IE 5441 1

Chapter 13. Models of Asset Dynamics

Shuzhong Zhang

IE 5441 2

The binomial lattice model

At each step, the stock price S either increases to uS or decreases to

dS. The probability of moving up is p. This will form a recombining

binomial tree:

Su4

Su3

Su2 Su3d

Su Su2d

S Sud1 Su2d2

Sd Sud2

Sd2 Sud3

Sd3

Sd4

Shuzhong Zhang

IE 5441 3

How to determine u, d and p?

Suppose that we wish to match

ν = E [ln(ST /S0)]

σ2 = var [ln(ST /S0)]

The choice of parameters will be

u = eσ√∆t

d = e−σ√∆t

p = 12 + ν

2σ

√∆t.

We will come back to discuss this point later.

Shuzhong Zhang

IE 5441 4

Example 13.1. Consider a stock with the parameter ν = 15% and

σ = 30%. We wish to make a binomial model based on weekly periods.

We compute that u = e0.3/√52 ≈ 1.04248, d = 1/u ≈ 0.95925, and

p = (1 + 1530

√1/52)/2 ≈ 0.534669. The binomial lattice is as follows:

118.11

113.29

108.67 108.67

104.25 104.25

100 100 100

95.93 95.93

92.02 92.02

88.27

84.67

Shuzhong Zhang

IE 5441 5

The additive model

The simplest model is

S(k + 1) = aS(k) + u(k),

for k = 0, 1, 2, ..., N , where u(k)’s are random.

Clearly,

S(k) = akS(0) + ak−1u(0) + ak−2u(1) + · · ·+ u(k − 1).

Suppose that all the u(k)’s have zero-mean. Then E[S(k)] = akS(0).

Shuzhong Zhang

IE 5441 6

The multiplicative model

Consider

S(k + 1) = u(k)S(k)

or,

lnS(k + 1) = lnS(k) + lnu(k).

Let w(k) = lnu(k), and suppose that w(k) follows normal distribution.

In this case, u(k) = ew(k) is said to follow the lognormal distribution.

Lognormal random variables always take positive values.

We have

lnS(k) = lnS(0) +k−1∑i=0

w(i).

Therefore, if w(i)’s are independent and have mean ν and variance σ2,

then

E[lnS(k)] = lnS(0) + kν

var (lnS(k)) = kσ2.

Shuzhong Zhang

IE 5441 7

How well does lognormal distribution fit the real stock returns?

• Overall, it fits quite well.

• The extremes of the distribution typically do not fit well, though.

The histograms of stock data typically display ‘fat-tails’.

• The observed histograms show that one tail is heavier than the

other: the density is skewed. The skewness is defined as

E[(lnS − E[lnS])3

]/σ3.

• It is more often negatively skewed, meaning that the stock prices

typically drop fast but recover slowly.

Shuzhong Zhang

IE 5441 8

One may use historical data to estimate the sample mean and variance:

ν̂ =1

N

N−1∑i=0

ln

[S(k + 1)

S(k)

]=

1

Nln

[S(N)

S(0)

]

σ̂2 =1

N − 1

N−1∑i=0

[ln

[S(k + 1)

S(k)

]− ν̂

]2.

In case w(k)’s are iid normal, we have

var (ν̂) = σ2/N

var (σ̂2) = 2σ4/(N − 1).

The mean is more difficult to get estimated accurately.

Shuzhong Zhang

IE 5441 9

Lognormal random variable

Suppose that w is a normally distributed random variable, with mean

µ and standard deviation σ.

Consider u = ew. The statistical properties of u are:

pdf: 1x√2πσ

e−(ln x−µ)2

2σ2 ;

mean: eµ+σ2/2;

variance: (eσ2 − 1)e2µ+σ2

;

skewness: (eσ2

+ 2)√eσ2 − 1.

If the volatility of w is small, then E[u] ≈ eE[w]. For instance, if

µ = 10% and σ = 15%, then E[u] = 1.105 while eE[w] = 1.117. But in

case the volatility is high compared to the mean, then this

approximation may be poor.

Shuzhong Zhang

IE 5441 10

Random walks and Wiener processes

As a first step, let us consider a discrete model where the time is

subdivided into N steps (each with length ∆t). Consider an additive

process

z(tk+1) = z(tk) + ϵ(tk)√∆t, k = 0, 1, 2, ..., N − 1.

In the above ϵ(tk)’s are assumed to be iid standard normal random

variables. The process is known as random walk.

For any j < k we have

z(tk)− z(tj) =

k−1∑i=j

ϵ(ti)√∆t.

We see that

E[z(tk)− z(tj)] = 0

var (z(tk)− z(tj)) = tk − tj .

Shuzhong Zhang

IE 5441 11

If we take the random walk process to limit by letting ∆t → 0, then

the limit process is known as the Wiener process, or Brownian motion.

In other words, the Wiener process is a continuous time stochastic

process z(t), satisfying

• z(t0) = 0.

• For any s < t, z(t)− z(s) is a normal random variable with mean

zero and variance t− s.

• For any 0 ≤ t1 < t2 ≤ t3 < t4, the random variables z(t2)− z(t1)

and z(t4)− z(t3) are uncorrelated.

