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CHAPTER 4
RANDOM WALK
&
BROWNIAN MOTION
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4.1 Introduction
A stochastic process in continuous time
𝑋𝑡 : 𝑡 ≥ 0
consist of
chronologically ordered random variables, but here the variable t is continuous
and it is a positive real number. Stock prices continuous time process which is
easier to handle analytically.
One of the well-known and long studied stochastic model is known as
Brownian motion and is named after the English botanist Robert Brown, in
1827. Brown described the unusual motion exhibited by a small particle that is
totally immersed in a liquid or gas. It is important in the modeling and analysis
of the price moments of stock and commodities in financial mathematics. A
formal mathematical description of Brownian motion and its properties was first
given by the great mathematician Norbort Winer beginning in 1918.
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Figure 4.1. Brownian motion of a molecule can be described as a random walk
where collisions with other molecules cause random direction changes.
Botanist Robert Brown
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Mathematician Norbert Wiener
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4.2 Random Walk
 Random walk is a process, a model or a rule to generate path sequence of
random motion.
 A random walk is a mathematical object, known as a stochastic or random
process, that describes a path that consists of a succession of random steps on
some mathematical space such as the integers.
Many natural phenomena can be modelled as random walk
 The path traced by a molecule as it travels in a liquid or gas,
 The search path of a foraging animal
 The price of a fluctuating stock
 The financial status of a gambler
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Figure 4.2: Simulation of Normally Distributed Random walk
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Figure 4.3: Simulation on Higher Dimension random walk
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Definition 4.1. (Random Walk)
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Consider a trial whose outcomes are success with probability 𝑝 or failure with
probability 1 − 𝑝. Repeat the trial infinitely many times (e.g., toss a fair coin
infinitely). The successive outcomes are denoted as 𝜔 = (𝜔1 , 𝜔2 , 𝜔3 , … ) e.g.,
𝜔 = (𝜔1 , 𝜔2 , 𝜔3 , … ) = (𝐻, 𝑇, 𝑇, … ) 𝑜𝑟 (𝑇, 𝐻, 𝑇, … ). Define
1 𝑖𝑓 𝜔𝑗 = 𝐻
𝑋𝑗 =
−1 𝑖𝑓 𝜔𝑗 = 𝑇
The probability mass function of 𝑋𝑗 is given by
𝑃 𝑋𝑗 = 1 = 𝑝;
𝑃 𝑋𝑗 = −1 = 1 − 𝑝;
𝑝 + 𝑞 = 1.
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Set 𝑆0 = 0 and define
𝑘
𝑆𝑘 = 𝑋1 + 𝑋2 + ⋯ … . +𝑋𝑘 =
𝑋𝑗
𝑘 = 1,2, … … . .
𝑗=1
where 𝑋𝑗 ’s are 𝑖. 𝑖. 𝑑. (independent and identically distributed) random variables.
Then, {𝑆𝑘 , 𝑘 = 0,1, … } is known as a random walk.
Sample path of the random walk is shown in the figure 4.1 for a outcome 𝜔 where
𝜔𝑖 = 𝑇, 𝑖 = 1,2, . . with 𝑝 = 0.45.
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Figure 4.4: A sample path of the random walk
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Properties of Random Walk
(a). Choose non-negative integers 0 = 𝑘0 < 𝑘1 < … < 𝑘𝑛 . Then
𝑘𝑙
𝑆𝑘𝑙 − 𝑆𝑘𝑖 =
𝑘𝑖
𝑋𝑗 −
𝑗=1
𝑘𝑙
𝑋𝑗 =
𝑗=1
𝑋𝑗
𝑗=𝑘𝑖 +1
Since 𝑋𝑗 are 𝑖. 𝑖. 𝑑 . random variables, 𝑆𝑘1 − 𝑆𝑘0 , 𝑆𝑘2 − 𝑆𝑘1 , … , 𝑆𝑘𝑛 − 𝑆𝑘𝑛−1 are
mutually independent variables. Hence , {𝑆𝑛 , 𝑛 = 0,1, … } has the independent
increment property.
For example if 𝑘𝑙 = 10 𝑎𝑛𝑑 𝑘𝑖 = 4 𝑡ℎ𝑒𝑛
10
𝑆10 − 𝑆4 =
4
𝑋𝑗 −
𝑗=1
10
𝑋𝑗 =
𝑗=1
𝑋𝑗 = 𝑋5 + 𝑋6 + ⋯ … . +𝑋10
𝑗=5
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(b). Similarly, for 1 ≤ 𝑖 < 𝑗 ≤ 𝑘 < 𝑙 𝑎𝑛𝑑 𝑙 − 𝑘 = 𝑗 − 𝑖 𝑡ℎ𝑒𝑛 𝑆𝑗 − 𝑆𝑖 ≡
𝑆𝑘 − 𝑆𝑙 . Hence, {𝑆𝑛 , 𝑛 = 0,1, … } has stationary increment property.
