# 7.0 Ito's Lemma(1) (1) ```7.0
Brownian Motion,
Itô’s Lemma and
Differentials
Aimi Syairah
7.1 Brownian Motion
 Brownian Motion
 Arithmetic Brownian Motion
 Geometric Brownian Motion
7.1.1 Introduction
 We will now study the theoretical background for BlackScholes pricing. In order to price options, we need:
1. A model for the price movement of the underlying asset
2. A way to calculate the price movement of a claim on the
asset as a function of the price movement of the asset
• Brownian motion is a model for price movements
• Itô’s lemma is a way to relate changes in values of functions
of assets to changes in values of assets
7.1.2 Discrete Random Walks
 A one-dimensional discrete random walk models somebody
starting out on the x axis at the point 0 and then moving left
or right at the rate of 1 per unit time, with the direction being
chosen randomly at every point
 Let 𝑋(𝑡) be the position at time 𝑡. For a random walk:
1. 𝑋(0) = 0
2. For 𝑡 > 0, if 𝑋 𝑡 − 1 = 𝑘, then
𝑘 + 1 with probability 1/2
𝑋 𝑡 =
𝑘 − 1 with probability 1/2
7.1.2 Discrete Random Walks
 To calculate the probability that 𝑋 3 = 1, we note that
for 𝑋 to move 1 at time 3, it would have to move up
twice and down once from time 0
 There are 23 = 8 possibilities for 3 movements and of
these 8, 3 of them would end up at 1, since we would
need 2 ups and 1 down
 Hence Pr 𝑋 3 = 1 =
3
8
7.1.2 Discrete Random Walks
 Suppose that instead of moving 1 per unit of time, we
moved ℎ per ℎ units of time and took the limit as ℎ →
0. We then have a continuous random walk
 The binomial random variable will converge to a normal
random variable and hence become Brownian motion
7.1.3 Brownian Motion
 Brownian motion is denoted by 𝑍(𝑡)
 It is a random process, a collection of random variables
indexed by time 𝑡, defined by the following properties:
1. 𝑍 0 = 0
2. 𝑍 𝑡 + 𝑠 |𝑍 𝑡 has a normal distribution with 𝜇 =
𝑍 𝑡 and 𝜎 2 = 𝑠
3. Increments are independent: 𝑍 𝑡 + 𝑠1 − 𝑍(𝑡)
independent of 𝑍 𝑡 − 𝑍(𝑡 − 𝑠2 )
4. 𝑍(𝑡) is continuous in 𝑡
is
Example 7.1.3
 The price of a stock follows a Brownian motion. The
price of the stock at time 3 is 52. Determine the
probability that the price of the stock is at least 55 at
time 12
Solution 7.1.3
 Pr 𝑍 12 > 55 𝑍 3 = 52 = Pr(𝑌 > 55)
 𝑍 12 |𝑍(3) has a normal distribution with mean
𝑍 3 = 52 and variance = 9
 Hence,
0.1587
Pr 𝑌 > 55 = 1 − 𝑁
55−52
9
=1−𝑁 1 =
7.1.4 Arithmetic Brownian Motion
 Arithmetic Brownian motion consists of a Brownian
motion scaled by multiplication and shifted by addition
 If 𝑋(𝑡) is an arithmetic Brownian motion (where 𝑍 𝑡 is
a Brownian motion), then
𝑋 𝑡 = 𝑋 0 + 𝛼𝑡 + 𝜎𝑍(𝑡)
 𝑋 𝑡 + 𝑠 − 𝑋(𝑡) has a normal distribution with mean 𝛼𝑠
and variance 𝜎 2 𝑠
 The parameter 𝛼 is called the drift of the process
7.1.4 Arithmetic Brownian Motion
 For the distribution of an arithmetic Brownian motion
𝑋 𝑢 :
 Let 𝑡 be the latest time for which we have the value 𝑋(𝑡)
 𝑋 𝑡 + 𝑠 |𝑋 𝑡 , 𝑠 > 0 is normally distributed with mean
𝑋 𝑡 + 𝛼𝑠 and variance 𝜎 2 𝑠
Example 7.1.4
 The price of a stock follows arithmetic Brownian motion of
the form 𝑋 𝑡 = 𝑋 0 + 𝑡 + 0.2𝑍(𝑡). The current price of
the stock is 40. Determine the probability that the price of
the stock at time 4 is less than 43
Solution 7.1.4
 Given 𝑋 𝑡 = 𝑋 0 + 𝑡 + 0.2𝑍(𝑡) or 𝛼 = 1, 𝜎 = 0.2
 We want to evaluate the distribution at time 𝑠 = 4
 𝑋 4 − 𝑋(0) is a normal random variable with mean
𝛼𝑠 = 1 4 = 4 and variance 𝜎 2 𝑠 = 0.22 (4)