Shuzhong Zhang

IE 5441 12

Basic properties of the Wiener process

• For given t > 0, the probability density function of z(t) is

ρz(t)(x) =1√2π

e−x2

2t .

• For given t > 0, we have E[z(t)] = 0 and var (z(t)) = t.

• Since E[z(t)2 − t] = 0 and

var (z(t)2 − t) = E[z(t)4 − 2tz(t)2 + t2] = 2t2,

we conclude that z(t)2 is almost constant (≈ t) for small t > 0.

• For given t, s > 0, cov (z(s), z(t)) = min{s, t}.

A Wiener process is nowhere differentiable. For any t < s,

E

[z(s)− z(t)

s− t

]2=

s− t

(s− t)2=

1

s− t→ ∞, as t → s.

Shuzhong Zhang

IE 5441 13

A sample path of Wiener process

Shuzhong Zhang

IE 5441 14

Generalized Wiener processes and Ito Processes

Generalized Wiener process

dx(t) = adt+ bdz(t),

where z(t) is a Wiener process, which has a solution

x(t) = x(0) + at+ bz(t).

An Ito process is an even more general extension

dx(t) = a(x, t)dt+ b(x, t)dz(t),

where z(t) is a Wiener process.

Ito processes do not have explicit solutions in general.

Shuzhong Zhang

IE 5441 15

A stock price process

As a basic assumption, we assume

d lnS(t) = νdt+ σdz(t).

The above is a generalized Wiener process; it is known as the

geometric Brownian motion.

Interestingly, we can show that

dS(t)

S(t)=

(ν + σ2/2

)dt+ σdz(t),

which follows from a very important result known as Ito’s lemma,

which we will discuss shortly.

Shuzhong Zhang

IE 5441 16

Since lnS(t) is a Wiener process, we have

lnS(t) = lnS(0) + νt+ σz(t).

Hence, the geometric Brownian motion is also a lognormal process:

For any given t, lnS(t) ∼ N(lnS(0) + νt, σ2t), or

S(t) = exp(lnS(t)) = S(0) exp(νt+ σz(t)).

We have

E[S(t)] = S(0)e(ν+12σ

2)t,

and

stdev(S(t)) = S(0)eνt+12σ

2t(eσ

2t − 1)1/2

.

Shuzhong Zhang

IE 5441 17

Ito’s lemma: Consider an Ito process x(t) defined by

dx(t) = a(x, t)dt+ b(x, t)dz(t).

Suppose that y(t) = F (x, t). Then, y(t) satisfies the Ito equation

dy(t) =

(∂F

∂xa+

∂F

∂t+

1

2

∂2F

∂x2b2)dt+

∂F

∂xbdz(t).

Proof.

y +∆y ≈ F (x, t) +∂F

∂x∆x+

∂F

∂t∆t+

1

2

∂2F

∂x2(∆x)2

= F (x, t) +∂F

∂x(a∆t+ b∆z) +

∂F

∂t∆t+

1

2

∂2F

∂x2(a∆t+ b∆z)2

= y(t) +

[(∂F

∂xa+

∂F

∂t

)∆t+

1

2

∂2F

∂x2b2(∆z)2

]+

∂F

∂xb∆z + o(∆t).

Shuzhong Zhang

IE 5441 18

According to a basic property of the Wiener process, we have

(∆z)2 ∼ ∆t.

Therefore,

y +∆y = y(t) +

(∂F

∂xa+

∂F

∂t+

1

2

∂2F

∂x2b2)∆t+

∂F

∂xb∆z + o(∆t).

This leads to

dy(t) =

(∂F

∂xa+

∂F

∂t+

1

2

∂2F

∂x2b2)dt+

∂F

∂xbdz(t).

2

Shuzhong Zhang

IE 5441 19

Now, let us see what Ito’s lemma tells us about the sock process. We

assumed that

d lnS(t) = νdt+ σdz(t).

We have S(t) = elnS(t). This suggests that we shall let x(t) = lnS(t)

and F (x, t) = ex, and then apply Ito’s lemma:

dS(t) = delnS(t) =

(νex(t) +

σ2

2ex(t)

)dt+ ex(t)σdz(t)

= S(t)

(ν +

σ2

2

)dt+ S(t)σdz(t).

Shuzhong Zhang

IE 5441 20

The parameters for the binomial lattice

For the binomial lattice, we have

E[lnSt] = p lnu+ (1− p) ln d

var (lnSt) = p(lnu)2 + (1− p)(ln d)2 − (p lnu+ (1− p) ln d)2

= p(1− p)(lnu− ln d)2.

Let U = lnu and D = ln d, and d = 1/u. Then we have

(2p− 1)U = ν∆t

4p(1− p)U2 = σ2∆t.

Shuzhong Zhang

IE 5441 21

This gives

p =1

2+

1

2√

σ2/(ν2∆t) + 1

U =√σ2∆t+ (ν∆t)2.

For small ∆t, this is approximately

p =1

2+

1

2

(νσ

)√∆t

u = eσ√∆t

d = e−σ√∆t.

Shuzhong Zhang