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Unformatted text preview: A Discussion of Financial Economics in Actuarial Models
A Preparation for the Actuarial Exam MFE/3F
Marcel B. Finan
Arkansas Tech University
c All Rights Reserved
Preliminary Draft
October 10, 2011 2 To Pallavi and Amin Preface
This is the third of a series of books intended to help individuals to pass actuarial exams. The
present manuscript covers the ﬁnancial economics segment of Exam M referred to by MFE/3F.
The ﬂow of topics in the book follows very closely that of McDonald’s Derivatives Markets. The
book covers designated sections from this book as suggested by the 2009 SOA Syllabus.
The recommended approach for using this book is to read each section, work on the embedded
examples, and then try the problems. Answer keys are provided so that you check your numerical
answers against the correct ones.
Problems taken from previous SOA/CAS exams will be indicated by the symbol ‡.
This manuscript can be used for personal use or class use, but not for commercial purposes. If you
ﬁnd any errors, I would appreciate hearing from you: mﬁ[email protected] Marcel B. Finan
Russellville, Arkansas
May 2010 3 4 PREFACE Contents
Preface 3 Parity and Other Price Options Properties
1 A Review of Options . . . . . . . . . . . . . . . . . . . . . . . . . .
2 PutCall Parity for European Options . . . . . . . . . . . . . . . .
3 PutCall Parity of Stock Options . . . . . . . . . . . . . . . . . . .
4 Conversions and Reverse Conversions . . . . . . . . . . . . . . . . .
5 Parity for Currency Options . . . . . . . . . . . . . . . . . . . . . .
6 Parity of European Options on Bonds . . . . . . . . . . . . . . . .
7 PutCall Parity Generalization . . . . . . . . . . . . . . . . . . . .
8 Labeling Options: Currency Options . . . . . . . . . . . . . . . . .
9 NoArbitrage Bounds on Option Prices . . . . . . . . . . . . . . . .
10 General Rules of Early Exercise on American Options . . . . . . .
11 Eﬀect of Maturity Time Growth on Option Prices . . . . . . . . .
12 Options with Diﬀerent Strike Prices but Same Time to Expiration
13 Convexity Properties of the Option Price Functions . . . . . . . . .
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. 9
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90 Option Pricing in Binomial Models
14 SinglePeriod Binomial Model Pricing of European Call Options .
15 RiskNeutral Option Pricing in the Binomial Model: A First Look
16 Binomial Trees and Volatility . . . . . . . . . . . . . . . . . . . . .
17 MultiPeriod Binomial Option Pricing Model . . . . . . . . . . . .
18 Binomial Option Pricing for European Puts . . . . . . . . . . . . .
19 Binomial Option Pricing for American Options . . . . . . . . . . .
20 Binomial Option Pricing on Currency Options . . . . . . . . . . .
21 Binomial Pricing of Futures Options . . . . . . . . . . . . . . . . .
22 Further Discussion of Early Exercising . . . . . . . . . . . . . . . .
23 RiskNeutral Probability Versus Real Probability . . . . . . . . . . .
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. 99
100
109
114
120
125
131
137
143
149
155 5 6 CONTENTS
24 Random Walk and the Binomial Model . . . . . . . . . . . . . . . . . . . . . . . . . . 165
25 Alternative Binomial Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
26 Estimating (Historical) Volatility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 The BlackScholes Model
27 The BlackScholes Formulas for European Options . .
28 Applying the BlackScholes Formula To Other Assets
29 Option Greeks: Delta, Gamma, and Vega . . . . . . .
30 Option Greeks: Theta, Rho, and Psi . . . . . . . . . .
31 Option Elasticity and Option Volatility . . . . . . . .
32 The Risk Premium and Sharpe Ratio of an Option . .
33 Proﬁt Before Maturity: Calendar Spreads . . . . . . .
34 Implied Volatility . . . . . . . . . . . . . . . . . . . . .
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. Option Hedging
35 DeltaHedging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
36 Option Price Approximations: Delta and DeltaGamma Approximations .
37 The DeltaGammaTheta Approximation and the MarketMaker’s Proﬁt .
38 The BlackScholes Analysis . . . . . . . . . . . . . . . . . . . . . . . . . .
39 DeltaGamma Hedging . . . . . . . . . . . . . . . . . . . . . . . . . . . .
An Introduction to Exotic Options
40 Asian Options . . . . . . . . . . . .
41 European Barrier Options . . . . .
42 Compound European Options . . .
43 Chooser and Forward Start Options
44 Gap Options . . . . . . . . . . . . .
45 Exchange Options . . . . . . . . . . .
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. The Lognormal Stock Pricing Model
46 The Normal Distribution . . . . . . . . . . .
47 The Lognormal Distribution . . . . . . . . .
48 A Lognormal Model of Stock Prices . . . . .
49 Lognormal Probability Calculations . . . . .
50 Conditional Expected Price and a Derivation .
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...............
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of BlackScholes Formula .
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. 183
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. 243
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308 .
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340
347 CONTENTS
Option Pricing Via Monte Carlo Simulation
51 Option Valuation as a Discounted Expected Value .
52 Computing Normal Random Numbers . . . . . . . .
53 Simulating Lognormal Stock Prices . . . . . . . . .
54 Monte Carlo Valuation for European Options . . . .
55 Monte Carlo Valuation of Asian Options . . . . . .
56 Control Variate Method . . . . . . . . . . . . . . . .
57 Antithetic Variate Method and Stratiﬁed Sampling . 7 .
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. Brownian Motion
58 Brownian Motion . . . . . . . . . . . . . . . . . . . . . . . . . .
59 Arithmetic Brownian Motion . . . . . . . . . . . . . . . . . . . .
60 Geometric Brownian Motion . . . . . . . . . . . . . . . . . . . .
61 Ito Process Multiplication Rules . . . . . . . . . . . . . . . . . .
62 Sharpe ratios of Assets that Follow Geometric Brownian Motions
63 The RiskNeutral Measure and Girsanov’s Theorem . . . . . . .
64 Single Variate Itˆ’s Lemma . . . . . . . . . . . . . . . . . . . . .
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65 Valuing a Claim on S a . . . . . . . . . . . . . . . . . . . . . . .
The BlackScholes Partial Diﬀerential Equation
66 Diﬀerential Equations for Riskless Assets . . . . . . . . .
67 Derivation of the BlackScholes PDE . . . . . . . . . . .
68 The BlackScholes PDE and Equilibrium Returns . . . .
69 The BlackScholes Equation and the Risk Neutral Pricing .
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. 353
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. 391
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429 .
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. 435
436
440
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450 Binary Options
453
70 CashorNothing Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 454
71 AssetorNothing Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 460
72 Supershares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 466
Interest Rates Models
73 Bond Pricing Model with Arbitrage Opportunity . . . . .
74 A BlackScholes Analogue for Pricing ZeroCoupon Bonds
75 ZeroCoupon Bond Pricing: RiskNeutral Process . . . .
76 The RendlemanBartter ShortTerm Model . . . . . . . .
77 The Vasicek ShortTerm Model . . . . . . . . . . . . . .
78 The CoxIngersollRoss ShortTerm Model . . . . . . . .
79 The Black Formula for Pricing Options on Bonds . . . . .
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. 469
470
475
481
486
489
495
502 8 CONTENTS
80 The Black Formula for Pricing FRA, Caplets, and Caps . . . . . . . . . . . . . . . . . 507 Binomial Models for Interest Rates
515
81 Binomial Model for Pricing of a ZeroCoupon Bond . . . . . . . . . . . . . . . . . . . 516
82 The Basics of the BlackDermanToy Model . . . . . . . . . . . . . . . . . . . . . . . 521
83 BlackDermanToy Pricing of Caplets and Caps . . . . . . . . . . . . . . . . . . . . . 529
Supplement
535
84 Jensen’s Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 536
85 Utility Theory and RiskNeutral Pricing . . . . . . . . . . . . . . . . . . . . . . . . . . 541
Answer Key 547 BIBLIOGRAPHY 548 Parity and Other Price Options
Properties
Parity is one of the most important relations in option pricing. In this chapter we discuss diﬀerent
versions of parity for diﬀerent underlying assets. The main parity relation for European options
can be rearranged to create synthetic securities. Also, options where the underlying asset and the
strike asset can be anything are discussed. Bounds of option prices for European and American
options are also discussed.
Since many of the discussions in this book are based on the noarbitrage principle, we will remind
the reader of this concept.
The concept of noarbitrage:
No arbitrage principle assumes there are no transaction costs such as tax and commissions.
Arbitrage is possible when one of three conditions is met:
1. The same asset does not trade at the same price on all markets (“the law of one price”).
2. Two assets with identical cash ﬂows do not trade at the same price.
3. An asset with a known price in the future does not today trade at its future price discounted at
the riskfree interest rate. 9 10 PARITY AND OTHER PRICE OPTIONS PROPERTIES 1 A Review of Options
Derivative securities are ﬁnancial instruments that derive their value from the value of other
assets. Examples of derivatives are forwards, options, and swaps. In this section we discuss brieﬂy
the basic vocabulary of options.
A forward contract is the obligation to buy or sell something at a prespeciﬁed time (called the
expiration date, the delivery date, or the maturity date) and at a prespeciﬁed price, known
as the forward price or delivery price. A forward contract requires no initial premium. In
contrast, an option is a contract that gives the owner the right, but not the obligation, to buy or
sell a speciﬁed asset at a speciﬁed price, on or by a speciﬁed date. The underlying assets include
stocks, major currencies, and bonds. The majority of options are traded on an exchange (such as
Chicago Board of Exchange) or in the overthecounter market.
There are two types of options: A call option gives the right to the owner to buy the asset. A
put option gives the right to the owner to sell the asset.
Option trading involves two parties: a buyer and a seller. The buyer or owner of a call (put)
option obtains the right to buy (sell) an asset at a speciﬁed price by paying a premium to the
writer or seller of the option, who assumes the collateral obligation to sell (buy) the asset, should
the owner of the option choose to exercise it. The buyer of an option is said to take a long position
in the option whereas the seller is said to take a short position in the option. A shortsale of
an asset (or shorting the asset) entails borrowing the asset and then selling it, receiving the cash.
Some time later, we buy back the asset, paying cash for it, and return it to the lender.
Additional terms are needed in understanding option contracts. The strike price or exercise
price (denoted by K ) is the ﬁxed price speciﬁed in the option contract for which the holder can
buy or sell the underlying asset. The expiration date, exercise date, or maturity (denoted by
T with T = 0 for “today”) is the last date on which the contract is still valid. After this date the
contract no longer exists. By exercising the option we mean enforcing the contract,i.e., buy or
sell the underlying asset using the option. Two types of exercising options (also known as style)
are: An American option may be exercised at any time up to the expiration date. A European
option on the other hand, may be exercised only on the expiration date. Unless otherwise stated,
options are considered to be Europeans.
Example 1.1
A call option on ABC Corp stock currently trades for $6. The expiration date is December 17,
2005. The strike price of the option is $95.
(a) If this is an American option, on what dates can the option be exercised?
(b) If this is a European option, on what dates can the option be exercised?
Solution.
(a) Any date before and including the expiration date, December 17, 2005. 1 A REVIEW OF OPTIONS 11 (b) Only on Dec 17, 2005
The payoﬀ or intrinsic value from a call option at the expiration date is a function of the
strike price K and the spot (or market) price ST of the underlying asset on the delivery date. It
is given by max{0, ST − K }. From this deﬁnition we conclude that if a call option is held until
expiration (which must be so for a European option, but not an American option) then the option
will be exercised if, and only if, ST > K, in which case the owner of the option will realize a net
payoﬀ ST − K > 0 and the writer of the option will realize a net payoﬀ(loss) K − ST < 0.
Likewise, the payoﬀ from a put option at the expiration date is a function of the strike price K
and the spot price ST of the underlying asset on the delivery date. It is given by max{0, K − ST }.
From this deﬁnition we conclude that if a put option is held until expiration (which must be so for
a European option, but not an American option) then the option will be exercised if, and only if,
ST < K, in which case the owner of the option will realize a net payoﬀ K − ST > 0 and the writer
of the option will realize a net payoﬀ(loss) ST − K < 0.
Example 1.2
Suppose you buy a 6month call option with a strike price of $50. What is the payoﬀ in 6 months
for prices $45, $50, $55, and $60?
Solution.
The payoﬀ to a purchased call option at expiration is:
Payoﬀ to call option = max{0, spot price at expiration − strike price}
The strike is given: It is $50. Therefore, we can construct the following table:
Price of assets in 6 months
45
50
55
60 Strike price
50
50
50
50 Payoﬀ to the call option
0
0
5
10 Payoﬀ does not take into consideration the premium which is paid to acquire an option. Thus, the
payoﬀ of an option is not the money earned (or lost). For a call option we deﬁne the proﬁt earned
by the owner of the option by
Buyer’s call proﬁt = Buyer’s call payoﬀ − future value of premium.
Likewise, we deﬁne the proﬁt of a put option by
Buyer’s put proﬁt = Buyer’s put payoﬀ − future value of premium. 12 PARITY AND OTHER PRICE OPTIONS PROPERTIES Payoﬀ and proﬁt diagrams of a call option and a put option are shown in the ﬁgure below. Pc and
Pp will denote the future value of the premium for a call and put option respectively. Example 1.3
You hold a European call option on 100 shares of CocaCola stock. The exercise price of the call is
$50. The option will expire in moments. Assume there are no transactions costs or taxes associated
with this contract.
(a) What is your proﬁt on this contract if the stock is selling for $51?
(b) If CocaCola stock is selling for $49, what will you do?
Solution.
(a) The proﬁt is 1 × 100 = $100.
(b) Do not exercise. The option is thus expired without exercise
Example 1.4
One can use options to insure assets we own (or purchase) or assets we short sale. An investor 1 A REVIEW OF OPTIONS 13 who owns an asset (i.e. being long an asset) and wants to be protected from the fall of the asset’s
value can insure his asset by buying a put option with a desired strike price. This combination of
owning an asset and owning a put option on that asset is called a ﬂoor. The put option guarantees
a minimum sale price of the asset equals the strike price of the put.
Show that buying a stock at a price S and buying a put option on the stock with strike price K and
time to expiration T has payoﬀ equals to the payoﬀ of buying a zerocoupon bond with parvalue
K and buying a call on the stock with strike price K and expiration time T.
Solution.
We have the following payoﬀ tables.
Transaction
Buy a stock
Buy a put
Total Payoﬀ at Time 0
−S
−P
−S − P Payoﬀ at Time T
ST ≤ K
ST
K − ST
K ST > K
ST
0
ST Transaction
Buy a Bond
Buy a Call
Total Payoﬀ at Time 0
−P V0,T (K )
−C
−P V0,T (K ) − C Payoﬀ at Time T
ST ≤ K
K
0
K ST > K
K
ST − K
ST Both positions guarantee a payoﬀ of max{K, ST }. By the noarbitrage principle they must have
same payoﬀ at time t = 0. Thus,
P + S = C + P V0,T (K )
An option is said to be inthemoney if its immediate exercise would produce positive cash ﬂow.
Thus, a put option is in the money if the strike price exceeds the spot price or (market price) of the
underlying asset and a call option is in the money if the spot price of the underlying asset exceeds
the strike price. An option that is not in the money is said to be outofthemoney. An option is
said to be atthemoney if its immediate exercise produces zero cash ﬂow. Letting ST denote the
market price at time T we have the following chart.
ST > K
ST = K
ST < K Call
Inthemoney
Atthemoney
Outofthemoney Put
outofthemoney
Atofthemoney
Inthemoney 14 PARITY AND OTHER PRICE OPTIONS PROPERTIES An option with an exercise price signiﬁcantly below (for a call option) or above (for a put option)
the market price of the underlying asset is said to be deep inthemoney. An option with an
exercise price signiﬁcantly above (for a call option) or below (for a put option) the market price of
the underlying asset is said to be deep outofthe money.
Example 1.5
If the underlying stock price is $25, indicate whether each of the options below is inthemoney,
atthemoney, or outofthemoney.
Strike
$20
$25
$30 Call Put Solution.
Strike Call
$20
Inthemoney
$25
Atthemoney
$30
Outofthemoney Put
outofthemoney
Atthemoney
Inthemoney 1 A REVIEW OF OPTIONS 15 Practice Problems
Problem 1.1
When you short an asset, you borrow the asset and sell, hoping to replace them at a lower price
and proﬁt from the decline. Thus, a short seller will experience loss if the price rises. He can insure
his position by purchasing a call option to protect against a higher price of repurchasing the asset.
This combination of short sale and call option purchase is called a cap.
Show that a shortsale of a stock and buying a call has a payoﬀ equals to that of buying a put and
selling a zerocoupon bond with parvalue the strike price of the options. Both options have the
same strike price and time to expiration.
Problem 1.2
Writing an option backed or covered by the underlying asset (such as owning the asset in the case
of a call or shorting the asset in the case of a put) is referred to as covered writing or option
overwriting. The most common motivation for covered writing is to generate additional income
by means of premium.
A covered call is a call option which is sold by an investor who owns the underlying assets. An
investor’s risk is limited when selling a covered call since the investor already owns the underlying
asset to cover the option if the covered call is exercised. By selling a covered call an investor is
attempting to capitalize on a neutral or declining price in the underlying stock. When a covered
call expires without being exercised (as would be the case in a declining or neutral market), the
investor keeps the premium generated by selling the covered call. The opposite of a covered call is
a naked call, where a call is written without owned assets to cover the call if it is exercised.
Show that the payoﬀ of buying a stock and selling a call on the stock is equal to the payoﬀ of buying
a zerocoupon bond with parvalue the strike price of the options and selling a put. Both options
have a strike price K and time to expiration T.
Problem 1.3
A covered put is a put option which is sold by an investor and which is covered (backed) by a
short sale of the underlying assets. A covered put may also be covered by deposited cash or cash
equivalent equal to the exercise price of the covered put. The opposite of a covered put would be a
naked put.
Show that the payoﬀ of shorting a stock and selling a put on the stock is equal to the payoﬀ of
selling a zerocoupon bond with parvalue the strike price of the options and selling a call. Both
options have a strike price K and time to expiration T.
Problem 1.4
A position that consists of buying a call with strike price K1 and expiration T and selling a call with 16 PARITY AND OTHER PRICE OPTIONS PROPERTIES strike price K2 > K1 and same expiration date is called a bull call spread. In contrast, buying
a put with strike price K1 and expiration T and selling a put with strike price K2 > K1 and same
expiration date is called a bull put spread. An investor who enters a bull spread is speculating
that the stock price will increase.
(a) Find formulas for the payoﬀ and proﬁt functions of a bull call spread. Draw the diagrams.
(b) Find formulas for the payoﬀ and proﬁt functions of a bull put spread. Draw the diagrams.
Problem 1.5
A bear spread is precisely the opposite of a bull spread. An investor who enters a bull spread
is hoping that the stock price will increase. By contrast, an investor who enters a bear spread is
hoping that the stock price will decline. Let 0 < K1 < K2 . A bear spread can be created by either
selling a K1 −strike call and buying a K2 −strike call, both with the same expiration date (bear call
spread), or by selling a K1 −strike put and buying a K2 −strike put, both with the same expiration
date (bear put spread).
Find formulas for the payoﬀ and the proﬁt of a bear spread created by selling a K1 −strike call and
buying a K2 −strike call, both with the same expiration date. Draw the diagram.
Problem 1.6
A (call) ratio spread is achieved by buying a certain number of calls with a high strike and selling
a diﬀerent number of calls at a lower strike. By replacing the calls with puts one gets a (put) ratio
spread. All options under considerations have the same expiration date and same underlying asset.
If m calls were bought and n calls were sold we say that the ratio is m .
n
An investor buys one $70strike call and sells two $85strike call of a stock. All the calls have
expiration date one year from now. The risk free annual eﬀective rate of interest is 5%. The
premiums of the $70strike and $85strike calls are $10.76 and $3.68 respectively. Draw the proﬁt
diagram.
Problem 1.7
A collar is achieved with the purchase of a put option with strike price K1 and expiration date T
and the selling of a call option with strike price K2 > K1 and expiration date T. Both options use
the same underlying asset. A collar can be used to speculate on the decrease of price of an asset.
The diﬀerence K2 − K1 is called the collar width. A written collar is the reverse of collar (sale
of a put and purchase of a call).
Find the proﬁt function of a collar as a function of the spot price ST and draw its graph.
Problem 1.8
Collars can be used to insure assets we own. This position is called a collared stock. A collared
stock involves buying the index, buy an atthemoney K1 −strike put option (which insures the 1 A REVIEW OF OPTIONS 17 index) and selling an outofthemoney K2 −strike call option (to reduce the cost of the insurance),
where K1 < K2 .
Find the proﬁt function of a collared stock as a function of ST .
Problem 1.9
A long straddle or simply a straddle is an option strategy that is achieved by buying a K −strike
call and a K −strike put with the same expiration time T and same underlying asset. Find the
payoﬀ and proﬁt functions of a straddle. Draw the diagrams.
Problem 1.10
A strangle is a position that consists of buying a K1 −strike put and a K2 −strike call with the
same expiration date and same underlying asset and such that K1 < K2 . Find the proﬁt formula
for a strangle and draw its diagram.
Problem 1.11
A butterﬂy spread is created by using the following combination:
(1) Create a written straddle by selling a K2 −strike call and a K2 −strike put.
(2) Create a long strangle by buying a K1 strike call and a K3 strike put.
Find the initial cost and the proﬁt function of the position. Draw the proﬁt diagram.
Problem 1.12
Given 0 < K1 < K2 < K3 . When a symmetric butterﬂy is created using these strike prices the
number K2 is the midpoint of the interval with endpoints K1 and K3 . What if K2 is not midway
between K1 and K3 ? In this case, one can create a butterﬂylike spread with the peak tilted either
to the left or to the right as follows: Deﬁne a number λ by the formula
λ= K3 − K2
.
K3 − K1 Then λ satisﬁes the equation
K2 = λK1 + (1 − λ)K3 .
Thus, for every written K2 −strike call, a butterﬂylike spread can be constructed by buying λ
K1 −strike calls and (1 − λ) K3 −strike calls. The resulting spread is an example of an asymmetric
butterﬂy spread.
Construct an asymmetric butterﬂy spread using the 35strike call (with premium $6.13), 43strike
call (with premium $1.525 ) and 45strike call (with premium $0.97). All options expire 3 months
from now. The risk free annual eﬀective rate of interest is 8.33%. Draw the proﬁt diagram. 18 PARITY AND OTHER PRICE OPTIONS PROPERTIES Problem 1.13
Suppose the stock price is $40 and the eﬀective annual interest rate is 8%.
(a) Draw a single graph payoﬀ diagram for the options below.
(b) Draw a single graph proﬁt diagram for the options below.
(i) 35strike call with a premium of $9.12.
(ii) 40strike call with a premium of $6.22.
(iii) 45strike call with a premium of $4.08.
Problem 1.14
A trader buys a European call option and sells a European put option. The options have the same
underlying asset, strike price, and maturity date. Show that this position is equivalent to a long
forward.
Problem 1.15
Suppose the stock price is $40 and the eﬀective annual interest rate is 8%.
(a) Draw the payoﬀ diagrams on the same window for the options below.
(b) Draw the proﬁt diagrams on the same window for the options below.
(i) 35strike put with a premium of $1.53.
(ii) 40strike put with a premium of $3.26.
(iii) 45strike put with a premium of $5.75.
Problem 1.16
Complete the following table.
Derivative
Maximum
Position
Loss
Long Forward
Short Forward
Long Call
Short Call
Long Put
Short Put Minimum
Gain Position with Respect
to Underlying Asset Strategy Problem 1.17 ‡
An insurance company sells single premium deferred annuity contracts with return linked to a stock
index, the time−t value of one unit of which is denoted by S (t). The contracts oﬀer a minimum
guarantee return rate of g %. At time 0, a single premium of amount π is paid by the policyholder, 1 A REVIEW OF OPTIONS 19 and π × y % is deducted by the insurance company. Thus, at the contract maturity date, T, the
insurance company will pay the policyholder
π × (1 − y %) × Max[S (T )/S (0), (1 + g %)T ].
You are given the following information:
(i) The contract will mature in one year.
(ii) The minimum guarantee rate of return, g%, is 3%.
(iii) Dividends are incorporated in the stock index. That is, the stock index is constructed with all
stock dividends reinvested.
(iv) S (0) = 100.
(v) The price of a oneyear European put option, with strike price of $103, on the stock index is
$15.21.
Determine y%, so that the insurance company does not make or lose money on this contract (i.e.,
the company breaks even.)
Problem 1.18 ‡
You are given the following information about a securities market:
(i) There are two nondividendpaying stocks, X and Y.
(ii) The current prices for X and Y are both $100.
(iii) The continuously compounded riskfree interest rate is 10%.
(iv) There are three possible outcomes for the prices of X and Y one year from now:
Outcome
1
2
3 X
$200
$50
$0 Y
$0
$0
$300 Consider the following two portfolios:
• Portfolio A consists of selling a European call on X and buying a European put on Y. Both
options expire in one year and have a strike price of $95. The cost of this portfolio at time 0 is then
CX − P Y .
• Portfolio B consists of buying M shares of X , N shares of Y and lending $P. The time 0 cost of
this portfolio is −100M − 100N − P.
Determine PY − CX if the time1 payoﬀ of Portfolio B is equal to the time1 payoﬀ of Portfolio A.
Problem 1.19 ‡
The following four charts are proﬁt diagrams for four option strategies: Bull Spread, Collar, Straddle, and Strangle. Each strategy is constructed with the purchase or sale of two 1year European 20 PARITY AND OTHER PRICE OPTIONS PROPERTIES options. Match the charts with the option strategies. 1 A REVIEW OF OPTIONS 21 22 PARITY AND OTHER PRICE OPTIONS PROPERTIES 2 PutCall Parity for European Options
Putcall parity is an important principle in options pricing.1 It states that the premium of a call
option implies a certain fair price for the corresponding put option having the same strike price and
expiration date, and vice versa.
To begin understanding how the putcall parity is established, we consider the following position:
An investor buys a call option with strike price K and expiration date T for the price of C (K, T )
and sells a put option with the same strike price and expiration date for the price of P (K, T ). If the
spot price at expiration is greater than K , the put will not be exercised (and thus expires worthless)
but the investor will exercise the call. So the investor will buy the asset for the price of K. If the
spot price at expiration is smaller than K , the call will not be exercised and the investor will be
assigned on the short put, if the owner of the put wishes to sell (which is likely the case) then the
investor is obliged to buy the asset for K. Either way, the investor is obliged to buy the asset for
K and the long callshort put combination induces a long forward contract that is synthetic since
it was fabricated from options.
Now, the portfolio consists of buying a call and selling a put results in buying the stock. The payoﬀ
of this portfolio at time 0 is P (K, T ) − C (K, T ). The portfolio of buying the stock through a long
forward contract and borrowing P V0,T (K ) has a payoﬀ at time 0 of P V0,T (K ) − P V0,T (F0,T ), where
F0,T is the forward price. Using the noarbitrage pricing theory2 , the net cost of owning the asset
must be the same whether through options or forward contracts, that is,
P (K, T ) − C (K, T ) = P V0,T (K − F0,T ). (2.1) Equation (2.1) is referred to as the putcall parity. We point out here that P V0,T (F0,T ) is the
prepaid forward price for the underlying asset and P V0,T (K ) is the prepaid forward price of the
strike.
It follows from (2.1), that selling a put and buying a call with the same strike price K and maturity
date create a synthetic long forward contract with forward price K and net premium P (K, T ) −
C (K, T ). To mimic an actual forward contract the put and call premiums must be equal.
Since American style options allow early exercise, putcall parity will not hold for American options
unless they are held to expiration. Note also that if the forward price is higher than the strike price
of the options, call is more expensive than put, and vice versa.
Remark 2.1
We will use the sign“+” for selling and borrowing positions and the sign “−” for buying and lending
positions.
1 Option pricing model is a mathemtical formula that uses the factors that determine an option’s price as inputs
to produce the theoretical fair value of an option.
2
The noarbitrage pricing theory asserts that acquiring an asset at time T should cost the same, no matter
how you achieve it. 2 PUTCALL PARITY FOR EUROPEAN OPTIONS 23 Remark 2.2
In the literature, (2.1) is usually given in the form
C (K, T ) − P (K, T ) = P V0,T (F0,T − K ).
Thus, we will use this form of parity in the rest of the book.
Remark 2.3
Note that for a nondividend paying stock we have P V0,T (F0,T ) = S0 , the stock price at time 0.
Example 2.1
The current price of a nondividendpaying stock is $80. A European call option that expires in six
months with strike price of $90 sells for $12. Assume a continuously compounded riskfree interest
rate of 8%, ﬁnd the value to the nearest penny of a European put option that expires in six months
and with strike price of $90.
Solution.
Using the putcall parity (2.1) with C = 12, T = 6 months= 0.5 years, K = 90, S0 = 80, and
r = 0.08 we ﬁnd
P = C − S0 + e−rT K = 12 − 80 + 90e−0.08(0.5) = $18.47
Example 2.2
Suppose that the current price of a nondividendpaying stock is S0 . A European put option that
expires in six months with strike price of $90 is $6.47 more expensive than the corresponding
European call option. Assume a continuously compounded interest rate of 15%, ﬁnd S0 to the
nearest dollar.
Solution.
The putcall parity
C (K, T ) − P (K, T ) = S0 − Ke−rT
gives
−6.47 = S0 − 90e−0.15×0.5 .
Solving this equation we ﬁnd S0 = $77
Example 2.3
Consider a European call option and a European put option on a nondividendpaying stock. You
are given:
(i) The current price of the stock is $60. 24 PARITY AND OTHER PRICE OPTIONS PROPERTIES (ii) The call option currently sells for $0.15 more than the put option.
(iii) Both the call option and put option will expire in T years.
(iv) Both the call option and put option have a strike price of $70.
(v) The continuously compounded riskfree interest rate is 3.9%.
Calculate the time of expiration T.
Solution.
Using the putcall parity
C (K, T ) − P (K, T ) = S0 − Ke−rT
we have
0.15 = 60 − 70e−0.039T .
Solving this equation for T we ﬁnd T ≈ 4 years 2 PUTCALL PARITY FOR EUROPEAN OPTIONS 25 Practice Problems
Problem 2.1
If a synthetic forward contract has no initial premium then
(A) The premium you pay for the call is larger than the premium you receive from the put.
(B) The premium you pay for the call is smaller than the premium you receive from the put.
(C) The premium you pay for the call is equal to the premium you receive from the put.
(D) None of the above.
Problem 2.2
In words, the PutCall parity equation says that
(A) The cost of buying the asset using options must equal the cost of buying the asset using a
forward.
(B) The cost of buying the asset using options must be greater than the cost of buying the asset
using a forward.
(C) The cost of buying the asset using options must be smaller than the cost of buying the asset
using a forward.
(D) None of the above.
Problem 2.3
State two features that diﬀerentiate a synthetic forward contract from a noarbitrage (actual) forward contract.
Problem 2.4
Recall that a covered call is a call option which is sold by an investor who owns the underlying
asset. Show that buying an asset plus selling a call option on the asset with strike price K (i.e. selling
a covered call) is equivalent to selling a put option with strike price K and buying a zerocoupon
bond with par value K.
Problem 2.5
Recall that a covered put is a put option which is sold by an investor and which is backed by a
short sale of the underlying asset(s). Show that short selling an asset plus selling a put option with
strike price K (i.e. selling a covered put) is equivalent to selling a call option with strike price K
and taking out a loan with maturity value of K.
Problem 2.6
The current price of a nondividendpaying stock is $40. A European call option on the stock with
strike price $40 and expiration date in three months sells for $2.78 whereas a 40strike European put
with the same expiration date sells for $1.99. Find the annual continuously compounded interest
rate r. Round your answer to two decimal places. 26 PARITY AND OTHER PRICE OPTIONS PROPERTIES Problem 2.7
Suppose that the current price of a nondividendpaying stock is S0 . A European call option that
expires in six months with strike price of $90 sells for $12. A European put option with strike
price of $90 and expiration date in six months sells for $18.47. Assume a continuously compounded
interest rate of 15%, ﬁnd S0 to the nearest dollar.
Problem 2.8
Consider a European call option and a European put option on a nondividendpaying stock. You
are given:
(i) The current price of the stock is $60.
(ii) The call option currently sells for $0.15 more than the put option.
(iii) Both the call option and put option will expire in 4 years.
(iv) Both the call option and put option have a strike price of $70.
Calculate the continuously compounded riskfree interest rate.
Problem 2.9
The putcall parity relationship
C (K, T ) − P (K, T ) = P V (F0,T − K )
can be rearranged to show the equivalence of the prices (and payoﬀs and proﬁts) of a variety of
diﬀerent combinations of positions.
(a) Show that buying an index plus a put option with strike price K is equivalent to buying a call
option with strike price K and a zerocoupon bond with par value of K.
(b) Show that shorting an index plus buying a call option with strike price K is equivalent to buying
a put option with strike price K and taking out a loan with maturity value of K.
Problem 2.10
A call option on XYZ stock with an exercise price of $75 and an expiration date one year from now
is worth $5.00 today. A put option on XYZ stock with an exercise price of $75 and an expiration
date one year from now is worth $2.75 today. The annual eﬀective riskfree rate of return is 8% and
XYZ stock pays no dividends. Find the current price of the stock.
Problem 2.11 ‡
You are given the following information:
• The current price to buy one share of XYZ stock is 500.
• The stock does not pay dividends.
• The riskfree interest rate, compounded continuously, is 6%.
• A European call option on one share of XYZ stock with a strike price of K that expires in one 2 PUTCALL PARITY FOR EUROPEAN OPTIONS 27 year costs $66.59.
• A European put option on one share of XYZ stock with a strike price of K that expires in one
year costs $18.64.
Using putcall parity, determine the strike price K.
Problem 2.12
The current price of a stock is $1000 and the stock pays no dividends in the coming year. The
premium for a oneyear European call is $93.809 and the premium for the corresponding put is
$74.201. The annual eﬀective riskfree interest rate is 4%. Determine the forward price of the
synthetic forward. Round your answer to the nearest dollar.
Problem 2.13
The actual forward price of a stock is $1020 and the stock pays no dividends in the coming year.
The premium for a oneyear European call is $93.809 and the premium for the corresponding put
is $74.201. The forward price of the synthetic forward contract is $1000. Determine the annual
riskfree eﬀective rate r.
Problem 2.14 ‡
You are given the following information:
• One share of the PS index currently sells for 1,000.
• The PS index does not pay dividends.
• The eﬀective annual riskfree interest rate is 5%.
You want to lock in the ability to buy this index in one year for a price of 1,025. You can do this by
buying or selling European put and call options with a strike price of 1,025. Which of the following
will achieve your objective and also gives the cost today of establishing this position.
(A) Buy the put and sell the call, receive 23.81.
(B) Buy the put and sell the call, spend 23.81.
(C) Buy the put and sell the call, no cost.
(D) Buy the call and sell the put, receive 23.81.
(E) Buy the call and sell the put, spend 23.81. 28 PARITY AND OTHER PRICE OPTIONS PROPERTIES 3 PutCall Parity of Stock Options
Recall that a prepaid forward contract on a stock is a forward contract with payment made at time
0. The forward price is the future value of the prepaid forward price and this is true regardless of
whether there are discrete dividends, continuous dividends, or no dividends.
If the expiration time is T then for a stock with no dividends the future value of the prepaid forward
price is given by the formula
F0,T = F V0,T (S0 ),
where S0 is the stock’s price at time t = 0.
In the case of a stock with discrete dividends, the prepaid forward price is
P
F0,T = S0 − P V0,T (Div ) so that
P
F0,T = F V0,T (F0,T ) = F V0,T (S0 − P V0,T (Div )). In particular, if the dividends D1 , D2 , · · · , Dn made at time t1 , t2 , · · · , tn prior to the maturity date
T , then
n F0,T =F V0,T (S0 − P V0,ti (Di ))
i=1
n =F V0,T (S0 ) − F V0,T ( P V0,ti Di )
i=1 In this case, Equation (2.1) takes the form
C (K, T ) − P (K, T ) = S0 − PV0,T (Div) − PV0,T (K)
or in the case of a continuously compounded riskfree interest rate r
n Di e−rti − e−rT K. C (K, T ) − P (K, T ) = S0 −
i=1 Example 3.1
Suppose ABC stock costs $75 today and is expected to pay semiannual dividend of $1 with the
ﬁrst coming in 4 months from today and the last just prior to the delivery of the stock. Suppose
that the annual continuously compounded riskfree rate is 8%. Find the cost of a 1year prepaid
forward contract. 3 PUTCALL PARITY OF STOCK OPTIONS 29 Solution.
The cost is
4 10 e−rT F0,T = S0 − PV0,T (Div) = 75 − e−0.08× 12 − e−0.08× 12 = $73.09
Example 3.2
A one year European call option with strike price of K on a share of XYZ stock costs $11.71 while
a one year European put with the same strike price costs $5.31. The share of stock pays dividend
valued at $3 six months from now and another dividend valued at $5 one year from now. The current
share price is $99 and the continuouslycompounded riskfree rate of interest is 5.6%. Determine
K.
Solution.
The putcall parity of stock options with discrete dividends is
n Di e−rti − e−rT K. C − P = S0 −
i=1 We are given C = 11.71, P = 5.31, T = 1, S0 = 99, D1 = 3, D2 = 5, t1 =
Solving for K we ﬁnd K = $89.85 6
,t
12 2 = 1 and r = 0.056. For a stock with continuous dividends we have
F0,T = F V0,T (S0 e−δT )
where δ is the continuously compounded dividend yield (which is deﬁned to be the annualized
divident payment divided by the stock price). In this case, Equation (2.1) becomes
C (K, T ) − P (K, T ) = S0 e−δT − P V0,T (K ).
Example 3.3
A European call option and a European put option on a share of XYZ stock have a strike price of
$100 and expiration date in nine months. They sell for $11.71 and $5.31 respectively. The price
of XYZ stock is currently $99, and the stock pays a continuous dividend yield of 2%. Find the
continuously compounded riskfree interest rate r.
Solution.
The putcall parity of stock options with continuous dividends is
C − P = S0 e−δT − e−rT K. 30 PARITY AND OTHER PRICE OPTIONS PROPERTIES We are given C = 11.71, P = 5.31, K = 100, T =
r = 12.39% 9
,
12 S0 = 99, and δ = 0.02. Solving for r we ﬁnd Parity provides a way for the creation of a synthetic stock. Using the putcall parity, we can
write
−S0 = −C (K, T ) + P (K, T ) − PV0,T (Div) − PV0,T (K).
This equation says that buying a call, selling a put, and lending the present value of the strike and
dividends to be paid over the life of the option is equivalent to the outright purchase1 of the stock.
Example 3.4
A European call option and a European put option on a share of XYZ stock have a strike price of
$100 and expiration date in nine months. They sell for $11.71 and $5.31 respectively. The price of
XYZ stock is currently $99 and the continuously compounded riskfree rate of interest is 12.39%.
A share of XYZ stock pays $2 dividends in six months. Determine the amount of cash that must
be lent at the given riskfree rate of return in order to replicate the stock.
Solution.
By the previous paragraph, in replicating the stock, we must lend PV0,T (Div) + PV0,T (K) to be
9
paid over the life of the option. We are given C = 11.71, P = 5.31, K = 100, T = 12 , S0 = 99,
and r = 0.1239. Thus,
PV0,T (Div) + PV0,T (K) = 2e−0.1239(0.5) + 100e−0.1239(0.75) = $93.01
The putcall parity provides a simple test of option pricing models. Any pricing model that produces
option prices which violate the putcall parity is considered ﬂawed and leads to arbitrage.
Example 3.5
A call option and a put option on the same nondividendpaying stock both expire in three months,
both have a strike price of 20 and both sell for the price of 3. If the continuously compounded
riskfree interest rate is 10% and the stock price is currently 25, identify an arbitrage.
Solution.
3
We are given that C = P = 3, r = 0.10, T = 12 , K = 20, and S0 = 25. Thus, C − P =
−rT
−0.10×0.25
0 and S0 − Ke
= 25 − 20e
> 0 so that the putcall parity fails and therefore an
arbitrage opportunity exists. An arbitrage can be exploited as follows: Borrow 3 at the continuously
compounded annual interest rate of 10%. Buy the call option. Shortsell a stock and invest that
25 at the riskfree rate. Three months later purchase the stock at strike price 20 and pay the bank
3e0.10×0.25 = 3.08. You should have 25e0.10×0.25 − 20 − 3.08 = $2.55 in your pocket
1 outright purchase occurs when an investor simultaneously pays S0 in cash and owns the stock. 3 PUTCALL PARITY OF STOCK OPTIONS 31 Remark 3.1
Consider the putcall parity for a nondividendpaying stock. Suppose that the option is atthemoney at expiration, that is, S0 = K. If we buy a call and sell a put then we will diﬀer the payment
for owning the stock until expiration. In this case, P − C = Ke−rT − K < 0 is the present value of
the interest on K that we pay for deferring the payment of K until expiration. If we sell a call and
buy a put then we are synthetically shortselling the stock. In this case, C − P = K − Ke−rT > 0
is the compensation we receive for deferring receipt of the stock price. 32 PARITY AND OTHER PRICE OPTIONS PROPERTIES Practice Problems
Problem 3.1
According to the putcall parity, the payoﬀs associated with ownership of a call option can be
replicated by
(A) shorting the underlying stock, borrowing the present value of the exercise price, and writing a
put on the same underlying stock and with the same exercise price
(B) buying the underlying stock, borrowing the present value of the exercise price, and buying a
put on the same underlying stock and with the same exercise price
(C) buying the underlying stock, borrowing the present value of the exercise price, and writing a
put on the same underlying stock and with the same exercise price
(D) none of the above.
Problem 3.2
A threeyear European call option with a strike price of $50 on a share of ABC stock costs $10.80.
The price of ABC stock is currently $42. The continuously compounded riskfree rate of interest is
10%. Find the price of a threeyear European put option with a strike price of $50 on a share of
ABC stock.
Problem 3.3
Suppose ABC stock costs $X today. It is expected that 4 quarterly dividends of $1.25 each will be
paid on the stock with the ﬁrst coming 3 months from now. The 4th dividend will be paid one day
before expiration of the forward contract. Suppose the annual continuously compounded riskfree
rate is 10%. Find X if the cost of the forward contract is $95.30. Round your answer to the nearest
dollar.
Problem 3.4
Suppose ABC stock costs $75 today and is expected to pay semiannual dividend of $X with the
ﬁrst coming in 4 months from today and the last just prior to the delivery of the stock. Suppose that
the annual continuously compounded riskfree rate is 8%. Find X if the cost of a 1year prepaid
forward contract is $73.09. Round your answer to the nearest dollar.
Problem 3.5
Suppose XYZ stock costs $50 today and is expected to pay quarterly dividend of $1 with the ﬁrst
coming in 3 months from today and the last just prior to the delivery of the stock. Suppose that
the annual continuously compounded riskfree rate is 6%. What is the price of a prepaid forward
contract that expires 1 year from today, immediately after the fourthquarter dividend? 3 PUTCALL PARITY OF STOCK OPTIONS 33 Problem 3.6
An investor is interested in buying XYZ stock. The current price of stock is $45 per share. This
stock pays dividends at an annual continuous rate of 5%. Calculate the price of a prepaid forward
contract which expires in 18 months.
Problem 3.7
A ninemonth European call option with a strike price of $100 on a share of XYZ stock costs $11.71.
A ninemonth European put option with a strike price of $100 on a share of XYZ stock costs $5.31.
The price of XYZ stock is currently $99, and the stock pays a continuous dividend yield of δ. The
continuously compounded riskfree rate of interest is 12.39%. Find δ.
Problem 3.8
Suppose XYZ stock costs $50 today and is expected to pay 8% continuous dividend. What is the
price of a prepaid forward contract that expires 1 year from today, immediately after the fourthquarter dividend?
Problem 3.9
Suppose that annual dividend of 30 on the stocks of an index is valued at $1500. What is the
continuously compounded dividend yield?
Problem 3.10
You buy one share of Ford stock and hold it for 2 years. The dividends are paid at the annualized
daily continuously compounded rate of 3.98% and you reinvest in the stocks all the dividends3 when
they are paid. How many shares do you have at the end of two years?
Problem 3.11
A 6month European call option on a share of XYZ stock with strike price of $35.00 sells for $2.27.
A 6month European put option on a share of XYZ stock with the same strike price sells for $P.
The current price of a share is $32.00. Assuming a 4% continuously compounded riskfree rate and
a 6% continuous dividend yield, ﬁnd P.
Problem 3.12
A ninemonth European call option with a strike price of $100 on a share of XYZ stock costs $11.71.
A ninemonth European put option with a strike price of $100 on a share of XYZ stock costs $5.31.
The price of XYZ stock is currently $99. The continuouslycompounded riskfree rate of interest is
12.39%. What is the present value of dividends payable over the next nine months?
3 In this case, one share today will grow to eδT at time T. See [2] Section 70. 34 PARITY AND OTHER PRICE OPTIONS PROPERTIES Problem 3.13
A European call option and a European put option on a share of XYZ stock have a strike price of
$100 and expiration date in T months. They sell for $11.71 and $5.31 respectively. The price of
XYZ stock is currently $99. A share of Stock XYZ pays $2 dividends in six months. Determine the
amount of cash that must be lent in order to replicate the purchased stock.
Problem 3.14 ‡
On April 30, 2007, a common stock is priced at $52.00. You are given the following:
(i) Dividends of equal amounts will be paid on June 30, 2007 and September 30, 2007.
(ii) A European call option on the stock with strike price of $50.00 expiring in six months sells for
$4.50.
(iii) A European put option on the stock with strike price of $50.00 expiring in six months sells for
$2.45.
(iv) The continuously compounded riskfree interest rate is 6%.
Calculate the amount of each dividend.
Problem 3.15 ‡
Given the following information about a European call option about a stock Z.
• The call price is 5.50
• The strike price is 47.
• The call expires in two years.
• The current stock price is 45.
• The continuously compounded riskfree is 5%.
• Stock Z pays a dividend of 1.50 in one year.
Calculate the price of a European put option on stock Z with strike price 47 that expires in two
years.
Problem 3.16 ‡
An investor has been quoted a price on European options on the same nondividendpaying stock.
The stock is currently valued at 80 and the continuously compounded risk free interest rate is 3%.
The details of the options are:
Option 1
Type
Put
Strike
82
Maturity 180 days Option 2
Call
82
180 days Based on his analysis, the investor has decided that the prices of the two options do not present
any arbitrage opportunities. He decides to buy 100 calls and sell 100 puts. Calculate the net cost
of his transaction. 3 PUTCALL PARITY OF STOCK OPTIONS 35 Problem 3.17 ‡
For a dividend paying stock and European options on this stock, you are given the following information:
• The current stock price is $49.70.
• The strike price of the option is $50.00.
• The time to expiration is 6 months.
• The continuously compunded riskfree interest rate is 3%.
• The continuous dividend yield is 2%.
• The call price is $2.00.
• The put price is $2.35.
Using putcall parity, calculate the present value arbitrage proﬁt per share that could be generated,
given these conditions.
Problem 3.18 ‡
The price of a European call that expires in six months and has a strike price of $30 is $2. The
underlying stock price is $29, and a dividend of $0.50 is expected in two months and again in
ﬁve months. The continuously compounded riskfree interest rate is 10%. What is the price of a
European put option that expires in six months and has a strike price of $30? 36 PARITY AND OTHER PRICE OPTIONS PROPERTIES 4 Conversions and Reverse Conversions
In a forward conversion or simply a conversion a trader buys a stock, buys a put option and
sells a call option with the same strike price and expiration date. The stock price is usually as close
as the options’ strike price. If the stock price at expiration is above the strike price, the short call
is exercised against the trader, which automatically oﬀsets his long position in the stock; the long
put option expires unexercised.
If the stock price at expiration is below the strike price, the long put option is exercised by the
trader, which automatically oﬀsets his long position in the stock; the short call option expires
unexercised.
Parity shows us
−S0 − P (K, T ) + C (K, T ) = −(PV0,T (K) + PV0,T (Div)).
We have thus created a position that costs PV0,T (K) + PV0,T (Div) and that pays K + FV0,T (Div)
at expiration. This is a synthetic long T −bill.1 You can think of a conversion as a synthetic
short stock (Long put/Short call) hedged with a long stock position.
Example 4.1
Suppose that the price of a stock is $52. The stock pays dividends at the continuous yield rate of
7%. Options have 9 months to expiration. A 50strike European call option sells for $6.56 and a
50strike put option sells for $3.61. Calculate the cost of buying a conversion T −bill that matures
for $1,000 in nine months.
Solution.
To create an asset that matures for the strike price of $50, we must buy e−δT shares of the stock,
sell a call option, and buy a put option. Using the putcall parity, the cost today is
PV0,T (K) = S0 e−δT + P(K, T) − C(K, T).
We are given that C = 6.56, P = 3.61, T = 0.75, K = 50, S0 = 52, and δ = 0.07. Substituting
we ﬁnd P V (K ) = $46.39. Thus, the cost of a long synthetic T −bill that matures for $1000 in nine
months is 20 × 46.39 = $927.80
Example 4.2
Suppose the S&R index is 800, the continuously compounded riskfree rate is 5%, and the dividend
yield is 0%. A 1year European call with a strike of $815 costs $75 and a 1year European put with
1 Tbills are purchased for a price that is less than their par (face) value; when they mature, the issuer pays the
holder the full par value. Eﬀectively, the interest earned is the diﬀerence between the purchase price of the security
and what you get at maturity. 4 CONVERSIONS AND REVERSE CONVERSIONS 37 a strike of $815 costs $45. Consider the strategy of buying the stock, selling the call, and buying
the put.
(a) What is the rate of return on this position held until the expiration of the options?
(b) What is the arbitrage implied by your answer to (a)?
(c) What diﬀerence between the call and put will eliminate arbitrage?
Solution.
(a) By buying the stock, selling the 815strike call, and buying the 815 strike put, your cost will be
−800 + 75 − 45 = −$770.
After one year, you will have for sure $815 because either the sold call commitment or the bought
put cancel out the stock price (create a payoﬀ table). Thus, the continuously compounded rate of
return r satisﬁes the equation 770er = 815. Solving this equation we ﬁnd r = 0.0568.
(b) The conversion position pays more interest than the riskfree interest rate. Therefore, we
should borrow money at 5%, and buy a large amount of the aggregate position of (a), yielding a
sure return of 0.68%. To elaborate, you borrow $770 from a bank to buy one position. After one
year, you owe the bank 770e0.05 = 809.48. Thus, from one single position you make a proﬁt of
815 − 809.40 = $5.52.
(c) To eliminate arbitrage, the putcall parity must hold. In this case C − P = S0 − Ke−rT =
800 − 815e−0.05×1 = $24.748
A reverse conversion or simply a reversal is the opposite to conversion. In this strategy, a
trader short selling a stock2 , sells a put option and buys a call option with the same strike price
and expiration date. The stock price is usually as close as the options’ strike price. If the stock
price at expiration is above the strike price, the long call option is exercised by the trader, which
automatically oﬀsets his short position in the stock; the short put option expires unexercised.
If the stock price at expiration is below the strike price, the short put option is exercised against
the trader, which automatically oﬀsets his short position in the stock; the long call option expires
unexercised.
Parity shows us
−C (K, T ) + P (K, T ) + S0 = PV0,T (K) + PV0,T (Div).
Thus, a reversal creates a synthetic short T −bill position that sells for PV0,T (K) + PV0,T (Div).
Also, you can think of a reversal as selling a bond with parvalue K + FV0,T (Div) for the price of
PV0,T (K) + PV0,T (Div).
2 Remember that shorting an asset or short selling an asset is selling an asset that have been borrowed from a third
party (usually a broker) with the intention of buying identical assets back at a later date to return to the lender. 38 PARITY AND OTHER PRICE OPTIONS PROPERTIES Example 4.3
A reversal was created by selling a nondividendpaying stock for $42, buying a threeyear European
call option with a strike price of $50 on a share of stock for $10.80, and selling a threeyear European
put option with a strike price of $50 for $P. The continuously compounded riskfree rate of interest
is 10%. Find P.
Solution.
We have
P (K, T ) = −S0 + C (K, T ) + PV0,T (K) = −42 + 10.80 + 50e−0.1×3 = $5.84
From the discussion of this and the previous sections, we have noticed that the putcall parity can
be rearranged to create synthetic securities. Summarizing, we have
• Synthetic long stock: buy call, sell put, lend present value of the strike and dividends:
−S0 = −C (K, T ) + P (K, T ) − (PV0,T (Div) + PV0,T (K)).
• Synthetic long Tbill: buy stock, sell call, buy put (conversion):
−(PV0,T (Div) + PV0,T (K)) = −S0 − P(K, T) + C(K, T).
• Synthetic short Tbill: shortsell stock, buy call, sell put (reversal):
PV0,T (Div) + PV0,T (K) = S0 + P(K, T) − C(K, T).
• Synthetic short call: sell stock and put, lend present value of strike and dividends:
C (K, T ) = S0 + P (K, T ) − (PV0,T (Div) + PV0,T (K)).
• Synthetic short put: buy stock, sell call, borrow present value of strike and dividends:
P (K, T ) = C (K, T ) − S0 + PV0,T (Div) + PV0,T (K). 4 CONVERSIONS AND REVERSE CONVERSIONS 39 Practice Problems
Problem 4.1
Which of the following applies to a conversion?
(A) Selling a call, selling a put, buying the stock
(B) Selling a call, buying a put, shortselling the stock
(C) Buying a call, buying a put, shortselling the stock
(D) Selling a call, buying a put, buying the stock
(E) Buying a call, selling a put, shortselling the stock
(F) Buying a call, selling a put, buying the stock
Problem 4.2
Which of the following applies to a reverse conversion?
(A) Selling a call, selling a put, buying the stock
(B) Selling a call, buying a put, shortselling the stock
(C) Buying a call, buying a put, shortselling the stock
(D) Selling a call, buying a put, buying the stock
(E) Buying a call, selling a put, shortselling the stock
(F) Buying a call, selling a put, buying the stock
Problem 4.3
A conversion creates
(A) A synthetic stock
(B) A synthetic short stock hedged with a long stock position
(C) A synthetic long stock hedged with a short stock position
(D) A synthetic long T −bill
Problem 4.4
Suppose that the price of a nondividendpaying stock is $41. A European call option with strike
price of $40 sells for $2.78. A European put option with strike price $40 sells for $1.09. Both options
expire in 3 months. Calculate the annual continuously compounded riskfree rate on a synthetic
Tbill created using these options.
Problem 4.5
The current price of a nondividendpaying stock is $99. A European call option with strike price
$100 sells for $11.71 and a European put option with strike price $100 sells for $5.31. Both options
expire in nine months. The continuously compounded riskfree interest rate is 12.39%. What is the
cost of a synthetic T −bill created using these options? 40 PARITY AND OTHER PRICE OPTIONS PROPERTIES Problem 4.6
A share of stock pays $3 dividends three months from now and nine months from now. The price
of the stock is $99. A ninemonth European call option on the stock costs $11.71. A nine month
European put option costs $5.31. Both options have a strike price of $100. The continuously
compounded riskfree interest rate is 12.39%.
(a) What is the cost of a synthetic T −bill created using these options?
(b) How much will this synthetic position pay at maturity?
Problem 4.7
A synthetic T −bill was created by buying a share of XYZ stock for $20, selling a call option with
underlying asset the share of XYZ stock for $5.43, and buying a put option with strike price of
$20 for $2.35. Both options expire in one year. The current stock price is $23. What is the
continuously compounded riskfree interest involved in this transaction? Assume that the stock
pays no dividends.
Problem 4.8
Using the putcall parity relationship, what should be done to replicate a short sale of a nondividendpaying stock?
Problem 4.9 (Synthetic Options)
Using the putcall parity relationship how would you replicate a long call option with underlying
asset a share of stock that pays dividends?
Problem 4.10 (Synthetic Options)
Using the putcall parity relationship how would you replicate a long put option with underlying
asset a share of stock that pays dividends?
Problem 4.11
A reversal was created by selling a nondividendpaying stock for $99, buying a 5month European
call option with a strike price of $96 on a share of stock for $6.57, and selling a 5month European
put option with a strike price of $96 for $P. The annual eﬀective rate of interest is 3%. Find P.
Problem 4.12 ‡
The price of a nondividendpaying stock is $85. The price of a European call with a strike price of
$80 is $6.70 and the price of a European put with a strike price of $80 is $1.60. Both options expire
in three months.
Calculate the annual continuously compounded riskfree rate on a synthetic T −Bill created using
these options. 5 PARITY FOR CURRENCY OPTIONS 41 5 Parity for Currency Options
In the previous section we studied the parity relationship that involves stock options. In this section
we consider the parity relationship applied to options with currencies as underlying assets.
Currencies are in general exchanged at an exchange rate. For example, the market today lists the
exchange rate between dollar and euro as e1 = $ 1.25 and $1 = e0.80. Exchange rates between
currencies ﬂuctuate over time. Therefore, it is useful for businesses involved in international markets
to be involved in currency related securities in order to be secured from exchange rate ﬂuctuation.
One such security is the currency forward contract.
Currency contracts are widely used to hedge against changes in exchange rates. A Forward currency contract is an agreement to buy or sell a speciﬁed amount of currency at a set price on
a future date. The contract does not require any payments in advance. In contrary, a currency
prepaid forward is a contract that allows you to pay today in order to acquire the currency at
a later time. What is the prepaid forward price? Suppose at a time T in the future you want to
acquire e1. Let re be the eurodenominated interest rate (to be deﬁned below) and let x0 be the
exchange rate today in $/e, that is e1 = $x0 . If you want e1 at time T you must have e−re T in
euros today. To obtain that many euros today, you must exchange x0 e−re T dollars into euros. Thus,
the prepaid forward price, in dollars, for a euro is
P
F0,T = x0 e−re T . Example 5.1
Suppose that re = 4% and x0 = $1.25/e. How much should be invested in dollars today to have
e1 in one year?
Solution.
We must invest today
1.25e−0.04 = $1.2
Now, the prepaid forward price is the dollar cost of obtaining e1 in the future. Thus, the forward
price in dollars for one euro in the future is
F0,T = x0 e(r−re )T
where r is the dollardenominated interest rate.
Example 5.2
Suppose that re = 4%, r = 6%, T = 1 and x0 = $1.25/e. Find the forward exchange rate after
one year. 42 PARITY AND OTHER PRICE OPTIONS PROPERTIES Solution.
The forward exchange rate is 1.25e0.06−0.04 = $1.275 per one euro after one year
Another currency related security that hedges against changes in exchange rates is the currency
option. A currency option is an option which gives the owner the right to sell or buy a speciﬁed
amount of foreign currency at a speciﬁed price and on a speciﬁed date. Currency options can be
either dollardenominated or foreigncurrency denominated. A dollardenominated option on a
foreign currency would give one the option to sell or buy the foreign currency at some time in the
future for a speciﬁed number of dollars. For example, a dollardenominated call on yen would give
one the option to obtain yen at some time in the future for a speciﬁed number of dollars. Thus,
a 3year, $0.008 strike call on yen would give its owner the option in 3 years to buy one yen for
$0.008. The owner would exercise this call if 3 years from now the exchange rate is greater than
$0.008 per yen.
A dollardenominated put on yen would give one the option to sell yen at some time in the future
for a speciﬁed number of dollars. Thus, a 3year, $0.008 strike put on yen would give its owner the
option in 3 years to sell one yen for $0.008. The owner would exercise this call if 3 years from now
the exchange rate is smaller than $0.008 per yen.
The payoﬀ on a call on currency has the same mathematical expression as for a call on stocks, with
ST being replaced by xT where xT is the (spot) exchange rate at the expiration time. Thus, the payoﬀ of a currency call option is max{0, xT − K } and that for a currency put option is max{0, K − xT }.
Consider again the options of buying euros by paying dollars. Since
F0,T = x0 e(r−re )T
the putcall parity
P (K, T ) − C (K, T ) = e−rT K − e−rT F0,T
reduces to
P (K, T ) − C (K, T ) = e−rT K − x0 e−re T .
Thus, buying a euro call and selling a euro put has the same payoﬀ to lending euros and borrowing
dollars. We will use the following form of the previous equation:
C (K, T ) − P (K, T ) = x0 e−re T − e−rT K.
Example 5.3
Suppose the (spot) exchange rate is $1.25/e, the eurodenominated continuously compounded interest rate is 4%, the dollardenominated continuously compounded interest rate is 6%, and the price
of 2year $1.20strike European put on the euro is $0.10. What is the price of a 2year $1.20strike
European call on the Euro? 5 PARITY FOR CURRENCY OPTIONS 43 Solution.
Using the putcall parity for currency options we ﬁnd
C = 0.10 + 1.25e−0.04×2 − 1.20e−0.06×2 = $0.19
A foreign currencydenominated option on the dollar is deﬁned in a similar way as above by
interchanging dollar with the foreign currency. Thus, a putcall parity of options to buy dollars by
paying euros takes the form
C (K, T ) − P (K, T ) = x0 e−rT − e−re T K
where x0 is the exchange rate in e/$.
Example 5.4
Suppose the (spot) exchange rate is e0.8/$, the eurodenominated continuously compounded interest rate is 4%, the dollardenominated continuously compounded interest rate is 6%, and the
price of 2year e0.833strike European put on the dollar is e0.08. What is the price of a 2year
e0.833strike European call on the dollar?
Solution.
Using the putcall parity for currency options we ﬁnd
C = 0.08 + 0.8e−0.06×2 − 0.833e−0.04×2 =e 0.0206
Remark 5.1
Note that in a dollardenominated option, the strike price and the premium are in dollars. In
contrast, in a foreign currencydenominated option, the strike price and the premium are in the
foreign currency. 44 PARITY AND OTHER PRICE OPTIONS PROPERTIES Practice Problems
Problem 5.1
A U.S. based company wants to buy some goods from a ﬁrm in Switzerland and the cost of the goods
is 62,500 SF. The ﬁrm must pay for the goods in 120 days. The exchange rate is x0 = $0.7032/SF.
Given that rSF = 4.5% and r$ = 3.25%, ﬁnd the forward exchange price.
Problem 5.2
Suppose the exchange rate is $1.25/e, the eurodenominated continuously compounded interest
rate is 4%, the dollardenominated continuously compounded interest rate is 6%, and the price of
2year $1.20strike European call on the euro is $0.19. What is the price of a 2year $1.20strike
European put on the Euro?
Problem 5.3
Suppose the exchange rate is £0.4/SF, the Swiss Francdenominated continuously compounded
interest rate is 5%, the British pounddenominated continuously compounded interest rate is 3%,
and the price of 3year £0.5strike European call on the SF is £0.05. What is the price of a 3year
£0.5strike European put on the SF?
Problem 5.4
Suppose the current $/eexchange rate is 0.95 $/e, the eurodenominated continuously compounded
interest rate is 4%, the dollardenominated continuously compounded interest rate is r. The price
of 1year $0.93strike European call on the euro is $0.0571. The price of a 1year $0.93strike put
on the euro is $0.0202. Find r.
Problem 5.5
A sixmonth dollardenominated call option on euros with a strike price of $1.30 is valued at $0.06.
A sixmonth dollardenominated put option on euros with the same strike price is valued at $0.18.
The dollardenominated continuously compounded interest rate is 5%.
(a) What is the 6month dollareuro forward price?
(b) If the eurodenominated continuously compounded interest rate is 3.5%, what is the spot exchange rate?
Problem 5.6
A ninemonth dollardenominated call option on euros with a strike price of $1.30 is valued at $0.06.
A ninemonth dollardenominated put option on euros with the same strike price is valued at $0.18.
The current exchange rate is $1.2/e and the dollardenominated continuously compounded interest
rate is 7%. What is the continuously compounded interest rate on euros? 5 PARITY FOR CURRENCY OPTIONS 45 Problem 5.7
Currently one can buy one Swiss Franc for $0.80. The continuouslycompounded interest rate for
Swiss Francs is 4%. The continuouslycompounded interest rate for dollars is 6%. The price of a
SF 1.15strike 1year call option is SF 0.127. The spot exchange rate is SF 1.25/$. Find the cost in
dollar of a 1year SF 1.15strike put option.
Problem 5.8
The price of a $0.02strike 1 year call option on an Indian Rupee is $0.00565. The price of a $0.02
strike 1 year put option on an Indian Rupee is $0.00342. Dollar and rupee interest rates are 4.0%
and 7.0%, respectively. How many dollars does it currently take to purchase one rupee?
Problem 5.9
Suppose the (spot) exchange rate is $0.009/ , the yendenominated continuously compounded
interest rate is 1%, the dollardenominated continuously compounded interest rate is 5%, and the
price of 1year $0.009strike European call on the yen is $0.0006. What is the dollardenominated
European yen put price such that there is no arbitrage opportunity?
Problem 5.10
Suppose the (spot) exchange rate is $0.009/ , the yendenominated continuously compounded
interest rate is 1%, the dollardenominated continuously compounded interest rate is 5%, and the
price of 1year $0.009strike European call on the yen is $0.0006. The price of a 1year dollardenominated European yen put with a strike of $0.009 has a premium of $0.0004. Demonstrate the
existence of an arbitrage opportunity.
Problem 5.11 ‡
You are given:
(i) The current exchange rate is $0.011/ .
(ii) A fouryear dollardenominated European put option on yen with a strike price of $0.008 sells
for $0.0005.
(iii) The continuously compounded riskfree interest rate on dollars is 3%.
(iv) The continuously compounded riskfree interest rate on yen is 1.5%.
Calculate the price of a fouryear yendenominated European put option on dollars with a strike
price of 125. 46 PARITY AND OTHER PRICE OPTIONS PROPERTIES 6 Parity of European Options on Bonds
In this section we construct the putcall parity for options on bonds. We start the section by
recalling some vocabulary about bonds. A bond is an interest bearing security which promises
to pay a stated amount (or amounts) of money at some future date (or dates). The company or
government branch which is issuing the bond outlines how much money it would like to borrow and
speciﬁes a length of time, along with an interest rate it is willing to pay. Investors who then lend
the requested money to the issuer become the issuer’s creditors through the bonds that they hold.
The term of the bond is the length of time from the date of issue until the date of ﬁnal payment.
The date of the ﬁnal payment is called the maturity date. Any date prior to, or including, the
maturity date on which a bond may be redeemed is termed a redemption date.
The par value or face value of a bond is the amount that the issuer agrees to repay the bondholder
by the maturity date.
Bonds with coupons are periodic payments of interest made by the issuer of the bond prior to
redemption. Zero coupon bonds are bonds that pay no periodic interest payments. It just pays
a lump sum at the redemption date.
Like loans, the price of a bond is deﬁned to be the present value of all future payments. The basic
formula for price is given by
B0 = F ran i + C (1 + i)−n
where F is the par value of the bond, r is the rate per coupon payment period, C is the amount of
money paid at a redemption date to the holder of the bond.
Example 6.1
A 15year 1000 par value bond bearing a 12% coupon rate payable annually is bought to yield 10%
convertible continuously. Find the price of the bond.
Solution.
The price of the bond is
B0 =120a15 + 1000e−0.10×15
1 − e−0.10×15
=120
+ 1000e−0.10×15
e0.10 − 1
=$1109.54
Example 6.2
A 10year 100 par value bond bearing a 10% coupon rate payable semiannually, and redeemable
at 105, is bought to yield 8% convertible semiannually. Find the price of the bond. 6 PARITY OF EUROPEAN OPTIONS ON BONDS 47 Solution.
The semiannual coupon payment is F r = 100 × 0.05 = $5. The price of the bond is
B0 = 5a20 0.04 + 105(1.04)−20 = $115.87
A bond option is an option that allows the owner of the option to buy or sell a particular bond
at a certain date for a certain price. Because of coupon payments (which act like stock dividends),
the prepaid forward price diﬀers from the bond’s price. Thus, if we let B0 be the bond’s price then
the putcall parity for bond options is given by
C (K, T ) = P (K, T ) + [B0 − PV0,T (Coupons)] − PV0,T (K).
Notice that for a noncoupon bond, the parity relationship is the same as that for a nondividendpaying stock.
Example 6.3
A 15year 1000 par value bond bearing a 12% coupon rate payable annually is bought to yield 10%
convertible continuously. A $1000strike European call on the bond sells for $150 and expires in 15
months. Find the value of a $1000strike European put on the bond that expires in 15 months.
Solution.
The price of the bond was found in Example 6.1. Using the putcall parity for options on bonds,
we have
C (K, T ) = P (K, T ) + [B0 − PV0,T (Coupons)] − PV0,T (K).
or
150 = P (1000, 1.25) + [1109.54 − 120e−0.1 ] − 1000e−0.1×1.25 .
Solving for P we ﬁnd P (1000, 1.25) = $31.54
Example 6.4
A 15year 1000 par value bond pays annual coupon of X dollars. The annual continuously compounded interest rate is 10%. The bond currently sells for $1109.54. A $1000strike European call
on the bond sells for $150 and expires in 15 months. A $1000strike European put on the bond sells
for $31.54 and expires in 15 months. Determine the value of X.
Solution.
Using the putcall parity for options on bonds we have
PV0,T (Coupons) = P(K, T) − C(K, T) + B0 − Ke−rT 48 PARITY AND OTHER PRICE OPTIONS PROPERTIES or
PV0,T (Coupons) = 31.54 − 150 + 1109.54 − 1000e−0.1×1.25 = 108.583.
The present value of the coupon payments is the present value of an annuity that pays X dollars a
year for 15 years at the interest rate of 10%. Thus,
108.583 = X 1 − e−0.10×15
e0.10 − 1 . Solving this equation we ﬁnd X = $14.621
Example 6.5
A 23year bond pays annual coupon of $2 and is currently priced at $100. The annual continuously
compounded interest rate is 4%. A $Kstrike European call on the bond sells for $3 and expires in
23 years. A $Kstrike European put on the bond sells for $5 and expires in 23 years.
(a) Find the present value of the coupon payments.
(b) Determine the value of K.
Solution.
(a) The present value of the coupon payments is the present value of an annuity that pays $2 a year
for 23 years at the continuous interest rate of 4%. Thus,
PV0,T (Coupons) = 2 1 − e−0.04×23
e0.04 − 1 = $29.476. (b) Using the putcall parity of options on bonds we have
Ke−rT = P (K, T ) − C (K, T ) + B0 − PV0,T (Coupons)
or
Ke−0.04×23 = 5 − 3 + 100 − 29.476.
Solving for K we ﬁnd K = $181.984 6 PARITY OF EUROPEAN OPTIONS ON BONDS 49 Practice Problems
Problem 6.1
Find the price of a $1000 par value 10year bond maturing at par which has coupons at 8% convertible semiannually and is bought to yield 6% convertible quarterly.
Problem 6.2
Find the price of a $1000 par value 10year bond maturing at par which has coupons at 8% convertible quarterly and is bought to yield 6% convertible semiannually.
Problem 6.3
Find the price of a 1000 par value 10year bond with coupons of 8.4% convertible semiannually,
which will be redeemed at 1050. The bond yield rate is 10% convertible semiannually for the ﬁrst
ﬁve years and 9% convertible semiannually for the second ﬁve years.
Problem 6.4
A zerocoupon bond currently sells for $67. The annual eﬀective interest rate is 4%. A $80strike
European call on the bond costs $11.56. Find the price of a $80strike put option on the bond.
Both options expire in 9 years.
Problem 6.5
The annual coupon payment of a 10year bond is $10. A $200strike European call expiring in
10 years costs $20. A $200strike European put expiring in 10 years costs $3. Assume an annual
eﬀective interest rate of 3%.
(a) Calculate the present value of the coupons.
(b) Find the price of the bond.
Problem 6.6
A 76year bond pays annual coupon of X dollars. The annual continuously compounded interest
rate is 5%. The bond currently sells for $87. A $200strike European call on the bond sells for
$5 and expires in 76 years. A $200strike European put on the bond sells for $2 and expires in 76
years. Determine the value of X.
Problem 6.7
A 1year bond is currently priced at $90. The bond pays one dividend of $5 at the end of one
year. A $100strike European call on the bond sells for $6 and expires in one year. A $100strike
European put on the bond sells for $5 and expires in one year. Let i be the annual eﬀective rate of
interest. Determine i. 50 PARITY AND OTHER PRICE OPTIONS PROPERTIES Problem 6.8
A 15year 1000 par value bond bearing a 12% coupon rate payable semiannually is bought to yield
10% continuously compounded interest. One month after the bond is issued, a $950strike European
put option on the bond has a premium of $25 and expiration date of 1 year. Calculate the value of
a $950strike European call option on the bond expiring in one year, one month after the bond is
issued.
Problem 6.9
A 10year bond with par value $1000 pays semiannual coupon of $35 and is currently priced at
$1000. The annual interest rate is 7% convertible semiannually. A $Kstrike European call on the
bond that expires in nine months has a premium $58.43 greater than a $Kstrike European put on
the bond that expires in nine months. Determine the value of K.