 We want the probability that it is less than 43 −
40 = 3
 Then,
3−4
Pr 𝑋 4 − 𝑋 0 < 3 = 𝑁
= 𝑁(−2.5)
0.2 4
7.1.5 Geometric Brownian Motion
 Arithmetic Brownian motion is not a good model for
stock price movement because it can go negative and
does not scale with stock price
 Just like we transform a normal distribution to a
lognormal distribution, we transform arithmetic
Brownian motion to geometric Brownian motion
7.1.5 Geometric Brownian Motion
 𝑋(𝑡) follows geometric Brownian motion if ln 𝑋(𝑡)
follows arithmetic Brownian motion
 If ln 𝑋 𝑡 is a normal random variable, then 𝑋(𝑡) is a
lognormal random variable
 If ln 𝑋𝑡 𝑋0 is normal with mean 𝜇𝑡 and variance 𝜎 2 𝑡,
then 𝑋𝑡 𝑋0 is lognormal, and its mean and variance are
𝐸[𝑋(𝑡) 𝑋 0 ] =
𝑉𝑎𝑟[𝑋(𝑡) 𝑋(0)] =
2𝑡
𝜇𝑡+0.5𝜎
𝑒
2𝑡
2𝑡
2𝜇𝑡+𝜎
𝜎
𝑒
(𝑒
− 1)
7.1.5 Geometric Brownian Motion
 Let 𝑆 𝑡 be the time- 𝑡 stock’s price. Suppose the
volatility of the stock is 𝜎, hence
𝑉𝑎𝑟[ln 𝑆 𝑡 𝑆 0 = 𝜎 2 𝑡
 Suppose the stock pays no dividends, then the only
return from the stock is its growth in price
 Suppose 𝐸 𝑆 𝑡 = 𝑆(0)𝑒 𝛼𝑡 , i.e., the continuously
compounded expected rate of return on the stock is 𝛼
7.1.5 Geometric Brownian Motion
 Now assume that the stock pays dividends at a
continuously compounded rate of 𝛿
 There are two sources of earnings on the stock: the
dividends, which return 𝛿 and the growth in price
 The sum of these returns is 𝛼, so the rate at which the
stock price increases is 𝛼 − 𝛿
 The expected stock price increase is
𝐸𝑆 𝑡
=𝑆 0 𝑒
𝛼−𝛿 𝑡
𝐸[𝑆(𝑡) 𝑆(0)] = 𝑒
and
𝛼−𝛿 𝑡
7.1.5 Geometric Brownian Motion
 But, for a lognormal distribution, the expected value is
(𝜇+0.5𝜎 2 )𝑡
𝐸[𝑋 𝑡 𝑋(0)] = 𝑒
 It follows that α − 𝛿 = 𝜇 + 0.5𝜎 2
 In other words, to go from a geometric Brownian motion
to the associated arithmetic Brownian motion, we
subtract 0.5𝜎 2
 We must go from the geometric Brownian motion to the
associated arithmetic Brownian motion if we wish to use
the normal tables to look up probabilities or percentiles
7.1.5 Geometric Brownian Motion
 If we want to calculate probabilities or percentiles for a
stock whose return is 𝛼 and pays dividends 𝛿 with
volatility 𝜎, set
𝑚 = 𝜇𝑡 = (𝛼 − 𝛿 − 0.5𝜎 2 )𝑡 and 𝑣 = 𝜎 𝑡
and use those parameters to look up the normal
distribution in the table
Example 7.1.5
 The time 𝑡 price of a stock is 𝑆(𝑡). The stock price is
modeled as following geometric Brownian motion. Your
are given:
 The stock’s continuously compounded expected rate of
return is 0.15
 The stock’s continuously compounded dividend yield is
0.04
 The stock’s volatility is 0.3
 𝑆(0) = 45
 𝑆(0.6) = 47
Calculate Pr(𝑆 1 < 45) given the above facts
Solution 7.1.5
The price of the stock at time 0 is irrelevant
𝜎 = 0.3
𝑆(1)
ln
𝑆 0.6
is normally distributed with parameters
𝑚 = 𝛼 − 𝛿 − 0.5𝜎 2 𝑡 = (0.15 − 0.04 − 0.5 0.32 1 − 0.6 = 0.026
and 𝑣 2 = 𝜎 2 𝑡 = 0.32 1 − 0.6 = 0.036
ln 𝑆(1) 𝑆(0) − 𝑚
Pr( 𝑆 1 < 45) = 𝑁
𝑣
=𝑁
ln 45 47 − 0.026
0.036
= 𝑁 −0.3662
7.2 Differentials and Itô’s Lemma
 Stochastic Differential Equations
 Itô’s Lemma
 Martingales
7.2.1 Differentials
 In ordinary calculus:
𝑦 = 𝑒 𝐽𝑡
𝑑𝑦
= 𝐽𝑒 𝐽𝑡
𝑑𝑡
 Multiplying both sides by 𝑑𝑡 gives us a differential expression:
𝑑𝑦 = 𝐽𝑒 𝐽𝑡 𝑑𝑡 = 𝐽𝑦𝑑𝑡
 The last expression says that the change of 𝑦 is proportional
to 𝐽𝑦 times the change of 𝑡
7.2.1 Differentials
 Suppose there were some uncertainty in this rate of change.
We could add 𝜎𝑑𝑍(𝑡) to the right hand side and get
𝑑𝑦
= 𝐽𝑑𝑡 + 𝜎𝑑𝑍(𝑡)
𝑦
 𝑑𝑍(𝑡) is a differential of a Brownian motion.