(Stationary increments is for any 𝑠 < 𝑡, 𝑆𝑡 − 𝑆𝑠 is equal in distribution to 𝑆𝑡−𝑠 .
This means that the probability distribution of any increment 𝑆𝑡 − 𝑆𝑠 depends
only on the length of time interval 𝑠 − 𝑡, the increments on same time interval
will result in the same probability distribution.)
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(c). Also, suppose 𝑆𝑛+ℎ − 𝑆ℎ = 𝑋ℎ+1 + 𝑋ℎ+2 + … + 𝑋ℎ+𝑛 , one can observe that,
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𝐸 𝑋𝑗 = 1 × 𝑝 + −1 × 𝑞 = 𝑝 − 𝑞
𝐸 𝑋𝑗2 = 12 × 𝑝 + −1
𝑉𝑎𝑟 𝑋𝑗 = 𝐸
𝑋𝑗2
2
×𝑞 =𝑝+𝑞
− 𝐸 𝑋𝑗
2
2
= 𝑝+𝑞
− 𝑝−𝑞
= 𝑝+𝑞
− 𝑝2 − 𝑞 2 + 2𝑝𝑞
= 𝑝 1 − 𝑝 + 𝑞 1 − 𝑞 + 2𝑝𝑞 = 4𝑝𝑞
Therefore
𝑛+ℎ
𝐸 𝑆𝑛+ℎ − 𝑆ℎ = 𝐸
𝑛+ℎ
𝑋𝑗
=
𝑗=ℎ+1
𝐸(𝑋𝑗 ) = 𝑛 𝑝 − 𝑞 𝑎𝑛𝑑
𝑗=ℎ+1
𝑛+ℎ
𝑉 𝑎𝑟 𝑆𝑛+ℎ − 𝑆ℎ = 𝑉𝑎𝑟
𝑛+ℎ
𝑋𝑗
=
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𝑉𝑎𝑟(𝑋𝑗 ) = 4𝑛𝑝𝑞 .
Result 4.1: For p ∈ 0,1 , 𝑛 ∈ ℕ 𝑎𝑛𝑑 𝑗 ∈ ℤ 𝑤𝑖𝑡ℎ 𝑗 ≤ 𝑛, given, the probability to
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end up in the integer 𝑗 after a random walk of 𝑛 steps, which starts at 0, is
Pr 𝑆𝑛 = 𝑗 =
𝑛
𝑛+𝑗
𝐶𝑛+𝑗 𝑝 2
2
1−𝑝
𝑛−𝑗
2
If 𝑗 and 𝑛 have same parity, and probability to end up in the integer number 𝑗 is
Pr 𝑆𝑛 = 𝑗 = 0, if 𝑗 and 𝑛 have opposite parity.
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Proof: Denote by 𝑛𝑅 the number of steps taken by the random walk right-ward
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by 𝑛𝐿 the number of steps taken left-ward. Obviously 𝑛𝑅 , 𝑛𝐿 ∈ 0,1,2 … … . 𝑛
and we have 𝑛𝑅 + 𝑛𝐿 = 𝑛 𝑎𝑛𝑑 𝑛𝑅 − 𝑛𝐿 = 𝑗.
If j and n have same parity, we have 𝑛𝑅 =
𝑛+𝑗
2
𝑎𝑛𝑑 𝑛𝐿 =
𝑛−𝑗
2
. . (𝑖).
If 𝑗 𝑎𝑛𝑑 𝑛 do not have the same parity, there are no integer numbers 𝑛𝑅 𝑎𝑛𝑑 𝑛𝐿
satisfying (𝑖), in fact there exist no random walk of n steps that ends up in 𝑗.
Every random walk of 𝑛 steps can be identified with 𝑛 Bernoulli trials. Hence, the
probability to make 𝑛𝑅 steps rightward and 𝑛𝐿 = 𝑛 − 𝑛𝑅 steps leftward is
Pr 𝑆𝑛 = 𝑛𝑅 =
𝑛
𝐶𝑛𝑅 𝑝𝑛𝑅 1 − 𝑝
𝑛𝐿 .
Therefore, if 𝑗 𝑎𝑛𝑑 𝑛 have the same parity, then the probability to end up in 𝑗 is
Pr 𝑆𝑛 = 𝑗 =
𝑛
𝑛+𝑗
𝐶𝑛+𝑗 𝑝 2
2
1−𝑝
𝑛−𝑗
2
𝑎𝑛𝑑
if 𝑗 𝑎𝑛𝑑 𝑛 have the opposite parity, then the probability to end up in 𝑗 𝑖𝑠 0.
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Example 4.1:
For a random walk on the integer axis, which starts at 0, the compute:
(i). 𝑃𝑟 𝑆6 = 𝑆10 = 𝑃𝑟 𝑋7 + 𝑋8 +𝑋9 +𝑋10 = 0 = 𝑃𝑟 𝑆6 − 𝑆10 = 2
4
=
𝐶2 𝑝2 1 − 𝑝
(4−2)
= 6𝑝2 1 − 𝑝
2
(ii).
