Chapter Five
Option Pricing
Answers to Problems and Questions
1. Arbitrage is the existence of a riskless profit, especially when you do not have
to commit funds to earn the profit.
2. Market prices move continually, and frequently move slightly away from an
equilibrium price. It is the actions of arbitrageurs that move them back to their
proper level.
3. Given the €/$ and €/AuD exchange rates, an Australian dollar should cost
$1.15/$1.63 = $0.7055. If the Australian dollar sells for less than this, the
appropriate arbitrage activity would be to buy the Australian dollar and
exchange it for another currency.
4. Differences in trading fees and taxes may make a situation profitable for one
person, but not for another. For instance, a trade in the tax-exempt account of
an endowment fund may be much less attractive from the perspective of a
taxable investor.
5. The law of one price states that equivalent assets should sell for the same price.
You should be able to buy just as much with a twenty dollar bill as with two
tens.
6. You can create a synthetic put by selling stock short and adding a long call
position. The profit and loss diagram is similar to that of a put option. More
formally, from put/call parity, P = C + K/(1+R)t – S. This means a package
comprised of a long call, short stock, and the investment of the present value of
the striking price is equivalent to a long put.
7. If interest rates are near zero, then C – P ≈ 0, meaning for a given expiration,
an at-the-money call should sell for approximately the same price as an at-themoney put.
8. 1) the put might have a longer expiration, 2) the stock price might be below the
striking price, 3) non-synchronous trading caused the put and call prices to be
determined at different points in time, and 4) the underlying stock may be
expected to pay a large cash dividend during the life of the option.
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Chapter Five. Option Pricing
9. If the branch probability is 100%, then the outcome is known for certain and is
riskless. A riskless outcome should earn the riskless rate of return. The option
premium, then, would be the present value of the option’s intrinsic value at
option expiration.
10. P = C – S + K/(1+R)t
P = $1 – 24.25 + 25/(1.0565).25 = $1.41
11. P = $0.50 – 23 + 25/(1.0615)86/365 = $2.15
12. The aggregate value of the calls should not change because of a stock split. In
this example, the investor would own six July $47.50 options post-split. They
would be worth 6 x 100 x .5 x $8.25 = $2,475.
13. Use the data on stock XYZ to get the implied interest rate. Assume there is
one year until expiration:
C – P = S – K/(1 + R)
$4.13 – 2.75 = 50 – 50/(1 + R)
R = 2.84%
Now use this rate and put/call parity to solve for the premium for the ABC put:
P = C – S + K/(1+R)t
P = $5 – 80 + 80/(1.0284) = $2.79
Note that the assumption of one year until expiration is not critical. You
can assume the XYZ option has any remaining life you wish. The implied
interest rate associated with the time period will still lead you to an ABC put
value of $2.79.
14. The difference between the call premium and the put premium (C – P) should
equal the difference between the stock price and the present value of the
striking price, S – K/(1 + R)t . In this example, the latter term equals $77 –
75/(1.0625)1/12 = $2.38. The difference between the call and put prices is
$1.12; this is too little. This means the quantity C – P must get larger, which
happens if the price of the call goes up and/or the price of the put goes down.
It would make sense to buy the call and write the put.
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Chapter Five. Option Pricing
Putting together a put/call parity arbitrage table like Table 5-3,
Activity
Buy 75 call
Short Stock
Write 75 put
Lend
Sum
Cash Flow
- 4.50
+ 77.00
+ 3.38
-75.88
0
Value at Option Expiration
$50
$75
$100
0
0
$25
-50
-75
-100
-25
0
0
76.26
76.26
76.26
+ $1.26
+ $1.26
+ $1.26
Regardless of what happens to the stock price, these trades would generate a
riskless profit.
15. 55 – 5N = 45
 N =2
(50 – 2C)(1 + .065/2) = 45  C = $3.21
16. 55 – 7N = 45
 N =1.429
(50 – 1.429C)(1 + .065/2) = 45  C = $4.49
17. 100 – 15N = 70  N =2
(85 – 2C)(1 + .065) = 70  C = $9.64
18. 100 – 22N = 70  N =1.364
(85 – 1.364C)(1 + .065) = 70  C = $14.13
19. 50 –5N = 42
 N = 1.6
(45 – 1.6C)(1 + .065/4) = 42  C = $2.29
20. 50 –3N = 42
 N = 2.667
(45 – 2.667C)(1 + .065/4) = 42  C = $1.38
21. 110 –10N = 97  N = 1.3
(95 – 1.3C)(1 + .065) = 97
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 C = $3.02
Chapter Five. Option Pricing
22. You can do this using put/call parity or directly via binomial pricing. Using
the latter method, find a portfolio of stock and long puts that will equal the
higher stock price at expiration:
45 + 5N = 55  N = 2
(50 + 2P)(1.065) = 55
 P = $0.82
Via put/call parity, first solve for the value of the $50 call:
55 – 5N = 45  N = 2
(50 – 2C)(1.065) = 45  C = $3.87
Then use the put/call parity model:
P = C – S + K/(1 + R)t
= 3.87 – 50 + 50/1.065 = $0.82
The penny difference comes from rounding.
23. 45 + 6N = 55  N = 1.667
(50 + 1.667P)(1.065) = 55  P = $0.99
24. 100 – 10N = 80  N = 2
(S – 2 x 4)(1.065) = 80  S = $83.12
25. 72 – 7N = 56  N = 2.286
(S – 2.286 x 5)(1 + .065/2) = 56  S = $65.67
26. 70 – 10N = 48  N = 2.2
(58 – 2.2 x 4.5)(1 + R) = 48
R = -0.21%
27. 65 – 15N = 35  N = 2
(50 – 2C)(1.0625)2 = 35
 C = $9.50
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Chapter Five. Option Pricing
value of 10,000 shares = $9.50 x 10,000 = $95,000
28. 65 – 10N = 35
N=3
(50 – 3 x 9.5)(1.0625)t = 35
 t = 8.04 years
29. Substitute the new striking price in the equation on page 118:
6.66 x (0.5557) – 7.5 e–0.5 (15/36) = $0.73
30. 6.66 x (0.5557) – 7.5 e–(0.5)(0.5) (15/36) = $0.65
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