Solution to Homework HW 10

ACTS 4302
Instructor: Natalia A. Humphreys
SOLUTION TO HOMEWORK 10
Lesson 16: Brownian Motion.
Lesson 17: Itˆo’s Lemma, Black-Scholes Equation.
Problem 1
Stock I and stock II open the trading day at the same price. Let X(t) denote the dollar amount
by which stock I’s price exceeds stock II’s price when 100t percent of the trading day has elapsed.
{X(t), 0 ≤ t ≤ 1} is modeled as an arithmetic Brownian motion process with drift 0.1 and variance
parameter σ = 0.3.
After 5/12 of the trading day has elapsed, the price of stock I is 51.75 and the price of stock II is 51.65.
Calculate the probability that X(11/12) ≥ 0
Solution. Let SI and SII be the prices of stocks I and II. Then X(t) = SI − SII
By definition of the Arithmetic Brownian motion,
X(t) = X(0) + αt + σZ(t).
If t is the latest time for which X(t) is known, then X(t + s)|X(t) has a normal distribution with
5
,
m = X(t) + αs and v 2 = σ 2 s. At time t = 12
5
5
5
11
5
X
= SI
− SII
= 51.75 − 51.65 = 0.1, and s =
−
= 0.5
12
12
12
12 12
5
2
2
2
Hence, X 11
12 )|X( 12 ∈ N m = 0.1 + 0.1 · 0.5 = 0.15, v = 0.3 · 0.5 = 0.045 = 0.2121 . Therefore,
0 − 0.15
11
5
= N (0.7071) ≈ N (0.71) = 0.7611 P r X( ) ≥ 0|X( ) = 0.1 = 1 − N
12
12
0.2121
Problem 2
S(t) denotes the price of the stock at time t. A stock’s price follows geometric Brownian motion. The
process has a drift α = 0.15 and volatility σ = 40%. S(0), the initial price of the stock, is $60.
Determine P r (55 < S(3) < 70).
Solution.
P r (55 < S(3) < 70|S(0) = 60) = P r (S(5) < 70|S(0) = 60) − P r (S(5) ≤ 55|S(0) = 60) = P1 − P2
By definition of the Geometric Brownian motion, if X(t) follows Geometric Brownian motion then
ln X(t) follows Arithmetic Brownian motion, i.e. if t is the latest time for which X(t) is known, then
ln X(t + s)/X(t)|X(t) has a normal distribution with parameters m = (α − δ − 0.5σ 2 )s and v 2 = σ 2 s.
Hence,
S(3)
ln
∈ N m = (0.15 − 0.5 · 0.42 )3 = 0.21, v 2 = 0.42 · 3 = 0.48 = 0.69282
S(0)
Therefore,
S(3)
7
ln 1.25 − 0.21
P1 = P r ln
< ln
=N
= N (−0.08) = 0.4681
S(0)
6
0.6928
S(3)
5.5
ln 0.9167 − 0.21
P2 = P r ln
< ln
=N
= N (−0.4284) ≈ N (−0.43) = 0.3336
S(0)
6
0.6928
Therefore,
P r (55 < S(3) < 70|S(0) = 60) = 0.4681 − 0.3336 = 0.1345
Copyright ©Natalia A. Humphreys, 2014
ACTS 4302. AU 2014. SOLUTION TO HOMEWORK 10.
Problem 3
A stock price is governed by a geometric Brownian motion model with the following parameters:
(i) Current stock price is $45
(ii) Volatility is 60% per annum
(iii) Expected return is 15% per annum, continuously compounded
(iv) The stock pays no dividends
(v) There are no transaction costs when it is bought or sold
Calculate the probability that a two and a half-year European call option purchased today with a strike
price of $65.00 will be exercised.
Solution. By definition of the Geometric Brownian motion, if X(t) follows Geometric Brownian motion
then ln X(t) follows Arithmetic Brownian motion, i.e. if t is the latest time for which X(t) is known,
then ln X(t + s)/X(t)|X(t) has a normal distribution with m = (α − δ − 0.5σ 2 )s and v 2 = σ 2 s.