𝑃𝑟 𝑆5 = −1 𝑎𝑛𝑑 𝑆11 = 3 = 𝑃𝑟 𝑆5 = −1 ∩ 𝑆11 = 3
= 𝑃𝑟 𝑆5 = −1)Pr(𝑆11 = 3 𝑆5 = −1
= 𝑃𝑟 𝑆5 = −1)Pr(𝑆11 − 𝑆5 = 4
= 𝑃𝑟 𝑋1 + 𝑋2 +𝑋3 +𝑋4 + 𝑋5 = −1)Pr(𝑋6 + 𝑋7 + 𝑋8 +𝑋9 +𝑋10 + 𝑋11 = 4
=
5
𝐶2 𝑝2 1 − 𝑝 3 .
= 60𝑝7 1 − 𝑝
6
𝐶5 𝑝5 1 − 𝑝
4
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(iii). 𝑃𝑟 𝑆6 = 3 = 𝑃𝑟 𝑋1 + 𝑋2 +𝑋3 +𝑋4 +𝑋5 +𝑋6 = 3 = 0 Since 𝑗 = 3 and
𝑛=6
have opposite parity.
(iv). 𝑃𝑟 𝑆6 − 𝑆2 = 2 𝑆2 = 2 = 𝑃𝑟 𝑋6 + 𝑋5 +𝑋4 +𝑋3 = 2 =
=
4
𝐶3 𝑝3 1 − 𝑝
(4−3)
= 4𝑝3 1 − 𝑝
(v). 𝑃𝑟 𝑆6 = 2 𝑆2 = 2 = 𝑃𝑟 𝑋6 + 𝑋5 +𝑋4 +𝑋3 = 0
=
4
𝐶2 𝑝2 1 − 𝑝
(4−2)
= 6𝑝2 1 − 𝑝
2
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4.3 Brownian Motion
Consider a particle performs a random walk such that in a small interval of time
of duration Δ𝑡, the displacement of the particle to the right or to the left is also of
small magnitude Δ𝑥. Let 𝑆(𝑡) be denote the total displacement of the particle in
time 𝑡 being 𝑥. Let 𝑋𝑗 be denote the length of the 𝑗th step taken by the particle in
a small interval of time Δ𝑡 with pmf
𝑃(𝑋𝑗 = Δ𝑥) = 𝑝; 𝑃(𝑋𝑗 = −Δ𝑥) = 1 − 𝑝; 𝑝 + 𝑞 = 1
0 < 𝑝 < 1, where 𝑝 is independent of 𝑥 and 𝑡. Partition the interval of length
𝑡 into n equal subintervals of length Δ𝑡. Then 𝑛Δ𝑡 = 𝑡 and the total
displacement 𝑆(𝑡) is the sum of 𝑛 i.i.d. random variables 𝑋𝑗 , i.e.,
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𝑛(𝑡)
𝑆 𝑡 =
𝑡
𝑛=𝑛 𝑡 =
Δ𝑡
𝑋𝑗
𝑗=1
𝐸 𝑋𝑗 = Δ𝑥 × 𝑝 + −Δ𝑥 × 𝑞 = Δ𝑥 𝑝 − 𝑞
𝐸 𝑋𝑗2 = Δ𝑥 2 × 𝑝 + −Δ𝑥
2
× 𝑞 = Δ𝑥
2
𝑝+𝑞
𝑣𝑎𝑟 𝑋𝑗 = 𝐸 𝑋𝑗2 − 𝐸(𝑋𝑗 )2
= Δ𝑥
= Δ𝑥
2
2
𝑝 + 𝑞 − Δ𝑥 𝑝 − 𝑞
𝑝+𝑞 − 𝑝−𝑞
= 4𝑝𝑞 Δ𝑥
2
2
2
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Using 𝐸 𝑋𝑗 = 𝑝 − 𝑞 Δ𝑥 𝑎𝑛𝑑 𝑣𝑎𝑟 𝑋𝑗 = 4𝑝𝑞 Δ𝑥
𝐸 𝑆 𝑡
= 𝑛𝐸 𝑋𝑗 =
𝑉𝑎𝑟 𝑆(𝑡) = 𝑛𝑉𝑎𝑟 𝑋𝑗
2
, we get
𝑡
𝑝 − 𝑞 Δ𝑥
Δ𝑡
𝑡
= 4𝑝𝑞 Δ𝑥 2 .
Δ𝑡
To get a meaningful result, as Δ𝑥 → 0, Δ𝑡 → 0, we must have
Δ𝑥 2
→ 𝑎 𝑙𝑖𝑚𝑖𝑡, 𝑝 − 𝑞 → a multiple of Δ𝑥 .