Problem 6.10
Amorphous Industries issues a bond with price 100 and annual coupons of 2, paid for 23 years. The
annual eﬀective interest rate is 0.04. A put option with a certain strike price and expiring in 23
years has price 5, whereas a call option with the same strike price and time to expiration has price
3. Find the strike price of both options. 7 PUTCALL PARITY GENERALIZATION 51 7 PutCall Parity Generalization
The options considered thus far have a strike asset consisting of cash. That is, the option’s underlying asset is exchanged for cash. Options can be designed to exchange any asset for any other
asset not just asset for cash. In this section we derive a putcall parity of such options. This parity
version includes all previous versions as special cases. Options where the underlying assets and the
strike assets can be anything are called exchange options.
Let the underlying asset be called asset A, and the strike asset, asset B. Let St be the price at time
P
t of asset A and Qt that of asset B. Let Ft,T (S ) denote the time t price of the prepaid forward on
P
asset A, paying ST at time T. Let Ft,T (Q) denote the time t price of the prepaid forward on asset
B , paying QT at time T.
Example 7.1
P
If the underlying asset is a share of stock, ﬁnd a formula for Ft,T (S ).
Solution.
We consider the following cases:
P
• If the stock pays no dividends then Ft,T (S ) = St .
P
• If the stock pays discrete dividends then Ft,T (S ) = St − P Vt,T (Div ) where P Vt,T stands for the
present value with T − t periods to expiration.
P
• If the stock pays continuous dividends then Ft,T (S ) = e−δ(T −t) St
Now, let C (St , Qt , T − t) be the time t price of an option with T − t periods to expiration, which
gives us the privilege to give asset B and get asset A at the expiration time T. Let P (St , Qt , T − t)
be time t price of the corresponding put option which gives us the privilege to give asset A and get
asset B. The payoﬀ of the call option at the expiration date T is
C (ST , QT , 0) = max{0, ST − QT }
and that of the put option is
P (ST , QT , 0) = max{0, QT − ST }.
Now, to understand how the generalized putcall parity is established, we consider the following
portfolio:
1) buying a call at the time t price of C (St , Qt , T − t);
2) selling a put at the time t price of P (St , Qt , T − t);
P
3) selling a prepaid forward on A at the price Ft,T (S );
P
4) buying a prepaid forward on B at the price Ft,T (Q).
The payoﬀ table of this portfolio is given next: 52 PARITY AND OTHER PRICE OPTIONS PROPERTIES Transaction
Buy Call
Sell Put
Sell Prepaid
Forward on A
Buy Prepaid
Forward on B
Net cash ﬂow Cost at Time t
−C (St , Qt , T − t)
P (St , Qt , T − t)
P
Ft,T (S ) Expiration
ST ≤ QT
0
ST − QT
−ST ST > QT
ST − QT
0
−ST P
−Ft,T (Q) QT QT P
−C (St , Qt , T − t) + Ft,T (S )
P
P (St , Qt , T − t) − Ft,T (Q) 0 0 These transactions, as a whole, create a portfolio that provides a sure payoﬀ of zero at time T. In
order to avoid arbitrage, the net cost of these transactions at time t must be zero, implying
P
P
C (St , Qt , T − t) − P (St , Qt , T − t) = Ft,T (S ) − Ft,T (Q). (7.1) This equation is referred to as the generalized putcall parity of options. It follows from this
equation that relative call and put premiums are determined by prices of prepaid forwards on the
underlying asset and the strike asset.
Remark 7.1
All European options satisfy equation (7.1) whatever the underlying asset.
Example 7.2
A nondividendpaying stock A is currently selling for $34 a share whereas a nondividendpaying
stock B is selling for $56 a share. Suppose A is the underlying asset and B is the strike asset.
P
P
(a) Find Ft,T (S ) and Ft,T (Q).
(b) Show that the put option is more expensive than the call for any time to expiration T of the
options.
Solution.
P
P
(a) Since the stocks pay no dividends, we have Ft,T (S ) = $34 and Ft,T (Q) = $56.
(b) Using the putcall parity we have
P
P
C (At , Bt , T − t) − P (At , Bt , T − t) = Ft,T (S ) − Ft,T (Q) = 34 − 56 = −$22. Thus, the put is $22 more expensive than the call
Example 7.3
A share of Stock A pays quarterly dividend of $1.2 and is currently selling for $55 per share. A 7 PUTCALL PARITY GENERALIZATION 53 share of Stock B pays dividends at the continuously compounded yield of 8% and is currently selling
for $72 per share. The continuously compounded riskfree interest rate is 6% per year. A European
exchange call option with stock B as the underlying asset and stock A as the strike asset sells for
$27.64. The option expires in one year. Find the premium of the corresponding put option.
Solution.
The quarterly eﬀective rate of interest is i = e0.06×0.25 − 1 = 1.5113%. We have
P
Ft,T (S ) = 72e−0.08 = 66.46437694 and
P
Ft,T (Q) = 55 − 1.2a4 i = $50.38. Now, using the generalized putcall parity we ﬁnd
27.64 − P (St , Qt , T − t) = 66.46437694 − 50.38
or
P (St , Qt , T − t) = 11.55562306
Example 7.4
A share of stock A pays dividends at the continuously compounded yield of 5% and is currently
selling for $110. A share of a nondividendpaying stock B is currently selling for $30. A 10month
European call option with underlying asset four shares of Stock B and strike asset one share of stock
A sells for $35.95. Find the price of a 10month put option with underlying asset twelve shares of
stock B and strike asset three shares of stock A.
Solution.
The current value of the four shares of stock B is 4 × 30 = $120. Using the putcall parity for
exchange options to ﬁnd the price of a 10month put option with underlying asset four shares of
stock B and strike asset one share of stock A to ﬁnd
10
10
P
P
C (120, 110, ) − P (120, 110, ) = F0, 10 (4B ) − F0, 10 (A)
12
12
12
12
or
10
10
35.95 − P (120, 110, ) = 120 − 110e−0.05× 12 .
12
Solving this equation we ﬁnd
10
P (120, 110, ) = 21.4608.
12
Hence, the price of a 10month put option with underlying asset twelve shares of stock B and strike
asset three shares of stock A is
4 × 21.4608 = $85.8433 54 PARITY AND OTHER PRICE OPTIONS PROPERTIES Example 7.5 ‡
Consider a model with two stocks. Each stock pays dividends continuously at a rate proportional
to its price. Let Sj (t) denotes the price of one share of stock j at time t where j = 1, 2.
Consider a claim maturing at time 3. The payoﬀ of the claim is
Maximum{S1 (3), S2 (3)}.
You are given:
(i) S1 (0) = $100
(ii)S2 (0) = $200
(iii) Stock 1 pays dividends of amount 0.05S1 (t)dt between time t and time t + dt.
(iv) Stock 2 pays dividends of amount 0.1S2 (t)dt between time t and time t + dt.
(v) The price of a European option to exchange Stock 2 for Stock 1 at time 3 is $10.
Calculate the price of the claim.
Hint: For each share of stock, the amount of dividends paid between t and t + dt is S (t)δdt.
Solution.
Note ﬁrst that the continuously compounded divided yield for stock 1 is 5% and that for stock 2
is 10%. Consider a portfolio consisting of one share of stock 2 and a European option that allows
you to exchange stock 2 for stock 1. The payoﬀ of this portfolio is Maximum{S1 (3), S2 (3)}. If at
time 3, S1 (3) < S2 (3) the owner will keep stock 2 and allows the option to expire unexercised. If
S1 (3) > S2 (3), then the owner will exercise the option by giving up stock 2 for stock 1.
The cost now for a share of stock at time 3 is
P
F0,3 (S2 ) = 200e−0.10×3 = $148.16. The cost of an exchange option allowing the owner to exchange stock 2 for stock 1 at time 3 is $10.
Thus, the price of the claim is 148.16 + 10 = $158.16 7 PUTCALL PARITY GENERALIZATION 55 Practice Problems
Problem 7.1
A share of stock A pays dividends at the continuous compounded dividend yield of 3%. Currently
a share of stock A sells for $65. A nondividendpaying stock B sells currently for $85 a share. A
9month call option with stock A as underlying asset and stock B as strike asset costs $40. Find
the price of the corresponding 9month put option.
Problem 7.2
A nondividendpaying stock A is currently selling for $34 a share whereas a nondividendpaying
stock B is selling for $56 a share. Suppose A is the underlying asset and B is the strike asset.
Suppose that a put option on A costs $25. Find the price of the corresponding call option.
Problem 7.3
A share of stock A pays dividends at the continuous compounded dividend yield of δ. Currently
a share of stock A sells for $67. A nondividendpaying stock B sells currently for $95 a share. A
13month call option with stock A as underlying asset and stock B as strike asset costs $45. The
corresponding 13month put option costs $74.44. Determine δ.
Problem 7.4
A share of stock A pays dividends at the continuously compounded yield of 10%. A share of stock
B pays dividends at the continuously compounded yield of 25% . A share of stock A currently sells
for $2000 and that of stock B sells for $D. A 12year call option with underlying asset A and strike
asset B costs $543. A 12year put option with underlying asset A and strike asset B costs $324.
Determine the value of D.
Problem 7.5
A share of stock A pays dividends at the continuously compounded yield of 4.88%. Currently a
share of stock A costs $356. A share of stock B pays no dividends and currently costs $567. A
T −year call option with underlying asset A and strike asset B costs $34. A T −year put option
with underlying asset A and strike asset B costs $290. Determine the time of expiration T.
Problem 7.6
A share of stock A pays no dividends and is currently selling for $100. Stock B pays dividends at
the continuously compounded yield of 4% and is selling for $100. A 7month call option with stock
A as underlying asset and stock B as strike asset costs $10.22. Find the price of the corresponding
7month put option. 56 PARITY AND OTHER PRICE OPTIONS PROPERTIES Problem 7.7
A share of stock A pays $3 dividends every two months at the continuously compounded interest
rate of 6% and is currently selling for $50. Stock B pays dividends at the continuously compounded
yield of 3% and is selling for $51. A 6month call option with stock A as underlying asset and stock
B as strike asset costs $2.70. Find the price of the corresponding 6month put option.
Problem 7.8
A share of stock A pays no dividends and is currently selling for $40. A share of Stock B pays
no dividends and is selling for $45. A 3month European exchange call option with stock A as
underlying asset and stock B as strike asset costs $6. Find the price of the corresponding 3month
exchange put option.
Problem 7.9
A share of stock A pays dividends at the continuously compounded yield of 4%. A share of stock
B pays dividends at the continuously compounded yield of 4% . A share of stock A currently sells
for $40 and that of stock B sells for $D. A 5month put option with underlying asset A and strike
asset B is $7.76 more expensive than the call option. Determine the value of D.
Problem 7.10
A share of stock B pays dividends at the continuously compounded yield of 4% and is currently
selling for $100. A share of a nondividendpaying stock A is currently selling for $100. A 7month
American call option with underlying asset one share of stock A and strike asset one share of stock
B sells for $10.22. Find the price of the corresponding 7month European put option. Hint: For a
stock with no dividends the price of an American call option is equal to the price of a European
call option. 8 LABELING OPTIONS: CURRENCY OPTIONS 57 8 Labeling Options: Currency Options
Any option can be labeled as a put (the privilege to sell) or a call (the privilege to buy). The
labeling process is a matter of perspective depending upon what we label as the underlying asset
and what we label as the strike asset. Thus, a call option can be viewed as a put option and vice
versa. To elaborate, you purchase an option that gives you the privilege of receiving one share of
stock A by surrendering one share of stock B in one year. This option can be viewed as either a
call or a put. If you view stock A as the underlying asset and stock B as the strike asset then the
option is a call option. This option gives you the privilege to buy, at T = 1, one share of stock A
by paying one share of stock B. If you view stock B as the underlying asset and stock A as the
strike asset then the option is a put option. This option gives you the privilege to sell, at T = 1,
one share of stock B for the price of one share of stock A.
Example 8.1
You purchase an option that gives you the privilege to sell one share of Intel stock for $35. Explain
how this option can be viewed either as a call option or as a put option.
Solution.
If you view the share of Intel stock as the underlying asset, the option is a put. The option gives
you the privilege to sell the stock for the price of $35. If you view $35 as the underlying asset then
the option is a call option. This option gives you the right to buy $35 by selling one share of the
stock
Options on Currencies
The idea that calls can be relabeled as put is used frequently by currency traders. Let xt be the value
in dollars of one foreign currency (f) at time t. That is, 1 f= $xt . A K −strike dollardenominated
call option on a foreign currency that expires at time T gives the owner the right to receive one (f)
in exchange of K dollars. This option has a payoﬀ at expiration, in dollars, given by
max{0, xT − K }.
In the foreign currency, this payoﬀ at the expiration time T would be
K
1
1
1
max{0, xT − K } = max{0, 1 −
} = K max{0, −
}.
xT
xT
K xT
Now, the payoﬀ at expiration of a
currency 1
−strike
K foreigndenominated put on dollars is in that foreign max{0, 1
1
−
}.
K xT 58 PARITY AND OTHER PRICE OPTIONS PROPERTIES It follows that on the expiration time T , the payoﬀ in the foreign currency of one K −strike dollar1
denominated call on foreign currency is equal to the payoﬀ in foreign currency of K K −strike
foreigndenominated puts on dollars. Thus, the two positions must cost the same at time t = 0, or
else there is an arbitrage opportunity. Hence,
1
K ×(premium in foreign currency of K −strike foreigndenominated put on dollars) = (premium in
foreign currency of K −strike dollardenominated call on foreign currency) = (premium in dollars
of K −strike dollardenominated call on foreign currency)/x0 . In symbols,
C$ (x0 , K, T )
= KPf
x0 11
, ,T
x0 K C$ (x0 , K, T ) = x0 KPf 11
, ,T
x0 K . P$ (x0 , K, T ) = x0 KCf 11
, ,T
x0 K . or Likewise, Example 8.2
A 1year dollardenominated call option on euros with a strike price of $0.92 has a payoﬀ, in
dollars, of max{0, x1 − 0.92}, where x1 is the exchange rate in dollars per euro one year from now.
Determine the payoﬀ, in euros, of a 1year eurodenominated put option on dollars with strike price
1
= e1.0870.
0.92
Solution.
The payoﬀ in euros is given by
max{0, 1
1
−}
0.92 x1 Example 8.3
The premium of a 1year 100strike yendenominated put on the euro is
8.763. The current
exchange rate is 95/e. What is the strike of the corresponding eurodenominated yen call, and
what is its premium?
Solution.
The strike price of the eurodenominated yen call is
eurodenominated yen call satisﬁes the equation 1
100 Pyen (95, 100, 1) = 95 × 100Ce = e0.01. The premium, in euros, of the
11
,
,1 .
95 100 8 LABELING OPTIONS: CURRENCY OPTIONS 59 Thus,
Ce 11
,
,1
95 100 = 1
× 0.01 × 8.763 = e0.00092242
95 Example 8.4
Suppose the (spot) exchange rate is $0.009/ , the yendenominated continuously compounded
interest rate is 1%,the dollardenominated continuously compounded interest rate is 5%, and the
price of 1year $0.009strike dollardenominated European call on the yen is $0.0006. What is the
price of a yendenominated dollar call?
Solution.
The dollardenominated call option is related to the yendenominated put option by the equation
C$ (x0 , K, T ) = x0 KP 11
, ,T
x0 K . Thus,
P 1
1
,
,1
0.009 0.009 = 0.0006
1
×
=
0.009
0.009 7.4074. Using the putcall parity of currency options we have
P
1
/$.
0.009 where x0 =
C 1
1
,
,1 − C
0.009 0.009 1
1
,
,1
0.009 0.009 = e−r T K − x0 e−r$ T Thus, 1
1
,
,1
0.009 0.009 = 7.4074 − 1 −0.01
1 −0.05
e
+
e
=
0.009
0.009 3.093907 60 PARITY AND OTHER PRICE OPTIONS PROPERTIES Practice Problems
Problem 8.1
You purchase an option that gives you the privilege to buy one share of Intel stock for $25. Explain
how this option can be viewed either as a call option or a put option.
Problem 8.2
A 1year dollardenominated call option on euros with a strike price of $0.92 costs $0.0337. The
current exchange rate is x0 = $0.90/e. What is the premium, in euros, of this call option?
Problem 8.3
A 1year dollardenominated call option on euros with a strike price of $0.92 costs $0.0337. The
current exchange rate is x0 = $0.90/e. What is the strike of the corresponding eurodenominated
dollar put, and what is its premium?
Problem 8.4
The premium on a 1year dollardenominated call option on euro with a strike price $1.50 is $0.04.
The current exchange rate is $1.52/e. Calculate the premium in euros of the corresponding 1year
eurodenominated put option on dollars.
Problem 8.5
Suppose the (spot) exchange rate is £0.6/$, the pounddenominated continuously compounded
interest rate is 8%,the dollardenominated continuously compounded interest rate is 7%, and the
price of 1year £0.58strike pounddenominated European put on the dollar is £0.0133. What is
the price of a dollardenominated pounds put?
Problem 8.6
Suppose the (spot) exchange rate is $1.20/e. The price of a 6month $1.25strike dollardenominated
European call on the euro is $0.083. What is the premium in euro of the eurodenominated dollar
put?
Problem 8.7
Suppose the (spot) exchange rate is e0.85/$. The price of a 6month e0.80strike eurodenominated
European call on the dollar is $0.0898. What is the premium in dollars of the dollardenominated
euro put?
Problem 8.8
Suppose the (spot) exchange rate is franc f 1.65/e. The price of a 1year f 1.60strike francdenominated European put on the euro is f 0.0918. What is the premium in euros of the eurodenominated franc call? 8 LABELING OPTIONS: CURRENCY OPTIONS 61 Problem 8.9
Let $ denote the Australian dollars. Suppose the (spot) exchange rate is £0.42/$, the pounddenominated continuously compounded interest rate is 8%,the dollardenominated continuously
compounded interest rate is 7%, and the price of 1year £0.40strike pounddenominated European
put on the dollar is £0.0133. What is the price of a dollardenominated pounds put?
Problem 8.10
Suppose the (spot) exchange rate is e0.70/$, the eurodenominated continuously compounded interest rate is 8%,the dollardenominated continuously compounded interest rate is 7%, and the price
of 6month e0.625strike eurodenominated European call on the dollar is e0.08. What is the price,
in euros, of a dollardenominated euro call?
Problem 8.11
Suppose the (spot) exchange rate is £0.38/$. The price of a 6month £1.40strike pounddenominated
European put on the dollar is £0.03. What is the premium in dollars of the dollardenominated
pounds call? 62 PARITY AND OTHER PRICE OPTIONS PROPERTIES 9 NoArbitrage Bounds on Option Prices
In this section lower and upper bounds of option prices are determined. Recall that American
options are options that can be exercised at any time up to (and including) the maturity date. On
the contrary, European options can be exercised only at the maturity. Since an American option
can duplicate a European option by exercising the American option at the maturity date, it follows
that an American option is always worth at least as much as a European option on the same asset
with the same strike price and maturity date. In symbols,
CAmer (K, T ) ≥ CEur (K, T )
and
PAmer (K, T ) ≥ PEur (K, T ).
Example 9.1
Show that for a nondividendpaying stock one has CAmer (K, T ) = CEur (K, T ).
Solution.
Suppose that CAmer (K, T ) > CEur (K, T ). We will show that this creates an arbitrage opportunity.
Consider the position of selling the American call and buying the European call. The payoﬀ for
this position is given next.
Exercise or Expiration
Transaction
Time 0
St ≤ K
St > K
Buy European call −CEur (K, T )
0
St − K
Sell American call CAmer
0
K − St
Total
CAmer − CEur (K, T ) > 0 0
0
The position has a sure payoﬀ of 0 and positive time0 payoﬀ. Thus, an arbitrage occurs
We next establish some bounds on the option prices. We ﬁrst consider call options:
• Since the best one can do with a call stock option is to own the stock so the call price cannot
exceed the current stock price. Thus,
S0 ≥ CAmer (K, T ) ≥ CEur (K, T ).
Example 9.2
Use a noarbitrage argument to establish that CAmer (K, T ) ≤ S0 . 9 NOARBITRAGE BOUNDS ON OPTION PRICES 63 Solution.
Suppose that CAmer (K, T ) > S0 . Consider the position of buying a stock and selling an American
call on the stock. The payoﬀ table for this position is given next.
Exercise or Expiration
Transaction Time 0
St ≤ K
Buy a stock −S0
St
Sell a call
CAmer
0
Total
CAmer − S0 > 0 St St > K
St
K − St
K Every entry in the row “Total” is nonnegative with CAmer − S0 > 0. Thus, an arbitrage occurs
• The price of a call or a put option has to be nonnegative because with these options you are
oﬀered the possibility for a future gain with no liability. In symbols, we have
0 ≤ CEur (K, T ) ≤ CAmer (K, T ).
• The price of a European call option must satisfy the putcall parity. Thus, for a nondividendpaying stock we have
CEur (K, T ) = PEur (K, T ) + S0 − P V0,T (K ) ≥ S0 − P V0,T (K ).
Combining, all of the above results we can write
S0 ≥ CAmer (K, T ) = CEur (K, T ) ≥ max{0, S0 − P V0,T (K )}.
Likewise, for a discretedividend paying stock we have
S0 ≥ CAmer (K, T ) ≥ CEur (K, T ) ≥ max{0, S0 − P V0,T (Div ) − P V0,T (K )}
and for a continuousdividend paying stock we have
S0 ≥ CAmer (K, T ) ≥ CEur (K, T ) ≥ max{0, S0 e−δT − P V0,T (K )}.
Next, we consider bounds on put options:
• Put options are nonnegative: 0 ≤ PEur (K, T ) ≤ PAmer (K, T ).
• The best one can do with a European put option is to get the strike price K at the maturity date
T. So a European put cannot be worth more than the present value of the strike price. That is,
PEur (K, T ) ≤ P V0,T (K ). 64 PARITY AND OTHER PRICE OPTIONS PROPERTIES • The best one can do with an American put option is to exercise it immediately after time zero
and receive the strike price K. So an American put cannot be worth more than the strike price. In
symbols,
PAmer (K, T ) ≤ K.
Combining the above results we ﬁnd
0 ≤ PEur (K, T ) ≤ PAmer (K, T ) ≤ K.
• The price of a European put option must obey the putcall parity. For a nondividendpaying
stock we have
PEur (K, T ) = CEur (K, T ) + P V0,T (K ) − S0 ≥ P V0,T (K ) − S0 .
Hence,
K ≥ PAmer (K, T ) ≥ PEur (K, T ) ≥ max{0, P V0,T (K ) − S0 }.
For a discretedividend paying stock we have
K ≥ PAmer (K, T ) ≥ PEur (K, T ) ≥ max{0, P V0,T (K ) + P V0,T (Div ) − S0 }.
For a continuousdividend paying stock we have
K ≥ PAmer (K, T ) ≥ PEur (K, T ) ≥ max{0, P V0,T (K ) − S0 e−δT }.
Example 9.3
The price of a K −strike European put option on a share of stock A that expires at time T is
exactly K. What is the price of an American put option on the same stock with the same strike
and expiration date?
Solution.
We have K = PEur (K, T ) ≤ PAmer (K, T ) ≤ K. Thus, PAmer (K, T ) = K
Example 9.4
What is a lower bound for the price of a 1month European put option on a nondividendpaying
stock when the stock price is $12, the strike price is $15, and the riskfree interest rate is 6% per
annum?
Solution.
A lower bound is given by
1 max{0, P V0,T (K ) − S0 } = max{0, 15e−0.06× 12 − 12} = $2.93 9 NOARBITRAGE BOUNDS ON OPTION PRICES 65 Example 9.5
A fourmonth European call option on a dividendpaying stock has a strike price of $60. The stock
price is $64 and a dividend of $0.80 is expected in one month. The continuously compounded
riskfree interest rate is 12%. Find a lower bound for the price of the call.
Solution.
A lower bound is given by
1 4 max{0, S0 − P V0,T (Div ) − P V0,T (K )} = 64 − 0.80e−0.12× 12 − 60e−0.12× 12 = $5.56
Example 9.6
Show that for a nondividendpaying stock we have
CAmer (K, T ) ≥ CEur (K, T ) ≥ S0 − K.
Solution.
Using the putcall parity we have
CEur (K, T ) = S0 − K + PEur (K, T ) + K (1 − e−rT ) ≥ S0 − K
where we used the fact that PEur (K, T ) ≥ 0 and K (1 − e−rT ) ≥ 0 66 PARITY AND OTHER PRICE OPTIONS PROPERTIES Practice Problems
Problem 9.1
The maximum value of a call stock option is equal to:
(A) the strike price minus the initial cost of the option
(B) the exercise price plus the price of the underlying stock
(C) the strike price
(D) the price of the underlying stock.
Problem 9.2
The lower bound of a call option:
(A) can be a negative value regardless of the stock or exercise price
(B) can be a negative value but only when the exercise price exceeds the stock price
(C) can be a negative value but only when the stock price exceeds the exercise price
(D) must be greater than zero
(E) can be equal to zero.
Problem 9.3
What is a lower bound for the price of a 4month call option on a nondividendpaying stock when
the stock price is $28, the strike price is $25, and the continuously compounded riskfree interest
rate is 8%?
Problem 9.4
A sixmonth European put option on a nondividendpaying stock has a strike price of $40. The
stock price is $37. The continuously compounded riskfree interest rate is 5%. Find a lower bound
for the price of the put.
Problem 9.5
A share of stock currently sells for $96. The stock pays two dividends of $8 per share, one six
months from now and the other one year from now. The annual continuously compounded rate of
interest is 5.6%. What is a lower bound for a 1year put option with strike price of $86?
Problem 9.6
A share of stock is currently selling for $99. The stock pays a continuous dividend yield of 2%. The
continuously compounded riskfree interest rate is 12.39%. Find a lower bound for a ninemonth
European call on the stock with strike price of $100.
Problem 9.7 ‡
Which of the following eﬀects are correct on the price of a stock option? 9 NOARBITRAGE BOUNDS ON OPTION PRICES 67 I. The premiums would not decrease if the options were American rather than European.
II. For European put, the premiums increase when the stock price increases.
III. For American call, the premiums increase when the strike price increases.
Problem 9.8
Show that the assumption
CAmer (K, T ) < CEur (K, T )
creates an arbitrage opportunity.
Problem 9.9
Use a noarbitrage argument to establish that PEur ≤ P V0,T (K ).
Problem 9.10
Use a noarbitrage argument to establish that CEur ≥ S0 − K.
Problem 9.11
Use a noarbitrage argument to establish that CEur (K, T ) + P V0,T (K ) ≥ S0 .
Problem 9.12 ‡
Consider European and American options on a nondividendpaying stock. You are given:
(i) All options have the same strike price of 100.
(ii) All options expire in six months.
(iii) The continuously compounded riskfree interest rate is 10%.
You are interested in the graph for the price of an option as a function of the current stock price.
In each of the following four charts IIV, the horizontal axis, S, represents the current stock price,
and the vertical axis, π, represents the price of an option.
Match the option (i.e. American, European calls and puts) with the shaded region in which its
graph lies. If there are two or more possibilities, choose the chart with the smallest shaded region. 68 PARITY AND OTHER PRICE OPTIONS PROPERTIES 10 GENERAL RULES OF EARLY EXERCISE ON AMERICAN OPTIONS 69 10 General Rules of Early Exercise on American Options
American options are contracts that may be exercised early, prior to expiration date. In this section,
we derive general sets of rules about when early exercise is not optimal, or under what conditions it
may be optimal. Further discussions about early exercising that require option pricing models will
be discussed in future sections. In this section the word option stands for American options unless
indicated otherwise.
First, we consider calls on a nondividendpaying stock. Suppose you buy a call option at time t = 0.
Let t denote today’s date so that 0 ≤ t < T. Let St denote today’s stock price and the stock price at
expiration be denoted by ST . Let r denote the annual continuously compounded riskfree interest
rate. We state our ﬁrst result in the form of a proposition.
Proposition 10.1
It is NEVER optimal to early exercise an American call option on a nondividendpaying stock.
Proof.
Using the generalized putcall parity for European options and the fact that PEur (St , K, T − t) ≥ 0
we can write
CEur (St , K, T − t) =PEur (St , K, T − t) + St − Ke−r(T −t)
≥St − Ke−r(T −t)
>St − K
where we use the fact that for 0 ≤ t < T we have 0 < e−r(T −t) < 1. Now, since CAmer (St , K, T − t) ≥
CEur (St , K, T − t), we have
CAmer (St , K, T − t) > St − K.
(10.1)
This shows that the cash ﬂow from selling the option at a time 0 ≤ t ≤ T is larger than the cash
ﬂow from exercising it. Indeed, if you sell the call option you receive CAmer while if you exercise the
option you will receive St − K and therefore you would lose money since CAmer (St , K, T − t) > St − K.
Hence, it is never optimal to early exercise an American call option on a nondividendpaying stock
Remark 10.1
The above proposition does not say that you must hold the option until expiration. It says that if
no longer you wish to hold the call, you should sell it rather than exercise it. Also, it follows from
the previous proposition that an American call on a nondividendpaying stock is like a European
call so that we can write CAmer = CEur . 70 PARITY AND OTHER PRICE OPTIONS PROPERTIES Remark 10.2
One of the eﬀects of early exercising is the time value of money: When you exercise the call option
at time t you pay the strike price K and you own the stock. However, you lose the interest you
could have earned during the time interval [t, T ] had you put K in a savings account, since owning
the stock has no gain during [t, T ] (stock pays no dividends).
The case CAmer (St , K, T − t) < St − K leads to an arbitrage opportunity as shown next.
Example 10.1
Suppose an American call option is selling for CAmer (St , K, T −t) < St −K. Demonstrate an arbitrage
opportunity.
Solution.
First note that CAmer (St , K, T − t) < St − K implies CAmer (St , K, T − t) < St − Ke−r(T −t) . Consider
the position of buying the call option for the price of CAmer (St , K, T − t), short selling the stock,
and lending Ke−r(T −t) . The payoﬀ table of this position is given next.
Transaction
Short sell a stock
Buy a call
Lend Ke−r(T −t)
Total Time t
St
−CAmer
−Ke−r(T −t)
St − CAmer − Ke−r(T −t) Exercise or Expiration
ST ≤ K
ST > K
−ST
−ST
0
ST − K
K
K
> 0 K − ST
0 Every entry in the row “Total” is nonnegative with St − CAmer − Ke−r(T −t) > 0. Thus, an arbitrage
occurs
Now, what if you can not sell the call option, should you ever exercise early then? By shortselling the stock we receive St that grows to St er(T −t) at time T. If at expiration, ST > K , you
exercise the call paying K and receiving the stock. You return the stock to the broker. In this case,
your proﬁt is St er(T −t) − K > St − K. If ST ≤ K, you let your call option expire worthless, purchase
a stock in the market, and return it to the broker. Your proﬁt is St er(T −t) − ST > St − K.
In conclusion, you never want to exercise a call when the underlying asset does not pay dividends.
Example 10.2
Suppose you own an American call option on a nondividendpaying stock with strike price of $100
and due in three months from now. The continuously compounded riskfree interest rate is 1% and
the stock is currently selling for $105.
(a) What is the payoﬀ from exercising the option now? 10 GENERAL RULES OF EARLY EXERCISE ON AMERICAN OPTIONS 71 (b) What is the least payoﬀ from selling the option now?
Suppose that you can not sell your call.
(c) What happens if you exercise now?
(d) What advantage do you have by delaying the exercise until maturity?
(e) What if you know the stock price is going to fall? Shouldn’t you exercise now and take your
proﬁts (by selling the stock), rather than wait and have the option expire worthless?
Solution.
(a) The payoﬀ from exercising the option now is 105 − 100 = $5.
(b) We know that CAmer ≥ max{0, 105 − 100e−0.01×0.25 } so that the least payoﬀ from selling the
option now is
105 − 100e−0.01×0.25 = $5.25.
(c) If you exercise now, you pay $100 and owns the stock which will be worth ST in three months.
(d) You can deposit the $100 into a savings account which will accumulate to 100e0.01×0.25 = $100.25.
At the maturity, if ST > 100 you exercise the call and pay $100 and own the stock. In this case,
you have an extra $0.25 as opposed to exercising now. If ST < 100, you don’t exercise the call and
you have in your pocket $100.25 as opposed to exercising it now (a loss of ST − 100.)
(e) Say you exercise now. You pay $100 for the stock, sell it for $105 and deposit the $5 in a
savings account that will accumulate to 5e0.01×0.25 = $5.0125. If you delay exercising until maturity, you can short the stock now, deposit $105 into a savings account that accumulates to
105e0.01×0.25 = $105.263. At maturity you cover the short by exercising the call option and in this
case your proﬁt is $5.263 > $5.0125
We next consider calls on a dividend paying stock. We ﬁrst establish a condition under which
early exercise is never optimal.
Proposition 10.2
Early exercise for calls on dividend paying stock can not be optimal if
K − P Vt,T (K ) > P Vt,T (Div ).
Proof.
Using putcall parity relationship for stocks with dividends given by
CEur (St , K, T − t) = PEur (St , K, T − t) + St − P Vt,T (Div ) − P Vt,T (K )
which can be written as
CEur (St , K, T − t) =St − K + PEur (St , K, T − t) − P Vt,T (Div ) + K − P Vt,T (K )
≥St − K − [P Vt,T (Div ) − K + P Vt,T (K )]. 72 PARITY AND OTHER PRICE OPTIONS PROPERTIES Thus,
CAmer ≥ St − K − [P Vt,T (Div ) − K + P Vt,T (K )].
To avoid early exercise we expect that selling the call (getting CAmer ) to be more proﬁtable than
exercising the call (getting St − K ). Thus, in order to achieve that we require
K − P Vt,T (K ) > P Vt,T (Div ). (10.2) In this case, we obtain
CAmer (St , K, T − t) > St − K (10.3) As in the discussion for call options with no dividends, selling the call is better than earlyexercising
it. The condition (10.2) says that early exercise should not occur if the interest on the strike price
exceeds the value of the dividends through early exercise. If (10.2) is violated, this does not tell us
that we will exercise, only that we cannot rule it out.
When does early exercising optimal then? If it’s optimal to exercise an American call early, then
the best time to exercise the call is immediately before the dividend payment. To further elaborate,
consider the picture below. We notice that by exercising at T1 instead of T2 we lose the interest that can be earned on K during
[T1 , T2 ], we lose the remaining call option on [T1 , T ], and gain nothing since there are no dividends
between T1 and T2 . Thus, exercising at T1 cannot be optimal. On the other hand, if we exercise
at T3 , we lose the dividend, the remaining call option, lose the interest that can be earned on K
during [T3 , T ],and gain just a tiny interest of the dividend on [T2 , T3 ]. Again, exercising at time T3
is not optimal.
We conclude that for a dividend paying stock, if it’s ever worthwhile to exercise an American call
early, you should exercise the call immediately before the dividend payment, no sooner or later.
Remark 10.3
It follows from our discussion that early exercise on American calls with dividends has its advantages
and disadvantages. Namely,
(+) You gain the dividends between t and T and the interest on the dividends.
(−) You lose the time value of money on the strike. That is, we lose the interest on K from time t
to T. 10 GENERAL RULES OF EARLY EXERCISE ON AMERICAN OPTIONS 73 (−) You lose the remaining call option on the time interval [t, T ].
(−) We pay K for a stock that might be worth less than K at T.
The second half of this section concerns early exercise of American put options. Consider an American put option on a nondividendpaying stock. Contrary to American call options for nondividend
paying stocks, an American put option on a nondividend paying stock may be exercised early.
To avoid early exercise, selling the put (getting PAmer ) should be more proﬁtable than exercising
(getting K − St ), that is, PAmer > K − St . Now, the putcall parity for European options says
PEur = CEur + (K − St ) − (K − P Vt,T (K )).
It follows from this equation that if
CEur > K − P Vt,T (K ) (10.4) then PAmer > K − St and therefore selling is better than exercising. This means that there is no
early exercise if the European call price is high (high asset price compared to strike price), the
strike price is low, or if the discounting until expiration is low (low interest rate or small time to
expiration).
It should be noted that if condition (10.4) is not satisﬁed, we will not necessarily exercise, but we
cannot rule it out.
Example 10.3
Consider an American put option on a stock. When the stock is bankrupt then St = 0 and it is
known that it will stay St = 0.
(a) What is the payoﬀ from early exercising?
(b) What is the present value at time t if put is exercised at maturity?
(c) Is early exercising optimal?
Solution.
Note that CEur = 0 < K − P V0,T (K ).
(a) The payoﬀ will be K.
(b) If the put is exercised at maturity, the present value of K at time t is P Vt,T (K ) < K.
(c) From (a) and (b) we see that early exercise is optimal
Finally, using an argument similar to the one considered for call options with dividends, it is
easy to establish (left as an exercise) that is not optimal to early exercise an American put option
with dividends satisfying
K − P Vt,T (K ) < P Vt,T (Div )
(10.5)
If condition (10.5) is not satisﬁed, we will not necessarily exercise, but we cannot rule it out. 74 PARITY AND OTHER PRICE OPTIONS PROPERTIES Example 10.4
For call options on a stock that pays no dividends, early exercise is never optimal. However, this is
not true in general for put options. Why not?
Solution.
Delaying exercise of a call gains interest on the strike, but delaying exercise of a put loses interest
on the strike 10 GENERAL RULES OF EARLY EXERCISE ON AMERICAN OPTIONS 75 Practice Problems
Problem 10.1
For which of the following options it is never optimal to early exercise?
(A) American put on a dividend paying stock
(B) American put on a nondividendpaying stock
(C) American call on a dividend paying stock
(D) American call on a nondividendpaying stock.
Problem 10.2
XYZ stock pays annual dividends of $5 per share of stock, starting one year from now. For which
of these strike prices of American call options on XYZ stocks might early exercise be optimal? All
of the call options expire in 5 years. The annual eﬀective riskfree interest rate is 3%.
(A) $756
(B) $554
(C) $396
(D) $256
(E) $43
(F) $10
Problem 10.3
American put options on XYZ stocks currently cost $56 per option. We know that it is never
optimal to exercise these options early. Which of these are possible values of the strike price K on
these options and the stock price St of XYZ stock?
(A) S = 123, K = 124
(B) S = 430, K = 234
(C) S = 234, K = 430
(D) S = 1234, K = 1275
(E) S = 500, K = 600
(F) S = 850, K = 800
Problem 10.4
A 1year European call option on a nondividendpaying stock with a strike price of $38 sells for $5.5
and is due six months from now. A 1year European put option with the same underlying stock
and same strike sells for $0.56. The continuously compounded riskfree interest rate is 5%. The
stock is currently selling for $42. Would exercising a 1year American put option with strike $38
be optimal if exercised now? 76 PARITY AND OTHER PRICE OPTIONS PROPERTIES Problem 10.5
A 1year European call option on a dividendpaying stock with a strike price of $38 sells for $5.5
now and is due six months from now. A 1year European put option with the same underlying
stock and same strike sells for $0.56. The continuously compounded riskfree interest rate is 5%. A
dividend of $0.95 is due at t = 10 months. Would it be optimal to exercise a 1year American call
with the same stock and strike price now?
Problem 10.6
A 1year European call option on a dividendpaying stock with a strike price of $38 sells for $5.5
now and is due six months. A 1year European put option with the same underlying stock and
same strike sells for $0.56. The continuously compounded riskfree interest rate is 5%. A dividend
of $0.95 is due at t = 6.1 months. Would it be optimal to exercise a 1year American call with the
same stock and strike price now?
Problem 10.7
A share of stock pays monthly dividends of $4, starting one month from now. 1year American call
options with the underlying stock are issued for a strike price of $23. The annual eﬀective interest
rate is 3%. Is it optimal to early exercise such American call options now?
Problem 10.8
A nondividendpaying stock is currently trading for $96 per share. American put options are written
with the underlying stock for a strike price of $100. What is the maximum price of the put options
at which early exercise might be optimal?
Problem 10.9
The following table list several American call options with a dividend paying stock. The stock is
currently selling fo $58. The ﬁrst dividend payment is $3 and due now whereas the second payment
due in nine months. Which of the options might early exercise be optimal? Assume a continuously
compounded riskfree interest rate of 5%.
Option
A
B
C
D
E Strike
40
50
50
52
59 Expiration(in years)
1.5
1.5
1.0
1.0
0.75 Problem 10.10
Suppose that you have an American call option that permits you to receive one share of stock A by
giving up one share of stock B. Neither stock pays a dividend. In what circumstances might you
early exercise this call? 10 GENERAL RULES OF EARLY EXERCISE ON AMERICAN OPTIONS 77 Problem 10.11
Suppose that you have an American put option that permits you to give up one share of stock A
by receiving one share of stock B. Neither stock pays a dividend. In what circumstances might you
early exercise this put? Would there be a loss from not earlyexercising if stock A price goes to
zero?
Problem 10.12 ‡
For a stock, you are given:
(i) The current stock price is $50.00.
(ii) δ = 0.08
(iii) The continuously compounded riskfree interest rate is r = 0.04.
(iv) The prices for oneyear European calls (C) under various strike prices (K) are shown below:
K
40
50
60
70 C
9.12
4.91
0.71
0.00 You own four special put options each with one of the strike prices listed in (iv). Each of these put
options can only be exercised immediately or one year from now.
Determine the lowest strike price for which it is optimal to exercise these special put option(s)
immediately. 78 PARITY AND OTHER PRICE OPTIONS PROPERTIES 11 Eﬀect of Maturity Time Growth on Option Prices
Recall that the time of expiration of an option is the time after which the option is worthless. In
this section, we discuss how time to expiration aﬀects the prices of American and European options.
Consider two American calls with the same underlying asset and strike price K but with expiration
time T1 < T2 . The call with expiration time T2 can be converted into a call with expiration time
T1 by voluntarily exercising it at time T1 . Thus, the price of the call with expiration time T2 is at
least as valuable as the call with expiration time T1 . In symbols, we have
CAmer (K, T1 ) ≤ CAmer (K, T2 ).
A similar result holds for American put options
PAmer (K, T1 ) ≤ PAmer (K, T2 ).
The above results are valid for all American options regardless whether the underlying asset pays
dividends or not.
Next, we consider European call options. If the underlying asset pays no dividends, then the
European call has the same price as an American call with the same underlying asset, same strike,
and same expiration date (See Example 9.1). Thus, for T1 < T2 , we have
CEur (K, T1 ) ≤ CEur (K, T2 )
and
PEur (K, T1 ) ≤ PEur (Ker(T2 −T1 ) , T2 ).
The above inequalities for European options may not be valid for European options with dividendpaying underlying asset. We illustrate these cases in the following two examples.
Example 11.1
Consider a stock that will pay a liquidating dividend two weeks from today. This means that the
stock is worthless after the dividend payment. Show that a 1week European call on the stock is
more valuable than a 3week European call.
Solution.
For T > 2 the stock is worthless, so that CEur (K, T ) = 0. If T ≤ 2, the call might be worth
something depending on how high the strike price K is. That is, CEur (K, T ) > 0. Let T1 = 1 and
T2 = 3. Then T1 < T2 but CEur (K, T1 ) > 0 = CEur (K, T2 )
Example 11.2
Show that when a company goes bankrupt the price of a longlived European put on the stock is
less valuable than a shorterlived European put. 11 EFFECT OF MATURITY TIME GROWTH ON OPTION PRICES 79 Solution.
When a company goes bankrupt, the underlying stock’s price is zero and stays like that. Thus, the
only dividend received will be the strike price. In this case, the value of a European put option
is just the present value of the strike price. Hence, the longlived European put will make the
present value smaller. In other words, the price of a longlived European put is less valuable than
a shorterlived European put
Next, let’s consider European options on nondividend paying stock with strike price growing over
time. Let K be the original (t = 0) strike price. Let C (t) denote the time 0 price for a European
call maturing at time t and with strike price Kt = Kert . Suppose t < T. Suppose that C (t) > C (T ).
Then we buy the call with T years to expiration and sell the call with t years to expiration. The
payoﬀ of the longerlived call at time T is max{0, ST − KT }. The payoﬀ of the shorterlived call at
time T is − max{0, St − Kt } accumulated from t to T.
• Suppose St < Kt . Then the payoﬀ of the shorterlived call is 0 and accumulates to 0 at time T.
If ST < KT then the payoﬀ of the longerlived call is 0 at time T. Thus, the total payoﬀ of the
position is 0 at time T.
• Suppose St < Kt . Then the payoﬀ of the shorterlived call is 0 and accumulates to 0 at time T.
If ST ≥ KT then the payoﬀ of the longerlived call is ST − KT at time T. Thus, the total payoﬀ of
the position is ST − KT at time T.
• Suppose St ≥ Kt . Borrow the stock and deliver it and collect the strike price Kt . Invest this
amount at the riskfree rate, so you will have Kt er(T −t) at time T. If ST ≤ KT , you buy back the
stock give it to the owner so that the total payoﬀ of this position is KT − ST at time T.
• Suppose St ≥ Kt . Borrow the stock and deliver it and collect the strike price Kt . Invest this
amount at the riskfree rate, so you will have Kt er(T −t) at time T. If ST > KT , exercise the option
and receive ST − KT . Buy the stock. For this position, the total payoﬀ is 0 at time T.
It follows that in order to avoid arbitrage, the longerlived call can’t sell for less than the shorterlived call. That is,
If T > t then CEur (KT , T ) ≥ CEur (Kt , t).
Likewise,
If T > t then PEur (KT , T ) ≥ PEur (Kt , t).
Example 11.3
The premium of a 6month European call with strike price $100 is $22.50. The premium of a
9month European call with strike price $102.53 is $20.00. The continuously compounded riskfree
interest rate is 10%.
(a) Demonstrate an arbitarge opportunity. 80 PARITY AND OTHER PRICE OPTIONS PROPERTIES (b) Given S0.5 = $98 and S0.75 = $101. Find the value of the accumulated arbitrage strategy after
9 months?
(c) Given S0.5 = $98 and S0.75 = $103. Find the value of the accumulated arbitrage strategy after
9 months? Solution.
(a) First, notice that 102.53 = 100e0.1×0.25 . We have C (0.5) = 22.50 > C (0.75). We buy one call
option with strike price $102.53 and time to expiration nine months from now. We sell one call
option with strike price of $100 and time to expiration six months from now. The payoﬀ table is
given next.
Payoﬀ at Time 9 months
S0.75 ≤ 102.53
S0.75 > 102.53
Payoﬀ at Time 6 months
Transaction
Time 0
S0.5 ≤ 100
S0.5 > 100
S0.5 ≤ 100
S0.5 > 100
Sell C 0.5)
C (0.5)
S0.75 − 102.53
0
S0.75 − 102.53
0
Buy C (0.75)
−C 0.75)
102.53 − S0.75 102.53 − S0.75
0
0
Total Payoﬀ C (0.5) − C (0.75)
0
102.53 − S0.75 S0.75 − 102.53
0
(b) The cash ﬂow of the two options nets to zero after 9 months. Thus, the accumulated value of
the arbitrage strategy at the end of nine months is
(22.50 − 20.00)e0.10×0.75 = $2.6947
(c) The cash ﬂow of the two options nets to S0.75 − 102.53 = 103 − 102.53 = 0.47. The accumulated
value of the arbitrage strategy at the end of nine months is
2.6947 + 0.47 = $3.1647 11 EFFECT OF MATURITY TIME GROWTH ON OPTION PRICES 81 Practice Problems
Problem 11.1
True or false:
(A) American call and put options (with same strike price) become more valuable as the time of
expiration increases.
(B) European call and put options (with same strike price) become more valuable as the time of
expiration increases.
(C) A longlived K −strike European call on a dividend paying stock is at least as valuable as a
shortlived K −strike European call.
(D) A longlived K −strike European call on a nondividendpaying stock is at least as valuable as
a shortlived K −strike European call.
Problem 11.2
Which of these statements about call and put options on a share of stock is always true?
(A) An American call with a strike price of $43 expiring in 2 years is worth at least as much as an
American call with a strike price of $52 expiring in 1 year.
(B) An American call with a strike price of $43 expiring in 2 years is worth at least as much as an
American call with a strike price of $43 expiring in 1 year.
(C) An American call with a strike price of $43 expiring in 2 years is worth at least as much as a
European call with a strike price of $43 expiring in 2 years.
(D) An American put option with a strike price of $56 expiring in 3 years is worth at least as much
as an American put option with a strike price of $56 expiring in 4 years.
(E) A European put option with a strike price of $56 expiring in 3 years is worth at least as much
as a European put option with a strike price of $56 expiring in 2 years.
Problem 11.3
Gloom and Doom, Inc. will be bankrupt in 2 years, at which time it will pay a liquidating dividend
of $30 per share. You own a European call option on Gloom and Doom, Inc. stock expiring in 2
years with a strike price of $20. The annual continuously compounded interest rate is 2%. How
much is the option currently worth? (Assume that, for T = 2, the option can be exercised just
before the dividend is paid.)
Problem 11.4
European put options are written on the stock of a bankrupt company. That is, the stock price is
$0 per share. What is the price of a put option that has strike price $92 and expires 2 years from
now? The annual continuously compounded interest rate is 21%. 82 PARITY AND OTHER PRICE OPTIONS PROPERTIES Problem 11.5
Imprudent Industries plans to pay a liquidating dividend of $20 in one year. Currently, European
calls and puts on Imprudent Industries are traded with the following strike prices (K ) and times
to expiration (T ). Which of the following European options will have the highest value? (Assume
that, for T = 1, the options can be exercised just before the dividend is paid.) The annual eﬀective
interest rate is 3%.
(A) Call; K = 3; T = 1.1
(B) Call; K = 19; T = 1
(C) Call; K = 19; T = 1.1
(D) Put; K = 10; T = 2
(E) Put; K = 10; T = 0.4.
Problem 11.6
European call options on Oblivious Co. are issued with a certain strike price that grows with
time. Which of these times to maturity will result in the highest call premium? (A higher number
indicates later time to maturity.)
(A) 2 months
(B) 4 months
(C) 1 year
(D) 2 years
(E) 2 weeks
(F) 366 days
Problem 11.7
Let P (t) denote the time 0 price for a European put maturing at time t and with strike price
Kt = Kert . Suppose t < T. Suppose that P (t) ≥ P (T ). Show that this leads to an arbitrage
opportunity.
Problem 11.8
The premium of a 6month European put with strike price $198 is $4.05. The premium of a 9month
European put with strike price $206.60 is $3.35. The continuously compounded riskfree interest
rate is 17%.
(a) Demonstrate an arbitrage opportunity.
(b) Given that S0.5 = $196 and S0.75 = $205. What is the value of the accumulated arbitrage
strategy after 9 months?
Problem 11.9
Two American call options on the same stock both having a striking price of $100. The ﬁrst option
has a time to expiration of 6 months and trades for $8. The second option has a time to expiration
of 3 months and trades for $10. Demonstrate an arbitrage. 11 EFFECT OF MATURITY TIME GROWTH ON OPTION PRICES 83 Problem 11.10
The premium of a 6month European call with strike price $198 is $23.50. The premium of a
9month European call with strike price $207.63 is $22.00. The continuously compounded riskfree
interest rate is 19%.
(a) Demonstrate an arbitrage opportunity.
(b) Given that S0.5 = $278 and S0.75 = $205. What is the value of the accumulated arbitrage
strategy after 9 months? 84 PARITY AND OTHER PRICE OPTIONS PROPERTIES 12 Options with Diﬀerent Strike Prices but Same Time to
Expiration
In this section we explore properties of options with diﬀerent strike prices but the same time to
expiration.
The ﬁrst result shows that the call premium decreases as the strike price increases.
Proposition 12.1
Suppose K1 < K2 with corresponding call option (American or European) prices C (K1 ) and C (K2 )
then
C (K1 ) ≥ C (K2 ).
(12.1)
Moreover,
C (K1 ) − C (K2 ) ≤ K2 − K1 . (12.2) Proof.
We will show (12.1) using the usual strategy of a noarbitrage argument. Let’s assume that the
inequality above does not hold, that is, C (K1 ) < C (K2 ). We want to set up a strategy that pays
us money today. We can do this by selling the highstrike call option and buying the lowstrike call
option (this is a bull spread). We then need to check if this strategy ever has a negative payoﬀ in
the future. Consider the following payoﬀ table:
Transaction
Sell C (K2 )
Buy C (K1 )
Total Expiration or Exercise
Time 0
St < K1 K1 ≤ St ≤ K2
C ( K2 )
0
0
−C (K1 )
0
S t − K1
C (K2 ) − C (K1 ) 0
S t − K1 St > K2
K2 − St
S t − K1
K2 − K1 Every entry in the row labeled “Total” is nonnegative. Thus, by selling the highstrike call and
buying the lowstrike call we are guaranteed not to lose money. This is an arbitrage. Hence, to
prevent arbitrage, (12.1) must be satisﬁed. If the options are Americans then we have to take in
consideration the possibility of early exercise of the written call. If that happens at time t < T, we
can simply exercise the purchased option, earning the payoﬀs in the table. If it is not optimal to
exercise the purchased option, we can sell it, and the payoﬀ table becomes
Transaction
Sell C (K2 )
Buy C (K1 )
Total Expiration or Exercise
Time 0
St < K1 K1 ≤ St ≤ K2
C ( K2 )
0
0
−C (K1 )
0
St − K1
C (K2 ) − C (K1 ) 0
St − K1 St > K2
K2 − St
C (K1 )
C (K1 ) + K2 − St 12 OPTIONS WITH DIFFERENT STRIKE PRICES BUT SAME TIME TO EXPIRATION 85
Since C (K1 ) > St − K1 we ﬁnd that C (K1 ) + K2 − St > K2 − K1 . That is, we get even higher
payoﬀs than exercising the purchased option
Example 12.1
Establish the relationship (12.2). That is, the call premium changes by less than the change in the
strike price.
Solution.
We will use the strategy of a noarbitrage argument. Assume C (K1 ) − C (K2 ) − (K2 − K1 ) > 0.
We want to set up a strategy that pays us money today. We can do this by selling the lowstrike
call option, buying the highstrike call option (this is a call bear spread), and lending the amount
K2 − K1 . We then need to check if this strategy ever has a negative payoﬀ in the future. Consider
the following payoﬀ table:
Transaction
Sell C (K1 )
Buy C (K2 )
Lend K2 − K1
Total Expiration or Exercise
Time 0
St < K1
K1 ≤ St ≤ K2
C (K1 )
0
K1 − S t
−C (K2 )
0
0
rt
K1 − K2
e (K2 − K1 ) ert (K2 − K1 )
C (K2 ) − C (K1 ) ert (K2 − K1 ) ert (K2 − K1 )
−(K2 − K1 )
−(St − K1 ) St > K2
K1 − St
St − K2
ert (K2 − K1 )
ert (K2 − K1 )
−(K2 − K1 ) Every entry in the row labeled “Total” is nonnegative. Thus, by selling the lowstrike call, buying
the highstrike call and lending K2 − K1 we are guaranteed not to lose money at time T. This is an
arbitrage. Hence, to prevent arbitrage, (12.2) must be satisﬁed. In the case of American options, if
the written call is exercised, we can duplicate the payoﬀs in the table by throwing our option away
(if K1 ≤ St ≤ K2 ) or exercising it (if St > K2 ). Since it never makes sense to discard an unexpired
option, and since exercise may not be optimal, we can do at least as well as the payoﬀ in the table
if the options are American
Remark 12.1
If the options are European, we can put a tighter restriction on the diﬀerence in call premiums,
namely CEur (K1 ) − CEur (K2 ) ≤ P V0,T (K2 − K1 ). We would show this by lending P V0,T (K2 − K1 )
instead of K2 − K1 . This strategy does not work if the options are American, since we don’t know
how long it will be before the options are exercised, and, hence, we don’t know what time to use in
computing the present value.
We can derive similar relationships for (American or European) puts as we did for calls. Namely,
we have 86 PARITY AND OTHER PRICE OPTIONS PROPERTIES Proposition 12.2
Suppose K1 < K2 with corresponding put option prices P (K1 ) and P (K2 ) then
P (K2 ) ≥ P (K1 ). (12.3) P (K2 ) − P (K1 ) ≤ K2 − K1 (12.4) Moreover,
and
PEur (K2 ) − PEur (K1 ) ≤ P V0,T (K2 − K1 ).
Example 12.2
The premium of a 50strike call option is 9 and that of a 55strike call option is 10. Both options
have the same time to expiration.
(a) What noarbitrage property is violated?
(b) What spread position would you use to eﬀect arbitrage?
(c) Demonstrate that the spread position is an arbitrage.
Solution.
(a) We are given that C (50) = 9 and C (55) = 10. This violates (12.1).
(b) Since C (55) − C (50) = 1 > 0, to proﬁt from an arbitrage we sell the 55strike call option and
buy the 50strike call option. This is a call bull spread position.
(c) We have the following payoﬀ table. Transaction
Sell C (55)
Buy C (50)
Total Time 0
10
−9
+1 Expiration or Exercise
St < 50 50 ≤ St ≤ 55
0
0
0
St − 50
0
St − 50 St > 55
55 − St
St − 50
5 Note that we initially receive money, and that at expiration the proﬁt is nonnegative. We have
found arbitrage opportunities
Example 12.3
The premium of a 50strike put option is 7 and that of a 55strike option is 14. Both options have
the same time of expiration.
(a) What noarbitrage property is violated?
(b) What spread position would you use to eﬀect arbitrage?
(c) Demonstrate that the spread position is an arbitrage. 12 OPTIONS WITH DIFFERENT STRIKE PRICES BUT SAME TIME TO EXPIRATION 87
Solution.
(a) We are given that P (50) = 7 and P (55) = 14. This violates (12.4).
(b) Since P (55) − P (50) = 4 > 0, to proﬁt from an arbitrage we sell the 55strike put option, buy
the 50strike call option. This is a put bull spread position. This positive cash ﬂow can be lent
resulting in the following table
(c) We have the following payoﬀ table.
Transaction
Sell P (55)
Buy P (50)
Total Time 0
14
−7
7 Expiration or Exercise
St < 50 50 ≤ St ≤ 55
St − 55 St − 55
50 − St 0
−5
St − 55 ≥ −5 St > 55
0
0
0 Note that we initially receive more money than our biggest possible exposure in the future
Example 12.4
A 90strike European call with maturity date of 2 years sells for $10 and a 95strike European
call with the same underlying asset and same expiration date sells for $5.25. The continuously
compounded freerisk interest rate is 10%. Demonstrate an arbitrage opportunity.
Solution.
We are given C (90) = 10 and C (95) = 5.25. We sell the 90strike call, buy the 95strike call (this
is a call bear spread), and loan $4.75. The payoﬀ table is shown next.
Transaction
Sell C (90)
Buy C (95)
Lend 4.75
Total Time 0
10
−5.25
−4.75
0 Expiration or Exercise
ST < 90 90 ≤ St ≤ 95
0
90 − ST
0
0
5.80
5.80
5.80
95.80 − ST > 0 St > 95
90 − ST
ST − 95
5.80
0.8 In all possible future states, we have a strictly positive payoﬀ. We demonstrated an arbitrage 88 PARITY AND OTHER PRICE OPTIONS PROPERTIES Practice Problems
Problem 12.1
Given two options with strike prices $77 and $89 respectively and with the same time to expiration.
Which of these statements are true about the following American call and put options?
(A) P (77) ≤ P (89)
(B) P (89) − P (77) ≥ 12
(C) C (77) ≥ C (89)
(D) P (77) ≥ P (89)
(E) C (89) − C (77) ≥ 12
(F) C (77) − C (89) ≥ 12
Problem 12.2
54strike Call options on a share of stock have a premium of $19. Which of the following are possible
values for call options on the stock with a strike price of $32 and the same time to expiration?
(A) 16
(B) 20
(C) 34
(D) 25
(E) 45
(F) It is impossible to have call options for this stock with a strike price of $32.
Problem 12.3
Suppose K1 < K2 with corresponding put option prices P (K1 ) and P (K2 ). Show that
P (K2 ) ≥ P (K1 )
Problem 12.4
Establish the relationship P (K2 ) − P (K1 ) ≤ K2 − K1 .
Problem 12.5
Establish the relationship PEur (K2 ) − PEur (K1 ) ≤ P V0,T (K2 − K1 ).
Problem 12.6
The premium of a 50strike put option is 7 and that of a 55strike put option is 6. Both options
have the same time to expiration.
(a) What noarbitrage property is violated?
(b) What spread position would you use to eﬀect arbitrage?
(c) Demonstrate that the spread position is an arbitrage. 12 OPTIONS WITH DIFFERENT STRIKE PRICES BUT SAME TIME TO EXPIRATION 89
Problem 12.7
The premium of a 50strike call option is 16 and that of a 55strike call option is 10. Both options
have the same time to expiration.
(a) What noarbitrage property is violated?
(b) What spread position would you use to eﬀect arbitrage?
(c) Demonstrate that the spread position is an arbitrage.
Problem 12.8 ‡
Given the following chart about call options on a particular dividend paying stock, which options
has the highest value?
Option
A
B
C
D
E Option Style
European
American
European
American
American Maturity
1 year
1 year
2 years
2 years
2 years Strike Price
50
50
50
50
55 Stock Price
42
42
42
42
42 Problem 12.9
A 70strike European put with maturity date of 1 year sells for $8.75 and a 75strike European
put with the same underlying asset and same expiration date sells for $13.50. The continuously
compounded freerisk interest rate is 6.9%.
(a) Demonstrate an arbitrage opportunity.
(b) What are the accumulated arbitrage proﬁts if the ﬁnal stock price is $68?
(c) What are the accumulated arbitrage proﬁts if the ﬁnal stock price is $73?
Problem 12.10 ‡
You are given:
(i) C (K, T ) denotes the current price of a K −strike T −year European call option on a nondividendpaying stock.
(ii) P (K, T ) denotes the current price of a K −strike T −year European put option on the same
stock.
(iii) S denotes the current price of the stock.
(iv) The continuously compounded riskfree interest rate is r.
Which of the following is (are) correct?
(I) 0 ≤ C (50, T ) − C (55, T ) ≤ 5e−rT
(II) 50e−rT ≤ P (45, T ) − C (50, T ) + S ≤ 55e−rT
(III) 45e−rT ≤ P (45, T ) − C (50, T ) + S ≤ 50e−rT . 90 PARITY AND OTHER PRICE OPTIONS PROPERTIES 13 Convexity Properties of the Option Price Functions
In this section, we consider the call option price and the put option price as functions of the strike
price while keeping the time to expiration ﬁxed. We will show that these functions are convex. The
results of this section are true for both American and European options.
We recall the reader that a function f with domain D is said to be convex if for all x, y in D and
0 < λ < 1 such that λx + (1 − λ)y ∈ D we must have
f (λx + (1 − λ)y ) ≤ λf (x) + (1 − λ)f (y ). Example 13.1
Let f be a convex function with domain D. Let x, y be in D such that f x+y
2 ≤ x+y
2 is in D. Show that f ( x) + f ( y )
.
2 Solution.
This follows from the deﬁnition of convexity with λ = 1
2
We next show that the call premium is a convex function of the strike price K. Proposition 13.1
Consider two call options with strike prices K1 < K2 and the same time to expiration. Let 0 < λ < 1.
Then
C (λK1 + (1 − λ)K2 ) ≤ λC (K1 ) + (1 − λ)C (K2 ). Proof.
We will show the required inequality by using the usual strategy of a noarbitrage argument. Let’s
assume that the inequality above does not hold, that is, C (λK1 +(1−λ)K2 ) > λC (K1 )+(1−λ)C (K2 )
or C (K3 ) > λC (K1 )+(1 − λ)C (K2 ) where K3 = λK1 +(1 − λ)K2 . Note that K3 = λ(K1 − K2 )+ K2 <
K2 and K1 = (1 − λ)K1 + λK1 < (1 − λ)K2 + λK1 < K3 so that K1 < K3 < K2 . We want to set
up a strategy that pays us money today. We can do this by selling one call option with strike price
K3 , buying λ call options with strike price K1 , and buying (1 − λ) call options with strike K2 . The
payoﬀ table of this position is given next. 13 CONVEXITY PROPERTIES OF THE OPTION PRICE FUNCTIONS Transaction
Sell 1 K3 −
strike call
Buy λ
K1 −strike calls
Buy (1 − λ)
K2 −strike calls
Total 91 Time 0
C ( K3 ) St < K1
0 Expiration or exercise
K1 ≤ St ≤ K3 K3 ≤ St ≤ K2
0
K3 − S t −λC (K1 ) 0 λ(St − K1 ) λ(St − K1 ) λ(St − K1 ) −(1 − λ)C (K2 ) 0 0 0 (1 − λ)(St − K2 ) C (K3 ) − λC (K1 )
−(1 − λ)C (K2 ) 0 λ(St − K1 ) (1 − λ)(K2 − St ) 0 St > K2
K3 − St Note that
K3 − St + λ(St − K1 ) =K3 − λK1 − (1 − λ)St
=(1 − λ)K2 − (1 − λ)St
=(1 − λ)(K2 − St )
and
K3 − St + λ(St − K1 ) + (1 − λ)(St − K2 ) = λK1 + (1 − λ)K2 − St + λ(St − K1 ) + (1 − λ)(St − K2 ) = 0.
The entries in the row “Total” are all nonnegative. In order to avoid arbitrage, the initial cost must
be nonnegative. That is
C (λK1 + (1 − λ)K2 ) ≤ λC (K1 ) + (1 − λ)C (K2 )
Example 13.2
Consider three call options with prices C (K1 ), C (K2 ), and C (K3 ) where K1 < K2 < K3 . Show that
C (K1 ) − C (K2 )
C (K2 ) − C (K3 )
≥
.
K2 − K1
K 3 − K2 (13.1) Solution.
Let λ = K3 −K2 . We have K1 < K2 so that K3 − K2 < K3 − K1 and hence 0 < λ < 1. Also, we note
K3 −K1
that
K3 − K 2
K2 − K 1
K2 =
K1 +
K3 = λK1 + (1 − λ)K3 .
K3 − K1
K3 − K 1
Using convexity we can write
C (K2 ) ≤ K3 − K2
K2 − K1
C ( K1 ) +
C (K3 )
K3 − K1
K3 − K1 92 PARITY AND OTHER PRICE OPTIONS PROPERTIES which is equivalent to
(K3 − K1 )C (K2 ) ≤ (K3 − K2 )C (K1 ) + (K2 − K1 )C (K3 )
or
(K3 − K1 )C (K2 ) − (K3 − K2 )C (K2 ) − (K2 − K1 )C (K3 ) ≤ (K3 − K2 )C (K1 ) − (K3 − K2 )C (K2 ).
Hence,
(K2 − K1 )[C (K2 ) − C (K3 )] ≤ (K3 − K2 )[C (K1 ) − C (K2 )]
and the result follows by dividing through by the product (K2 − K1 )(K3 − K2 )
Example 13.3
Consider three call options with the same time to expiration and with strikes: $50, $55, and $60.
The premiums are $18, $14, and $9.50 respectively.
(a) Show that the convexity property (13.1) is violated.
(b) What spread position would you use to eﬀect arbitrage?
(c) Demonstrate that the spread position is an arbitrage.
Solution.
(a) We are given K1 = 50, K2 = 55, and K3 = 60. Since
C (K3 ) − C (K2 )
C (K1 ) − C (K2 )
= 0.8 <
= 0.9
K2 − K1
K3 − K2
the convexity property (13.1) is violated.
(b) We ﬁnd
60 − 55
K3 − K2
λ=
=
= 0.5.
K3 − K1
60 − 50
Thus, a call butterﬂy spread1 is constructed by selling one 55strike call, buying 0.5 units of 50strike
calls and 0.5 units of 60strike calls. To buy and sell round lots, we multiply all the option trades
by 2.
(c) The payoﬀ table is given next.
1 A call butterﬂy spread is an option strategy that involves selling several call options and at the same time
buying several call options with the same underlying asset and diﬀerent strike prices. This strategy involves three
diﬀerent strike prices. 13 CONVEXITY PROPERTIES OF THE OPTION PRICE FUNCTIONS Transaction
Sell 2 55−
strike calls
Buy one
50−strike call
Buy one
60−strike call
Total 93 Time 0
28 Expiration or exercise
St < 50 50 ≤ St ≤ 55 55 ≤ St ≤ 60 St > 60
0
0
110 − 2St
110 − 2St −18 0 St − 50 St − 50 St − 50 −9.50 0 0 0 St − 60 0.50 0 St − 50 60 − St 0 Note that we initially receive money and have nonnegative future payoﬀs. Therefore, we have
found an arbitrage possibility, independent of the prevailing interest rate
Similar convexity results hold for put options. Namely, we have
Proposition 13.2
(a) For 0 < λ < 1 and K1 < K2 we have
P (λK1 + (1 − λ)K2 ) ≤ λP (K1 ) + (1 − λ)P (K2 ).
(b) For K1 < K2 < K3 we have
P (K2 ) − P (K1 )
P ( K3 ) − P ( K2 )
≤
.
K2 − K1
K3 − K2
(c) For λ = K3 −K2
K3 −K1 (13.2) we have
P (K2 ) ≤ λP (K1 ) + (1 − λ)P (K3 ). Example 13.4
Consider three put options with the same time to expiration and with strikes: $50, $55, and $60.
The premiums are $7, $10.75, and $14.45 respectively.
(a) Show that the convexity property (13.2) is violated.
(b) What spread position would you use to eﬀect arbitrage?
(c) Demonstrate that the spread position is an arbitrage.
Solution.
(a) We are given K1 = 50, K2 = 55, and K3 = 60. Since
P (K2 ) − P (K1 )
P ( K3 ) − P ( K2 )
= 0.75 >
= 0.74
K 2 − K1
K3 − K2 94 PARITY AND OTHER PRICE OPTIONS PROPERTIES the convexity property for puts is violated.
(b) We ﬁnd
K3 − K2
60 − 55
λ=
=
= 0.5.
K3 − K1
60 − 50
Thus, a put butterﬂy spread is constructed by selling one 55strike put, buying 0.5 units of 50strike
puts and 0.5 units of 60strike puts. To buy and sell round lots, we multiply all the option trades
by 2.
(c) The payoﬀ table is given next.
Transaction
Sell 2 55−
strike puts
Buy one
50−strike put
Buy one
60−strike put
Total Time 0
21.50 Expiration or exercise
St < 50
50 ≤ St ≤ 55 55 ≤ St ≤ 60
2St − 110 2St − 110
0 St > 60
0 −7 50 − St 0 0 0 −14.45 60 − St 60 − St 60 − St 0 0.05 0 St − 50 60 − St 0 Note that we initially receive money and have nonnegative future payoﬀs. Therefore, we have
found an arbitrage possibility, independent of the prevailing interest rate
Example 13.5 ‡
Near market closing time on a given day, you lose access to stock prices, but some European call
and put prices for a stock are available as follows:
Strike Price
$40
$50
$55 Call Price
$11
$6
$3 Put Price
$3
$8
$11 All six options have the same expiration date.
After reviewing the information above, John tells Mary and Peter that no arbitrage opportunities
can arise from these prices.
Mary disagrees with John. She argues that one could use the following portfolio to obtain arbitrage
proﬁt: Long one call option with strike price 40; short three call options with strike price 50; lend
$1; and long some calls with strike price 55.
Peter also disagrees with John. He claims that the following portfolio, which is diﬀerent from
Mary’s, can produce arbitrage proﬁt: Long 2 calls and short 2 puts with strike price 55; long 1 call 13 CONVEXITY PROPERTIES OF THE OPTION PRICE FUNCTIONS 95 and short 1 put with strike price 40; lend $2; and short some calls and long the same number of
puts with strike price 50.
Who is correct?
Solution.
One of the requirements for an arbitrage position is that it costs nothing on net to enter into. The
second requirement is that it will make the owner a proﬁt, irrespective of future price movements.
First consider Mary’s proposal. The call option prices do not satisfy the convexity condition
C (K1 ) − C (K2 )
C (K2 ) − C (K3 )
≥
K2 − K1
K2 − K3
so there is an arbitrage opportunity: Purchasing one 40strike call costs $11, while selling 3 50strike
call options gives her 3 × 6 = $18. She also buys X $55strike calls at a price of $3 each and lends
out $1. So her net cost is −11 + 18 − 1 − 3X or 6 − 3X. In order for 6 − 3X to be 0, X must equal
2. In this case, we have the following payoﬀ table for Mary’s position
Transaction
Buy 1 40Strike call
Sell 3 50strike calls
Lend $
Buy 2 55strike calls
Total Time 0
−11
18
−1
−6
0 ST < 40
0
0
erT
0
erT > 0 40 ≤ ST ≤ 50
ST − 40
0
erT
0
erT + ST − 40 > 0 50 ≤ ST ≤ 55
ST − 40
−3(ST − 50)
erT
0
erT + 2(55 − ST ) > 0 ST > 55
ST − 40
−3(ST − 50)
erT
2(ST − 55)
erT > 0 Thus, Mary’s portofolio yields an arbitrage proﬁt.
