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ACTS 4302
Instructor: Natalia A. Humphreys
SOLUTION TO HOMEWORK 4
Lesson 6: Binomial trees: miscellaneous topics.
Lesson 7: Modeling stock prices with the lognormal distribution.
Problem 1
For a 1-year American put option on a stock with 6 months left to expiry, you are given:
(i) The strike price is 42.
(ii) The continuous dividend rate for the stock is 0.03.
(iii) 6 months before expiry, the stock price is 33.
(iv) The value of a 6-month call option on the stock with a strike price of 42 is 0.155.
Calculate the lowest possible continuously compounded risk-free rate so that exercising the option at
6 months before expiry is optimal.
Solution. Since early exercise is optimal, the present value of interest on the strike price must exceed
the present value of dividends plus the value of the implicit call option:
K(1 − e−rt ) ≥ S(1 − e−δt ) + C
The present value of dividends is:
33 1 − e−0.5·0.03 = 0.4913
The value of the call option is given as 0.155. The present value of interest on the strike price is
42 1 − e−0.5r
Hence,
42 1 − e−0.5r ≥ 0.4913 + 0.155 = 0.6463
e−0.5r = 0.9846
r = 0.031
Problem 2
The Jarrow-Rudd binomial tree is used to price an option on a non-dividend paying stock. You are
given:
(i) The risk-free rate is 0.045.
(ii) The annual volatility is 29%.
(iii) The period of the binomial tree is 7 months.
Determine the risk-neutral probability of an up move.
Solution. Recall that for a Lognormal (Jarrow-Rudd) tree:
u = e(r−δ−0.5σ
2 )h+σ
√
h
, d = e(r−δ−0.5σ
2 )h−σ
√
h
and p∗ =
e(r−δ)h − d
u−d
Hence,
2
√
u = e0.5833[0.045−0.5(0.29 )]+0.29√ 0.5833 = e0.2232 = 1.2501
2
d = e0.5833[0.045−0.5(0.29 )]−0.29 0.5833 = e−0.2198 = 0.8027
e0.045·0.5833 − 0.8027
1.0266 − 0.8027
p∗ =
=
= 0.5005
1.2501 − 0.8027
0.4474
Copyright ©Natalia A. Humphreys, 2014
ACTS 4302. AU 2014. SOLUTION TO HOMEWORK 4.
Problem 3
A Cox-Ross-Rubinstein binomial tree is used to model an option. You are given:
(i) The continuously compounded risk-free rate is 5%.
(ii) σ = 0.06.
(iii) δ = 0.
Determine the largest period for which this tree can be used without violating arbitrage conditions
on the nodes.
Solution. To avoid arbitrage, u and d must satisfy:
d < e(r−δ)h < u
For a Cox-Ross-Rubinstein tree:
u = eσ
√
h
, d = e−σ
√
h
Thus, to avoid arbitrage, we need e(r−δ)h < u ⇔ e(r−δ)h < eσ
eσ
√
√
h.
Since δ = 0,
√
0.06 h
h
> erh ⇔ e
> e0.05h ⇔
√
√
6
0.06 h > 0.05h ⇔ h < ⇔ h < 1.44
5
Problem 4
An American company expects to receive £450,000 from sales in England at the end of 9 months.
The current exchange rate is $1.5/£. The company would like to guarantee that it will get at least
this rate when it receives the pounds, so that it will receive at least $675,000.
You are given:
(i)
(ii)
(iii)
(iv)
The continuously compounded risk-free rate in dollars is 4.3%.
The continuously compounded risk-free rate in pounds is 3.5%.
Relative volatility of the currencies is 0.2.
A two-period Cox-Ross-Rubinstein binomial tree is used to determine the price of options.
Determine the cost of an option, in dollars, which will guarantee the current exchange rate at the
end of 9 months.
Solution. The company should purchase a European put on pounds with strike price 1.50. For this
option, the domestic currency is dollars and the foreign currency is pounds. Thus,
r = rd = r$ = 0.043
δ = rf = r£ = 0.035
Also, note that h = 0.75/2 = 0.375.
In a Cox-Ross-Rubinstein binomial tree the up and down movements, and the risk-neutral probability
of an up move, are
√
√
h
= e0.2 0.375