Δ𝑡
𝐴𝑠 Δ𝑥 → 0, Δ𝑡 → 0, 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑡𝑖𝑚𝑒 𝐸 𝑆 𝑡
→ 𝜇 𝑎𝑛𝑑 𝑉𝑎𝑟 𝑆(𝑡) → 𝜎 2 . 𝐻𝑒𝑛𝑐𝑒,
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We get
Δ𝑥 = 𝜎 ∆𝑡
1/2
1
𝑝 = 1 + 𝜇 ∆𝑡
2
,
1
𝑞 = 1 − 𝜇 ∆𝑡
2
1/2 /𝜎
1/2 /𝜎
Then
𝐸 𝑆 𝑡
= 𝑛𝐸 𝑋𝑗 =
𝑡
𝑡 1
1
𝑝 − 𝑞 Δ𝑥 =
1 + 𝜇 ∆𝑡 1/2 /𝜎 − 1 − 𝜇 ∆𝑡
Δ𝑡
Δ𝑡 2
2
1
𝑡
𝜇 ∆𝑡 2
1
Δ𝑡
=
𝜎 ∆𝑡 2 = 𝜇𝑡
𝜎
𝑡
𝑉𝑎𝑟 𝑆(𝑡) = 4𝑝𝑞 Δ𝑥
Δ𝑡
2
𝑡
1
= 4
1 + 𝜇 ∆𝑡
Δ𝑡 2
1/2
/𝜎
1
1 − 𝜇 ∆𝑡
2
1/2
/𝜎
1/2
/𝜎
𝜎 ∆𝑡
𝜎 ∆𝑡
1/2
1/2 2
= 𝑡𝜎
Now, we present the limiting distribution of symmetric random walk is
Brownian motion using central limit theorem.
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Limiting Case of Random Walk
For large 𝑛(𝑡)(= 𝑛), 𝑆 𝑡 =
𝑛(𝑡)
𝑗=1 𝑋𝑗
converges in distribution to 𝑁(𝜇𝑡, 𝜎 2 𝑡). Since
𝑡 represents the length of the interval of time during which the displacement, we
conclude for 0 < 𝑠 < 𝑡,
𝑆 𝑡 − 𝑆 𝑠
Further,
the
increments
𝑖𝑠 𝑁 𝜇 𝑡 − 𝑠 , 𝜎 2 𝑡 − 𝑠 .
{𝑆(𝑠) − 𝑆(0)} and
{𝑆(𝑡) − 𝑆(𝑠)} are
mutually
independent.
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Definition 4.2:Brownian Motion
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A Gaussian random process {𝑊(𝑡), 𝑡 ∈ [0, ∞)} is called a Brownian motion or
a Wiener process if
1. W(0) = 0.
2. For all 0 ≤ 𝑡1 < 𝑡2 , 𝑊(𝑡2 ) − 𝑊(𝑡1 ) ∼ 𝑁(0, 𝜎 2 (𝑡2 −𝑡1 )).
3. 𝑊(𝑡) has independent increments. That is, for all 0 ≤ 𝑡1 < 𝑡2 < 𝑡3 ⋯ < 𝑡𝑛 ,
the
random variables 𝑊 𝑡2 − 𝑊 𝑡1 , 𝑊 𝑡3 − 𝑊 𝑡2 , … … … , 𝑊(𝑡𝑛 ) − 𝑊(𝑡𝑛−1 ) are
independent.
4. W(t) has continuous sample paths.
If 𝜎 2 = 1 is called a standard Wiener process or standard Brownian motion.
Note: The wiener process has no jumps. But fluctuations heavily the paths are
continuous but highly erratic. Those paths are not differentiable with probability 1.
Sample path of Brownian motion is shown in the figure as follows.
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Figure 4.5: A sample path of a Brownian motion
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Example 4.2:
Let 𝑊(𝑡) be a standard Brownian motion. For all 𝑠, 𝑡 ∈ [0, ∞), find
𝐶𝑜𝑣(𝑊(𝑠), 𝑊(𝑡)).
Solution: Let's assume 𝑠 ≤ 𝑡. Then, we have
𝐶𝑜𝑣 𝑊 𝑠 , 𝑊 𝑡
= 𝐶𝑜𝑣 𝑊(𝑠), 𝑊(𝑠) + 𝑊(𝑡) − 𝑊(𝑠)
= 𝐶𝑜𝑣[𝑊 𝑠 , 𝑊 𝑠 ] + 𝐶𝑜𝑣[𝑊(𝑠), 𝑊(𝑡) − 𝑊(𝑠)]
= 𝑉𝑎𝑟(𝑊(𝑠)) + 𝐶𝑜𝑣(𝑊(𝑠), 𝑊(𝑡) − 𝑊(𝑠))
= 𝑠 + 𝐶𝑜𝑣 𝑊 𝑠 , 𝑊 𝑡 − 𝑊 𝑠
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Brownian
motion
has
independent
variables 𝑊(𝑠) = 𝑊(𝑠) − 𝑊(0)
increments,
so
and 𝑊(𝑡) − 𝑊(𝑠)
the
two
random
are independent.
Therefore, 𝐶𝑜𝑣(𝑊(𝑠), 𝑊(𝑡) − 𝑊(𝑠)) = 0.
We conclude 𝐶𝑜𝑣(𝑊(𝑠), 𝑊(𝑡)) = 𝑠.
