6.4 Geometric Brownian Motion.
When we model a stock price X(t) using Brownian motion we have to take into account the fact that the
magnitude of a stock's price has an effect on how it changes. For example a 25¢ increase in the price of a
stock whose current price is $5.00 is similar to a $2.50 increase in the price of a stock whose current price is
$50.00. In order to take this into account we model the logarithm Y(t) of the stock's price as a Brownian
motion rather than the price itself. When we do this we say the stock price itself is being modeled by a
geometric Brownian motion. The Brownian motion parameters  and  for Y(t) are called the drift and
volatility of the stock price.
Example 1. Let X(t) be the price of FMC stock at time t years from the present. Assume that X(t) is a
geometric Brownian motion with drift  = – 0.05 / yr and volatility  = 0.4 / yr1/2. If the current price of
FMC stock is $2.50, what is the probability that the price will be at least $2.60 one year from now.
Since X(t) is a geometric Brownian motion, log(X(t)) is a regular Brownian motion with drift rate
 = - 0.05 / yr and  = 0.4 / yr1/2. We want to know the probablity that log(X(1))  log(2.60) given that
log(X(0))  log(2.50). This means
2.60
log(X(1)) - log(X(0))  log(2.60) - log(2.50) = log
2.50 = log(1.04)  0.0392
In this case Z = (log(X(1)) - log(X(0)) – t)/( t) = (log(X(1)) - log(X(0)) - (- 0.05)(1))/(0.4)( 1)) =
(log(X(1)) - log(X(0)) +0.05)/0.4 is a standard normal random variable. So
Pr{log(X(1)) - log(X(0)) > 0.0392} = Pr{log(X(1)) - log(X(0)) + 0.05 > 0.0392 + 0.05}
= Pr{
log(X(1)) - log(X(0)) + 0.05 0.0392 + 0.05
>
} = Pr{ Z > 0.223}
0.4
0.4
= 1 - Pr{ Z  0.293} = 1 - (0.293) = 1 – 0.5884 = 0.4116  41%
z
1
 e-u2/2 du is the distribution function of a standard normal random variable.
Here (z) = Pr{Z  z} =
2 
-
Example 2. Let X(t) be the price of JetCo stock at time t years from the present. Assume that
X(t) is a geometric Brownian motion with zero drift and volatility  = 0.4 / yr1/2. If the current
price of JetCo stock is $8.00, what is the probability that the price will be at least $8.40 six
months from now.
Since X(t) is a geometric Brownian motion, log(X(t)) is a regular Brownian motion with zero
drift and  = 0.4 / yr1/2. We want to know the probablity that log(X(1/2))  log(8.40) given
that log(X(0))  log(8.00). This means
8.40
log(X(1/2)) - log(X(0))  log(8.40) - log(8.00) = log
8.00 = log(1.05)  0.0488
In this case Z = (log(X(1/2)) - log(X(0)) – (0)(1/2))/(0.4 0.5) =
(log(X(1/2)) - log(X(0)))/0.283 is a standard normal random variable. So
6.4.1 - 1
Pr{log(X(1/2)) - log(X(0)) > 0.0488} = Pr{
log(X(1/2)) - log(X(0)) 0.0488
>
}
0.283
0.283
= Pr{ Z > 0.172} = 1 - Pr{ Z  0.172} = 1 - (0.172) = 1 – 0.568 = 0.432  43%
Example 3. Redo Example 2 if the drift rate is – 0.02 / yr instead of zero.
Now Z = (log(X(1/2)) - log(X(0)) – (- 0.02)(1/2))/(0.4 0.5) =
(log(X(1/2)) - log(X(0)) + 0.01)/0.283 is a standard normal random variable. So
log(X(1/2)) - log(X(0)) + 0.01 0.0488 + 0.01
Pr{log(X(1/2))-log(X(0)) > 0.0488} = Pr{
>
}
0.283
0.283
= Pr{ Z > 0.21} = 1 - Pr{ Z  0.21} = 1 - (0.21) = 1 – 0.5832 = 0.4168  42%
Expected Values: Geometric Brownian motion has a little quirk, namely its expected value is higher than
one might think at first. If X(t) is a regular Brownian motion with zero drift then E{X(t) – X(0)} = 0 for all
t. However, this is not true for geometric Brownian motion with zero drift.
Proposition 1. Suppose X(t) is a geometric Brownian motion with drift  and volatility . Then
E{X(t)} = E{X(0)}e( + 
2/2)t
.
Proof. Since X(t) is a geometric Brownian motion with the given drift and volatility, Y(t) = log(X(t)) is a
regular Brownian motion and Y(t) – Y(0) =  tZ(t) + t where Z(t) is a standard normal random variable.
1/2Z(t) + t
So X(t) = eY(t) = eY(t) – Y(0) + Y(0) = eY(0)eY(t) – Y(0) = X(0)et
t
independent of Y(0) one has E{X(t)} = E{X(0)}e E{e
t1/2Z(t)
1/2Z(t)
= X(0)etet
. Since Y(t) – Y(0) is
}. In order to complete the proof it suffices to
2
show that if Z is a standard normal random variable then E{eZ} = e /2. One has



z  1  -z2/2
 1   e-(z2 - 2)/2 dz =  1  
 e-(z2 - 2 + 2 - 2)/2 dz
E{e } = 
 e  2 e dz =  2 
 2 -