Now consider Peter’s proposal. Purchasing one 40strike call and selling one 40strike put option
cost −11 + 3 = −$8. Purchasing 2 55strike puts and selling 2 55strike calls cost 2(−3 + 11) = $16.
Lending $2 costs −$2 at time = 0. Selling X 50strike calls and purchasing X 50strike puts cost
X (6 − 8) = −2X. Thus, her net cost at time t = 0 is −8 + 16 − 2 − 2X = 6 − 2X. In order for
6 − 2X to be zero we must have X = 3. In this case, we have the following payoﬀ table for Peter’s
position
Transaction
Buy 2 55strike calls
Sell 2 55strike puts
Buy 1 40strike call
Sell 1 40strike put
Lend $
Buy 3 50strike puts
Sell 3 50strike calls
Total Time 0
16 ST < 40
40 ≤ ST ≤ 50 50 ≤ ST ≤ 55 ST > 55
2(ST − 55) 2(ST − 55)
2(ST − 55)
2(ST − 55) −8 ST − 40 −2
−6
0 ST − 40 ST − 40 ST − 40 2erT
2erT
3(50 − ST ) 3(50 − ST ) 2erT
3(50 − ST ) 2erT
3(50 − ST ) 2erT > 0 2erT > 0 2erT > 0 2erT > 0 96 PARITY AND OTHER PRICE OPTIONS PROPERTIES Thus, Peter’s portofolio yields an arbitrage proﬁt. So Mary and Peter are right while John is
wrong
Remark 13.1
We conclude from this and the previous section that a call(put) price function is decreasing (increasing) and concave up. A graph is given below. 13 CONVEXITY PROPERTIES OF THE OPTION PRICE FUNCTIONS 97 Practice Problems
Problem 13.1
Three call options on a stock with the same time to expiration trade for three strike prices: $46,
$32, and $90. Determine λ so that C (32) ≤ λC (46) + (1 − λ)C (90).
Problem 13.2
Call options on a stock trade for three strike prices: $43, $102, and $231. The price of the $43strike
option is $56. The price of the $231strike option is $23. What is the maximum possible price of
the $102strike option? All options have the same time to expiration.
Problem 13.3
Call options on a stock trade for three strike prices, $32, $34, and $23. The $32strike call currently
costs $10, while the $34strike call costs $7. What is the least cost for the $23strike call option?
Problem 13.4
Put options on a stock trade for three strike prices: $102, $105, and K3 > 105. Suppose that
λ = 0.5. The $102strike put is worth $20, the $105strike put is worth $22, and the K3 −strike put
is worth $24. Find the value of K3 .
Problem 13.5
Consider two put options with strike prices K1 < K2 and the same time to expiration. Let 0 < λ < 1.
Show that
P (λK1 + (1 − λ)K2 ) ≤ λP (K1 ) + (1 − λ)P (K2 ).
Problem 13.6
Consider three put options with prices P (K1 ), P (K2 ), and P (K3 ) where K1 < K2 < K3 . Show that
P (K2 ) ≤ λP (K1 ) + (1 − λ)P (K3 ).
Problem 13.7
Consider three put options with prices P (K1 ), P (K2 ), and P (K3 ) where K1 < K2 < K3 . Show that
P ( K3 ) − P ( K2 )
P (K2 ) − P (K1 )
≤
.
K2 − K1
K3 − K2
Problem 13.8
Consider three call options with the same time to expiration and with strikes:$80, $100, and $105.
The premiums are $22, $9, and $5 respectively.
(a) Show that the convexity property is violated.
(b) What spread position would you use to eﬀect arbitrage?
(c) Demonstrate that the spread position is an arbitrage. 98 PARITY AND OTHER PRICE OPTIONS PROPERTIES Problem 13.9
Consider three put options with the same time to expiration and with strikes: $80, $100, and $105.
The premiums are $4, $21, and $24.80 respectively.
(a) Show that the convexity property is violated.
(b) What spread position would you use to eﬀect arbitrage?
(c) Demonstrate that the spread position is an arbitrage.
Problem 13.10
Consider three put options that expire in 6 months and with strikes: $50, $55, and $60. The
premiums are $7, $10.75, and $14.45 respectively. The continuously compounded risk free interest
rate is 10%. Complete the following table
Stock at Expiration
48
53
59
63 Accumulated Strategy Proﬁts Option Pricing in Binomial Models
The binomial option pricing model is a model founded by Cox, Ross, and Rubenstein in 1979. This
model computes the noarbitrage price of an option when the price of the underlying asset has
exactly twostate prices at the end of a period. Binomial pricing model displays the option pricing
in a simple setting that involves only simple algebra. In this chapter, we introduce the binomial
model and use it in computing European and American option prices for a variety of underlying
assets. 99 100 OPTION PRICING IN BINOMIAL MODELS 14 SinglePeriod Binomial Model Pricing of European Call
Options
In this section we develop the one period binomial model to exactly price a long call option. For
simplicity, we will assume that the underlying asset is a share of a stock that pays continuous
dividends. We will make the assumption that for a given period the stock price either goes up (up
state) or down (down state) in value. No other outcomes are possible for this stock’s price. The
restriction to only two possible values justiﬁes the use of the word “binomial”.
We will introduce the following notation: Let h be the length of one period. Let S be the stock
price at the beginning of the period. We deﬁne the up ratio by u = 1 + g = Sh where g is the rate
S
of capital gain when the stock goes up and Sh is the stock price at the end of the period. We deﬁne
the down ratio by d = 1 + l = Sh where l is the rate of capital loss when the stock goes down.
S
For example, if S = 100 and the stock price at the end of the period is 175 then g = 75%. If at the
end of the period the stock drops to 75 then l = −25%. When the stock goes up we shall write uS
and when it goes down we shall write dS. These two states can be described by a tree known as a
binomial tree as shown in Figure 14.1. Figure 14.1
Now, let K be the strike price of a call option on the stock that pays continuous dividends and that
matures at the end of the period. At the end of the period, we let Cu = max{0, uS − K } denote
the value of the call option in the up state and Cd = max{0, dS − K } the value of the call option
in the down state. Let C denote the price of the option at the beginning of the period. Let r be
the continuously compounded riskfree annual rate. Thus, the periodic rate is rh. Let the dividend
yield be δ. We will assume that the dividends are reinvested in the stock so that one share at the
beginning of the period grows to eδh shares at the end of the period.1
The major question is to determine the current price C of the option. For that, we are going to use
two diﬀerent approaches: The replicating portfolio approach and the riskneutral approach. Either
approach will yield the same answer. In this section, we will discuss the former one leaving the
latter one to the next section.
The replicating portfolio approach consists of creating a portfolio that replicates the actual call
option. Let Portfolio A consists of buying a call option on the stock and Portfolio B consists of
1 See [2] Section 70. 14 SINGLEPERIOD BINOMIAL MODEL PRICING OF EUROPEAN CALL OPTIONS 101 buying ∆ shares of the stock and borrowing $B > 0. The payoﬀ tables of these positions are shown
below.
Transaction
Buy a Call
Total Time 0 Up State Down State
−C
Cu
Cd
−C
Cu
Cd Transaction
Time 0
Buy ∆ Shares −∆S
Borrow $B
B
Total
−∆S + B Up State
∆eδh uS
−Berh
∆eδh uS − Berh Down State
∆eδh dS
−Berh
∆eδh dS − Berh If Portfolio B is to replicate portfolio A we must have
C = ∆S − B
Cu = ∆eδh uS − Berh
Cd = ∆eδh dS − Berh .
Solving the last two equations we ﬁnd
Cu − Cd
S (u − d) (14.1) uCd − dCu
.
d−u (14.2) ∆ = e−δh
and
B = e−rh
Thus,
C = ∆S − B = e−rh Cu e(r−δ)h − d
u − e(r−δ)h
+ Cd
u−d
u−d . (14.3) Remark 14.1
Note that ∆S is not a change in S. It is monetary value of the shares of stock in the portfolio. Note
that
Cu − Cd
∆=
Su − Sd
where Su = uS and Sd = dS. As h approaches zero, Su − Sd approaches zero so that we can write
∆ = ∂C . Hence, ∆ measures the sensitivity of the option to a change in the stock price. Thus, if
∂S
the stock price increases by $1, then the option price, ∆S − B , changes by ∆. 102 OPTION PRICING IN BINOMIAL MODELS Example 14.1
Consider a European call option on the stock of XYZ with strike $95 and six months to expiration. XYZ stock does not pay dividends and is currently worth $100. The annual continuously
compounded riskfree interest rate is 8%. In six months the price is expected to be either $130 or
$80. Using the singleperiod binomial option pricing model, ﬁnd ∆, B and C.
Solution.
We are given: S = 100, K = 95, r = 0.08, δ = 0, h = 0.5, u =
Cu = 35 and Cd = 0. Hence,
Cu − Cd
∆ = e−δh
= 0.7
S (u − d)
B = e−rh 130
100 = 1.3, and d = 80
100 = 0.8. Thus, uCd − dCu
= $53.8042
d−u and the price of the call option
C=e −rh e(r−δ)h − d
u − e(r−δ)h
+ Cd
Cu
u−d
u−d = $16.1958 The noarbitrage principle was a key factor in establishing formulas (14.1)  (14.3). The following
result exhibits a condition for which the principle can be valid.
Proposition 14.1 (noarbitrage proﬁts condition)
The noarbitrage condition is
d < e(r−δ)h < u. (14.4) Proof.
Suppose ﬁrst that e(r−δ)h > u (i.e. Se(r−δ)h > Su). Consider the following position: Shortsell the
stock and collect S. Now, invest the proceeds for a length of time h while paying dividends to the
original owner of the stock, should they occur. At the end of the period you will get Se(r−δ)h . At the
end of the period, if the stock ends up at uS, buy back the stock and return it to its original owner.
Now, pocket the diﬀerence Se(r−δ)h − Su > 0. If the stock goes down to dS instead, the payoﬀ is
even larger: Se(r−δ)h − Sd > Se(r−δ)h − Su > 0. Thus, the condition e(r−δ)h > u demonstrates an
arbitrage.
Next, suppose that d > e(r−δ)h (i.e., Sdeδh > Serh .) Consider the following position: Borrow $S at
the riskfree rate and use it to buy one share of stock. Now, hold the stock for a period of h. If
the stock goes down, collect Sdeδh (sale of stock + dividends received.) Next, repay the loan plus
interest, i.e., Serh . In this case, you pocket Sdeδh − Serh > 0. If the stock goes up to uS instead the
payoﬀ is even larger Sueδh − Serh > Sdeδh − Serh > 0. Again, the condition d > e(r−δ)h demonstrates 14 SINGLEPERIOD BINOMIAL MODEL PRICING OF EUROPEAN CALL OPTIONS 103 an arbitrage opportunity
Arbitrage opportunities arise if the options are mispriced, that is, if the actual option price is
diﬀerent from the theoretical option price:
• If an option is overpriced, that is, the actual price is greater than the theoretical price, then we
can sell the option. However, the risk is that the option will be in the money at expiration, and we
will be required to deliver the stock. To hedge this risk, we can buy a synthetic option at the same
time we sell the actual option.
• If the option is underpriced, that is, the actual price is smaller than the theoretical price, then
we buy the option. To hedge the risk associated with the possibility of the risk price falling at
expiration, we sell a synthetic option at the same time we buy the actual option.
We illustrate these results in the next example.
Example 14.2
Consider the option of Example 14.1.
(a) Suppose you observe a call price of $17 (i.e. option is overpriced). What is the arbitrage?
(b) Suppose you observe a call price of $15.50 (i.e. option is underpriced). What is the arbitrage?
Solution.
(a) The observed price is larger than the theoretical price. We sell the actual call option for $17
and synthetically create a call option by buying 0.7 of one share and borrowing $53.8042. This
synthetic option hedges the written call as shown in the payoﬀ table.
Transaction
Written Call
0.7 Purchased shares
Repay Loan
Total Stock Price in Six Months
$80
$130
0
$35
$56
$91
−$56
−$56
$0
$0 Now, the initial cash ﬂow is
17.00 − 0.7 × 100 + 53.8042 = $0.8042.
Thus, we earn $0.8042 which is the amount the option is mispriced.
(b) The observed price is smaller than the theoretical price. We buy the option and synthetically
create a short position in an option. In order to do so, we sell 0.7 units of the share and lend
$53.8042. The initial cash ﬂow is
−15.50 + 0.7 × 100 − 53.8042 = $0.6958 104 OPTION PRICING IN BINOMIAL MODELS Thus, we earn $0.6958 which is the amount the option is mispriced
Graphical Interpretation of the Binomial Formula
Consider graphing the price function after one period for a nondividendpaying stock. Let Ch be
the option price after one period with corresponding stock price Sh . Then
Ch = ∆Sh − erh B.
The graph is the straight line going through the points A, E, D as shown below with slope ∆ and
vertical intercept −erh B. Thus, any line replicating a call must have a positive slope and a negative vertical intercept.
Example 14.3
Consider a nondividend paying stock and a 1year call on the stock. The stock is currently trading
for $60. Suppose that Ch = −32.38 when Sh = 0 and Ch = 26.62 when Sh = 88.62. Find the
current price of the call option if the continuously compounded riskfree rate is 9%.
Solution.
The current price of the call is given by
C = ∆S − B = 60∆ − B
where
∆= 26.62 − (−32.38)
= 0.6658
88.62 − 0 14 SINGLEPERIOD BINOMIAL MODEL PRICING OF EUROPEAN CALL OPTIONS
and
−32.38 = 0.6658(0) − e0.09 B.
Solving for B we ﬁnd B = 29.5931. Thus, the ﬁnal answer is
C = 0.6658(60) − 29.5931 = $10.35 105 106 OPTION PRICING IN BINOMIAL MODELS Practice Problems
Problem 14.1
XYZ currently has a stock price of $555 per share. A replicating portfolio for a particular call
option on XYZ stock involves borrowing $B and buying 3 of one share. The price of the call option
4
is $360.25. Calculate B using the oneperiod binomial option pricing model.
Problem 14.2
Show that ∆ ≤ 1.
Problem 14.3
The graph of the value of a replicating portfolio of a nondividendpaying stock is given below. Determine ∆ and B given that h = 1 and the continuously compounded riskfree interest rate
is 8%.
Problem 14.4
XYZ currently has a stock price of $41 per share. A replicating portfolio for a particular call option
2
on XYZ stock involves borrowing $18.462 and buying 3 of one share. Calculate the price of the call
option using the oneperiod binomial option pricing model.
Problem 14.5
XYZ currently has a stock price of $555 per share. A replicating portfolio for a particular call option
on XYZ stock involves borrowing $56 and buying ∆ of one share. The price of the call option is
$360.25. Calculate ∆ using the oneperiod binomial option pricing model. 14 SINGLEPERIOD BINOMIAL MODEL PRICING OF EUROPEAN CALL OPTIONS 107 Problem 14.6
Consider a European call option on the stock of XYZ with strike $110 and 1 year to expiration. XYZ
stock does not pay dividends and is currently worth $100. The annual continuously compounded
riskfree interest rate is 5%. In one year the price is expected to be either $120 or $90. Using the
singleperiod binomial option pricing model, ﬁnd ∆ and B.
Problem 14.7
A call option on a stock currently trades for $45. The stock itself is worth $900 per share. Using
the oneperiod binomial option pricing model, a replicating portfolio for the call option is equal to
buying 1 of one share of stock and borrowing $B. Calculate B.
5
Problem 14.8
A share of stock XYZ pays continuous dividends at the annual yield rate of δ. The stock currently
trades for $65. A European call option on the stock has a strike price of $64 and expiration time
of one year. Suppose that in one year, the stock will be worth either $45 or $85. Assume that
9
the portfolio replicating the call consists of 20 of one share. Using the oneperiod binomial option
pricing model, what is the annual continuouslycompounded dividend yield?
Problem 14.9
Consider a European call option on the stock of XYZ, with a strike price of $25 and one year
to expiration. The stock pays continuous dividends at the annual yield rate of 5%. The annual
continuously compounded risk free interst rate is 11%. The stock currently trades for $23 per share.
Suppose that in two months, the stock will trade for either $18 per share or $29 per share. Use the
oneperiod binomial option pricing to ﬁnd the today’s price of the call.
Problem 14.10 ‡
A nondividendpaying stock S is modeled by the tree shown below. A European call option on S expires at t = 1 with strike price K = 12. Calculate the number
of shares of stock in the replicating portfolio for this option. 108 OPTION PRICING IN BINOMIAL MODELS Problem 14.11 ‡
You are given the following information:
• A particular stock is currently worth 100
• In one year, the stock will be worth either 120 or 90
• The annual continuouslycompounded riskfree rate is 5%
Calculate the delta for a call option that expires in one year and with strike price of 105.
Problem 14.12
Which of the following binomial models with the given parameters represent an arbitrage?
(A) u = 1.176, d = 0.872, h = 1, r = 6.3%, δ = 5%
(B) u = 1.230, d = 0.805, h = 1, r = 8%, δ = 8.5%
(C) u = 1.008, d = 0.996, h = 1, r = 7%, δ = 6.8%
(D) u = 1.278, d = 0.783, h = 1, r = 5%, δ = 5%
(E) u = 1.100, d = 0.982, h = 1, r = 4%, δ = 6%. 15 RISKNEUTRAL OPTION PRICING IN THE BINOMIAL MODEL: A FIRST LOOK 109 15 RiskNeutral Option Pricing in the Binomial Model: A
First Look
An alternative to the replicating portfolio approach to valuing options using the binomial model is
the riskneutral approach. The basic argument in the risk neutral approach is that investors are
riskneutral, i.e., assets expected return is the riskfree rate.1 In this world of riskneutral investors,
one computes riskneutral probabilities associated with the stock. These probabilities are used to
compute the expected option payoﬀ, and this payoﬀ is discounted back to the present at the riskfree
rate in order to get today’s option price.
Now, recall the option price derived with the replicating portfolio approach:
C = e−rh Cu u − e(r−δ)h
e(r−δ)h − d
+ Cd
u−d
u−d (15.1) Let
pu = e(r−δ)h −d
u−d and pd = u−e(r−δ)h
u−d Example 15.1
Using the condition d < e(r−δ)h < u, show that pu > 0, pd > 0, and pu + pd = 1.
Solution.
We have e(r−δ)h − d > 0, and u − e(r−δ)h > 0, and u − d > 0. Hence, pu > 0 and pd > 0. Now, adding
pu and pd we ﬁnd
u−d
e(r−δ)h − d u − e(r−δ)h
+
=
= 1.
pu + pd =
u−d
u−d
u−d
Thus, pu and pd can be interpreted as probabilities
We will call pu the riskneutral probabiltiy2 of an increase in the stock price. In terms of
pu we can write
C = e−rh (pu Cu + (1 − pu )Cd ) .
(15.2)
Let X be the discrete random variable representing the call option price. Thus, the range of X is
the set {Cu , Cd } with Cu occurring with a probability pu and Cd occurring with a probability 1 − pu .
Thus, the expected future price of the option is
pu Cu + (1 − pu )Cd .
It follows from (15.2) that the current option price can be viewed as a discounted expected future
price of the option.
1
2 Riskneutral will be discussed in more details in Section 23.
pu and pd look like probabilites but they are not in general. See Section 23. 110 OPTION PRICING IN BINOMIAL MODELS Example 15.2
Let X be the discrete random variable representing the stock price. Show that the expected value
of X (i.e., the expected stock price) is just the forward price of the stock from time t to time t + h.
Solution.
The range of X is the set {uS, dS }. The expected value of X is
u − e(r−δ)h
e(r−δ)h − d
uS +
dS
u−d
u−d
e(r−δ)h uS − duS + udS − e(r−δ)h dS
=
u−d
e(r−δ)h (u − d)S
=
= e(r−δ)h S = Ft,t+h
u−d
Thus, we can think of pu as the probability for which the expected stock price is the forward price
pu uS + (1 − pu )dS = In the following example, we ﬁnd the price of the option in Example 14.1 using the riskneutral
approach.
Example 15.3
Consider a European call option on the stock of XYZ with strike $95 and six months to expiration. XYZ stock does not pay dividends and is currently worth $100. The annual continuously
compounded riskfree interest rate is 8%. In six months the price is expected to be either $130 or
$80. Find C using the riskneutral approach as discussed in this section.
Solution.
We have e(0.08−0)×0.5 − 0.8
= 0.4816215
1.3 − 0.8
This is the riskneutral probability of the stock price increasing to $130 at the end of six months.
The probability of it going down to $80 is pd = 1 − pu = 0.5183785. Now given that if the stock
price goes up to $130, a call option with an exercise price of $95 will have a payoﬀ of $35 and
$0 if the stock price goes to $80, a riskneutral individual would assess a 0.4816215 probability of
receiving $35 and a 0.5183785 probability of receiving $0 from owning the call option. As such, the
risk neutral value would be:
pu = C = e−0.08×0.5 [0.4816215 × 35 + 0.5183785 × 0] = $16.1958
which agrees with the answer obtained by using the replicating portfolio method3
3 It is not easy but it can be shown that while the two approaches appear to be diﬀerent, they are the same. As
such, either approach can be used. 15 RISKNEUTRAL OPTION PRICING IN THE BINOMIAL MODEL: A FIRST LOOK 111 Practice Problems
Problem 15.1
Consider a European call option on the stock of XYZ with strike $65 and one month to expiration.
XYZ stock does not pay dividends and is currently worth $75. The annual continuously compounded
riskfree interest rate is 6%. In one month the price is expected to be either $95 or $63. Find C
using the riskneutral approach as discussed in this section.
Problem 15.2
Stock XYZ price is expected to be either $75 or $40 in one year. The stock is currently valued at
$51. The stock pays continuous dividends at the yield rate of 9%. The continuously compounded
riskfree interest rate is 12%. Find the riskneutral probability of an increase in the stock price.
Problem 15.3
GS Inc pays continuous dividends on its stock at an annual continuouslycompounded yield of 9%.
The stock is currently selling for $10. In one year, its stock price could either be $15 or $7. The
riskneutral probability of the increase in the stock price is 41.3%. Using the oneperiod binomial
option pricing model, what is the annual continuouslycompounded riskfree interest rate?
Problem 15.4
The stock of GS, which current value of $51, will sell for either $75 or $40 one year from now. The
annual continuously compounded interest rate is 7%. The riskneutral probability of an increase in
the stock price (to $75) is 0.41994. Using the oneperiod binomial option pricing model, ﬁnd the
current price of a call option on GS stock with a strike price of $50.
Problem 15.5
The probability that the stock of GS Inc will be $555 one year from now is 0.6. The probability
that the stock will be $521 one year from now is 0.4. Using the oneperiod binomial option pricing
model, what is the forward price of a oneyear forward contract on the stock?
Problem 15.6
Consider a share of nondividendpaying stock in a oneyear binomial framework with annual price
changes, with the current price of the stock being 55, and the price of the stock one year from
now being either 40 or 70. Let the annual eﬀective riskfree interest rate be 12%, continuously
compounded. Calculate riskneutral probability that the price of the stock will go up.
Problem 15.7
A nondividendpaying stock, currently priced at $120 per share, can either go up $25 or down
$25 in any year. Consider a oneyear European call option with an exercise price of $130. The
continuouslycompounded riskfree interest rate is 10%. Use a oneperiod binomial model and a
riskneutral probability approach to determine the current price of the call option. 112 OPTION PRICING IN BINOMIAL MODELS Problem 15.8 ‡
For a oneyear straddle on a nondividendpaying stock, you are given:
(i) The straddle can only be exercised at the end of one year.
(ii) The payoﬀ of the straddle is the absolute value of the diﬀerence between the strike price and
the stock price at expiration date.
(iii) The stock currently sells for $60.00.
(iv) The continuously compounded riskfree interest rate is 8%.
(v) In one year, the stock will either sell for $70.00 or $45.00.
(vi) The option has a strike price of $50.00.
Calculate the current price of the straddle.
Problem 15.9 ‡
You are given the following regarding stock of Widget World Wide (WWW):
(i) The stock is currently selling for $50.
(ii) One year from now the stock will sell for either $40 or $55.
(iii) The stock pays dividends continuously at a rate proportional to its price. The dividend yield
is 10%.
(iv) The continuously compounded riskfree interest rate is 5%.
While reading the Financial Post, Michael notices that a oneyear atthemoney European call
written on stock WWW is selling for $1.90. Michael wonders whether this call is fairly priced. He
uses the binomial option pricing model to determine if an arbitrage opportunity exists.
What transactions should Michael enter into to exploit the arbitrage opportunity (if one exists)?
Problem 15.10
A binomial tree can be constructed using the equations
u = e(r−δ)h+σ
and
d = e(r−δ)h−σ √ √ h h . Where σ denotes the volatility of the stock to be discussed in the next section. Show that
pu = eσ √ 1
h +1 . Problem 15.11 ‡
On January 1 2007, the Florida Property Company purchases a oneyear property insurance policy
with a deductible of $50,000. In the event of a hurricane, the insurance company will pay the
Florida Property Company for losses in excess of the deductible. Payment occurs on December 31 15 RISKNEUTRAL OPTION PRICING IN THE BINOMIAL MODEL: A FIRST LOOK 113 2007. For the last three months of 2007, there is a 20% chance that a single hurricane occurs and
an 80% chance that no hurricane occurs. If a hurricane occurs, then the Florida Property company
will experience $1000000 in losses. The continuously compounded risk free rate is 5%.
On October 1 2007, what is the risk neutral expected value of the insurance policy to the Florida
Property Company? 114 OPTION PRICING IN BINOMIAL MODELS 16 Binomial Trees and Volatility
The goal of a binomial tree is to characterize future uncertainty about the stock price movement.
In the absence of uncertainty (i.e. stock’s return is certain at the end of the period), a stock must
appreciate at the riskfree rate less the dividend yield. Thus, from time t to time t + h we must
have
St+h = Ft,t+h = St e(r−δ)h .
In other words, under certainty, the price next period is just the forward price.
What happens in the presence of uncertainty (i.e. stock’s return at the end of the period is uncertain)? First, a measure of uncertainty about a stock’s return is its volatility which is deﬁned
as the annualized standard deviation of the return of the stock when the return is expressed using
continuous compounding. Thus, few facts about continuously compounded returns are in place.
Let St and St+h be the stock prices at times t and t + h. The continuously compounded rate
of return in the interval [t, t + h] is deﬁned by
rt,t+h = ln St+h
.
St Example 16.1
Suppose that the stock price on three consecutive days are $100, $103, $97. Find the daily continuously compounded returns on the stock.
Solution.
The daily continuously compounded returns on the stock are
103
97
ln 100 = 0.02956 and ln 103 = −0.06002 Now, if we are given St and rt,t+h we can ﬁnd St+h using the formula
St+h = St ert,t+h .
Example 16.2
Suppose that the stock price today is $100 and that over 1 year the continuously compounded
return is −500%. Find the stock price at the end of the year.
Solution.
The answer is
S1 = 100e−5 = $0.6738 16 BINOMIAL TREES AND VOLATILITY 115 Now, suppose rt+(i−1)h,t+ih , 1 ≤ i ≤ n, is the continuously compounded rate of return over the time
interval [t + (i − 1)h, t + ih]. Then the continuously compounded return over the interval [t, t + nh]
is
n rt,t+nh = rt+(i−1)h,t+ih . (16.1) i=1 Example 16.3
Suppose that the stock price on three consecutive days are $100, $103, $97. Find the continuously
compounded returns from day 1 to day 3.
Solution.
The answer is
0.02956 − 0.06002 == −0.03046
The rate of returns on the sample space of stock prices are random variables. Suppose a year is
splitted into n periods each of length h. Let r(i−1)h,ih be the rate of return on the time interval
[(i − 1)h, ih]. Deﬁne the random variable rAnnual to be the annual continuously compounded rate of
return. Thus, we can write
n rAnnual = r(i−1)h,ih .
i=1 The variance of the annual return is therefore
n
2 σ =Var(rAnnual ) = Var r(i−1)h,ih
i=1
n n = 2
σh,i Var(r(i−1)h,ih ) =
i=1 i=1 where we assume that the return in one period does not aﬀect the expected returns in subsequent
periods. That is, periodic rates are independent. If we assume that each period has the same
variance of return σh then we can write
2
σ 2 = nσh = 2
σh
.
h Thus, the standard deviation of the period of length h is
√
σh = σ h. 116 OPTION PRICING IN BINOMIAL MODELS Now, one way to incorporate uncertainty in the future stock price is by using the model
uSt =Ft,t+h eσ √ dSt =Ft,t+h e−σ h √ h Note that in the absence of uncertainty, σ = 0 and therefore uSt = dSt = Ft,t+h . Now using the fact
that Ft,t+h = St e(r−δ)h we obtain
u = e(r−δ)h+σ
d = e(r−δ)h−σ √ √ h (16.2) h (16.3) We will refer to a tree constructed using equations (16.2)(16.3) as a forward tree.
Remark 16.1
From the relation
St+h = St e(r−δ)h±σ √ h we ﬁnd that the continuously compound return to be equal to
√
(r − δ )h ± σ h.
That is, the continuously compound return consists of two parts, one of which is certain (r − δ )h,
√
and the other of which is uncertain and generates the up and down stock moves ±σ h.
Example 16.4
The current price of a stock is $41. The annual continuously compounded interest rate is 0.08, and
the stock pays no dividends. The annualized standard deviation of the continuously compounded
stock return is 0.3. Find the price of a European call option on the stock with strike price $40 and
that matures in one year.
Solution.
Using equations (16.1)(16.2) we ﬁnd
√ uS =41e(0.08−0)×1+0.3×
dS =41e 1 √
(0.08−0)×1−0.3× 1 = $59.954
= $32.903 16 BINOMIAL TREES AND VOLATILITY 117 It follows that
59.954
= 1.4623
41
32.903
d=
= 0.8025
41
Cu =59.954 − 40 = $19.954
Cd =0
19.954 − 0
∆=
= 0.7376
41 × (1.4623 − 0.8025)
1.4623 × 0 − 0.8025 × 19.954
B =e−0.08
= $22.405.
0.8025 − 1.4623
u= Hence, the option price is given by
C = ∆S − B = 0.7376 × 41 − 22.405 = $7.839.
Forward trees for the stock price and the call price are shown below. Remark 16.2
Volatility measures how sure we are that the stock return will be close to the expected return.
Stocks with a larger volatility will have a greater chance of return far from the expected return.
Remark 16.3
A word of caution of the use of volatility when the underlying asset pays dividends. For a paying
dividend stock, volatility is for the prepaid forward price S − P V0,T (Div ) and not for the stock
price. Thus, for nondividendpaying stock the stock price volatility is just the prepaid forward price
volatility. Mcdonald gives the following relationship between stock price volatility and prepaid
forward price volatility
S
σF = σstock × P .
F
P
Note that when the stock pays no dividends then F = S and so σF = σstock 118 OPTION PRICING IN BINOMIAL MODELS Practice Problems
Problem 16.1
Given the daily continuously compound returns on four consecutive days: 0.02956,−0.06002, and
r2,3 . The threeday continuously compounded return is −0.0202. Determine r2,3 .
Problem 16.2
Suppose that the stock price today is $S and at the end of the year it is expected to be $0.678. The
annual continuously compounded rate of return is −500%. Find today’s stock price.
Problem 16.3
Establish equality (16.1).
Problem 16.4
Given that the volatility of a prepaid forward price on the stock is 90%, the annual continuously
compounded interest rate is 7%. The stock pays dividends at an annual continuously compounded
yield of 5%. Find the factors by which the price of the stock might increase or decrease in 4 years?
Problem 16.5
(a) Find an expression of pu in a forward tree.
(b) Show that pu decreases as h increases. Moreover, pu approaches 0.5 as h → 0.
Problem 16.6
The forward price on a 10year forward contract on GS stock is $567. The annualized standard
deviation of the continuously compounded stock return is currently 0.02. Find the price of the stock
after 10 years if we know that it is going to decrease.
Problem 16.7
GS stock may increase or decline in 1 year under the assumptions of the oneperiod binomial
option pricing model. The stock pays no dividends and the annualized standard deviation of the
continuously compounded stock return is 81%. A 1year forward contract on the stock currently
sells for $100. GS stock currently sells for $90. What is the annual continuously compounded
riskfree interest rate?
Problem 16.8
A stock currently sell for $41. Under the assumptions of the oneperiod binomial option pricing
model the stock is expected to go up to $59.954 in one year. The annual continuously compounded
return is 8% and the stock pays no dividends. Determine the annualized standard deviation of the
continuously compounded stock return. 16 BINOMIAL TREES AND VOLATILITY 119 Problem 16.9 ‡
Consider the following information about a European call option on stock ABC:
• The strike price is $100
• The current stock price is $110
• The time to expiration is one year
• The annual continuouslycompounded riskfree rate is 5%
• The continuous dividend yield is 3.5%
• Volatility is 30%
• The length of period is 4 months.
Find the riskneutral probability that the stock price will increase over one time period.
Problem 16.10 ‡
A three month European call on a stock is modeled by a single period binomial tree using the
following parameters
• The annual continuouslycompounded riskfree rate is 4%
• Stock pays no dividends
• Annual volatility is 15%
• Current stock price is 10
• Strike price is 10.5
Calculate the value of the option. 120 OPTION PRICING IN BINOMIAL MODELS 17 MultiPeriod Binomial Option Pricing Model
The singleperiod binomial model extends easily to a multiperiod model. In this section, we examine
the special cases of the two and threeperiod models.
The binomial trees of the stock prices as well as the call prices of the twoperiod model are shown
in Figure 17.1. Figure 17.1
Note that an up move in the stock price for one period followed by a down move in the stock price
in the next period generates the same stock price to a down move in the ﬁrst period followed by an
up move in the next. A binomial tree with this property is called a recombining tree.
Note that we work backward when it comes to pricing the option since formula (14.3) requires
knowing the option prices resulting from up and down moves in the subsequent periods. At the
outset, the only period where we know the option price is at expiration.
Knowing the price at expiration, we can determine the price in period 1. Having determined that
price, we can work back to period 0. We illustrate this process in the following example.
Example 17.1
A stock is currently worth $56. Every year, it can increase by 30% or decrease by 10%. The stock
pays no dividends, and the annual continuouslycompounded riskfree interest rate is 4%. Find the
price today of one twoyear European call option on the stock with a strike price of $70.
Solution.
We are given u = 1.3, d = 0.9, r = 0.04, h = 1, and K = 70. In one year the stock is worth 17 MULTIPERIOD BINOMIAL OPTION PRICING MODEL 121 either uS = 1.3 × 56 = $72.8 or dS = 0.9 × 56 = $50.4. In two years, the stock is worth either
u2 S = 72.8 × 1.3 = $94.64 or udS = 1.3 × 50.4 = $65.52 or d2 S = 0.9 × 50.4 = 45.36.
Year 2, Stock Price = $94.64 Since we are at expiration, the option value is Cuu = 94.64 − 70 =
$24.64.
Year 2, Stock Price = $65.52 Again we are at expiration and the option is out of the money so
that Cud = 0.
Year 2, Stock Price = $45.36.52 So at expiration we have Cdd = 0.
Year 1, Stock Price = $72.8 At this node we use (14.3) to compute the option value:
1.3 − e0.04
e0.04 − 0.9
+0×
1.3 − 0.9
1.3 − 0.9 Cu = e−0.04 24.64 × = $8.3338. Year 1, Stock Price = $50.4 At this node we use (14.3) to compute the option value:
Cd = e−0.04 0 × 1.3 − e0.04
e0.04 − 0.9
+0×
1.3 − 0.9
1.3 − 0.9 = $0. Year 0, Stock Price = $56 At this node we use (14.3) to compute the option value:
1.3 − e0.04
e0.04 − 0.9
+0×
1.3 − 0.9
1.3 − 0.9 C = e−0.04 8.3338 × = $2.8187 The content of this section can be extended to any number of periods.
Example 17.2
Find the current price of a 60strike 1.5year (18month) European call option on one share of an
underlying dividendpaying stock. Let S = 60, r = 0.03, σ = 0.25, δ = 0.03, and h = 0.50 (the
binomial interval is 6 months − thus, you need a threestep tree).
Solution.
We ﬁrst ﬁnd u and d. We have
u = e(r−δ)h+σ
and
d = e(r−δ)h−σ
and
pu = √ √ √ h = e(0.03−0.03)×0.5+0.25 h = e(0.03−0.03)×0.5−0.25 √ 0.5 = 1.1934 0.5 = 0.8380 e(r−δ)h − d
1 − 0.8380
=
= 0.4558.
u−d
1.1934 − 0.8380 122 OPTION PRICING IN BINOMIAL MODELS We have
Period 3, Stock Price = u3 S = 101.9787. Since we are at expiration, the option value is Cuuu =
101.9787 − 60 = $41.9787.
Period 3, Stock Price = u2 dS = 71.6090 and Cuud = 11.6090.
Period 3, Stock Price = ud2 S = 50.2835 and Cudd = Cddu = 0.
Period 3, Stock Price = d3 S = 35.3088 and Cddd = 0.
Period 2, Stock Price = u2 S = 85.4522 and Cuu = e−0.03×0.5 [0.4588 × 41.9787 + (1 − 0.4588) ×
11.6090] = 25.1623.
Period 2, Stock Price = udS = 60.0042 and Cud = e−0.03×0.5 [0.4588 × 11.6090 + (1 − 0.4588) × 0] =
5.2469.
Period 2, Stock Price = d2 S = 42.1346 and Cdd = e−0.03×0.5 [0.4588 × 0 + (1 − 0.4588) × 0] = 0.
Period 1, Stock Price = uS = 71.604 and Cu = e−0.03×0.5 [0.4588 × 25.1623+(1 − 0.4588) × 5.2469] =
14.1699.
Period 1, Stock Price = dS = 50.28 and Cd = e−0.03×0.5 [0.4588 × 5.2469+(1 − 0.4588) × 0] = 2.3714.
Period 0, Stock Price = $60 and the current option value is:
C = e−0.03×0.5 [0.4588 × 14.1699 + (1 − 0.4588) × 2.3714] = 7.6687
Example 17.3
Consider a twoperiod binomial model. Show that the current price of a call option is given by the
formula
C = e−2rh [p2 Cuu + 2pu (1 − pu )Cud + (1 − pu )2 Cdd ].
u
Solution.
We have
C =e−rh [pu Cu + (1 − pu )Cd ]
=e−rh pu e−rh [pu Cuu + (1 − pu )Cud ] + (1 − pu )e−rh [pu Cdu + (1 − pu )Cdd ]
=e−2rh (p2 Cuu + pu (1 − pu )Cud + pu (1 − pu )Cdu + (1 − pu )2 Cdd )
u
=e−2rh [p2 Cuu + 2pu (1 − pu )Cud + (1 − pu )2 Cdd ]
u 17 MULTIPERIOD BINOMIAL OPTION PRICING MODEL 123 Practice Problems
Problem 17.1
A stock is currently worth $41. Every year, it can increase by 46.2% or decrease by 19.7%. The
stock pays no dividends, and the annual continuouslycompounded riskfree interest rate is 8%.
Find the price today of one twoyear European call option on the stock with a strike price of $40.
Problem 17.2
The annualized standard deviation of the continuously compounded stock return on GS Inc is
23%. The annual continuously compounded rate of interest is 12%, and the annual continuously
compounded dividend yield on GS Inc. is 7%. The current price of GS stock is $35 per share. Using
a twoperiod binomial model, ﬁnd the price of GS Inc., stock if it moves up twice over the course
of 7 years.
Problem 17.3 ‡
Consider the following information about a European call option on stock ABC:
• The strike price is $95
• The current stock price is $100
• The time to expiration is two years
• The annual continuouslycompounded riskfree rate is 5%
• The stock pays non dividends
• The price is calculated using twostep binomial model where each step is one year in length.
The stock price tree is Calculate the price of a European call on the stock of ABC. 124 OPTION PRICING IN BINOMIAL MODELS Problem 17.4
Find the current price of a 120strike sixmonth European call option on one share of an underlying
nondividendpaying stock. Let S = 120, r = 0.08, σ = 0.30, and h = 0.25 (the binomial interval is
3 months).
Problem 17.5
GS Inc., pays dividends on its stock at an annual continuously compounded yield of 6%. The annual
eﬀective interest rate is 9%. GS stock is currently worth $100. Every two years, it can change by a
factor of 0.7 or 1.5. Using a twoperiod binomial option pricing model, ﬁnd the price today of one
fouryear European call option on GS, Inc., stock with a strike price of $80.
Problem 17.6
Find the current price of a 95strike 3year European call option on one share of an underlying stock
that pays continuous dividends. Let S = 100, r = 0.05, σ = 0.3, δ = 0.03 and h = 1.
Problem 17.7
Given the following information about a European call option: S = $40, r = 3%, δ = 5%, u =
1.20, d = 0.90, K = $33, and T = 3months. Using a threeperiod binomial tree, ﬁnd the price of
the call option.
Problem 17.8
A European call option on a stock has a strike price $247 and expires in eight months. The annual
continuously compounded riskfree rate is 7% and the compounded continuously dividend yield is
2%. The current price of the stock is $130. The price volatility is 35%. Use a 8period binomial
model to ﬁnd the price of the call.
Problem 17.9
A European put option on a stock has a strike price $247 and expires in eight months. The annual
continuously compounded riskfree rate is 7% and the compounded continuously dividend yield is
2%. The current price of the stock is $130. The price volatility is 35%. Use a 8period binomial
model to ﬁnd the price of the put. Hint: Putcall parity and the previous problem. 18 BINOMIAL OPTION PRICING FOR EUROPEAN PUTS 125 18 Binomial Option Pricing for European Puts
Binomial option pricing with puts can be done using the exact same formulas and conceptual tools
developed for European call options except of one diﬀerence that occurs at expiration: Instead of
computing the price as max{0, S − K }, we use max{0, K − S }. The objective of this section is to
establish the conceptual approach to binomial option pricing with puts.
The replicating portfolio approach consists of creating a portfolio that replicates the actual
short put option. Let Portfolio A consists of selling a put option on the stock and Portfolio B
consists of selling ∆ shares of the stock and lending $B < 0. The payoﬀ tables of these positions
are shown below.
Transaction
Sell a Put
Total Time 0 Up State Down State
P
− Pu
−Pd
P
− Pu
−Pd Transaction
Time 0
Sell ∆ Shares ∆S
Lend $B
−B
Total
∆S − B Up State
−∆eδh uS
Berh
−∆eδh uS + Berh Down State
−∆eδh dS
Berh
−∆eδh dS + Berh If Portfolio B is to replicate portfolio A we must have
P = ∆S − B
Pu = ∆eδh uS − Berh
Pd = ∆eδh dS − Berh .
Solving the last two equations we ﬁnd
Pu − Pd
S (u − d) (18.1) uPd − dPu
.
d−u (18.2) ∆ = e−δh
and
B = e−rh
Thus,
P = ∆S − B = e−rh Pu e(r−δ)h − d
u − e(r−δ)h
+ Pd
u−d
u−d Using riskneutral probability we can write
P = e−rh [pu Pu + (1 − pu )Pd ]. . (18.3) 126 OPTION PRICING IN BINOMIAL MODELS Example 18.1
Consider a European put option on the stock of XYZ with strike $50 and one year to expiration.
XYZ stock does not pay dividends and is currently worth $51. The annual continuously compounded
riskfree interest rate is 7%. In one year the price is expected to be either $75 or $40.
(a) Using the singleperiod binomial option pricing model, ﬁnd ∆, B and P.
(b) Find P using the riskneutral approach.
Solution.
(a) We have: u = 75
51 and d = 40
.
51 Also, Pd = 10 and Pu = 0. Hence, ∆ = e−δh P u − Pd
0 − 10
=
= −0.2857.
40
S (u − d)
51 75 − 51
51 Also,
B = e−rh 75
× 10 − 0
uPd − dPu
= e−0.07×1 51 40 75 = −19.9799.
d−u
− 51
51 The current price of the put is
P = ∆S − B = −0.2857 × 51 + 19.9799 = $5.41.
(b) We have
pu = e(r−δ)h − d
= 0.41994
u−d and
P = e−rh [pu Pu + (1 − pu )Pd ] = $5.41
Example 18.2
Consider a European put option on the stock of XYZ with strike $95 and six months to expiration. XYZ stock does not pay dividends and is currently worth $100. The annual continuously
compounded riskfree interest rate is 8%. In six months the price is expected to be either $130 or
$80.
(a) Using the singleperiod binomial option pricing model, ﬁnd ∆, B and P.
(b) Suppose you observe a put price of $8 (i.e. option is overpriced). What is the arbitrage?
(c) Suppose you observe a put price of $6 (i.e. option is underpriced). What is the arbitrage?
Solution.
(a) We are given: S = 100, K = 95, r = 0.08, δ = 0, h = 0.5, u =
Thus, Pu = 0 and Pd = 95 − 80 = 15. Hence,
∆ = e−δh Cu − Cd
= −0.3
S (u − d) 130
100 = 1.3, and d = 80
100 = 0.8. 18 BINOMIAL OPTION PRICING FOR EUROPEAN PUTS
B = e−rh 127 uPd − dPu
= −$37.470788
d−u and the price of the put option
P = ∆S − B = $7.4707.
(b) The observed price is larger than the theoretical price. We sell the actual put option for $8 and
synthetically create a long put option by selling 0.3 units of one share and lending $37.471. This
synthetic option hedges the written put as shown in the payoﬀ table.
Stock Price in Six Months
Transaction
$80
$130
Written Put
−$15
0
0.3 Written shares −$24
−$39
Lending
$39
$39
Total
$0
$0
Now, the initial cash ﬂow is
8.00 + 0.3 × 100 − 37.471 = $0.529.
Thus, we earn $0.529 which is the amount the option is mispriced.
(c) The observed price is smaller than the theoretical price. We buy the option and synthetically
create a short put option. In order to do so, we buy 0.3 units of the share and borrow $37.471. The
initial cash ﬂow is
−6.00 − 0.3 × 100 + 37.471 = $1.471
Thus, we earn $1.471 which is the amount the option is mispriced
Example 18.3
The current price of a stock is $65. The annual continuously compounded interest rate is 0.08, and
the stock pays dividends at the continuously compounded rate of 0.05. The annualized standard
deviation of the continuously compounded stock return is 0.27. Find the price of a European put
option on the stock with strike price $63 and that matures in one year.
Solution.
Using equations (16.1)(16.2) we ﬁnd
√ uS =65e(0.08−0.05)×1+0.27× 1 √
(0.08−0.05)×1−0.27× 1 dS =65e = $87.7408
= $51.1308 128 OPTION PRICING IN BINOMIAL MODELS It follows that
87.7408
= 1.3499
65
51.1308
d=
= 0.7866
65
Pu =$0
Pd =63 − 51.1308 = 11.8692
Pu − Pd
∆ =e−δh
= −0.3084
S (u − d)
uPd − dPu
1.3499 × 11.8692 − 0.7866 × 0
B =e−rh
= e−0.08
= −$26.2567.
d−u
0.7866 − 1.3499
u= Hence, the option price is given by
P = ∆S − B = −0.3084 × 65 + 26.2567 = $6.2107 18 BINOMIAL OPTION PRICING FOR EUROPEAN PUTS 129 Practice Problems
Problem 18.1
Consider a European put option on the stock of XYZ with strike $130 and one year to expiration.
XYZ stock does not pay dividends and is currently worth $100. The annual continuously compounded riskfree interest rate is 5%. In one year the price is expected to be either $120 or $90.
Using, the oneperiod binomial option pricing model, ﬁnd the price today of one such put option.
Problem 18.2
Consider a European put option on the stock of XYZ, with a strike price of $30 and two months
to expiration. The stock pays continuous dividends at the annual continuously compounded yield
rate of 5%. The annual continuously compounded risk free interst rate is 11%. The stock currently
trades for $23 per share. Suppose that in two months, the stock will trade for either $18 per share
or $29 per share. Use the oneperiod binomial option pricing to ﬁnd the today’s price of the put.
Problem 18.3
One year from today, GS stock will sell for either $130 or $124. The annual continuously compounded interest rate is 11%. The riskneutral probability of an increase in the stock price (to
$130) is 0.77. Using the oneperiod binomial option pricing model, ﬁnd the current price of a
oneyear European put option on the stock with a strike price of $160.
Problem 18.4
Consider a twoperiod binomial model. Show that the current price of a put option is given by the
formula
P = e−2rh [p2 Puu + 2pu (1 − pu )Pud + (1 − pu )2 Pdd ].
u
Problem 18.5
GS stock is currently worth $56. Every year, it can increase by 30% or decrease by 10%. The stock
pays no dividends, and the annual continuouslycompounded riskfree interest rate is 4%. Using a
twoperiod binomial option pricing model, ﬁnd the price today of one twoyear European put option
on the stock with a strike price of $120
Problem 18.6
GS stock pays dividends at an annual continuously compounded yield of 6%. The annual continuosuly compounded interest rate is 9%. The stock is currently worth $100. Every two years, it can
increase by 50% or decrease by 30%. Using a twoperiod binomial option pricing model, ﬁnd the
price today of one fouryear European put option on the stock with a strike price of $130. 130 OPTION PRICING IN BINOMIAL MODELS Problem 18.7
Given the following information about a stock: S = $100, σ = 0.3, r = 0.05, δ = 0.03, K = 95.
Using a threeperiod binomial tree, ﬁnd the current price of a European put option with strike $95
and expiring in three years.
Problem 18.8 ‡
For a twoyear European put option, you are given the following information:
• The stock price is $35
• The strike price is $32
• The continuously compounded riskfree interest rate is 5%
• The stock price volatility is 35%
• The stock pays no dividends.
Find the price of the put option using a twoperiod binomial pricing.
Problem 18.9
A European put option on a nondividendpaying stock has a strike price $88 and expires in seven
months. The annual continuously compounded riskfree rate is 8%. The current price of the stock
is $130. The price volatility is 30%. Use a 7period binomial model to ﬁnd the price of the put.
Hint: Compare d7 S and ud6 S.
Problem 18.10
A European call option on a nondividendpaying stock has a strike price $88 and expires in seven
months. The annual continuously compounded riskfree rate is 8%. The current price of the stock
is $130. The price volatility is 30%. Use a 7period binomial model to ﬁnd the price of the call. 19 BINOMIAL OPTION PRICING FOR AMERICAN OPTIONS 131 19 Binomial Option Pricing for American Options
Binomial trees are widely used within ﬁnance to price American type options. The binomial method
constructs a tree lattice which represents the movements of the stock and prices the option relative
to the stock price by working backward through the tree. At each node, we compare the value of
the option if it is held to expiration to the gain that could be realized upon immediate exercise.
The higher of these is the American option price.
Thus, for an American call the value of the option at a node is given by
C (St , K, t) = max{St − K, e−rh [pu C (uSt , K, t + h) + (1 − pu )C (dSt , K, t + h)]}
and for an American put it is given by
P (St , K, t) = max{K − St , e−rh [pu P (uSt , K, t + h) + (1 − pu )P (dSt , K, t + h)]}
where
• t is the time equivalent to some node in the tree
• St is the stock price at time t
• h the length of a period
• r is the continuously compounded riskfree interest rate
• K is the strick price
• pu is the riskneutral probability on an increase in the stock
• P (St , K, t) is the price of an American put with strike price K and underlying stock price St
• C (St , K, t) is the price of an American call with strike price K and underlying stock price St .
Example 19.1
Given the following information about a stock: S = $100, σ = 0.3, r = 0.05, δ = 0.03, K = 95. Using
a threeperiod binomial tree, ﬁnd the current price of an American call option with strike $95 and
expiring in three years.
Solution.
We ﬁrst ﬁnd u and d. We have
u = e(r−δ)h+σ
and
d = e(r−δ)h−σ √ √ h = e(0.05−0.03)+0.3 = 1.3771 h = e(0.05−0.03)−0.3 = 0.7558. Thus,
pu = e(r−δ)h − d
e0.02 − 0.7558
=
= 0.4256
u−d
1.3771 − 0.7558 132 OPTION PRICING IN BINOMIAL MODELS Now, we have
Year 3, Stock Price = u3 S = 261.1539 Since we are at expiration, the option value is Cuuu =
261.1539 − 95 = $166.1539.
Year 3, Stock Price = u2 dS = 143.3302 and Cuud = 143.3302 − 95 = 48.3302.
Year 3, Stock Price = ud2 S = 78.6646 and Cudd = Cddu = 0.
Year 3, Stock Price = d3 S = 43.1738 and Cddd = 0.
Year 2, Stock Price = u2 S = 189.6404
Cuu = max{189.6404 − 95, e−0.05 [0.4256 × 166.1539 + (1 − 0.4256) × 48.3302]}
=189.6404 − 95
=94.6404.
Year 2, Stock Price = udS = 104.0812
Cud = max{104.0812 − 95, e−0.05 [0.4256 × 48.3302 + (1 − 0.4256) × 0]}
=e−0.05 [0.4256 × 48.3302 + (1 − 0.4256) × 0]
=19.5662.
Year 2, Stock Price = d2 S = 57.1234
Cdd = max{57.1234 − 95, e−0.05 [0.4256 × 0 + (1 − 0.4256) × 0]}
=e−0.05 [0.4256 × 0 + (1 − 0.4256) × 0] = 0
Year 1, Stock Price = uS = 137.71
Cu = max{137.71 − 95, e−0.05 [0.4256 × 94.6404 + (1 − 0.4256) × 19.5662]}
=e−0.05 [0.4256 × 94.6404 + (1 − 0.4256) × 19.5662]
=49.0052
Year 1, Stock Price = dS = 75.58
Cd = max{75.58 − 95, e−0.05 [0.4256 × 19.5662 + (1 − 0.4256) × 0]}
=e−0.05 [0.4256 × 19.5662 + (1 − 0.4256) × 0]
=7.9212
Year 0, Stock Price = $100 and the current option value is:
C = max{100 − 95, e−0.05 [0.4256 × 49.0052 + (1 − 0.4256) × 7.9212]}
=e−0.05 [0.4256 × 51.578 + (1 − 0.4256) × 7.9212]
=25.167 19 BINOMIAL OPTION PRICING FOR AMERICAN OPTIONS 133 Example 19.2
Given the following information about a stock: S = $100, σ = 0.3, r = 0.05, δ = 0.03, K = 95. Using
a threeperiod binomial tree, ﬁnd the current price of an American put option with strike $95 and
expiring in three years.
Solution.
The values of u, d, and pu are the same as in the previous example. We have
Year 3, Stock Price = u3 S = 261.1539 Since we are at expiration, the option value is Puuu = 0.
Year 3, Stock Price = u2 dS = 143.3302 and Puud = 0.
Year 3, Stock Price = ud2 S = 78.6646 and Pudd = Pddu = 95 − 78.6646 = 16.3354.
Year 3, Stock Price = d3 S = 43.1738 and Pddd = 95 − 43.1738 = 51.8262.
Year 2, Stock Price = u2 S = 189.6404 and
Puu = max{95 − 189.6404, e−0.05 [0.4256 × 0 + (1 − 0.4256) × 0]}
=0.
Year 2, Stock Price = udS = 104.0812 and
Pud = max{95 − 104.08212, e−0.05 [0.4256 × 0 + (1 − 0.4256) × 16.3554]}
=e−0.05 [0.4256 × 0 + (1 − 0.4256) × 16.3354] = 9.3831
Year 2, Stock Price = d2 S = 57.1234 and
Pdd = max{95 − 57.123, e−0.05 [0.4256 × 16.3354 + (1 − 0.4256) × 51.8262]}
=95 − 57.123 = 37.877
Year 1, Stock Price = uS = 137.71 and
Pu = max{95 − 137.71, e−0.05 [0.4256 × 0 + (1 − 0.4256) × 9.3831]}
=e−0.05 [0.4256 × 0 + (1 − 0.4256) × 9.3831] = 5.1268
Year 1, Stock Price = dS = 75.58 and
Pd = max{95 − 75.58, e−0.05 [0.4256 × 9.3831 + (1 − 0.4256) × 37.877]}
=e−0.05 [0.4256 × 9.3831 + (1 − 0.4256) × 37.877] = 24.4942
Year 0, Stock Price = $100 and the current option value is:
P = max{95 − 100, e−0.05 [0.4256 × 5.1268 + (1 − 0.4256) × 24.4942]}
=e−0.05 [0.4256 × 5.1268 + (1 − 0.4256) × 24.4942] = 15.4588 134 OPTION PRICING IN BINOMIAL MODELS Practice Problems
Problem 19.1
A stock is currently worth $100. In one year the stock will go up to $120 or down to $90. The stock
pays no dividends, and the annual continuouslycompounded riskfree interest rate is 5%. Find the
price today of a one year America put option on the stock with a strike price of $130.
Problem 19.2
One year from now, GS stock is expected to sell for either $130 or $124. The annual continuously
compounded interest rate is 11%. The riskneutral probability of the stock price being $130 in one
year is 0.77. What is the current stock price for which a oneyear American put option on the stock
with a strike price of $160 will have the same value whether calculated by means of the binomial
option pricing model or by taking the diﬀerence between the stock price and the strike price?
Problem 19.3
GS Inc., pays dividends on its stock at an annual continuously compounded yield of 6%. The annual
eﬀective interest rate is 9%. GS Inc., stock is currently worth $100. Every two years, it can go
up by 50% or down by 30%. Using a twoperiod binomial option pricing model, ﬁnd the price two
years from today of one fouryear American call option on the stock with a strike price of $80 in
the event that the stock price increases two years from today.
Problem 19.4
Repeat the previous problem in the event that the stock price decreases two years from today.
Problem 19.5
The current price of a stock is $110. The annual continuously compounded interest rate is 0.10, and
the stock pays continuous dividends at the continuously compounded yield 0.08. The annualized
standard deviation of the continuously compounded stock return is 0.32. Using a threeperiod
binomial pricing model, ﬁnd the price of an American call option on the stock with strike price
$100 and that matures in nine months.
Problem 19.6
Repeat the previous problem with an American put.
Problem 19.7
Given the following: S = $72, r = 8%, δ = 3%, σ = 23%, h = 1, K = 74. Use twoperiod binomial
pricing to ﬁnd the current price of an American put. 19 BINOMIAL OPTION PRICING FOR AMERICAN OPTIONS 135 Problem 19.8 ‡
You are given the following information about American options:
• The current stock price is 72
• The strike price is 80
• The continuouslycompounded riskfree rate is 5%
• Time to expiration is one year
• Every six months, the stock price either increases by 25% or decreases by 15%.
Using a twoperiod binomial tree, calculate the price of an American put option.
Problem 19.9 ‡
For a twoperiod binomial model, you are given:
(i) Each period is one year.
(ii) The current price for a nondividendpaying stock is $20.
(iii) u = 1.2840, where u is one plus the rate of capital gain on the stock per period if the stock
price goes up.
(iv) d = 0.8607, where d is one plus the rate of capital loss on the stock per period if the stock price
goes down.
(v) The continuously compounded riskfree interest rate is 5%.
Calculate the price of an American call option on the stock with a strike price of $22.
Problem 19.10 ‡
Consider the following threeperiod binomial tree model for a stock that pays dividends continuously
at a rate proportional to its price. The length of each period is 1 year, the continuously compounded
riskfree interest rate is 10%, and the continuous dividend yield on the stock is 6.5%.
Calculate the price of a 3year atthemoney American put option on the stock. 136 OPTION PRICING IN BINOMIAL MODELS Problem 19.11 ‡
For a twoperiod binomial model for stock prices, you are given:
(i) Each period is 6 months.
(ii) The current price for a nondividendpaying stock is $70.00.
(iii) u = 1.181, where u is one plus the rate of capital gain on the stock per period if the price goes
up.
(iv) d = 0.890, where d is one plus the rate of capital loss on the stock per period if the price goes
down.
(v) The continuously compounded riskfree interest rate is 5%.
Calculate the current price of a oneyear American put option on the stock with a strike price of
$80.00.
Problem 19.12 ‡
Given the following information for constructing a binomial tree for modeling the price movements
of a stock. (This tree is sometimes called a forward tree.)
(i) The length of each period is one year.
(ii) The current stock price is 100.
(iii) Volatility is 30%.
(iv) The stock pays dividends continuously at a rate proportional to its price. The dividend yield
is 5%.
(v) The continuously compounded riskfree interest rate is 5%.
Calculate the price of a twoyear 100strike American call option on the stock.
Problem 19.13 ‡
Given the following information for constructing a binomial tree for modeling the price movements
of a stock:
(i) The period is 3 months.
(ii) The initial stock price is $100.
(iii) The stock’s volatility is 30%.
(iv) The continuously compounded riskfree interest rate is 4%.
At the beginning of the period, an investor owns an American put option on the stock. The option
expires at the end of the period.
Determine the smallest integervalued strike price for which an investor will exercise the put option
at the beginning of the period. 20 BINOMIAL OPTION PRICING ON CURRENCY OPTIONS 137 20 Binomial Option Pricing on Currency Options
In the case of currency options with underlying asset a foreign currency and strike asset dollars, the
replicating portfolio consists of buying ∆ units of the foreign currency and borrowing cash in the
amount of B dollars. Letting rf denote the foreign currency denominated interest rate, one unit of
the foreign currency now increases to erf h units at the end of one period. Thus, after one period, the
payoﬀ of the portfolio is ∆ × ux × erf h − erh B in the up state and ∆ × dx × erf h − erh B in the down
state where x is the exchange rate at the beginning of the period and r is the dollardenominated
interest rate. Now, if this portfolio is to replicate the call option then by the noarbitrage principle
we must have
∆ × ux × erf h − erh B = Cu and ∆ × dx × erf h − erh B = Cd .
This is a system of two equations in the unknowns ∆ and B. Solving the system we ﬁnd
∆ = e−rf h C u − Cd
x(u − d) (20.1) uCd − dCu
(20.2)
d−u
Now, the call price at the beginning of the period is deﬁned to be the total amount invested in the
equivalent portfolio. That is,
B = e−rh C = ∆x − B = e−rh Cu u − e(r−rf )h
e(r−rf )h − d
+ Cd
u−d
u−d . (20.3) These equations look similar to the ones for stock with the dividend yield being replaced by the
foreign denominated interet rate.
Now, from equation (20.3) we can deﬁne the riskneutral probability of an up move by the formula
pu = e(r−rf )h − d
.
u−d By Section 16, we have the following equations
ux =Ft,t+h eσ √ dx =Ft,t+h e−σ h √ h where Ft,t+h = xe(r−rf )h . Hence u and d are found using the equations
u =e(r−rf )h+σ
d =e(r−rf )h−σ √
√ h
h 138 OPTION PRICING IN BINOMIAL MODELS Example 20.1
You are given the following information: x =$1.20/e, r = 5%, re = 9%, σ = 15%. Using a threeperiod binomial tree, calculate the price of a nine months European call on the euro, denominated
in dollars, with a strike price of $1.10.
Solution.
We ﬁrst ﬁnd u and v given by
u = e(r−rf )h+σ
and
d = e(r−rf )h−σ √ √ √ h = e(0.05−0.09)×0.25+0.15 h = e(0.05−0.09)×0.25−0.15 √ 0.25 = 1.06716 0.25 = 0.91851 The riskneutral probability of an up move is
pu = e−0.04×0.25 − 0.918151
e(r−rf )h − d
=
= 0.48126.
u−d
1.06716 − 0.91851 We have
9month, Exchange rate = u3 x = $1.4584/e Since we are at expiration, the option value is
Cuuu = 1.4584 − 1.1 = $0.3584.
9month, Exchange rate = u2 dx = $1.2552/e and Cuud = 1.2552 − 1.10 = $0.1552.
9month, Exchange rate = ud2 x = $1.0804/e and Cudd = Cddu = 0.
9month, Exchange rate = d3 x = $0.9299/e and Cddd = 0.
6month, Exchange rate = u2 x = $1.3666/e and
Cuu = e−0.05×0.25 [0.48126 × 0.3584 + (1 − 0.48126) × 0.1552] = $0.2499.
6month, Exchange rate = udx = $1.1762/e and
Cud = e−0.05×0.25 [0.48126 × 0.1552 + (1 − 0.48126) × 0] = $0.0738.
6month, Exchange rate = d2 x = $1.0124/e and
Cdd = e−0.05×0.25 [0.48126 × 0 + (1 − 0.48126) × 0] = 0.
3month, Exchange rate = ux = $1.2806/e and
Cu = e−0.05×0.25 [0.48126 × 0.2499 + (1 − 0.48216) × 0.0738] = $0.1565.
3month, Exchange rate = dx = $1.1022/e and
Cd = e−0.05×0.25 [0.48216 × 0.0738 + (1 − 0.48216) × 0] = $0.0351.
The current option value is:
C = e−0.05×0.25 [0.48216 × 0.1565 + (1 − 0.48216) × 0.0351] = $0.0924 20 BINOMIAL OPTION PRICING ON CURRENCY OPTIONS 139 Example 20.2
You are given the following information: x =$1.20/e, r = 5%, re = 9%, σ = 15%. Using a threeperiod binomial tree, calculate the price of a nine months American call on the euro, denominated
in dollars, with a strike price of $1.10.
Solution.
From the previous example, we have u = 1.06716, d = 0.91851, and pu = 0.48126. Now, we have
9month, Exchange rate = u3 x = $1.4584/e Since we are at expiration, the option value is
Cuuu = 1.4584 − 1.1 = $0.3584.
9month, Exchange rate = u2 dx = $1.2552/e and Cuud = 1.2552 − 1.10 = $0.1552.
9month, Exchange rate = ud2 x = $1.0804/e and Cudd = Cddu = 0.
9month, Exchange rate = d3 x = $0.9299/e and Cddd = 0.
6month, Exchange rate = u2 x = $1.3666/e and
Cuu = max{1.3666 − 1.10, e−0.05×25 [0.48216 × 0.3584 + (1 − 0.48126) × 0.1552]}
=1.3666 − 1.10
=0.2530.
6month, Exchange rate = udx = $1.1762/e and
Cud = max{1.1762 − 1.10, e−0.05×0.25 [0.48126 × 0.1552 + (1 − 0.48126) × 0]}
=1.1762 − 1.10
=0.0752.
6month, Exchange rate = d2 x = $1.0124/e and
Cdd = max{1.0124 − 1.1, e−0.05×0.25 [0.48126 × 0 + (1 − 0.4826) × 0]}
=e−0.05×0.25 [0.48126 × 0 + (1 − 0.48216) × 0] = 0
3month, Exchange rate = ux = $1.2806/e and
Cu = max{1.2806 − 1.1, e−0.05×0.25 [0.48126 × 0.2530 + (1 − 0.48126) × 0.0752]}
=1.2806 − 1.1
=0.1806
3month, Exchange rate = dx = $1.1022/e and
Cd = max{1.1022 − 1.10, e−0.05×0.25 [0.48126 × 0.0752 + (1 − 0.48126) × 0]}
=e−0.05×0.25 [0.48126 × 0.0752 + (1 − 0.48126) × 0]
=0.0357 140 OPTION PRICING IN BINOMIAL MODELS The current option value is:
C = max{1.20 − 1.10, e−0.05×0.25 [0.48126 × 0.1806 + (1 − 0.48126) × 0.0357]}
=e−0.05×0.25 [0.48126 × 0.1806 + (1 − 0.48126) × 0.0357]
=0.1041 20 BINOMIAL OPTION PRICING ON CURRENCY OPTIONS 141 Practice Problems
Problem 20.1
One euro currently trades for $1.56. The dollardenominated annual continuouslycompounded
riskfree interest rate is 2%, and the eurodenominated annual continuouslycompounded riskfree
interest rate is 9%. Calculate the price of a 10year forward contract on euros, denominated in
dollars.
Problem 20.2
One euro currently trades for $1.56. The dollardenominated annual continuouslycompounded
riskfree interest rate is 0.02, and the eurodenominated annual continuouslycompounded riskfree
interest rate is 0.09. The annualized standard deviation of the continuously compounded return on
the euro is 0.54. Using a oneperiod binomial model, calculate what the euro price in dollars will
be in two years if the euro’s price increases.
Problem 20.3
One dollar is currently trading for FC 45. The dollardenominated annual continuously compounded
riskfree interest rate is 0.13, while the FCdenominated annual continuously compounded riskfree
interest rate is 0.05. The annualized standard deviation of the continuously compounded return
on dollars is 0.81. Using the oneperiod binomial option pricing model, what is the riskneutral
probability that the price of a dollar will increase in two months?
Problem 20.4
You are given the following information: x =$0.92/e, r = 4%, re = 3%, u = 1.2, d = 0.9, and
h = 0.25. Using a threeperiod binomial tree, calculate the price of a nine months European put on
the euro, denominated in dollars, with a strike price of $0.85.
Problem 20.5
You are given the following information: x =$0.92/e, r = 4%, re = 3%, u = 1.2, d = 0.9, and
h = 0.25. Using a threeperiod binomial tree, calculate the price of a nine months American put on
the euro, denominated in dollars, with a strike price of $1.00.
Problem 20.6
One yen currently trades for $0.0083. The dollardenominated annual continuouslycompounded
riskfree interest rate is 0.05, and the yendenominated annual continuouslycompounded riskfree
interest rate is 0.01. The annualized standard deviation of the continuously compounded return on
the dollar is 0.10. Using a oneperiod binomial model, calculate what the yen price in dollars will
be in two years if the yen’s price increases. 142 OPTION PRICING IN BINOMIAL MODELS Problem 20.7
Given the following information: r = 0.05, rf = 0.01, σ = 0.10, h =
probability of an up move. 1
.
3 Find the riskneutral Problem 20.8
One dollar is currently trading FC 45. The dollardenominated annual continuouslycompounded
riskfree interest rate is 13%, while the FCdenominated annual continuouslycompounded riskfree
interest rate is 5%. The annualized standard deviation of the continuously compounded return
on dollars is 0.81. For a certain twomonth European call option on one dollar, the replicating
portfolio involves buying 3 dollars and borrowing FC 23. Using the oneperiod binomial option
4
pricing model, what would the price of this call option (in FC) be in two months in the case there
is a decrease in the dollar value?
Problem 20.9
A dollar is currently selling for 118. The dollardenominated annual continuouslycompounded
riskfree interest rate is 0.06, and the yendenominated annual continuouslycompounded riskfree
interest rate is 0.01. The annualized standard deviation of the continuously compounded return on
the dollar is 0.11. Using a threeperiod binomial tree, ﬁnd the current price of a yendenominated
1year American call on the dollar with strike price of 118.
Problem 20.10 ‡
Consider a 9month dollardenominated American put option on British pounds. You are given
that:
(i) The current exchange rate is 1.43 US dollars per pound.
(ii) The strike price of the put is 1.56 US dollars per pound.
(iii) The volatility of the exchange rate is σ = 0.3.
(iv) The US dollar continuously compounded riskfree interest rate is 8%.
(v) The British pound continuously compounded riskfree interest rate is 9%.
Using a threeperiod binomial model, calculate the price of the put. 21 BINOMIAL PRICING OF FUTURES OPTIONS 143 21 Binomial Pricing of Futures Options
In this section we consider applying the binomial model to price options with underlying assets
futures contracts.
Recall that a futures contract, or simply a futures, is a pledge to purchase at a future date a given
amount of an asset at a price agreed on today. A futures contract requires no initial payment.
At the maturity date, cash is exchanged for the asset. For example, a 3month futures contract
for 1000 tons of soybean at a forward price of $165/ton is a committment from the owner of the
contract to buy 1000 tons of soybean in three months for a price of $165 a ton.
The replicating portfolio consists of buying ∆ units of the futures contracts and lending cash in the
amount of B dollars. At the end of a given period, a futures’ contract payoﬀ is just the change in
the futures price1 . Thus, after one period, the payoﬀ of the portfolio is ∆ × (uF − F ) + erh B in the
up state and ∆ × (dF − F ) + erh B in the down state where F is the delivery price of the contract
at the beginning of the period and r is the continuously compounded riskfree interest rate. Now, if
this portfolio is to replicate a call option on the futures then by the noarbitrage principle we must
have
∆ × (uF − F ) + erh B = Cu and ∆ × (dF − F ) + erh B = Cd .
This is a system of two equations in the unknowns ∆ and B. Solving the system we ﬁnd
∆= Cu − Cd
F (u − d) (21.1) u−1
1−d
+ Cd
.
(21.2)
u−d
u−d
Now, the call price at the beginning of the period is time 0 value of the replicating portfolio. That
is,
u−1
1−d
+ Cd
C = B = e−rh Cu
(21.3)
u−d
u−d
since a futures requires no initial premium. Now, from equation (21.2) we can deﬁne the riskneutral
probability of an up move by the formula
B = e−rh Cu 1−d
.
u−d
By Section 16, the up and down move of the forward price are modeled by the following equations
pu = uF =Ft,t+h eσ √ dF =Ft,t+h e−σ
1 See [2], Section 59. h √ h 144 OPTION PRICING IN BINOMIAL MODELS where Ft,t+h = F is the forward price. Hence u and d are found by using the equations
u =eσ √ d =e−σ h √ h Example 21.1
An option has a gold futures contract as the underlying asset. The current 1year gold futures
price is $300/oz, the strike price is $290, the continuously compounded riskfree interest rate is 6%,
volatility is 10%, and the time to expiration is 1 year. Using a oneperiod binomial model, ﬁnd
∆, B, and the price of the call.