Similarly, if 𝑡 ≤ 𝑠 we obtain 𝐶𝑜𝑣(𝑊(𝑠), 𝑊(𝑡)) = 𝑡.
We conclude
𝑪𝒐𝒗(𝑾(𝒔), 𝑾(𝒕)) = 𝒎𝒊𝒏(𝒔, 𝒕), 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒔, 𝒕.
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Example 4.3:
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Let 𝑊(𝑡) be a standard Brownian motion.
a. Find 𝑃(1 < 𝑊(1) < 2).
b.Find 𝑃(𝑊(2) < 3 𝑊(1) = 1).
Solution
a. We have 𝑊(1) ∼ 𝑁(0,1). Thus,
𝑃 1 < 𝑊 1 < 2 = Φ 2 − Φ 1 ≈ 0.136
b. Note that 𝑊(2) = 𝑊(1) + 𝑊(2) − 𝑊(1). Also, note that 𝑊(1) and 𝑊(2) −
𝑊(1) are independent, and 𝑊(2) − 𝑊(1) ∼ 𝑁(0,1).
We conclude that
𝑊(2) 𝑊(1) = 1 ∼ 𝑁(1,1).
Thus,
𝑃(𝑊(2) < 3 𝑊(1) = 1) = Φ
3−1
= Φ(2) ≈ 0.98
1
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Example 4.4:
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Let 𝑊(𝑡) be a standard Brownian motion. Find 𝑃(𝑊(1) + 𝑊(2) > 2).
Solution
Let 𝑋 = 𝑊(1) + 𝑊(2). Since 𝑊(𝑡) is a Gaussian process, 𝑋 is a normal random
variable.
𝐸(𝑋) = 𝐸[𝑊(1)] + 𝐸[𝑊(2)] = 0,
𝑉𝑎𝑟(𝑋) = 𝑉𝑎𝑟(𝑊(1)) + 𝑉𝑎𝑟(𝑊(2)) + 2𝐶𝑜𝑣(𝑊(1), 𝑊(2)) = 1 + 2 + 2 ⋅ 1
We conclude
𝑋 ∼ 𝑁(0,5).
Thus,
𝑃(𝑋 > 2) = 1 − Φ
2−0
5
≈ 0.186
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28
Example 4.5:
If {𝐵(𝑡), 𝑡 ≥ 0} is a standard Brownian motion then the following are also
Brownian motion.
(i). 𝐵1 𝑡 = 𝑐𝐵 𝑡/𝑐 2
𝑐≠0
Solution:
Consider 𝐵1 0 = 𝑐𝐵 0/𝑐 2 = 𝑐𝐵 0 = 0 (Since 𝐵(𝑡) is a standard BM
from property (i), 𝐵(0) = 0)
Satisfy the first property.
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𝐸 𝐵1 𝑡 − 𝐵1 𝑠
= 𝑐𝐸 𝐵 𝑡/𝑐 2 − 𝑐𝐵 𝑠/𝑐 2
= 𝑐 𝐸 𝐵 𝑡/𝑐 2
− 𝐸 𝐵 𝑠/𝑐 2
=0−0=0
𝐸
𝐵1 𝑡 − 𝐵1 𝑠
= 𝑐2 𝐸
𝐵 𝑡/𝑐 2
2
2
= 𝑐2𝐸
+𝐸
𝐵 𝑡/𝑐 2 − 𝑐𝐵 𝑠/𝑐 2
𝐵 𝑠/𝑐 2
2
2
− 2𝐸 𝐵 𝑡/𝑐 2
𝐵 𝑠/𝑐 2
= 𝑐 2 𝑡/𝑐 2 + 𝑠/𝑐 2 − 2𝑚𝑖𝑛 𝑡/𝑐 2 , 𝑠/𝑐 2
= 𝑐 2 𝑡/𝑐 2 + 𝑠/𝑐 2 − 2𝑠/𝑐 2
=𝑡−𝑠
Therefore
𝐵1 𝑡 − 𝐵1 𝑠 ~𝑁 0, 𝑡 − 𝑠
Property two satisfied.
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30
Note that if 𝑡1 < 𝑡2 ≤ 𝑡3 < 𝑡4 then
increments 𝐵
𝑡4
𝑐2
−𝐵
𝑡3
𝑐2
and 𝐵
𝑡1
𝑐2
𝑡2
𝑐2
<
𝑡2
𝑐2
−𝐵
≤
𝑡3
𝑐2
𝑡1
𝑐2
<
𝑡4
𝑐2
and the corresponding
are independent. Then the
multiplies of each by c are independent and so 𝐵1 𝑡4 − 𝐵1 (𝑡3 ) and 𝐵1 𝑡2 −
𝐵1 (𝑡1 ) are independent.
Third property satisfied.
𝐵1 𝑡 = 𝑐𝐵 𝑡/𝑐 2
𝑐 ≠ 0 is a standard BM.