-
Z
-

= e
2/2

1 
2
2
2
2
 1 

e-(z - ) /2 dz= e /2 
e-u /2 dz = e /2


 2 -
 2 -
Example 4. Let X(t) be the price of FMC stock at time t years from the present. Assume that X(t) is a
geometric Brownian motion with drift  = – 0.05 / yr and volatility  = 0.4 / yr1/2. If the current price of
FMC stock is $2.50, what is the expected price six months from now. By Proposition 1 one has
E{X(t)} = E{X(0)}e( + 
2/2)t
= (2.5)e(- 0.05 + (0.4)
2/2)(1/2)
= (2.5)e(- 0.05 + 0.08)(1/2)
= (2.5)e0.015 = (2.5)(1.0151) = $2.538
Interest. There is one more factor that enters into the pricing of stock options. This is the effect of interest.
An alternative to investing in the stock market is to put one's money in an investment that is guaranteed to
pay a certain amount of interest such as a certificate of deposit. In certain situations putting one's money in
a guaranteed investment might by better than investing in stocks or stock options. Let's suppose
6.4.1 - 2
r = interest rate on guaranteed investments.
So an investment of M grows
Mt = M(1 + r)t
= amount M grows to after a time t.
Example 1. How much does $8.00 grow to after two years if the annual interest rate is 1%?
In this case M = $8.00, r = 0.01 and t = 2. After two years this grows to M2 = M(1 + r)t = (8)(1 + 0.01)2 =
(8)(1.01)2 = (8)(1.0201) = $8.1608.
When one is modeling investments in continuous time it is convenient to express the factor (1 + r)t in terms
of e. Let
 = log(1 + r) = instantaneous interest rate
so
(1 + r) = e
(1 + r)t = et
Mt = M(1 + r)t = Met = amount an investment of M grows to in t years
Example 2. What is the instantaneous interest rate if the regular interest rate is 1% per year?
In this case r = 0.01 so  = log(1 + r) = log(1 + 0.01) = log(1.01) = 0.00995 / yr.
Example 3. How much does $8.00 grow to after two years if the instantaneous annual interest rate is 1%?
In this case M = $8.00,  = 0.01 and t = 2. After two years this grows to M2 = Met = (8)e(0.01)(2) = (8)e0.02 =
(8)(1.0202) = $8.1616.
One effect of interest is that a future payment of a certain amount is not same as a payment of the same
amount right now. If we are to get a payment of an amount v at a time t in the future then this is the same as
a payment of M now where Met = v. So M = ve-t. Thus a payment of an amount v at a time t in the future
is the same as a payment of ve-t now. This is called the present value of the future payment and e-t is
called the discount factor, because it is the amount funture payments are discounted to get their present
value. So
vt = ve-t = present value of a payment of v at a future time v
e-t = discount factor of future payments at time t
Example 4. What is the present value of a payment of $8.00 to be made in two years if the instantaneous
annual interest rate is 1%?
6.4.1 - 3
In this case M = $8.00,  = 0.01 and t = 2. The present value is v2 = vet = (8)e-(0.01)(2) = (8)e-0.02 =
(8)(0.980199) = $7.84159.
In the previous section we saw that the expected value of a geometric Brownian motion at time t was
E{X(t)} = E{X(0)}e( + 
2/2)t
.
In order to get the present value of this we need to multiply by the discount factor e-t. So
E{e-tX(t)} = E{X(0)}e( + 
2/2 - )t
= present value of a geometric Brownian motion at time t
Example 4. Let X(t) be the price of FMC stock at time t years from the present. Assume that X(t) is a
geometric Brownian motion with drift  = – 0.05 / yr and volatility  = 0.4 / yr1/2. If the current price of
FMC stock is $2.50, what is the expected present value of its price six months from now if the annual
interest rate is 5%.
The instanteous interest rate is  = log(1.05) = 0.0487. Using the above formula one has
2
 + 2 -  = - 0.05 + (0.4)2/2 – 0.0487 = 0.08 – 0.0987 = - 0.0187
E{X(t)} = E{X(0)}e( + 
2/2 - )t
= (2.5)e(- 0.0187)(1/2) = (2.5)e(- 0.00935
= (2.5)(0.990694) = $2.47673
Arbitrage. If the value of the price of a certain stock follows a geometric Brownian motion and x0 is its
current price then by the above formula the present value of its expected price at time t is x0e( + 
2/2 - )t
.
Whether or not it is better to invest in the stock or put one's money in a guaranteed investment depends on
the factor  +
2
2
- .
2
 + 2 - > 0
Stock is better
2
 + 2 - < 0
Guaranteed investment is better
2
 + 2 - = 0
Stock and guaranteed investment are equal
Some people believe the stock's drift rate adjusts so that investing in the stock has the same expected
outcome as a guaranteed investment, i.e. so
2
 + 2 - = 0
When this happens an arbitrage is said to occur. We shall assume this happens below.
6.4.1 - 4
Problem 1. Suppose the current price of DTE stock is $49, the price has a volatility of
 = 0.15 / yr1/2 and the current interest rate is  = 0.05 / yr. Assume that arbitrage is working so
the drift  is  =  - 2/2.
a.
What is the probability that the stock price will be at least $50 in 3 months?
b.
What is the expected stock price in 3 months?
6.4.1 - 5