Solution.
We ﬁrst ﬁnd u and d. We have
u = eσ
and
d = e−σ √ √ h = e0.10 = 1.1052 h = e−0.10 = 0.9048. We also have
Cu = 1.1052 × 300 − 290 = $41.56
and
Cd = 0.
Thus,
∆= Cu − Cd
41.56 − 0
=
= 0.6913
F (u − d)
300(1.1052 − 0.9048) and
C = B = e−rh Cu 1−d
u−1
+ Cd
u−d
u−d = e−0.06 41.56 × 1 − 0.9048
= $18.5933
1.1052 − 0.9048 Alternatively, we could have found the riskneutral probability of an up move
pu = 1−d
1 − 0.9048
=
= 0.4751
u−d
1.1052 − 0.9048 and then ﬁnd the price of the call to be
C = B = e−rh [pu Cu + (1 − pu )Cd ] = e−0.06 × 0.4751 × 41.56 = $18.5933 21 BINOMIAL PRICING OF FUTURES OPTIONS 145 Example 21.2
An option has a futures contract as the underlying asset. The current 1year futures price is $1000,
the strike price is $1000, the continuously compounded riskfree interest rate is 7%, volatility is
30%, and the time to expiration is one year. Using a threeperiod binomial model, ﬁnd the price of
an American call.
Solution.
We have
u = eσ
and
d = e−σ h √1
= e0.30 3 = 1.18911 h √1
= e−0.30 3 = 0.84097. √ √ The riskneutral probability of an up move is
pu = 1 − 0.84097
1−d
=
= 0.45681.
u−d
1.18911 − 0.84097 Now, we have
12month, Futures Price = u3 F = 1681.3806 Since we are at expiration, the option value is
Cuuu = 1681.3806 − 1000 = $681.3806.
12month, Futures Price = u2 dF = 1189.1099 and Cuud = 1189.1099 − 1000 = 189.1099.
12month, Futures Price = ud2 F = 840.9651 and Cudd = Cddu = 0.
12month, Futures Price = d3 F = 594.7493 and Cddd = 0.
8month, Futures Price = u2 F = 1413.9825
1 Cuu = max{1413.9825 − 1000, e−0.07× 3 [0.45681 × 681.3806 + (1 − 0.45681) × 189.1099]}
=1413.9825 − 1000
=413.9825.
8month, Futures Price = udF = 1000
1 Cud = max{10000 − 1000, e−0.07× 3 [0.45681 × 189.1099 + (1 − 0.45681) × 0]}
1 =e−0.07× 3 [0.45681 × 189.1099 + (1 − 0.45681) × 0]
=84.3943.
8month, Futures Price = d2 F = 707.2224
1 Cdd = max{707.2224 − 1000, e−0.07× 3 [0.45681 × 0 + (1 − 0.4256) × 0]}
1 =e−0.07× 3 [0.45681 × 0 + (1 − 0.4256) × 0 = 0 146 OPTION PRICING IN BINOMIAL MODELS 4month, Futures Price = uF = 1189.1099
1 Cu = max{1189.1099 − 1000, e−0.07× 3 [0.45681 × 413.9825 + (1 − 0.45681) × 84.3943]}
1 =e−0.07× 3 [0.45681 × 413.9825 + (1 − 0.45681) × 84.3943]
=229.5336.
4month, Futures Price = dF = 840.9651
1 Cd = max{849.9651 − 1000, e−0.07× 3 [0.45681 × 84.3943 + (1 − 0.45681) × 0]}
1 =e−0.07× 3 [0.45681 × 84.3943 + (1 − 0.45681) × 0]
=37.6628.
0month, Futures Price = $1000 and the current option value is:
1 C = max{1000 − 1000, e−0.07× 3 [0.45681 × 229.5336 + (1 − 0.45681) × 37.6628]}
1 =e−0.07× 3 [0.45681 × 229.5336 + (1 − 0.45681) × 37.6628]
=122.4206 21 BINOMIAL PRICING OF FUTURES OPTIONS 147 Practice Problems
Problem 21.1
An option has a gold futures contract as the underlying asset. The current 1year gold futures
price is $600/oz, the strike price is $620, the continuously compounded riskfree interest rate is 7%,
volatility is 12%, and the time to expiration is 1 year. Using a oneperiod binomial model, ﬁnd the
price of the call.
Problem 21.2
An option has a futures contract as the underlying asset. The current 1year futures price is $1000,
the strike price is $1000, the continuously compounded riskfree interest rate is 7%, volatility is
30%, and the time to expiration is one year. Using a threeperiod binomial model, ﬁnd the time0
number of futures contract ∆ in the replicating portfolio of an American call on the futures.
Problem 21.3
Consider a European put option on a futures contract. Suppose that d = 3 u when using a two4
1
period binomial model and the riskneutral probability of an increase in the futures price is 3 .
Determine u and d.
Problem 21.4
Consider a European put option on a futures contract with expiration time of 1 year and strike
price of $80. The time0 futures price is $80. Suppose that u = 1.2, d = 0.9, r = 0.05. Using a
twoperiod binomial model, ﬁnd the current price of the put.
Problem 21.5
Consider an American put option on a futures contract with expiration time of 1 year and strike
price of $80. The time0 futures price is $80. Suppose that u = 1.2, d = 0.9, r = 0.05. Using a
twoperiod binomial model, ﬁnd the current price of the put.
Problem 21.6
Consider an option on a futures contract. The time0 futures price is $90. The annualized standard
deviation of the continuously compounded return on the futures contract is 34%. Using a threeperiod binomial option pricing model, ﬁnd the futures contract price after 6 years if the contract
always increases in price every time period.
Problem 21.7
Find the riskneutral probability of a down move in Problem 21.6. 148 OPTION PRICING IN BINOMIAL MODELS Problem 21.8
Consider an option on a futures contract. The time0 futures price is $90. The annualized standard
deviation of the continuously compounded return on the futures contract is 0.34. The annual
continuously compounded riskfree interest rate is 0.05. Using a oneperiod binomial option pricing
model, ﬁnd ∆ for a replicating portfolio equivalent to one twoyear European call option on futures
contract with a strike price of $30.
Problem 21.9
Consider an option on a futures fontract. The time0 futures price is $49. The annual continuously
compounded riskfree interest rate is 15%. The price today of one particular threemonth European
call option on the contract is $10. The value of ∆ in a replicating portfolio equivalent to one such
option is 0.4. If in three months, futures prices will be worth 0.85 of its present amount, what will
the price of the call option be? Use a oneperiod binomial option pricing model.
Problem 21.10 ‡
You are to price options on a futures contract. The movements of the futures price are modeled by
a binomial tree. You are given:
(i) Each period is 6 months.
(ii) u = 4 , where u is one plus the rate of gain on the futures price if it goes up, and d is one plus
d
3
the rate of loss if it goes down.
(iii) The riskneutral probability of an up move is 1 .
3
(iv) The initial futures price is 80.
(v) The continuously compounded riskfree interest rate is 5%.
Let CI be the price of a 1year 85strike European call option on the futures contract, and CII be
the price of an otherwise identical American call option. Determine CII − CI . 22 FURTHER DISCUSSION OF EARLY EXERCISING 149 22 Further Discussion of Early Exercising
As you have seen by now, when American options are valued using the binomial tree, one compares
the value of exercising immediately with the value of continuing holding the option and this is done
at each binomial node. In this section, we examine early exercise in more detail. More speciﬁcally,
we want to know when is it rational to early exercise.
The early exercise decision weighs three economic considerations:
• The dividends on the underlying asset.
• The interest on the strike price.
• The insurance value of keeping the option alive.
Example 22.1
Discuss the three considerations of early exercise for an American call option holder.
Solution.
By exercising, the option holder
• Receives the underlying asset and captures all related future dividends.
• Pays the strike price and therefore loses the interest from the time of exercising to the time of
expiration.
• Loses the insurance/ﬂexibility implicit in the call. The option holder is not protected anymore
when the underlying asset has a value less than the strike price at expiration
It follows that for a call, dividends encourage early exercising while interest and insurance weigh
against early exercise.
Example 22.2
Consider an American call option on a stock with strike price of $100. The stock pays continuous
dividends at the continuous yield rate of 5%. The annual continuously compounded riskfree interest
rate is 5%.
(a) Suppose that one year is left to expiration and that the stock price is currently $200. Compare
the amount of dividends earned from acquiring the stock by early exercise to the amount of interest
saved by not exercising.
(b) According to your answer to (a), what is the only economic consideration for an option holder
not to exercise early?
Solution.
(a) The amount of dividends earned from early exercise is 200e0.05 − 200 = $10.25. The amount
of interest saved from not exercising is 100e0.05 − 100 = $5.13. Thus, the dividends lost by not
exercising exceed interest saved by deferring exercise.
(b) The only reason for not exercising in this case is to keep the implicit insurance provided by the 150 OPTION PRICING IN BINOMIAL MODELS call against a drop in the stock price below the strike price
According to the previous example, one of the reasons an option holder may defer early exercising is the insurance feature of the option. However, with zero volatility this insurance has zero
value. In this case, what will be the optimal decision? Obviously, it is optimal to defer exercise as
long as the interest savings on the strike exceeds the dividends lost. In symbol, we want
er(T −t) K − K > eδ(T −t) St − St .
Example 22.3
Show that the condition rK > δSt implies er(T −t) K − K > eδ(T −t) St − St .
Solution.
Suppose that rK > δSt . Using the Taylor series expansion of the function et around zero we
can write er(T −t) K − K ≈ Kr(T − t) and eδ(T −t) St − St ≈ δSt (T − t). Thus, rK > δSt implies
er(T −t) K − K > eδ(T −t) St − St
It follows that for an American call option where the volatility is zero, it is optimal to defer exercise
as long as the following condition holds:
rK > δSt .
It is optimal to exercise whenever1
rK
.
δ
With American put options, the reverse holds. It is optimal to exercise early when
St > St < rK
δ and it is optimal to defer exercise when
rK < δSt .
Example 22.4
Consider an American call option with zero volatility. Suppose that r = 2δ. When it is optimal to
exercise?
1 This condition may be wrong in some instances. However, this condition is always correct for inﬁnitelylived
American options. 22 FURTHER DISCUSSION OF EARLY EXERCISING
Solution.
It is optimal to exercise whenever St >
strike price rK
δ 151 = 2K. That is, when the stock price is at least twice the When volatility is positive, the implicit insurance has value, and the value varies with time to
expiration. Figure 22.1 shows, for a ﬁxed time, the lowest stock price above which early exercise
is optimal for a 5year American call with strike $100, r = δ = 5%, for three diﬀerent volatilities.
Recall that if it is optimal to exercise a call at a given stock price then it is optimal to exercise at all
higher stock prices. The ﬁgure shows the eﬀect of volatility. The exercise bounds for lower volatility
are lower than the exercise bounds for higher volatility. This stem from the fact that the insurance
value lost by earlyexercise is greater when volatility is greater. Moreover, for a ﬁxed volatility, the
passage of time decreases the exercise bounds. This stems from the fact that the value of insurance
diminishes as the options approach expiration. Figure 22.1 In the case of an American put, the higher the volatility the lower the exercise bound. Moreover,
the passage of time increases the exercise bounds. See Figure 22.2. 152 OPTION PRICING IN BINOMIAL MODELS Figure 22.2
Example 22.5
Suppose that for a volatility of 10%, an early exercise for an American call option is optimal when
the lowest price of the underlying asset is $130. For a volatility of 30%, would the lowest price of
the underlying asset be larger or smaller than $130 in order for early exercise to be optimal?
Solution.
It has to be larger than $130
Example 22.6
You are given the following information about an American call on a stock:
• The current stock price is $70.
• The strike price is $68
• The continuously compounded riskfree interest rate is 7%.
• The continuously compounded dividend yield is 5%.
• Volatility is zero.
Find the time until which early exercise is optimal.
Solution.
Early exercise is optimal if and only of
0.07 × 68
rK
=
= $95.20.
δ
0.05
That is, when the stock price reaches $95.20. Now,
S> u = e(r−δ)h+σ √ h = e0.02h = d. Thus, early exercise is optimal when uS = 95.20 or 70e0.02T = 95.20. Solving this equation we ﬁnd
T = 15.37 22 FURTHER DISCUSSION OF EARLY EXERCISING 153 Practice Problems
Problem 22.1
Discuss the three considerations of early exercise for an American put option holder.
Problem 22.2
Suppose that for a volatility of 10%, an early exercise for an American put option is optimal when
the lowest price of the underlying asset is $80. For a volatility of 30%, would the highest price of
the underlying asset be larger or smaller than $80 in order for early exercise to be optimal?
Problem 22.3
Consider an American call option with strike K such that r = δ. If at time t, St > K, would it be
optimal to exercise?
Problem 22.4
Given the following information about an American call option on a stock:
• The current price of the stock is $456.
• The continuously compounded riskfree interest rate is 5%.
• The continuously compounded yield rate is 4%.
• Volatility is zero.
What is the least strike value so that to defer early exercise?
Problem 22.5
Given the following information about an American call option on a stock:
• The strike price is $30.
• The continuously compounded riskfree interest rate is 14%.
• The continuously compounded yield rate is 11%.
• Volatility is zero.
What is the lowest stock price for which early exercise is optimal?
Problem 22.6
You own an American put option on the nonvolatile GS Inc. stock. The strike price of the option is
23, and the stock’s annual continuously compounded dividend yield is 0.4. The annual continuously
compounded interest rate is 0.1. For which of these stock prices would it be optimal to exercise the
option?
(A) S = 20
(B) S = 15.65
(C) S = 13.54
(D) S = 12.34
(E) S = 6.78
(F) S = 3.45 154 OPTION PRICING IN BINOMIAL MODELS Problem 22.7
GS Co. stock prices currently have a volatility of σ = 0.3. For American call options with a strike
price of $120 and time to expiration 1 year, you know that the lowest stock price where exercise is
optimal is $160. If the stock price volatility changes, for which of these volatilities will exercise still
be optimal at a stock price of $160?
(A) σ = 0.1
(B) σ = 0.2
(C) σ = 0.4
(D) σ = 0.5
(E) σ = 0.6
Problem 22.8
GS Co. stock prices currently have a volatility of σ = 0.3. For American put options with a strike
price of $90 and time to expiration 1 year, you know that the highest exercise bound for optimal
exercise price is $40. If the stock price volatility changes, for which of these volatilities will exercise
still be optimal at a stock price of $40?
(A) σ = 0.1
(B) σ = 0.2
(C) σ = 0.4
(D) σ = 0.5
(E) σ = 0.6
Problem 22.9
You are given the following information about an American call on a stock:
• The current stock price is $50.
• The strike price is $48.
• The continuously compounded riskfree interest rate is 9%.
• The continuously compounded dividend yield is 6%.
• Volatility is zero.
Find the time until which early exercise is optimal. 23 RISKNEUTRAL PROBABILITY VERSUS REAL PROBABILITY 155 23 RiskNeutral Probability Versus Real Probability
The riskneutral pricing approach was ﬁrst introduced in Section 15. In this section, we examine
this approach in more detail and we compare the riskneutral approach with the pricing approach
using true probability.
We ﬁrst discuss the meaning of riskneutral. Suppose you are oﬀered the following two scenarios.
In the ﬁrst scenario you are guaranteed to receive $100. In the second scenario, a coin is ﬂipped
and you receive $200 if the coin shows head or $0 otherwise. The expected payoﬀ for both scenarios
is $100. Investors may have diﬀerent risk attitudes. A riskaverse investor prefers the sure thing
which is the ﬁrst scenario. A riskneutral investor is indiﬀerent between the bet and a certain
$100 payment since both have the same expected payoﬀ. That is, a riskneutral investor will be
equally happy with either scenarios.
In most of the future pricing calculations that will occur in the rest of the book, investors are
assumed to be riskaverse unless otherwise indicated. However, let’s consider a hypothetical world
of riskneutral investors only. In such a world, investors would only be concerned with expected
returns, and not about the level of risk. Hence, investors will not charge or require a premium for
risky securities. Therefore, risky securities would have the same expected rate of return as riskless
securities. In other words, investors hold assets with an expected return equal to the riskfree rate.
In this imaginary riskneutral world, we let pu denote the probability of the stock going up such that
the stock is expected to earn the riskfree rate. In the binomial model, pu for one period satisﬁes
the equation
pu uSeδh + (1 − pu )dSeδh = erh S.
Solving for pu we ﬁnd e(r−δ)h − d
.
u−d
This is why we referred to pu introduced in Section 15 as a riskneutral probability. It is the
probability that the stock price would increase in a riskneutral world. Also, in the hypothetical
world, the option price valuation for a call is given by
pu = C = e−rh [pu Cu + (1 − pu )Cd ].
Example 23.1
You are given the following information about a stock:
• The stock pays continuous dividend at the continuously compounded yield of 6%.
• The continuously compounded riskfree interest rate is 9%.
• Every h years the stock price increases by 90% or decreases by 80%.
• The riskneutral probability of the stock price’s increase in h years is 0.72.
Find h. 156 OPTION PRICING IN BINOMIAL MODELS Solution.
We know that
pu = e(r−δ)h − d
u−d or e(0.09−0.06)h − 0.2
.
1.9 − 0.2
Solving this equation we ﬁnd h = 11.7823 years
0.72 = What is the option pricing in the riskaverse world? In this world, we let p denote the real probability of the stock going up. Let α be the continuously compounded expected return on the stock.
Then p satisﬁes the equation
puSeδh + (1 − p)dSeδh = eαh S.
Solving for p we ﬁnd
e(α−δ)h − d
.
p=
u−d
The real probability for the stock to go down is then
1−p= u − e(α−δ)h
.
u−d Imposing the condition d < e(α−δ)h < u we obtain 0 < p < 1. Now, using p we can ﬁnd the actual
expected payoﬀ at the end of the period:
pCu + (1 − p)Cd .
Example 23.2
Consider a nondividend payingstock. Every two years, the stock price either increases by 5% or
decreases by 6%. Find an upper constraint on the annual continuously compounded expected return
on the stock.
Solution.
We use the condition d < eαh < u where we write u = eah . Thus, 1.05 = e2a → a =
0.024395 ln 1.05
2 = Example 23.3
Consider a nondividend payingstock. Every two years, the stock price either increases by 5% or
decreases by 6%. The annual continuously compounded expected return on the stock is 5%. Find
the real probability that the stock will increase in price in two years. 23 RISKNEUTRAL PROBABILITY VERSUS REAL PROBABILITY 157 Solution.
We have eαh − d
e0.05×2 − 0.94
=
= 0.01817
u−d
1.05 − 0.94
For riskneutral probabilities we discounted the expected payoﬀ at the riskfree rate in order to
obtain the current option price. At what rate do we discount the actual expected payoﬀ? Deﬁnitely
not at the rate α since the option is a type of leveraged investment in the stock so that it is riskier
than the stock.
Let γ denote the appropriate perperiod discount rate.1 We use the result of Brealey and Meyer
which states that the return on any portfolio is the weighted average of the returns on the assets
in the portfolio. We apply this result to the portfolio consisting of ∆ shares of nondividendpaying
stock and B bonds that mimic the payoﬀ of the call option to obtain
p= eγh = B
S∆
eαh −
erh .
S∆ − B
S∆ − B Since an option is equivalent to holding a portfolio consisting of ∆ shares and B bonds, the denominator of the previous relation is just the option price. Thus, discounted cash ﬂow is not used
in practice to price options: It is necessary to compute the option price in order to compute the
correct discount rate.
We can now compute the option price as the discounted expected payoﬀ at the rate γ to obtain
C = e−γh [pCu + (1 − p)Cd ].
Example 23.4
Show that the option price obtained with real probabilities is the same as the one with riskneutral
probabilities.
Solution.
We have
e−γh [pCu + (1 − p)Cd ] = But and 1 erh − d
S∆ − B
u − erh
eαh − erh
Cu +
Cd +
(Cu − Cd ) .
u−d
u−d
S ∆eαh − Berh u − d erh − d
u − erh
Cu +
Cd = erh (S ∆ − B )
u−d
u−d
eαh − erh
(Cu − Cd ) = (eαh − erh )∆S.
u−d This means that in a multiperiodbinomial model, the perperiod discount rate is diﬀerent at each node. 158
Thus OPTION PRICING IN BINOMIAL MODELS erh − d
u − erh
eαh − erh
Cu +
Cd +
(Cu − Cd ) = S ∆eαh − Berh .
u−d
u−d
u−d Hence,
e−γh [pCu + (1 − p)Cu ] = S ∆ − B
which is the same result as the one obtained using riskneutral probabilities
Example 23.5
Given the following information about a 1year European call option on a stock:
• The strike price is $40.
• The current price of the stock is $41.
• The expected rate of return is 15%.
• The stock pays no dividends.
• The continuously compounded riskfree rate is 8%.
• Volatility is 30%.
Use a oneperiod binomial model to compute the price of the call
(a) Using true probabilities on the stock.
(b) Using riskneutral probabilities.
Solution.
We ﬁrst compoute u and d. We have
u = e(r−δ)h+σ
and
d = e(r−δ)h−σ √ √ √ h = e(0.08−0)×1+0.30 h = e(0.08−0)×1−0.30 √ 1 = 1.4623 1 = 0.8025. The number of shares in the replicating portfolio is
∆= [1.4623 × 41 − 40] − 0
Cu − Cd
=
= 0.738.
S ( u − d)
41(1.4623 − 0.8025) The amount of money borrowed is
B = e−rh uCd − dCu
−0.8025 × 19.854
= e−0.08 ×
= $22.405.
d−u
0.8023 − 1.4623 (a) The true probability of the stock going up in value is
p= eαh − d
e0.15 − 0.8025
=
= 0.5446.
u−d
1.4623 − 0.8025 23 RISKNEUTRAL PROBABILITY VERSUS REAL PROBABILITY 159 Now,
S∆
B
eαh −
erh
S∆ − B
S∆ − B
41 × 0.738
22.405
=
e0.15 −
0.738 × 41 − 22.405
0.738 × 41 − 22.405
=1.386 eγh = Thus,
γ = ln 1.386 = 0.3264.
The price of the option is
C = e−γh [pCu + (1 − p)Cd ] = e−0.3264 [0.5446 × 19.954 + (1 − 0.5446) × 0] = $7.839.
(b) The riskneutral probability of a move up state is
pu = e0.08 − 0.8025
e(r−δ)h − d
=
= 0.4256.
u−d
1.4625 − 0.8025 Thus, the price of the call is
C = e−rh [pu Cu + (1 − pu )Cd ] = e−0.08 [0.4256 × 19.954 + (1 − 0.4256) × 0] = $7.839
Remark 23.1
(1) Notice that in order to ﬁnd γ we had to ﬁnd ∆ and B. But then the option price is just
C = S ∆ + B and there is no need for any further computations. It can be helpful to know how to
ﬁnd the actual expected return, but for valuation it is pointless.
(2) In general, expected rate of returns α are hard to estimate, thus by Example 23.4, riskneutral
approach is easier to use in price valuation.
Example 23.6
Consider the following information about a 2year American call on a stock:
• Stock pays dividends at the continuously compounded yield rate of 10%.
• The continuously compounded riskfree rate is 11%.
• The continuously compounded rate of return is 24%.
• Volatility is 40%.
• The current price of the stock is $50.
• The strike price is $40.
Using a twoperiod binomial model, ﬁnd the continuously compounded discount rate at each node. 160 OPTION PRICING IN BINOMIAL MODELS Solution.
We ﬁrst compute u and d. We have
u = e(r−δ)h+σ
and
d = e(r−δ)h−σ √ √ h h √ = e0.11−0.10+0.4 √ = e(0.11−0.10−0.4 1 = 1.5068 1 = 0.6771. The binomial trees are shown below. The riskneutral probability is
pu = e(r−δ)h − d
e(0.11−0.10)×1 − 0.6771
=
= 0.4013.
u−d
1.5068 − 0.6771 The true probability for an upward movement is
p=
Node
Node
Node
Node with
with
with
with Stock
Stock
Stock
Stock price
price
price
price =
=
=
= e(α−δ)h − d
e0.24−0.10 − 0.6771
=
= 0.5703.
u−d
1.5068 − 0.6771
$113.5223 γ = N/A.
$51.0127 γ = N/A.
$22.9232 γ = N/A.
$75.34 We have
e−γ [pCuu + (1 − p)Cud ] = 35.34 23 RISKNEUTRAL PROBABILITY VERSUS REAL PROBABILITY
or
e−γ [0.5703 × 73.5223 + (1 − 0.5703) × 11.0127] = 35.34.
Solving this equation we ﬁnd γ = 0.2779.
Node with Stock price = $33.855 We have
e−γ [pCud + (1 − p)Cdd ] = 3.9590
or
e−γ [0.5703 × 11.0127 + (1 − 0.5703) × 0] = 3.9590.
Solving this equation we ﬁnd γ = 0.4615.
Node with Stock price = $50 We have
e−γ [pCu + (1 − p)Cd ] = 14.8280
or
e−γ [0.5703 × 35.34 + (1 − 0.5703) × 3.9590] = 14.8280.
Solving this equation we ﬁnd γ = 0.3879
Remark 23.2
A discussion of why the riskneutral pricing works is covered in Section 85. 161 162 OPTION PRICING IN BINOMIAL MODELS Practice Problems
Problem 23.1
Consider the following two scenarios:
Scenario 1: You have $1000 which you deposit into a savings account that pays annual continuously
compounded riskfree interest rate of r.
Scenario 2: You purchase a stock for $1000. The stock pays dividends with an annual continuously
compounded yield of 12%. You hold the stock for 13 years, at the end of which it will be either 3.4
or 0.1 of its present price. The riskneutral probability of a stock price increase is 0.82.
Find the balance in the savings account, 13 years from now.
Problem 23.2
Given the following information about a stock:
• After one period, the stock price will either increase by 30% or decrease by 20%.
• The annual compounded continuously expected rate of return is 15%.
• The true probability of increase in the stock price is 0.5446.
Determine the length of the period in years.
Problem 23.3
Given the following information of a 1year European call on a stock:
• Stock does not pay dividends.
• The continuously compounded riskfree rate is 8%.
• Volatility is 30%.
• The current price of the stock is $50.
• The strike price is $48.
• The true probability of an upward movement is 0.46.
Using a oneperiod binomial model, ﬁnd the continuously compounded discount rate γ.
Problem 23.4
Consider the following information about a 1year European put on a stock:
• Stock does not pay dividends.
• The continuously compounded riskfree rate is 8%.
• Volatility is 30%.
• The current price of the stock is $50.
• The strike price is $48.
• The true probability of an upward movement is 0.46.
Using a oneperiod binomial model, ﬁnd the continuously compounded discount rate γ. 23 RISKNEUTRAL PROBABILITY VERSUS REAL PROBABILITY 163 Problem 23.5
Consider the following information about a 1year European put on a stock:
• Stock pays dividends at the continuously compounded yield rate of 4%.
• The continuously compounded riskfree rate is 8%.
• Volatility is 24%.
• The current price of the stock is $62.
• The strike price is $64.
• The continuously compounded expected rate of return is 12%.
Using a oneperiod binomial model, ﬁnd the continuously compounded expected rate of discount γ.
Problem 23.6
Consider the following information about a 1year European call on a stock:
• Stock pays dividends at the continuously compounded yield rate of 2%.
• The continuously compounded riskfree rate is 7%.
• Volatility is 27%.
• The current price of the stock is $38.
• The strike price is $40.
• The continuously compounded discount rate of return is 34.836%.
Using a oneperiod binomial model, ﬁnd
(a) the true probability of a downward movement.
(b) the expected return on the stock α.
Problem 23.7
Consider the following information about a 1year American call on a stock:
• Stock pays dividends at the continuously compounded yield rate of 5%.
• The continuously compounded riskfree rate is 6%.
• The continuously compounded rate of return is 10%.
• Volatility is 30%.
• The current price of the stock is $50.
• The strike price is $47.
Using a twoperiod binomial model, ﬁnd the continuously compounded discount rate after one
upward movement.
Problem 23.8 ‡
For a oneperiod binomial model for the price of a stock, you are given:
(i) The period is one year.
(ii) The stock pays no dividends.
(iii) u = 1.433, where u is one plus the rate of capital gain on the stock if the price goes up. 164 OPTION PRICING IN BINOMIAL MODELS (iv) d = 0.756, where d is one plus the rate of capital loss on the stock if the price goes down.
(v) The continuously compounded annual expected return on the stock is 10%.
Calculate the true probability of the stock price going up.
Problem 23.9
√
For a oneperiod binomial model, the up and down moves are modeled by the equations u = eσ h
√
and d = e−σ h . Given that the period is 6 months, the continuously compounded expected return
on the stock is 15%, the continuous compounded yield is 5%, and the stock price volatility is 30%.
Find the true probability of the stock going up in price. 24 RANDOM WALK AND THE BINOMIAL MODEL 165 24 Random Walk and the Binomial Model
Stock market price movements cannot be predicted since they move randomly. It is conjectured
that stock prices follow a random walk model. In this section we introduce the random walk model
and we show that the binomial model is a variant of a random walk and that it approximates a
lognormal distribution.
Symmetric Random Walk
The onedimensional random walk is constructed as follows: Starting from location 0, you randomly
want to walk along a line either forward or backward, each pace being the same length. To decide
whether you want to move forward or backward, you ﬂip a coin. That is, the walk is simulated with
a coin ﬂip. If it’s a head, you take one step forward( add 1 to your current position). If it’s a tail,
you take one step back(subtract 1 from your current position). The coin is unbiased, so the chances
of heads or tails are equal. The problem is to ﬁnd the probability of landing at a given spot after a
given number of steps, and, in particular, to ﬁnd how far away you are on average from where you
started.
Example 24.1
Suppose that the ﬂip of the coin after 10 steps shows the following outcomes: HT HT T T HHHT.
What is your current position?
Solution.
We have the following chart
Time 1
23
Coin H
TH
Step 1 −1 1
Position 1
01 4
5
T
T
−1 −1
0 −1 6
7 8 9 10
T
HHH
T
−1
1 1 1 −1
−2 −1 0 1
0 Thus, you are at the initial starting point
Consider the case of n steps. For the ith step we deﬁne the random variable Yi by
Yi = +1 if coin displays head
−1 if coin displays tail We deﬁne the cumulative total to be the sum
n Zn = Yi .
i=1 166 OPTION PRICING IN BINOMIAL MODELS So Zn gives your position from the starting point after n steps of the walk. The random walk model
states that the larger the number of steps, the likelier it is that we will be farther away from where
we started (either from the left or the right of the starting point).
Example 24.2
Suppose that Zn = 5. What are the possible values for Zn+1 ?
Solution.
If Zn = 5, then Zn+1 = 4 or 6 with probability of 1
2 each Example 24.3
You ﬂip a coin ﬁve times and you get the following outcomes: Head, Head, Tail, Head, Tail. What
is the Value of Z5 ?
Solution.
We have Y1 = 1, Y2 = 1, Y3 = −1, Y4 = 1, and Y5 = −1. Thus, Z5 = 1 + 1 − 1 + 1 − 1 = 1
How does the random walk model relate to asset price movements? In “eﬃcient markets”, an
asset price should reﬂect all available information. Thus, in response to new information, the asset
price should move up or down with equal probability, as with the coin ﬂip. The asset price after
a period must be the initial price plus the cumulative up and down movements during the period
which are the results of new information. For example, under the oneperiod binomial random walk
model, the asset price after one period is either uS = (1 + g )S or dS = (1 + l) with probability pu
and 1 − pu , respectively.
Modeling Stock Prices as a Random Walk
The random model described above is a bad model for stock price movements. We name three
problems with this model:
(I) Negative stock prices are possible: If by chance we get enough cumulative down movements, the
stock price becomes negative, something that does not happen. A stock price is either positive or
zero (case of a bankrupt ﬁrm).
(II) The magnitude of the move stays constant regardless of the stock price (ﬁxed amount not
proportional to stock price).
(III) The expected return on the stock is always zero. The stock on average should have a positive
return.
The binomial model is a variant of the the random model that solves all of the above problems.
The binomial model assumes that continuously compounded returns1 can be modeled by a random
1 See Section 16 for a discusssion of the continuously compounded returns. 24 RANDOM WALK AND THE BINOMIAL MODEL 167 walk.
Recall that the binomial model for stock pricing is given by
St,t+h = St e(r−δ)h±σ √ h . Taking logs, we obtain
rt,t+h = ln St,t+h
St √
= (r − δ )h ± σ h (24.1) That is, the continuously compounded returns consist of two parts, one of which is certain (r − δ )h,
√
and the other of which is uncertain and generates the up and down stock moves ±σ h. Thus, the
binomial model is a particular way to model the continuously compounded return.
Equation (24.1) solves the three problems mentioned above:
(I) The stock price cannot be negative since the price at the beginning of the period is multiplied
√
by the exponential function e(r−δ)h±σ h > 0.
√
(II) As stock price √
moves occur more frequently, h gets smaller, therefore the up move σ h and
the down move −σ h get smaller. Thus, the moves are proportional to the stock price.
(III) There is a (r − δ )h term so we can choose the probability of an up move, so we can guarantee
that the expected return on the stock is positive.
Lognormality and the Binomial Model
What is a lognormal distribution2 ? In probability theory, a random variable X is said to have the
lognormal distribution, with parameters µ and σ, if ln X has the normal distribution with mean
µ and standard deviation σ. Equivalently,
X = eY
where Y is normally distributed with mean µ and standard deviation σ.
It is known that the continuously compounded returns at the nodes of a binomial tree have a normal
distribution3 . It follows that the stock prices in the binomial model tree approximate a lognormal
distribution. Figure 24.1 compares the probability distribution for a 25period binomial tree with
the corresponding lognormal distribution.
2
3 See Section 47 for a further discussion of lognormal distributions.
See Section 48 168 OPTION PRICING IN BINOMIAL MODELS Figure 24.1
The binomial model implicitly assigns probabilities to the various nodes. Let pu be the riskneutral
probability of an upward movement in the stock’s price. For a binomial tree with three periods we
ﬁnd the following tree of probabilities Example 24.4
Find the probability assigned to the node uud in a threeperiod binomial tree.
Solution.
There are three paths to reach the node: uud, udu, and duu. At each node the probability is
p2 (1 − pu ). Thus, the answer is 3p2 (1 − pu )
u
u
Remark 24.1
We have seen that the binomial model requires that the volatility be constant, large stock price
movements do not suddenly occur, and the periodic stock returns are independent of each other.
These assumptions are not considered realistic. 24 RANDOM WALK AND THE BINOMIAL MODEL 169 Practice Problems
Problem 24.1
Find the expected value and the standard deviation of the random variable Yi .
Problem 24.2
Find Var(Zn ).
Problem 24.3
Show that Zn+1 = Zn + Yn .
Problem 24.4
Let Sn represent the price of the stock on day n with S0 representing the initial stock price. Find
a relation between Sn and Yn .
Problem 24.5
Suppose that the annualized standard deviation of returns on a stock is 0.67. What is the standard
deviation of returns on the same stock after 12 years?
Problem 24.6
The standard deviation of returns on a stock over 10 years is 0.02. The standard deviation of
returns on the same stock over Z years is 0.15. Find Z.
Problem 24.7
The standard deviation of returns on a stock over 10 years is 0.02; the annual continuouslycompounded interest rate is 0.03, and the stock pays dividends at an annual continuouslycompounded
rate of 0.01. The stock price is currently $120/share. If the stock price increases in 10 years, what
will it be?
Problem 24.8
A coin was ﬂipped 13 times, and you know that Z13 = 6, Y12 = −1, Y11 = −1, Y10 = 1, and Y9 = 1.
Find Z8 .
Problem 24.9
Construct a threeperiod binomial tree depicting stock price paths, along with riskneutral probabilities of reaching the various terminal prices. 170 OPTION PRICING IN BINOMIAL MODELS Problem 24.10
For an nperiod binomial tree, the probability for reaching the ith node from the top in the terminal
prices column (i.e. the last column in the tree) is given by the formula
n
pu−i (1 − pu )i n!
.
(n − i)!i! Use a 15period binomial tree to model the price movements of a certain stock. For each time
period, the riskneutral probability of an upward movement in the stock price is 0.54. Find the
probability that stock price will be at the 8th node of the binomial tree at the end of 15 periods. 25 ALTERNATIVE BINOMIAL TREES 171 25 Alternative Binomial Trees
Up to this point, a binomial tree is constructed based on the formulas
u = e(r−δ)h+σ √ h √ d = e(r−δ)h−σ h
e(r−δ)h − d
.
pu =
u−d
In this section, we discuss two more additional ways for constructing binomial trees that approximate a lognormal distribution.
The CoxRossRubinstein Binomial Tree
The CoxRossRubinstein binomial tree is constructed based on the formulas
u = eσ √ h √ d = e−σ h
e(r−δ)h − d
.
pu =
u−d
The CRR approach is often used in practice. However, the approach breaks down for any choice of
√
h and σ such that erh > eσ h . In practice, h is usually small so that such a problem does not occur.
Example 25.1
The current price of a stock is $1230. The price volatility 30%. Use a CRR threeperiod binomial
tree to ﬁnd the price of the stock at the node uud. A period consists of two months.
Solution.
We have
u = eσ √ h √1
= e0.30 6 = 1.1303 and d = 1.1303−1 = 0.8847. Thus, the price of the stock at node uud is u2 dS = 1.13032 × 0.8847 × 1230 = $1390.24
Example 25.2
During 54 periods in a binomial model, the stock price of GS LLC has gone up 33 times and gone
down 21 times at the end of which the price of the stock is $32/share. The price volatility is 0.2,
and one time period in the binomial model is 6 months. Using a CoxRubinstein binomial tree,
calculate the original price of the stock. 172 OPTION PRICING IN BINOMIAL MODELS Solution.
We are asked to ﬁnd S such that u33 d21 S = 32. Since ud = 1, we just need to solve the equation
32
u12 S = 32 or S = u12 = e12×032×√0.5 = $5.86
.2
Example 25.3
You are given the following information about a 1year European call on a stock:
• The current price of the stock is $38.
• The strike price is $40.
• The stock price volatility is 30%
• The continuously compounded riskfree rate is 7%.
• The stock pays no dividends.
Use a oneperiod CRR binomial tree to ﬁnd the current price of the call.
Solution.
We have
u = eσ
and
d = e−σ √ √ √ h = e0.30 h 1 = e−0.30 √ = 1.3498
1 = 0.7408. The riskneutral probability is
e(r−δ)h − d
e0.07 − 0.7408
pu =
=
= 0.5447.
u−d
1.3498 − 0.7408
Now, Cu = 1.3498 × 38 − 40 = 11.2924 and Cd = 0. Thus,
C = e−rh [pu Cu + (1 − pu )Cd ] = e−0.07 × 0.5447 × 11.2924 = $5.735
The Jarrow and Rudd (lognormal) Binomial Tree
This tree is constructed based on the formulas
u = e(r−δ−0.5σ 2 )h+σ 2 √ h √
h d = e(r−δ−0.5σ )h−σ
e(r−δ)h − d
pu =
.
u−d Example 25.4
The stock prices of GS LLC can be modeled via a lognormal tree with 6 years as one time period.
The annual continuously compounded riskfree interest rate is 0.08, and the stock pays dividends
with an annual continuously compounded yield of 0.01. The prepaid forward price volatility is 0.43.
The stock price is currently $454 per share. Find the stock price in 72 years if it goes up 7 times
and down 5 times. 25 ALTERNATIVE BINOMIAL TREES 173 Solution.
√
√
2
2
σ
We want to ﬁnd u7 d5 S. But u = e(r−δ−0.5σ )h+√ h = e(0.08−0.01−0.5×0.43 )×6+0.43 6 = 2.5057 and
√
2
2
d = e(r−δ−0.5σ )h−σ h = e(0.08−0.01−0.5×0.43 )×6−0.43 6 = 0.3048. Thus, u7 d5 S = 2.50577 × 0.30485 ×
454 = $741.19
Example 25.5
You are given the following information about a 1year European call on a stock:
• The current price of the stock is $38.
• The strike price is $40.
• The stock price volatility is 30%
The continuously compounded riskfree rate is 7%.
• The stock pays no dividends.
Use a oneperiod lognormal binomial tree to ﬁnd the current price of the call.
Solution.
We have
u = e(r−δ−0.5σ 2 )h+σ and
d = e(r−δ−0.5σ 2 )h−σ √ √ 2 )+0.30 h = e(0.07−0.5(0.30) h = e(0.07−0.5(0.30) 2 )−0.30 = 1.3840
= 0.7596 The riskneutral probability is
pu = e(r−δ)h − d
e0.07 − 0.7596
=
= 0.5011.
u−d
1.3840 − 0.7596 Now, Cu = 1.3840 × 38 − 40 = 12.592 and Cd = 0. Thus,
C = e−rh [pu Cu + (1 − pu )Cd ] = e−0.07 × 0.5011 × 12.592 = $5.8833 174 OPTION PRICING IN BINOMIAL MODELS Practice Problems
Problem 25.1
During the course of 34 time periods of 1 year each in the binomial model, two stocks that were
identically priced at the beginning diverged in their prices. Stock A went up consistently, while
Stock B went down consistently. The volatility of both stocks’ prices is 0.09. Using CRR binomial
tree, ﬁnd the ratio of the price of Stock A to the price of Stock B.
Problem 25.2
Suppose that, in a CRR tree, you are given h = 1 and u
d = 1.8221. Find σ. Problem 25.3
Consider a 1year European call on a nondividend paying stock. The strike price is $40. The current
price of the stock is $38 and the stock’s price volatility is 30%. Suppose that the true probability
of an upward movement is 0.661. Using a oneperiod CRR binomial model, ﬁnd the expected rate
of return α.
Problem 25.4
You are given the following information about a 1year European put on a stock:
• The current price of the stock is $38.
• The strike price is $40.
• The stock price volatility is 30%
• The continuously compounded riskfree rate is 7%.
• The stock pays no dividends.
Use a oneperiod CRR binomial tree to ﬁnd the current price of the put.
Problem 25.5
Given the following information about a threemonth European call on a stock:
• The current price of the stock is $100.
• The strike price is $95.
• The prepaid forward price volatility is 30%
• The continuously compounded riskfree rate is 8%.
• The stock dividend yield is 5%.
• The threemonth expected discount rate is 21.53%.
Use a oneperiod CRR binomial tree to ﬁnd the true probability for an upward movement.
Problem 25.6
You are given the following information about a 1year European put on a stock:
• The current price of the stock is $38. 25 ALTERNATIVE BINOMIAL TREES 175 • The strike price is $40.
• The stock price volatility is 30%
• The continuously compounded riskfree rate is 7%.
• The stock pays no dividends.
Use a oneperiod lognormal binomial tree to ﬁnd the current price of the put.
Problem 25.7
You are given the following information about a 3month European call on a stock:
• The current price of the stock is $100.
• The strike price is $95.
• The prepaid forward price volatility is 30%
• The continuously compounded riskfree rate is 8%.
• The continuously compounded dividend rate is 5%.
• The expected 3month discount rate is 52.81%
Use a oneperiod lognormal binomial tree to ﬁnd
(a) the true probability of an upward movement
(b) the expected rate of return α.
Problem 25.8
You are given the following information about a 6month European call on a stock:
• The current price of the stock is $58.
• The strike price is $60.
• The prepaid forward price volatility is 27%
• The continuously compounded riskfree rate is 12%.
• The continuously compounded dividend rate is 4.5%.
Use the oneperiod lognormal binomial tree to calculate ∆, the number of shares in the replicating
portfolio.
Problem 25.9
You are given the following information about a 6month European call on a stock:
• The current price of the stock is $100.
• The strike price is $95.
• The stock price volatility is 30%
• The continuously compounded riskfree rate is 8%.
• The stock pays dividends at the continuously compound yield rate 5%.
Use the oneperiod CRR model to ﬁnd the current price of the call.
Problem 25.10
You are given the following information about a 1year European call on a stock: 176 OPTION PRICING IN BINOMIAL MODELS • The current price of the stock is $38.
• The strike price is $40.
• The stock price volatility is 30%
• The continuously compounded riskfree rate is 7%.
• The stock pays no dividends.
Use a oneperiod lognormal binomial tree to ﬁnd the current price of the call. 26 ESTIMATING (HISTORICAL) VOLATILITY 177 26 Estimating (Historical) Volatility
Recall that volatility is deﬁned to be the annualized standard deviation of the continuously compounded stock returns. This parameter cannot usually be observed directly. In this section we
examine a procedure for estimating volatility.
Suppose we are given a periodic data about the price of a stock. As usual, we let St denote the
price of the stock at time t and St+h the price at time t + h. The continuously compounded return
for the interval [t, t + h] is
St+h
.
rt,t+h = ln
St
Next, we describe how to estimate volatility, known as historical volatility. Suppose that we
have data for n periods where each period is of length h. We compute the (historical) stock returns:
rt,t+h , rt+h,t+2h , · · · , rt+(n−1)h,t+nh . We assume that the returns are independent and identically distributed. The next step is to calculate the average of the returns
r= n
i=1 rt+(i−1)h,t+ih .
n
From probability theory, an estimate of the periodic standard deviation is given by the formula
σh = n
i=1 (rt+(i−1)h,t+ih n−1 − r)2 . Volatility, which is the annualized standard deviation, is then found using the formula
σh
σ=√ .
h
Volatility computed from historical stock returns is referred to as historical volatility. We illustrate the above procedure in the next example.
Example 26.1
The table below lists the closing price of a stock for 6 weeks. Estimate the annualized standard
deviation of the stock’s prices.
Date
03/05/03
03/12/03
03/19/03
03/26/03
04/02/03
04/09/03 Price
85
81
87
80
86
93 178 OPTION PRICING IN BINOMIAL MODELS Solution.
The table below shows the historical weekly returns, the average of the returns, the squared deviations, and the sum of the squared deviations.
Date
03/05/03
03/12/03
03/19/03
03/26/03
04/02/03
04/09/03 Price
85
81
87
80
86
93 (rt − r)2 rt = ln St /St−1
−0.048202
0.071459
−0.083881
0.072321
0.078252
r = 0.017990 0.004381
0.002859
0.010378
0.002952
0.003632
5
2
i=1 (ri − r ) = 0.024201 The estimate of the standard deviation of the weekly returns is
σ1 =
52 The historical volatility is 0.024201
= 0.07784.
4
σ1 σ= 52 = 0.561 1
52 Summarizing, the volatility needed for the binomial model can be estimated by computing the
standard deviation of periodic continuously compounded returns and annualizing the result. Once
the annualized standard deviation is found, we can use it to construct binomial trees. We multiply
√
σ by h to adapt the annual standard deviation to any size binomial step. 26 ESTIMATING (HISTORICAL) VOLATILITY 179 Practice Problems
Problem 26.1
Weekly prices of a stock are given from 04/02/03 to 05/28/03. Estimate the value of σ.
Date
04/02/03
04/09/03
04/16/03
04/23/03
04/30/03
05/07/03
05/14/03
05/21/03
05/28/03 Price
77.73
75.18
82.00
81.55
81.46
78.71
82.88
85.75
84.90 Problem 26.2
Monthly prices of a stock are shown in the table below. Estimate the volatility σ.
Month
1
2
3
4
5 Price
100
110
112
105
113 Problem 26.3
The data of a stock’s price for 7 months are given. Estimate the volatility σ.
Month
1
2
3
4
5
6
7 Price
85
81
87
93
102
104
100 Problem 26.4 ‡
You are to estimate a nondividendpaying stock’s annualized volatility using its prices in the past
nine months. 180 OPTION PRICING IN BINOMIAL MODELS
Month
1
2
3
4
5
6
7
8
9 Stock Price per share
80
64
80
64
80
100
80
64
80 Calculate the historical volatility for this stock over the period.
Problem 26.5
Suppose that the stock prices for six weeks were given and the estimate of the standard deviation
of the weekly returns is 0.07784. Find the value of the sum of the squared deviations.
Problem 26.6
Stock prices for n months were given. Suppose that the estimated standard deviation for monthly
returns is 0.059873 and that the sum of the squared deviations is 0.017924. Determine the value of
n.
Problem 26.7
Suppose that the historical volatility is 0.82636 and the periodic standard deviation is 0.23855.
Find the length of a period.
Problem 26.8
Monthly prices of a stock are shown in the table below.
Month
1
2
3
4
5 Price
100
110
112
105
X Suppose that the average of the returns is 0.042185. Determine the value of X.
Problem 26.9
The data of a stock’s price for 7 months are given. 26 ESTIMATING (HISTORICAL) VOLATILITY
Month
1
2
3
4
5
6
7 181
Price
85
81
87
93
102
X
X −4 Suppose that the average of the returns is 0.027086. Determine the value of X. 182 OPTION PRICING IN BINOMIAL MODELS The BlackScholes Model
In this chapter we present the BlackScholes formula for pricing European options and discuss
various topics related to it. 183 184 THE BLACKSCHOLES MODEL 27 The BlackScholes Formulas for European Options
In this section we present the BlackScholes formulas for European calls and puts options. Before
doing that we examine a function that will appear in the formula.
The Cumulative Normal Distribution Function
Let X denote the standard normal random variable, that is, the distribution of X is normal with
mean 0 and standard deviation 1. The probability density function1 of X is given by
x2
1
f ( x) = √ e − 2 .
2π Let N (x) denote the cumulative normal distribution function. That is,
x N (x) = P r(X ≤ x) =
−∞ t2
1
√ e− 2 dt.
2π N (x) is the area under the standard normal curve to the left of x. The SOA asks that you compute
N (x) using the standard normal distribution tables. However, in this book, we will ﬁnd the value
of N (x) using Excel’s NormSdist.
Example 27.1
Show that N (x) + N (−x) = 1.
Solution.
The function N (x) = f (x) is an even function so that N (−x) = N (x) for all real numbers x. Integrating both sides we obtain N (x) = −N (−x) + C. Letting x = 0 we ﬁnd C = 2N (0) = 2(0.5) = 1.
Hence, N (x) + N (−x) = 1
Before examining the BlackScholes formulas, we list the following assumptions that were required
in the derivation of the formulas:
• Continuously compounded returns on the stock are normally distributed and independent over
time.
• The volatility of continuously compounded returns is known and constant.
• Future dividends are known, either as a dollar amount or as a ﬁxed dividend yield.
• The riskfree interest rate is known and constant.
• There are no transaction costs or taxes.
• It is possible to shortsell costlessly and to borrow at the riskfree rate.
1 See Section 46 for a further discussion of normal distributions. 27 THE BLACKSCHOLES FORMULAS FOR EUROPEAN OPTIONS 185 BlackScholes Formula for European Calls
For a European call on a stock that pays continuous dividends, the price is given by
C (St , K, σ, r, T − t, δ ) = St e−δ(T −t) N (d1 ) − Ke−r(T −t) N (d2 )
where
d1 = ln (St /K ) + (r − δ + 0.5σ 2 )(T − t)
√
σ T −t and √
ln (St /K ) + (r − δ − 0.5σ 2 )(T − t)
√
= d1 − σ T − t.
σ T −t
As with the binomial model, the six inputs in the above formula are
• St is the stock price at time t.
• K is the strike price of the option.
• σ is the annual standard deviation of the rate of return on the stock or the prepaid forward price
volatility.
• r is the annual continuously compounded riskfree interest rate.
• T is the time to expiration.
• δ is the annual continuously compounded dividend yield.
d2 = Example 27.2
The stock of XYZ currently sells for $41 per share. The annual stock price volatility is 0.3, and the
annual continuously compounded riskfree interest rate is 0.08. The stock pays no dividends.
(a) Find the values of N (d1 ) and N (d2 ) in the BlackScholes formula for the price of a call option
on the stock with strike price $40 and time to expiration of 3 months.
(b) Find the BlackScholes price of the call option.
Solution.
(a) With t = 0 we have
ln (S/K ) + (r − δ + 0.5σ 2 )T
√
σT 41
32
ln 40 + 0.08 − 0 + 0.2 × 0.25 √
=N 0.3 0.25 N (d1 ) =N 1
=N (0.3729507506) = √
2π
=0.645407 0.3729507506 x2 e− 2 dx
−∞ 186 THE BLACKSCHOLES MODEL and
1
N (d2 ) = N (0.2229507506) = √
2π 0.2229507506 x2 e− 2 dx = 0.588213.
−∞ (b) The price of the call is
C (41, 40, 0.3, 0.08, 0.25, 0) = 41 × e−0×0.25 × 0.645407 − 40e−0.08×0.25 × 0.588213 = $3.399
BlackScholes Formula for European Puts
For a European put on a stock that pays continuous dividends, the price is given by
P (St , K, σ, r, T − t, δ ) = Ke−r(T −t) N (−d2 ) − St e−δ(T −t) N (−d1 )
Proposition 27.1
The BlackScholes formulas for calls and puts satisfy the putcall parity
C (St , K, σ, r, T − t, δ ) − P (St , K, σ, r, T − t, δ ) = St e−δ(T −t) − Ke−r(T −t) .
Proof.
We have
C (St , K, σ, r, T − t, δ ) − P (St , K, σ, r, T − t, δ ) =St e−δ(T −t) N (d1 ) − Ke−r(T −t) N (d2 ) − [Ke−r(T −t) N (−d2 )
−St e−δ(T −t) N (−d1 )]
=St e−δ(T −t) [(N (d1 ) + N (−d1 )]
−Ke−r(T −t) [N (d2 ) + N (−d2 )]
=St e−δ(T −t) − Ke−r(T −t)
Example 27.3
Using the same inputs as in Example 27.2, ﬁnd the BlackScholes price of a put option on the stock
with strike price $40 and time to expiration of 3 months.
Solution.
Using the putcall parity we ﬁnd
P (41, 40, 0.3, 0.08, 0.25, 0) = 3.373 + 40e−0.08×0.25 − 41e−0×0.25 = $1.581
Example 27.4 ‡
You are asked to determine the price of a European put option on a stock. Assuming the BlackScholes framework holds, you are given: 27 THE BLACKSCHOLES FORMULAS FOR EUROPEAN OPTIONS 187 (i) The stock price is $100.
(ii) The put option will expire in 6 months.
(iii) The strike price is $98.
(iv) The continuously compounded riskfree interest rate is r = 0.055.
(v) δ = 0.01
(vi) σ = 0.50
Calculate the price of this put option.
Solution.
With t = 0 we have
ln (S/K ) + (r − δ + 0.5σ 2 )T
√
σT 00
ln 198 + 0.055 − 0.01 +
√
=N −
0.5 0.5 N (−d1 ) =N 1
=N (−0.297558191) = √
2π
=0.38302 0.52
2 × 0.5 −0.297558191 x2 e− 2 dx
−∞ and
√ 1
N (−d2 ) = N (−d1 + σ T ) = N (0.0559951996) = √
2π 0.0559951996 x2 e− 2 dx = 0.522327.
−∞ The price of the put is
P (100, 98, 0.5, 0.055, 0.5, 0.01) = 98 × e−0.055×0.5 × 0.522327 − 100e−0.01×0.5 × 0.38302 = $11.688 188 THE BLACKSCHOLES MODEL Practice Problems
Problem 27.1
A European call option on XYZ stock has the following speciﬁcations: Strike price = $45, current
stock price = $46, time to expiration = 3 months, annual continuously compounded interest rate
= 0.08, dividend yield = 0.02, volatility=0.35. Calculate the BlackScholes price of the option.
Problem 27.2
Using the same inputs as the previous problem, calculate the BlackScholes price of a put option.
Problem 27.3
The stock of GS Co. currently sells for $1500 per share. The prepaid forward price volatility is
0.2, and the annual continuously compounded riskfree interest rate is 0.05. The stock’s annual
continuously compounded dividend yield is 0.03. Within the BlackScholes formula for the price of
a put option on GS Co. stock with strike price $1600 and time to expiration of 3 years, ﬁnd the
value of N (−d2 ).
Problem 27.4
You are asked to determine the price of a European call option on a stock. Assuming the BlackScholes framework holds, you are given:
(i) The stock price is $40.
(ii) The put option will expire in one year.
(iii) The strike price is $35.
(iv) The continuously compounded riskfree interest rate is r = 0.10.
(v) δ = 0.02
(vi) σ = 0.30
Calculate the price of this call option.
Problem 27.5
Show that d1 = ln S e−δT
Ke−rT
√ +0.5σ 2 T σT . Problem 27.6 ‡
Assume the BlackScholes framework.
Eight months ago, an investor borrowed money at the riskfree interest rate to purchase a oneyear
75strike European call option on a nondividendpaying stock. At that time, the price of the call
option was 8.
Today, the stock price is 85. You are given:
(i) The continuously compounded riskfree rate interest rate is 5%.
(ii) The stock’s volatility is 26%.
Find the current price of the call. 27 THE BLACKSCHOLES FORMULAS FOR EUROPEAN OPTIONS 189 Problem 27.7 ‡
For a European call option on a stock within the BlackScholes framework, you are given:
(i) The stock price is $85.
(ii) The strike price is $80.
(iii) The call option will expire in one year.
(iv) The continuously compound riskfree interest rate is 5.5%.
(v) σ = 0.50.
(vi) The stock pays no dividends.
Calculate the price of the call.
Problem 27.8 ‡
For a dividendpaying stock and a European option on the stock, you are given the following
information:
• The current stock price is $58.96.
• The strike price of the option is $60.00.
• The expected annual return on the stock is 10%.
• The volatility is 20%.
• The continuously compounded riskfree rate is 6%.
• The continuously dividend yield is 5%.
• The expiration time is three months.
Calculate the price of the call.
Problem 27.9 ‡
For a nondividendpaying stock and a European option on the stock, you are given the following
information:
• The current stock price is $9.67.
• The strike price of the option is $8.75.
• The volatility is 40%.
• The continuously compounded riskfree rate is 8%.
• The expiration time is three months.
Calculate the price of the put.
Problem 27.10 ‡
Assume the BlackScholes framework. Consider a 9month atthemoney European put option on
a futures contract. You are given:
(i) The continuously compounded riskfree interest rate is 10%.
(ii) The strike price of the option is 20.
(iii) The price of the put option is 1.625.
If three months later the futures price is 17.7, what is the price of the put option at that time? 190 THE BLACKSCHOLES MODEL 28 Applying the BlackScholes Formula To Other Assets
In the previous section the options under consideration had stocks with continuous dividends as
underlying assets. In this section we want to adapt the BlackScholes formula to options with underlying assets consisting of stocks with discrete dividends, futures contracts, and currency contracts.
According to Problem 27.5, we can write
ln S e−δT
Ke−rT + 0.5σ 2 T √
σT d1 = . But the term Se−δT is the prepaid forward price for the stock and Ke−rT is the prepaid forward
price for the strike. Thus,
ln
d1 = P
F0,T (S )
P
F0,T (K ) + 0.5σ 2 T √
σT . Now, the BlackScholes formulas can be written in terms of prepaid forward prices: for a call we
have
P
P
P
P
C (F0,T (S ), F0,T (K ), σ, T ) = F0,T (S )N (d1 ) − F0,T (K )N (d2 )
and for a put we have
P
P
P
K
P (F0,T (S ), F0,T (K ), σ, T ) = F0,T (S )N (−d2 ) − F0,T (K )N (−d1 ). These formulas when written in terms of prepaid forward prices are useful when pricing options
with underlying assets other than stocks with continuous dividends, namely, stocks with discrete
dividends, futures, or currencies.
Example 28.1
Show that
P
P
P
P
P
P
P (F0,T (S ), F0,T (K ), σ, T ) = C (F0,T (S ), F0,T (K ), σ, T ) + F0,T (K ) − F0,T (S ). Solution.
By the putcall parity we have
P
P
P
P
P
P
C (F0,T (S ), F0,T (K ), σ, T ) − P (F0,T (S ), F0,T (K ), σ, T ) = Se−δT − Ke−rT = F0,T (S ) − F0,T (K ).
P
P
Now the result follows by solving this equation for P (F0,T (S ), F0,T (K ), σ, T ) 28 APPLYING THE BLACKSCHOLES FORMULA TO OTHER ASSETS 191 Example 28.2 ‡
Let S (t) denote the price at time t of a stock that pays no dividends. The BlackScholes framework
holds. Consider a European call option with exercise date T, T > 0, and exercise price S (0)erT ,
where r is the continuously compounded riskfree interest rate. You are given:
(i) S (0) = $100
(ii) T = 10
(iii) Var[ln S (t)] = 0.4t, t > 0.
Determine the price of the call option.
Solution.
The variance over the interval [0, t] is given by
2
σt = Var[ln (St /S0 )] = Var[ln St ] − Var[ln 100] = Var[ln St ] = 0.4t. Thus,
σ = σ1 = √ 0.4. We also have
ln
d1 =
and P
F0,T (S )
P
F0,T (K ) + 0.5σ 2 T √ σT = ln (100/100) + 0.5 × 0.4 × 10
=1
2 √
d2 = d1 − σ T = 1 2 = −1. Hence,
1
N (d1 ) = √
2π
and
1
N (d2 ) = √
2π 1 x2 e− 2 dx = 0.841345
−∞
−1 x2 e− 2 dx = 0.158655.
−∞ The BlackScholes price of the call option is
P
P
C = F0,T (S )N (d1 ) − F0,T (K )N (d2 ) = 100 × 0.841345 − 100 × 0.158655 = $68.269 • If the underlying asset is a stock with discrete dividends, then the prepaid forward price is
P
F0,T (S ) = S0 − P V0,T (Div). 192 THE BLACKSCHOLES MODEL Example 28.3
Consider a stock that pays dividends of $40 in two years and $32 in six years. The stock currently
trades for $221 per share. The annual continuously compounded riskfree interest rate is 5%, and
the annual price volatility relevant for the BlackScholes equation is 30%. Find the BlackScholes
price of a call option with strike price $250 and expiration time of 8 years.
Solution.
The prepaid forward price of the stock is
P
F0,T (S ) = S0 − P V0,T (Div) = 221 − 40e−0.05×2 − 32e−0.05×6 = $161.1003. The prepaid forward on the strike is
P
F0,T (K ) = Ke−rT = 250e−0.05×8 = $167.58. The values of d1 and d2 are
ln
d1 =
and P
F0,T (S )
P
F0,T (K ) + 0.5σ 2 T √ σT = ln (161.1003/167.58) + 0.5 × 0.32 × 8
√
= 0.3778
0.3 8 √
√
d2 = d1 − σ T = 0.3778 − 0.3 8 = −0.4707. Thus,
1
N (d1 ) = √
2π
and
1
N (d2 ) = √
2π
The BlackScholes price of the call option is 0.3778 x2 e− 2 dx = 0.6472
−∞
−0.4707 x2 e− 2 dx = 0.3189.
−∞ P
P
P
P
C (F0,T (S ), F0,T (K ), σ, T ) =F0,T (S )N (d1 ) − F0,T (K )N (d2 ) =161.1003 × 0.6472 − 167.58 × 0.3189 = $50.8261
P
• If the underlying asset is a foreign currency, then F0,T (x) = x0 e−rf T where rf is the foreign
currency interest rate and x0 is the current exchange rate (expressed as domestic currency per unit
of foreign currency). In this case, the BlackScholes price of a call option is given by C (x0 , K, σ, r, T, rf ) = x0 e−rf T N (d1 ) − Ke−rT N (d2 ) 28 APPLYING THE BLACKSCHOLES FORMULA TO OTHER ASSETS
where
d1 =
and 193 ln (x0 /K ) + (r − rf + 0.5σ 2 )T
√
σT
√
d2 = d1 − σ T . The above formula is also known as the GarmanKohlhagan formula.
The BlackScholes for a European currency put option is given by
P (x0 , K, σ, r, T, rf ) = Ke−rT N (−d2 ) − x0 e−rf T N (−d1 ).
We can also get the put formula via putcall parity:
P (x0 , K, σ, r, T, rf ) = C (x0 , K, σ, r, T, rf ) + Ke−rT − x0 e−rf T .
The options prices are in the domestic currency.
Example 28.4
One euro is currently trading for $0.92. The dollardenominated continuously compounded interest
rate is 6% and the eurodenominated continuously compounded interest rate is 3.2%. Volatility is
10%.
(a) Find the BlackScholes price of a 1year dollardenominated euro call with strike price of $0.9/e.
(b) Find the BlackScholes price of a 1year dollardenominated euro put with strike price of $0.9/e.
Solution.
We ﬁrst ﬁnd d1 and d2 . We have
d1 =
and ln (x0 /K ) + (r − rf + 0.5σ 2 )T
ln (0.92/0.9) + (0.06 − 0.032 + 0.5 × 0.102 ) × 1
√
√
=
= 0.549789
0.10 1
σT
√
d2 = d1 − σ T = 0.549789 − 0.10 = 0.449789. Thus,
1
N (d1 ) = √
2π
and
1
N (d2 ) = √
2π 0.549789 x2 e− 2 dx = 0.708768
−∞
0.449789 x2 e− 2 dx = 0.673569.
−∞ 194 THE BLACKSCHOLES MODEL (a) The BlackScholes price of the call is
C = 0.92e−0.032 × 0.708768 − 0.9e−0.06 × 0.673569 = $0.0606.
(b) The BlackScholes price of the put is
P = C + Ke−rT − x0 e−rf T = 0.0606 + 0.9e−0.06 − 0.92e−0.032 = $0.0172
• The prepaid forward price for a futures contract is just the present value of the futures price.
P
Letting F denote the futures price, we have F0,T (F ) = F e−rT . The BlackScholes formula for a call,
also known as the Black formula is given by
C (F, K, σ, r, T, r) = F e−rT N (d1 ) − Ke−rT N (d2 )
where ln (F/K ) + 0.5σ 2 T
√
d1 =
σT and √
d2 = d1 − σ T . The put price is given by
P (F, K, σ, r, T, r) = Ke−rT N (−d2 ) − F e−rT N (−d1 )
which can be found via the putcall parity:
P (F, K, σ, r, T, r) = C (F, K, σ, r, T, r) + Ke−rT − F e−rT .
Example 28.5
Futures contracts on natural gas currently trade for $2.10 per MMBtu. The annual futures contract
price volatility is 0.25, and the annual continuously compounded currency riskfree interest rate is
0.055.
(a) Find the BlackScholes price of 1year European call on natural gas futures contracts with strike
price of $2.10.
(b) Find the BlackScholes price of 1year European put on natural gas futures contracts with strike
price of $2.10.
Solution.
We ﬁrst ﬁnd d1 and d2 . We have
d1 = ln (2.10/2.10) + 0.5 × 0.252 × 1
ln (F/K ) + 0.5σ 2 T
√
√
=
= 0.125
0.25 1
σT 28 APPLYING THE BLACKSCHOLES FORMULA TO OTHER ASSETS
and 195 √
d2 = d1 − σ T = 0.125 − 0.25 = −0.125. Thus,
1
N (d1 ) = √
2π
and
1
N (d2 ) = √
2π 0.125 x2 e− 2 dx = 0.549738
−∞
−0.125 x2 e− 2 dx = 0.450262
−∞ (a) The price of the call is
C = F e−rT N (d1 ) − Ke−rT N (d2 ) = 2.10e−0.055 × 0.549738 − 2.10e−0.055 × 0.450262 = $0.19772.
(b) The price of the put is
P = C + Ke−rT − F e−rT = C = $0.19772 196 THE BLACKSCHOLES MODEL Practice Problems
Problem 28.1 ‡
Which of the following is an assumption of the BlackScholes option pricing model?
(A) Stock prices are normally distributed
(B) Stock price volatility is a constant
(C) Changes in stock price are lognormally distributed
(D) All transaction cost are included in stock returns
(E) The riskfree interest rate is a random variable.
Problem 28.2
Consider a stock that pays dividends of $5 one month from now. The stock currently trades for
$46 per share. The annual continuously compounded riskfree interest rate is 8%, and the annual
price volatility relevant for the BlackScholes equation is 39.24%. Find the BlackScholes price of a
call option with strike price $45 and expiration time of three months.
Problem 28.3
Consider a stock that pays dividends of $40 in two years and $32 in six years. The stock currently
trades for $221 per share. The annual continuously compounded riskfree interest rate is 5%, and
the annual price volatility relevant for the BlackScholes equation is 30%. Find the BlackScholes
price of a put option with strike price $250 and expiration time of 8 years.
Problem 28.4
You are given the following information about a call option on a stock in the BlackScholes framework:
• The annual continuouslycompounded interest rate is 0.03.
• The annual price volatility is 0.03
• The current stock price is $56
• The option’s time to expiration is 2 years
• The price of the prepaid forward on the strike asset is $42
• d1 = 0.4
• The stock will pay a dividend in the amount of d dollars one year from today.
Find the size of the dividend.
Problem 28.5
Consider a stock that pays dividends of $7 in 3 months and $10 in 9 months. The stock currently
trades for $77 per share. The annual continuously compounded riskfree interest rate is 10%, and
the annual price volatility relevant for the BlackScholes equation is 27.43%. Find the BlackScholes
price of a put option with strike price $73 and expiration time of 6 months. 28 APPLYING THE BLACKSCHOLES FORMULA TO OTHER ASSETS 197 Problem 28.6
One euro is currently trading for $1.25. The dollardenominated continuously compounded interest
rate is 6% and the eurodenominated continuously compounded interest rate is 3.5%. Volatility
relevant for the BlackScholes equation is 11%.
(a) Find the BlackScholes price of a 6month dollardenominated euro call with strike price of
$1.3/e.
(b) Find the BlackScholes price of a 6month dollardenominated euro put with strike price of
$1.3/e.
Problem 28.7
One euro is currently trading for $1.25. The dollardenominated continuously compounded interest
rate is 8% and the eurodenominated continuously compounded interest rate is 5%. Volatility
relevant for the BlackScholes equation is 10%. Find the BlackScholes price of an atthemoney
1year eurodenominated dollar put.
Problem 28.8
Futures contracts on superwidgets currently trade for $444 per superwidget. The annual futures
contract price volatility relevant for the BlackScholes equation is 0.15, and the annual continuously
compounded currency riskfree interest rate is 0.03.
(a) Find the BlackScholes price of 2year European call on superwidget futures contracts with
strike price of $454.
(b) Find the BlackScholes price of 2year European put on superwidget futures contracts with
strike price of $454.
Problem 28.9
The dividend yield in the BlackScholes formula for stock option pricing is analogous to which of
these variables in other related formulas?
(A) The riskfree interest rate in the BlackScholes formula for stock option pricing.
(B) The riskfree interest rate in the Black formula for futures contract option pricing.
(C) The domestic riskfree interest rate in the GarmanKohlhagen formula for currency option
pricing.
(D) The foreign riskfree interest rate in the GarmanKohlhagen formula for currency option pricing.
(E) The volatility in the Black formula for futures contract option pricing.
Problem 28.10
A stock XYZ pays dividends at the continuously compounded rate of 5%. Currently the stock is
trading for $70. The continuously compounded riskfree interest rate is 9%. The volatility relevant
for the BlackScholes equation is 30%. Find the BlackScholes of a European call on futures contracts
on XYZ stock with strike price $65 and expiration of six months. 198 THE BLACKSCHOLES MODEL Problem 28.11
A stock XYZ pays no dividends. Currently the stock is trading for $100. The continuously compounded riskfree interest rate is 7%. The volatility relevant for the BlackScholes equation is 35%.
Find the BlackScholes of a European put on futures contracts on XYZ stock with strike price $105
and expiration of one year.
Problem 28.12 ‡
On January 1, 2007, the following currency information is given:
• Spot exchange rate: $0.82/e
• Dollar interest rate= 5% compounded continuously
• Euro interest rate = 2.5% compounded continuously
• Exchange rate volatility relevant for the BlackScholes equation = 0.10.
What is the price of 850 dollardenominated euro call options with a strike exchange rate $0.80/ethat
expire on January 1, 2008?
Problem 28.13 ‡
You are considering the purchase of 100 European call options on a stock, which pays dividends
continuously at a rate proportional to its price. Assume that the Black Scholes framework holds.
You are given:
(i) The strike price is $25.
(ii) The options expire in 3 months.
(iii) δ = 0.03.
(iv) The stock is currently selling for $20.
(v) The volatility relevant for the BlackScholes equation σ = 0.24.
(vi) The continuously compounded riskfree interest rate is 5%.
Calculate the price of the block of 100 options.
Problem 28.14 ‡
For a sixmonth European put option on a stock, you are given:
(i) The strike price is $50.00.
(ii) The current stock price is $50.00.
(iii) The only dividend during this time period is $1.50 to be paid in four months.
(iv) σS = 0.30
(v) The continuously compounded riskfree interest rate is 5%.
Under the BlackScholes framework, calculate the price of the put option.
Problem 28.15 ‡
Consider a oneyear 45strike European put option on a stock S. You are given: 28 APPLYING THE BLACKSCHOLES FORMULA TO OTHER ASSETS 199 (i) The current stock price, S (0), is 50.00.
(ii) The only dividend is 5.00 to be paid in nine months.
P
2
(iii) σh = Var[ln Fh,1 (S )] = 0.01h, 0 ≤ h ≤ 1.
(iv) The continuously compounded riskfree interest rate is 12%.