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(ii). 𝐵2 𝑡 = 𝐵 𝑡 cos 𝛼 + 𝐵′ 𝑡 sin ∝
∝≥0
Where {𝐵′ (𝑡), 𝑡 ≥ 0} is a standard Brownian motion independent of
{𝐵(𝑡), 𝑡 ≥
0} .
Solution:
Consider
𝐵2 0 = 𝐵 0 cos 𝛼 + 𝐵′ 0 sin ∝ = 0 (Since
𝐵 𝑡 𝑎𝑛𝑑 𝐵′ 𝑡
are
standard BM from property (i), 𝐵 0 = 𝐵′ 𝑡 = 0)
Satisfy the first property.
𝐸 𝐵2 𝑡 − 𝐵2 𝑠
= 𝐸 𝐵 𝑡 cos 𝛼 + 𝐵′ 𝑡 sin ∝ − 𝐵 𝑠 cos 𝛼 − 𝐵′ 𝑠 sin ∝
= 𝐸 cos 𝛼 𝐵 𝑡 − 𝐵(𝑠) + sin ∝ 𝐵′ 𝑡 − 𝐵′(𝑠)
= cos 𝛼 𝐸 𝐵 𝑡 − 𝐵(𝑠) + sin ∝ 𝐸 𝐵′ 𝑡 − 𝐵′(𝑠)
= 0−0=0
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𝐸
2
𝐵2 𝑡 − 𝐵2 𝑠
= cos 2 𝛼 𝐸 𝐵(𝑡)
2
2 cos 2 𝛼 𝐸 𝐵 𝑡 𝐵 𝑠
= 𝐸 𝐵 𝑡 cos 𝛼 + 𝐵′ 𝑡 sin ∝ − 𝐵 𝑠 cos 𝛼 − 𝐵′ 𝑠 sin ∝
+cos 2 𝛼 𝐸 𝐵(𝑠)
2
+sin2 𝛼 𝐸 𝐵′(𝑡)
− 2 sin2 𝛼 𝐸 𝐵′ 𝑡 𝐵′ 𝑠
2
+ sin2 𝛼 𝐸 𝐵′(𝑠)
2
2
−
+ cos 𝛼 sin 𝛼 𝐸 𝐵 𝑡 𝐵′ 𝑡 +….
= cos 2 𝛼 𝑡 + cos 2 𝛼 s+sin2 𝛼 t + sin2 𝛼 𝑠 − 2 cos 2 𝛼 min 𝑡, 𝑠 − 2 sin2 𝛼 min(𝑡, 𝑠)
= cos 2 𝛼 𝑡 + 𝑠 − 2𝑠 + sin2 𝛼 𝑡 + 𝑠 − 2𝑠
=𝑡−𝑠
Therefore
𝐵2 𝑡 − 𝐵2 𝑠 ~𝑁 0, 𝑡 − 𝑠
Satisfy the second property.
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33
Note that if 𝑡1 < 𝑡2 ≤ 𝑡3 < 𝑡4 then increments
𝐵2 𝑡4 − 𝐵2 𝑡3 = 𝐵 𝑡4 cos 𝛼 + 𝐵′ 𝑡4 sin ∝ − 𝐵 𝑡3 cos 𝛼 − 𝐵′ 𝑡3 sin ∝
= cos 𝛼 𝐵 𝑡4 − 𝐵 𝑡3
𝐵2 𝑡2 − 𝐵2 𝑡1
= 𝐵 𝑡2 cos 𝛼 + 𝐵′ 𝑡2 sin ∝ − 𝐵 𝑡1 cos 𝛼 − 𝐵′ 𝑡1 sin ∝
= cos 𝛼 𝐵 𝑡2 − 𝐵 𝑡1
Since
𝐵 𝑡4 − 𝐵 𝑡3
+ sin ∝ 𝐵′ 𝑡4 − 𝐵′ 𝑡3
and
𝐵′ 𝑡4 − 𝐵′ 𝑡3
+ sin ∝ 𝐵′ 𝑡2 − 𝐵′ 𝑡1
are independent from
𝐵 𝑡2 −
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34
Problem 4.1:
Let 𝑊(𝑡) be standard Brownian motion.
(a) Find the probability that 0 < 𝑊(1) < 1.
(b) Find the probability that 0 < 𝑊(1) < 1 and 1 < 𝑊(2) − 𝑊(1) < 3.
(c) Find the probability that 0 < 𝑊(1) < 1 and 1 < 𝑊(2) − 𝑊(1) <
3 and 0 < 𝑊(3) − 𝑊(2) < 1/2.
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Brownian Motion with Drift
35
This is a stochastic process of the form {𝐵1 𝑡 = 𝜇𝑡 + 𝐵(𝑡), 𝑡 ≥ 0} where 𝜇 is a
constant and {𝐵(𝑡), 𝑡 ≥ 0} is Brownian motion.
Definition 4.3:Brownian Motion with Drift
A stochastic process {𝐵(𝑡), 𝑡 ≥ 0} is called Brownian Motion (or Wiener
Process) with volatility 𝝈 and drift 𝝁 if
(i). 𝐵 0 = 0
(ii). 𝐵(𝑡) has independent stationary increments.
