σ2G]dt + σnGdz
This shows that G = Sn follows geometric Brownian motion where the
expected return is
µn+ ½ n(n-1) σ2 and the volatility is nσ.
The stock price S has an expected return of µ and the expected value of
ST is S0eµT.
The expected value of SnT is
1
[ n  n ( n1) 2 ]T
n
2
0
S e
Ref : John C Hull, Options, Futures and Other Derivatives
42
Ito’s lemma and Black Scholes
 The Ito’s lemma is very useful when it comes to framing the Black Scholes
differential equation.
 Let us assume that the stock price follows Geometric Brownian motion,
i.e.,
ΔS = μSΔt + SΔz
 Let f be the price of a call option written on the stock whose price is
modeled as S.
 f is a function of S and t.
 or ΔS = a (S,t) dt +b (S,t) dS.
 Applying Ito’s lemma, we can relate the change in f to the change in S .
 Comparing with the general expression for Ito’s Lemma, we get:
 G = f , a = μS and b = S, x = S,

or
 f
f 1  2 f 2 2 
f
f   s  

S

t

Sz

2

s

t
2

S

s


43
The Black Scholes differential equation
 Our aim is to create a risk free portfolio whose value does not depend on the S, the
stochastic variable.
 Suppose we create a portfolio with a long position of
one call option.
f
s
shares and a short position of
 The value of the portfolio will be

 = -f +fs
S
 (Value means the net positive investment made. So a purchase gets a plus sign and a
short sale gets a negative sign.)
f
 We will see later that s is nothing but “delta” and the technique used to create a risk
free portfolio is called delta hedging.
 Change in the value of the portfolio will be:
 Δ
=
- Δf +
f
s

=
-
 f
f 1  2 f 2 2 
f
f
 s  


S

t


S

z

s
2


s

t
2

s

s

s


Δs
44
The Black Scholes differential equation
 But Δs = μSΔt+SΔz
 or Δ =

=
-
f
f
1 2 f 2 2
f
f
st  t 

s

t


s

z

( st  sz )
s
t
2 s 2
s
s
-
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
 or Δ =
-

-
=
f
1 2 f 2 2
t 
 s t
t
2 s 2
 f 1  2 f 2 2 
 
 s t
2

t
2

s


 This equation does not have a Δs term.
 It is a riskless portfolio, with the stochastic or risky component having been
eliminated.
 The total return depends only on the time. That means the return on the portfolio is
the same as that on other short term risk free securities. Otherwise, arbitrage would
be possible.
 So we could write the change in value of the portfolio as:
 Δ = r  Δt where r is the risk free rate. (Because this is a risk free portfolio)
45
The Black Scholes differential equation
 But  = -f +
 and
 or  or
 or
f
s
s
f 1  2 f 2 2
  ( 
 s )t
t 2 s 2
 f 1  2 f 2 2 
 
 s t =
2

t
2

s


2
f 1  f 2 2
f
rf  
 s r s
2
t 2 s
s
f

r  f 
s


s t

f
f 1 2 2  2 f
rf 
 rs   s
t
s 2
s 2
 This is the Black Scholes differential equation.
 The portfolio used in deriving the Black Scholes differential equation is riskless
only for a very short period of time when f is constant.
s
 With change in stock price and passage of time,
f
s
can change.
 So the portfolio will have to be continuously rebalanced to achieve what is called a
perfectly hedged or zero delta position.
 This is also called dynamic hedging.
 Solving the equation with appropriate boundary conditions gives us the black
Scholes formula.
46
The Black Scholes formula
Let C be the value of the call, P that of the put, K the strike price
C
= S0 N(d1) – Ke-rT N(d2)
d1
= [ln(S0/k) + (r+2/2)T] / T
d2
= [ln(S0/k) + (r-2/2)T] / T = d1 - T
As per put call parity,
C – P=
S0 – Ke-rT
or P =
C – S0 + Ke-rT

