Download Derivation of the Black–Scholes Option Pricing Model

Chapter 27 Solutions
1.
Itˆo’s lemma is a useful result because it allows the computation of stochastic
differentials of arbitrary functions having as an argument a stochastic process
that itself is assumed to possess a stochastic differential. In this respect, Itˆo’s
formula is as useful as the chain rule of ordinary calculus. Given an Itˆo
stochastic process S(t,w) with respect to a given Wiener process Z(t,w), and
letting Y (t,w) = u[t, S(t,w)] be a new process, then based on Itˆo’s lemma and
Taylor’s theorem, we can obtain
dt × dt = 0, dZ× dZ = dt, dt × dZ = 0
1
dY = utdt + uSdS + 2 uSS (dS)2
A stochastic process is an Itˆo process if the random variable dS(t,w) can be
represented by
dS(t,w) = μ[t, S(t,w)]dt + σ[t, S(t,w)]dZ(t,w)
where The first term, μ[t, S(t,w)]dt, is the expected change in S(t,w) at time t.
The second term, σ[t, S(t,w)]dZ(t,w), reflects the uncertain term. dZ(t,w), is
called white noise followed standard normal distribution; it denotes an
infinitesimal change in the Wiener process, models financial uncertainty in
continuous time.
2. The stock price has a constant expected return μ and the volatility of its return
is constant σ times the Wiener process dZ(t,w) which follows normal
distribution with zero mean and variance, t. Therefore, by equation (27.12) in
1
section 27.3, the stock price follows a lognormal distribution with mean (μ-2
σ2)t and variance σ2t.
3. Assume the stock price follows the stochastic process as follows
dS(t,w) = μS(t,w)dt + σS(t,w)dZ(t,w)
The expected value of a call option is 𝐶 = 𝐸[𝑒 −𝑟𝑇 (𝑆𝑇 − 𝑘)+ ] where k is the
strike price in call option contract.
By equation (27.12), we can arrange the call option formula as
1 2
)𝑇+𝜎𝑍
𝐶 = 𝐸[𝑒 −𝑟𝑇 (𝑆𝑇 − 𝑘)+ ] = 𝐸[𝑒 −𝑟𝑇 (𝑆0 𝑒 (𝜇−2𝜎
− 𝑘)+ ]
(1)
Where z is the normal distribution N(0, T).
To derive the Black–Scholes call option model, first we should prove 𝜇 equal
to risk-free rater, r, in risk-neutral framework.
That is, 𝐸 𝑄 [𝑒 −𝑟𝑇 (𝑆𝑇 )] = 𝑆0 under the risk-neutral measure Q.
Proof:
1 2
)𝑇+𝜎𝑍
𝐸 𝑄 [𝑒 −𝑟𝑇 (𝑆𝑇 )] = 𝐸 𝑄 [𝑒 −𝑟𝑇 (𝑆0 𝑒 (𝜇−2𝜎
)] = 𝑆0 𝑒 (𝜇−𝑟)𝑇 = 𝑆0
Therefore, 𝜇 = 𝑟 and replace r into equation (1), we can obtain equation (2)
as follows:
1 2
)𝑇+𝜎𝑍
𝑆0 𝑒 (𝑟−2𝜎
− 𝑘 ≥ 0 (2)
Let standard normal distribution Y=