(iii). 𝐵 𝑡 has normal distribution with mean μ𝑡 and variance 𝜎 2 𝑡.
Note: Since 𝐵 𝑡 has normal distribution with mean μ𝑡 and variance 𝜎 2 𝑡 , 𝐵 𝑡 −
𝐵 𝑠 has normal distribution with mean μ 𝑡 − 𝑠 𝑎𝑛𝑑 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒𝜎 2 𝑡 − 𝑠 .
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36
4.4 Geometric Brownian Motion
Under the Brownian motion we think stock’s price as a particle and also
normal distribution seems like a reasonable choice to model a stock price. However
there are some obvious problems with the BM viewpoint for stock prices
themselves.
a. BM process can be negative where as stock prices are never negative.
b. In a BM process, the increments 𝐵 𝑡 − 𝐵(𝑠) have distribution , that depends
only on (𝑡 − 𝑠) then thus if a stock price were to have a BM process 𝐵(𝑡) then the
expected change
𝐸 𝐵 𝑡 − 𝐵(𝑠) in the stock price over a period of time could
be 𝜇 (𝑡 − 𝑠), which does not depends on the initial price 𝐵(𝑠) . This is not realistic.
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Example 4.6:
37
Consider change in price is 𝜇 𝑡 − 𝑠 = $10. A $10 expected change of price
might be quite reasonable if the stock priced (initially) at 𝐵 𝑠 = $100. But not
nearly as reasonable if the stock is initially priced at 𝐵 𝑠 = $1.
To resolve this issue, it would seem to make more sense to model the rate of return
of the stock price as a BM process.
Example 4.7:
Say that a stock’s price rate of return is 10% is to say that the price may grow
$100 to $110 or $1 to $1.10.
This can be handling, by assuming that stock price 𝑆 𝑡 at time t is given by
𝑆 𝑡 = 𝑆 0 𝑒 𝐻𝑡
Where 𝐻𝑡
Continuously compounded rate of return of the stock price over the
period of time [0, t]. Also 𝐻𝑡 refers to as the logarithmic growth of the stock price ,
satisfy 𝐻𝑡 = 𝑙𝑛
𝑆 𝑡
𝑆 0
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.
38
Definition 4.3: (Geometric Brownian Motion)
A stochastic process of the form 𝑒 𝐻𝑡 𝑡 ≥ 0 is a Brownian motion process called
a geometric Brownian Motion process.
If we assume that 𝐻𝑡 follows a Brownian motion process with drift then we can
write
𝑆 𝑡
𝐻𝑡 = 𝑙𝑛
𝑆 0
= 𝜇𝑡 + 𝜎𝐵(𝑡)
Where {𝐵(𝑡), 𝑡 ≥ 0} is standard Brownian motion. Therefore 𝐻𝑡 has a normal
distribution with
𝐸 𝐻𝑡 = 𝜇𝑡
𝑉𝑎𝑟 𝐻𝑡 = 𝜎 2 𝑡
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39
It is clear that the continuous time model is a limiting case of the Cox-RossRubistine model. The continuously compounded rate of return 𝐻𝑡 , over the time
period 0 , 𝑡 , which satisfied the equation
𝑆 𝑡 = 𝑆 0 𝑒 𝐻𝑡
𝑆 𝑡
ln
𝑆 0
= 𝐻𝑡 = 𝜇𝑡 + 𝜎𝐵(𝑡)
Differentiate with respect to t
1 𝑑 𝑆 𝑡
𝑆 𝑡 𝑑𝑡 𝑆 0
𝑆 0
𝑑
= 𝜇 + 𝜎 𝐵(𝑡)
𝑑𝑡
1 𝑑𝑆 𝑡
𝑑
= 𝜇 + 𝜎 𝐵(𝑡)
𝑆 𝑡 𝑑𝑡
𝑑𝑡
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40
The most common approach to the continuous time model of stock price assumes
that the instantaneous percentage return is a Brownian motion process more
specifically
𝑑𝑆 𝑡
𝑑𝐵(𝑡)
= 𝜇𝑑𝑡 + 𝜎
𝑆 𝑡
𝑑𝑡
𝑑𝑆(𝑡) = 𝑆(𝑡) 𝜇𝑑𝑡 + 𝜎𝑑𝐵(𝑡)
Where {𝐵(𝑡), 𝑡 ≥ 0} is standard Brownian motion and 𝜇 and 𝜎 are constants.
Further
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41
We can say that the stochastic process
𝑑𝑆(𝑡)
𝑆(𝑡)
𝑡 ≥ 0 is assume to follow Brownian motion process with drift 𝜇 and volatility 𝜎.
The formula can be written
𝑑𝑆 𝑡 = 𝜇𝑆 𝑡 𝑑𝑡 + 𝜎𝑆 𝑡 𝑑𝐵 𝑡
This is an instance of what is known as a stochastic differential equation.