=
S0N(d1) – Ke-rT N(d2) – S0 + Ke-rT

=
Ke-rT [1 – N(d2)] + S0 [N (d1) – 1]

=
Ke-rT N(-d2) – S0 N(-d1)
Problem
What is the price of a European call option on a non-dividendpaying stock when the stock price is $52, the strike price is $50,
the risk-free interest rate is 12% per annum, the volatility is 30%
per annum, and the time to maturity is 3 months?
In this case S0 = 52, K = 50, r = 0.12, σ = 0.30 and T = 0.25
ln( 52 / 50)  (0.12  0.302 / 2)0.25
d1 
 0.5365
0.30 0.25
d 2  d1  0.30 0.25  0.3865
The price of the European call is

52N(0.5365) – 50e-0.12x0.25N(0.3865)
=
52 x 0.7042 – 50e-0.03 x 0.6504
=
$ 5.06
48
Problem
A call has 91 days until it expires.The risk-free interest rate is 8
percent/year. The strike price is 60.The stock price is 64.The
standard deviation of the stock’s monthly return is 0.144.
Compute the value of the call. What is the delta of the call?
 The volatility of the stock’s annual returns is .144 x√12 .=.4988. t = 91/365 =
0249315.
ln( S / K )  (r   2 / 2)T ln( 64 / 60)  (0.08  12 0.248832)0.249315
d1 

 T
0.49833 0.249315
= 0.46374
N(d1) = 0.6786
d2 = d1 - σ√T = 0.46374 – 0.49883√0.249315 = 0.46374 – 0.2491 = 0.2146
N(d2) = 0.5850
Ke-rT = (60) (e-0.019945) = (60) (0.98025) = 58.8151
C = SN(d1) – Ke-rTN(d2) = 64(0.6786) – 58.8151 (0.5850) =
9.0235
Options And Financial Futures: by Dubofsky
49
Problem
Use Black Scholes to value the following call option:
Stock price =$200, Strike price=$210,
Time to expiration =156 days, Risk-free interest rate = 11%
Variance of monthly stock returns = 0.02
S
=
200, K = 210, T = 156/365 = 0.4274 year
r
=
0.11, σ= √[12x0.02] =
0.489898 / year
ln( S / K )  (r   2 / 2)T ln( 200 / 210)  (0.11  0.12)0.4274
d1 

0.320275
 T
= 0.1546
50
Options And Financial Futures: by Dubofsky
Solution
N(d1) = 0.5614
d2 = d1 - σ√T = 0.1546 – 0.3203 = - 0.1657
N(d2) = 0.4342
Ke-rT = 210e-(0.11)(0.4274) = (210)e-0.0470 =(210)(0.9541)

=
200.3556
C = SN(d1) - Ke-rTN(d2)

= (200) (0.5614) – (200.3556) (0.4342)

=
25.2902
The equivalent portfolio consists of long 0.5614 shares of
stock and borrowing $200.3556.
51
Consider an American call option on a stock. The stock price is $50, the time to
maturity is 15 months, the risk-free rate of interest is 8% per annum, the exercise
price is $55, and the volatility is 25%. Dividends of $1.50 are expected in 4
months and 10 months. Calculate the price of the option.
The present value of the dividends is
1.5e-0.3333x0.08 + 1.5e-0.8333x0.08 = 2.864
The option can be valued using the European pricing formula with:
S0 = 50 – 2.864 = 47.136, K = 55, σ = 0.25, r = 0.08, T = 1.25
 d1 = [{ln(47.136/55) + ( .08 + .0625/2)1.25}]/[.25√1.25] = -.0545
 d2 = d1 – 0.25 √1.25 = - 0.3340
 N(d1) = 0.4783,
N (d2) = 0.3692
 and the call price is 47.136 x 0.4783 – 55e-0.08x1.25 x 0.3692 = 4.17