Under the BlackScholes framework, calculate the price of 100 units of the put option.
Problem 28.16 ‡
Company A is a U.S. international company, and Company B is a Japanese local company. Company A is negotiating with Company B to sell its operation in Tokyo to Company B. The deal will
be settled in Japanese yen. To avoid a loss at the time when the deal is closed due to a sudden devaluation of yen relative to dollar, Company A has decided to buy atthemoney dollardenominated
yen put of the European type to hedge this risk.
You are given the following information:
(i) The deal will be closed 3 months from now.
(ii) The sale price of the Tokyo operation has been settled at 120 billion Japanese yen.
(iii) The continuously compounded riskfree interest rate in the U.S. is 3.5%.
(iv) The continuously compounded riskfree interest rate in Japan is 1.5%.
(v) The current exchange rate is 1 U.S. dollar = 120 Japanese yen.
(vi) The yen per dollar exchange rate and the dollar per yen exchange rate have the same daily
volatility 0.261712%.
(vii) 1 year = 365 days; 3 months = 1 year.
4
Assuming the BlackScholes pricing framework, calculate Company A’s option cost. 200 THE BLACKSCHOLES MODEL 29 Option Greeks: Delta, Gamma, and Vega
In mathematical ﬁnance, the Greeks are the quantities representing the sensitivities of derivatives
such as options to a change in underlying parameters on which the value of an instrument or portfolio of ﬁnancial instruments is dependent. The name is used because the most common of these
sensitivities are often denoted by Greek letters.
In practice, option values change on account of a number of factors: movements in the price of the
underlying asset, passage of time, changes in volatility of the underlying asset, changes in the rate
of interest, and changes in the dividend yield if the underlying asset pays dividends. There are
formulas that measure the changes in the option price when only one parameter is being changed
while leaving the remaining parameters ﬁxed. Letters from the Greek alphabet are used to represent
these derived measures:
• Delta(∆) measures the change in the option price when the stock price increases by $1.
• Gamma (Γ) measures the change in delta when the stock price increases by $1.
• Vega (V) measures the change in the option price for an increase by 1% in volatility.
• Theta(θ) measures the change in the option price when time to maturity decreases by 1 day.
• Rho (ρ) measures the change in the option price when the riskfree interest rate increases by 1%.
• Psi (ψ ) measures the change in the option price when the dividend yield increases by 1%.
Even though greek measures can be camputed for options with any kind of underlying asset, we
will focus our attention on stock options. We will be examining each greek measure in turn, for a
purchased option. The Greek for a written option is opposite in sign to that for the same purchased
option. In this section we will examine the ﬁrst three Greek measures: Delta, gamma, and vega.
In what follows, the BlackScholes value of a European call option at time t is given by the formula
Ct = Se−δ(T −t) N (d1 ) − Ke−r(T −δ) N (d2 ).
For a put option the formula is
Pt = Ke−r(T −t) N (−d2 ) − Se−δ(T −δ) N (−d1 )
where
d1 =
and ln (S/K ) + (r − δ + 0.5σ 2 )(T − t)
√
σ T −t
√
d2 = d1 − σ T − t. The Delta Measure
The option Greek Delta (∆) measures the option price change when the stock price increases by 29 OPTION GREEKS: DELTA, GAMMA, AND VEGA 201 $1. More formally, the delta of an option is deﬁned as the rate of change of the option value with
respect to stock price:
∂V
∆=
∂S
Proposition 29.1
We have
∆Call = e−δ(T −t) N (d1 )
and
∆Put = −e−δ(T −t) N (−d1 )
Proof.
Using Problem 29.1 we ﬁnd
∂N (d1 )
∂N (d2 )
∂Ct
= e−δ(T −t) N (d1 ) + Se−δ(T −t)
− Ke−r(T −t)
∂S
∂S
∂S
∂N (d1 ) ∂d1
∂N (d2 ) ∂d2
=e−δ(T −t) N (d1 ) + Se−δ(T −t)
− Ke−r(T −t)
∂d1 ∂S
∂d2 ∂S
d2
d2
1
S e(r−δ)(T −t)
1
1
1
1
√
·
=e−δ(T −t) N (d1 ) + e−δ(T −t) · √ e− 2 · √
− Ke−r(T −t) · √ e− 2 ·
K Sσ T − t
σ T −t
2π
2π
=e−δ(T −t) N (d1 ) ∆Call = A similar argument for the put
According to the previous result, the delta of a call option is positive. Thus, an increase in stock
price increases the option value. On the contrary, the delta of a put option is negative. Hence, an
increase in the stock price decreases the value of the option.
The BlackScholes formula tells us that a long call option on the stock can be replicated by buying
e−δ(T −t) N (d1 ) shares of the stock and borrowing Ke−r(T −t) N (d2 ). For a put option, the replicating
portfolio consists of selling e−δ(T −t) N (−d1 ) shares and lending Ke−r(T −t) N (−d2 ).
Next, we look at the range of values of delta:
• An option that is inthemoney will be more sensitive to price changes than an option that is
outofthemoney. For an option inthemoney with a stock price high relative to the strike price
(i.e. deepinthe money), the option is most likely to be exercised. In this case, the option exhibits
the price sensitivity of one full share. That is, ∆ approaches 1. If the option is outofthe money,
it is unlikely to be exercised and the option has a low price, behaving like a position with almost
no shares. That is, ∆ approaches 0. For an atthemoney option, the option may or may not be
exercised leading to 0 < ∆ < 1. 202 THE BLACKSCHOLES MODEL • As time of expiration increases, delta is less at high stock prices and greater at low stock prices.
See Figure 29.1.1 Indeed, for a greater time to expiration, the likelihood is greater for an outofthe
money option to become inthemoney option, and the likelihood is greater that an inthemoney
option to become outofthe money. Figure 29.1
Example 29.1
The BlackScholes price for a certain call option on GS stock is $50. The stock currently trades for
$1000 per share, and it is known that $452 must be borrowed in the replicating portfolio for this
option. Find the delta of the option.
Solution.
We have C = S ∆ − B with B > 0. Substituting we ﬁnd 50 = 1000∆ − 452. Solving this equation
we ﬁnd ∆ = 0.502
The Gamma Measure
The option Greek gamma (Γ) measures the change in delta when the stock price increases by $1.
More formally, the option Greek gamma is deﬁned as the rate of change in the delta value with
respect to stock price:
∂∆
∂ 2V
Γ=
=
.
∂S
∂S 2
1 See [1]. 29 OPTION GREEKS: DELTA, GAMMA, AND VEGA
Proposition 29.2
We have
ΓCall = 203 e−δ(T −t)
√
N (d1 )
Sσ T − t and
ΓPut = ΓCall
where
N (d1 ) = d2
∂N (d1 )
1
1
= √ e− 2 > 0.
∂d1
2π Proof.
We show the ﬁrst part. The second follows from the putcall parity. We have
∂C ΓCall = ∂S
∂S
∂ (e−δ(T −t) N (d1 ))
=
∂S
∂N (d1 ) ∂d1
=e−δ(T −t)
∂d1 ∂S
−δ (T −t)
e
∂N (d1 )
=√
Sσ T − t ∂d1
It follows from the above result that Γ > 0 for a purchased option. Thus, the call and put option
prices are convex functions. Also, delta increases as the stock price increases. Hence, for a call,
delta (∆ > 0) approaches 1 as the stock price increases. For a put, delta (∆ < 0) approaches 0 as
the stock price increases.
Recall that ∆ is close to 1 for deep inthemoney options. Thus, ∆ cannot change much as the
stock price increases. Therefore, Γ is close to 0. Similarly, for deep outofthe money options, Γ is
close to zero.
Example 29.2
Suppose that Γ = 0.02 and ∆ = 0.5. What is the new value of ∆ if the stock price increases by $3?
Solution.
The new value of ∆ is 0.05 + 0.02 × 3 = 0.56
The Vega Measure
Vega (the only greek that isn’t represented by a real Greek letter) measures the change in the option
price for an increase by 1% in volatility. More formally, it measures the change in the option price
to changes in volatility: 204 THE BLACKSCHOLES MODEL
V= Proposition 29.3
We have ∂Ct
∂σ or V = ∂Pt
∂σ √
VCall = Se−δ(T −t) T − tN (d1 ) > 0 and
VPut = VCall .
Proof.
We have
V= ∂N (d1 )
∂N (d2 )
∂N (d1 ) ∂d1
∂N (d2 ) ∂d2
∂Ct
= Se−δ(T −t)
− Ke−r(T −t)
= Se−δ(T −t)
− Ke−r(T −t)
∂σ
∂σ
∂σ
∂d1 ∂σ
∂d2 ∂σ
−δ (T −t) =Se d2
1
1
√ e− 2
2π −r(T −t) −Ke −δ (T −t) =Se 3 d2 S
1
1
√ e− 2 e(r−δ)(T −t)
K
2π d2
1
1
√ e− 2
2π 1 σ 2 (T − t) 2 − ln (S/K ) + (r − δ + 0.5σ 2 )(T − t) (T − t) 2
σ 2 (T − t) 3 σ 2 (T − t) 2
σ 2 (T − t) 1 − ln (S/K ) + (r − δ + 0.5σ 2 )(T − t) (T − t) 2
σ 2 (T − t)
√
= Se−δ(T −t) T − tN (d1 ) The second result follows from the putcall parity
It follows that higher volatility means higher option prices. Figure 29.2 shows that 40strike call
vegas tends to be greater for atthemoney options, and greater for options with moderate than
with short times to expiration. The opposite is true for very longlived options. Figure 29.2 29 OPTION GREEKS: DELTA, GAMMA, AND VEGA 205 Remark 29.1
It is common to report vega as the change in the option price per percentage point change in
volatility. This requires dividing the vega formula above by 100. We will follow this practice in the
problems.
Example 29.3
The price of a call option on XYZ is currently $2.00. Suppose that the vega is 0.20 with the (prepaid
forward) volatility of XYZ at 30%.
(a) If the volatility of XYZ rises to 31%, what will the price of the call option be?
(b) If the volatility of XYZ falls to 29%, what will the price of the call option be?
Solution.
(a) The value of the XYZ call will rise to $2.20.
(b) The value of the XYZ call will drop to $1.80 206 THE BLACKSCHOLES MODEL Practice Problems
Problem 29.1
Show that d2
1
∂N (d2 )
S
1
= √ e− 2 · e(r−δ)(T −t) .
∂d2
K
2π Problem 29.2
Show that ∆Put = ∆Call − e−δ(T −t) .
Problem 29.3
Consider a call option on an a nondividend paying stock. Suppose that for ∆ = 0.4 the option is
trading for $33 an option. What is the new price of the option if the stock price increases by $2?
Problem 29.4
A certain stock is currently trading for $95 per share. The annual continuously compounded riskfree interest rate is 6%, and the stock pays dividends with an annual continuously compounded
yield of 3%. The price volatility relevant for the BlackScholes formula is 32%.
(a) Find the delta of a call option on the stock with strike price of $101 and time to expiration of
3 years.
(b) Find the delta of a put option on the stock with strike price of $101 and time to expiration of
3 years.
Problem 29.5
For otherwise equivalent call options on a particular stock, for which of these values of strike price
(K) and time to expiration (T) would you expect delta to be the highest? The stock price at both
T = 0.3 and T = 0.2 is $50.
(A) K = $43, T = 0.3
(B) K = $43, T = 0.2
(C) K = $55, T = 0.3
(D) K = $55, T = 0.2
(E) K = $50, T = 0.3
(F) K = $50, T = 0.2
Problem 29.6
A certain stock is currently trading for $86 per share. The annual continuously compounded riskfree interest rate is 9.5%, and the stock pays dividends with an annual continuously compounded
yield of 3%. The price volatility relevant for the BlackScholes formula is 35%. Find the delta of a
put option on the stock with strike price of $90 and time to expiration of 9 months. 29 OPTION GREEKS: DELTA, GAMMA, AND VEGA 207 Problem 29.7
A stock currently trades for $60 per share. For which of these otherwise equivalent options and
strike prices (K) is the gamma the highest?
(A) Call, K = 2
(B) Put, K = 20
(C) Call, K = 45
(D) Put, K = 61
(E) Call, K = 98
(F) Put, K = 102
Problem 29.8
A call option on XYZ stock has a delta of 0.45, and a put option on XYZ stock with same strike
and date to expiration has a delta of −0.55. The stock is currently trading for $48.00. The gamma
for both the call and put is 0.07.
(a) What is the value of ∆ for the call and the put if the price of the stock moves up $1?
(b) What is the value of ∆ for the call and the put if the price of the stock drops $1?
Problem 29.9
A stock has a price of $567 and a volatility of 0.45. A certain put option on the stock has a price
of $78 and a vega of 0.23. Suddenly, volatility increases to 0.51. Find the new put option price.
Problem 29.10
The stock of GS Inc., has a price of $567. For which of these strike prices (K) and times to
expiration (T, in years) is the vega for one of these otherwise equivalent call options most likely to
be the highest?
(A) K = 564, T = 0.2
(B) K = 564, T = 1
(C) K = 564, T = 30
(D) K = 598, T = 0.2
(E) K = 598, T = 1
(F) K = 598, T = 30
Problem 29.11
A stock is currently selling for $40. The stock pays no dividends. Given that the volatility of the
stock relevant for the BlackScholes equation is 30% and the continuously compounded riskfree
interest rate is 8%. Consider a $45strike purchased call on the stock with time to expiration in 6
months. What are the delta, gamma, and vega? 208 THE BLACKSCHOLES MODEL Problem 29.12 ‡
You are considering the purchase of a 3month 41.5strike American call option on a nondividendpaying stock. You are given:
(i) The BlackScholes framework holds.
(ii) The stock is currently selling for 40.
(iii) The stock’s volatility is 30%.
(iv) The current call option delta is 0.5.
Which of these expressions represents the price of this option?
x2
0.15
(A) 20 − 20.453 −∞ e− 2 dx
0.15 − x2
e 2 dx
−∞
0.15 − x2
20 − 40.453 −∞ e 2 dx
x2
0.15
16.138 −∞ e− 2 dx − 20.453
x2
0.15
40.453 −∞ e− 2 dx − 20.453. (B) 20 − 16.138
(C)
(D)
(E) 30 OPTION GREEKS: THETA, RHO, AND PSI 209 30 Option Greeks: Theta, Rho, and Psi
In this section we examine the three remaining option Greeks: Theta, Rho and Psi.
The Theta Measure
The option Greek theta (θ) measures the change in the option price when there is a decrease in
the time to maturity of 1 day (also called time decay). More formally, it is deﬁned as the rate of
change of the option price with respect to the passage of time. We can write
θ= ∂V
∂t where V is the option value and t is the passage of time or the time with T − t periods to expiration.
Proposition 30.1
We have
θCall = δe−δ(T −t) N (d1 ) − rKe−r(T −t) N (d2 ) − Se−δ(T −t) N (d1 )σ and
θPut = rKe−r(T −t) N (−d2 ) − δSe−δ(T −t) N (−d1 ) − 2 (T − t)
Se−δ(T −t) N (d1 )σ
2 (T − t) Proof.
We consider the following version of the BlackScholes formula
Ct = Se−δ(T −t) N (d1 ) − Ke−r(T −t) N (d2 ).
We have
θCall = ∂C
∂t ∂N (d2 )
∂N (d1 )
− Ke−r(T −t) N (d2 ) − Ke−δ(T −t)
∂t
∂t
∂N (d1 ) ∂d1
∂N (d2 ) ∂d2
=Se−δ(T −t) N (d1 ) + Se−δ(T −t)
− Ke−r(T −t) N (d2 ) − Ke−δ(T −t)
∂d1 ∂t
∂t
∂t =Se−δ(T −t) N (d1 ) + Se−δ(T −t) d2
1
1
=Se−δ(T −t) N (d1 ) + Se−δ(T −t) √ e− 2
2π −Ke−r(T −t) N (d2 ) − Ke−δ(T −t) ln (S/K )
2σ (T − t) 3
2 − 1 S − d2 (r−δ)(T −t)
1
√
e 2e
2π K r − δ + 0.5σ 2 r − δ + 0.5σ 2
√
√
+
σ T −t
2σ T − t
ln (S/K )
3 2σ (T − t) 2 − r − δ − 0.5σ 2 r − δ − 0.5σ 2
√
√
+
σ T −t
2σ T − t 210 THE BLACKSCHOLES MODEL d2
1
1
=Se−δ(T −t) N (d1 ) − Ke(r(T −t) + Se−δ(T −t) √ e− 2
2π
−δ (Tt )
Se
σN (d1 )
√
=Se−δ(T −t) N (d1 ) − Ke(r(T −t) −
2 T −t σ
−√
2 T −t The result for the put can be shown in a similar way
Remark 30.1
If time to expiration is measured in years, theta will be the annualized change in the option value.
To obtain a perday theta, divide by 365. This is the case we will consider in the problems.
The value of an option is the combination of time value and stock value. When time passes, the
time value of the option decreases, that is, the option becomes less valuable. Thus, the rate of
change of the option price with respect to the passage of time, theta, is usually negative. There are
a few exceptions to this rule. Theta can be positive for deep inthemoney European calls when the
underlying asset has a high dividend yield or for deep inthemoney European puts.
The theta for a purchased call and put at the same strike price and the same expiration time are
not equal (See Problem 30.1). Data analysis shows that time decay is most rapid at expiration
and that the theta for a call is highest (i.e. largest in absolute value) for atthemoney shortlived
options, and is progressively lower (turns less and less negative) as options are inthemoney and
outofthe money.
The Measure Rho
The option Greek rho (ρ) measures the change in the option price when there is an increase in the
interest rate of 1 percentage point. More formally, The rho of an option is deﬁned as the rate of
change of the option price with respect to the interest rate:
ρ= ∂V
∂r where V is the option value.
Proposition 30.2
We have
ρCall = (T − t)Ke−r(T −t) N (d2 )
and
ρPut = −(T − t)Ke−r(T −t) N (−d2 ) 30 OPTION GREEKS: THETA, RHO, AND PSI 211 Proof.
We have
∂Ct
∂N (d1 )
∂N (d2 )
= Se−δ(T −t)
+ (T − t)Ke−r(T −t) N (d2 ) − Ke−r(T −t)
∂r
∂r
∂r
∂N (d1 ) ∂d1
∂N (d2 ) ∂d2
=Se−δ(T −t)
+ (T − t)Ke−r(T −t) N (d2 ) − Ke−r(T −t)
∂d1 ∂r
∂d2 ∂r
√
1 − d2
T −t
1
+ (T − t)Ke−r(T −t) N (d2 )
=Se−δ(T −t) √ e 2
σ
2π
√
1 − d2 S (r−δ)(T −t)
T −t
1
−r(T −t)
√ e2 e
−Ke
K
σ
2π
√
d2
d2
1
T −t
1
1
1
=Se−δ(T −t) √ e− 2
+ (T − t)Ke−r(T −t) N (d2 ) − Se−δ(T −t) √ e− 2
σ
2π
2π
−r(T −t)
=(T − t)Ke
N (d2 ) ρCall = √ T −t
σ A similar argument for the put option
Remark 30.2
We will use the above formula divided by 100. That is, the change of option value per 1% change
in interest rate.
The rho for an ordinary stock call option should be positive because higher interest rate reduces
the present value of the strike price which in turn increases the value of the call option. Similarly,
the rho of an ordinary put option should be negative by the same reasoning. Figure 30.1 shows
that for a European call, rho increases with time to maturity. This is also true for increases in the
stock price. Figure 30.1 212 THE BLACKSCHOLES MODEL The Measure Psi
The option Greek psi (ψ ) measures the change in the option price when there is an increase in the
continuous dividend yield of 1%. More formally, the psi of an option is deﬁned as the rate of change
of the option price with respect to the dividend yield:
ψ= ∂V
∂δ where V is the value of the option.
Proposition 30.3
We have
ψCall = −(T − t)Se−δ(T −t) N (d1 )
and
ψPut = (T − t)Se−δ(T −t) N (−d1 )
To interpret psi as a price change per percentage point change in the dividend yield, divide by
100. It follows that for call options, psi is negative. For put options, psi is positive. Data analysis
shows that for a European call, psi decreases as time to maturity increases. For a European put,
psi increases as time to maturity increases.
Remark 30.3
In the problems we will use the above formula divided by 100.
Example 30.1
Consider a stock with annual volatility of 30% . The stock pays no dividends and is currently selling
for $40. The strike price of a purchased call option is $45 and the time to maturity is six months.
The continuously compounded riskfree rate is 8%. What are θ, ρ, and ψ ?
Solution.
We ﬁrst ﬁnd d1 and d2 . We have
d1 =
and ln (40/45) + (0.08 − 0 + 0.5(0.3)2 )(0.5)
ln (S/K ) + (r − δ + 0.5σ 2 )T
√
√
=
= −0.260607
0.3 0.5
σT
√
√
d2 = d1 − σ T − t = −0.260607 − 0.3 0.5 = −0.47274. 30 OPTION GREEKS: THETA, RHO, AND PSI 213 Thus,
1
N (d1 ) = √
2π
and
1
N (d2 ) = √
2π −0.260607 x2 e− 2 dx = 0.397198
−∞
−0.47274 x2 e− 2 dx = 0.318199.
−∞ We have
θCall = 1
Se−δ(T −t) N (d1 )σ
[δe−δ(T −t) N (d1 ) − rKe−r(T −t) N (d2 ) −
365
2 (T − t)
0.2606072 ρCall
ψCall 1
40(0.3) e− 2
−0.08(0.5)
√
=
[−0.08(45)e
(0.318199) − √ = −0.0122
365
2 0.5
2π
1
1
[(T − t)Ke−r(T −t) N (d2 )] =
[0.5(45)e−0.08(0.5) (0.318199)] = 0.0688
=
100
100
1
1
[−(T − t)Se−δ(T −t) N (d1 )] =
[−(0.5)(40)(0.100396)] = −0.077
=
100
100 Greek Measures for Portfolios
The Greek measure of a portfolio is the sum of the Greeks of the individual portfolio components.
For a portfolio containing n options with a single underlying stock, where the quantity of each
option is given by ni , 1 ≤ i ≤ n, we have
n G= ni Gi
i=1 where the G and Gi ’s are the option Greeks. For example,
n ∆= ni ∆i
i=1 Example 30.2
A portfolio consists of 45 call options on Asset A (with ∆A = 0.22), 14 put options on Asset B
(with ∆B = −0.82), 44 put options on Asset C (with ∆C = −0.33), and 784 call options on Asset
D (with ∆D = 0.01). Find the delta of the entire portfolio.
Solution.
The delta of the entire portfolio is
∆ = (45)(0.22) + (14)(−0.82) + (44)(−0.33) + (784)(0.01) = 8.26 214 THE BLACKSCHOLES MODEL Practice Problems
Problem 30.1
Show that
θPut = θCall + rKe−r(T −t) − δSe−δ(T −t) .
Problem 30.2
Prove Proposition 30.3.
Problem 30.3
The stock of GS Co. pays dividends at an annual continuously compounded yield of 0.12. The
annual continuously compounded riskfree interest rate is 0.34. Certain call options on the stock of
GS Co. have time to expiration of 99 days. The option currently trades for $56.
(a) Suppose θ = −0.03(per day). Find the price of the call option 34 days from expiration, all other
things equal.
(b) Suppose ρ = 0.11. Find the price of the call option 34 days from expiration if the interest rate
suddenly increases to 0.66, all other things equal.
(c) Suppose ψ = −0.04. Find the price of the call option 34 days from expiration if the stock’s
dividend yield suddenly decreases to 0.02, all other things equal.
Problem 30.4
A stock is currently selling for $40. The stock pays no dividends. Given that the volatility relevant
for the BlackScholes equation is 30% and the continuously compounded riskfree interest rate is
8%. Consider a $40strike purchased call on the stock with time to expiration in 6 months. What
are the theta, rho, and psi?
Problem 30.5
Consider a bull spread where you buy a 40strike call and sell a 45strike call. Suppose S = $40, σ =
30%, r = 8%, δ = 0%, and T = 0.5. What are theta and rho for this portfolio?
Problem 30.6
A stock is currently selling for $40. The stock pays no dividends. Given that the volatility relevant
for the BlackScholes equation is 30% and the continuously compounded riskfree interest rate is
8%. Consider a $40strike purchased put on the stock with time to expiration in 6 months. What
are the delta, gamma, vega, theta, and rho?
Problem 30.7
A stock is currently selling for $40. The stock pays no dividends. Given that the volatility is relevant
for the BlackScholes equation 30% and the continuously compounded riskfree interest rate is 8%.
Consider a $45strike purchased put on the stock with time to expiration in 6 months. What are
the delta, gamma, vega, theta, and rho? 30 OPTION GREEKS: THETA, RHO, AND PSI 215 Problem 30.8
Consider a bull spread where you buy a 40strike put and sell a 45strike put. Suppose S = $40, σ =
30%, r = 8%, δ = 0, and T = 0.5. What are delta, gamma, vega, theta, and rho?
Problem 30.9
Show that the delta of a K1 − K2 call bull spread is equal to the K1 − K2 put call spread when the
underlying stock pays no dividends. Here K1 < K2 .
Problem 30.10 ‡
You compute the delta for a 5060 bull spread with the following information:
(i) The continuously compounded riskfree rate is 5%.
(ii) The underlying stock pays no dividends.
(iii) The current stock price is $50 per share.
(iv) The stock’s volatility relevant for the BlackScholes equation is 20%.
(v) The time to expiration is 3 months.
How much does the delta change after 1 month, if the stock price does not change? 216 THE BLACKSCHOLES MODEL 31 Option Elasticity and Option Volatility
Suppose a stock price changes from S to S . Let = S − S denote the change in the stock price.
Let CV = V − V denote the change in an option value. From the deﬁnition of the option Greek
∆, as the change in option value over the change in stock value we can write
CV = ∆.
That is, the change in option value is the change in stock value multiplied by delta.
The option elasticity Ω is deﬁned as the percentage change in the option price relative to the
precentage change in the stock price:
Ω= V −V
V
S −S
S = ∆
V = S S∆
.
V Thus, option elasticity gives the percentage change in the option value for a 1% change in the stock
value. It tells us the risk of the option relative to the stock in percentage form.
For a call option we shall use the notation
ΩCall = S∆
C and for a put option
S∆
P
For a call option we have S ∆ = Se−δT N (d1 ) ≥ Se−δT N (d1 ) − Ke−rT N (d2 ) = C. That is, ΩCall ≥ 1.
For a put option ∆ < 0 so that ΩPut ≤ 0.
ΩPut = Example 31.1
A certain stock is currently trading for $41 per share with a stock price volatility of 0.3. Certain
call options on the stock have a delta of 0.6991 and a price of $6.961. Find the elasticity of such a
call option.
Solution.
The call elasticity is
S∆
41 × 0.6911
=
= 4.071.
C
6.961
This says that if the stock price goes up to $41.41, the call option value goes up to 6.9661+0.04071 ×
6.9661 = $7.25
ΩC = 31 OPTION ELASTICITY AND OPTION VOLATILITY 217 Figure 31.1 displays the elasticity of a call option with strike price of $35 with diﬀerent maturities. Figure 31.1
The following observations are in place:
• ΩCall increases as stock price decreases.
• ΩCall decreases as time to maturity increases.
• ΩCall increases as the option becomes more outofthe money. ΩCall decreases as the option becomes
more inthemoney.
Example 31.2
A European call option on XYZ stock has the following speciﬁcations: Strike price = $45, current
stock price = $46, time to expiration = 3 months, annual continuously compounded interest rate
= 0.08, dividend yield = 0.02, prepaid forward price volatility=0.35. Calculate the elasticity of the
call.
Solution.
We ﬁrst calculate d1 and d2 . We have
ln
ln (S/K ) + (r − δ + 0.5σ 2 )T
√
=
d1 =
σT
and 46
45 + (0.08 − 0.02 + 0.352 × 0.5)(0.25)
√
= 0.29881
0.35 0.25 √
√
d2 = d1 − σ T = 0.29881 − 0.35 0.25 = 0.12381. 218 THE BLACKSCHOLES MODEL Thus,
1
N (d1 ) = √
2π
and
1
N (d2 ) = √
2π 0.29881 x2 e− 2 dx = 0.617457
−∞
0.12381 x2 e− 2 dx = 0.549267.
−∞ Thus, the price of the call is
C = 46e−0.02×0.25 × 0.617457 − 45e−0.08×0.25 × 0.549267 = $4.0338.
Now,
∆Call = e−δT N (d1 ) = e−0.02×0.25 × 0.617457 = 0.6144.
Thus,
46 × 0.6144
S∆
=
= 7.0064
C
4.0338
The volatility of an option can be found in terms of the option elasticity.
ΩCall = Proposition 31.1
The volatility of an option is the option elasticity times the volatility of the stock. That is,
σoption = σstock × Ω.
Proof.
Consider a hedge portfolio consisting of purchased ∆ shares of the stock and one short option. The
portfolio value is V − S ∆. One period later the portfolio will be worth
(V + δV ) − ∆(S + δS )
where δ means the change in S or V. If this portfolio is hedged, its value should grow at the riskfree
rate. Hence, the following condition must hold:
(V − ∆S )(1 + r) = (V + δV ) − ∆(S + δS ).
This leads to the equation
δV
δS
S
δS
=r+
−r ∆ =r+
− r Ω.
V
S
V
S 31 OPTION ELASTICITY AND OPTION VOLATILITY 219 Using the expression created above for the return on the option as a function of the return on the
stock, we take the variance of the option return:
δS
δV
=Var r +
−r Ω
Var
V
S
δS
=Var
Ω
S
δS
=Ω2 Var
S
Taking square root of both sides we obtain
σoption = σstock × Ω
It follows that the option’s risk is the risk of the stock times the risk of the option relative to the
stock.
Example 31.3
A certain stock is currently trading for $41 per share with a stock price volatility of 0.3. Certain
call options on the stock have a delta of 0.6991 and a price of $6.961. Find the volatility of such a
call option.
Solution.
The answer is
σCall = σstock × ΩCall  = 0.3 × 4.071 = 1.2213
Example 31.4
Consider a 3year call option on a stock that pays dividends at the continuously compounded yield
of 3%. Suppose that the prepaid forward price volatility is 20% and the elasticity of the call is
3.9711. Find the volatility of the call option.
Solution.
We know from Remark 16.2 that
σF = σstock ×
But S
.
P
F0,T S
S
= −δT = eδT .
P
Se
F0,T Thus,
σstock = σF e−δT = 0.2w−0.03×3 = 0.1828.
Hence,
σCall = σstock × ΩCall  = 0.1828 × 3.9711 = 0.7259 220 THE BLACKSCHOLES MODEL Practice Problems
Problem 31.1
Since C = max{0, ST − K }, as the strike decreases the call becomes more in the money. Show that
the elasticity of a call option decreases as the strike price decreases.
Problem 31.2
Which of the following statements is true?
(A) ΩPut ≤ −1.
(B) ΩCall ≥ 0.
(C) The elasticity of a call increases as the call becomes more outofthe money.
(D) The elasticity of a call increases as stock price increases.
Problem 31.3
A certain stock is currently trading for $41 per share with a stock volatility of 0.3. Certain put
options on the stock have a delta of −0.3089 and a price of $2.886. Find the elasticity of such a
put option.
Problem 31.4
A European put option on XYZ stock has the following speciﬁcations: Strike price = $45, current
stock price = $46, time to expiration = 3 months, annual continuously compounded interest rate =
0.08, dividend yield = 0.02, prepaid forward price volatility=0.35. Calculate the elasticity of such
a put.
Problem 31.5
A European call option on XYZ stock has the following speciﬁcations: Strike price = $45, current
stock price = $46, time to expiration = 3 months, annual continuously compounded interest rate
= 0.08, dividend yield = 0.02, prepaid forward price volatility=0.35. Calculate the volatility of the
call.
Problem 31.6
Given the following information about a 3year call option on a certain stock:
• The current stock price is $550.
• The prepaid forward price volatility (the volatility relevant for the BlackScholes formula) is 0.2.
• The strike price is $523.
• The stock pays dividends at an annual continuously compounded yield of 0.03.
• The annual continuously compounded interest rate is 0.07.
Find the elasticity of such a call option. 31 OPTION ELASTICITY AND OPTION VOLATILITY 221 Problem 31.7
Find the call option volatility in the previous problem.
Problem 31.8
Given the following information about a 3year put option on a certain stock:
• The current stock price is $550.
• The prepaid forward price volatility (the volatility relevant for the BlackScholes formula) is 0.2.
• The strike price is $523.
• The stock pays dividends at an annual continuously compounded yield of 0.03.
• The annual continuously compounded interest rate is 0.07.
Find the elasticity of such a put option.
Problem 31.9 ‡
A call option is modeled using the BlackScholes formula with the following parameters:
• S = 25
• K = 24
• r = 4%
• δ = 0%
• σ = 20%
• T = 1.
Calculate the call option elasticity.
Problem 31.10 ‡
For a European call option on a stock within the BlackScholes framework, you are given:
(i) The stock price is $85.
(ii) The strike price is $80.
(iii) The call option will expire in one year.
(iv) The continuously compound riskfree interest rate is 5.5%.
(v) σ = 0.50
(vi) The stock pays no dividends.
Calculate the volatility of this call option.
Problem 31.11
For a European put option on a stock within the BlackScholes framework, you are given:
(i) The stock price is $50.
(ii) The strike price is $55.
(iii) The put option will expire in one year.
(iv) The continuously compound riskfree interest rate is 3%. 222
(v) σ = 0.35
(vi) The stock pays no dividends.
Calculate the volatility of this put option. THE BLACKSCHOLES MODEL 32 THE RISK PREMIUM AND SHARPE RATIO OF AN OPTION 223 32 The Risk Premium and Sharpe Ratio of an Option
When buying an asset you lose cash that can be invested at riskfree interest rate and you acquire
the risk of owning the stock. On average you expect to earn interest as compensation for the time
value of money and additional return as compensation for the risk of the stock. This additional
return in known as the risk premium.
In ﬁnance, the risk premium on an asset is deﬁned to be the excess return of the asset over the
riskfree rate. Thus, if r denotes the riskfree rate and α the expected return on the asset, then the
risk premium on the asset is the diﬀerence
α − r.
Now, consider a call option on a stock. We know that the call can be replicated by buying ∆ units
of a share of the stock and borrowing B. A result in ﬁnance due to Brealey and Meyer states that
the return on any portfolio is the weighted average of the returns on the assets in the portfolio. We
apply this result to the above portfolio. Let γ be the expected return on the call, α the expected
return on the stock and r the riskfree rate of return. Then
B
S∆
α−
r.
S∆ − B
S∆ − B γ= But C = S ∆ − B so that −B = C − S ∆. Thus, the above equation can be written as
γ=
Since Ω = S∆
,
C S∆
S∆
α+ 1−
C
C r. (32.1) the above equation reduces to
γ = Ωα + (1 − Ω)r or
γ − r = (α − r) × Ω.
This says that the risk premium of the option is the risk premium of the stock multiplied by the
option elasticity.
Remark 32.1
∆
Note that in terms of the replicating portfolio, in Equation (32.1), SC is the percentage of the value
∆
of the option invested in the stock and 1 − SC is the percentage of the value of the option for
borrowing. 224 THE BLACKSCHOLES MODEL Remark 32.2
Suppose α − r > 0. Since ΩCall ≥ 1, we have γ − r ≥ α − r or γ ≥ α. For a put, we know that
ΩPut ≤ 0 which leads to γ − r = (α − r) × Ω ≤ 0 < α − r. That is, γ < α.
We know from Section 31 that ΩCall increases if either the stock price goes down or the strike
price goes up. Therefore, the expected return on a call option increases if either the stock price
goes down or the strike price goes up. That is, the expected return increases if the option is more
outofthemoney and decreases if the option is more inthemoney.
Example 32.1
You are given the following information of a call option on a stock:
• The expected return on the stock is 15% compounded continuously.
• The continuously compounded riskfree interest rate is 7%.
• The call elasticity is 4.5.
(a) Find the risk premium on the option.
(b) Find the expected annual continuously compounded return on the call option.
Solution.
(a) The risk premium on the option is γ − r = (α − r) × Ω = (0.15 − 0.07) × 4.5 = 0.36.
(b) The expected return on the option is γ = 0.36 + 0.07 = 0.43
The Sharpe Ratio of an Option
The Sharpe ratio of an asset is the risk premium per unit of risk in an investment asset (deﬁned
as the standard deviation of the asset’s excess return.) It is given by
Sharpe ratio = α−r
.
σ The Sharpe ratio is used to characterize how well the return of an asset compensates the investor
for the risk taken.
Applying this deﬁnition to a call option on a stock, we can deﬁne the Sharpe ratio of a call by
Sharpe ratio for call = α−r
(α − r)Ω
=
.
σΩ
σ Thus, the Sharpe ratio of a call option is equal to the Sharpe ratio of the underlying stock.
The Sharpe ratio for a put option on a stock is
Sharpe ratio for put = (α − r)Ω
(α − r)Ω
r−α
=
=
.
σ × Ω
σ × (−Ω)
σ Thus, the Sharpe ratio of a put is the opposite of the Sharpe ratio of the stock. 32 THE RISK PREMIUM AND SHARPE RATIO OF AN OPTION 225 Example 32.2
You expect to get an annual continuously compounded return of 0.3 on the stock of GS Co. The
stock has annual price volatility of 0.22. The annual continuously compounded riskfree interest
rate is 0.02. A certain call option on GS Co. stock has elasticity of 2.3. Find the Sharpe ratio of
the call option.
Solution.
The Sharpe ratio is 0.3 − 0.02
α−r
=
= 1.2727
σ
0.22
The Elasticity and Risk Premium of a Portfolio of Options
Consider a portfolio comprising of n calls with the same underlying stock. For 1 ≤ i ≤ n, we let
Ci denote the value of the ith call and ∆i the change in the value of the ith call for a $1 change in
the stock. Suppose that there are ni units of the ith call in the portfolio. We deﬁne ωi to be the
percentage of the portfolio value invested in the ith call. That is
ni C i
,
n
j =1 nj Cj ωi = where n=1 nj Cj is the portfolio value.
j
Now, a $1 change in the stock results in a change of the portfolio value by
n nj ∆j .
j =1 Now, the elasticity of the portfolio is the percentage change in the portfolio (of calls) price divided
by the percentage change in the stock price, or
Ωportfolio = n
i=1 ni ∆i
n
j =1 nj Cj 1
S n =
i=1 ni Ci
n
j =1 nj Cj S ∆i
=
Ci n ωi Ωi .
i=1 Like the case of a single option, it is easy to establish that the risk premium of the portfolio is the
elasticity times the risk premium of the underlying stock.
Example 32.3
Given the following information:
Call
A
B
C Price
$9.986
$7.985
$6.307 Elasticity
4.367
4.794
5.227 226 THE BLACKSCHOLES MODEL Find the elasticity of the portfolio consisting of buying one call option on stock A, one call option
on stock B and selling one call option on stock C.
Solution.
The cost of the portfolio is
9.986 + 7.985 − 6.307 = $11.664.
The elasticity of the portfolio is
3 Ωportfolio = ωi Ωi =
i=1 7.985
6.307
9.986
× 4.367 +
× 4.794 +
× 5.227 = 4.1943
11.664
11.664
11.664 32 THE RISK PREMIUM AND SHARPE RATIO OF AN OPTION 227 Practice Problems
Problem 32.1
You are given the following information of a call option on a stock:
• The expected return on the stock is 30% compounded continuously.
• The continuously compounded riskfree interest rate is 2%.
• The call elasticity is 2.3.
(a) Find the risk premium on the option.
(b) Find the expected annual continuously compounded return on the call option.
Problem 32.2
A call option on a stock has elasticity of 3.4. The continuously compounded riskfree rate is 3.3%
and the BlackScholes price volatility of the stock is 0.11. Suppose that the risk premium on the
stock is 77% of the stock volatility. Find the expected annual continuously compounded return on
the option.
Problem 32.3
You expect an annual continuously compounded return of 0.145 on the stock of GS, Inc., and an
annual continuously compounded return of 0.33 on a certain call option on that stock. The option
elasticity is 4.44. Find the annual continuously compounded riskfree interest rate.
Problem 32.4
The price of a put option on a stock is $4.15. The elasticity of the put is −5.35. Consider the
replicating portfolio.
(a) What is the amount of money invested in the stock?
(b) What is the amount of money to be lent at the riskfree interest rate?
Problem 32.5
The stock of GS Co. has a Sharpe ratio of 0.77 and annual price volatility of 0.11. The annual
continuously compounded riskfree interest rate is 0.033. For a certain call option on GS stock, the
elasticity is 3.4. Find the expected annual continuously compounded return on the option.
Problem 32.6
A call option on a stock has option volatility of 1.45 and elasticity 4.377. The annual continuously
compounded riskfree is 6% and the annual continuously compounded return on the stock is 15%.
Find the Sharpe ratio of such a call option.
Problem 32.7
A stock has volatility of 25%. The annual continuously compounded riskfree is 4% and the annual
continuously compounded return on the stock is 8%. What is the Sharpe ratio of a put on the
stock? 228 THE BLACKSCHOLES MODEL Problem 32.8
Consider the following information:
Call
A
B
C Price
$9.986
$7.985
$6.307 Elasticity
4.367
4.794
5.227 Find the elasticity of a portfolio consisting of 444 options A, 334 options B, and 3434 options C.
Problem 32.9
Consider again the information of the previous problem. The expected annual continuously compounded return on the stock is 0.24, and the annual continuously compounded riskfree interest
rate is 0.05. Find the risk premium on this option portfolio.
Problem 32.10
Consider the following information:
Call
A
B
C Price
$9.986
$7.985
$6.307 Elasticity
4.367
4.794
5.227 A portfolio consists of 444 of options A, 334 of options B and X units of option C. The elasticity
of the portfolio is 5.0543. Find X to the nearest integer.
Problem 32.11
Consider the following information:
Call
A
B
C Price
$9.986
$7.985
$6.307 Elasticity
4.367
4.794
ΩC A portfolio consists of 444 of options A, 334 of options B and 3434 units of option C. The risk
premium of the portfolio is 0.960317. The expected annual continuously compounded return on the
stock is 0.24, and the annual continuously compounded riskfree interest rate is 0.05. Find ΩC .
Problem 32.12
Consider a portfolio that consists of buying a call option on a stock and selling a put option. The
stock pays continuous dividends at the yield rate of 5%. The options have a strike of $62 and expire
in six months. The current stock price is $60 and the continuously compounded riskfree interest
rate is 13%. Find the elasticity of this portfolio. 32 THE RISK PREMIUM AND SHARPE RATIO OF AN OPTION 229 Problem 32.13
Given the following information
Option Price Elasticity
Call
$10
ΩC
Put
$5
ΩP
Consider two portfolios: Portfolio A has 2 call options and one put option. The elasticity of this
portfolio is 2.82. Portfolio B consists of buying 4 call options and selling 5 put options. The delta
of this portfolio is 3.50. The current value of the stock is $86. Determine ΩC and ΩP .
Problem 32.14 ‡
An investor is deciding whether to buy a given stock, or European call options on the stock. The
value of the call option is modeled using the BlackScholes formula and the following assumptions:
• Continuously compounded riskfree rate = 4%
• Continuously compounded dividend = 0%
• Expected return on the stock = 8%
• Current stock price = 37
• Strike price = 41
• Estimated stock volatility = 25%
• Time to expiration = 1 year.
Calculate the sharp ratio of the option.
Problem 32.15 ‡
Assume the BlackScholes framework. Consider a stock, and a European call option and a European
put option on the stock. The current stock price, call price, and put price are 45.00, 4.45, and 1.90,
respectively.
Investor A purchases two calls and one put. Investor B purchases two calls and writes three puts.
The current elasticity of Investor A’s portfolio is 5.0. The current delta of Investor B’s portfolio is
3.4.
Calculate the current putoption elasticity.
Problem 32.16 ‡
Assume the BlackScholes framework. Consider a 1year European contingent claim on a stock.
You are given:
(i) The time0 stock price is 45.
(ii) The stock’s volatility is 25%.
(iii) The stock pays dividends continuously at a rate proportional to its price. The dividend yield
is 3%. 230
(iv) The continuously compounded riskfree interest rate is 7%.
(v) The time1 payoﬀ of the contingent claim is as follows: Calculate the time0 contingentclaim elasticity. THE BLACKSCHOLES MODEL 33 PROFIT BEFORE MATURITY: CALENDAR SPREADS 231 33 Proﬁt Before Maturity: Calendar Spreads
In this section we discuss the concepts of the holding period proﬁt and calendar spreads.
The holding period proﬁt of a position is deﬁned to be the current value of the position minus
the cost of the position, including interest. For example, consider a call option with maturity time
T and strike price K. Suppose that at time t = 0 the price of the option is C. Let Ct be the price
of the option t < T years later. We want to ﬁnd the holding period proﬁt at time t. Using the
BlackScholes formula we have
Ct = Se−δ(T −t) N (d1 ) − Ke−r(T −t) N (d2 ).
Thus, the holding period proﬁt is
Ct − Cert .
Example 33.1 ‡
Assume the BlackScholes framework.
Eight months ago, an investor borrowed money at the riskfree interest rate to purchase a oneyear
75strike European call option on a nondividendpaying stock. At that time, the price of the call
option was 8.
Today, the stock price is 85. The investor decides to close out all positions.
You are given:
(i) The continuously compounded riskfree rate interest rate is 5%.
(ii) The stock’s volatility is 26%.
Calculate the eightmonth holding proﬁt.
Solution.
We ﬁrst ﬁnd the current value (t = 8) of the option. For that we need to ﬁnd d1 and d2 . We have
d1 = ln (85/75) + (0.05 + 0.5(0.26)2 )(4/12)
ln (S/K ) + (r − δ + 0.5σ 2 )T
√
=
= 1.02
σT
0.26 4
12 and
√
d2 = d1 − σ T = 1.02 − 0.26 4
= 0.87.
12 Thus,
1
N (d1 ) = √
2π 1.02 x2 e− 2 dx = 0.846136
−∞ 232 THE BLACKSCHOLES MODEL and
1
N (d2 ) = √
2π 0.87 x2 e− 2 dx = 0.80785.
−∞ Thus,
4 Ct = Se−δ(T −t) N (d1 ) − Ke−r(T −t) N (d2 ) = 85 × 0.846136 − 75e−0.05× 12 × 0.80785 = 12.3342.
Thus, the eightmonth holding proﬁt is
8 12.3342 − 8e0.05× 12 = $4.0631
Calendar Spread
If an investor believes that stock prices will be stable for a foreseeable period of time, he or she can
attempt to proﬁt from the declining time value of options by setting a calendar spread.
In simplest terms, a calendar spread, also called a time spread or horizontal spread, involves
buying an option with a longer expiration and selling an option with the same strike price and a
shorter expiration. The long call usually has higher premium than the short call.
Suppose that the stock price is very low at the expiry of the shorterlived call. Then the call will
not be exercised and therefore it is worthless. But also, the value of the longerlived call is close to
zero. So the investor loss is the cost of setting up the spread. If the stock price is very high, the
shorterlived call will be exercised and this costs the investor ST − K and the longlived call has
value close to ST − K so that the investor again incurs a loss which is the cost of setting the spread.
If ST is close to K , the shortlived call will not be exercised so it is worthless and costs the investor
almost nothing but the longlived call is still valuable and thus the investor incurs a net gain.
Example 33.2
The current price of XYZ stock is $40. You sell a 40strike call with three months to expiration
for a premium of $2.78 and at the same time you buy a 40strike call with 1 year to expiration for
a premium of $6.28. The theta for the long call is −0.0104 and that for the short call is −0.0173.
Assume that the continuously compounded riskfree rate is 8%, ﬁnd the proﬁt once the sold option
has expired, if the stock price remains at $40 and nothing else has changed.
Solution.
Three months from now, at a stock price of $40, the written call option will be atthemoney and so
will not be exercised. You keep the entire premium on the option− i.e., $2.78, and this value could
have accumulated interest over three months. Thus, your proﬁt from selling the 3month option is
2.78e0.08×0.25 = $2.8362.
You also lose some money on your purchased 1year option because of time decay. Due to time
decay, the option price changes by 90θ = 90(−0.0104) = −0.936. 33 PROFIT BEFORE MATURITY: CALENDAR SPREADS 233 You further lose some money in the form of the interest that could have accumulated on the $6.28
premium you paid for the purchased call option, if instead you invested the money at the riskfree
interest rate. That interest is 6.28(e0.08×0.25 − 1) = $0.1269. So your net gain from entering into this
spread is
2.8362 − 0.936 − 0.1269 = $1.7733
Example 33.3
Suppose that the current month is April, 08 and the stock price is $80. You buy a call option
expiring on Dec, 08 and sell a call expiring on July, 08. Both calls have the same strike $80. The
premium for the long call is $12 and for the short call $7. On May, 08 the price of the stock drops
to $50 and the longlived call is selling for $1. In terms of premium only, what would be your proﬁt
in May?
Solution.
The cost on April, 08 from creating the spread is 12 − 7 = $5. On May 08, the short call is nearly
worthless while the longcall can be sold for $1. Thus, the loss from premiums is $4
A calendar spread may be implemented using either call options or put options, but never with
calls and puts used together in the same trade.
Example 33.4
The current price of XYZ stock is $40. You sell a 40strike put with three months to expiration
for a premium of $2.78 and at the same time you buy a 40strike put with 1 year to expiration for
a premium of $6.28. The theta for the long put is −0.002 and that for the short put is −0.008.
Assume that the continuously compounded riskfree rate is 8%, ﬁnd the proﬁt at the expiration
date of the shorter put, if the stock price remains at $40 and nothing else has changed.
Solution.
Three months from now, at a stock price of $40, the written put option will be atthemoney and so
will not be exercised. You keep the entire premium on the option− i.e., $2.78, and this value could
have accumulated interest over three months. Thus, your proﬁt from selling the 3month option is
2.78e0.08×0.25 = $2.8362.
You also lose some money on your purchased 1year option because of time decay. Due to time
decay, the option price changes by 90θ = 90(−0.002) = −0.18.
You further lose some money in the form of the interest that could have accumulated on the $6.28
premium you paid for the purchased put option, if instead you invested the money at the riskfree
interest rate. That interest is 6.28(e0.08×0.25 − 1) = $0.1269. So your net gain (after three months)
from entering into this spread is
2.8362 − 0.18 − 0.1269 = $2.53 234 THE BLACKSCHOLES MODEL Practice Problems
Problem 33.1
Assume the BlackScholes framework.
Eight months ago, an investor borrowed money at the riskfree interest rate to purchase a oneyear
90strike European put option on a nondividendpaying stock. At that time, the price of the put
option was 7.
Today, the stock price is 80. The investor decides to close out all positions.
You are given:
(i) The continuously compounded riskfree rate interest rate is 6%.
(ii) The stock’s volatility is 28%.
Calculate the eightmonth holding proﬁt.
Problem 33.2
Assume the BlackScholes framework.
Eight months ago, an investor borrowed money at the riskfree interest rate to purchase a oneyear
75strike European call option on a nondividendpaying stock. Today, the stock price is 85.
You are given:
(i) The continuously compounded riskfree rate interest rate is 5%.
(ii) The stock’s volatility is 26%.
(iii) The eightmonth holding proﬁt is $4.0631.
Calculate the initial cost of the call.
Problem 33.3
Eight months ago the price of a put option on a nondividendpaying stock was $7. Currently
the price of the put is $10.488. Find the continuously compounded riskfree interest rate if the
eightmonth holding period proﬁt is $3.7625.
Problem 33.4
You own a calendar spread on a the stock of GS Co., which you bought when the stock was priced
at $22. The spread consists of a written call option with a strike price of $22 and a longerlived
purchased call option with a strike price of $22. Upon the expiration of the shorterlived option, at
which stock price will you make the most money on the calendar spread?
(A) $3 (B) $12 (C) $22 (D) $23 (E) $33
Problem 33.5
The current price of XYZ stock is $60. You sell a 60strike call with two months to expiration for
a premium of $3.45 and at the same time you buy a 60strike call with 1 year to expiration for
a premium of $18.88. The theta for the long call is −0.05 and that for the short call is −0.04. 33 PROFIT BEFORE MATURITY: CALENDAR SPREADS 235 Assume that the continuously compounded riskfree rate is 6%, ﬁnd the proﬁt once the sold option
has expired, if the stock price remains at $60 and nothing else has changed.
Problem 33.6
A reverse calendar spread is constructed by selling a longterm option and simultaneously buying
a shortterm option with the same strike price. Would you expect to proﬁt if the stock price moves
away in either direction from the strike price by a great deal?
Problem 33.7
Suppose that the current month is April, 08 and the stock price is $80. You sell a call option
expiring on Dec, 08 and buy a call price expiring on July, 08. Both calls have the same strike $80.
The premium for the long call is $12 and for the short call $7. On May, 08 the price of the stock
drops to $50 and the longlived call is selling for $1. In terms of premium only, what would be your
proﬁt in May?
Problem 33.8
Currently stock XYZ is selling for $40. You believe that in the next three months the stock price
will be almost the same. Assume that the continuously compounded riskfree rate is 8%. Create a
calendar spread from the following options with the largest proﬁt at the time of expiration of the
shortlived call.
Call option A: K = $40, θ = −0.017, T =3months, Premium $2.78.
Call option B: K = $40, θ = −0.01, T =1 year, Premium $3.75.
call Option C: K = $40, θ = −0.006, T =3years, Premium $7.05
Problem 33.9
Currently, stock XYZ is selling for $40. You believe that in the next three months the stock price
will be almost the same. Assume that the continuously compounded riskfree rate is 8%. Create a
calendar spread from the following options with the largest proﬁt at the time of expiration of the
shortlived call.
Put option A: K = $40, θ = −0.008, T =3months, Premium $2.78.
Put option B: K = $40, θ = −0.002, T =1 year, Premium $6.28.
Put option C: K = $40, θ = −0.0001, T =3years, Premium $9.75. 236 THE BLACKSCHOLES MODEL 34 Implied Volatility
In Section 26 we described how to estimate volatility of the underlying asset using previously known
returns. We called such a volatility historical volatility. The problem with this notion of volatility
is that it uses the past to estimte future volatility and thus cannot be considered reliable. In this
section we introduce a diﬀerent type of volatility that depends on the observed market price of an
option.
Implied volatility of the underlying asset is the volatility that, when used in a particular pricing
model, yields a theoretical value for the option equal to the current market price of that option.
Thus, historical volatility tells us how volatile an asset has been in the past. Implied volatility is
the market’s view on how volatile an asset will be in the future.
Now, if we assume that the option price can be modeled by the BlackScholes formula, and the other
variables − stock price (S), strike price (K), annual continuously compounded riskfree interest rate
(r), time to expiration (T), and annual continuously compounded dividend yield (δ ) are known, the
implied volatility σ is then the solution to the equation
ˆ
Market option price = C (S, K, σ , r, T, δ ).
ˆ (34.1) There is no way to solve directly for implied volatility in (34.1). Instead, either iterative methods
(such as Newton’s method) or ﬁnancial softwares are used.
Example 34.1
Suppose we observe a 40strike 3month European call option with a premium of $2.78. The stock
price is currently $45, the interest rate is 8%, and the stock pays no dividends. Find the implied
volatility.
Solution.
We want to ﬁnd σ that satisﬁes the equation
ˆ
2.78 = C (45, 40, σ , 0.08, 0.25, 0).
ˆ
Using an electronic device, we ﬁnd σ ≈ 36%
ˆ
When using the BlackScholes or the binomial model, it is possible to conﬁne implied volatility
within particular boundaries by calculating option prices for two diﬀerent implied volatilities. If
one of these prices is greater than the observed option price and the other is less than the observed
option price, then we know that the implied volatility is somewhere between the two values for
which calculations were made. 34 IMPLIED VOLATILITY 237 Remark 34.1
On the actuarial exam, you will be given several ranges within which implied volatility might fall.
Test the extreme values of those ranges and see if the observed option price falls somewhere in
between the prices calculated by considering each of these extreme values. If it does, then you have
obtained the correct value of implied volatility.
Example 34.2
A oneyear European call option is currently valued at $60. The following parameters are given:
• The current stock price is $300
• The continuously compounded riskfree rate is 9%
• The continuously compounded dividend yield is 3%
• The strike price is $300.
Using a singleperiod binomial model and assuming the implied volatility of the stock to be at least
6%, determine the interval containing σ.
(A) less than 10%
(B) At least 10% but less than 20%
(C) At least 20% but less than 30%
(D) At least 30% but less than 40%
(E) At least 40%.
Solution.
We test the option prices that would be generated by volatilities of 0.06, 0.10, 0.20, 0.30, and 0.40.
We have
u = e(r−δ)h+σ √ h = e0.06+σ and d = e(r−δ)h−σ √ h = e0.06−σ . Because we know that σ > 0.06, we know that d < e0.06−0.06 = 1, so that Cd = 0. Thus, in calculating
the option price we need to only apply the formula
C =e−rh [pu Cu + (1 − pu )Cd ] = e−0.06 [pu Cu + (1 − pu )0]
=e−0.09 pu [300u − 300]
e0.06 − d
=e−0.09
(300u − 300)
u−d
e0.06 − e0.06−σ
=3000e−0.09 0.06+σ
(e0.06+σ − 1)
e
− e0.06−σ 238 THE BLACKSCHOLES MODEL Now we can insert various values of σ to ﬁnd corresponding values of C (σ ).
C (0.06) =16.95
C (0.10) =22.60
C (0.20) =36.65
C (0.30) =50.56
C (0.40) =64.27
Thus, the implied volatility is between 0.30 and 0.40 so that the correct answer is (D)
European puts and calls on the same asset, strike, and maturity should have the same implied
volatility to prevent arbitrage. That is, the call and put prices satisfy the putparity relationship.
A Typical pattern for volatility with regards to strike price and maturity time occurs1 :
• For ﬁxed time to maturity, implied volatility tends to decrease as strike price increases.
• For ﬁxed strike price, implied volatility tends to decrease as time to expiry increases.
Consider the graph of the implied volatility against the strike price:
• If the graph is a skew curvature then we refer to the graph as volatility skew. If the graph
a rightskewed curve then the volatility is higher as the option goes more inthemoney and lower
as the option goes outofthe money. A similar observation for leftskewed curve. A rightskewed
curve is also called volatility smirk.
• When both the inthemoney and outofthe money options have a higher implied volatility than
atthemoney options, we have a volatility smile.
• Rarely, both the inthemoney and outofthe money options have a lower implied volatility than
atthemoney options, and we have a volatility frown.
Implied volatility is important for a number of reasons. Namely,
(1) It allows pricing other options on the same stock for which there may not be observed market
price. This is done by setting the BlackScholes formula to the price of a traded option on the same
stock and solving the equation for the volatility. This volatility can be used in conjunction with
the BlackScholes formula to price options on the same stock with no observed market price.
(2) It is a quick way to describe the level of option prices. That is, a range of option prices for the
same asset can be assigned one single volatility.
(3) Volatility skew provides a measure of how good the BlackScholes pricing model is. Any as1 Since implied volatility changes with respect to time and expiry, the BlackScholes model does not fully describe
option prices. Nevertheless, it is widely used Benchmark 34 IMPLIED VOLATILITY 239 sets with options displaying a Volatility Skew is displaying inconsistency with the theories of the
BlackScholes Model in terms of constant volatility. 240 THE BLACKSCHOLES MODEL Practice Problems
Problem 34.1
A stock currently costs $41 per share. In one year, it might increase to $60. The annual continuously
compounded riskfree interest rate is 0.08, and the stock pays dividends at an annual continuously
compounded yield of 0.03. Find the implied volatility of this stock using a oneperiod binomial
model.
Problem 34.2
The stock of GS LLC currently costs $228 per share. The annual continuously compounded riskfree
interest rate is 0.11, and the stock pays dividends at an annual continuously compounded yield of
0.03. The stock price will be $281 next year if it increase. What will the stock price be next year
if it decreases? Use a oneperiod binomial model.
Problem 34.3
For a 1year European call, the following information are given:
• The current stock price is $40
• The strike price is $45
• The continuously compounded riskfree interest rate is 5%
• ∆ = 0.5. That is, for every increase of $1 in the stock price, the option price increases by $0.50.
Determine the implied volatility.
Problem 34.4
Consider a nondividendpaying stock with a current price of S, and which can go up to 160, or
down to 120, one year from now. Find the implied volatility, σ, of the stock, assuming the binomial
model framework.
Problem 34.5
Given the following information about a 1year European put on a stock:
• The current price of the stock $35
• The strike price is $35
• The continuously compounded riskfree interest rate is 8%
• The continuously compounded dividend yield is 6%.
• The observed put option price is $3.58.
Is σ = 20% the implied volatility of the stock under the BlackScholes framework?
Problem 34.6 ‡
A oneyear European call option is currently valued at 0.9645. The following parameters are given:
• The current stock price is 10 34 IMPLIED VOLATILITY 241 • The continuously compounded riskfree rate is 6%
• The continuously compounded dividend yield is 1%
• The strike price is 10.
Using a singleperiod binomial model and assuming the implied volatility of the stock to be at least
5%, determine the interval containing σ.
(A) less than 10%
(B) At least 10% but less than 20%
(C) At least 20% but less than 30%
(D) At least 30% but less than 40%
(E) At least 40%.
Problem 34.7 ‡
Assume the BlackScholes framework. Consider a oneyear atthemoney European put option on
a nondividendpaying stock.
You are given:
(i) The ratio of the put option price to the stock price is less than 5%.
(ii) Delta of the put option is −0.4364.
(iii) The continuously compounded riskfree interest rate is 1.2%.
Determine the stock’s volatility. 242 THE BLACKSCHOLES MODEL Option Hedging
In this chapter we explore the BlackScholes framework by considering the marketmaker perspective
on options. More speciﬁcally, we look at the issues that a market professional encounters. In deriving
the BlackScholes formula, it is assumed that the marketmakers are proﬁtmaximizers who want
to hedge (i.e., minimize) the risk of their option positions. On average, a competetive marketmaker should expect to break even by hedging. It turns out that the break even price is just the
BlackScholes option price. Also, we will see that the hedging position can be expressed in terms
of Greeks: delta, gamma, and theta.
We have been using the term “marketmaker” in the introduction of this chapter. What do we
mean by that term? By a marketmaker we mean a market trader who sells assets or contracts to
buyers and buys them from sellers. In other words, he/she is an intermediary between the buyers
and sellers. Marketmakers buy at the bid and sell at the ask and thus proﬁt from the askbid
spread. They do not speculate! A marketmaker generates inventory as needed by shortselling.
In contrast to marketmakers, proprietary traders are traders who buy and sell based on view of
the market, if that view is correct, trading is proﬁtable, if not leads to losses. Mostly speculators
engage in proprietary trading. 243 244 OPTION HEDGING 35 DeltaHedging
Marketmakers have positions generated by fulﬁlling customer orders. They need to hedge the risk
of these positions for otherwise an adverse price move has the potential to bankrupt the marketmaker.
One way to control risk is by deltahedging which is an option strategy that aims to reduce
(hedge) the risk associated with price movements in the underlying asset by oﬀsetting long and short
positions. This requires computing the option Greek delta and this explains the use of the term
“deltahedged.” For example, a long call position may be deltahedged by shorting the underlying
stock. For a written call, the position is hedged by buying shares of stock. The appropriate number
of shares is determined by the number delta. The delta is chosen so that the portfolio is deltaneutral, that is, the sum of the deltas of the portfolio components is zero.
There is a cost for a deltahedged position: The costs from the long and short positions are not
the same and therefore the marketmaker must invest capital to maintain a deltahedged position.
Deltahedged positions should expect to earn the riskfree rate: the money invested to maintain a
deltahedged position is tied up and should earn a return on it and moreover it is riskless.
Next, we examine the eﬀect of an unhedged position.
Example 35.1
Suppose that the current price of a stock is $40. The stock has a volatility of 30% and pays no
dividends. The continuously compounded riskfree interest rate is 8%. A customer buys a call
option from the marketmaker with strike price $40 and time to maturity of 91 days. The call is
written on 100 shares of the stock.
(a) Using the BlackScholes framework, ﬁnd C and ∆ at day the time of the transaction.
(b) What is the risk for the marketmaker?
(c) Suppose that the marketmaker leaves his position unhedged. What is the realized proﬁt if the
stock increases to $40.50 the next day?
(d) Suppose that the market makerhedges his position by buying 0.58240 shares of the stock. What
is his proﬁt/loss in the next day when the price increases to $40.50?
(e) The next day the value of delta has increased. What is the cost of keeping the portfolio hedged?
(f) Now on Day 2, the stock price goes down to $39.25. What is the marketmaker gain/loss on
that day?
Solution.
(a) Using the BlackScholes formula we ﬁnd
C = Se−δT N (d1 ) − Ke−rT N (d2 ) 35 DELTAHEDGING 245 where
d1 = ln (40/40) + (0.08 − 0 + 0.5(0.3)2 ) ×
ln (S/K ) + (r − δ + 0.5σ 2 )T
√
=
σT
0.3 91 91
365 = 0.2080477495. 365 and √
d2 = d1 − σ T = 0.2080477495 − 0.3 91
= 0.0582533699
365 Thus,
1
N (d1 ) = √
2π
and
1
N (d2 ) = √
2π 0.2080477495 x2 e− 2 dx = 0.582404
−∞
0.0582533699 x2 e− 2 dx = 0.523227.
−∞ Hence, 91 C = 40 × 0.582404 − 40e−0.08× 365 × 0.523227 = $2.7804
From the marketmaker perspective we have ∆ = −e−δT N (d1 ) = −0.5824.
(b) The risk for the marketmaker who has written the call option is that the stock price will rise.
(c) Suppose the stock rises to $40.50 the next day. We now have
d1 = ln (40.50/40) + (0.08 − 0 + 0.5(0.3)2 ) ×
ln (S/K ) + (r − δ + 0.5σ 2 )T
√
=
σT
0.3 90 90
365 = 0.290291 365 and √
d2 = d1 − σ T = 0.290291 − 0.3 90
= 0.141322.
365 Thus,
1
N (d1 ) = √
2π
and
1
N (d2 ) = √
2π
Hence, 0.290291 x2 e− 2 dx = 0.614203
−∞
0.141322 x2 e− 2 dx = 0.556192.
−∞
90 C = 40.5 × 0.614203 − 40e−0.08× 365 × 0.556192 = $3.0621.
1 Thus, there is a one day gain on the premium so the marketmaker proﬁt is 2.7804e0.08× 365 − 3.0621 =
−$0.281. That is a loss of about 28 cents a share or $28 for 100 shares. 246 OPTION HEDGING (d) On day 0, the hedged portfolio consists of buying 58.24 shares at $40 for a total cost of
58.24 × 40 = $2329.60. Since the marketmaker received only 2.7804 × 100 = $278.04, he/she
must borrow 2329.60 − 278.04 = $2051.56. Thus, the initial position from the marketmaker perspective is −2329.60 + 2051.56 + 278.04 = $0.
The ﬁnance charge for a day on the loan is
1 2051.56(e0.08× 365 − 1) = $0.45
If the price of the stock goes up to $40.50 the next day, then there is a gain on the stock, a loss on
the option and a oneday ﬁnance charge on the loan. Thus, the marketmaker realizes a proﬁt of
58.24(40.50 − 40) + (278.04 − 306.21) − 0.45 = $0.50.
This process is referred to as markingthemarket. Thus, there is an overnight proﬁt of $0.50.
(e) The delta for the price of $40.50 is ∆ = −N (d1 ) = −0.6142. In order, to deltaneutralize
the portfolio, the marketmaker needs to buy additional 61.42 − 58.24 = 3.18 shares at a cost of
3.18 × 40.50 = $128.79. This process is known as rebalancing the portfolio.
(f) Markingthemarket: The new call option price is $2.3282 per share. In this case, there
is a loss on the shares of 61.42 × (39.25 − 40.50) = −$76.78 and a gain on the option of about
306.21 − 232.82 = $73.39, plus the ﬁnance charge. Thus, the Day 2 proﬁt is
1 73.39 − 76.78 − (2051.56 + 128.79)(e0.08× 365 − 1) = −$3.87.
That is, a loss of $3.87
The daily proﬁt calculation over 5 days for the hedged portfolio is given next. Table 35.1
From this table, we notice that the return on a deltahedged position does not depend on the
direction in which the stock price moves, but it does depend on the magnitude of the stock price 35 DELTAHEDGING 247 move.
The three sources of cash ﬂow into and out of the portfolio in the previous example are:
• Borrowing: The marketmaker capacity to borrowing is limited by the market value of the
securities in the portfolio. In practice, the marketmaker can borrow only part of the funds required
to buy the securities so he/she must have capital to make up the diﬀerence.
•Purchase or Sale of shares The marketmaker must buysell shares in order to oﬀset changes
in the option price.
•Interest: The ﬁnance charges paid on borrowed money.
Remark 35.1
The calculation of the proﬁts in the above example is referred to as marktomarket proﬁts. With
positive marktomarket proﬁt the marketmaker is allowed to take money out of the portfolio. In
the case of negative marktomarket proﬁt the investor must put money into the portfolio. A hedged
portfolio that never requires additional cash investments to remain hedged is called a selfﬁnancing
portfolio. It can be shown that any portfolio for which the stock moves according to the binomial
model is approximately selfﬁnancing. More speciﬁcally, the deltahedged portfolio breaks even if
the stock moves one standard deviation. See Problem 35.6.
Example 35.2
A stock is currently trading for $40 per share. The stock will pay no dividends. The annual
continuously compounded riskfree interest rate is 0.08, and the stock price volatility is 0.3. Suppose
you sell a 40strike put on 100 shares with 91 days to expiration.
(a) What is delta?
(b) What investment is required for a deltahedged portfolio?
(c) What is your proﬁt the next day if the stock falls to $39?
(d) What if the stock goes up to $40.50 instead?
Solution.
(a) Using the BlackScholes formula we ﬁnd
P = Ke−rT N (−d2 ) − Se−δT N (−d1 )
where
d1 = ln (40/40) + (0.08 − 0 + 0.5(0.3)2 ) ×
ln (S/K ) + (r − δ + 0.5σ 2 )T
√
=
σT
0.3 91
365 and √
d2 = d1 − σ T = 0.20805 − 0.3 91
= 0.0582.
365 91
365 = 0.20805 248 OPTION HEDGING Thus,
1
N (d1 ) = √
2π
and
1
N (d2 ) = √
2π 0.20805 x2 e− 2 dx = 0.582405
−∞
0.0582 x2 e− 2 dx = 0.523205.
−∞ Hence,
91 P = 40e−0.08× 365 (1 − 0.523205) − 40(1 − 0.582405) = $1.991
and
∆ = −e−δT N (−d1 ) = −(1 − 0.582405) = −0.4176.
The value of delta from the marketmaker perspective is ∆ = 0.4176.
(b) The marketmaker sells 41.76 shares and deposit the amount
41.76 × 40 + 199.1 = $1869.50
at a savings account earning the riskfree interest rate. The initial position, from the perspective of
the marketmaker is
41.76 × 40 + 199.1 − 1869.50 = $0.
(c) Suppose the stock falls to $39.00 the next day. We now have
ln (39/40) + (0.08 − 0 + 0.5(0.3)2 ) ×
ln (S/K ) + (r − δ + 0.5σ 2 )T
√
d1 =
=
σT
0.3 90 90
365 365 and √
d2 = d1 − σ T = 0.03695 − 0.3 90
= −0.1121.
365 Thus,
1
N (d1 ) = √
2π
and
1
N (d2 ) = √
2π 0.03695 x2 e− 2 dx = 0.514738
−∞
−0.1121 x2 e− 2 dx = 0.455372.
−∞ Hence,
90 P = 40e−0.08× 365 (1 − 0.455372) − 39(1 − 0.514738) = $2.434. = 0.03695 35 DELTAHEDGING 249 Hence, there is a gain on the shares but a loss on the option and gain from the interest. The
marketmaker overnight proﬁt is
1 41.76(40 − 39) − (243.40 − 199.10) + 1869.50(e0.08× 365 − 1) = −$2.130.
(d) Suppose the stock rises to $40.50 the next day. We now have
ln (40.50/40) + (0.08 − 0 + 0.5(0.3)2 ) ×
ln (S/K ) + (r − δ + 0.5σ 2 )T
√
=
d1 =
σT
0.3 90 90
365 = 0.290291 365 and √
d2 = d1 − σ T = 0.290291 − 0.3 90
= 0.141322.
365 Thus,
1
N (d1 ) = √
2π
and
1
N (d2 ) = √
2π
Hence, 0.290291 x2 e− 2 dx = 0.614203
−∞
0.141322 x2 e− 2 dx = 0.556192.
−∞ 90 P = 40e−0.08× 365 (1 − 0.556192) − 40.50(1 − 0.614203) = $1.7808.
Thus, there is a gain on the option but a loss on the shares and interest gain. The overnight proﬁt
is
1
(199.10 − 178.08) − 41.76 × 0.50 + 1869.50(e0.08× 365 − 1) = $0.55 250 OPTION HEDGING Practice Problems
Problem 35.1
A stock is currently trading for $40 per share. The stock will pay no dividends. The annual
continuously compounded riskfree interest rate is 0.08, and the stock price volatility is 0.3. Suppose
you sell a 45strike call on 100 shares with 91 days to expiration.