= 1.1303
u = eσ √
√
−σ h
−0.2 0.375
d=e
=e
= 0.8847
u − d = 0.2456
e(r−δ)h − d
e(0.043−0.035)0.375 − 0.8847
p∗ =
=
= 0.4817
u−d
0.2456
1 − p∗ = 0.5183
The spot exchange rate tree:
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Copyright ©Natalia A. Humphreys, 2014
ACTS 4302. AU 2014. SOLUTION TO HOMEWORK 4.
u2 x0 = 1.9163
ux0 = 1.6954
x0 = 1.5
udx0 = 1.5
dx0 = 1.3271
ddx0 = 1.1741
The put tree:
Puu = 0
Pu
Pud = 0
P
Pd
Pdd = 0.3259
Pd = e−rh (p∗ Pud + (1 − p∗ )Pdd ) = e−0.043·0.375 · 0.5183 · 0.3259 = 0.1662
Since the put is the European put, we continue with this value of Pd :
P = e−0.043·0.375 · 0.5183 · 0.1662 = 0.0848$/£
This is the cost of a put to buy £1. The total price of a put on £450,000 is:
450,000 · 0.0848 = 38,143.61
Note that if the put were an American put, the answer would be slightly different:
If exercised, 1.5 − 1.3271 = 0.1729. Since 0.1729 > 0.1662, we use 0.1729 in further calculations:
P = e−0.043·0.375 · 0.5183 · 0.1729 = 0.0882$/£
This is the cost of a put to buy £1. The total price of a put on £450,000 is:
450,000 · 0.0882 = 39,684.74
Problem 5
You are given the following information on the price of a stock:
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Copyright ©Natalia A. Humphreys, 2014
ACTS 4302. AU 2014. SOLUTION TO HOMEWORK 4.
Date
Stock price
Jul. 1, 2007
35.30
Aug. 1, 2007
33.90
Sep. 1, 2007
41.20
Oct. 1, 2007
31.95
Nov. 1, 2007
38.25
Dec. 1, 2007
46.18
Estimate the annual volatility of continuously compounded return on the stock.
(A) 0.20
(B) 0.62
(C) 0.68
(D) 0.69
(E) 0.75
Key: D
Solution. To estimate volatility from historical data:
s
P
P 2
√
xt
n
St
xt
2
−x
¯ , where xt = ln
, x
¯=
σ
ˆ= N
n−1
n
St−1
n
N is the number of periods per year, n is the number one less than the number of observations of
stock price.
In our problem N=12 and n=5. We calculate xt = ln(St /St−1 ) and x2t :
St
xt
x2t
35.30
33.90 -0.0405 0.0016
41.20
0.1950
0.0380
31.95 -0.2543 0.0647
38.25
0.1800
0.0324
46.18
0.1884
0.0355
Summing up the third column and its squares,
5
X
5
X
0.2687
= 0.05373,
x2t = 0.1722
5
t=1
t=1
√
5
0.1722
s2 =
− 0.053732 = 0.03944, s = 0.03944 = 0.1986
4
5
That is the monthly volatility. The annual volatility is
√
0.1986 12 = 0.688 ≈ 0.69 xt = 0.2687, x
¯=
Problem 6
A stock’s price follows a lognormal model. You are given:
(i) The current price of the stock is 105.
(ii) The probability that the stock’s price will be less than 98 at the end of 6 months is 0.3483.
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Copyright ©Natalia A. Humphreys, 2014
ACTS 4302. AU 2014. SOLUTION TO HOMEWORK 4.
(iii) The probability that the stock’s price will be less than 115 at the end of 9 months is 0.7123.
Calculate the expected price of the stock at the end of one year.
Solution. Recall from Lesson 7 that probabilities of payoffs of stock prices are:
P r(St < K) = N (−dˆ2 ), and P r(St > K) = N (dˆ2 ), where
ln SK0 + (α − δ + 0.5σ 2 )t
ln SK0 + (µ + σ 2 )t
ˆ
√
√
d1 =
=
σ t
σ t
ln SK0 + µt
ln SK0 + (α − δ − 0.5σ 2 )t
ˆ
√
√
=
d2 =
σ t
σ t
√
dˆ2 = dˆ1 − σ t
Hence, our statements could be written as:
P r (S0.5 < 98) = N (−dˆ2 (0.5)) = 0.3483
P r (S0.75 < 115) = N (−dˆ2 (0.75)) = 0.7123
Calculating dˆ2 (0.5) and dˆ2 (0.75), we obtain:
105
98 √+
ln
dˆ2 (0.5) =
ln
dˆ2 (0.75) =
µ · 0.5
σ 0.5
105
115 √+
=
µ · 0.75
0.069 + 0.5µ
0.7071σ
=
−0.091 + 0.75µ
0.866σ
σ 0.75
On the other hand, using the standard normal probability table for N (z) = 0.3483 and N (z) = 0.7123,
we obtain:
−dˆ2 (0.5) = −0.39
−dˆ2 (0.75) = 0.56
Thus, we solve the system:


0.069+0.5µ
0.7071σ
= 0.39
⇔


0.069 + 0.5µ = 0.2758σ

 −0.091+0.75µ
−0.091 + 0.75µ = −0.485σ
= −0.56
0.866σ


 0.5µ − 0.2758σ = −0.069
 µ − 0.5517σ = −0.138
⇔


0.75µ + 0.485σ = 0.091
µ + 0.6467σ = 0.1213
to obtain: σ = 0.2164, µ = −0.01864. Therefore, m = µ · t = −0.01864t, v 2 = σ 2 t = 0.0468t and the
expected value of the stock after one year is
2
E[S1 ] = S0 em+0.5v = 105e−0.01864+0.5·0.0468 = 105.5025
This problem could be solved slightly differently, using the theory of percentiles.
The probability statements result in these two equations expressing the percentiles in terms of standard normal coefficients of the 100pth percentile, or zp :
√
105e0.5µ+ √0.5σz0.3483 = 98
105e0.75µ+ 0.75σz0.7123 = 115
The normal percentiles are: z0.3483 = −0.39 and z0.7123 = 0.56. Hence,
0.5µ − 0.2758σ = −0.069
0.75µ + 0.485σ = 0.091
This is the same system as in the first approach above. Page 5 of 7
Copyright ©Natalia A. Humphreys, 2014
ACTS 4302. AU 2014. SOLUTION TO HOMEWORK 4.
Problem 7
A stock’s price follows a lognormal model. You are given:
(i) The initial price is 95.
(ii) α = 0.14.
(iii) δ = 0.07.
(iv) σ = 0.4.
Construct a 95% confidence interval for the price of the stock at the end of five years.
Solution. The lognormal parameter m for the distribution of the stock price after five years is
(α − δ − 0.5σ 2 )t = (0.14 − 0.07 − 0.5 · 0.42 )5 = −0.05
√
The lognormal parameter v is 0.4 5 = 0.8944. The confidence interval is
−0.05 ± 1.96(0.8944) = (−1.8031, 1.7031)
Exponentiating, we get
e−1.8031 , e1.7031 = (0.1648, 5.4909)
Multiplying by the initial stock price of 95, the answer is (15.66, 521.64).
Problem 8
A stock’s price follows a lognormal model. The current price of the stock is 50. A 95% confidence
interval for the price of the stock at the end of one year is (41.20, 73.05).
Construct a 90% confidence interval for the price of the stock at the end of two years.
Solution. We know that
41.20 = 50eµ−1.96σ and 73.05 = 50eµ+1.96σ
so the quotient is:
73.05
= 1.7731
41.20
ln 1.7731
σ=
= 0.1461
3.92
e3.92σ =
and the product is
e2µ =
We need
41.20 · 73.05
= 1.2039
502
√
2σ
50e2µ−1.645
√
and 50e2µ+1.645
2σ
Calculating:
√
√
e1.645 2σ =√e1.645 2·0.1461 = 1.4049
50e2µ−1.645 2σ = 50(1.2039)/1.4049 = 42.85
√
50e2µ+1.645
2σ
= 50(1.2039)(1.4049) = 84.56
Thus the 90% confidence interval for the price of the stock at the end of two years is (42.85, 84.56).
Problem 9
A stock’s price follows a lognormal model. You are given:
(i) The stock’s initial price is 60.
(ii) The stock’s continuously compounded rate of return is 0.06.
(iii) The stock’s continuously compounded dividend rate is 0.03.
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ACTS 4302. AU 2014. SOLUTION TO HOMEWORK 4.
Copyright ©Natalia A. Humphreys, 2014
(iv) Volatility is 0.25.
Calculate the conditional expected value of the stock after 1 year given that it is greater than 70.
Solution. We use the formula
E[St |St > K] =
S0 e(α−δ)t N (dˆ1 )
N (dˆ2 )
Calculating dˆ1 and dˆ2 :
ln SK0 + (α − δ + 0.5σ 2 )t
ln (60/70) + 0.06 − 0.03 + 0.5(0.252 )
ˆ
√
d1 =
=
= −0.3716
0.25
σ t
√
dˆ2 = dˆ1 − σ t = −0.3716 − 0.25 = −0.6216
Then
N (−0.3716)
=
N (−0.6216)
1 − N (0.3716)
1 − N (0.3716)
= 61.8273
= 61.8273
≈
1 − N (0.6216)
1 − N (0.6216)
1 − N (0.37)
1 − 0.6443
≈ 61.8273
= 61.8273
= 82.1822
1 − N (0.62)
1 − 0.7324
E[St |St > 70] = 60e0.03
Problem 10
A stock’s price follows a lognormal model. You are given:
(i) The stock’s initial price is 42.
(ii) The stock’s continuously compounded rate of return is 0.156.
(iii) The stock’s continuously compounded dividend rate is 0.03.
(iv) Volatility is 0.36.
A European call option on the stock expires in 6 months and has strike price 47.
Calculate the expected payoff for the call option.
Solution. We use the formula
E[max(0, St − K)] = S0 e(α−δ)t N (dˆ1 ) − KN (dˆ2 )
Calculating dˆ1 and dˆ2 :
S0
ln
+ (α − δ + 0.5σ 2 )t
ln (42/47) + (0.156 − 0.03 + 0.5(0.362 ))0.5
K
√
√
dˆ1 =
=
= −0.0671 ≈ −0.07
σ t
0.36 0.5
√
√
dˆ2 = dˆ1 − σ t = −0.0671 − 0.36 0.5 = −0.3217 ≈ −0.32
N (dˆ1 ) = N (−0.07) = 1 − N (0.07) = 1 − 0.5279 = 0.4721
N (dˆ2 ) = N (0.32) = 1 − N (0.32) = 1 − 0.6255 = 0.3745
The expected payoff for the call option is
E[max(0, St − K)] = 42e0.063 (0.4721) − 47(0.3745) = 3.5161
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