1
𝑌 < 𝑑2 =
𝜎√𝑇
𝑘
1
𝑆0
2
𝑍
√𝑇
, then we can rearrange equation (2) as
[ln ( ) + (𝑟 − 𝜎 2 ) 𝑇] (3)
Therefore, equation (2) holds if and only if equation (3) holds. Then equation
(1) can be written as
𝑑
2
𝐶 = 𝐸[(𝑆𝑇 − 𝑘)+ ] = ∫−∞
𝑑
2
= ∫−∞
1
√2𝜋
𝑑
1
√2𝜋
𝑑
2
) 𝑑𝑦 − 𝑒 −𝑟𝑇 𝑘 ∫−∞
1 2
− 𝑒 −𝑟𝑇 𝑘) 𝑒 −2𝑦 𝑑𝑦
1
√2𝜋
1 2
𝑒 −2𝑦 𝑑𝑦
2
1
𝑒 −2(𝑦+𝜎√𝑇) 𝑑𝑦 − 𝑒 −𝑟𝑇 𝑘𝑁(𝑑2 )
𝑑 +𝜎√𝑇 1
2
= 𝑆0 ∫−∞
√2𝜋
1 2
)𝑇+𝜎√𝑇𝑦
(𝑒 −𝑟𝑇 𝑆0 𝑒 (𝑟−2𝜎
1 2
1
−𝜎√𝑇𝑦− 𝜎2 𝑇)
2
(𝑆0 𝑒 (−2𝑦
2
= 𝑆0 ∫−∞
1
√2𝜋
1
𝑒 −2(𝑦
′ )2
𝑑𝑦 − 𝑒 −𝑟𝑇 𝑘𝑁(𝑑2 ) where y ' =y+σ√T
= 𝑆0 𝑁(𝑑1 ) − 𝑒 −𝑟𝑇 𝑘𝑁(𝑑2 ) where 𝑑1 = 𝑑2 + 𝜎√𝑇
4. let
𝑌 = 𝑆 𝑛 (𝑡, 𝑤) and use Itˆo’s lemma to find 𝑑𝑌
𝑑𝑆
𝑛 (𝑡,
𝜕𝑌
𝜕𝑌
1 𝜕 2𝑌
𝑤) = 𝑑𝑌 =
𝑑𝑡 +
𝑑𝑆 +
(𝑑𝑆)2
𝜕𝑡
𝜕𝑆
2 (𝜕𝑆)2
1
𝑑𝑌 = 0𝑑𝑡 + 𝑛𝑆 𝑛−1 (𝑡, 𝑤)𝑑𝑆 + 2 (𝑛)(𝑛 − 1)𝑆 𝑛−2 (𝑡, 𝑤)(𝑑𝑆)2
= 𝑛𝑆 𝑛−1 (𝑡, 𝑤)[𝜇𝑆(𝑡, 𝑤)𝑑𝑡 + 𝜎𝑆(𝑡, 𝑤)𝑑𝑍(𝑡, 𝑤)]
1
+ 2 (𝑛)(𝑛 − 1)𝑆 𝑛−2 (𝑡, 𝑤)(𝜎 2 𝑆 2 (𝑡, 𝑤)𝑑𝑡)
1
= (𝑛𝜇 + 2 (𝑛)(𝑛 − 1)𝜎 2 ) 𝑆 𝑛 (𝑡, 𝑤)𝑑𝑡 + 𝑛𝜎𝑆 𝑛 (𝑡, 𝑤)𝑑𝑍(𝑡, 𝑤)
1 2
)𝑡+𝜎𝑍
5. The stock price 𝑆𝑇 = 𝑆0 𝑒 (𝜇−2𝜎
follows lognormal distribution where Z
follows normal distribution with zero mean and variance, T.
a. Given the information 𝜇 = 0.15, 𝜎 = 0.2 and 𝑆0 = $50 in the first
two years, then the stock price at the end of two years is
1
𝑆2𝑦 = 50𝑒 (0.15−2
(0.2)2 )(2)+(0.2)𝑍
= 50𝑒 0.26+(0.2)𝑍
Where Z follows normal distribution with mean zero and variance, 2.
Therefore, the stock price follows lognormal distribution with mean 4.172
(ln50+0.26 = 4.172) and variance 0.8 (0.22(2) = 0.8).
b. At the end of three years, given the information 𝜇 = 0.28, 𝜎 = 0.35
in the last year, the stock price can be written in terms of the stock
price at the end of two years as
1
𝑆3𝑦 = 𝑆2𝑦 𝑒 (0.28−2
(0.35)2 )+(0.35)𝑍 ′
= 𝑆2𝑦 𝑒 0.4375+(0.35)𝑍
′
Where Z’ is the normal distribution with mean zero and variance, 1.
From the part (a), we can obtain
′
𝑆3𝑦 = 𝑆2𝑦 𝑒 0.4375+(0.35)𝑍 = 50𝑒 0.26+(0.2)𝑍 𝑒 0.4375+(0.35)𝑍
′
′
= 50𝑒 0.6975+(0.2)𝑍+(0.35)𝑍 = 𝑒 ln(50)+0.6975+(0.2)𝑍+(0.35)𝑍
′
′
= 𝑒 4.6095+(0.2)𝑍+(0.35)𝑍
Since Z and Z’ are two independent normal distributions, therefore the
sum of these two independent normal distribution Z”= (0.2)𝑍 +
(0.35)𝑍 ′ is following a normal distribution with mean zero and
variance 0.2025 (0.22(2)+0.352 = 0.2025).
Then the stock price at the end of three years follows lognormal
distribution with mean 4.6095 and variance 0.2025 (𝜎 = 0.45).