(a) What is delta?
(b) What investment is required for a deltahedged portfolio?
(c) What is your proﬁt the next day if the stock falls to $39?
(d) What if the stock goes up to $40.50?
Problem 35.2
A stock is currently trading for $40 per share. The stock will pay no dividends. The annual
continuously compounded riskfree interest rate is 0.08, and the stock price volatility is 0.3. Suppose
you buy a 40strike call and sell a 45strike call both expiring in 91 days. Both calls are written on
100 shares.
(a) What investment is required for a deltahedged portfolio?
(b) What is your proﬁt the next day if the stock falls to $39?
(c) What if the stock goes up to $40.50 instead?
Problem 35.3
A stock is currently trading for $60 per share. The stock will pay no dividends. The annual continuously compounded riskfree interest rate is 0.08, and the stock price volatility is 0.25. Suppose
you sell a 60strike call expiring in 91 days. What is you overnight proﬁt if the stock goes up to
$60.35?
Problem 35.4
A stock is currently trading for $40 per share. The stock will pay no dividends. The annual
continuously compounded riskfree interest rate is 0.08, and the stock price volatility is 0.3. Suppose
you buy a 45strike put and sell two 40strike puts both expiring in 91 days. That is, you enter into
a put ratio spread. Both puts are written on 100 shares.
(a) What investment is required for a deltahedged portfolio?
(b) What is your proﬁt the next day if the stock falls to $39?
(c) What if the stock goes up to $40.50 instead?
Problem 35.5
1
Suppose that the stock moves up and down according to the binomial model with h = 365 . Assume
that on Days 1 and 5 the stock price goes up and on Days 2, 3, and 4 it goes down. Find the
√
magnitude move σ √h. Complete the following table. Assume σ = 0.3 and S = $40. Hint: Recall
that St+h = St erh±σ h . 35 DELTAHEDGING 251
Day
Stock Price 01
40 2 3 4 5 Problem 35.6
Construct a table similar to Table 35.1 using the daily change in stock prices found in the previous
problem.
Problem 35.7 ‡
A marketmaker has sold 100 call options, each covering 100 shares of a dividendpaying stock, and
has deltahedged by purchasing the underlying stock.
You are given the following information about the marketmaker’s investment:
• The current stock price is $40.
• The continuously compounded riskfree rate is 9%.
• The continuous dividend yield of the stock is 7%.
• The time to expiration of the options is 12 months.
• N (d1 ) = 0.5793
• N (d2 ) = 0.5000
The price of the stock quickly jumps to $41 before the marketmaker can react. This change the price
of one call option to increase by $56.08. Calculate the net proﬁt on the marketmaker investment
associated with this price move. Note: Because of sudden jump interest on borrowing or lending is
ignored.
Problem 35.8 ‡
Several months ago, an investor sold 100 units of a oneyear European call option on a nondividendpaying stock. She immediately deltahedged the commitment with shares of the stock, but has not
ever rebalanced her portfolio. She now decides to close out all positions.
You are given the following information:
(i) The riskfree interest rate is constant.
(ii)
Several Months Ago
Stock Price
$40
Call Option Price $8.88
Put Option Price $1.63
Call Option Delta 0.794 Now
$50
$14.42
$0.26 The put option in the table above is a European option on the same stock and with the same strike
price and expiration date as the call option.
Calculate her proﬁt. 252 OPTION HEDGING 36 Option Price Approximations: Delta and DeltaGamma
Approximations
In this section we use ﬁnite Taylor approximations to estimate option price movements when the
underlying stock price changes, particularly ﬁrst and second order (delta and gamma.)
We start by recalling from calculus the Taylor series expansion given by
f (x + h) = f (x) + h df
h2 d2 f
h3 d3 f
( x) +
+
(x) + higher order terms.
dx
2 dx2
6 dx3 Let V (St ) denote the option price when the stock price is St . Recall that
∆= ∂V
.
∂S A linear approximation of the value of the option when the stock price is St+h can be found using
Taylor series of ﬁrst order given by
V (St+h ) = V (St ) + ∆(St ) + higher order terms
where = St+h − St is the stock price change over a time interval of length h. We call the approximation
V (St+h ) ≈ V (St ) + ∆(St )
the delta approximation. Notice that the delta approximation uses the value of delta at St .
Recall that the price functions (for a purchased option) are convex functions of the stock price (See
Section 29). The delta approximation is a tangent line to the graph of the option price. Hence, the
delta approximation is always an underestimate of the option price. See Figure 36.1.
Example 36.1
Consider a nondividend paying stock. The annual continuously compounded riskfree interest rate
is 0.08, and the stock price volatility is 0.3. Consider a 40strike call on 100 shares with 91 days to
expiration.
(a) What is the option price today if the current stock price is $40?
(b) What is the option price today if the stock price is $40.75?
(c) Estimate the option price found in (b) using the delta approximation.
Solution.
(a) Using the BlackScholes formula we ﬁnd
C = Se−δT N (d1 ) − Ke−rT N (d2 ) 36 OPTION PRICE APPROXIMATIONS: DELTA AND DELTAGAMMA APPROXIMATIONS253
where
d1 = ln (40/40) + (0.08 − 0 + 0.5(0.3)2 ) ×
ln (S/K ) + (r − δ + 0.5σ 2 )T
√
=
σT
0.3 91 91
365 = 0.2080477495. 365 and √
d2 = d1 − σ T = 0.2080477495 − 0.3 91
= 0.0582533699
365 Thus,
1
N (d1 ) = √
2π
and
1
N (d2 ) = √
2π 0.2080477495 x2 e− 2 dx = 0.582404
−∞
0.0582533699 x2 e− 2 dx = 0.523227.
−∞ Hence,
91 C = 40 × 0.582404 − 40e−0.08× 365 × 0.523227 = $2.7804
and
∆ = e−δT N (d1 ) = 0.5824.
(b) Suppose that the option price goes up to $40.75. Then
d1 = ln (40.75/40) + (0.08 − 0 + 0.5(0.3)2 ) ×
ln (S/K ) + (r − δ + 0.5σ 2 )T
√
=
σT
0.3 91 90
365 = 0.3320603167 365 and √
d2 = d1 − σ T = 0.3320603167 − 0.3 91
= 0.1822659371.
365 Thus,
1
N (d1 ) = √
2π
and
1
N (d2 ) = √
2π 0.3320603167 x2 e− 2 dx = 0.630078
−∞
0.1822659371 x2 e− 2 dx = 0.572313.
−∞ Hence,
91 C ($40.75) = 40.75 × 0.630078 − 40e−0.08× 365 × 0.572313 = $3.2352 254 OPTION HEDGING (c) Using the delta approximation we ﬁnd
C ($40.75) = C ($40) + 0.75 × 0.5824 = 2.7804 + 0.75 × 0.5824 = $3.2172.
The error is 3.2172 − 3.2352 = −0.018
A second method of approximation involves both delta and gamma and uses Taylor approximation of order two:
V (St+h ) = V (St ) + ∆(St ) + 12
Γ(St ) + higher order terms.
2 We deﬁne the deltagamma approximation by
V (St+h ) ≈ V (St ) + ∆(St ) + 12
Γ(St ).
2 Example 36.2
Consider the information of the previous example.
(a) Find the option Greek gamma.
(b) Estimate the value of C ($40.75) using the deltagamma approximation.
Solution.
(a) We have
0.20804774952 Γcall 2
e−
e−δ(T −t) N (d1 )
√
=
=
√
Sσ T − t
40 × 0.3 2π × = 0.0652.
91
365 (b) We have the estimate
C ($40.75) = 2.7804 + 0.5824(0.75) + 0.5 × 0.752 × 0.0652 = $3.2355.
In this case, the error is 3.2355 − 3.2352 = 0.0003. Thus, the approximation is signiﬁcantly closer
to the true option price at $40.75 than the delta approximation
Figure 36.1 shows the result of approximating the option price using the delta and deltagamma
approximations. 36 OPTION PRICE APPROXIMATIONS: DELTA AND DELTAGAMMA APPROXIMATIONS255 Figure 36.1 256 OPTION HEDGING Practice Problems
Problem 36.1
For a stock price of $40 the price of call option on the stock is $2.7804. When the stock goes up to
$40.75 the price of the option estimated by the delta approximations is $3.2172. Find ∆(40).
Problem 36.2
A stock is currently trading for $45 per share. The stock will pay no dividends. The annual
continuously compounded riskfree interest rate is 0.08, and the stock price volatility is 0.3. Consider
a 45strike call on 100 shares with 91 days to expiration. Use delta approximation to estimate
C ($44.75).
Problem 36.3
A stock is currently trading for $45 per share. The stock will pay no dividends. The annual
continuously compounded riskfree interest rate is 0.08, and the stock price volatility is 0.3. Consider
a 45strike call on 100 shares with 91 days to expiration. Use deltagamma approximation to
estimate C ($44.75).
Problem 36.4
A stock is currently trading for $40 per share. The stock will pay no dividends. The annual
continuously compounded riskfree interest rate is 0.08, and the stock price volatility is 0.3. Consider
a 40strike put on 100 shares with 91 days to expiration. Using the delta approximation estimate
P ($40.55).
Problem 36.5
A stock is currently trading for $40 per share. The stock will pay no dividends. The annual
continuously compounded riskfree interest rate is 0.08, and the stock price volatility is 0.3. Consider
a 40strike put on 100 shares with 91 days to expiration. Using the deltagamma approximation
estimate P ($40.55).
Problem 36.6
A stock is currently trading for $1200 per share. A certain call option on the stock has a price of
$35, a delta of 0.72, and a certain value of gamma. When the stock price suddenly falls to $1178,
the call price falls to $23. Using the deltagamma approximation, what is the gamma of this call
option?
Problem 36.7
A stock is currently trading for $13 per share. A certain call option on the stock has a price of
$1.34, a gamma of 0.025, and a certain value of delta. When the stock price suddenly rises to $19
per share, the call option price increases to $5.67. Using the deltagamma approximation, what is
the original delta of this call option? 36 OPTION PRICE APPROXIMATIONS: DELTA AND DELTAGAMMA APPROXIMATIONS257
Problem 36.8
The stock of GS Co. currently trades for $657 per share. A certain call option on the stock has a
price of $120, a delta of 0.47, and a gamma of 0.01. Use a deltagamma approximation to ﬁnd the
price of the call option if, after 1 second, the stock of GS Co. suddenly begins trading at $699 per
share.
Problem 36.9
When the stock of GS Co. suddenly decreased in price by $6 per share, a certain put option on
the stock increased in price to $5.99. The put option had an original ∆ of −0.49 and a gamma of
0.002. Find the original put option price using the deltagamma approximation.
Problem 36.10
A stock is currently trading for a price greater than $75 per share. A certain put option on the stock
has a price of $5.92, a delta of −0.323, and a gamma of 0.015. Use a deltagamma approximation
to ﬁnd the current price of the stock if, after 1 second, the put is valued at $6.08 when the stock
price is $86.
Problem 36.11 ‡
Assume that the BlackScholes framework holds. The price of a nondividendpaying stock is $30.00.
The price of a put option on this stock is $4.00.
You are given:
(i) ∆ = −0.28
(ii) Γ = 0.10.
Using the deltagamma approximation, determine the price of the put option if the stock price
changes to $31.50.
Problem 36.12 ‡
Assume that the BlackScholes framework holds. Consider an option on a stock.
You are given the following information at time 0:
(i) The stock price is S (0), which is greater than 80.
(ii) The option price is 2.34.
(iii) The option delta is −0.181.
(iv) The option gamma is 0.035.
The stock price changes to 86.00. Using the deltagamma approximation, you ﬁnd that the option
price changes to 2.21.
Determine S (0). 258 OPTION HEDGING 37 The DeltaGammaTheta Approximation and the MarketMaker’s Proﬁt
The deltagamma approximation discussed in the previous section does not take into consideration
the sensitivity of the option price change due to time passing. Recall that the Greek option theta
measures the option’s change in price due to time passing. Thus, the change in price over a period
of length h is approximately hθ(St ). Adding this term to the deltagamma approximation formula
we obtain
V (St , T − t − h) ≈ V (St , T − t) + ∆(St , T − t) + 12
Γ(St , T − t) + hθ(St , T − t).
2 (37.1) This formula is known as the deltagammatheta approximation.
Example 37.1
Yesterday, a nondividend paying stock was selling for $40 and a call option on the stock was selling
for $2.7804 and has 91 days left to expiration. The call option on the stock had a delta of 0.5824,
a gamma of 0.0652, and an annual theta of −6.3145. Today, the stock trades for $40.75. The
annual continuously compounded riskfree interest rate is 0.08. Find the new option price using the
deltagammatheta approximation.
Solution.
Using the deltagammatheta approximation we have
C (40, 91
91
1
91
91
90
) ≈C (40,
) + ∆(40,
) + 2 Γ(40,
) + hθ(40,
)
365
365
365
2
365
365
1
1
=2.7804 + 0.75(0.5824) + × 0.752 +
(−6.3145) = $3.2182
2
365 We next examine the marketmaker proﬁt. Consider the case of long delta shares and short a call.
The value of the marketmaker investment is the cost of the stock plus the proceeds received from
selling the call, that is,
C (St ) − St ∆(St ) < 0.
This amount is borrowed at the riskfree interest rate r. Now, suppose that over the time interval
h, the stock price changes from St to St+h . In this case, the marketmaker proﬁt is
(St+h − St )∆(St ) − (C (St+h ) − C (St )) − rh[St ∆ − C (St )].
But from equation (37.1), using = St+h − St , we have
C (St+h ) − C (St ) = ∆(St ) + 12
Γ(St ) + hθ.
2 37 THE DELTAGAMMATHETA APPROXIMATION AND THE MARKETMAKER’S PROFIT259
Thus, the marketmaker proﬁt can be expressed in the form
MarketMaker Proﬁt = ∆(St ) −
=− ∆(St ) + 12
Γ(St ) + hθ − rh[St ∆(St ) − C (St )]
2 (37.2) 12
Γ(St ) + hθ(St ) + rh[St ∆(St ) − C (St )] .
2 From this equation we observe the following:
• Gamma factor: Because the gamma of a purchased call is positive (so a stock price increase will
increase the value of the call) and the marketmaker is the writer of the call then the marketmaker
will lose money in proportion to the square of the stock price change . Thus, the larger the stock
move, the greater is the loss.
• Theta factor: The theta for a purchased call is negative which means that the call value decreases
with time passing. Since the marketmaker is the writer of the call, he/she beneﬁts from theta.
• Interest cost: The marketmaker being the seller has a net investment which requires borrowing
since the ∆(St ) shares are more expensive than one short option. The net cost of borrowing for a
time interval h is the term −rh[St ∆(St ) − C (St )].
It follows that time passing beneﬁts the marketmaker, whereas interest and gamma work against
him or her.
Example 37.2
22 days ago, GS stock traded for $511 per share. A certain call option on the stock had a delta
of 0.66, a gamma of 0.001, and an annual theta of −10.95. The option used to trade for $59. Now
the stock trades for $556. The annual continuously compounded riskfree interest rate is 0.08.
A hypothetical marketmaker has purchased delta shares and shortsold one call. Find what a
marketmaker’s proﬁt on one such option would be using (37.2).
Solution.
Using (37.2) we ﬁnd
12
Γ(St ) + hθ(St ) + rh[St ∆(St ) − C (St )]
2
1
22
22
=−
× 452 × 0.001 + (−10.95) ×
+ 0.08 ×
[0.66 × 511 − 59]
2
365
365
= − $1.694246849 MarketMaker proﬁt = − Now, we examine the eﬀect of 2 on the proﬁt. Recall from Section 35 that the magnitude of 2 −
and not the direction of the stock price move− aﬀects the marketmaker proﬁt. Also, recall that 260 OPTION HEDGING the marketmaker approximately breaks even for a onestandard deviation move in the stock. So if
σ is measured annually, then a onestandarddeviation move over a period of length h is
√
√
= St+h − St = St (1 + σ h) − St = σSt h.
With this expression for , the marketmaker proﬁt when the stock moves one standard deviation
is
122
σ St Γ(St ) + θ(St ) + r[St ∆(St ) − C (St )] h.
MarketMaker proﬁt = −
2
Example 37.3
22 days ago, the stock of GS traded for $511 per share. A certain call option on the stock had
a delta of 0.66, a gamma of 0.001, and a daily theta of −0.03. The option used to trade for $59.
Now the stock trades for $556. The annual continuously compounded riskfree interest rate is 0.08.
A hypothetical market maker has purchased delta shares and shortsold the call. Assume a onestandarddeviation of stock price move, what is the annual standard deviation of the stock price
movement?
Solution.
We have
2 σ= St2 h 1
2 = 452
5112 × 1
2 22
365 = 0.3586961419 Greeks in the Binomial Model
We will use some of the relations of this section to compute the binomial Greeks. Consider a
binomial model with period of length h. At time t = 0 we have
∆(S, 0) = e−δh Cu − Cd
.
uS − dS That’s the only Greek we can compute at that time. After one period, we have
∆(uS, h) = e−δh Cuu − Cud
uuS − udS Cud − Cdd
udS − ddS
∆(uS, h) − ∆(dS, h)
Γ(S, 0) ≈ Γ(Sh , h) =
.
uS − dS
∆(dS, h) = e−δh Now, letting
= udS − S 37 THE DELTAGAMMATHETA APPROXIMATION AND THE MARKETMAKER’S PROFIT261
we can use the deltagammatheta approximation to write
C (udS, 2h) = C (S, 0) + ∆(S, 0) + 12
Γ(S, 0) + 2hθ(S, 0).
2 Solving this equation for θ we ﬁnd
θ(S, 0) = C (udS, 2h) − ∆(S, 0) − 0.5 2 Γ(S, 0) − C (S, 0)
.
2h Here S is the stock price at time t = 0.
Example 37.4 ‡
Consider the following threeperiod binomial tree model for a stock that pays dividends continuously at a rate proportional to its price. The length of each period is 1 year, the continuously
compounded riskfree interest rate is 10%, and the continuous dividend yield on the stock is 6.5%. Approximate the value of gamma at time 0 for the 3year atthemoney American put on the
stock
Solution.
We ﬁrst ﬁnd u and d. We have u =
pu = 375
300 = 1.25 and s = 210
300 = 0.70. Thus, e(r−δ)h − d
300e0.10−0.065 − 210
=
= 0.61022.
u−d
375 − 210 we have
Year 3, Stock Price = u3 S = 585.9375 Since we are at expiration, the option value is Puuu = 0. 262
Year
Year
Year
Year OPTION HEDGING
3,
3,
3,
2, Stock
Stock
Stock
Stock Price
Price
Price
Price = u2 dS = 328.125 and Puud = 0.
= ud2 S = 183.75 and Pudd = Pddu = 300 − 183.75 = 116.25.
= d3 S = 102.90 and Pddd = 300 − 102.90 = 197.10.
= u2 S = 468.75 Puu = max{300 − 468.75, e−0.10 [0.61022 × 0 + (1 − 0.61022) × 0]}
=0.
Year 2, Stock Price = udS = 262.50
Pud = max{300 − 262.50, e−0.10 [0.61022 × 0 + (1 − 0.61022) × 116.25]}
=e−0.10 [0.61022 × 0 + (1 − 0.61022) × 116.25]
=41.00
Year 2, Stock Price = d2 S = 147
Pdd = max{300 − 147, e−0.10 [0.61022 × 116.25 + (1 − 0.61022) × 197.10]}
=300 − 147
=153.
Hence,
∆(uS, h) = e−δh Puu − Pud
0 − 41
= e−0.065×1
= −0.1863
uuS − udS
468.75 − 262.50 and
∆(dS, h) = e−δh Pud − Pdd
41 − 153
= e−0.065×1
= −0.9087.
udS − ddS
262.50 − 147 Thus,
Γ(S, 0) ≈ Γ(Sh , h) = ∆(uS, h) − ∆(dS, h)
−0.1863 − (−0.9087)
=
= 0.004378
uS − dS
375 − 210 37 THE DELTAGAMMATHETA APPROXIMATION AND THE MARKETMAKER’S PROFIT263 Practice Problems
Problem 37.1
Yesterday, a nondividend paying stock was selling for $40 and a call option on the stock was selling
for $2.7804 and has 91 days left to expiration. The call option on the stock had a delta of 0.5824,
a gamma of 0.0652, and a daily theta of −0.0173. Today, the stock trades for $39.25. The annual
continuously compounded riskfree interest rate is 0.08. Find the new option price using the deltagammatheta approximation.
Problem 37.2
A stock is currently trading for $678 per share. 98 days ago, it traded for $450 per share and a
call option on the stock was selling for $56. Now the call option trades for $100. The option has a
delta of 0.33 and a gamma of 0.006. What is the daily option theta? Use the deltagammatheta
approximation.
Problem 37.3
A call option on a nondividendpaying stock was valued $6.13 yesterday when the stock price was
$54. Today, the stock price is $56. Estimate the price of option using the deltagammatheta
approximation. The option has a delta of 0.5910, a gamma of 0.0296 and an annualized theta of
−14.0137.
Problem 37.4
GS stock has a price volatility of 0.55. A certain call option on the stock today costs $71.80. The
option has a delta of 0.32, a gamma of 0.001, and a daily theta of −0.06. The stock price today is
$3000 per share, and the annual continuously compounded riskfree interest rate is 0.1. Find what
a marketmaker’s proﬁt on one such option would be after 1 year using the deltagammatheta
approximation. Assume stock price moves one standard deviation.
Problem 37.5 ‡
Which of the following are correct for option Greeks?
(I) If the gamma of a call is positive, by writing the call the marketmaker will lose money in
proportion to the stock price change.
(II) If the theta for a call is negative, the option writer beneﬁts from theta.
(III) If, in order to hedge, the marketmaker must purchase stock, then the net carrying cost is a
component of the overall cost.
Problem 37.6
Yesterday, a stock traded for $75 per share. A certain call option on the stock had a delta of
0.5910, a gamma of 0.0296, and an annual theta of −14.0317. The option used to trade for $6.13. 264 OPTION HEDGING Now the stock trades for $77. The annual continuously compounded riskfree interest rate is 0.10.
A hypothetical marketmaker has purchased delta shares and shortsold one call. Find what a
marketmaker’s proﬁt on one such option would be using (37.2).
Problem 37.7
Consider the following threeperiod binomial tree model for a stock that pays dividends continuously at a rate proportional to its price. The length of each period is 1 year, the continuously
compounded riskfree interest rate is 10%, and the continuous dividend yield on the stock is 6.5%. Compute the value of delta at time 0 for the 3year atthemoney American put on the stock
Problem 37.8
Consider the same threeperiod binomial tree model as above for a stock that pays dividends
continuously at a rate proportional to its price. The length of each period is 1 year, the continuously
compounded riskfree interest rate is 10%, and the continuous dividend yield on the stock is 6.5%.
Estimate the value of theta at time 0 for the 3year atthemoney American put on the stock 38 THE BLACKSCHOLES ANALYSIS 265 38 The BlackScholes Analysis
In this section we describe the BlackScholes analysis for pricing options which incorporate deltahedging with pricing. We ﬁrst start by listing the assumptions under which the BlackScholes
equation holds for options:
• The riskfree interest rate r and the stock volatility σ are constant over the life of the option.
• Both the stock and the option do not pay dividends during the life of the option.
• The stock price moves one standard deviation over a small time interval (i.e., binomial model is
used).
Black and Scholes argued that the marketmaker proﬁt is zero for a onestandarddeviation of the
price of a stock. But we know that the marketmaker proﬁt in this case is given by the formula
MarketMaker proﬁt = − 122
σ St Γ(St ) + θ(St ) + r[St ∆(St ) − C (St )] h.
2 Thus, setting this equation to 0 and rearranging terms we ﬁnd the diﬀerential equation
122
σ St Γ(St ) + rSt ∆(St ) + θ = rC (St ).
2 (38.1) Since the Greeks ∆, Γ, and θ are partial derivatives, the previous equation is a partial diﬀerential
equation of order 2. We call (38.1) the BlackScholes partial diﬀerential equation. A similar
equation holds for put options.
Example 38.1
A stock has a current price of $30 per share, and the annual standard deviation of its price is 0.3.
A certain call option on this stock has a delta of 0.4118, a gamma of 0.0866, and an annual theta
of −4.3974. The annual continuously compounded riskfree interest rate is 0.08. What is the price
of this call option, as found using the BlackScholes equation?
Solution.
Using equation (38.1) we can write
122
σ St Γ(St ) + rSt ∆(St ) + θ
2
1
1
× 0.32 × 0.0866 × 302 + 0.08 × 0.4118 × 30 − 4.3974
=
0.08 2
=$1.22775 C (St ) = 1
r Equation (38.1) holds for European options. The BlackScholes equation applies to American
options only when immediate exercise is not optimal. We examine this issue more closely. Consider 266 OPTION HEDGING an American call option where early exercise is optimal. Then the option price is C = S − K.
2
Hence, ∆ = ∂C = 1, Γ = ∂ C = 0, and θ = ∂C = 0. In this case, equation (38.1) becomes
∂S
∂S 2
∂t
122
σ St × 0 + rSt × (1) + 0 = r(S − K ).
2
This leads to rK = 0 which is impossible since r > 0 and K > 0.
Rehedging
Up to this point, we have assumed that marketmakers maintain their hedged position every time
there is a change in stock price. Because of transaction costs, this process becomes expensive. So
what would the optimal frequency of rehedging be?
One approach to answer this question is to consider hedging at discrete intervals, rather than every
time the stock price changes. For that purpose we introduce the BoyleEmanuel framework:
• The marketmaker shorts a call and deltahedge it.
• Each discrete interval has length h with h in years. That is, rehedging occurring every h years.
• Let xi be a standard random variable deﬁned to be the number of standard deviations the stock
price moves during time interval i. We assume that the xi s are independent random variables, that
is, they are uncorrelated across time.
• Let Rh,i denote the period−i return (not rate of return!). For xi standard deviations the marketmaker proﬁt is given by
MarketMaker proﬁt = − 1222
σ S xi Γ + θ + r[S ∆ − C ] h.
2 Thus,
122
σ S Γ + θ + r [S ∆ − C ] h
2
1222
−−
σ S xi Γ + θ + r [ S ∆ − C ] h
2
1
= σ 2 S 2 (x2 − 1)Γh
i
2 Rh,i = − Thus, we can write
Var(Rh,i ) = 122
σ S Γh
2 2 Var(x2 − 1).
i But x2 is a gamma distribution with α = 0.5 and θ = 2 so that Var(x2 − 1) = Var(x2 ) = αθ2 = 2.
i
i
i
Hence,
1 22
2
Var(Rh,i ) =
σ S Γh .
2 38 THE BLACKSCHOLES ANALYSIS 267 The annual variance of return is
Annual variance of return = 1
1 22
Var(Rh,i ) =
σSΓ
h
2 2 h. This says that the annual variance of return is proportional to h. Thus, frequent hedging implies a
smaller h and thus reduces the variance of the return. As an example, let’s compare hedging once
a day against hedging hourly. The daily variance of the return earned by the marketmaker who
hedges once a day is given by
2
1
Γ
Var(R 1 ,1 ) =
S 2σ2
.
365
2
365
Let R 1 ,i be the variance of the return of the marketmaker who hedges hourly. But the daily
365×24
return of the marketmaker who hedges hourly is the sum of the hourly returns. Then, using
independence, we have
24 Var 24 Rh,i = i=1 Var(R
i=1
24 =
i=1 = 1
,i
365×24 ) 1 22
[S σ Γ/(24 × 365)]2
2 1
Var(R 1 ,1 )
365
24 Thus, by hedging hourly instead of daily the marketmaker’s variance daily return is reduced by a
factor of 24.
Example 38.2
A deltahedged marketmaker has a short position in a call option on a certain stock. Readjusted
hedges occur every 2 months. The stock has a price of $45; the standard deviation of this price is
0.33. The gamma of the call option is 0.02. During a particular 2month period, the stock price
moves by 0.77 standard deviations. Find the variance of the return to the marketmaker during
this time period.
Solution.
Using BoyleEmanuel formula we have
1
Var(R 1 ,1 ) = (S 2 σ 2 Γ/6)2
6
2
1
= (452 × 0.332 × 0.02/6)2 = 0.2701676278
2 268 OPTION HEDGING Practice Problems
Problem 38.1
Show that the BlackScholes equation does not hold for a put option with optimal early exercise.
Problem 38.2
A stock has a current price of $30 per share, and the annual standard deviation of its price is 0.3.
A certain put option on this stock has a delta of −0.5882, a gamma of 0.0866, and an annual theta
of −1.8880. The annual continuously compounded riskfree interest rate is 0.08. What is the price
of this put option, as found using the BlackScholes equation?
Problem 38.3
A stock has a current price of $50 per share, and the annual standard deviation of its price is 0.32.
A certain European call option on this stock has a delta of 0.5901, a gamma of 0.0243, and an
annual theta of −4.9231. The annual continuously compounded riskfree interest rate is 0.08. What
is the price of a European put option on the stock? Both options have a strike price of $53 and a
1year maturity.
Problem 38.4
Assume that the BlackScholes framework holds. The stock of GS has a price of $93 per share, and
the price has an annual standard deviation of 0.53. A certain European call option on the stock has
a price of $4, a delta of 0.53, and a gamma of 0.01. The annual continuously compounded riskfree
interest rate is 0.02. What is the theta for this call option?
Problem 38.5
Assume that the BlackScholes framework holds. The price of a stock has a standard deviation of
0.32. A certain put option on this stock has a price of $2.59525, a delta of −0.5882, a gamma of
0.0866, and a theta of −1.8880. The annual continuously compounded riskfree interest rate is 0.08.
What is the current price of one share of the stock? Round your answer to the nearest dollar.
Problem 38.6
Assume that the BlackScholes framework holds. The stock of GS has a price of $93 per share, and
the price has an annual standard deviation of 0.53. A certain European call option on the stock
has a price of $4, a delta of 0.53, and an annual theta of −13.0533205. The annual continuously
compounded riskfree interest rate is 0.02. What is the gamma for this call option?
Problem 38.7
A stock has a price of $30, and this price has a standard deviation of 0.3. A certain call option
in this stock has a price of $1.22775, a gamma of 0.0866, and a theta of −4.3974. The annual
continuously compounded riskfree interest rate is 0.08. Find the delta of this call option. 38 THE BLACKSCHOLES ANALYSIS 269 Problem 38.8
A deltahedged marketmaker has a short position in a call option on a certain stock. Readjusted
hedges occur every hour. The stock has a price of $50; the standard deviation of this price is 0.30.
The gamma of the call option is 0.0521. During a particular 1hour period, the stock price moves by
0.57 standard deviations. Find the standard deviation of the hourly returns to the marketmaker
during this time period.
Problem 38.9
A deltahedged marketmaker has a long position in a call option on a certain stock. Readjusted
hedges occur every 5 months. The stock has a price of $500; the standard deviation of this price is
0.03. The gamma of the call option is 0.001. During a particular 5month period, the stock price
moves by 0.23 standard deviations. Find the return for the marketmaker during this time period.
Problem 38.10
A deltahedged marketmaker has a short position in a call option on a certain stock. Readjusted
hedges occur every 2 days. The stock has a price of $80; the standard deviation of this price is 0.3.
The gamma of the call option is 0.02058. Find the variance of the 2day return if the marketmaker
hedges daily instead of every two days.
Problem 38.11 ‡
Consider the BlackScholes framework. A marketmaker, who deltahedges, sells a threemonth
atthemoney European call option on a nondividendpaying stock.
You are given:
(i) The continuously compounded riskfree interest rate is 10%.
(ii) The current stock price is 50.
(iii) The current call option delta is 0.6179.
(iv) There are 365 days in the year.
If, after one day, the marketmaker has zero proﬁt or loss, determine the stock price move over the
day. 270 OPTION HEDGING 39 DeltaGamma Hedging
One of the major risk that a marketmaker faces is the extreme moves of prices. There are at least
four ways for a delta hedging marketmaker to protect against extreme price moves.
(1)GammaNeutrality: Recall from the previous section that a deltahedged portfolio with a
negative gamma results in large losses for increasing move in the stock price. Thus, gamma is
a factor that a deltahedged marketmaker needs to worry about. Since a deltahedged portfolio
maintains a deltaneutral position, it makes sense for the marketmaker to adopt a gammaneutral
position. But the gamma of a stock is zero so the marketmaker has to buy/sell options so as to
oﬀset the gamma of the existing position.
Example 39.1
Suppose that the current price of a stock is $40. The stock has a volatility of 30% and pays no
dividends. The continuously compounded riskfree interest rate is 8%. A customer buys a call
option from the marketmaker with strike price $40 and time to maturity of three months. The call
is written on 100 shares of the stock.
(a) Using the BlackScholes framework, ﬁnd C, ∆ and Γ.
(b) Find C, ∆ and Γ for a purchased 45strike call with 4 months to expiration.
(c) Find the number of purchased 45strike call needed for maintaining a gammahedged position.
(d) Find the number of shares of stock needed for maintaining a deltahedged position.
Solution.
(a) Using the BlackScholes formula we ﬁnd
C = Se−δ(T −t) N (d1 ) − Ke−r(T −t) N (d2 )
where
ln (40/40) + (0.08 − 0 + 0.5(0.3)2 ) ×
ln (S/K ) + (r − δ + 0.5σ 2 )(T − t)
√
=
d1 =
σ T −t
0.3 3
12 and √
d2 = d1 − σ T − t = 0.208333 − 0.3 3
= 0.058333
12 Thus,
1
N (d1 ) = √
2π 0.208333 x2 e− 2 dx = 0.582516
−∞ 3
12 = 0.208333. 39 DELTAGAMMA HEDGING 271 and
0.058333 1
N (d2 ) = √
2π x2 e− 2 dx = 0.523258.
−∞ Hence,
3 C = 40 × 0.582516 − 40e−0.08× 12 × 0.523258 = $2.7847,
∆ = e−δ(T −t) N (d1 ) = 0.5825,
and
d2
1 1
e− 2
√
= 0.0651.
Γ= √ ×
2π Sσ T − t
From the marketmaker perspective we have ∆ = −0.5825 and Γ = −0.0651.
(b) Using the BlackScholes formula as in (a), we ﬁnd C = $1.3584, ∆ = 0.3285, and Γ = 0.0524.
(c) Let n1 be the number of K1 −strike written options and n2 the number of K2 −strike purchased
options such that the total gamma is 0. Then, n1 Γ1 + n2 Γ2 = 0 so that
n2 = − n1 Γ1
.
Γ2 In our case, n1 = 1, Γ1 = −0.0651 and Γ2 = 0.0524. Thus,
n2 = 0.0651
= 1.2408.
0.0524 Hence, to maintain a gammahedged position, the marketmaker must buy 1.2408 of the 45strike
4month call. Indeed, we have
Γ40 + 1.2408Γ45 = −0.0651 + 1.2408 × 0.3285 = 0.
(d) Since ∆40 + 1.2408∆45 = −0.5825 + 1.2408 × 0.3285 = −0.1749, the marketmaker needs to buy
17.49 shares of stock to be deltahedged (i.e., deltaneutral)
Figure 39.1 compares a deltahedged position to a deltagamma hedged position. Note that the loses
from a large drop in stock prices for a deltagamma hedged position is relatively small compared to
the loses from a deltahedged position. Moreover, for a large stock price increase, the deltahedge
causes loss whereas the deltagamma hedged position causes gains. 272 OPTION HEDGING Figure 39.1
(2) Static option replication strategy: This strategy uses options to hedge options. It requires
little if any rebalancing and thus the word “static” compared to “dynamic”. For example, suppose
that the marketmaker sells a call on a share of stock. By the putcall parity relation we can write
C − P − S + Ke−rT = 0.
To create a hedge that is both delta and gammaneutral, the marketmaker purchases a put option
with the same strike price and expiration, a share of stock, and borrow money to fund this position.
(3) Buy out of the money options as insurance− deep out of the money options are inexpensive but
have positive gammas.
(4) Create a new ﬁnancial product such as a variance swap in which a marketmaker makes a
payment if the stock movement in either direction is small, but receives a payment if the stock
movement in either direction is large. 39 DELTAGAMMA HEDGING 273 Practice Problems
Problem 39.1
A hedged portfolio consists of selling 100 50strike calls on a share of stock. The gamma of the call
is −0.0521. Find the number of 55strike call options that a marketmaker must purchase in order
to bring the hedged portfolio gamma to zero. The gamma of the purchased call is 0.0441.
Problem 39.2
A hedged portfolio consists of selling 100 50strike calls on a stock. The gamma of the call is −0.0521
and the delta is −0.5824. A gammahedged portfolio is created by purchasing 118.14 55strike call
options. The gamma of the purchased call is 0.0441 and the delta is 0.3769. Find the number of
shares of stock needed for maintaining a deltahedged position.
Problem 39.3
A stock is currently trading for $50. A hedged portfolio consists of selling 100 50strike calls on a
stock. The gamma of the call is −0.0521 and the delta is −0.5824. The price of the call is $3.48.
A gammahedged portfolio is created by purchasing 118.14 units of a 55strike call option. The
gamma of the purchased call is 0.0441 and the delta is 0.3769. The price of this call is $2.05. Find
the cost of establishing the deltahedged position.
Problem 39.4
Consider the hedged portfolio of the previous problem. Suppose that in the next day the stock
price goes up to $51. For such a move, the price of the 50strike call option goes up to $4.06 while
the 55strike call goes up to $2.43. Assume a continuously compounded riskfree rate of 8%, ﬁnd
the overnight proﬁt of this position.
Problem 39.5 ‡
An investor has a portfolio consisting of 100 put options on stock A, with a strike price of 40, and 5
shares of stock A. The investor can write put options on stock A with strike price of 35. The deltas
and the gammas of the options are listed below
40strike put
Delta
−0.05
Gamma
0.25 35strike put
−0.10
0.50 What should be done to delta and gammaneutralize the investor’s portfolio?
Problem 39.6 ‡
For two European call options, CallI and CallII, on a stock, you are given: 274 OPTION HEDGING
Greek
Delta
Gamma
Vega CallI
0.5825
0.0651
0.0781 CallII
0.7773
0.0746
0.0596 Suppose you just sold 1000 units of CallI. Determine the numbers of units of CallII and stock you
should buy or sell in order to both deltahedge and gammahedge your position in CallI. An Introduction to Exotic Options
In this chapter we discuss exotic options. By an exotic option we mean an option which is created
by altering the contractual terms of a standard option such as European and American options
(i.e. Vanilla option). They permit hedging solutions tailored to speciﬁc problems and speculation
tailored to particular views. The types of exotic options that we will discuss in this chapter are:
Asian, barrier, compound, gap, and exchange options. 275 276 AN INTRODUCTION TO EXOTIC OPTIONS 40 Asian Options
An Asian option is an option where the payoﬀ is not determined by the underlying price at
maturity but by the average underlying price over some preset period of time. Averages can be
either arithmetic or geometric.
T
Suppose that a time interval [0, T ] is partitioned into N equal subintervals each of length h = N .
Let Sih denote the stock price at the end of the ith interval. The arithmetic average of the stock
price is deﬁned by
1
A(T ) =
N N Sih .
i=1 This type of average is typically used. However, there are no simple pricing formulas for options
based on the arithmetic average. A diﬀerent type of average that is computationally easier than
the arithmetic average, but less common in practice, is the geometric average deﬁned as
1 G(T ) = (S1h × S2h × · · · × SN h ) N .
Simple pricing formulas exist for geometric average options.
Proposition 40.1
The geometric average is less than or equal to the arithmetic average. That is, G(T ) ≤ A(T ).
Proof.
We know that ex ≥ 1+ x for all real number x. For 1 ≤ i ≤ N, let xi =
Multiplying these inequalities we ﬁnd Sih
− 1.
A(T ) Thus, Sih
A(T ) Sih ≤ e A(T ) −1 . S1h +S2h +···+SN h
S1h S2h
SN h
−N
A(T )
···
≤e
= 1.
A(T ) A(T )
A(T ) Hence,
S1h S2h · · · SN h ≤ (A(T ))N
or
1 G(T ) = (S1h S2h · · · SN h ) N ≤ A(T )
Example 40.1
Suppose you observe the following prices {345, 435, 534, 354}. What are the arithmetic and geometric
averages? 40 ASIAN OPTIONS 277 Solution.
We have
A(T ) = 345 + 435 + 534 + 354
= 417
4 and 1 G(T ) = (345.435.534.354) 4 = 410.405543
The payoﬀ at maturity of an Asian option can be computed using the average stock price either
as the underlying asset price (an average price option) or as strike price (an average strike
option). Hence, there are eight basic kinds of Asian options with payoﬀs listed next:
Type
Arithmetic average price call
Arithmetic average price put
Arithmetic average strike call
Arithmetic average strike put
Geometric average price call
Geometric average price put
Geometric average strike call
Geometric average strike put Payoﬀ
max{0, A(T ) − K }
max{0, K − A(T )}
max{0, ST − A(T )}
max{0, A(T ) − ST }
max{0, G(T ) − K }
max{0, K − G(T )}
max{0, ST − G(T )}
max{0, G(T ) − ST } Example 40.2
At the end of each of the past four months, Stock GS had the following prices: $345, $435, $534,
and $354. A certain 4month Asian geometric average price call on this stock has a strike price of
$400. It expires today, and its payoﬀ is computed on the basis of the geometric average of the stock
prices given above. What is the payoﬀ of this option?
Solution.
The payoﬀ is
max{0, G(T ) − K } = max{0, 410.405543 − 400} = $10.405543
Example 40.3
Consider the following information about a European call on a stock:
• The strike price is $100
• The current stock price is $100
• The time to expiration is one year
• The stock price volatility is 30%
• The annual continuouslycompounded riskfree rate is 8%
• The stock pays no dividends 278 AN INTRODUCTION TO EXOTIC OPTIONS • The price is calculated using twostep binomial model where each step is 6 months in length.
(a) Construct the binomial stock price tree including all possible arithmetic and geometric averages
after one year?
(b) What is the price of an Asian arithmetic average price call?
(c) What is the price of an Asian geometric average price call?
Solution.
(a) We ﬁrst ﬁnd u and d. We have
u = e(r−δ)h+σ
and
d = e(r−δ)h−σ √ √ h h √ = e0.08×0.5+0.3 = e0.08×0.5−0.3 √ 0.5 0.5 = 1.2868 = 0.84187. Thus, the riskneutral probability for an up move is
e(r−δ)h − d
e0.08(0.5) − 0.84187
=
= 0.44716.
u−d
1.2868 − 0.84187
The binomial stock price tree with all the possible arithmetic and geometric averages is shown in
the tree below.
pu = (b) The price of an Asian arithmetic average price call is
e−0.08(0.5) pu (e−0.08(0.5) [pu (47.13) + (1 − pu )(18.50)]) = $12.921.
(c) The price of an Asian geometric average price call is
e−0.08(0.5) [pu (e−0.05(0.5) [pu (45.97) + (1 − pu )(18.07)])] = $12.607 40 ASIAN OPTIONS 279 Remark 40.1
An Asian option is an example of pathdependent option, which means that the payoﬀ at
expiration depends upon the path by which the stock reached its ﬁnal price. In the previous
example, the payoﬀ of an Asian call at expiration for the stock price $108.33, using arithmetic
average price, is either $18.50 or $0 depending on the path leading to the price $108.33.
Example 40.4
Show that the price of an arithmetic average price Asian call is greater or equal to a geometric
average price Asian call.
Solution.
By Proposition 40.1 we have G(T ) ≤ A(T ) so that G(T )−K ≤ A(T )−K. Thus, max{0, G(T )−K } ≤
max{0, A(T ) − K }
Next, consider geometric average price and geometric average strike call and put options with
S = K = $40, r = 0.08, σ = 0.3, δ = 0 and T = 1. The table below gives the premiums of these
Asian options. We observe the following:
• The value of an average price Asian option decreases as N increases (because the more samples,
the less volatile the average). For Asian options that average the stock price, averaging reduces the
volatility of the value of the underlying asset. Thus, the price of the option at issuance is less than
otherwise equivalent standard option.
• The value of an average strike Asian option increases as N increases. 280 AN INTRODUCTION TO EXOTIC OPTIONS Practice Problems
Problem 40.1
At the end of each of the past four months, a certain stock had the following prices: $345, $435,
$534, and $354. A certain 4month Asian arithmetic average price call on this stock has a strike
price of $400. It expires today, and its payoﬀ is computed on the basis of the arithmetic average of
the stock prices given above. What is the payoﬀ of this option?
Problem 40.2
A stock price is $1 at the end of month 1 and increases by $1 every month without exception. A
99month Asian arithmetic average price put on the stock has a strike price of $56 and a payoﬀ
that is computed based on an average of monthly prices. The option expires at the end of month
99. What is the payoﬀ of this option? Recall that 1 + 2 + · · · + n = n(n2+1) .
Problem 40.3
A stock price is $1 at the end of month 1 and increases by $1 every month without exception. A
10month Asian geometric average price call on the stock has a strike price of $2 and a payoﬀ that
is computed based on an average of monthly prices. The option expires at the end of month 10.
What is the payoﬀ of this option?
Problem 40.4
Consider the information of Example 40.3.
(a) What is the price of an Asian arithmetic average strike call?
(b) What is the price of an Asian geometric average strike call?
Problem 40.5
Consider the information of Example 40.3.
(a) What is the price of an Asian arithmetic average price put?
(b) What is the price of an Asian geometric average price put?
Problem 40.6
Consider the information of Example 40.3.
(a) What is the price of an Asian arithmetic average strike put?
(b) What is the price of an Asian geometric average strike put?
Problem 40.7
(a) Show that the price of an arithmetic average price Asian put is less than or equal to a geometric
average price Asian put.
(b) Show that the price of an arithmetic average strike Asian call is less than or equal to a geometric 40 ASIAN OPTIONS 281 average strike Asian call.
(c) Show that the price of an arithmetic average strike Asian put is greater than or equal to a
geometric average strike Asian put.
Problem 40.8 ‡
You have observed the following monthly closing prices for stock XYZ:
Date
January 31, 2008
February 29, 2008
March 31, 2008
April 30, 2008
May 31, 2008
June 30, 2008
July 31, 2008
August 31, 2008
September 30, 2008
October 31, 2008
November 30, 2008
December 31, 2008 Stock Price
105
120
115
110
115
110
100
90
105
125
110
115 Calculate the payoﬀ of an arithmetic average Asian call option (the average is calculated based on
monthly closing stock prices) with a strike of 100 and expiration of 1 year.
Problem 40.9 ‡
At the beginning of the year, a speculator purchases a sixmonth geometric average price call option
on a company’s stock. The strike price is 3.5. The payoﬀ is based on an evaluation of the stock
price at each month end.
Date
January 31
February 28
March 31
Apricl
May 31
June 30 Stock Price
1.27
4.11
5.10
5.50
5.13
4.70 Based on the above stock prices, calculate the payoﬀ of the option. 282 AN INTRODUCTION TO EXOTIC OPTIONS 41 European Barrier Options
The second type of exotic options is the socalled barrier option. By a barrier option we mean an
option whose payoﬀ depends on whether the underlying asset price ever reaches a speciﬁed level−
or barrier− during the life of the option. Thus, a barrier option is another example of a pathdependent option. If the barrier’s level is the strike price, then the barrier option is equivalent to a
standard option.
Barrier options either come to existence or go out of existence the ﬁrst time the asset price reaches
the barrier. Thus, three are three types of barrier options:
• Knockout options: Options go out of existence when the asset price reaches the barrier before
option maturity. These options are called downandout when the asset price has to decline to
reach the barrier. They are called upandout when the price has to increase to reach the barrier.
“Out” options start their lives active and become null and void when the barrier price is reached.
• Knockin options: Options come into existence when asset price reaches the barrier before
option maturity. These options are called downandin when the asset price has to decline to
reach the barrier. They are called upandin when the price has to increase to reach the barrier.
“In” options start their lives worthless and only become active when barrier price is reached.
• Rebate options: Options make a ﬁxed payment if the asset price reaches the barrier. The
payment can be made at the time the barrier is reached or at the time the option expires. If the
latter is true, then the option is a deferred rebate. Up rebate options occur when the barrier is
above the current asset price. Down rebate options occur when the barrier is below the current
price.
The inout parity of barrier options is given by
“knockin” option + “knockout” option = standard option
For example, for otherwise equivalent options, we can write
Cupandin + Cupandout
Cdownandin + Cdownandout
Pupandin put + Pupandout
Pdownandin + Pdownandout =C
=C
=P
=P Since barrier options premiums are nonnegative, the inout parity shows that barrier options have
lower premiums than standard options.
Example 41.1
An ordinary call option on a certain stock has a strike price of $50 and time to expiration of 1 year 41 EUROPEAN BARRIER OPTIONS 283 trades for $4. An otherwise identical upandin call option on the same stock with a barrier of $55
trades for $2.77. Find the price of an upandout call option on the stock with a barrier of $55,a
strike price of $50, and time to expiration of 1 year.
Solution.
Using the inout parity relation we can write
2.77 + upandout call = 4.
Thus, upandout call premium = 4 − 2.77 = $1.23
Example 41.2
The stock of Tradable Co. once traded for $100 per share. Several barrier option contracts were
then written on the stock. Suddenly, the stock price increased to $130 per share − which is the
barrier for the options. Find the payoﬀ of
(a) An upandout call option with a strike of $120.
(b) An upandin call option with a strike of $120.
(c) An upandout put option with a strike of $120.
(d) An upandin put option with a strike of $120.
(e) A rebate option that pays a rebate of $12
Solution.
(a) The barrier 130 is reached so the option is knockedout. Therefore, the payoﬀ of the option is
zero.
(b) The barrier 130 is reached so the option is knockedin. Therefore, the payoﬀ of the option is
max{0, 130 − 120} = $10.
(c) The barrier 130 is reached so the option is knockedout. Therefore, the payoﬀ of the option is
zero.
(d) The barrier 130 is reached so the option is knockedin. Therefore, the payoﬀ of the option is
max{0, 120 − 130} = $0.
(e) Payoﬀ is $12
Example 41.3
When is a barrier option worthless? Assume that the stock price is greater than the barrier at
issuance for a downandout and less than the barrier for upandout. 284 AN INTRODUCTION TO EXOTIC OPTIONS Solution.
Consider an upandout call option with barrier level H. For this option to have a positive payoﬀ
the ﬁnal stock price must exceed the strike price. If H < K then the ﬁnal stock price exceeds also
the barrier H and hence is knockedout. Consider a downandout put. For this option to have a
positive payoﬀ, the ﬁnal stock price must be less than the strike price. Suppose H > K , since the
stock price at issuance is greater than H, for the ﬁnal stock price to be less than K means that the
stock price exceeded H and hence the option is knockedout
Example 41.4
You are given the following information about a European call option: S = $75, K = $75, σ =
0.30, r = 0.08, T = 1, and δ = 0.
(a) Using the BlackScholes framework, ﬁnd the price of the call.
(b) What is the price of a knockin call with a barrier of $74?
(c) What is the price of a knockout call with a barrier of $74?
Solution.
(a) We have
d1 =
and ln (75/75) + (0.08 + 0.5 × 0.32 )
ln (S/K ) + (r − δ + 0.5σ )T
√
=
= 0.4167
0.3
σT
√
d2 = d1 − σ T = 0.4167 − 0.3 = 0.1167. Thus,
1
N (d1 ) = √
2π
and
1
N (d2 ) = √
2π 0.4167 x2 e− 2 dx = 0.661551
−∞
0.1167 x2 e− 2 dx = 0.546541.
−∞ Hence,
C = Se−δT N (d1 ) − Ke−rT N (d2 ) = 75 × 0.661551 − 75e−0.08 × 0.546541 = $11.777.
(b) For a standard call to ever be in the money, it must pass through the barrier. Therefore, the
knockin call option has the same price as the standard call.
(c) Using the putcall parity for barrier options we ﬁnd that the price of a knockout call is zero 41 EUROPEAN BARRIER OPTIONS 285 Practice problems
Problem 41.1
When is a barrier option the same as an ordinary option? Assume that the stock price has not
reached the barrier.
Problem 41.2
Consider a European upandout barrier call option with strike price $75 and barrier $72. What is
the payoﬀ of such a call option?
Problem 41.3
Consider a European call option on a stock with strike price of $50 and time to expiration of 1 year.
An otherwise identical knockin and knockout call options with a barrier of $57 trade for $10.35
and $5.15 respectively. Find the price of the standard call option.
Problem 41.4
Consider a European call option on a stock with strike price of $50 and time to expiration of 1 year.
An otherwise identical knockin and knockout call options with a barrier of $57 trade for $10.35
and $5.15 respectively. An otherwise identical knockin call option with a barrier of $54 has a price
of $6.14. Find the price of an otherwise identical knockout option with a barrier of $54.
Problem 41.5
GS owns a portfolio of the following options on the stock of XYZ:
• An upandin call with strike of $63 and barrier of $66
• An upandout put with strike of $78 and barrier of $65
• An up rebate option with rebate of $14 and barrier of $61
• An upandout call with strike of $35 and barrier of $61
Originally, the stock traded at $59 per share. Right before the options expired, the stock began to
trade at $67 per share, its record high. What is GS’s total payoﬀ on this portfolio?
Problem 41.6
The stock of GS pays no dividends. The stock currently trades at $54 per share. An upandin
call with strike $55 and barrier $60 has a price of $3.04, and an upandout call with strike $55
and barrier $60 has a price of $1.32. The options expire in 2 months, and the annual continuously
compounded interest rate is 0.03. Find the price of one ordinary put option on the stock of GS
with strike price of $55 and time to expiration of 2 months.
Problem 41.7
You are given the following information about a call option: S = $40, K = $45, σ = 0.30, r =
0.08, T = 1,and δ = 0. Find the standard price of the option using oneperiod binomial model. 286 AN INTRODUCTION TO EXOTIC OPTIONS Problem 41.8 ‡
You have observed the following monthly closing prices for stock XYZ:
Date
January 31, 2008
February 29, 2008
March 31, 2008
April 30, 2008
May 31, 2008
June 30, 2008
July 31, 2008
August 31, 2008
September 30, 2008
October 31, 2008
November 30, 2008
December 31, 2008 Stock Price
105
120
115
110
115
110
100
90
105
125
110
115 The following are oneyear European options on stock XYZ. The options were issued on December
31, 2007.
(i) An arithmetic average Asian call option (the average is calculated based on monthly closing
stock prices) with a strike of 100.
(ii) An upandout call option with a barrier of 125 and a strike of 120.
(iii) An upandin call option with a barrier of 120 and a strike of 110.
Calculate the diﬀerence in payoﬀs between the option with the largest payoﬀ and the option with
the smallest payoﬀ.
Problem 41.9 ‡
Barrier option prices are shown in the table below. Each option has the same underlying asset and
the same strike price.
Type of option
downandout
upandout
downandin
upandin
down rebate
up rebate
Calculate X, the price of the upandin option. Price
$25
$15
$30
$X
$25
$20 Barrier
30,000
50,000
30,000
50,000
30,000
50,000 41 EUROPEAN BARRIER OPTIONS 287 Problem 41.10 ‡
Prices for 6month 60strike European upandout call options on a stock S are available. Below is
a table of option prices with respect to various H, the level of the barrier. Here, S (0) = 50.
H Price of upandout call
60
0
70
0.1294
80
0.7583
90
1.6616
∞
4.0861
Consider a special 6month 60strike European “knockin, partial knockout” call option that knocks
in at H1 = 70, and “partially” knocks out at H2 = 80. The strike price of the option is 60. The
following table summarizes the payoﬀ at the exercise date:
H1 Hit
H1 Not Hit
H2 Not Hit
H2 Hit
0
2 max{S (0.5) − 60, 0} max{S (0.5) − 60, 0}
Calculate the price of the option. 288 AN INTRODUCTION TO EXOTIC OPTIONS 42 Compound European Options
In this section we discuss a third type of exotic options. A compound option is an option with
underyling asset another option. Thus, a compound option has two strikes and two expirations
associated with it− one for the underlying option and one for the compound option itself.
There are four types of compound options: a call on a call (option to buy a call), a call on a put
(option to buy a put), a put on a call (option to sell a call), and a put on a put (option to sell
a put). For example, consider a call on a call. The owner of the compound option has until the
expiration date of the compound call to decide whether to exercise the compound option. If so,
he/she will receive the underyling call option with its own strike and expiration. If that call option
is exercised, he/she will receive the underlying asset.
Let Ku be the strike price of the underlying call option and Tu be the maturity date. Let Kc be the
strike price of the compound call option and Tc < Tu be the maturity time. Because we are dealing
with European options, the owner of the compound option cannot exercise the option before time
Tc . At time Tu , either the owner will pay Kc and receives the underlying call or do nothing and let
the option expires worthless. The owner will exercise if the price of the underlying call is greater
than Kc . In this case, the payoﬀ is
max{C (STc , Ku , Tu − Tc ) − Kc , 0}
where C (STc , Ku , Tu − Tc ) is the price of the underlying call at time Tc .
Now, let S ∗ be the underlying asset price for which the price of the underlying call is the cost of
acquiring it. That is, C (S ∗ , Ku , Tu − Tc ) = Kc . Then for STc > S ∗ we expect that C (STc , Ku , Tu −
Tc ) > Ku so that the compound option is exercised. It follows that two conditions must exist
for a compound CallOnCall option to be valuable ultimately: The ﬁrst is that STc > S ∗ and the
second is that STu > Ku . Thus, the pricing of a CallOnCall requires a bivariate normal cummulative
probability distribution as opposed to the univariate distribution in the BlackScholes formula.
Even though the CallOnCall option looks conceptually like an ordinary option on a stock, it is
mathematically diﬀerent. That is, the BlackScholes pricing formula cannot be applied. Even
though the underyling asset of the call is lognormally distributed, the price of the underyling asset
is not.
It can be shown that the pricing of a call on a call satisﬁes the following parity relationship:
CallOnCall − PutOnCall + Kc e−rTc = BSCall
where BSCall is the price of the underlying option according to the BlackScholes model. For a put,
the parity relationship is
CallOnPut − PutOnPut + Kc e−rTc = BSPut. 42 COMPOUND EUROPEAN OPTIONS 289 Example 42.1
The BlackScholes price of Call Option Q− which expires is 2 years− is $44. The annual continuouslycompounded riskfree interest rate is 0.03. The price of a PutOnCall option on Option Q with a
strike price of $50 and expiring in 1 year is $10. What is the price of a CallOnCall option on Option
Q with a strike price of $50 and expiring in 1 year?
Solution.
Using the putcall parity for compound options we can write
CallOnCall − PutOnCall + Kc e−rTc = BSCall.
We are given PutOnCall = 10, BSCall = 44, r = 0.03, Tc = 1, Kc = 50. Thus,
CallOnCall − 10 + 50e−0.03 = 44.
Solving this equation we ﬁnd CallOnCall = $5.477723323
Valuing Options on DividendPaying Stocks
Next, we examine the use of the compound option model to price an American call option on a
stock where the stock pays a single dividend of D at time Tc (the expiration time of the compound
option). We can either exercise the option at the cumdivident price1 STc + D, or we can hold the
option until expiration. In the latter case, the underlying option will be priced on the basis of the
stock price STc after the dividend is paid. The call option payoﬀ at Tc is
max{C (STc , Tu − Tc ), STc + D − Ku }.
By the putcall parity, the value of the unexercised call is
C (STc , Tu − Tc ) = P (STc , Tu − Tc ) − Ku e−r(Tu −Tc ) .
The call option payoﬀ at time Tc can thus be written as
STc + D − Ku + max{P (STc , Tu − Tc ) − [D − Ku (1 − e−r(Tu −Tc ) )], 0}. (42.1) The current value of the American call option is the present value of this amount. Note that
max{P (STc , Tu − Tc ) − [D − Ku (1 − e−r(Tu −Tc ) )], 0} is the payoﬀ of a call on a put option with strike
price D − Ku (1 − e−r(Tu −Tc ) ) and maturity date Tc permitting the owner to buy a put option with
strike Ku and maturity date Tu . Letting S0 = P V0,Tc (STc + D) we obtain
CAmer = S0 − Ku e−rTc + CallOnPut.
1 The exdividend date is the ﬁrst date when buying a stock does not entitle the new buyer to the declared
dividend, as the transfer of stock ownership cannot be completed before the company initiates dividend payment.
Before this date, the stock trades cumdividend. 290 AN INTRODUCTION TO EXOTIC OPTIONS Remark 42.1
Note that in the above compound option we assume that the strike D − K (1 − e−r(Tu −Tc ) ) is positive
for otherwise the interest on the strike (over the life of the option from the exdividend date to
expiration) would exceed the value of the dividend and early exercise would never be optimal.
Remark 42.2
The compound option that is implicit in the early exercise decision gives the right to acquire a
put option after the dividend is paid. (The put option is acquired if we do not exercise the call.
If the call is unexercised after the dividend, all subsequent valuation will be with respect to the
exdividend value of the stock.) Thus, for purposes of valuing the compound option, the underlying
asset is really the stock without the dividend, which is the prepaid forward. In obtaining this put,
the call owner gives up the dividend and earns interest on the strike. Remember that exercising the
compound option is equivalent to keeping the option on the stock unexercised.
Example 42.2
Suppose that a stock will pay a dividend of $2 in 4 months. A call option on the stock has a strike
price of $33 and will expire in 6 months. Four months from now, the stock price is expected to be
$35 and the price of the call will be $3 after the dividend is paid. What will be the value of the call
option in four months?
Solution.
The value of the call option in 4 months is given by
max{C (STc , Tu − Tc ), STc + D − Ku }.
We are given that STc = 35, Tc = 1 , Tu = 0.5, Ku = 33, D = 2, and C (STc , Tu − Tc ) = 3. Thus,
3
max{3, 35 + 2 − 33} = $4
Example 42.3
You have perfect knowledge that 3 months from now, a stock will pay a dividend of $5 per share.
Right after it pays the dividend, the stock will be worth $80 per share. A certain put option
on this stock has a strike price of $83, time to expiration of 6 months. The annual continuously
compounded riskfree interest rate r is 0.07.
(a) What is the strike price of the CallOnPut compound option (after the dividend is paid)?
(b) Suppose that three months from now, the CallOnPut compound option has a price of $4.04.
What is the price of the American call? 42 COMPOUND EUROPEAN OPTIONS
Solution.
(a) The strike price of the CallOnPut compound option is
D − Ku (1 − e−r(Tu −Tc ) ) = 5 − 83(1 − e−0.07(0.5−0.25) ) = $3.56.
(b) The price of the American call today is
CAmer = S0 − Ku e−rTc + CallOnPut = 80 − 83e−0.07×0.25 + 4.04 = $2.48 291 292 AN INTRODUCTION TO EXOTIC OPTIONS Practice Problems
Problem 42.1
The annual continuouslycompounded riskfree interest rate is 0.12. A CallOnCall option on Call
Option A has a price of $53. A PutOnCall option on Option A has a price of $44. Both compound
options have a strike price of $100 and time to expiration of 2 years. Find the BlackScholes price
of the underlying Option.
Problem 42.2
The BlackScholes price of Call Option A is $12. A CallOnCall option on Call Option A has a price
of $5. A PutOnCall option on Option A has a price of $3.33. Both compound options have a strike
price of $14 and expire in 3 years. Find the annual continuouslycompounded interest rate.
Problem 42.3
The BlackScholes price of put Option Q− which expires is 2 years− is $6.51. The annual continuouslycompounded riskfree interest rate is 0.05. The price of a CallOnPut option on Option Q with a
strike price of $3 and expiring in 1 year is $4.71. What is the price of a PutOnPut option on Option
Q with a strike price of $3 and expiring in 1 year?
Problem 42.4 ‡
You are given the following information on a CallOnPut option:
• The continuously compounded riskfree rate is 5%
• The strike price of the underlying option is 43
• The strike price of the compound option is 3
• The compound option expires in 6 month
• The underlying option expires six months after te compound option.
• The underlying option is American. 42 COMPOUND EUROPEAN OPTIONS 293 Based on the above binomial stock price tree, calculate the value of the compound option.
Problem 42.5
Suppose that a stock will pay a dividend of $2 in 4 months. A call option on the stock has a strike
price of $33 and will expire in 6 months. Four months from now, the stock price is expected to
be $35 and the price of the call will be $3 after the dividend is paid. The annual continuously
compounded riskfree interest rate r is 0.03. If the call option is unexercised, what will be the value
4 months from now of a put option on the stock with a strike price of $33 and time to expiration
of 6 months?
Problem 42.6
You have perfect knowledge that 1 year from now, the stock of GS will pay a dividend of $10 per
share. Right after it pays the dividend, the stock will be worth $100 per share. A certain put option
on this stock has a strike price of $102, time to expiration of 2 years, and will have a price of $12 in
1 year if unexercised. The annual continuously compounded riskfree interest rate r is 0.08. What
will be the price in 1 year (after the dividend is paid) of a call option on this stock with a strike
price of $102 and time to expiration of 2 years?
Problem 42.7
You have perfect knowledge that 1 year from now, the stock of GS will pay a dividend of $10 per
share. Right after it pays the dividend, the stock will be worth $100 per share. A certain put option
on this stock has a strike price of $102, time to expiration of 2 years, and will have a price of $12
in 1 year if unexercised. The annual continuously compounded riskfree interest rate r is 0.08. You
can use a compound CallOnPut option on the GS put to determine the price in 1 year (after the
dividend is paid) of a call option on this stock with a strike price of $102 and time to expiration of
2 years. What is the strike price of such a CallOnPut option?
Problem 42.8
You have perfect knowledge that 3 months from now, a stock will pay a dividend of $6 per share.
Right after it pays the dividend, the stock will be worth $65 per share. A certain put option on this
stock has a strike price of $65, time to expiration of one. The annual continuously compounded
riskfree interest rate r is 0.08.
(a) What is the strike price of the CallOnPut compound option (after the dividend is paid)?
(b) Suppose that the current American call price is $4.49. What is the price of the CallOnPut
compound option? 294 AN INTRODUCTION TO EXOTIC OPTIONS 43 Chooser and Forward Start Options
A chooser option is exactly what its name suggests− it is an option which allows the holder to
choose whether his/her option is a call or a put at a predetermined time t. Both options have the
same strike price K.
The payoﬀ at the choice time t is given by:
Payoﬀ = max{C (St , t, T ), P (St , t, T )}
where C (St , t, T ) and P (St , t, T ) denote the respective timet price of a European call and a put at
the choice date t.
Note that if T = t, that is the chooser option and the underlying options expire simultaneously,
then the chooser payoﬀ is the payoﬀ of a call if ST > K and the payoﬀ of a put if ST < K. That is,
the chooser option is equivalent to a straddle with strike price K and expiration time T.
Now, suppose that the chooser must be exercised at choice time t, using callput parity at t, the
value of a chooser option at t can be expressed as:
max{C (St , t, T ), P (St , t, T )} =C (St , t, T ) + max{0, P (St , t, T ) − C (St , t, T )}
=C (St , t, T ) + max{0, Ke−r(T −t) − St e−δ(T −t) }
=C (St , t, T ) + e−δ(T −t) max{0, Ke−(r−δ)(T −t) − St }
This says that the chooser option is equivalent to a call option with strike price K and maturity T
and e−δ(T −t) put options with strike price Ke−(r−δ)(T −t) and maturity t.
Example 43.1
Consider a chooser option (also known as an asyoulikeit option) on a nondividendpaying stock.
At time 2, its holder will choose whether it becomes a European call option or a European put
option, each of which will expire at time 3 with a strike price of $90. The time0 price of the call
option expiring at time 3 is $30. The chooser option price is $42 at time t = 0. The stock price is
$100 at time t = 0. Find the time0 price of a European call option with maturity at time 2 and
strike price of $90. The riskfree interest rate is 0.
Solution.
The time0 price of the chooser is given by
chooser option price = C (St , t, T ) + e−δ(T −t) max{0, Ke−(r−δ)(T −t) − St }.
We are given that r = δ = 0, S0 = 100, C (100, 2, 3) = 30, t = 2, T = 3, K = 90. Thus, at time t = 0
we have
42 = 30 + P (90, 2). 43 CHOOSER AND FORWARD START OPTIONS 295 Solving this equation we ﬁnd P (90, 2) = 12. Using the putcall parity of European options we can
write
C (90, 2) = P (90, 2) + S0 e−2δ − Ke−2r = 12 + 100 − 90 = $22
Example 43.2
Consider a chooser option on a stock. The stock currently trades for $50 and pays dividend at the
continuously compounded yield of 4%. The choice date is six months from now. The underlying
European options expire 9 months from now and have a strike price of $55. The time0 price of the
European option is $4. The continuously compounded riskfree rate is 10% and the volatility of the
prepaid forward price of the stock is 30%. Find the time0 chooser option price.
Solution.
The chooser option time0 price is the time0 price of a European call with strike $55 and maturity
date of 9 months and the time0 price of e−δ(T −t) = e−0.04×0.25 put options with strike price of
Ke−(r−δ)(T −t) = 55e−(0.10−0.04)×0.25 = $54.1812 and expiration of 6 months. To ﬁnd the price of such
a put option, we use the BlackScholes framework. We have
d1 =
and ln (50/54.1812) + (0.10 − 0.04 + 0.5 × 0.32 ) × 0.5
ln (S/K ) + (r − δ + 0.5σ 2 )T
√
√
=
= −0.1311
0.3 0.5
σT
√
√
d2 = d1 − σ T = −0.1311 − 0.3 0.5 = −0.3432. Thus,
1
N (d1 ) = √
2π
and
1
N (d2 ) = √
2π −0.1311 x2 e− 2 dx = 0.447848
−∞
−0.3432 x2 e− 2 dx = 0.365724.
−∞ The price of the European put with strike $54.1812 and maturity of six months is
P (54.1812, 0.5) = 54.1812e−0.10×0.5 (1 − 0.365724) − 50e−0.04×0.5 (1 − 0.447848) = $5.6289.
The time0 price of the chooser option is
4 + e−0.04×0.25 × 5.6289 = $9.5729
Forward Start Option
A Forward start option is an option whose strike will be determined at some future date. Like a
standard option, a forwardstart option is paid for in the present; however, the strike price is not 296 AN INTRODUCTION TO EXOTIC OPTIONS fully determined until an intermediate date before expiration.
Consider a forward start option with maturity T. Suppose that at time 0 < t < T the forward start
option will give you one call with T − t periods to expiration and strike αSt . The timet price of
the forward start option is
C = St e−δ(T −t) N (d1 ) − αSt e−r(T −t) N (d2 )
where
d1 =
and − ln α + (r − δ + 0.5σ 2 )(T − t)
√
σ T −t
√
d2 = d1 − σ T − t. Now, the time0 price of the forward start option is e−rt C.
Example 43.3
A nondividendpaying stock currently trades for $100. A forward start option with maturity of nine
months will give you, six months from today, a 3month atthemoney call option. The expected
stock volatility is 30% and the continuously compounded riskfree rate is 8%. Assume that r, σ,
and δ will remain the same over the next six months.
(a) Find the value of the forward start option six months from today. What fraction of the stock
price does the option cost?
(b) What investment today would guarantee that you had the money in 6 months to purchase an
atthemoney call option?
(c) Find the time0 price of the forward start option today.
Solution.
(a) After six months, the price of forward start option is found as follows:
d1 =
and − ln 1 + (0.08 − 0 + 0.5(0.3)2 )(0.25)
− ln α + (r − δ + 0.5σ 2 )(T − t)
√
√
=
= 0.2083
σ T −t
0.3 0.25
√
√
d2 = d1 − σ T − t = 0.2083 − 0.3 0.25 = 0.0583. Thus,
1
N (d1 ) = √
2π
and
1
N (d2 ) = √
2π 0.2083 x2 e− 2 dx = 0.582503
−∞
0.0583 x2 e− 2 dx = 0.523245.
−∞ 43 CHOOSER AND FORWARD START OPTIONS 297 Hence, the price of the forward start six months from now is
C = St e−δ(T −t) N (d1 ) − St e−r(T −t) N (d2 ) = St [0.582503 − e−0.08×0.25 × 0.523245] = 0.069618St .
(b) In six months we will need 0.069618S0.5 dollars to buy the call. Thus, we must have 0.069618
shares of the stock today.
(c) The time0 price of the forward start option must be 0.069618 multiplied by the time0 price of
a security that gives St as payoﬀ at time t = 0.5 years, i.e., multiplied by 100 or $6.9618
Example 43.4
A dividendpaying stock currently trades for $50. The continuouslycompound dividend yield is
2%. A forward start option with maturity of two years will give you, one year from today, a 1year
atthemoney put option. The expected stock volatility is 30% and the continuously compounded
riskfree rate is 10%. Assume that r, σ, and δ will remain the same over the year.
(a) Find the value of the forward start option one year from today. What fraction of the stock price
does the option cost?
(b) What investment today would guarantee that you had the money in one year to purchase an
atthemoney put option?