It is the limit, as ℎ goes to 0, of a random variable equal to ℎ
with probability 0.5 and – ℎ with probability 0.5
 𝑡 will represent time
 𝑍(𝑡) is independent of 𝑡
7.2.1 Differentials
 The value at time 𝑡 of an arithmetic Brownian motion
𝑋(𝑡) with drift 𝜇 and coefficient of 𝑍(𝑡) equal to 𝜎 can
be expressed as:
𝑋 𝑡 = 𝑋 0 + 𝜇𝑡 + 𝜎𝑍(𝑡)
 The differential of this arithmetic Brownian motion
𝑋(𝑡) is
𝑑𝑋 𝑡 = 𝜇𝑑𝑡 + 𝜎𝑑𝑍(𝑡)
 This says that a small change in 𝑋 equals
 𝜇 times a small change in time, or 𝑑𝑡, plus
 𝜎 times a small change in a Brownian motion, or 𝑑𝑍(𝑡)
7.2.1 Differentials
 The Brownian motion itself, 𝑍(𝑡), is a r.v. normally
distributed with mean 0 and variance 𝑡
 A small change in a Brownian motion is a r.v. normally
distributed with mean 0 and variance 𝑑𝑡
 The time value of a GBM 𝑋(𝑡) can be expressed in terms
of its logarithm
ln 𝑋(𝑡) = ln 𝑋(0) + 𝛼 − 𝛿 − 0.5𝜎 2 𝑡 + 𝜎𝑍(𝑡)
(1)
 The differential of this expression is
𝑑(ln 𝑋(𝑡)) = 𝛼 − 𝛿 − 0.5𝜎 2 𝑑𝑡 + 𝜎𝑑𝑍(𝑡)
(2)
7.2.1 Differentials
 The same GBM can be expressed directly as
𝛼−𝛿−0.5𝜎 2 𝑡+𝜎𝑍(𝑡)
𝑋 𝑡 = 𝑋(0)𝑒
(3)
 The differential of this GBM 𝑋(𝑡) is
𝑑𝑋 𝑡 = 𝛼 − 𝛿 𝑋 𝑡 𝑑𝑡 + 𝜎𝑋 𝑡 𝑑𝑍(𝑡)
 If 𝑋(𝑡) satisfies any of equation (1) through (3), then
𝑋(𝑡) follows a GBM with coefficients 𝛼 − 𝛿 𝑋 𝑡 for 𝑑𝑡
and 𝜎𝑋 𝑡 for 𝑑𝑍(𝑡)
7.2.1 Differentials
 Any process of the form
𝑑𝑆 𝑡 = (𝛼 − 𝛿) 𝑆 𝑡 , 𝑡 𝑑𝑡 + 𝜎 𝑆 𝑡 , 𝑡 𝑑𝑍(𝑡)
where (𝛼 − 𝛿) 𝑆, 𝑡 and 𝜎 𝑆, 𝑡 are functions of 𝑆 and 𝑡
is called Itô process
 The coefficient of 𝑑𝑡 is called the drift process
Example 7.2.1A
You are given an Itô process of the form
𝑑𝑆 𝑡 = 0.25𝑆 𝑡 𝑑𝑡 + 0.10𝑆 𝑡 𝑑𝑍(𝑡)
Calculate the probability that 𝑆(𝑡) is at least 5% higher
than 𝑆(0):
i.
At time 𝑡 = 0.1
ii.
At time 𝑡 = 1
Solution 7.2.1A
 The Itô process is a GBM with 𝛼 − 𝛿 = 0.25 and σ =
0.10
 We calculate 𝜇 = 𝛼 − 𝛿 − 0.5𝜎 2 to obtain the 𝜇 of the
corresponding arithmetic Brownian motion
d ln S (t )  (0.25  0.5(0.10 2 ))dt  0.10dZ (t )  0.245dt  0.10dZ (t )
 Then
Pr S (t ) / S (0)  1.05  Pr ln S (t )  ln S (0)  ln 1.05
Solution 7.2.1A
1. For 𝑡 = 0.1, 𝑚 = 𝜇 0.1 = 0.0245, 𝑣 = 0.10 0.1
 S (1)