Hence,
S(2.5)
ln
∈ N m = (0.15 − 0.5 · 0.62 )2.5 = −0.075, v 2 = 0.62 · 2.5 = 0.9 = 0.94872
S(0)
Therefore,
S(2.5)
65
>
S(0)
45
S(2.5)
= P r ln
> ln 1.4444 =
S(0)
P = P r (S(2.5) ≥ 65|S(0) = 45) = P r
ln 1.4444 + 0.075
=1−N
= 1 − N (0.4667) ≈ N (−0.47) = 0.3192
0.9487
Problem 4
Consider a European call option on a stock with nine months to expiry and strike price 80. The
underlying stock follows the Itˆ
o process
dS(t)
= 0.17dt + 0.45dZ(t)
S(t)
You are given:
(i) S(0) = 70
(ii) The continuously compounded dividend rate is 3%.
(iii) The Sharpe ratio of the stock is 0.35.
Calculate the price of the option.
Solution. We are given that
α−r
= 0.35, α − δ = 0.17, σ = 0.45.
σ
Since δ = 0.03, it follows that α = 0.2 and, hence, r = α − φ · σ = 0.2 − 0.35 · 0.45 = 0.0425.
Thus, the price of the European call is:
φ=
C = Se−δt N (d1 (K)) − Ke−rt N (d2 (K)), where
ln 70
+ 0.0425 − 0.03 + 0.5 · 0.452 · 0.75
ln (S/K) + (r − δ + 0.5σ 2 )t
80
√
√
d1 =
=
=
σ t
0.45 · 0.75
0.0482
=−
= −0.1237
0.3897
N (d1 ) = N (−0.12) = 0.4522
√
d2 = d1 − σ t = 0.1237 − 0.3897 = −0.5134
N (d2 ) = N (−0.51) = 0.3050
C = 70 · e−0.03·0.75 · 0.4522 − 80e−0.0425·0.75 · 0.3050 = 30.9497 − 23.6345 = 7.3152
Page 2 of ??
Copyright ©Natalia A. Humphreys, 2014
ACTS 4302. AU 2014. SOLUTION TO HOMEWORK 10.
Problem 5
Consider a call option C(S, t) on a non-dividend paying stock. The stock follows the process
dS(t)
= 0.15dt + 0.35dZ(t)
S(t)
where {Z(t)} is a Brownian motion. The call option follows the process
dC(S, t)
= 0.37dt + σC dZ(t),
C(S, t)
The risk free rate is 0.05.
Determine σC
Solution. We are given that:
α − δ = 0.15, σ = 0.35, γ = 0.37
We need to find σC .
By the connection between the risk premium of an option and the risk premium of the underlying stock
γ − r = Ω(α − r)
Therefore,
0.37 − 0.05
0.32
γ−r
=
=
= 3.2
α−r
0.15 − 0.05
0.1
By the connection between the volatility of an option and the volatility of the underlying stock
Ω=
σoption = σstock |Ω|
Therefore,
σC = 0.35 · 3.2 = 1.12
Problem 6
A contract will give you one share of stock at the end of a year if the price of the stock is less than 45
at that time.
You are given:
(i) The stock follows the process
dS(t)
= 0.14 dt + 0.18 dZ(t)
S(t)
(ii) S(0) = 55
(iii) The continuously compounded dividend rate is 3%.
(iv) The continuously compounded risk free rate is 5%.
Calculate the value of the contract.
Solution. We are given:
α − δ = 0.14, σ = 0.18
The contract is the asset-or-nothing put option. It’s value is
SO = Se−δT N (−d1 )
Calculating
ln 55
+ 0.05 − 0.03 + 0.5 · 0.182 · 1
ln (S/K) + (r − δ + 0.5σ 2 )t
45
√
d1 =
=
=
0.18 · 1
σ t
0.2369
= 1.3159
=
0.18
N (−d1 ) = N (−1.32) = 0.0934
SO = 55 · e−0.03 · 0.0934 = 4.9852 ≈ 4.99
Page 3 of ??
Copyright ©Natalia A. Humphreys, 2014
ACTS 4302. AU 2014. SOLUTION TO HOMEWORK 10.