(c) Find the time0 price of the forward start option today.
Solution.
(a) After one year, the price of forward start option is found as follows:
− ln 1 + (0.10 − 0.02 + 0.5(0.3)2 )(1)
− ln α + (r − δ + 0.5σ 2 )(T − t)
√
√
=
= 0.41667
d1 =
σ T −t
0.3 1
and √
√
d2 = d1 − σ T − t = 0.41667 − 0.3 1 = 0.11667. Thus,
1
N (d1 ) = √
2π
and
1
N (d2 ) = √
2π 0.41667 x2 e− 2 dx = 0.66154
−∞
0.11667 x2 e− 2 dx = 0.546439.
−∞ Hence, the price of the forward start one year from now is
P = St e−r(T −t) N (−d2 )−St e−δ(T −t) N (−d1 ) = St [(1−0.546439)e−0.10 −e−0.02 ×(1−0.66154)] = 0.0786St . 298 AN INTRODUCTION TO EXOTIC OPTIONS (b) In one year we will need 0.0786S1 dollars to buy the put. Thus, we must have 0.0786 shares of
the stock today.
(c) The time0 price of the forward start option must be 0.0786 multiplied by the time0 price of
a security that gives St as payoﬀ at time t = 1 year, i.e., multiplied by the prepaid forward price
P
F0,1 (St ). Hence, the time0 price of the forward start option is
P
0.0786F0,1 (St ) = 0.0786e−0.02 × 50 = 3.85128 43 CHOOSER AND FORWARD START OPTIONS 299 Practice Problems
Problem 43.1
Consider a chooser option on a stock. The stock currently trades for $50 and pays dividend at the
continuously compounded yield of 8%. The choice date is two years from now. The underlying
European options expire in four years from now and have a strike price of $45. The continuously
compounded riskfree rate is 5% and the volatility of the prepaid forward price of the stock is 30%.
Find the delta of the European call with strike price of $45 and maturity of 4 years.
Problem 43.2
Consider a chooser option on a stock. The stock currently trades for $50 and pays dividend at the
continuously compounded yield of 8%. The choice date is two years from now. The underlying
European options expire in four years from now and have a strike price of $45. The continuously
compounded riskfree rate is 5% and the volatility of the prepaid forward price of the stock is 30%.
Find the delta of the European put with strike price of $47.7826 and maturity of 2 years.
Problem 43.3
Consider a chooser option on a stock. The stock currently trades for $50 and pays dividend at the
continuously compounded yield of 8%. The choice date is two years from now. The underlying
European options expire in four years from now and have a strike price of $45. The continuously
compounded riskfree rate is 5% and the volatility of the prepaid forward price of the stock is 30%.
Find the delta of the chooser option.
Problem 43.4
Consider a chooser option on a stock that pays dividend at the continuously compounded yield of
5%. After one year, its holder will choose whether it becomes a European call option or a European
put option, each of which will expire in 3 years with a strike price of $100. The time0 price of a
call option expiring in one year is $9.20. The stock price is $95 at time t = 0. The time0 price of
the chooser option is $28.32. The continuously compounded riskfree interest rate is 5%. Find the
time0 price of the European option with strike price of $100 and maturity of 3 years.
Problem 43.5 ‡
Consider a chooser option (also known as an asyoulikeit option) on a nondividendpaying stock.
At time 1, its holder will choose whether it becomes a European call option or a European put
option, each of which will expire at time 3 with a strike price of $100. The chooser option price is
$20 at time t = 0. The stock price is $95 at time t = 0. Let C (T ) denote the price of a European
call option at time t = 0 on the stock expiring at time T, T > 0, with a strike price of $100. You
are given:
(i) The riskfree interest rate is 0. 300 AN INTRODUCTION TO EXOTIC OPTIONS (ii) C (1) = $4
Determine C (3).
Problem 43.6
A dividendpaying stock currently trades for $50. The continuouslycompound dividend yield is
2%. A forward start option with maturity of two years will give you, one year from today, a 1year
atthemoney call option. The expected stock volatility is 30% and the continuously compounded
riskfree rate is 10%. Assume that r, σ, and δ will remain the same over the year.
(a) Find the value of the forward start option one year from today. What fraction of the stock price
does the option cost?
(b) What investment today would guarantee that you had the money in one year to purchase an
atthemoney call option?
(c) Find the time0 price of the forward start option today.
Problem 43.7
A nondividendpaying stock currently trades for $100. A forward start option with maturity of
nine months will give you, six months from today, a 3month call option with strike price 105%
of the current stock price. The expected stock volatility is 30% and the continuously compounded
riskfree rate is 8%. Assume that r, σ, and δ will remain the same over the next six months.
(a) Find the value of the forward start option six months from today. What fraction of the stock
price does the option cost?
(b) What investment today would guarantee that you had the money in 6 months to purchase an
atthemoney call option?
(c) Find the time0 price of the forward start option today.
Problem 43.8 ‡
Consider a forward start option which, 1 year from today, will give its owner a 1year European
call option with a strike price equal to the stock price at that time.
You are given:
(i) The European call option is on a stock that pays no dividends.
(ii) The stock’s volatility is 30%.
(iii) The forward price for delivery of 1 share of the stock 1 year from today is $100.
(iv) The continuously compounded riskfree interest rate is 8%.
Under the BlackScholes framework, determine the price today of the forward start option.
Problem 43.9 ‡
You own one share of a nondividendpaying stock. Because you worry that its price may drop
over the next year, you decide to employ a rolling insurance strategy, which entails obtaining one 43 CHOOSER AND FORWARD START OPTIONS 301 3month European put option on the stock every three months, with the ﬁrst one being bought
immediately.
You are given:
(i) The continuously compounded riskfree interest rate is 8%.
(ii) The stock’s volatility is 30%.
(iii) The current stock price is 45.
(iv) The strike price for each option is 90% of the thencurrent stock price.
Your broker will sell you the four options but will charge you for their total cost now.
Under the BlackScholes framework, how much do you now pay your broker? 302 AN INTRODUCTION TO EXOTIC OPTIONS 44 Gap Options
We know by now that the payoﬀ of a standard option is the result of comparing the stock price with
the strike price. In the case of a call, we obtain a nonzero payoﬀ when the stock price is greater
than the strike price. In the case of a put, we obtain a nonzero payoﬀ when the stock price is less
than the strike price. For a gap option, we obtain exactly these payoﬀs but by comparing the stock
price to a price diﬀerent than the strike price.
A gap option involves a strike price K1 and a trigger price K2 . We assume that K1 and K2 are
diﬀerent.1 The trigger price determines whether the option has a nonnegative payoﬀ. For example,
the payoﬀ of a gap call option is given by
Gap call option payoﬀ = ST − K1 if ST > K2
0
if ST ≤ K2 The diagram of the gap call option’s payoﬀ is shown in Figure 44.1. Figure 44.1
Note that the payoﬀs of a gap call can be either positive or negative. Also, note the gap in the
graph when ST = K2 .
Likewise, we deﬁne the payoﬀ of a gap put option by
Gap put option payoﬀ =
1 If K1 = K2 a gap option is just an ordinary option. K1 − ST if ST < K2
0
if ST ≥ K2 44 GAP OPTIONS 303 The diagram of the gap put option’s payoﬀ is shown in Figure 44.2. Figure 44.2
The pricing formulas of gap options are a modiﬁcation of the BlackScholes formulas where the
strike price K in the primary formulas is being replaced by K1 and the strike price in d1 is being
replaced by the trigger price K2 . Thus, the price of a gap call option is given by
C = Se−δT N (d1 ) − K1 e−rT N (d2 )
and that for a gap put option is
P = K1 e−rT N (−d2 ) − Se−rT N (−d1 )
where and ln (S/K2 ) + (r − δ + 0.5σ 2 )T
√
d1 =
σT
√
d2 = d1 − σ T . It follows that the deltas of an ordinary option and otherwise an identical gap option are equal.
Example 44.1
Assume that the BlackScholes framework holds. A gap call option on a stock has a trigger price
of $55, a strike price of $50, and a time to expiration of 2 years. The stock currently trades for $53 304 AN INTRODUCTION TO EXOTIC OPTIONS per share and pays dividends with a continuously compounded annual yield of 0.03. The annual
continuously compounded riskfree interest rate is 0.09, and the relevant price volatility for the
BlackScholes formula is 0.33. Find the BlackScholes price of this gap call.
Solution.
The values of d1 and d2 are
d1 =
and ln (53/55) + (0.09 − 0.03 + 0.5(0.33)2 × 2
ln (S/K2 ) + (r − δ + 0.5σ 2 )T
√
√
=
= −0.3603
0.33 2
σT
√
√
d2 = d1 − σ T = −0.3603 − 0.33 2 = −0.8269. Thus,
1
N (d1 ) = √
2π
and
1
N (d2 ) = √
2π −0.3603 x2 e− 2 dx = 0.359311
−∞
−0.8269 x2 e− 2 dx = 0.204147
−∞ Hence,
C = Se−δT N (d1 ) − K1 e−rT N (d2 ) = 53e−0.03×2 × 0.359311 − 50e−0.09×2 × 0.204147 = $9.409
Because negative payoﬀs are possible, gap options can have negative premiums.
Example 44.2
Assume that the BlackScholes framework holds. A gap call option on a stock has a trigger price
of $52, a strike price of $65, and a time to expiration of 3 months. The stock currently trades
for $50 per share and pays dividends with a continuously compounded annual yield of 0.04. The
annual continuously compounded riskfree interest rate is 0.07, and the relevant price volatility for
the BlackScholes formula is 0.40. Find the BlackScholes price of this gap call.
Solution.
The values of d1 and d2 are
d1 =
and ln (50/52) + (0.07 − 0.04 + 0.5(0.40)2 × 0.25
ln (S/K2 ) + (r − δ + 0.5σ 2 )T
√
√
=
= −0.05860
0.40 0.252
σT
√
√
d2 = d1 − σ T = −0.05860 − 0.40 0.252 = −0.25860. 44 GAP OPTIONS 305 Thus,
1
N (d1 ) = √
2π
and
1
N (d2 ) = √
2π −0.05860 x2 e− 2 dx = 0.476635
−∞
−0.25860 x2 e− 2 dx = 0.397972
−∞ Hence,
C = Se−δT N (d1 ) − K1 e−rT N (d2 ) = 50e−0.04×0.25 × 0.476635 − 65e−0.07×0.25 × 0.397972 = −$1.825 306 AN INTRODUCTION TO EXOTIC OPTIONS Practice Problems
Problem 44.1
True or False: When the strike equal the trigger, the premium for a gap option is the same as for
an ordinary option with the same strike.
Problem 44.2
True or False: For a ﬁxed K1 , if K2 > K1 for a gap put option then increasing the trigger price
reduces the premium.
Problem 44.3
True or False: For a ﬁxed K2 , increasing K1 for a gap put option then increasing will reduce the
premium.
Problem 44.4
True or False: For any strike K1 > 0, there is a K2 such that PGap = 0.
Problem 44.5
Assume that the BlackScholes framework holds. A gap call option on a stock has a trigger price
of $31, a strike price of $29, and a time to expiration of 6 months. The stock currently trades
for $30 per share and pays dividends with a continuously compounded annual yield of 0.03. The
annual continuously compounded riskfree interest rate is 0.06, and the relevant price volatility for
the BlackScholes formula is 0.30. Find the BlackScholes price of this gap call.
Problem 44.6 ‡
Let S (t) denote the price at time t of a stock that pays dividends continuously at a rate proportional
to its price. Consider a European gap option with expiration date T, T > 0.
If the stock price at time T is greater than $100, the payoﬀ is
S (T ) − 90;
otherwise, the payoﬀ is zero.
You are given:
(i) S (0) = $80
(ii) The price of a European call option with expiration date T and strike price $100 is $4.
(iii) The delta of the call option in (ii) is 0.2.
Calculate the price of the gap option. 44 GAP OPTIONS 307 Problem 44.7 ‡
Which one of the following statements is true about exotic options?
(A) Asian options are worth more than European options
(B) Barrier options have a lower premium than standard options
(C) Gap options cannot be priced with the BlackScholes formula
(D) Compound options can be priced with the BlackScholes formula
(E) Asian options are pathindependent options.
Problem 44.8
Consider three gap calls A, B, and C with underlying asset a certain stock. The strike price of call
A is $48, that of call B is $56 and call C is $50. Suppose that CA = $4.98, CB = $1.90. All three
options have the same trigger price and time to expiration. Find CC .
Problem 44.9 ‡
A marketmaker sells 1,000 1year European gap call options, and deltahedges the position with
shares.
You are given:
(i) Each gap call option is written on 1 share of a nondividendpaying stock.
(ii) The current price of the stock is 100.
(iii) The stock’s volatility is 100%.
(iv) Each gap call option has a strike price of 130.
(v) Each gap call option has a payment trigger of 100.
(vi) The riskfree interest rate is 0%.
Under the BlackScholes framework, determine the initial number of shares in the deltahedge. 308 AN INTRODUCTION TO EXOTIC OPTIONS 45 Exchange Options
A exchange option, also known as an outperformance option, is an option that gives the owner
the right to exchange an underlying asset with a strike asset known as the benchmark. Such an
option pays oﬀ only when the underyling asset outperforms the strike asset. For example, a call
option on a stock is an exchange option which entails the owner to exchange the stock with cash
when the stock outperforms cash.
The pricing formula for an exchange option is a variant of the BlackScholes formula. The exchange
call option price is
ExchangeCallPrice = Se−δS T N (d1 ) − Ke−δK T N (d2 )
and that for a put is
ExchangePutPrice = Ke−δK T N (−d2 ) − Se−δS T N (−d1 )
where
ln
d1 =
and S e−δS T
Ke−δK T + 0.5σ 2 T √
σT √
d2 = d1 − σ T . Here, σ denotes the volatility of ln (S/K ) over the life of the option. Hence,
Var[ln (S/K )] =Var[ln (S )] + Var[ln (K )] − 2Cov[ln (S ), ln (K )]
2
2
=σS + σK − 2ρσS σK
where ρ is the correlation between the continuously compounded returns on the two assets.
Example 45.1
One share of Stock A is used as the underlying asset on an exchange option, for which the benchmark asset is one share of a Stock B. Currently, Stock A trades for $321 per share, and Stock B
trades for $300 per share. Stock A has an annual price volatility of 0.34 and pays dividends at an
annual continuously compounded yield of 0.22. Stock B has an annual price volatility of 0.66 and
pays dividends at an annual continuously compounded yield of 0.02. The correlation between the
continuously compounded returns on the two assets is 0.84. The exchange option expires in 4 years.
Find the BlackScholes price of this option.
Solution.
The underlying asset is a share of Stock A and the strike asset is one share of Stock B. Therefore,
we have S = 321 and K = 300. The volatility of ln (S/K ) is
√
2
2
σ = σS + σK − 2ρσS σK = 0.342 + 0.662 − 2 × 0.84 × 0.34 × 0.66 = 0.4173823187. 45 EXCHANGE OPTIONS 309 The values of d1 and d2 are
ln + 0.5σ 2 T √
σT d1 =
ln
= S e−δS T
Ke−δK T 321e−0.22×4
300e−0.02×4 + 0.5(0.4173823187)2 (4) 0.4173823187 × 2
= − 0.4599204785
and √
d2 = d1 − σ T = −0.4599204785 − 0.4173823187 × 2 = −1.294685116. Thus,
1
N (d1 ) = √
2π
and
1
N (d2 ) = √
2π
The value of the exchange call is −0.4599204785 x2 e− 2 dx = 0.32278665
−∞
−1.294685116 x2 e− 2 dx = 0.097714438.
−∞ ExchangeCallPrice =Se−δS T N (d1 ) − Ke−δK T N (d2 )
=321e−0.22×4 × 0.32278665 − 300e−0.02×4 × 0.097714438 = $15.91699158
Example 45.2
An exchange call option with expiration of nine months allows the owner to acquire one share of
a stock A for three shares of a stock B. The price of the option is $17.97. Stock A pays dividends
at the continuously compounded yield of 4% and that of stock B is 3%. Stock A currently trades
for $100 and stock B trades for $30. Find the value of an exchange option that allows the owner to
acquire one share of stock B by giving up 1 shares of stock A.
3
Solution.
Let P (St , Kt , T − t) be the price of an option that allows the owner to give up one share of stock
A for three shares of stock B. Then by the putcall parity we have
P
P
C (St , Kt , T − t) − P (St , Kt , T − t) = Ft,T (S ) − Ft,T (K ) or
17.97 − P (St , Kt , T − t) = 100e−0.04×0.75 − 3(30)e−0.03×0.75 .
Thus, P (St , Kt , T − t) = $8.923. Hence, the value of an option allowing its owner to give up
of stock A in exchange of 1 share of stock B is 8.923/3 = $2.97 1
3 shares 310 AN INTRODUCTION TO EXOTIC OPTIONS Example 45.3
Assume the BlackScholes framework. Consider two nondividendpaying stocks whose time−t prices
are denoted by S1 (t) and S2 (t), respectively.
You are given:
(i) S1 (0) = 20 and S2 (0) = 11.
(ii) Stock 1’s volatility is 0.15.
(iii) Stock 2’s volatility is 0.20.
(iv) The correlation between the continuously compounded returns of the two stocks is −0.30.
(v) The continuously compounded riskfree interest rate is 6%.
(vi) A oneyear European option with payoﬀ max{22 − max{S1 (1), 2S2 (1)}, 0} has a current (time0) price of 0.29.
Consider a European option that gives its holder the right to buy either one share of Stock 1 or two
shares of Stock 2 at a price of 22 one year from now. Calculate the current (time0) price of this
option.
Solution.
Suppose that X = max{S1 (1), 2S2 (1)} is the payoﬀ of a certain asset A. Using (vi) we have the
payoﬀ of a European put option on asset A
max{22 − max{S1 (1), 2S2 (1)}, 0} = max{22 − X, 0} = 0.29.
We are asked to ﬁnd the time0 price of a European call option with payoﬀ, one year from today,
given by
max{max{S1 (1), 2S2 (1)} − 22, 0} = max{X − 22, 0}.
Using, the putcall parity
C − P = S − Ke−rT
we ﬁnd C = 0.29 + A0 − 22e−0.06×1 where A0 is the time0 price of asset A. It follows that in order
to ﬁnd C we must ﬁnd the value of A0 . Since
max{S1 (1), 2S2 (1)} = 2S2 (1) + max{S1 (1) − 2S2 (1), 0}
the time0 value of A is twice the time0 value of stock 2 plus the time0 value of an exchange call
option that allows the owner to give two shares of stock 2 (strike asset) for one share of stock 1
(underlying asset). Hence,
A0 = 2S2 (0) + ExchangeCallPrice
We next ﬁnd the time0 value of the exchange call option. We have
σ= 2
2
σS + σK − 2ρσS σK = 0.152 + 0.202 − 2(−0.30)(0.15)(0.20) = 0.28373 45 EXCHANGE OPTIONS 311
ln + 0.5σ 2 T √
σT d1 =
ln
= S e−δS T
Ke−δK T (20)e−0×1
2×11e−0×1 + 0.5(0.28373)2 (1) 0.28373 × 1 = − 0.19406
and √
d2 = d1 − σ T = −0.19406 − 0.28373 × 1 = −0.47779. Thus,
1
N (d1 ) = √
2π
and
1
N (d2 ) = √
2π
The value of the exchange call is −0.19406 x2 e− 2 dx = 0.423064
−∞
−0.47779 x2 e− 2 dx = 0.3164.
−∞ ExchangeCallPrice =Se−δS T N (d1 ) − Ke−δK T N (d2 )
=20 × 0.423064 − 2 × 11 × 0.3164 = 1.5$
Hence,
A0 = 2 × 11 + 1.5 = $23.5
and
C = 0.29 + 23.5 − 22e−0.06 = 3.07 312 AN INTRODUCTION TO EXOTIC OPTIONS Practice Problems
Problem 45.1
One share of Stock A is used as the underlying asset on an exchange option, for which the benchmark
asset is four shares of a Stock B. Currently, Stock A trades for $42 per share, and Stock B trades
for $10 per share. Stock A has an annual price volatility of 0.3 and pays no dividends. Stock B has
an annual price volatility of 0.5 and pays dividends at an annual continuously compounded yield of
0.04. The correlation between the continuously compounded returns on the two assets is 0.4. The
exchange option expires in 1 year. Find the BlackScholes price of this call option.
Problem 45.2
One share of Stock A is used as the underlying asset on an exchange option, for which the benchmark
asset is one share of a Stock B. Currently, Stock A trades for $75 per share, and Stock B trades for
$75 per share. Stock A has an annual price volatility of 0.3 and pays no dividends. Stock B has
an annual price volatility of 0.25 and pays no dividends. The correlation between the continuously
compounded returns on the two assets is 1. The exchange option expires in 1 year. Find the
BlackScholes price of this call option.
Problem 45.3
An exchange call option with expiration of one year allows the owner to acquire one share of a
stock A for one share of a stock B. The price of the option is $2.16. Stock A pays dividends at the
continuously compounded yield of 7%. Stock B pays no dividends. Stock A currently trades for
$50 and stock B trades for $55. Find the value of an exchange option that allows the owner to give
up one share of stock A for one share of stock B.
Problem 45.4
One share of Stock A is used as the underlying asset on an exchange option, for which the benchmark asset is four shares of a Stock B. Currently, Stock A trades for $42 per share, and Stock B
trades for $10 per share. Stock A has an annual price volatility of 0.4 and pays dividends at the
continuously compounded yield rate of 2%. Stock B has an annual price volatility of 0.3 and pays
no dividends. The correlation between the continuously compounded returns on the two assets is
0.5. The exchange option expires in 1 year. Find the BlackScholes price of this call option.
Problem 45.5 ‡
Assume the BlackScholes framework. Consider two nondividendpaying stocks whose time−t prices
are denoted by S1 (t) and S2 (t), respectively.
You are given:
(i) S1 (0) = 10 and S2 (0) = 20.
(ii) Stock 1’s volatility is 0.18. 45 EXCHANGE OPTIONS 313 (iii) Stock 2’s volatility is 0.25.
(iv) The correlation between the continuously compounded returns of the two stocks is −0.40.
(v) The continuously compounded riskfree interest rate is 5%.
(vi) A oneyear European option with payoﬀ max{min{2S1 (1), S2 (1)}− 17, 0} has a current (time0)
price of 1.632.
Consider a European option that gives its holder the right to sell either two shares of Stock 1 or
one share of Stock 2 at a price of 17 one year from now. Calculate the current (time0) price of this
option. 314 AN INTRODUCTION TO EXOTIC OPTIONS The Lognormal Stock Pricing Model
The BlackScholes pricing model assumes that the asset prices are lognormally distributed. The
purpose of this chapter is to examine in more details the lognormal distribution. 315 316 THE LOGNORMAL STOCK PRICING MODEL 46 The Normal Distribution
A normal random variable with parameters µ and σ 2 has a pdf
f (x) = √ (x−µ)2
1
e − 2σ 2 ,
2πσ − ∞ < x < ∞. This density function is a bellshaped curve that is symmetric about µ (See Figure 46.1). Figure 46.1
To prove that the given f (x) is indeed a pdf we must show that the area under the normal curve is
1. That is,
∞
(x−µ)2
1
√
e− 2σ2 dx = 1.
2πσ
−∞
First note that using the substitution y =
∞ √
−∞ x− µ
σ we have (x−µ)2
1
1
e− 2σ2 dx = √
2πσ
2π ∞ y2 e− 2 dy.
−∞ 2 Toward this end, let I = y
∞
e− 2
−∞ ∞ dy. Then
∞ y2 e− 2 dy I2 = ∞ x2 e− −∞ −∞
∞ 2π −∞
∞ r2 e− 2 rdrdθ = 2π =
0 0 ∞ e− 2 dx = x2 +y 2
2 dxdy −∞
r2 re− 2 dr = 2π
0 √
Thus, I = 2π and the result is proved. Note that in the process above, we used the polar
substitution x = r cos θ, y = r sin θ, and dydx = rdrdθ. 46 THE NORMAL DISTRIBUTION 317 Example 46.1
The width of a bolt of fabric is normally distributed with mean 950mm and standard deviation 10
mm. What is the probability that a randomly chosen bolt has a width between 947 and 950 mm?
Solution.
Let X be the width (in mm) of a randomly chosen bolt. Then X is normally distributed with mean
950 mm and variation 100 mm. Thus,
P r(947 ≤ X ≤ 950) = 1
√
10 2π 950 e− (x−950)2
200 dx ≈ 0.118 947 Theorem 46.1
If X is a normal distribution with parameters (µ, σ 2 ) then Y = aX + b is a normal distribution
with parmaters (aµ + b, a2 σ 2 ).
Proof.
We prove the result when a > 0. The proof is similar for a < 0. Let FY denote the cdf of Y. Then
FY (x) =P r(Y ≤ x) = P r(aX + b ≤ x)
x−b
x−b
=P r X ≤
= FX
a
a
Diﬀerentiating both sides to obtain
1
x−b
fY (x) = fX
a
a
1
x−b
=√
exp −(
− µ)2 /(2σ 2 )
a
2πaσ
1
=√
exp −(x − (aµ + b))2 /2(aσ )2
2πaσ
which shows that Y is normal with parameters (aµ + b, a2 σ 2 )
−
Note that if Z = Xσ µ then this is a normal distribution with parameters (0,1). Such a random
variable is called the standard normal random variable.
Theorem 46.2
If X is a normal random variable with parameters (µ, σ 2 ) then
(a) E (X ) = µ
(b) Var(X ) = σ 2 . 318 THE LOGNORMAL STOCK PRICING MODEL Proof.
(a) Let Z =
E (Z ) = ∞
−∞ X −µ
σ be the standard normal distribution. Then xfZ (x)dx = √1
2π ∞
−∞ x2 x2 xe− 2 dx = − √1 π e− 2
2 ∞
−∞ =0 Thus,
E (X ) = E (σZ + µ) = σE (Z ) + µ = µ.
(b) ∞ 1
Var(Z ) = E (Z 2 ) = √
2π
Using integration by parts with u = x and dv = xe
Var(Z) = √1
2π x2 −xe− 2 ∞
−∞ + x2
∞
e− 2 dx
−∞ = √1
2π x2 x2 e− 2 dx.
−∞ 2
−x
2 we ﬁnd x2
∞
e− 2 dx
−∞ = 1. Thus,
Var(X ) = Var(σZ + µ) = σ 2 Var(Z ) = σ 2
Figure 46.2 shows diﬀerent normal curves with the same µ and diﬀerent σ. Figure 46.2
It is traditional to denote the cdf of Z by Φ(x). However, to be consistent with the BlackScholes
notation, we will use the letter N. That is,
1
N ( x) = √
2π
x2 x y2 e− 2 dy.
−∞ Now, since fZ (x) = N (x) = √1 π e− 2 , fZ (x) is an even function. This implies that N (−x) = N (x).
2
Integrating we ﬁnd that N (x) = −N (−x) + C. Letting x = 0 we ﬁnd that C = 2N (0) = 2(0.5) = 1. 46 THE NORMAL DISTRIBUTION 319 Thus,
N (x) = 1 − N (−x), − ∞ < x < ∞. (46.1) This implies that
P r(Z ≤ −x) = P r(Z > x).
Now, N (x) is the area under the standard curve to the left of x. The values of N (x) for x ≥ 0 are
given in the table at the end of the section. Equation 46.1 is used for x < 0.
Example 46.2
On May 5, in a certain city, temperatures have been found to be normally distributed with mean
µ = 24◦ C and variance σ 2 = 9. The record temperature on that day is 27◦ C.
(a) What is the probability that the record of 27◦ C will be broken next May 5 ?
(b) What is the probability that the record of 27◦ C will be broken at least 3 times during the
next 5 years on May 5 ? (Assume that the temperatures during the next 5 years on May 5 are
independent.)
(c) How high must the temperature be to place it among the top 5% of all temperatures recorded
on May 5?
Solution.
(a) Let X be the temperature on May 5. Then X has a normal distribution with µ = 24 and σ = 3.
The desired probability is given by
X − 24
27 − 24
>
= P r(Z > 1)
3
3
=1 − P r(Z ≤ 1) = 1 − N (1) = 1 − 0.8413 = 0.1587 P r(X > 27) =P r (b) Let Y be the number of times with broken records during the next 5 years on May 5. Then, Y
has a binomial distribution with n = 5 and p = 0.1587. So, the desired probability is
P r(Y ≥ 3) =P r(Y = 3) + P r(Y = 4) + P r(Y = 5)
=C (5, 3)(0.1587)3 (0.8413)2 + C (5, 4)(0.1587)4 (0.8413)1
+C (5, 5)(0.1587)5 (0.8413)0
≈0.03106
(c) Let x be the desired temperature. We must have P r(X > x) = 0.05 or equivalently P r(X ≤
x) = 0.95. Note that
P r(X ≤ x) = P r X − 24
x − 24
<
3
3 = Pr Z < x − 24
3 = 0.95 320 THE LOGNORMAL STOCK PRICING MODEL From the Zscore Table (at the end of this section) we ﬁnd P r(Z ≤ 1.65) = 0.95. Thus, we set
x−24
= 1.65 and solve for x we ﬁnd x = 28.95◦ C
3
Next, we point out that probabilities involving normal random variables are reduced to the ones
involving standard normal variable. For example
P r(X ≤ a) = P r a−µ
X −µ
≤
σ
σ =N a−µ
σ . Example 46.3
Let X be a normal random variable with parameters µ and σ 2 . Find
(a)P r(µ − σ ≤ X ≤ µ + σ ).
(b)P r(µ − 2σ ≤ X ≤ µ + 2σ ).
(c)P r(µ − 3σ ≤ X ≤ µ + 3σ ).
Solution.
(a) We have
P r(µ − σ ≤ X ≤ µ + σ ) =P r(−1 ≤ Z ≤ 1)
=N (1) − N (−1)
=2(0.8413) − 1 = 0.6826.
Thus, 68.26% of all possible observations lie within one standard deviation to either side of the
mean.
(b) We have
P r(µ − 2σ ≤ X ≤ µ + 2σ ) =P r(−2 ≤ Z ≤ 2)
=N (2) − N (−2)
=2(0.9772) − 1 = 0.9544.
Thus, 95.44% of all possible observations lie within two standard deviations to either side of the
mean.
(c) We have
P r(µ − 3σ ≤ X ≤ µ + 3σ ) =P r(−3 ≤ Z ≤ 3)
=N (3) − N (−3)
=2(0.9987) − 1 = 0.9974.
Thus, 99.74% of all possible observations lie within three standard deviations to either side of the
mean 46 THE NORMAL DISTRIBUTION 321 Sum of Normal Random Variables1 and the Central Limit Theorem
Suppose that X and Y are two jointly normally distributed2 random variables. For any real numbers
a and b, the sum aX + bY is also a normal random variable3 with variance
Var(aX + bY ) =Cov(aX + bY, aX + bY ) = a2 Cov(X, X ) + 2abCov(X, Y ) + b2 Cov(Y, Y )
2
2
=a2 σX + b2 σY + 2abρσX σY
where
ρ= Cov(X, Y )
Var(X )Var(Y ) is the correlation coeﬃcient that measures the degree of linearity between X and Y and Cov(X, Y )
is the covariance of X and Y.
Now, by the linearity of the mean we have E (aX + bY ) = aE (X ) + bE (Y ).
The normal distribution arises so frequently in applications due the amazing result known as the
central limit theorem which states that the sum of large number of independent identically
distributed random variables is wellapproximated by a normal random variable.
Example 46.4
Let X have a standard normal distribution. Let Z be a random variable independent from X such
1
that P r(Z = 1) = P r(Z = −1) = 2 . Deﬁne Y = XZ.
(a) Show that Y has the standard normal distribution.
(b) Show that X + Y is not normally distributed.
1 See Section 36 of [3]
See Section 29 of [3].
3
In general, the sum of two normal random variables is not normal. However, the sum of two independent normal
random variables is normal.
2 322 THE LOGNORMAL STOCK PRICING MODEL Solution.
(a) We have
FY (x) = P r(Y ≤ x) = P r(XZ ≤ x) = FX (x)
if Z = 1
1 − FX (−x) if Z = −1 Taking the derivative of both sides we ﬁnd fY (x) = fX (x).
(b) This follows from the fact that P r(X + Y = 0) = P r(Z = −1) = 1
2 =1 Example 46.5
Suppose that X1 is a random variable with mean 1 and variance 5 and X2 is a normal random
variable independent from X1 with mean −2 and variance 2. The covariance between X1 and X2
is 1.3. What is the distribution X1 − X2 ?
Solution.
Since the random variables are normal and independent, the diﬀerence is a random variable with
mean E (X1 − X2 ) = E (X2 ) − E (X2 ) = 1 − (−2) = 3 and variance
Var(X1 − X2 ) = 52 + 22 − 2(1.3) = 26.4
Remark 46.1
In general, the product of two normal random variables needs not be normal. 46 THE NORMAL DISTRIBUTION 323 Area under the Standard Normal Curve from −∞ to x
x
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
3.1
3.2
3.3
3.4 0.00
0.01
0.02
0.03
0.5000 0.5040 0.5080 0.5120
0.5398 0.5438 0.5478 0.5517
0.5793 0.5832 0.5871 0.5910
0.6179 0.6217 0.6255 0.6293
0.6554 0.6591 0.6628 0.6664
0.6915 0.6950 0.6985 0.7019
0.7257 0.7291 0.7324 0.7357
0.7580 0.7611 0.7642 0.7673
0.7881 0.7910 0.7939 0.7967
0.8159 0.8186 0.8212 0.8238
0.8413 0.8438 0.8461 0.8485
0.8643 0.8665 0.8686 0.8708
0.8849 0.8869 0.8888 0.8907
0.9032 0.9049 0.9066 0.9082
0.9192 0.9207 0.9222 0.9236
0.9332 0.9345 0.9357 0.9370
0.9452 0.9463 0.9474 0.9484
0.9554 0.9564 0.9573 0.9582
0.9641 0.9649 0.9656 0.9664
0.9713 0.9719 0.9726 0.9732
0.9772 0.9778 0.9783 0.9788
0.9821 0.9826 0.9830 0.9834
0.9861 0.9864 0.9868 0.9871
0.9893 0.9896 0.9898 0.9901
0.9918 0.9920 0.9922 0.9925
0.9938 0.9940 0.9941 0.9943
0.9953 0.9955 0.9956 0.9957
0.9965 0.9966 0.9967 0.9968
0.9974 0.9975 0.9976 0.9977
0.9981 0.9982 0.9982 0.9983
0.9987 0.9987 0.9987 0.9988
0.9990 0.9991 0.9991 0.9991
0.9993 0.9993 0.9994 0.9994
0.9995 0.9995 0.9995 0.9996
0.9997 0.9997 0.9997 0.9997 0.04
0.05
0.06
0.07
0.5160 0.5199 0.5239 0.5279
0.5557 0.5596 0.5636 0.5675
0.5948 0.5987 0.6026 0.6064
0.6331 0.6368 0.6406 0.6443
0.6700 0.6736 0.6772 0.6808
0.7054 0.7088 0.7123 0.7157
0.7389 0.7422 0.7454 0.7486
0.7704 0.7734 0.7764 0.7794
0.7995 0.8023 0.8051 0.8078
0.8264 0.8289 0.8315 0.8340
0.8508 0.8531 0.8554 0.8577
0.8729 0.8749 0.8770 0.8790
0.8925 0.8944 0.8962 0.8980
0.9099 0.9115 0.9131 0.9147
0.9251 0.9265 0.9279 0.9292
0.9382 0.9394 0.9406 0.9418
0.9495 0.9505 0.9515 0.9525
0.9591 0.9599 0.9608 0.9616
0.9671 0.9678 0.9686 0.9693
0.9738 0.9744 0.9750 0.9756
0.9793 0.9798 0.9803 0.9808
0.9838 0.9842 0.9846 0.9850
0.9875 0.9878 0.9881 0.9884
0.9904 0.9906 0.9909 0.9911
0.9927 0.9929 0.9931 0.9932
0.9945 0.9946 0.9948 0.9949
0.9959 0.9960 0.9961 0.9962
0.9969 0.9970 0.9971 0.9972
0.9977 0.9978 0.9979 0.9979
0.9984 0.9984 0.9985 0.9985
0.9988 0.9989 0.9989 0.9989
0.9992 0.9992 0.9992 0.9992
0.9994 0.9994 0.9994 0.9995
0.9996 0.9996 0.9996 0.9996
0.9997 0.9997 0.9997 0.9997 0.08
0.5319
0.5714
0.6103
0.6480
0.6844
0.7190
0.7517
0.7823
0.8106
0.8365
0.8599
0.8810
0.8997
0.9162
0.9306
0.9429
0.9535
0.9625
0.9699
0.9761
0.9812
0.9854
0.9887
0.9913
0.9934
0.9951
0.9963
0.9973
0.9980
0.9986
0.9990
0.9993
0.9995
0.9996
0.9997 0.09
0.5359
0.5753
0.6141
0.6517
0.6879
0.7224
0.7549
0.7852
0.8133
0.8389
0.8621
0.8830
0.9015
0.9177
0.9319
0.9441
0.9545
0.9633
0.9706
0.9767
0.9817
0.9857
0.9890
0.9916
0.9936
0.9952
0.9964
0.9974
0.9981
0.9986
0.9990
0.9993
0.9995
0.9997
0.9998 324 THE LOGNORMAL STOCK PRICING MODEL Practice Problems
Problem 46.1
Scores for a particular standardized test are normally distributed with a mean of 80 and a standard
deviation of 14. Find the probability that a randomly chosen score is
(a) no greater than 70
(b) at least 95
(c) between 70 and 95.
(d) A student was told that her percentile score on this exam is 72%. Approximately what is her
raw score?
Problem 46.2
Suppose that egg shell thickness is normally distributed with a mean of 0.381 mm and a standard
deviation of 0.031 mm.
(a) Find the proportion of eggs with shell thickness more than 0.36 mm.
(b) Find the proportion of eggs with shell thickness within 0.05 mm of the mean.
(c) Find the proportion of eggs with shell thickness more than 0.07 mm from the mean.
Problem 46.3
Assume the time required for a certain distance runner to run a mile follows a normal distribution
with mean 4 minutes and variance 4 seconds.
(a) What is the probability that this athlete will run the mile in less than 4 minutes?
(b) What is the probability that this athlete will run the mile in between 3 min55sec and 4min5sec?
Problem 46.4
You work in Quality Control for GE. Light bulb life has a normal distribution with µ = 2000 hours
and σ = 200 hours. What’s the probability that a bulb will last
(a) between 2000 and 2400 hours?
(b) less than 1470 hours?
Problem 46.5
Human intelligence (IQ) has been shown to be normally distributed with mean 100 and standard
deviation 15. What fraction of people have IQ greater than 130 (“the gifted cutoﬀ”), given that
Φ(2) = .9772?
Problem 46.6
Let X represent the lifetime of a randomly chosen battery. Suppose X is a normal random variable
with parameters (50, 25).
(a) Find the probability that the battery lasts at least 42 hours.
(b) Find the probability that the battery will lasts between 45 to 60 hours. 46 THE NORMAL DISTRIBUTION 325 Problem 46.7
For Company A there is a 60% chance that no claim is made during the coming year. If one or more
claims are made, the total claim amount is normally distributed with mean 10,000 and standard
deviation 2,000 .
For Company B there is a 70% chance that no claim is made during the coming year. If one or
more claims are made, the total claim amount is normally distributed with mean 9,000 and standard
deviation 2,000 .
Assuming that the total claim amounts of the two companies are independent, what is the probability that, in the coming year, Company B’s total claim amount will exceed Company A’s total
claim amount?
Problem 46.8
If for a certain normal random variable X, P r(X < 500) = 0.5 and P r(X > 650) = 0.0227, ﬁnd
the standard deviation of X.
Problem 46.9
Suppose that X is a normal random variable with parameters µ = 5, σ 2 = 49. Using the table of
the normal distribution , compute: (a) P r(X > 5.5), (b) P r(4 < X < 6.5), (c) P r(X < 8), (d)
P r(X − 7 ≥ 4).
Problem 46.10
A company wants to buy boards of length 2 meters and is willing to accept lengths that are oﬀ by
as much as 0.04 meters. The board manufacturer produces boards of length normally distributed
with mean 2.01 meters and standard deviation σ.
If the probability that a board is too long is 0.01, what is σ ?
Problem 46.11
Let X be a normal random variable with mean 1 and variance 4. Find P r(X 2 − 2X ≤ 8).
Problem 46.12
Scores on a standardized exam are normally distributed with mean 1000 and standard deviation
160.
(a) What proportion of students score under 850 on the exam?
(b) They wish to calibrate the exam so that 1400 represents the 98th percentile. What should they
set the mean to? (without changing the standard deviation)
Problem 46.13
The daily number of arrivals to a rural emergency room is a Poisson random variable with a mean
of 100 people per day. Use the normal approximation to the Poisson distribution to obtain the
approximate probability that 112 or more people arrive in a day. 326 THE LOGNORMAL STOCK PRICING MODEL Problem 46.14
A machine is used to automatically ﬁll 355ml pop bottles. The actual amount put into each bottle
is a normal random variable with mean 360ml and standard deviation of 4ml.
(a) What proportion of bottles are ﬁlled with less than 355ml of pop?
(b) Suppose that the mean ﬁll can be adjusted. To what value should it be set so that only 2.5%
of bottles are ﬁlled with less than 355ml?
Problem 46.15
Suppose that your journey time from home to campus is normally distributed with mean equal to
30 minutes and standard deviation equal to 5 minutes. What is the latest time that you should
leave home if you want to be over 99% sure of arriving in time for a class at 2pm?
Problem 46.16
Suppose that the current measurements in a strip of wire are assumed to follow a normal distribution
with mean of 12 milliamperes and a standard deviation of 3 (milliamperes).
(a) What is the probability that a measurement will exceed 14 milliamperes?
(b) What is the probability that a current measurement is between 9 and 16 milliamperes?
(c) Determine the value for which the probability that a current measurement is below this value
is 0.95.
Problem 46.17
Suppose that X1 is a random variable with mean 2 and variance 0.5 and X2 is a normal random
variable independent from X1 with mean 8 and variance 14. The correlation coeﬃcient is −0.3.
What is the distribution X1 + X2 ? 47 THE LOGNORMAL DISTRIBUTION 327 47 The Lognormal Distribution
If X is a normal random variable with parameters µ and σ then the random variable Y = eX is
called lognormal random variable with parameters µ and σ. In other words, Y is lognormal if and
only if ln Y is normal. Equivalently, we can write X = ln Y.
The lognormal density function is given by
fY (x) = 1 ln x−µ 2
1
√ e− 2 ( σ ) , x > 0
xσ 2π and 0 otherwise. To see this, we have for x > 0
FY (x) = P r(Y ≤ x) = P r(X ≤ ln x) = FX (ln x).
Diﬀerentiating both sides with respect to x we ﬁnd
fY (x) = fX (ln x) · 1 ln x−µ 2
1
1
√ e− 2 ( σ )
=
x
xσ 2π where F stands for the cumulative distribution function. Notice that the graph of a lognormal
distribution is nonnegative and skewed to the right as shown in Figure 47.1. Figure 47.1
The mean of Y is 1 2 E (Y ) = eµ+ 2 σ .
Indeed, we have
∞
X ∞
x E (Y ) =E (e ) = e fX (x)dx =
−∞ 1
µ+ 2 σ 2 ∞ =e −∞
1 =eµ+ 2 σ 2 −1
1
√ e2
σ 2π −∞
x−(µ+σ 2 )
σ 1
2
2
1
√ e− 2σ2 [(x−µ) −2σ x] dx
σ 2π 2 dx 328 THE LOGNORMAL STOCK PRICING MODEL since the integral on the right is the total area under a normal density with mean µ + σ 2 and variance
σ 2 so its value is 1.
Similarly,
∞ ∞ e2x fX (x)dx = E (Y 2 ) =E (e2X ) = −∞ −∞ =e 2µ+2σ 2 ∞
−∞ =e2µ+2σ 1
√e
σ 2π −1
2 x−(µ+2σ 2 )
σ 1
2
2
1
√ e− 2σ2 [(x−µ) −4σ x] dx
σ 2π 2 dx 2 Hence, the variance of Y is
2 2 Var(Y ) = E (Y 2 ) − [E (Y )]2 = e2µ+2σ − e2µ+σ .
Example 47.1
Show that the product of two independent lognormal variables is lognormal.1
Solution.
Suppose that Y1 = eX1 and Y2 = eX2 are two independent lognormal random variables. Then X1
and X2 are independent normal random variables. Hence, by the previous section, X1 + X2 is
normal. But Y1 Y2 = eX1 +X2 so that Y1 Y2 is lognormal
Example 47.2
Lifetimes of a certain component are lognormally distributed with parameters µ = 1 day and σ = 0.5
days. Find the probability that a component lasts longer than four days.
Solution.
Let Y represent the lifetime of a randomly chosen component. We need to ﬁnd P r(Y > 4). We
have
P r(Y > 4) = P r(eX > 4) = P r(X > ln 4) = P r(X > 1.386)
where X is a normal random variable with mean 1 and variance 0.52 . Using the zscore, we have
z= 1.386 − 1
= 0.77.
0.5 Using the ztable from the previous section we ﬁnd P r(X < 1.386) = P r( X −1 < 0.77) = 0.7794.
0.5
Hence,
P r(Y > 4) = 1 − P r(X < 1.386) = 1 − 0.7794 = 0.2206
1 The sum and the product of any two lognormally distributed random variables need not be lognormally distributed. 47 THE LOGNORMAL DISTRIBUTION 329 Example 47.3
Suppose that Y is lognormal with parameters µ and σ 2 . Show that for a > 0, aY is lognormal with
parameters ln a + µ and σ 2 .
Solution.
We have for x > 0
FaY (x) = P r(aY ≤ x) = P r(Y ≤ x
) = FY (x/a).
a Taking the derivative of both sides we ﬁnd
1 ln x−(ln a+µ) 2
1
1
) , x>0
σ
√ e− 2 (
faY (x) = fY (x/a) =
a
xσ 2π and 0 otherwise. Thus, aY is lognormal with parameters ln a + µ and σ 2
Example 47.4
Find the median of a lognormal random variable Y with parameters µ and σ 2 .
Solution.
The median is a number α > 0 such that P r(Y ≤ α) = 0.5. That is N (Z ≤
zscore table we ﬁnd that
ln α − µ
=0
σ
Solving for α we ﬁnd α = eµ ln α−µ
)
σ = 0.5. From the Example 47.5
Find the mode of a lognormal random variable Y with parameters µ and σ 2 .
Solution.
The mode is the point of global maximum of the pdf function. It is found by solving the equation
(fY (x)) = 0. But
1 ln x−µ 2
−1
ln x − µ
dfY
√ e− 2 ( σ ) 1 +
(x) =
2 σ 2π
dx
σ2
x
Setting this equation to 0 and solving for x we ﬁnd
x = eµ−σ 2 330 THE LOGNORMAL STOCK PRICING MODEL Practice Problems
Problem 47.1
Which of the following statements best characterizes the relationship between normal and lognormal
distribution?
(A) The lognormal distribution is logarithm of the normal distribution
(B) If ln X is lognormally distributed then X is normally distributed
(C) If X is lognormally distributed then ln X is normally distributed
(D) The two distributions have nothing in common.
Problem 47.2
Which of the following statements are true?
(A) The sum of two independent normal random variables is normal
(B) The product of two normal random variables is normal
(C) The sum of two lognormal random variables is lognormal
(D) The product of two independent lognormal random variables is lognormal.
Problem 47.3
Suppose that X is a normal random variable with mean 1 and standard variation 0.5. Find the
mean and the variance of the random variable Y = eX .
Problem 47.4
For a lognormal variable X , we know that ln X has a normal distribution with mean 0 and standard
deviation of 0.2. What is the expected value of X ?
Problem 47.5
Suppose Y is lognormal with parameters µ and σ 2 . Show that
and σ 2 . 1
Y is lognormal with parameters −µ Problem 47.6
The lifetime of a semiconductor laser has a lognormal distribution with paramters µ = 10 hours
and σ = 1.5 hours. What is the probability that the lifetime exceeds 10,000 hours?
Problem 47.7
The lifetime of a semiconductor laser has a lognormal distribution with paramters µ = 10 hours
and σ = 1.5 hours. What lifetime is exceeded by 99% of lasers?
Problem 47.8
The lifetime of a semiconductor laser has a lognormal distribution with paramters µ = 10 hours
and σ = 1.5 hours. Determine the mean and the standard deviation of the lifetime of the laser. 47 THE LOGNORMAL DISTRIBUTION 331 Problem 47.9
The time (in hours) to fail for 4 light bulbs are:115, 155, 183, and 217. It is known that the time
to fail of the light bulbs is lognormally distributed. Find the parameters of this random variable.
Problem 47.10
The time (in hours) to fail for 4 light bulbs are:115, 155, 183, and 217. It is known that the time
to fail of the light bulbs is lognormally distributed. Find the probability that the time to fail is
greater than 100 hours.
Problem 47.11
Concentration of pollution produced by chemical plants historically are known to exhibit behavior
that resembles a lognormal distribution. Suppose that it is assumed that the concentration of a
certain pollutant, in parts per million, has a lognormal distribution with parameters µ = 3.2 and
σ 2 = 1. What is the probability that the concentration exceeds 8 parts per million?
Problem 47.12
The life, in thousands per mile, of a certain electronic control for locomotives has a lognormal
distribution with parameters µ = 5.149 and σ = 0.737. Find the 5th percentile of the life of such a
locomotive.
Problem 47.13
Let X be a normal random variable with mean 2 and variance 5. What is E (eX )? What is the
median of eX ? 332 THE LOGNORMAL STOCK PRICING MODEL 48 A Lognormal Model of Stock Prices
Let St denote the price of a stock at time t. Consider the relative price of an asset between periods
0 and t deﬁned by
St
= 1 + R0,t
S0
where R0,t is the holding period return. For example, if S0 = $30 and S1 = $34.5 then
so that R0,t = 15%.
We deﬁne the continuously compounded return from time 0 to t by
r0,t = ln St
S0 S1
S0 = 1.15 = ln (1 + R0,t ). For the above example, r0,1 = ln 1.5 = 13.97% which is lower than the holding period return. Note
that
St
St−1
S1
St
=
×
× ··· ×
−→ r0,t = rt−1,t + · · · + r0,1 .
S0
St−1 St−2
S0
Thus, the continuously compounded return from 0 to t is the sum of oneperiod continuously compounded returns.
Example 48.1
Given the following: S0 = $100, r0,1 = 15%, r1,2 = 3%. Find the stock price after 2 year.
Solution.
We have S2 = S0 er0,2 = S0 er0,1 +r1,2 = 100e0.15+0.03 = 100e0.18 = $119.7222
It is commonly assumed in investments that return are independent and identically distributed
over time. This means that investors can not predict future returns based on past returns and
the distribution of returns is stationary. It follows that if the oneperiod continuously compounded
returns are normally distributed, their sum will also be normal. Even if they are not, then by the
Central Limit Theorem, their sum will be normal. From this, we conclude that the relative price
of the stock is lognormally distributed random variable.
t
Now, take the period of time from 0 to t and divide it into n equal subintervals each of length h = n .
Since the continuously compounded returns are identically distributed, we can let E (r(i−1)h,ih ) = µh
2
and Var(r(i−1)h,ih ) = σh for 1 ≤ i ≤ n. Thus,
n E (r0,t ) = E (r(i−1)h,ih ) = nµh =
i=1 t
µh
h 48 A LOGNORMAL MODEL OF STOCK PRICES
and n
2
Var(r(i−1)h,ih ) = nσh = Var(r0,t ) =
i=1 333 t2
σ.
hh Hence, the mean and the variance of the continuously compounded returns are proportional to time.
In what follows t will be expressed in years. For t = 1, we let E (r0,1 ) = α and Var(r0,1 ) = σ 2 . Let δ
be the continuously compounded annual yield on the stock. We examine a lognormal model of the
stock price as follows: We assume that the continuously compounded return from time 0 to time
t is normally distributed with mean (α − δ − 0.5σ 2 )t and variance σ 2 t. Using the Z −score we can
write
√
St
= (α − δ − 0.5σ 2 )t + σ tZ
ln
S0
and solving for St we ﬁnd
√
2
St = S0 e(α−δ−0.5σ )t+σ tZ .
(48.1)
Example 48.2
Find the expected stock price St .
Solution.
12
First recall that if X is a normal random variable with parameters µ and σ then E (eX ) = eµ+ 2 σ .
St
Let X = ln S0 . Then X is a normal random variable with parameters (α − δ − 0.5σ 2 )t and σ 2 t.
Hence,
12
2
E (eX ) = e(α−δ−0.5σ )t+ 2 σ t .
But eX = St
S0 so that
12
1
2
E (St ) = e(α−δ−0.5σ )t+ 2 σ t .
S0 Hence,
E (St ) = S0 e(α−δ)t .
We call the diﬀerence α − δ the expected continuously rate of appreciation on the stock
Example 48.3
Find the median stock price. That is the price where 50% of the time prices will be above or below
that value.
Solution.
For the median we want to ﬁnd the stock price such that
Pr Z < ln (St /S0 ) − (α − δ − 0.5σ 2 )t
√
σt = 0.5. 334 THE LOGNORMAL STOCK PRICING MODEL This happens when
ln (St /S0 ) − (α − δ − 0.5σ 2 )t
√
= 0.
σt
Solving we ﬁnd
St = S0 e(α−δ−0.5σ
1 2 )t 1 = E (St )e− 2 σ 2t 2 Now, since e− 2 σ t < 1, we use the last equation to write St < E (St ). That is, the median is below
the mean. This says that more than 50% of the time, a lognormally distributed stock price will earn
below its expected return. Note that if σ is such that (α − δ − 0.5σ 2 )t < 0 then St < S0 < E (St ).
In this case, a lognormally distributed stock will lose money more than half the time.
Example 48.4
A nondividendpaying stock is currently trading for $100. The annual expected rate of return on
the stock is α = 10% and the annual standard deviation is σ = 30%. Assume the stock price is
lognormally distributed.
(a) Using the above model, ﬁnd an expression for the stock price after 2 years.
(b) What is the expected stock price two years from now?
(c) What is the median stock price?
Solution.
(a) The stock price is
S2 = S0 e(α−δ−0.5σ 2 )×2+σ √ 2Z 1 √ 2 )×2+0.3 = 100e(0.1− 2 ×0.3 2Z . (b) We have
E (S2 ) = 100e0.1×2 = $122.14.
(c) The median stock price is
1 1 2 E (S2 )e− 2 σ t = 122.14e− 2 (0.3) 2 (2) 2 = 122.14e−0.3 = $111.63 Now, since
Z= ln (St /S0 ) − E (ln (St /S0 ))
Var (ln (St /S0 )) we deﬁne a one standard deviation move up in the stock price by letting Z = 1 and a one
standard deviation down by letting Z = −1
Example 48.5
Using the same information in the prvious example, ﬁnd the stock price over 2 years for a one
standard deviation up and a one standard deviation down. 48 A LOGNORMAL MODEL OF STOCK PRICES 335 Solution.
A one standard deviation move up over 2 years is given by
1 S2 = 100e(0.1− 2 (0.3) √ 2 )×2+0.3 2 = $170.62 and a one standard deviation move down over 2 years is given by
1 2 )×2−0.3 S2 = 100e(0.1− 2 (0.3) √ 2 = $73.03 Estimating the Parameters of the Lognormal Pricing Model
We next see how to estimate the mean and the variance of lognormally distributed price data.
When stocks are lognormally distributed, if St−h is the stock price at the beginning of a period then
the price at the end of the period is found using the formula
St = St−h e(α−δ−0.5σ 2 )h+σ √ hZ . From this it follows that ln (St /St−h ) is normally distributed with mean (α − δ − 0.5σ 2 )h and
variance σ 2 h. That is, by using the log of ratio prices at adjacent points in time, we can compute
the continuously compounded mean and variance.
Example 48.6
The table below contains seven weekly stock price observations. The stock pays no dividends.
Week
Price 1
2
3
4
5
6
7
100 105.04 105.76 108.93 102.50 104.80 104.13 (a) Estimate the continuously compounded mean and variance.
(b) Estimate the annual expected rate of return.
Solution.
(a) We compute the weekly continuous compounded returns
Week
1
2
3
4
5
6
7
Price
100 105.04 105.76 108.93 102.50
104.80 104.13
ln (St /St−1 ) −
0.0492 0.0068 0.0295 −0.0608 0.0222 −0.0064
The weekly continuously compounded mean is
X= 0.0492 + 0.0068 + 0.0295 − 0.0608 + 0.0222 − 0.0064
= 0.00675.
6 336 THE LOGNORMAL STOCK PRICING MODEL The annual continuously compounded mean is 0.0675 × 52 = 3.51.
The weekly continuously compounded variance is
6
t=1 (ln (St /St−1 ) − X )2 5 = 0.03822 . The annualized standard deviation is
σAnnual = 0.0382 × √ 52 = 0.274. (b) The annual expected rate of return is the annual mean of onehalf the variance. That is,
α = 0.00675 × 52 + 0.5 × 0.2742 = 0.3885
Remark 48.1
The above data are hypothetical data. Statistical theory tells us that the observed mean is determined by the diﬀerence between the beginning and ending stock price. What happens in between
is irrelevant. Thus, having frequent observations is not helpful in estimating the mean returns. Observations taken over a longer period of time are more likely to reﬂect the true mean and therefore
they increase the precision of the estimate.
Unlike the mean, one can increase the precision of the estimate of the standard deviation by making
more frequent observations. 48 A LOGNORMAL MODEL OF STOCK PRICES 337 Practice Problems
Problem 48.1
Suppose that S0 = $100 and S2 = $170.62. Find the relative stock price from time 0 to time 2.
What is the holding period return over the 2 years?
Problem 48.2
The continuously compounded return from time 0 to time 2 is 18%. Find the holding period return
over the 2year period.
Problem 48.3
Suppose the stock price is initially $100 and the continuously compounded return on a stock is 15%
in one year and r1,2 the next year. What is r1,2 if the stock price after 2 years is $119.722?
Problem 48.4
Suppose the stock price is initially $100 and the continuously compounded return on a stock is 15%
in one year and r1,2 the next year. What is R1,2 if the stock price after 2 years is $119.722?
Problem 48.5
Suppose that the continuously compounded return from time 0 to time t is normal with mean
1
(α − δ − 2 σ 2 )t and variance σ 2 t. The stock pays dividend at the annual continuously compounded
yield of 3%. The annual variance is 0.09. Initially, the stock was trading for $75. Suppose that the
mean for the continuously compounded 2year return is 0.07. Find the the annual expected rate of
return α.
Problem 48.6
Suppose that the continuously compounded return from time 0 to time t is normal with mean
(α − δ − 1 σ 2 )t and variance σ 2 t. The stock pays dividend at the annual continuously compounded
2
yield of 3%. The annual variance is 0.09. Initially, the stock was trading for $75. Suppose that the
mean for the continuously compounded 2year return is 0.07. Find the expected stock price after 4
years.
Problem 48.7
Suppose that the continuously compounded return from time 0 to time t is normal with mean
(α − δ − 1 σ 2 )t and variance σ 2 t. The stock pays dividend at the annual continuously compounded
2
yield of 3%. The annual variance is 0.09. Initially, the stock was trading for $75. Suppose that
the mean for the continuously compounded 2year return is 0.07. Find the median price at time 4
years. 338 THE LOGNORMAL STOCK PRICING MODEL Problem 48.8
Suppose that the continuously compounded return from time 0 to time t is normal with mean
1
(α − δ − 2 σ 2 )t and variance σ 2 t. The stock pays dividend at the annual continuously compounded
yield of 3%. The annual variance is 0.09. Initially, the stock was trading for $75. Suppose that the
mean for the continuously compounded 2year return is 0.07. Also, suppose that the dividends are
reinvested in the stock. Find the median of the investor’s position at time 4 years.
Problem 48.9
A nondividendpaying stock is currently trading for $100. The annual expected rate of return
α = 10% and the annual standard deviation is σ = 60%. Assume the stock price is lognormally
distributed.
(a) Using the above model, ﬁnd an expression for the stock price after 2 years.
(b) What is the expected stock price two years from now?
(c) What is the median stock price?
Problem 48.10
Consider a stock with annual variance of 0.16. Suppose that the expected stock price in ﬁve years
is $135. Find the median of the stock price at time 5 years.
Problem 48.11
A nondividendpaying stock is currently trading for $100. The annual expected rate of return on
the stock is α = 13% and the annual standard deviation is σ = 25%. Assume the stock price is
lognormally distributed. Find the stock price over 6 months for a one standard deviation up and a
one standard deviation down.
Problem 48.12
A stock is currently trading for $100. The annual expected rate of return on the stock is α = 13%
and the annual standard deviation is σ = 25%. The stock pays dividends at the continuously
compounded yield of 2%. Assume the stock price is lognormally distributed. Find the stock price
over 6 months for a two standard deviation up.
Problem 48.13
The table below contains seven monthly stock price observations. The stock pays no dividends.
Week
Price 1
2
345
100 104 97 95 103 (a) Estimate the annual continuously compounded mean and standard deviation.
(b) Estimate the annual expected rate of return. 48 A LOGNORMAL MODEL OF STOCK PRICES
Problem 48.14 ‡
Assume the Black−Scholes framework.
The price of a nondividendpaying stock in seven consecutive months is:
Month
1
2
3
4
5
6
7 Price
54
56
48
55
60
58
62 Estimate the continuously compounded expected rate of return on the stock. 339 340 THE LOGNORMAL STOCK PRICING MODEL 49 Lognormal Probability Calculations
In this section we use the lognormal stock pricing model discussed in the previous section to compute
a number of probabilities and expectations. Among the questions we consider are the question of
the probability of an option to expire in the money, and given that the option expires in the money,
what will be the expected stock price.
Let K be an arbitrary positive number. K can be the strike price of an option. We would like to
ﬁnd the probability that a stock price is less than K. Since
−1
1
√ e2
xσ 2π ln (St /S0 )−(α−δ −0.5σ 2 )t
√
σt 2 −1
1
√ e2
=
xσ 2π ln St −ln S0 −(α−δ −0.5σ 2 )t
√
σt 2 the random variable ln St is normally distributed with parameters ln S0 + (α − δ − 0.5σ 2 )t and σ 2 t.
Now, we have
P r(St < K ) =P r(ln St < ln K )
ln St − ln S0 − (α − δ − 0.5σ 2 )t
ln K − ln S0 − (α − δ − 0.5σ 2 )t
√
√
=P r
<
σt
σt
ln K − ln S0 − (α − δ − 0.5σ 2 )t
√
=P r Z <
σt
ˆ
=N (−d2 )
where ln S0 − ln K + (α − δ − 0.5σ 2 )t
ˆ
√
d2 =
.
σt
ˆ
ˆ
Now, using the property that N (d2 ) + N (−d2 ) = 1 we obtain
ˆ
P r(St ≥ K ) = 1 − P r(St < K ) = N (d2 ).
Remark 49.1
ˆ
We have found that the probability for a call option to expire in the money is N (d2 ) and that for
ˆ
a put option is N (−d2 ). These probabilities involve the true expected return on the stock. If α is
being replaced by the riskfree rate r then we obtain the riskneutral probability that an option
expires in the money.
Example 49.1
You are given the following information:S0 = 50, α = 0.10, σ = 0.25, and δ = 0. What is P r(S2 >
$65)? 49 LOGNORMAL PROBABILITY CALCULATIONS 341 Solution.
ˆ
We are asked to ﬁnd P r(S2 > 65) = N (d2 ) where
ln S0 − ln K + (α − δ − 0.5σ 2 )t
ln 50 − ln 65 + (0.10 − 0 − 0.5 × 0.252 )(2)
ˆ
√
√
d2 =
=
= −0.3532.
σt
0.25 2
Thus,
ˆ
P r(St > 65) = N (d2 ) = N (−0.3532) = 0.361969
Next, suppose we are interested in knowing the range of prices such that there is 95% probability
that the stock price will be in that range after a certain time. This requires generating a 95%
conﬁdence interval. Using the lognormal distribution, we will discuss how to generate conﬁdence
intervals.
Suppose that we want to be (1 − p)% conﬁdent that a stock price is in the range [StL , StU ]. This is
equivalent to saying that we want to be p% conﬁdent that either St < StL or St > StU . We split this
probability into two, half for each tail. In other words, we want
P r(St < StL ) =
But P r(St < StL ) = p
2 p
2 and P r(St > StU ) = p .
2 ˆ
ˆ
implies N (−d2 ) = p/2 so that d2 = −N −1 (p/2). Hence,
N −1 (p/2) = − ln StL − ln S0 + (α − δ − 0.5σ 2 )t
√
.
σt Solving this equation for StL we ﬁnd
StL = S0 e(α−δ−0.5σ 2 )t+σ √ tN −1 (p/2) . Likewise, we solve for the StU such that
ˆ
N −1 (1 − p/2) = d2
This gives us
StU = S0 e(α−δ−0.5σ 2 )t+σ √ tN −1 (1−p/2) . Example 49.2
Use the Zscore table to derive the 95% conﬁdence interval if S0 = $100, t = 2, δ = 0, α = 0.10, and
σ = 0.30. 342 THE LOGNORMAL STOCK PRICING MODEL Solution.
ˆ
ˆ
We have p = 5% so that N (−d2 ) = 0.025 and N (d2 ) = 0.975. Using the Z −score table we ﬁnd
N (0.025) = −1.96 and N (0.975) = 1.96. Hence,
StL = S0 e(α−δ−0.5σ 2 )t+σ and
StU = S0 e(α−δ−0.5σ 2 )t+σ √ √ tN −1 (p/2) √ 2 )×2−0.30 = 100e(0.10−0.5×0.3 tN −1 (1−p/2) 2×1.96 √ 2 )×2+0.30 = 100e(0.10−0.5×0.3 = $48.599 2×1.96 = $256.40 There is 95% probability that in two years the stock price will be between $48.599 and $256.40
Portfolio Insurance in the Long Run
Consider a portfolio of stocks. Historical return statistics has indicated that the rate of return from
investing in stocks over a long time has outperformed that from investing in riskfree bonds. That
is, one is led to believe that if a stock is held for suﬃciently long period of time, the investment is
as safe as the freerisk bond.
To guarantee that investing in stocks is safe in the long run, it was suggested that a put option is
to be bought today insuring that after T years the stock portfolio would worth at least as much as
if one had instead invested in a zerocoupon bond. If the initial stock price is S0 then one has to
set the strike price of the put option to KT = S0 erT . Using the BlackScholes model, it was shown
that in the absence of arbitrage the premium of the insurance increases with time to maturity T.
We can use the results of this section to show that the probability of the payoﬀ of the put option
increases with the time to maturity. Suppose that the stock price is lognormally distributed using
the model of the previous section. Assume that the stock pays no dividends. Then the probability
that the bond will outperform the stock is given by
P r(ST < KT ) = N 12
σ
2 − (α − r) √
T
σ . In the BlackScholes analysis, the put price depends in part on the riskneutral probability that the
stock will underperform bonds. We can obtain this probability by setting α = r in the previous
equation obtaining
1√
P r∗ (ST < KT ) = N
σT.
2
This says that the riskneutral probability that the put will pay oﬀ is increasing with time.
Example 49.3
Consider a European put option on a stock with initial price S0 . Suppose that the riskpremium
is greater than 0.5σ 2 . Show that the probability that the option expires in the money is less than
50%. 49 LOGNORMAL PROBABILITY CALCULATIONS 343 Solution.
The probability that the option expires in the money is
ˆ
P r(ST < S0 erT ) = N (−d2 )
where √
2
2
ˆ2 = ln S0 − ln K + (α − δ − 0.5σ )t = [(α − r) − 0.5σ ] T .
√
d
σ
σt Hence,
P r(ST < S0 erT ) = N √
[0.5σ 2 − (α − r)] T
σ ˆ
Since α − r > 0.5σ 2 , we ﬁnd 0.5σ 2 − (α − r) < 0 and therefore −d2 < 0. This shows that P r(ST <
rT
S0 e ) < 50% 344 THE LOGNORMAL STOCK PRICING MODEL Practice Problems
Problem 49.1
A nondividendpaying stock that is lognormally distributed is currently selling for $50. Given that
the annual expected rate of return is 15% and the annual standard deviation is 30%. Find the
monthly continuously compounded mean return and the monthly standard deviation.
Problem 49.2
A nondividendpaying stock that is lognormally distributed is currently selling for $50. Given that
the annual expected rate of return is 15% and the annual standard deviation is 30%. It is known
that for the standard normal distribution, there is 68.27% probability of drawing a number in the
interval (−1, 1) and a 95.45% probability of drawing a number in the interval (−2, 2).
(a) Find rL and rU so that there is 68.37% that the monthly continuously compounded return on
the stock will be between rL and rU .
(b) Find rL and rU so that there is 95.45% that the monthly continuously compounded return on
the stock will be between rL and rU .
Problem 49.3
A nondividendpaying stock that is lognormally distributed is currently selling for $50. Given that
the annual expected rate of return is 15% and the annual standard deviation is 30%. It is known
that for the standard normal distribution, there is 68.27% probability of drawing a number in the
interval (−1, 1) and a 95.45% probability of drawing a number in the interval (−2, 2).
(a) Find S L and S U so that there is 68.37% that the monthly stock price will be between S L and
SU .
(b) Find S L and S U so that there is 95.45% that the monthly stock price will be between S L and
SU .
Problem 49.4
You are given the following information:S0 = 50, α = 0.10, σ = 0.25, and δ = 0. What is P (S2 <
$65)?
Problem 49.5
You are given: S0 = 75, α = 0.16, δ = 0, σ = 0.30, t = 0.5 and p = 0.10. Find StU .
Problem 49.6
Let P (St < S0 ) be the probability that the stock price at the end of t years is less than the original
stock price. You are given:α = 0.10, δ = 0, and σ = 0.4 Show P (St < S0 ) < 50%. Hint: Assume
the contrary and get a contradiction. 49 LOGNORMAL PROBABILITY CALCULATIONS 345 Problem 49.7
Consider a European put option on a stock with initial price S0 . Suppose that the riskpremium is
greater than 0.5σ 2 . Show that the probability that the option expires in the money decreases as the
time to expiration increases.
Problem 49.8
A stock is currently selling for $100. Given that annual rate of return on the stock is 10%. The annual continuously compounded riskfree rate of 8%. Suppose that T years from now the probability
that the stock will outperform a zerocoupon bond is 50%. Find the volatility σ.
Problem 49.9
A nondividendpaying stock that is lognormally distributed is currently selling for $75. Given that
the annual expected rate of return is 12% and the annual standard deviation is 25%. Find the
biannual continuously compounded mean return and the biannual standard deviation.
Problem 49.10
A nondividendpaying stock that is lognormally distributed is currently selling for $75. Given that
the annual expected rate of return is 12% and the annual standard deviation is 25%. It is known
that for the standard normal distribution, there is 68.27% probability of drawing a number in the
interval (−1, 1) and a 95.45% probability of drawing a number in the interval (−2, 2).
(a) Find rL and rU so that there is 68.37% that the biannual continuously compounded return on
the stock will be between rL and rU .
(b) Find rL and rU so that there is 95.45% that the biannual continuously compounded return on
the stock will be between rL and rU .
Problem 49.11
A nondividendpaying stock that is lognormally distributed is currently selling for $75. Given that
the annual expected rate of return is 12% and the annual standard deviation is 25%. It is known
that for the standard normal distribution, there is 68.27% probability of drawing a number in the
interval (−1, 1) and a 95.45% probability of drawing a number in the interval (−2, 2).
(a) Find S L and S U so that there is 68.37% that the biannual stock price will be between S L and
SU .
(b) Find S L and S U so that there is 95.45% that the biannual stock price will be between S L and
SU .
Problem 49.12
A stock is currently selling for $100. The stock pays dividends at the annual continuously compounded yield of 3%. The annual rate of return on the stock is 15% and the annual standard
deviation is 30%. Find the probability that a call on the stock with strike price of $125 will expire
in the money 2 years from today. 346 THE LOGNORMAL STOCK PRICING MODEL Problem 49.13 ‡
You are given the following information about a nondividendpaying stock:
(i) The current stock price is 100.
(ii) The stockprice is lognormally distributed.
(iii) The continuously compounded expected return on the stock is 10%.
(iv) The stock’s volatility is 30%.
Consider a ninemonth 125strike European call option on the stock. Calculate the probability that
the call will be exercised.
Problem 49.14 ‡
Assume the BlackScholes framework.
You are given the following information for a stock that pays dividends continuously at a rate
proportional to its price.
(i) The current stock price is 0.25.
(ii) The stock’s volatility is 0.35.
(iii) The continuously compounded expected rate of stockprice appreciation is 15%.