 ln 1.05  0.0245 


Pr  ln
 ln 1.05   1  N 
  0.22121
 0.10 0.1 
 S ( 0)

2. For 𝑡 = 1, 𝑚 = 0.245, 𝑣 = 0.10
 ln 1.05  0.245 
1 N
  0.97512
0.10


Example 7.2.1B
 You are given that 𝑆(𝑡) follows an Ito process of the form
dS (t )
 0.15dt  0.20dZ (t )
S (t )
 Given that 𝑆(9) = 40, calculate the probability that
40 < 𝑆(13) < 50
Solution 7.2.1B
 𝑆(𝑡) follows GBM, implying 𝑆(𝑡) is lognormally
distributed
 The associated normal distribution has parameters
𝜇𝑡 = 4 0.15 − 0.5(0.22 ) = 0.52 and
𝜎 𝑡 = 0.2 4 = 0.4
Solution 7.2.1B
 The probability that 40 40 < 𝑆(13) 𝑆(9) < 50 40 or
that 0 < ln[𝑆(13) 𝑆(9)] < ln 1.25 is
 ln 1.25  0.52 
  0.52 
N
  N
  0.13220
0 .4


 0.4 
7.2.2 Language of Brownian Motion

𝑑𝑆
𝑆
= 𝛼 − 𝛿 𝑑𝑡 + 𝜎𝑑𝑍(𝑡) is a form of an Itô process
where:
 𝛼 − 𝛿 and 𝜎 are constants
 The stock follows the Black-Scholes framework
 The continuous rate of increase in the stock price is the
continuously compounded rate of return on the stock (𝛼)
minus the continuous dividend rate (𝛿)
 The volatility is 𝜎
 To actually apply the Black-Scholes formula, we will need the
dividend rate and the risk-free rate
7.2.3 Itô’s Lemma
 An Itô process is a random process 𝑋(𝑡) whose
differential can be expressed as
dX (t )   t , X (t ) dt   t , X (t ) dZ (t )
 It generalizes arithmetic and geometric Brownian
motions
 For arithmetic Brownian motion, 𝜉 𝑡, 𝑋 𝑡
𝜎 𝑡, 𝑋 𝑡 = 𝜎 with 𝛼 and 𝜎 constants
 For geometric Brownian motion, 𝜉 𝑡, 𝑋 𝑡
𝜎 𝑡, 𝑋 𝑡 = 𝜎𝑋(𝑡)
= 𝛼 and
= 𝜉𝑋(𝑡) and
7.2.3 Itô’s Lemma
 Itô’s lemma is a formula for evaluating 𝑑𝐶(𝑆, 𝑡) if 𝐶(𝑆, 𝑡)
is a function of 𝑆 and 𝑡.
 From ordinary calculus’ chain rule:
dC  Cs dS  Ct dt
 where 𝐶𝑠 =
𝑆(𝑡) and
 𝐶𝑡 =
𝜕𝐶
𝜕𝑡
𝜕𝐶
𝜕𝑆
is the partial derivative of 𝐶(𝑡) wrt
is the partial derivative of 𝐶(𝑡) wrt to 𝑡
7.2.3 Itô’s Lemma
 Itô’s lemma: 𝑑𝐶 = 𝐶𝑠 𝑑𝑆 + 0.5𝐶𝑠𝑠 𝑑𝑆
 Where 𝐶𝑠𝑠 =
𝐶(𝑡) wrt 𝑆(𝑡)
𝜕2 𝐶
𝜕𝑆 2
2
+ 𝐶𝑡 𝑑𝑡
, the second partial derivative of