Thus, Geometric BM in differential form is,
𝑑𝑆 𝑡 = 𝜇 𝑆 𝑡 𝑑𝑡 + 𝜎𝑆 𝑡 𝑑𝐵(𝑡)
And
Geometric BM in integral form is,
𝑡
𝑡
𝑆 𝑡 = 𝑆 0 + 0 𝜇 𝑆 𝑢 𝑑𝑢 + 0 𝜎𝑆 𝑢 𝑑𝐵(𝑢)
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42
Figure 4.6: Two sample paths of Geometric Brownian motion, with different parameters.
The blue line has larger drift, the green line has larger variance.
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Example 4.8. Suppose that stock price {𝑆(𝑡), 𝑡 ≥ 0} follows geometric Brownian
43
motion with drift μ = 8% per year and variance 𝜎 2 = 8.5% per annum. Assume
that, the current price of the stock is 𝑆(0) = 60. find
(i). 𝐸[𝑆(3)]
According to the geometric Brownian motion we have
𝑆 𝑡
𝑙𝑛
𝑆 0
= 𝜇𝑡 + 𝜎𝐵(𝑡)
ln 𝑆 𝑡 − ln 𝑆(0) = 𝜇𝑡 + 𝜎𝐵(𝑡)
ln 𝑆 𝑡 = 𝜇𝑡 + ln 𝑆(0) + 𝜎𝐵(𝑡)
Since 𝐵(𝑡)~𝑁 0 , 𝑡
ln 𝑆 𝑡 ~𝑁 𝜇𝑡 + ln 𝑆(0) , 𝜎 2 𝑡
Therefore
𝑆 𝑡 ~𝑙𝑜𝑔𝑛𝑜𝑟𝑚𝑎𝑙 𝜇𝑡 + ln 𝑆(0) , 𝜎 2 𝑡
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Result 2.1: Suppose that X has the lognormal distribution with parameters 𝜇 𝑎𝑛𝑑 𝜎,
then
1
𝐸 𝑋 𝑛 = 𝑒𝑥𝑝 𝑛 𝜇 + 𝑛2 𝜎 2
2
𝑛∈𝑁
Now
𝑆 𝑡 ~𝑙𝑜𝑔𝑛𝑜𝑟𝑚𝑎𝑙 𝜇𝑡 + ln 𝑆(0) , 𝜎 2 𝑡
Therefore
𝐸 𝑆 𝑡
𝐸 𝑆 3
1 2
= 𝑒𝑥𝑝 𝜇𝑡 + ln 𝑆(0) + 𝜎 𝑡
2
1
= 𝑒𝑥𝑝 0.08 × 3 + ln 60 + × 0.085 × 3
2
= 84.65
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(ii). 𝑃 𝑆 3 > 100 = 𝑃
𝑆 3
60
>
100
60
= 𝑃 log
𝑆 3
60
> log
100
60
100
log 60 − 0.08 ∗ 3
=𝑃 𝑍>
0.085 ∗ 3
= 𝑃𝑟 𝑍 > 0.54 = 1 − 0.7054
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Example 4.9. The price of a stock follows a geometric Brownian motion with
parameters μ = 0.12 and 𝜎 = 0.24. If the present price of the stock is 𝑅𝑠. 40,
what is the probability that a call option having four months to exercise time and
with a strike price 𝐾 = 𝑅𝑠. 42, will be exercised?
Solution
A call option on a stock with exercise time 𝑇 and strike 𝐾 is the right, but not the
obligation, to buy a certain number of shares at time T for the fixed price 𝐾. If 𝑆(𝑡)
denotes the market price of this number of shares then the value of the option at
time 𝑇 is 𝑆(𝑇) − 𝐾 𝑖𝑓 𝑆(𝑇) > 𝐾, since the shares can be bought for 𝐾 and
immediately sold in the market for 𝑆(𝑇). On the other hand the option is worthless,
and will not be exercised, if 𝑆(𝑇) ≤ 𝐾. The required probability is
𝑃𝑟{𝑆(1/3) > 42}.
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Since
𝑆(𝑇)
~𝑙𝑜𝑔𝑛𝑜𝑟𝑚𝑎𝑙
𝑆(0)
𝜇𝑇 , 𝜎 2 𝑇
4
For the problem μ = 0.12 𝑎𝑛𝑑 𝜎 = 0.24 , 𝑇 = 12 = 1/3 𝑎𝑛𝑑 𝑆 0 = 40
𝑆(1/3)
1
~𝑙𝑜𝑔𝑛𝑜𝑟𝑚𝑎𝑙 0.12 × , 0.242 × 1/3
40
3
𝑆(1/3)
42
𝑃𝑟 𝑆 1/3 > 42 = 𝑃𝑟
>
40
40
𝑆(1/3)
42
= 𝑃𝑟 𝑙𝑛
> ln
40
40
42
1
ln 40 − 0.12 × 3
= 𝑃𝑟 𝑍 >
0.242 × 1/3
= 𝑃𝑟 𝑍 > 0.063)
= 1 − 0.5239
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