Problem 7
The time-t price of a non-dividend paying stock is S(t). S(t) follows an arithmetic Brownian motion
with α = 0.15, σ = 0.25.
The time-t price of an exotic option on the stock is C(t). The partial derivatives of C(t) are:
∂C
∂2C
∂C
= 0.6,
= 0.08,
= 0.12
2
∂S
∂S
∂t
Determine the drift and the volatility of the Itˆo process followed by C.
Solution. By definition of the Arithmetic Brownian Motion and corresponding differential:
S(t) = S(0) + 0.15t + 0.25Z(t)
dS = 0.15dt + 0.25dZ
By Itˆo’s Lemma
dC(S, t) = CS dS + 0.5CSS (dS)2 + Ct dt = 0.6dS + 0.5 · 0.08(dS)2 + 0.12dt =
= 0.6 (0.15dt + 0.25dZ) + 0.04 · 0.0625dt + 0.12dt = 0.2109dt + 0.15dZ
Hence,
αC = 0.2125, σC = 0.15
Problem 8
The stochastic process X(t) can be expressed as
X(t) = (tZ(t))2 + 4 t3 Z(t)
Using Itˆo’s lemma, express dX(t) in terms of dt and dZ(t).
Solution. By Itˆ
o’s Lemma
dX(Z, t) = XZ dZ + 0.5XZZ (dZ)2 + Xt dt
XZ = 2t2 Z + 4t3
XZZ = 2t2
Xt = 2tZ 2 + 12t2 Z
dX = (2t2 Z + 4t3 )dZ + 0.5 · 2t2 dt + (2tZ 2 + 12t2 Z)dt = t2 (2Z + 4t)dZ + t(t + 2Z2 + 12tZ)dt
Problem 9
For an option on a stock, you are given
(i) The price of the option is 15.2
(ii) ∆ = 0.85
(iii) Γ = 0.02
(iv) θ = −1.96 per year.
(v) The stock’s volatility is 30% per year.
(vi) The continuously compounded risk-free rate is 4%.
Determine the price of the stock.
(A) 18.21
(B) 26.56
(C) 33.34
(D) 37.77
(E) 44.80
Key: E
Solution. By Black-Scholes Equation
∆S(r − δ) + 0.5ΓS 2 σ 2 + θ = rC
Page 4 of ??
Copyright ©Natalia A. Humphreys, 2014
ACTS 4302. AU 2014. SOLUTION TO HOMEWORK 10.
Using the given ∆, Γ, θ, r and C, we have:
0.5 · 0.02 · 0.32 S 2 + 0.85(0.04 − 0.02)S − 0.04 · 15.2 = 0
0.0009S 2 + 0.017S − 2.568 = 0
D = 0.0172 + 4 · 0.0009 · 2.568 = 0.0095 = 0.04982
−0.017 + 0.0498
S=
= 44.8007 ≈ 44.80 2 · 0.0009
Problem 10
For two options on a non-dividend paying stock following the Black-Scholes framework, you are given:
Option
∆
Γ
θ
Option Premium
1
0.5500 0.0700 -0.015
2.95
2
0.3200
1.08
0.030
-0.007
θ is expressed per day. The stock’s price is 51.
Determine r, the continuously compounded risk-free rate.
Solution. By Black-Scholes Equation
∆S(r − δ) + 0.5ΓS 2 σ 2 + θ = rC
Since δ = 0, the above equation becomes:
∆Sr + 0.5ΓS 2 σ 2 + θ = rC
Given the information about the two options, we have the following system of equations:
(0.55 · 51 − 2.95)r + 0.5 · 0.07 · 512 · σ 2 − 0.015 · 365 = 0
⇔
(0.32 · 51 − 1.08)r + 0.5 · 0.03 · 512 · σ 2 − 0.007 · 365 = 0
25.1r + 91.035σ 2 = 5.475
r + 3.6269σ 2 = 0.2181
⇔
⇔
2
15.24r + 39.015σ = 2.555
r + 2.56σ 2 = 0.1677
1.0669σ 2 = 0.0504
σ 2 = 0.0472
⇔
2
r = 0.0468
r + 2.56σ = 0.1677
Page 5 of ??

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