 In ordinary calculus, second order terms such as 𝑑𝑡 2 are
zero
 In stochastic calculus 𝑑𝑍 𝑡 × 𝑑𝑍 𝑡 = 𝑑𝑡 ≠ 0
 To multiply differentials, we use the following table:
Multiplication table
𝑑𝑡
𝑑𝑍(𝑡)
𝑑𝑡
0
0
𝑑𝑍(𝑡)
0
𝑑𝑡
Example 7.2.3A
 You are given that 𝑋 𝑡 = 𝛼𝑡 + 𝜎𝑍(𝑡). Calculate 𝑑𝑋 𝑡
using Itô’s lemma
 Solution:
𝑋(𝑡) is a function of 𝑍(𝑡) so, 𝑋 will play the role of 𝐶 and 𝑍
will play the role of 𝑆 in Itô’s lemma
𝜕𝑋
𝜕𝑍
=𝜎
𝜕2 𝑋
𝜕𝑍 2
𝜕𝑋
1 𝜕2𝑋
𝑑𝑋 =
𝑑𝑍 +
𝑑𝑍
2
𝜕𝑍
2 𝜕𝑍
𝜕𝑋
𝜕𝑡
2
=0
𝜕𝑋
𝜕𝑡
=𝛼
𝜕𝑋
+
𝑑𝑡 = 𝜎𝑑𝑍 + 0 + 𝛼𝑑𝑡
𝜕𝑡
does not involve 𝑍(𝑡) as 𝑍(𝑡) in Itô’s lemma is treated as an
independent variable from t
Example 7.2.3B
 You are given that 𝑋 𝑡 = 𝑋(0)𝑒
𝑑𝑋(𝑡) using Itô’s lemma
 Solution:
2 +𝜎𝑍 𝑡
𝜕𝑋
𝛼−0.5𝜎
= 𝜎𝑋 0 𝑒
𝜕𝑍
𝜕2 𝑋
2
=
𝜎
𝑋(𝑡)
𝜕𝑍 2
𝜕𝑋
= 𝛼 − 0.5𝜎 2 𝑋(𝑡)
𝜕𝑡
𝛼−0.5𝜎 2 𝑡+𝜎𝑍(𝑡) .
Calculate
= 𝜎𝑋(𝑡)
𝑑𝑋 𝑡
= 𝜎𝑋 𝑡 𝑑𝑍 𝑡 + 0.5𝜎 2 𝑋 𝑡 𝑑𝑍(𝑡) 2 + 𝛼 − 0.5𝜎 2 𝑋 𝑡 𝑑𝑡
= 𝜎𝑋 𝑡 𝑑𝑍 𝑡 + 0.5𝜎 2 𝑋 𝑡 𝑑𝑡 + 𝛼 − 0.5𝜎 2 𝑋 𝑡 𝑑𝑡
= 𝛼𝑋 𝑡 𝑑𝑡 + 𝜎𝑋 𝑡 𝑑𝑍(𝑡)
Example 7.2.3C
𝑑𝑋(𝑡)
𝑋(𝑡)
 You are given
using Itô’s lemma.
 Solution:
= 𝜉𝑑𝑡 + 𝜎𝑑𝑍(𝑡). Calculate 𝑑 ln 𝑋(𝑡)
Let 𝑌 𝑡 = ln 𝑋(𝑡)
𝑑𝑌(𝑡)
1
=
𝑑𝑋(𝑡) 𝑋(𝑡)
𝑑 2 𝑌(𝑡)
1
=−
2
𝑑𝑋(𝑡)
𝑋(𝑡)2
𝑑𝑌(𝑡)
=0
𝑑𝑡
Example 7.2.3C
 Hence:
1
1
𝑑 𝑌(𝑡) =
𝑑𝑋 𝑡 − 0.5
𝑋 𝑡
𝑋(𝑡)2
= 𝜉𝑑𝑡 + 𝜎𝑑𝑍 𝑡 − 0.5𝜎 2 𝑑𝑡
= 𝜉 − 0.5𝜎 2 𝑑𝑡 + 𝜎𝑑𝑍(𝑡)
𝑑𝑋(𝑡)2
Exercise 7.2.3
Given an Itô process 𝑑𝑋(𝑡) and 𝑑𝑌(𝑡). 𝑑𝑌(𝑡) is defined by
𝑑𝑌(𝑡)
= 0.1𝑑𝑡 + 0.5𝑑𝑍(𝑡)
𝑌(𝑡)
And 𝑋 = 𝑌𝑒 0.02𝑡
𝑑𝑋(𝑡) can be expressed as
𝑑𝑋 𝑡 = 𝛼𝑋 𝑡 𝑑𝑡 + 𝜎𝑋 𝑡 𝑑𝑍(𝑡)
Use Itô’s lemma to find 𝛼
7.2.4 Martingales (
 An Itô process is a martingale if and only if the
coefficient of 𝑑𝑡, the drift, is identically zero
 We can use Itô’s lemma to calculate the coefficient of 𝑑𝑡
Example 7.2.4
 The process 𝑋 𝑡 = 𝑍(𝑡)3 + 𝑐𝑡𝑍(𝑡) is a martingale.
Determine c. (Easier way to solve)
Solution:
dX (t )
 3Z (t ) 2  ct
dZ (t )
d 2 X (t )
 6 Z (t )
2
dZ (t )
dX (t )
 cZ (t )
dt
dX (t )  3Z (t ) 2  ct dZ (t )  0.56 Z (t ) dt  cZ (t )dt