John C Hull, Options, Futures and Other Derivatives
52
Problem
A company can buy an option for the delivery of 1 million barrels of
oil in 3 years at $25 per barrel. The 3-year futures price of oil is
$24 per barrel. The risk-free interest rate is 5% per annum with
continuous compounding and the volatility of the futures price is
20% per annum. How much is the option worth?.
The option can be valued using Black’s model. We use futures/forward
price, F0 instead of spot price, S0 of underlying.
F0 = 24, K = 25, r = 0.05, σ = 0.2, and T = 3. The value of a option to
purchase one barrel of oil at $25 is
2
ln( F0 / K )   2T / 2
ln(
F
/
K
)


T /2
0
d2 
e-rT [F0N(d1) – KN(d2)], d1 
 T
 T
d1 = 0.0554, d2 = -.2910, N(d1) = .52209
N(d2) = .38553
F0N(d1) – KN(d2) = 2.891, e-rT = 0.86
 C =0.86 X 2.891
= 2.489
The value of the option to purchase one million barrels is $2,489,000.
53
John C Hull, Options, Futures and Other Derivatives
Problem
Use the Black’s model to value a 1-year European put option on a
10-year bond. Assume that the current value of the bond is $125, the
strike price is $110, the 1-year interest rate is 10% per annum, the
bond’s forward price volatility is 8% per annum, and the present value
of the coupons to be paid during the life of the option is $10.
In this case, F0 = (125 – 10)e0.1x1 = 127.09, K = 110, σ= 0.08, and T = 1.0
 d1 = {[ln(127.09/110) + .0064/2]}/.08 = 1.8456
d2 = d1 – 0.08 = 1.7656
The value of the option is

110e-0.1x1 N (-1.7656) – 115N (-1.8456) = 0.12
Or
$0.12
John C Hull, Options, Futures and Other Derivatives
54
Problem
 Calculate the value of a 4-year European call option on a bond that will mature 5
years from today using Black’s model. The 5-year cash bond price is $105, the
cash price of a 4-year bond with the same coupon is $102, the strike price is
$100, the 4-year risk-free interest rate is 10% per annum with continuous
compounding, and the volatility for the bond price in 4 years is 2% per annum.
 We use Black’s formula.
 The present value of the principal in the four year bond is 100e-4x0.1 = 67.032.
 The present value of the coupons is, therefore, 102 – 67.032
 So forward price of the five-year bond is : (105 – 34.968)e4x0.1
= 34.968.
= 104.475
 F0 = 104.475, K = 100, r = 0.1, T = 4, and σ = 0.02.
 d1 = ln [(104.475/100) + ..0004X4/2]/ [.02x √4] = 1.1144
 d2 = d1 – 0.02√4 = 1.0744
 Price of the European call is e-0.1x4[104.475N(1.1144) – 100N(1.0744)] = 3.19
John C Hull, Options, Futures and Other Derivatives
55
Problem
A company’s stock price is $50. The company is considering giving
its employees at-the-money 5-year call options. The stock price
volatility is 25%, the 5-year risk-free rate is 5% and the company
does not pay dividends. Calculate the value of the option.
d1 = { ln(50/50) + [.05+.0625 x 5 /2]} /[.25X√5] = .3690,
N(d1 ) = .6439
d2 = .3690- . 25X√5 = -.1900,
N(d2 )= .4247
e-rt = .7789, r = .05, t =5.
C = 50 x .6439- 50 x.7789x .4247 = 15.66
56
Problem
A company’s stock price is $50 and 10 million shares are
outstanding. The company is considering giving its employees 3
million at-the-money 5-year call options. Option exercises will be
handled by issuing more shares. Estimate the cost to the company
of the employee stock option issue. Asssume the value of an option
is $ 15.66.
N = No. of existing shares, M, the no. of new options
 The cost to the company of the option is[ NS + MK]/[N+M] - K
 = [Nx (S-K)]/(N+M) where, N=10, M=3, S-K = option value = 15.66
= 10/[10+3} X 15.66 = $ 12.05 per option.
 The total cost is therefore 3 million times this or $36.15 million.
57
John C Hull, Options, Futures and Other Derivatives