6. The stock price change following the stochastic process in question 2 is more
appropriate than the stochastic process in question 6 because the parameters 𝜇
and 𝜎 are meaningful for the return of stock price and can be estimated by
the expected return and standard deviation of the stock’s returns, respectively.
In addition, investors prefer to know how much percent of return for their
investment rather than the change of the investment.
7.
When the stock price follows the stochastic process as follows:
dS(t,w) = μdt + σdZ(t,w), μ=1.5, σ=2 and S0 = $110
a) Then St follows a normal distribution N(S0+ μt, σ2t). Therefore, for next
year, S1 ~ N(110+1.5, 22) = N(111.5, 4). It implies that the stock price in
the next year follows a normal distribution with mean 111.5 and standard
deviation 4.
b) 95% confidence limits for the stock price in the next year are between the
mean±standard deviation, that is, [ 111.5- 4, 111.5+ 4] =[107.5, 115.5]
8. When the stock price follows the stochastic process as follows:
dS(t,w) = μ S(t,w) dt + σ S(t,w) dZ(t,w), μ=0.7, σ2=25% and S0 = $100
1 2
)𝑡+𝜎𝑍
Then, the stock price 𝑆𝑇 = 𝑆0 𝑒 (𝜇−2𝜎
follows lognormal distribution
where Z follows normal distribution with zero mean and variance, T
a) At the end of six months,
𝑆0.5𝑦 = 100𝑒
1
2
(0.7− (0.25))(0.5)+(0.5)𝑍1
Where Z1 follows normal distribution with zero mean and variance 0.5.
Therefore, the stock price follows lognormal distribution with mean 4.893
1
(ln100+(0.7 − 2 (0.25)) (0.5) = 4.893) and variance 0.125 (0.52(0.5) =
0.125)
The expected stock price at the end of six months is e 4.893=133.353
b) The standard deviation of the stock price at the end of six months
is e 0.125=1.133
c) In next year, 𝑆1𝑦 = 100𝑒
1
2
(0.7− (0.25))+(0.5)𝑍2
Where Z2 follows normal distribution with zero mean and variance 1.
Therefore, the expected stock price is 100𝑒
1
2
(0.7− (0.25))
= 177.713
9. Assume the stock price follows the stochastic process as follows:
dS(t,w) = μ S(t,w) dt + σ S(t,w) dZ(t,w)
(a) Y=3 S(t,w), then dY=3 dS(t,w). Therefore, Y follows the stochastic
process as dY=3dS(t,w) = 3μ S(t,w) dt + 3σ S(t,w) dZ(t,w)
(b) Y= Sn(t,w), then
1
dY= nSn-1(t,w) dS(t,w)+2 n(n-1)Sn-2(t,w) [dS(t,w)]2
1
= nSn-1(t,w) [μ S(t,w) dt + σ S(t,w) dZ(t,w)]+ 2 n(n-1)Sn-2(t,w) (σ2 ) S2(t,w)dt
1
= (nμ+2 n(n-1) σ2 )Sn(t,w) dt + (n σ) Sn(t,w) dZ(t,w)
(c) Y=e r(T-t) Sn(t,w), then
1
dY= -e r(T-t) Sn(t,w)dt+ e r(T-t) {nSn-1(t,w) dS(t,w)+2 n(n-1)Sn-2(t,w) [dS(t,w)]2}
1
=(nμ+2 n(n-1) σ2 -1) e r(T-t) Sn(t,w) dt + e r(T-t) (n σ) Sn(t,w) dZ(t,w)
1 2
)𝑡+𝜎𝑍
10. The stock price 𝑆𝑇 = 𝑆0 𝑒 (𝜇−2𝜎
follows lognormal distribution where
Z follows normal distribution with zero mean and variance, T. The stock price
in two years in equation 5 is
𝑆2𝑦 = 50𝑒
1
2
(0.15− (0.2)2 )(2)+(0.2)𝑍
= 50𝑒 0.26+(0.2)𝑍
Therefore, the stock price follows lognormal distribution with mean 4.172
(ln50+0.26 = 4.172) and variance 0.8 (0.22(2) = 0.8).
The probability that stock price larger than $130 in two year can be calculated
as
P(𝑆2𝑦 > $130)=P(50𝑒 0.26+(0.2)𝑍 > 130)
= 𝑃 (𝑍 >
ln(
130
)−0.26
50
0.2
) = 𝑃(𝑋 >
ln(
130
)−0.26
50
0.2√0.2
)=P(X>7.7761)=0
Where X is the standard normal distribution and Z is the normal distribution
with zero mean and variance, 2.