 3Z (t ) 2  ct dZ (t )  (3  c) Z (t )dt
To make (3  c) Z (t )  0, we need c  3
7.2.5 Sharpe Ratio
 If an Itô process is expressed as
dS (t )
  t , S (t )    t , S (t ) dt   t , S (t ) dZ (t )
S (t )
We use 𝛼 𝑡, 𝑆(𝑡) and 𝜎 𝑡, 𝑆(𝑡) to obtain the Sharpe ratio
of 𝑆. The dividend rate is not subtracted from 𝛼 (Follow
geometric)
 If an unusual asset followed an arithmetic Brownian
motion, we have to express the process as the equation
above, by dividing the coefficients of 𝑑𝑡 and 𝑑𝑍(𝑡) by
𝑆(𝑡) because 𝑑𝑆(𝑡) is divided by 𝑆(𝑡)
Example 7.2.5A
The price of a nondividend paying stock 𝑆 satisfies the
following stochastic differential equation:
d ln S (t )   0.1dt  0.3dZ (t )
The continuously compounded risk-free rate is 0.04.
Calculate the Sharpe ratio for 𝑆 (𝐹𝑜𝑙𝑙𝑜𝑤 𝐴𝑟𝑖𝑡ℎ𝑚𝑒𝑡𝑖𝑐)
Solution:


dS (t )
 0.1  0.5(0.32 ) dt  0.3dZ (t )  0.145dt  0.3dZ (t )
S (t )
  r 0.145  0.04
Sharpe ratio  

 0.35

0 .3
7.2.5 Sharpe Ratio
 The Sharpe ratio may vary with time (𝑡), the risk-free
rate 𝑟(𝑡) which itself may vary with time, or with the
Brownian motion 𝑍(𝑡) that is part of the 𝑆(𝑡)
 At any time 𝑡, for two Itô processes depending on the
same 𝑍(𝑡), the Sharpe ratios are equal
Example 7.2.5B (This Ito process
You are given two nondividend paying assets 𝑋(𝑡) and 𝑌(𝑡)
satisfying the following stochastic differential equations:
dX (t )  0.15 X (t )dt  0.30 X (t )dZ (t )
dY (t )  AY (t )dt  0.20Y (t )dZ (t )
The continuously compounded risk-free rate is 0.03.
Determine A.
Solution:
0.15  0.03
 0 .4
0 .3
A  0.03
For Y (t ) :  
 0.4, A  0.11
0 .2
From X (t ) :  
Next Agenda
 Bond Pricing and Binomial Interest Rates
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