Calculate the upper limit of the 90% lognormal conﬁdence interval for the price of the stock in 6
months. 50 CONDITIONAL EXPECTED PRICE AND A DERIVATION OF BLACKSCHOLES FORMULA347 50 Conditional Expected Price and a Derivation of BlackScholes Formula
Given that an option expires in the money, what is the expected stock price? More speciﬁcally,
suppose that a call option with a strike K expires in the money, the expected stock price is the
conditional stock price E (ST ST > K ) which is deﬁned to be the ratio of the partial expectation
over the probability that ST > K. That is,
E (ST ST > K ) = P E (St St > K )
P r(St > K ) where P E stands for partial expectation which is given by
∞ P E (St St > K ) = St g (St , S0 )dSt
K where
g (St , S0 ) = 1 −1
2 √
e
St 2πσ 2 t ln St −ln S0 −(α−δ −0.5σ 2 )t
√
σt 2 . Now, we have
∞ P E (St St > K ) = St g (St , S0 )dSt
K =√ 1 ∞ −1
2 2 ln St −ln S0 −(α−δ −0.5σ 2 )t
√
σt e
dSt
2πσ 2 t K
∞
√
1
2
2
=√
eln S0 +(α−δ−0.5σ )t+σ twt −0.5wt dwt
2 )t
2π ln K −ln S0 −(αt−δ−0.5σ
√
σ
=eln S0 +(α−δ−0.5σ
1
=S0 e(α−δ)t √
2π 2 )t+0.5σ 2 t 1
√
2π ∞ 1 ln K −ln S0 −(α−δ −0.5σ 2 )t
√
σt ∞
ln K −ln S0 −(α−δ +0.5σ 2 )t
√
σt 1 2 e− 2 wt dwt ln S0 − ln K + (α − δ + 0.5σ 2 )t
√
σt
ˆ
=S0 e(α−δ)t N (d1 ) =S0 e(α−δ)t N where ln S0 − ln K + (α − δ + 0.5σ 2 )t
ˆ
√
d1 =
.
σt e− 2 (wt −σ √ t)2 dwt 348 THE LOGNORMAL STOCK PRICING MODEL Hence,
E (St St > K ) = ˆ
ˆ
N (d1 )
S0 e(α−δ)t N (d1 )
= S0 e(α−δ)t
.
ˆ
P (St > K )
N (d2 ) For a put, See Problem 50.1, we ﬁnd
E (St St < K ) = S0 e(α−δ)t ˆ
N (−d1 )
.
ˆ
N (−d2 ) Example 50.1
You are given the following information:S0 = 50, α = 0.10, σ = 0.25, and δ = 0. What is E (S2 S2 >
$65)?
Solution.
ˆ
N (d )
We are asked to ﬁnd E (S2 S2 > 65) = S0 e(α−δ)t N (dˆ1 ) where
2 ln 50 − ln 65 + (0.10 − 0 + 0.5 × 0.252 )(2)
ln S0 − ln K + (α − δ + 0.5σ 2 )t
ˆ
√
√
=
d1 =
= 0.0004
σt
0.25 2
and
ln 50 − ln 65 + (0.10 − 0 − 0.5 × 0.252 )(2)
ln S0 − ln K + (α − δ − 0.5σ 2 )t
ˆ
√
√
=
= −0.3532.
d2 =
σt
0.25 2
Thus, using Excel spreadsheet we ﬁnd
ˆ
ˆ
N (d1 ) = N (0.0004) = 0.50016 and N (d2 ) = N (−0.3532) = 0.361969.
Hence,
E (S2 S2 > 65) = 50e0.10×2 × 0.50016
= $84.385
0.361969 Example 50.2
A stock is currently selling for $100. The stock pays dividends at the continuously compounded
yield of 3%. Consider a European call on the stock with strike price of $100 and time to maturity T.
Suppose that the annual rate of return is 3%. Show that the partial expectation of ST conditioned
on ST > 100 increases as the time to maturity increases.
Solution.
We have √
ˆ
P E (St St > K ) = S0 e(α−δ)t N (d1 ) = 100N (0.5σ T ). 50 CONDITIONAL EXPECTED PRICE AND A DERIVATION OF BLACKSCHOLES FORMULA349
√
It follows that as T increases N (0.5σ T ) increases and therefore the partial expectation increases
Derivation of the BlackScholes Formula
We next examine a derivation of the BlackScholes formula using the results established above. We
use riskneutral probability where α = r in the formulas above. We let g ∗ denote the riskneutral
pdf, E ∗ denote the expectation with respect to riskneutral probabilities, and P r∗ the riskneutral
probability.
For a European call option with strike K , the price is deﬁned as the expected value of e−rt max{0, St −
K }. That is,
C (S, K, σ, r, t, δ ) = E [e−rt max{0, St − K }]
Now we proceed to establish the BlackScholes formula based on riskneutral probability:
C (S, K, σ, r, t, δ ) =E [e−rt max{0, St − K }]
∞ =e−rt (St − K )g ∗ (St , S0 )dSt
K =e−rt E ∗ (St − K St > K )P r∗ (St > K )
=e−rt E ∗ (St St > K )P r∗ (St > K ) − e−rt E ∗ (K St > K )P r∗ (St > K )
=e−δt S0 N (d1 ) − Ke−rt N (d2 )
which is the celebrated BlackScholes formula.
A similar argument, shows that the price of a put option is given by
P (S, K, σ, r, t, δ ) = Ke−rt N (−d2 ) − S0 e−δt N (−d1 ). 350 THE LOGNORMAL STOCK PRICING MODEL Practice Problems
Problem 50.1
The partial expectation of St , conditioned on St < K, is given by
K P E (St St < K ) = St g (St , S0 )dSt
0 where
1 1
−2 ln St −ln S0 −(α−δ −0.5σ 2 )t
√
σt 2 √
e
.
St 2πσ 2 t
What is the expected stock price conditioned on a put option with strike price K expiring in the
money?
g (St , S0 ) = Problem 50.2
Derive the BlackScholes formula for a put:
P (S, K, σ, r, t, δ ) = Ke−rt N (−d2 ) − S0 e−δt N (−d1 ).
Problem 50.3
A stock is currently selling for $100. The stock pays dividends at the continuously compounded
yield of 3%. Consider a European put on the stock with strike price of $100 and time to maturity
T. Suppose that the annual rate of return is 3%. Show that the expectation of ST conditioned on
ST < 100 decreases as the time to maturity increases.
Problem 50.4
A stock is currently selling for $100. The stock pays dividends at the continuously compounded
yield of 3%. Consider a European put on the stock with strike price of $100 and time to maturity T.
Suppose that the annual rate of return is 3%. Show that the partial expectation of ST conditioned
on ST < 100 decreases as the time to maturity increases.
Problem 50.5
A stock is currently selling for $100. The stock pays no dividends and has annual standard deviation
of 25%. Consider a European call on the stock with strike price of K and time to maturity in 2
years. Suppose that the annual rate of return is 10%. It is found that the partial expectation of St ,
conditioned on St > K is 50e0.2 . Determine the value of K.
Problem 50.6
A stock is currently selling for $50. The stock pays dividends at the continuously compounded yield
of 4%. The annual standard deviation is 30%. Consider a European call on the stock with strike
price of $48 and time to maturity in six months. Suppose that the annual rate of return is 13%.
Find the partial expectation of St if the call expires in the money. 50 CONDITIONAL EXPECTED PRICE AND A DERIVATION OF BLACKSCHOLES FORMULA351
Problem 50.7
A stock is currently selling for $50. The stock pays dividends at the continuously compounded yield
of 4%. The annual standard deviation is 30%. Consider a European call on the stock with strike
price of $48 and time to maturity in six months. Suppose that the annual rate of return is 13%.
Find the conditional expectation that the call expires in the money.
Problem 50.8
A stock is currently selling for S0 . The stock pays dividends at the continuously compounded yield
of 3%. Consider a European call on the stock with strike price S0 and time to maturity of 3 years.
The annual standard deviation on the stock price is 25%. The annual return on the stock is 9%.
The partial expectation that the stock will expire in the money in three years is 85.88. That is,
P E (S3 S3 > S0 ) = 85.88. Find the value of S0 .
Problem 50.9
Consider a binomial model in which a put strike price is $50, and the stock price at expiration can
be $20,$40,$60, and $80 with probabilities 1/8,3/8,3/8, and 1/8.
(a) Find the partial expectation for a put to be in the money.
(b) Find the conditonal expectation. 352 THE LOGNORMAL STOCK PRICING MODEL Option Pricing Via Monte Carlo
Simulation
A Monte Carlo option model uses Monte Carlo methods to calculate the price of an option
with complicated features. In general, the technique is to generate several thousand possible (but
random) price paths for the underlying asset via simulation, and to then calculate the associated
payoﬀ of the option for each path. These payoﬀs are then averaged and discounted to today, and
this result is the value of the option today. 353 354 OPTION PRICING VIA MONTE CARLO SIMULATION 51 Option Valuation as a Discounted Expected Value
In this section, we examine an option pricing that is performed by discounting expected payoﬀ
values at the riskfree rate. We will see that Monte Carlo valuation model exploits the ideas of this
section.
Consider a stock that is currently trading for $41. A European call option on the stock has a strike
price of $40 and expiration of one year. The stock pays no dividends and its price volatility is 30%.
The continuously compounded riskfree interest rate is 8%. Figure 51.1
We consider a threeperiod binomial tree. The values of u and d are
√1
√
1
u = e(r−δ)h+σ h = e0.08× 3 +0.3 3 = 1.221
and √
(r−δ )h−σ h d=e √1 0.08× 1 −0.3
3 =e 3 = 0.864. The riskneutral probability of an up movement is
1 e(r−δ)h − d
e0.08× 3 − 0.864
pu =
=
= 0.4566.
u−d
1.221 − 0.864 51 OPTION VALUATION AS A DISCOUNTED EXPECTED VALUE 355 The tree of stock prices, option payoﬀs, the associated probabilities, and the expected payoﬀ values
are shown in Figure 51.1. The option price can be found be discounting the expected payoﬀ values
obtaining:
C = e−0.08 (3.290 + 4.356) = $7.058
which is the same price obtained by using the binomial pricing model as discussed in Section 17.
We can use Figure 51.1 to illustrate Monte Carlo estimation of the option price. Imagine a gambling
wheel divided into four unequal sections: A (corresponds to the payoﬀ $34.362), B ($12.811),
C ($0.00), and D ($0.00). Suppose we spin the wheel 100 times with the following number of
occurrence: A (30), B (22), C (38), and D (10). Then an estimate of the expected payoﬀ is
30 × 34.362 + 22 × 12.811 + 38 × 0 + 10 × 0
= $13.13.
100
An estimate to the call option price is
C ≈ e−0.08 × 13.13 = $12.12.
The procedure we used to price the call option discussed earlier does not work if riskneutral
probabilities are replaced by true probabilities. The reason is that the discount rate is not the same
at all nodes of the tree. Suppose that the expected rate of return is 15%. Then the true probability
of an upward move is
1 e(0.15−0)× 3 − 0.864
e(α−δ)h − d
=
= 0.5246.
p=
u−d
1.221 − 0.864
We next ﬁnd the discount
Node with Stock price
Node with Stock price
Node with Stock price
Node with Stock price
Node with Stock price rate γ at each node of the tree:
= $74.632 γ = N/A.
= $52.811 γ = N/A.
= $37.371 γ = N/A.
= $26.444 γ = N/A.
= $61.124 We have e−γ ×h [pCuuu + (1 − p)Cuud ] = e−rh [pu Cuuu + (1 − pu )Cuud
or
1 1 e−γ × 3 [0.5246 × 34.632 + (1 − 0.5246) × 12.811] = e−0.08× 3 [0.4566 × 34.632 + (1 − 0.4566) × 12.811].
Solving this equation we ﬁnd γ = 0.269. Likewise, we ﬁnd
Node with Stock price = $43.253 γ = 0.495 356 OPTION PRICING VIA MONTE CARLO SIMULATION Node with Stock price = $30.606 γ =N/A
Node with Stock price = $50.061 γ = 0.324
Node with Stock price = $35.424 γ = 0.496
Node with Stock price = $41.00 γ = 0.357.
With true probabilities, the unique discount rate we used for the riskneutral probability is now
being replaced by the discount rate for each path. The discount rate of each path is deﬁned to
be the average of the discount rate at each node along the path. In Figure 51.1, there are eight
possible paths for the stock price but only four have a positive option payﬀ, namely the paths:
uuu, uud, udu, and duu. We next ﬁnd the discounting payoﬀ for each path:
• Along the uuu path, the path discount rate is (0.357 + 0.324 + 0.269)/3 = 31.67%. The probability
of the path is p3 = 0.1444 with payoﬀ of $34.632. Hence, the discounted expected value of the path
is e−0.3167 × 0.1444 × 34.632 = $3.643.
• Along the uud path, the path discount rate is (0.357 + 0.324 + 0.269)/3 = 31.67%. The probability
of the path is p2 (1 − p) = 0.1308 with payoﬀ of $12.811. Hence, the discounted expected value of
the path is e−0.3167 × 0.1308 × 12.811 = $1.221.
• Along the udu path, the path discount rate is (0.357 + 0.324 + 0.496)/3 = 39.23%. The probability
of the path is p2 (1 − p) = 0.1308 with payoﬀ of $12.811. Hence, the discounted expected value of
the path is e−0.3923 × 0.1308 × 12.811 = $1.132.
• Along the duu path, the path discount rate is (0.357 + 0.496 + 0.495)/3 = 44.93%. The probability
of the path is p2 (1 − p) = 0.1308 with payoﬀ of $12.811. Hence, the discounted expected value of
the path is e−0.4493 × 0.1308 × 12.811 = $1.069.
It follows that the call option price is
3.643 + 1.221 + 1.132 + 1.069 = $7.065
Remark 51.1
In remainder of the chapter it is assumed that the world is riskneutral. 51 OPTION VALUATION AS A DISCOUNTED EXPECTED VALUE 357 Practice Problems
Problem 51.1
Consider a stock that is currently trading for $50. A European put option on the stock has a
strike price of $47 and expiration of one year. The continuously compounded dividend yield is 0.05.
The price volatility is 30%. The continuously compounded riskfree interest rate is 6%. Using a
twoperiod binomial model, ﬁnd the riskneutral probability of two ups, one up and one down, and
two downs.
Problem 51.2
Using the information of the previous problem, construct a ﬁgure similar to Figure 51.1
Problem 51.3
Using a gambling wheel with unequal sections, where each section has a probability to one of the
option payoﬀs in the previous problem, the following occurrence are recorded for 140 spins of the
wheel: 31 (for the uu path), 70 (for the ud or du path) and 39 (for the dd path). Estimate the put
option price.
Problem 51.4
Consider a stock that is currently trading for $50. A European call option on the stock has a
strike price of $47 and expiration of one year. The continuously compounded dividend yield is 0.05.
The price volatility is 30%. The continuously compounded riskfree interest rate is 6%. Using a
twoperiod binomial model, construct a tree showing stock prices, option payoﬀs, the associated
probabilities, and the expected payoﬀ values.
Problem 51.5
Using a gambling wheel with unequal sections, where each section has a probability to one of the
option payoﬀs in the previous problem, the following occurrence are recorded for 140 spins of the
wheel: 31 (for the uu path), 70 (for the ud or du path) and 39 (for the dd path). Estimate the call
option price.
Problem 51.6
Given the following information about a 1year European call option on a stock:
• The strike price is $47.
• The current price of the stock is $50.
• The expected rate of return is 10%.
• The continuously compounded yield is 5%.
• The continuously compounded riskfree rate is 6• Volatility is 30%.
Find the discount rate at each node in a twoperiod binomial tree using actual probabilities. 358 OPTION PRICING VIA MONTE CARLO SIMULATION Problem 51.7
Estimate the price of the option in the previous exercise using the discounted expected value approach. 52 COMPUTING NORMAL RANDOM NUMBERS 359 52 Computing Normal Random Numbers
In this section we discuss a couple of techniques to compute normally distributed random numbers
required for Monte Carlo valuation. Since both techniques involve uniformly distributed random
variables, we start by a discussion of this concept.
A continuous random variable X is said to be uniformly distributed over the interval a ≤ x ≤ b if
its pdf is given by
1
if a ≤ x ≤ b
b−a
f (x) =
0
otherwise
Since F (x) = x
−∞ f (t)dt, the cdf is given by F (x) = 0 x−a
b−a 1 if x ≤ a
if a < x < b
if x ≥ b If a = 0 and b = 1 then X is called the standard uniform random variable.
Remark 52.1
The values at the two boundaries a and b are usually unimportant because they do not alter
the value of the integral of f (x) over any interval. Sometimes they are chosen to be zero, and
1
1
sometimes chosen to be b−a . Our deﬁnition above assumes that f (a) = f (b) = f (x) = b−a . In the
case f (a) = f (b) = 0 then the pdf becomes
f (x) = 1
b−a 0 if a < x < b
otherwise Example 52.1
You are the production manager of a soft drink bottling company. You believe that when a machine
is set to dispense 12 oz., it really dispenses 11.5 to 12.5 oz. inclusive. Suppose the amount dispensed
has a uniform distribution. What is the probability that less than 11.8 oz. is dispensed?
Solution.
Since f (x) = 1
12.5−11.5 = 1, P r(11.5 ≤ X ≤ 11.8) = area of rectangle of base 0.3 and height 1 = 0.3
Example 52.2
Suppose that X has a uniform distribution on the interval (0, a), where a > 0. Find P (X > X 2 ). 360 OPTION PRICING VIA MONTE CARLO SIMULATION Solution.
If a ≤ 1 then P r(X > X 2 ) =
1
P r(X > X 2 ) = min{1, a } a1
dx
0a = 1. If a > 1 then P r(X > X 2 ) = 11
dx
0a = 1
.
a Thus, The expected value of X is
b xf (x) E (X ) =
a
b =
a x
dx
b−a
b x2
=
2(b − a) a
b 2 − a2
a+b
=
=
2(b − a)
2
and so the expected value of a uniform random variable is halfway between a and b. Because
b E (X 2 ) =
a x2
dx
b−a
b x3
3(b − a) a
b 3 − a3
=
3(b − a)
a2 + b2 + ab
=
3
= then
Var(X ) = E (X 2 ) − (E (X ))2 = a2 + b2 + ab (a + b)2
(b − a)2
−
=
.
3
4
12 Example 52.3
You arrive at a bus stop 10:00 am, knowing that the bus will arrive at some time uniformly distributed between 10:00 and 10:30 am. Let X be your wait time for the bus. Then X is uniformly
distributed in (0, 30). Find E (X ) and Var(X ).
Solution.
We have a = 0 and b = 30. Thus, E (X ) = a+b
2 = 15 and Var(X ) = (b−a)2
12 = 302
12 = 75 52 COMPUTING NORMAL RANDOM NUMBERS 361 The ﬁrst technique to compute normally distributed random variables consists of summing 12 uniformly distributed random variables on (0, 1) and subtracting 6. That is, considering the random
variable
n ˆ
Z= Ui − 6
i=1 where the Ui are uniformly distributed random variables on (0, 1). The variance of each Ui is 1/12
ˆ
and the mean is 1/2. Thus, the mean of Z is
12 ˆ
E (Z ) = E (Ui ) − 6 = 12 ×
i=1 1
−6=0
2 and the variance is
12 ˆ
Var(Z ) = 12 Var Ui − Var(Ui ) =
i=1 i=1 1
2 12 = Var(Ui ) = 1.
i=1 What we obtain is technically not normally distributed but is symmetric with mean zero and
standard deviation of 1.0, which are three properties associated with the normal distribution. We
ˆ
can use the central limit theorem and look at Z as a close approximation to a standard normal
random variable.
Example 52.4
Using the function RAND() that generates unifromly distributed random numbers on (0,1) we draw
the set of numbers {0.123, 0.456, 0.013, 0.222, 0.781, 0.345, 0.908, 0.111, 0.415, 0.567, 0.768, 0.777}. Find
the standard normal estimate of this draw.
Solution.
The sum of these numbers is
0.123 + 0.456 + 0.013 + 0.222 + 0.781 + 0.345 + 0.908 + 0.111 + 0.415 + 0.567 + 0.768 + 0.777 = 5.486.
Hence,
ˆ
Z = 5.486 − 6 = −0.514
The second technique that we consider uses the inverse of the cumulative standard normal distribution function N (x). The idea is to convert a single uniformly distributed random number to a
normally distributed random number. Let u be a uniformly distributed random number in (0, 1). 362 OPTION PRICING VIA MONTE CARLO SIMULATION The idea is to interpret F (u) as a quantile.1 Thus, if F (u) = 0.5, we interpret it as 50% quantile.
We then use the inverse distribution function, N −1 (u), to ﬁnd the value from the standard normal
distribution corresponding to that quantile.
Remark 52.2
The above process simulate draws from the standard normal distribution. Exponentiating these
draws we simulate a lognormal random variable. The above procedure of using the inverse cumulative distribution function works for any distribution for which the cdf has an inverse. That is,
suppose that D is a CDF such that D−1 exists. Let u be a number from the uniform distribution on
(0, 1). To ﬁnd the estimate of this draw in the distribution of D we solve the equation D(d) = F (u)
for d or d = D−1 (F (u)).
Example 52.5
Find the 30% quantile of the standard normal random variable.
Solution.
We want to ﬁnd z such that P r(Z ≤ z ) ≥ 0.30 or equivalently N (z ) ≥ 0.30. Considering the
equation N (z ) = 0.30 we ﬁnd z = N −1 (0.30) = −0.524 using NormSInv in Excel. Using a table,
we have N (−z ) = 1 − N (z ) = 1 − 0.30 = 0.70 ≈ 0.7019 so that −z = 0.53 or z = −0.53
Example 52.6
A draw from a uniformly distributed random variable on (3, 5) is 4.6. Find the corresponding single
draw from the standard normal distribution.
Solution.
We have 4.6 − 3
= 0.8.
5−3
Next, we want to ﬁnd the 0.8 quantile of the standard normal variable. That is, we want to
ﬁnd z such that N (z ) = 0.8. Using the inverse cumulative standard normal distribution we ﬁnd
z = N −1 (0.8) = 0.842 using Excel spreadsheet. Using the table we will have z = N −1 (80) =
N −1 (0.7995) = 0.84
F (4.6) = 1 The q th quantile of a random variable X is the smallest number such that FX (x) = P (X ≤ x) ≥ q. Quantiles
for any distribution are uniformly distributed which means that any quantile is equally likely to be drawn. 52 COMPUTING NORMAL RANDOM NUMBERS 363 Practice Problems
Problem 52.1
The total time to process a loan application is uniformly distributed between 3 and 7 days.
(a) Let X denote the time to process a loan application. Give the mathematical expression for the
probability density function.
(b) What is the probability that a loan application will be processed in fewer than 3 days ?
(c) What is the probability that a loan application will be processed in 5 days or less ?
Problem 52.2
Customers at Santos are charged for the amount of salad they take. Sampling suggests that the
amount of salad taken is uniformly distributed between 5 ounces and 15 ounces. Let X = salad
plate ﬁlling weight
(a) Find the probability density function of X.
(b) What is the probability that a customer will take between 12 and 15 ounces of salad?
(c) Find E (X ) and Var(X).
Problem 52.3
Suppose thst X has a uniform distribution over the interval (0, 1). (a) Find F (x).
(b) Show that P (a ≤ X ≤ a + b) for a, b ≥ 0, a + b ≤ 1 depends only on b.
Problem 52.4
Using the function RAND() that generates unifromly distributed random numbers on (0,1) we draw
the set of numbers {0.126, 0.205, 0.080, 0.303, 0.992, 0.481, 0.162, 0.786, 0.279, 0.703, 0.752, 0.994}. Find
the standard normal estimate of this draw.
Problem 52.5
Find the 10%quantile of the standard normal random variable.
Problem 52.6
Consider an exponentially distributed random variable X with CDF given by F (x) = 1 − e−0.5x .
Let 0.21072 be drawn from this distribution. Find the corresponding single draw from the standard
normal distribution.
Problem 52.7
Consider the following three draws from the uniform distribution in (0, 1) : 0.209, 0.881, and 0.025.
Find the corresponding draws from the standard normal distribution. 364 OPTION PRICING VIA MONTE CARLO SIMULATION 53 Simulating Lognormal Stock Prices
Recall that for a standard normal distribution Z , a lognormal stock price is given by
St = S0 e(α−δ−0.5σ 2 )t+σ √ tZ . (53.1) Suppose we want to draw random stock prices for t years from today. We can do that by randomly
drawing a set of standard Z’s and substituting the results into the equation above.
Example 53.1
Given the following: S0 = 50, α = 0.12, δ = 0, σ = 0.30, and T = 3. Find the random set of
lognormally distributed stock prices over 3 years using the set of uniform random numbers on
(0, 1) : 0.209, 0.881, 0.025.
Solution.
We have N (Z1 ) = F (0.209) = 0.209 so that Z1 = N −1 (0.209) = −0.8099. Similarly, Z2 =
N −1 (0.881) = 1.18 and Z3 = N −1 (0.025) = −1.95996. Thus, the new stock prices are
S3 =50e(0.12−0−0.5×0.3 √ 2 )(3)+0.3 3×(−0.8099) √
(0.12−0−0.5×0.32 )(3)+0.3 3×(1.18) S3 =50e √ (0.12−0−0.5×0.32 )(3)+0.3 S3 =50e = $41.11 = $115.60 3×(−1.96) = $22.61 Now, if we want to simulate the path of the stock price over t years (which is useful for pricing
pathdependent options) then we can do so by splitting t into n equal intervals each of length h,
t
that is, n = h . In this case, we ﬁnd the following stock prices:
Sh =S0 e(α−δ−0.5σ 2 )h+σ S2h =Sh e(α−δ−0.5σ
..
. =.
.. 2 )h+σ Snh =S(n−1)h e(α−δ−0.5σ
Note that √
(α−δ −0.5σ 2 )t+σ t St = S0 e
n
i=1 √ hZ (1) √ hZ (2) 2 )h+σ 1
√
n √ hZ (n) n
i=1 Z (i) . 1
Since √n
Z (i)1 is a normal random variable on (0, 1), we get the same distribution at time t
with the above equation as if we had drawn a single normal random variable on (0, 1) as in equation
1 Note that the Zi s are independent. 53 SIMULATING LOGNORMAL STOCK PRICES 365 (53.1). The advantage of splitting up the problem into n draws is to create a simulation of the path
taken by the stock price.
Example 53.2
Given the following information about a stock: T = 1, S0 = 100, α = 0.10, δ = 0, and σ = 0.30. Find
the simulated stock price one year from today based on the drawing of the following two random
numbers from the uniform distribution on (0, 1) : 0.15 and 0.65.
Solution.
We ﬁrst ﬁnd the corresponding draws from the standard normal distribution. We have N (z1 ) =
F (0.15) = 0.15 so that z1 = N −1 (0.15) = −1.0363. Likewise, z2 = N −1 (0.65) = 0.38532. Hence,
√
(α−δ −0.5σ 2 )T +σ T ST =S0 e z1 +z2
√
2 √
√
(0.10−0−0.5×0.32 )×1+0.3 1× −1.0363+0.38532 =100e
=$112.16 2 Example 53.3
The price of a stock is to be estimated using simulation. It is known that:
(i) The timet stock price, St , follows the lognormal distribution: ln (St /St−1 ) is normal with mean
(α − 0.5σ 2 )t and variance σ 2 t.
(ii) S0 = 50, α = 0.12, and σ = 0.30.
The following are three uniform (0, 1) random numbers:
0.209, 0.881, 0.025.
Use each of these three numbers to simulate a time3 stock price. Calculate the mean of the three
simulated prices.
Solution.
We have Z (1) = N −1 (0.209) = −0.81, Z (2) = N −1 (0.881) = 1.18, and Z (3) = N −1 (0.025) = −1.96.
Thus,
2 )×3+0.3× 1
S1 =50e(0.12−0.5×0.3 2 )×3+0.3× 2
S1 =50e(0.12−0.5×0.3
3
S1 √
√ 3×(−0.81)
3×1.18 = 115.60 √
(0.12−0.5×0.32 )×3+0.3× 3×(−1.96) =50e = 41.11 = 22.61 Thus, the mean of the three simulated prices is
41.11 + 115.60 + 22.61
= $59.77
3 366 OPTION PRICING VIA MONTE CARLO SIMULATION Practice Problems
Problem 53.1
A stock is currently selling for $100. The continuously compounded riskfree rate is 11%. The
continuously compounded dividend yield is 3%. The volatility of the stock according to the BlackScholes framework is 30%. Find the new stock prices along a path based on the uniformly distributed
random numbers: 0.12, 0.87, and 0.50, given that T = 1 year. That is, ﬁnd S 1 , S 2 , and S1 .
3 3 Problem 53.2
A nondividendpaying stock is currently selling for $40. The continuously compounded riskfree
rate is 8%. The volatility of the stock is 30%. Find a simulation of the stock prices along a path
based on the uniformly distributed random numbers: 0.038, 0.424, 0.697, and 0.797, given that
T = 1 year.
Problem 53.3 ‡
The price of a stock is to be estimated using simulation. It is known that:
(i) The timet stock price, St , follows the lognormal distribution: ln (St /St−1 ) is normal with mean
(α − 0.5σ 2 )t and variance σ 2 t.
(ii) S0 = 50, α = 0.15, and σ = 0.30.
The following are three uniform (0, 1) random numbers:
0.9830, 0.0384, 0.7794.
Use each of these three numbers to simulate a time2 stock price. Calculate the mean of the three
simulated prices. 54 MONTE CARLO VALUATION FOR EUROPEAN OPTIONS 367 54 Monte Carlo Valuation for European Options
In this section we discuss the pricing of standard European options using Monte Carlo simulation.
1
2
n
Consider an option on a stock with strike price K and expiration time T. Let ST , ST , · · · , ST be n
randomly drawn timeT stock prices. Let V (S, T ) denote the timeT option payoﬀ. We deﬁne the
time0 Monte Carlo price of the option by
1
V (S0 , 0) = e−rT
n n
i
V (ST , T ) (54.1) i=1 i
i
where r is the riskfree interest rate. For the case of a call option we have V (ST , T ) = max{0, ST −
i
i
K }. For a put, we have V (ST , T ) = max{0, K − ST }. Note that equation (54.1) uses simulated
lognormal stock prices to approximate the lognormal stock price distribution. Also, note that the
Monte Carlo valuation uses the riskneutral probabilities so that the simulated stock price is given
by the formula
√
2
ST = S0 e(r−δ−0.5σ )T +σ T Z . In the next example, we price a European call option using both the BlackScholes pricing model
and the Monte Carlo pricing model so that we can assess the performance of Monte Carlo Valuation.
Example 54.1
Consider a nondividend paying stock. The annual continuously compounded riskfree interest rate
is 0.08, and the stock price volatility is 0.3. Consider a 40strike call on the stock with 91 days to
expiration.
(a) Within the BlackScholes framework, what is the option price today if the current stock price
is $40?
(b) Using the Monte Carlo, what is the option price today if the current stock price is $40?
Solution.
(a) Using the BlackScholes formula we ﬁnd
C = Se−δT N (d1 ) − Ke−rT N (d2 )
where
d1 = ln (40/40) + (0.08 − 0 + 0.5(0.3)2 ) ×
ln (S/K ) + (r − δ + 0.5σ 2 )T
√
=
σT
0.3 91 91
365 365 and √
d2 = d1 − σ T = 0.2080477495 − 0.3 91
= 0.0582533699
365 = 0.2080477495. 368 OPTION PRICING VIA MONTE CARLO SIMULATION Thus,
1
N (d1 ) = √
2π
and
1
N (d2 ) = √
2π 0.2080477495 x2 e− 2 dx = 0.582404
−∞
0.0582533699 x2 e− 2 dx = 0.523227.
−∞ Hence, 91 C = 40 × 0.582404 − 40e−0.08× 365 × 0.523227 = $2.7804
(b) We draw random 3month stock prices using the formula
S3months = 40e(0.08−0.5(0.3) 2 )×0.25+0.3 √ 0.25Z . For each stock price, we compute the option payoﬀ
Option payoﬀ = max{0, S3months − 40}.
The following table shows the results of Monte Carlo valuation of the call option using 5 trials where
each trial uses 500 random draws.
Trial
1
2
3
4
5 Computed Call Price
2.98
2.75
2.63
2.75
2.91 Averaging the prices in the table we ﬁnd $2.804 versus the BlackScholes price $2.7804
Example 54.2
A stock is currently selling for $100. The annual continuously compounded yield is 0.02. The
annual continuously compounded riskfree interest rate is 0.07, and the stock price volatility is 0.25.
Consider a $102strike put with one year to expiration. Using the four draws from the uniform
distribution on (0,1): 0.90, 0.74, 0.21, 0.48, compute the price of the put using the Monte Carlo
valuation.
Solution.
First we ﬁnd the corresponding draws from the normal distribution on (0,1):
N (Z1 ) = F (0.90) ⇒ Z1 = N −1 (0.90) = 1.282 54 MONTE CARLO VALUATION FOR EUROPEAN OPTIONS 369 N (Z2 ) = F (0.74) ⇒ Z1 = N −1 (0.74) = 0.64
N (Z3 ) = F (0.21) ⇒ Z1 = N −1 (0.21) = −0.806
N (Z4 ) = F (0.48) ⇒ Z1 = N −1 (0.48) = −0.05
The stock prices are found using the formula
ST = S0 e(r−δ−0.5σ 2 )T +σ √ TZ . The four simulated stock prices with their payoﬀs are shown in the table below
i
i
S1
V (S1 , 1)
140.39
0
119.57
0
83.30
18.70
100.63
1.37 Thus, the Monte Carlo time0 price of the put is
V (100, 0) = e−0.07
[0 + 0 + 18.70 + 1.37] = $4.68
4 Error Analysis of Monte Carlo Methods
The Monte Carlo valuation is simple but relatively ineﬃcient. A key question in error analysis is
how many simulated stock prices needed in order to achieve a certain desired degree of accuracy.
One approach is to run the simulation diﬀerent times and see how much variability there is in the
result. For example, the table of Example 54.1 shows the results from running ﬁve Monte Carlo
valuations, each consisting of 500 stock price draws. It took 2500 draws to get a close answer to
the correct answer of $2.7804.
The option price estimate obtained from a Monte Carlo simulation is a sample average. Thus an
elementary principle of statistics states that the standard deviation of the estimate is the standard
deviation of the sample (i.e., of the individual price estimates) divided by the square root of the
sample size. Consequently, the error reduces at the rate of 1 over the square root of the sample size.
Thus, the accuracy of the estimate is increased by increasing the sample size. To see this, suppose
σ is the standard deviation of the sample. We ﬁrst conduct a Monte Carlo simulation using n1
random drawings. Since the option value is a sample mean, the standard deviation of our estimate
√
of the option value is σ/ n1 . Now suppose we wanted to reduce that standard deviation of the
estimate in half. How much larger must the sample be? Let this new sample size be n2 . Then its
√
standard deviation of the estimate of the option price is σ/ n2 . But then
√σ
n2 = 1 √σ
2 n1 if and only if n2 = 4n1 . 370 OPTION PRICING VIA MONTE CARLO SIMULATION Thus, to achieve a 50% reduction in error, i.e., a 50% increase in accuracy, we must quadruple the
number of random drawings. That is, the standard error reduces only at the rate of the square root
of the sample size, not at the rate of the sample size itself.
Example 54.3
The standard deviation of the 2500 price estimates of Example 54.1 is $4.05. Find the standard
deviation of a sample of 500 draws. What percentage of the correct option price is that?
Solution.
For 500 draws, the standard deviation is
4.05
σ
√ =√
= $0.18
n
500
which is 0.18/2.78 = 6.5% of the correct option price 54 MONTE CARLO VALUATION FOR EUROPEAN OPTIONS 371 Practice Problems
Problem 54.1
A stock is currently selling for $100. The annual continuously compounded yield is 0.03. The
annual continuously compounded riskfree interest rate is 0.11, and the stock price volatility is 0.30.
Consider a $102strike call with six months to expiration. Using the two draws from the uniform
distribution on (0,1): 0.15 and 0.65 compute the price of the call using the Monte Carlo valuation.
Problem 54.2
A stock is currently selling for $100. The annual continuously compounded yield is 0.03. The
annual continuously compounded riskfree interest rate is 0.11, and the stock price volatility is 0.30.
Consider a $102strike put with one year to expiration. Using the three draws from the uniform
distribution on (0,1): 0.12, 0.87, and 0.50, compute the price of the put using the Monte Carlo
valuation.
Problem 54.3
How many draws are required in Example 54.1 so that the standard deviation of the estimate is 1%
of the correct price?
Problem 54.4
In a Monte Carlo estimate of a European call, the standard deviation of the individual price estimates is found to be 3.46. What is the standard deviation of the estimate in a sample of 10000
draws?
Problem 54.5
In a Monte Carlo estimate of a European call, the standard deviation of the individual price estimates is found to be 3.46. The correct price is $2.74. How many draws are required so that the
standard deviation of the estimate is 2% of the correct price? 372 OPTION PRICING VIA MONTE CARLO SIMULATION 55 Monte Carlo Valuation of Asian Options
In this section, we consider the Monte Carlo valuation of an average (either arithmetic or geometric)
price1 Asian option where the payoﬀ is based on an average stock price that replaces the actual
price at expiration. Similar discussion holds for average strike Asian options.
Consider an Asian call option with strike price K and time to expiration T. Split T into n equal
periods each of length h = T . Let Z (i), 1 ≤ i ≤ n, be random draws from the normal distribution
n
on (0, 1). We can ﬁnd the simulated stock prices as follows:
S1 =S0 e(r−δ−0.5σ 2 )h+σ S2 =S1 e(r−δ−0.5σ
..
. =.
.. 2 )h+σ Snh =S(n−1)h e(r−δ−0.5σ √
√ hZ (1)
hZ (1) 2 )h+σ √ hZ (1) For an average price Asian option the payoﬀ is given by
max{0, S − K }
where
S= S1 + S2 + · · · + Snh
n in the case of an arithmetic average or
1 S = (S1 × S2 × · · · × Snh ) n
for a geometric average.
Example 55.1
A stock price is currently selling for $100. The continuously compounded riskfree rate is 11%. The
continuously compounded yield is 3%. The stock price volatility is 30%. Consider an arithmetic
average price Asian put option with time to maturity of 3 years and strike price of $100. The
stock price at the end of years 1, 2 , and 3 are simulated using the following (0,1) uniform random
numbers: 0.12, 0.87, and 0.50. Find the payoﬀ of this put at expiration.
Solution.
Using NormSInv in Excel spreadsheet we ﬁnd Z (1) = N −1 (0.12) = −1.175, Z (2) = N −1 (0.87) =
1 The advantage of averaging stock prices is that it reduces the likelihood of large gains or losses. 55 MONTE CARLO VALUATION OF ASIAN OPTIONS 373 1.126, and Z (3) = N −1 (0.50) = 0.000. Thus the yearly stock prices are
S1 =S0 e(r−δ−0.5σ 2 )T +σ (r−δ −0.5σ 2 )T +σ S2 =S1 e (r−δ −0.5σ 2 )T +σ S3 =S2 e √
√
√ T Z (1)
T Z (2)
T Z (3) = 100e(0.11−0.03−0.5(0.3) √ 2 )+0.3 1×(−1.175)
√ (0.11−0.03−0.5(0.3)2 )+0.3 = 71.36e 1×(1.126) √
(0.11−0.03−0.5(0.3)2 )+0.3 1×(0) = 101.04e = $71.36
= $101.04 = $102.57 Thus, the payoﬀ of the put option is
max{0, 100 − (71.36 + 101.04 + 102.57)/3} = $8.34
Example 55.2
A stock price is currently selling for $100. The continuously compounded riskfree rate is 11%.
The continuously compounded yield is 3%. The stock price volatility is 30%. Consider a geometric
average strike Asian call option with time to maturity of 1 year. The stock price at the end of
four months, eight months, and 12 months are simulated using the following (0,1) uniform random
numbers: 0.12, 0.87, and 0.50. Find the payoﬀ of this call at expiration.
Solution.
Using NormSInv in Excel spreadsheet we ﬁnd Z (1) = N −1 (0.12) = −1.175, Z (2) = N −1 (0.87) =
1.126, and Z (3) = N −1 (0.50) = 0.000. Thus, the simulated stock prices are
√1
√
1
2
2
S1 =S0 e(r−δ−0.5σ )T +σ T Z (1) = 100e(0.11−0.03−0.5(0.3) )× 3 +0.3 3 ×(−1.175) = $82.54
√1
√
1
2
2
S2 =S1 e(r−δ−0.5σ )T +σ T Z (2) = 82.54e(0.11−0.03−0.5(0.3) )× 3 +0.3 3 ×(1.126) = $101.49
√1
√
1
2
2
S3 =S2 e(r−δ−0.5σ )T +σ T Z (3) = 101.49e(0.11−0.03−0.5(0.3) )× 3 +0.3 3 ×(0) = $102.68
Thus, the payoﬀ of the call option is
1 max{0, 102.68 − (82.54 × 101.49 × 102.68) 3 } = $7.58
Example 55.3
A stock price is currently selling for $100. The continuously compounded riskfree rate is 11%.
The continuously compounded yield is 3%. The stock price volatility is 30%. Consider a geometric
average strike Asian call option with time to maturity of 1 year. Find the payoﬀ of this call at
expiration if he stock price at the end of four months, eight months, and 12 months are simulated
using the following (0,1) uniform random numbers:
(a) 0.12, 0.87, and 0.50.
(b) 0.341, 0.7701, and 0.541
(c) 0.6751, 0.111, and 0.078.
Estimate the Monte Carlo valuation of this call option. 374 OPTION PRICING VIA MONTE CARLO SIMULATION Solution.
(a) From the previous example, we found the payoﬀ of the call option to be
1 max{0, 102.68 − (82.54 × 101.49 × 102.68) 3 } = $7.58.
(b) Using NormSInv in Excel spreadsheet we ﬁnd Z (1) = N −1 (0.341) = −0.41, Z (2) = N −1 (0.7701) =
0.739, and Z (3) = N −1 (0.541) = 0.103. Thus, the simulated stock prices are
√
√
(r−δ −0.5σ 2 )T +σ T Z (1)
(0.11−0.03−0.5(0.3)2 )× 1 +0.3 1 ×(−0.41)
3
3
S1 =S0 e
= 100e
= $94.24
√1
√
2 )T +σ T Z (2)
2 )× 1 +0.3
×(0.739)
3
S2 =S1 e(r−δ−0.5σ
= 94.24e(0.11−0.03−0.5(0.3) 3
= $108.37
√1
√
1
2
2
S3 =S2 e(r−δ−0.5σ )T +σ T Z (3) = 108.37e(0.11−0.03−0.5(0.3) )× 3 +0.3 3 ×(0.103) = $111.62
Thus, the payoﬀ of the call option is
1 max{0, 111.62 − (94.24 × 108.37 × 111.62) 3 } = $7.16.
(c) Using NormSInv in Excel spreadsheet we ﬁnd Z (1) = N −1 (0.6751) = 0.454, Z (2) = N −1 (0.111) =
−1.22, and Z (3) = N −1 (0.078) = −1.41. Thus, the simulated stock prices are
√
√
(0.11−0.03−0.5(0.3)2 )× 1 +0.3 1 ×(0.454)
(r−δ −0.5σ 2 )T +σ T Z (1)
3
3
= 100e
= $109.45
S1 =S0 e
√1
√
2 )T +σ T Z (2)
2 )× 1 +0.3
×(−1.22)
3
S2 =S1 e(r−δ−0.5σ
= 109.45e(0.11−0.03−0.5(0.3) 3
= $89.64
√1
√
1
2
2
S3 =S2 e(r−δ−0.5σ )T +σ T Z (3) = 89.64e(0.11−0.03−0.5(0.3) )× 3 +0.3 3 ×(−1.41) = $71.04
Thus, the payoﬀ of the call option is
1 max{0, 71.04 − (109.45 × 89.64 × 71.04) 3 } = $0.00
The Monte Carlo estimate of the call option price is
CAsian = e−0.11
[7.58 + 7.16 + 0.00] = $0.90
3 55 MONTE CARLO VALUATION OF ASIAN OPTIONS 375 Practice Problems
Problem 55.1
A stock price is currently selling for $50. The continuously compounded riskfree rate is 8%. The
stock pays no dividends. The stock price volatility is 30%. Consider an arithmetic average price
Asian call option with time to maturity of 3 years and strike price of $50. The stock price at the
end of years 1, 2 , and 3 are simulated using the following (0,1) uniform random numbers: 0.983,
0.0384, and 0.7794. Find the payoﬀ of this call at expiration.
Problem 55.2
The price of a stock is to be estimated using simulation. It is known that:
(i) The timet stock price, St , follows the lognormal distribution: ln (St /St−1 ) is normal with mean
(α − 0.5σ 2 )t and variance σ 2 t.
(ii) S0 = 50, α = 0.15, and σ = 0.30.
The following are three uniform (0, 1) random numbers:
0.9830, 0.0384, 0.7794.
Use each of these three numbers to simulate stock prices at the end of four months, eight months,
and 12 motnhs. Calculate the arithmetic mean and the geometric mean of the three simulated
prices.
Problem 55.3
You are given the following:
(i) The timet stock price, St , follows the lognormal distribution: ln (St /St−1 ) is normal with mean
(α − 0.5σ 2 )t and variance σ 2 t.
(ii) S0 = 50, α = 0.15, and σ = 0.30.
The following are three uniform (0, 1) random numbers:
0.9830, 0.0384, 0.7794.
Find the payoﬀ at expiration of an Asian call option with strike price of $50 based on the arithmetic
average of the simulated prices of Problem 55.2
Problem 55.4
You are given the following:
(i) The timet stock price, St , follows the lognormal distribution: ln (St /St−1 ) is normal with mean
(α − 0.5σ 2 )t and variance σ 2 t. 376 OPTION PRICING VIA MONTE CARLO SIMULATION (ii) S0 = 50, α = 0.15, and σ = 0.30.
The following are three uniform (0, 1) random numbers:
0.9830, 0.0384, 0.7794.
Find the payoﬀ at expiration of an Asian put option with strike price of $70 based on the arithmetic
average of the simulated prices of Problem 55.2
Problem 55.5
You are given the following:
(i) The timet stock price, St , follows the lognormal distribution: ln (St /St−1 ) is normal with mean
(α − 0.5σ 2 )t and variance σ 2 t.
(ii) S0 = 50, α = 0.15, and σ = 0.30.
The following are three uniform (0, 1) random numbers:
0.9830, 0.0384, 0.7794.
Find the payoﬀ at expiration of an arithmetic average strike Asian call option based on the simulated
prices of Problem 55.2
Problem 55.6
You are given the following:
(i) The timet stock price, St , follows the lognormal distribution: ln (St /St−1 ) is normal with mean
(α − 0.5σ 2 )t and variance σ 2 t.
(ii) S0 = 50, α = 0.15, and σ = 0.30.
The following are three uniform (0, 1) random numbers:
0.9830, 0.0384, 0.7794.
Find the payoﬀ at expiration of an arithmetic average strike Asian put option based on the simulated
prices of Problem 55.2
Problem 55.7
You are given the following:
(i) The timet stock price, St , follows the lognormal distribution: ln (St /St−1 ) is normal with mean
(α − 0.5σ 2 )t and variance σ 2 t.
(ii) S0 = 50, α = 0.15, and σ = 0.30.
The following are three uniform (0, 1) random numbers:
0.9830, 0.0384, 0.7794.
Find the payoﬀ at expiration of an Asian call option with strike price of $50 based on the geometric
average of the simulated prices of Problem 55.2 55 MONTE CARLO VALUATION OF ASIAN OPTIONS 377 Problem 55.8
You are given the following:
(i) The timet stock price, St , follows the lognormal distribution: ln (St /St−1 ) is normal with mean
(α − 0.5σ 2 )t and variance σ 2 t.
(ii) S0 = 50, α = 0.15, and σ = 0.30.
The following are three uniform (0, 1) random numbers:
0.9830, 0.0384, 0.7794.
Find the payoﬀ at expiration of an Asian put option with strike price of $70 based on the geometric
average of the simulated prices of Problem 55.2
Problem 55.9
You are given the following:
(i) The timet stock price, St , follows the lognormal distribution: ln (St /St−1 ) is normal with mean
(α − 0.5σ 2 )t and variance σ 2 t.
(ii) S0 = 50, α = 0.15, and σ = 0.30.
The following are three uniform (0, 1) random numbers:
0.9830, 0.0384, 0.7794.
Find the payoﬀ at expiration of a geometric average strike Asian call option based on the simulated
prices of Problem 55.2
Problem 55.10
You are given the following:
(i) The timet stock price, St , follows the lognormal distribution: ln (St /St−1 ) is normal with mean
(α − 0.5σ 2 )t and variance σ 2 t.
(ii) S0 = 50, α = 0.15, and σ = 0.30.
The following are three uniform (0, 1) random numbers:
0.9830, 0.0384, 0.7794.
Find the payoﬀ at expiration of a geometric average strike Asian put option based on the simulated
prices of Problem 55.2
Problem 55.11
The table below lists the arithemtic and geometric averages of simulated stock prices as well as the
stock price at expiration (i.e., end of one year). 378 OPTION PRICING VIA MONTE CARLO SIMULATION
Trial
1
2
3
4 Arithmetic Ave.
39.13
60.77
61.85
53.44 Geometric Ave.
37.42
59.68
61.35
53.43 Stock Price at Exp.
32.43
47.26
60.11
52.44 The continuously compounded riskfree rate is 10%.
(a) Find the Monte Carlo estimate of an arithmetic average strike of an Asian put option that
matures in one year.
(b) Find the Monte Carlo estimate of a geometric average strike of an Asian put option that matures
in one year. 56 CONTROL VARIATE METHOD 379 56 Control Variate Method
In this section, we discuss a method, the control variate method, that increases the accuracy of
Monte Carlo valuations.
A control variate in this context is a somewhat similar option to the option we are trying to price
but whose true value is known. We then obtain a simulated value of that option. The diﬀerence
between the true value of the control variate and its simulated value is then added to the simulated
value of the option we are trying to price. In this manner, the error in the control variate is added
to the simulated value of the option of interest. Let us see how this method works.
Let Cs be the simulated price of the option we are trying to price. Let Vt be the true value of another
similar option (the control variate) and Vs be its simulated value. Our control variate estimate is
then found as
C ∗ = Cs + (Vt − Vs ).
(56.1)
Since Monte Carlo valuation provides an unbiased estimate, we have E (Vs ) = Vt and E (Cs ) = Ct .
Thus, E (C ∗ ) = E (Cs ) = Ct . For the variance of C ∗ we have the following result from probability
theory
Var(C ∗ ) = Var(Cs − Vs ) = Var(Cs ) + Var(Vs ) − 2Cov(Cs , Vs ).
This will be less than Var(Cs ) if Var(Vs ) < 2Cov(Cs , Vs ), meaning that the control variate method
relies on the assumption of a large covariance between Cs and Vs . The control variate chosen should
be one that is very highly correlated with the option we are pricing.
The reduction in the variance is
Var(Cs ) − Var(C ∗ ).
Example 56.1
Consider the pricing of an arithmetic average strike Asian put. A control variate is a geometric
average strike Asian put since we have a formula for such an option.1 Suppose that for a given set
of random stock prices the following results were found
Trial
1
2
3
4 Arithmetic Geometric
Average
Average
39.13
37.42
60.77
59.68
61.85
61.35
53.44
53.43 Stock Price
at Exp.
32.43
47.26
60.11
52.44 Arithmetic Option
payoﬀ
6.70
13.51
1.74
1.00 Geometric Option
payoﬀ
4.99
12.42
1.24
0.99 Suppose that the continuously compounded riskfree rate is 10%. Find the control variate estimate
of the arithmetic average strike Asian put.
1 See Appendix 19.A of [1]. 380 OPTION PRICING VIA MONTE CARLO SIMULATION Solution.
The Monte Carlo estimate for the arithmetic average Asian put is
As = e−0.10
[6.70 + 13.51 + 1.74 + 1.00] = $5.19
4 The Monte Carlo estimate for the geometric average Asian put is
Gs = e−0.10
[4.99 + 12.42 + 1.24 + 0.99] = $4.44.
4 The true price of the geometric option is $2.02. Hence, the control variate estimate of the arithmetic
average strike option is
A∗ = As + (Gt − Gs ) = 5.19 + (2.02 − 4.44) = $2.77
A better estimate of the option price is to replace (56.1) with the equation
C ∗ = Cs + β (Vt − Vs ).
In this case, E (C ∗ ) = Ct and
Var(C ∗ ) =Var(Cs ) + β 2 Var(Vs ) − 2β Cov(Cs , Vs )
=Var(Cs ) − Cov2 (Cs , Vs )
Var(Vs ) Cov(Cs , Vs )
+ Var(Vs ) β −
Var(Vs ) The variance Var(C ∗ ) is minimized when
β= Cov(Cs , Vs )
.
Var(Vs ) The minimal variance is
Var(C ∗ ) = Var(Cs ) − Cov2 (Cs , Vs )
Var(Vs ) The variance reduction factor is
Var(C ∗ )
= 1 − ρ2 s ,Vs
C
Var(Cs )
where ρCs ,Vs is the correlation coeﬃcient between Cs and Vs . . 2 . 56 CONTROL VARIATE METHOD 381 Estimating Cov(X, Y )/Var(X )
Let {x1 , x2 , · · · , xn } be a set of observations drawn from the probability distribution of a random
variable X. We deﬁne the sample mean by
X= x1 + x2 + · · · + xn
n The sample variance estimation is given by
n
i=1 (xi − X )2
.
n−1 Var(X ) = If {y1 , y2 , · · · , yn } is a set of observations drawn from the probability distribution of a random
variable Y, then an estimation of the covariance is given by
n
i=1 (xi Cov(X, Y ) = − X )(yi − Y )
.
n−1 Thus, an estimation of β = Cov(X, Y )/Var(X ) is
β= Example 56.2
Show that β = n
i=1 xi yi −nX Y
2
n
2
i=1 xi −nX n
i=1 (xi − X )(yi −
n
2
i=1 (xi − X ) Y) . (56.2) . Soltution.
This follows from
n n
2 2 (x2 − 2Xxi + X )
i (xi − X ) =
i=1 i=1
n n x2
i =
i=1
n − 2X
i=1
2 i=1 2 x2 − 2nX + nX
i xi + nX = x2 − nX .
i = n
2 i=1 2 382 OPTION PRICING VIA MONTE CARLO SIMULATION and
n n (xi − X )(yi − Y ) =
i=1 n xi yi − Y
i=1
n n xi − X
i=1 y i + nX Y
i=1 xi yi − nX Y − nX Y + nX Y =
i=1
n xi yi − nX Y =
i=1 Example 56.3
Let C (K ) denote the BlackScholes price for a 3month Kstrike European call option on a nondividendpaying stock.
Let Cs (K ) denote the Monte Carlo price for a 3month Kstrike European call option on the stock,
calculated by using 5 random 3month stock prices simulated under the riskneutral probability
measure.
You are to estimate the price of a 3month 52strike European call option on the stock using the
formula
C ∗ (52) = Cs (52) + β [C (50) − Cs (50)],
where the coeﬃcient β is such that the variance of C ∗ (52) is minimized.
You are given: (i) The continuously compounded riskfree interest rate is 10%.
(ii) C (50) = $2.67.
(iii) Both Monte Carlo prices, C( 50) and Cs (52), are calculated using the following 5 random 3month stock prices:
43.30, 47.30, 52.00, 54.00, 58.90
(a) Based on the 5 simulated stock prices, estimate β.
(b) Compute C ∗ (52).
Solution.
(a) We know that
β= Cov(Cs , Vs )
.
Var(Vs ) We use the β of the previous example with xi and yi are simulated payoﬀs of the 50strike and
52strike calls respectively corresponding to the ith simulated price where 1 i ≤ 5. Note that the
50strike call is our control variate. We do not need to discount the payoﬀs because the eﬀect of
discounting is canceled in the formula of β. Thus, we have the following table: 56 CONTROL VARIATE METHOD 383
max{0, S0.25 − 50} max{0, S0.25 − 52}
0.00
0.00
0.00
0.00
2.00
0.00
4.00
2.00
8.90
6.90 Simulated S0.25
43.30
47.30
52.00
54.00
58.90
Hence, 2 + 4 + 8.90
= 2.98
5
2 + 6.90
Y=
= 1.78
5 X= 5 x2 = 22 + 42 + 8.902 = 99.21
i
i=1
5 xi yi = 4 × 2 + 8.90 × 6.90 = 69.41.
i=1 The estimate of β is
β= 69.41 − 5 × 2.98 × 1.78
= 0.78251.
99.21 − 5 × 2.982 (b) We have
Cs (50) = e−0.10×0.25
× (2 + 4 + 8.90) = $2.91
5 and
Cs (52) = e−0.10×0.25
× (2 + 6.90) = $1.736.
5 Hence,
C ∗ (52) = Cs (52) + β [C (50) − Cs (50)] = 1.736 + 0.78251(2.67 − 2.91) = $1.548 384 OPTION PRICING VIA MONTE CARLO SIMULATION Practice Problems
Problem 56.1
For the control variate method, you are given the following parameters: ρCs ,Vs = 0.8, Var(Cs ) = 3,
and Var(Vs ) = 5. Calculate
(a) The covariance of Cs and Vs .
(b) The variance of of the control variate estimate.
(c) The variance reduction.
Problem 56.2
For the control variate method with beta, you are given the following: Var(C ∗ ) = 0.875, Var(Cs ) = 2.
Find the variance reduction factor.
Problem 56.3
For the control variate method with beta, you are given the following: Var(C ∗ ) = 0.875, Var(Cs ) = 2.
Find the coeﬃcient of correlation between Cs and Vs .
Problem 56.4
For the control variate method with beta, you are given the following: Var(Cs ) = 2 and ρCs ,Vs = 0.75.
Claculate the variance of the control variate estimate.
Problem 56.5 ‡
Let C (K ) denote the BlackScholes price for a 3month Kstrike European call option on a nondividendpaying stock.
Let Cs (K ) denote the Monte Carlo price for a 3month Kstrike European call option on the stock,
calculated by using 5 random 3month stock prices simulated under the riskneutral probability
measure.
You are to estimate the price of a 3month 42strike European call option on the stock using the
formula
C ∗ (42) = Cs (42) + β [C (40) − Cs (40)],
where the coeﬃcient β is such that the variance of C ∗ (42) is minimized.
You are given:
(i) The continuously compounded riskfree interest rate is 8%.
(ii) C (40) = $2.7847.
(iii) Both Monte Carlo prices, Cs (40) and Cs (42), are calculated using the following 5 random 3month stock prices:
33.29, 37.30, 40.35, 43.65, 48.90 56 CONTROL VARIATE METHOD 385 (a) Based on the 5 simulated stock prices, estimate β.
(b) Based on the 5 simulated stock prices, compute C ∗ (42).
Problem 56.6
Let G denote the BlackScholes price for a 3month geometric average strike put option on a
nondividendpaying stock.
Let Gs denote the Monte Carlo price for a 3month geometric average strike put option on the
stock, calculated by using the following geoemtric average strike prices
37.423, 59.675, 61.353, 53.425
simulated under the riskneutral probability measure.
The corresponding simulated arithmetic average strike prices are
39.133, 60.767, 61.847, 53.440.
You are to estimate the price of a 3month arithmetic average strike put option on the stock using
the formula
A∗ = As + β [G − Gs ],
where the coeﬃcient β is such that the variance of A∗ is minimized.
You are given:
(i) The continuously compounded riskfree interest rate is 10%.
(ii) G = $2.02.
(iii) The 3month simulated stock prices are
32.43, 47.26, 70.11, 52.44
(a) Estimate β.
(b) Compute the control variate estimate A∗ . 386 OPTION PRICING VIA MONTE CARLO SIMULATION 57 Antithetic Variate Method and Stratiﬁed Sampling
A second variance reduction technique for improving the eﬃciency of Monte Carlo valuation is the
antithetic variate method. The fundamental idea behind the method is to bring in negative
correlation between two estimates.
Remember that in the standard Monte Carlo simulation, we are generating observations of a standard normal random variable. The standard normal random variable is distributed with a mean
of zero, a variance of 1.0, and is symmetric. Thus, for each value we draw, there is an equally
likely chance of having drawn the observed value times −1. Consequently, for each value of x we
draw, we can legitimately create an artiﬁcially observed companion observation of −x. This is the
antithetic variate. This procedure automatically doubles our sample size without having increased
the number of random drawings.
(1)
Let Z (1), Z (2), · · · , Z (n) be n standard normal numbers that simulate stock prices and let Cs be
the corresponding Monte Carlo estimate. Using, the standard normal numbers −Z (1), −Z (2), · · · , −Z (n),
we simulate an additional set of n stock prices and obtain a corresponding Monte Carlo estimate
(2)
Vs .
We next show that the variance of the antithetic estimate is reduced based on the fact that the
(1)
(2)
estimates Vs and Vs are negatively correlated:
Var(Vs(1) + Vs(2) ) =Var(Vs(1) ) + Var(Vs(2) ) + 2Cov(Vs(1) , Vs(2) )
=2Var(Vs(1) ) + 2Cov(Vs(1) , Vs(2) )
=2Var(Vs(1) )(1 + ρ)
(1) where ρ is the coeﬃcient of correlation between Vs
Var(Vs(1) + Vs(2) ) < (2) and Vs . Since ρ < 0 we conclude that 2Var(Vs(1) ) Var(V (1) )
=2
n where Var(V (1) ) is the variance of the sample {V (Z (1)), V (Z (2)), · · · , V (Z (n))}. Hence, for n ≥ 2,
(1)
(2)
(1)
we have Var(Vs + Vs ) < Var(Vs ).
Example 57.1
A stock is currently selling for $40. The continuously compounded riskfree rate is 0.08. The stock
pays no dividends. The stock price volatility is 0.35. Six months stock prices are simulated using
the standard normal numbers:
−0.52, 0.13, −0.25, −1.28.
Find the antithetic control estimate of a call option with strike price $40 and time to maturity of
six months. 57 ANTITHETIC VARIATE METHOD AND STRATIFIED SAMPLING 387 Solution.
The simulated prices are found by means of the following formula:
ST = S0 e(r−δ−0.5σ 2 )T +σ √ TZ . Thus, we obtain the following chart
Z
Z
S0.5 Option Payoﬀ
−0.52 35.50
0.00
0.13 41.70
1.70
−0.25 37.95
0.00
−1.28 29.41
0.00
0.52 45.92
5.92
−0.13 39.10
0.00
0.25 42.96
2.96
1.28 55.43
15.43 Hence, the antithetic variate estimate of the price of the option is
Cs = e−0.08×0.5
× (0 + 1.70 + 0 + 0 + 5.92 + 0 + 2.96 + 15.43) = $3.12
8 Stratiﬁed Sampling
A third variance reduction technique is the stratiﬁed sampling method. The idea behind this
method is to generate random numbers in such a way to improve the Monte Carlo estimate.
Given uniform random numbers U1 , U2 , · · · , Un on (0,1). Divide the interval (0, 1) into k ≤ n equal
subintervals. We can use the given uniform random numbers to generate uniform random in each
of the k intervals by using the formula
ˆ
Ui = i−1+Ui
k
(i−k)−1+Ui
k if i ≤ k
if i > k 1
12
ˆ
ˆ
Thus, U1 is a random number in the interval (0, k ), U2 belongs to the interval ( k , k ) and so on. More1
ˆˆ
ˆ
over, the numbers U1 , U1+k , U1+2k , · · · are uniformly distributed in the interval (0, k ), the numbers
12
ˆˆ
ˆ
U2 , U2+k , U2+2k , · · · are uniformly distributed in ( k , k ) and so on. Example 57.2
Consider the following 8 uniform random numbers in (0,1): 388 OPTION PRICING VIA MONTE CARLO SIMULATION
i
Ui 1
2
3
4
5
6
7
8
0.4880 0.7894 0.8628 0.4482 0.3172 0.8944 0.5013 0.3015 Use the given numbers, to generate uniform random numbers in each quartile, i.e. k = 4.
Solution.
We have
1 − 1 + U1
ˆ
= 0.122
U1 =
4
2 − 1 + U2
ˆ
U2 =
= 0.44735
4
3 − 1 + U3
ˆ
U3 =
= 0.7157
4
4 − 1 + U4
ˆ
U4 =
= 0.86205
4
(5 − 4) − 1 + U5
ˆ
= 0.0793
U5 =
4
(6 − 4) − 1 + U6
ˆ
U6 =
= 0.4736
4
(7 − 4) − 1 + U7
ˆ
= 0.625325
U7 =
4
(8 − 4) − 1 + U8
ˆ
U8 =
= 0.825375
4 57 ANTITHETIC VARIATE METHOD AND STRATIFIED SAMPLING 389 Practice Problems
Problem 57.1
A stock is currently selling for $40. The continuously compounded riskfree rate is 0.08. The stock
pays no dividends. The stock price volatility is 0.35. Six months stock prices are simulated using
the standard normal numbers:
−0.52, 0.13, −0.25, −1.28.
Find the antithetic control estimate of a put option with strike price $40 and time to maturity of
six months.
Problem 57.2
Consider the following 8 uniform random numbers in (0,1):
i
Ui 1
2
3
4
5
6
7
8
0.4880 0.7894 0.8628 0.4482 0.3172 0.8944 0.5013 0.3015 Find the diﬀerence between the largest and the smallest simulated uniform random variates generated by the stratiﬁed sampling method.
Problem 57.3
Consider the following 8 uniform random numbers in (0,1):
i
Ui 1
2
3
4
5
6
7
8
0.4880 0.7894 0.8628 0.4482 0.3172 0.8944 0.5013 0.3015 ˆ
Compute 8 standard normal random variates by Zi = N −1 (Ui ), where N −1 is the inverse of the
cumulative standard normal distribution function.
Problem 57.4 ‡
Michael uses the following method to simulate 8 standard normal random variates:
Step 1: Simulate 8 uniform (0, 1) random numbers U1 , U2 , · · · , U8 .
Step 2: Apply the stratiﬁed sampling method to the random numbers so that Ui and Ui+4 are
ˆ
ˆ
transformed to random numbers Ui and Ui+4 that are uniformly distributed over the interval ((i −
1)/4, i/4), i = 1, 2, 3, 4.
ˆ
Step 3: Compute 8 standard normal random variates by Zi = N −1 (Ui ), where N −1 is the inverse
of the cumulative standard normal distribution function.
Michael draws the following 8 uniform (0,1) random numbers:
i
Ui 1
2
3
4
5
6
7
8
0.4880 0.7894 0.8628 0.4482 0.3172 0.8944 0.5013 0.3015 390 OPTION PRICING VIA MONTE CARLO SIMULATION Find the diﬀerence between the largest and the smallest simulated normal random variates.
Problem 57.5
Given the following: S0 = 40, α = 0.08, δ = 0, σ = 0.30, and T = 1. Also given the following uniform
random numbers in (0, 1).
i
Ui 1
2
3
4
0.152 0.696 0.788 0.188 (a) Apply the stratiﬁed sampling method with k = 4 to ﬁnd the uniformly distributed numbers,
ˆˆˆˆ
U1 , U2 , U3 , U4 .
ˆ
(b) Compute thet 4 standard normal random numbers Zi = N −1 (Ui ), 1 ≤ i ≤ 4.
(c) Find the random set of lognormally distributed stock prices over the one year period.
Problem 57.6
Using the information of the previous problem, simulate the stock prices at the end of three months,
six months, nine months, and one year.
Problem 57.7
Using the simulated stock prices of the previous problem, ﬁnd the payoﬀ of an arithmetic average
price Asian call that expires in one year. Brownian Motion
Brownian motion is the basic building block for standard derivatives pricing models. For example,
the BlackScholes option pricing model assumes that the price of the underlying asset follows a
geometric Brownian motion. In this chapter we explain what this means. 391 392 BROWNIAN MOTION 58 Brownian Motion
A Brownian motion1 can be thought of as a random walk where a coin is ﬂipped inﬁnitely fast
and inﬁnitesimally small steps (backward or forward based on the ﬂip) are taken at each point.
That is, a Brownian motion is a random walk occurring in continuous time with movements that
are continuous rather than discrete. The motion is characterized by a family of random variables
Z = {Z (t)} indexed by time t where Z (t) represents the random walk− the cumulative sum of all
moves− after t periods. Such a family is called a stochastic process.
For a Brownian motion that starts from z, the process Z (t) has the following characteristics:
(1) Z (0) = z.
(2) Z (t + s) − Z (t) is normally distributed with mean 0 and variance s.
(3) Z (t + s1 ) − Z (t) is independent of Z (t) − Z (t − s2 ) where s1 , s2 > 0. That is, nonoverlapping
increments are independently distributed. In other words, Z (t + s1 ) − Z (t) and Z (t) − Z (t − s2 )
are independent random variables.
(4) Z (t) is a continuous function of time t. (Continuous means you can draw the motion without
lifting your pen from the paper.)
If z = 0 then the Brownian motion is called a standard or a pure Brownian motion. Also, a
standard Brownian motion is known as Wiener motion.
Example 58.1
What is the variance of Z (t) − Z (s)? Here, 0 ≤ s < t.
Solution.
We have Z (t) − Z (s) = Z (s + (t − s)) − Z (s) so that by (2), the variance is t − s
Example 58.2
Show that E [Z (t + s)Z (t)] = Z (t).
Solution.
For any random variables X, Y, and Z we know that E (X + Y Z ) = E (X Z ) + E (Y Z ) and
E (X X ) = X. Using these properties, We have
E [Z (t + s)Z (t)] =E [Z (t + s) − Z (t) + Z (t)Z (t)]
=E [Z (t + s) − Z (t)Z (t)] + E [Z (t)Z (t)]
=0 + Z (t) = Z (t)
where we used Property (2). The above result implies that Z (t) is a martingale2
1
2 Named after the Scottish botanist Robert Brown.
A stochastic process Z (t) for which E [Z (t + s)Z (t)] = Z (t) is called a martingale. 58 BROWNIAN MOTION 393 Example 58.3
Suppose that Z (2) = 4. Compute E [Z (5)Z (2)].
Solution.
We have E [Z (5)Z (2)] = E [Z (2 + 3)Z (2)] = Z (2) = 4
Example 58.4
Let Z be a standard Brownian motion. Show that E (Z (t)Z (s)) = min{t, s} where t, s ≥ 0.
Solution.
Assume t ≥ s. We have
E (Z (t)Z (s)) =E [(Z (s) + Z (t) − Z (s))Z (s))] = E [(Z (s))2 ] + E [(Z (t) − Z (s))Z (s)]
=s + E [Z (t) − Z (s)]E [Z (s)] = s + 0 = min{t, s}
since W (s) is normal with mean 0 and variance s and Z (t) − Z (s) and Z (s) are independent
We next show that a standard Brownian motion Z can be approximated by a sum of independent binomial random variables. By the continuity of Z, for a small time period h we can estimate
the change in Z from t to t + h by the equation
√
Z (t + h) − Z (t) = Y (t + h) h
where Y (t) is a random draw from a binomial distribution3 . Now, take the interval [0, T ]. Divide
[0, T ] into n equal subintervals each of length h = T . Then we have
n
n n [Z (ih) − Z ((i − 1)h)] = Z (T ) =
i=1 √
Y (ih) h i=1 √
1
=T√
n n Y (ih) (58.1) i=1 Since E (Y (ih)) = 0, we have
1
E√
n
3 n i=1 1
Y (ih) = √
n n E (Y (ih)) = 0.
i=1 Y (t) = ±1 with probability 0.5. Also, E (Y (t)) = 0 and Var(Y (t)) = 1. 394 BROWNIAN MOTION Likewise, since Var(Y (ih)) = 1 we have
1
Var √
n n i=1 1
Y (ih) =
n n 1 = 1.
i=1 Hence, we can think of a standard Brownian motion as being approximately generated from the
sum of independent binomial draws with mean 0 and variance h.
Now, by the Central Limit Theorem4 we have that the limit
1
lim √
n→∞
n n Y (ih)
i=1 approaches a standard normal distribution, say W. Hence,
√
Z (T ) = T W (T ).
It follows that Z (T ) is approximated by a normal random variable with mean 0 and variance T.
In Stochastic calculus, Z (t) can be represented in integral form
T Z (T ) = dZ (t).
0 The integral on the right is called a stochastic integral.
Example 58.5
Let Z be a standard Brownian motion. Calculate E (Z (4)Z (5)).
Solution.
Note that Z (4) = Z (4) − Z (0) and Z (5) − Z (4) are independent random variables. Thus,
E (Z (4)Z (5)) =E [Z (4){Z (5) − Z (4) + Z (4)}] = E [Z (4)2 ] + E [Z (4)(Z (5) − Z (4))]
=E [Z (4)2 ] + E [Z (4)]E [Z (5) − Z (4)]
=E [Z (4)2 ] − [E (Z (4))]2 + E [Z (4)]E [Z (5)]
=Var(Z (4)) = 4
Now, renaming h as dt and the change in Z as dZ (t) we can write the diﬀerential form5
√
dZ (t) = W (t) dt.
In words, this equation says, that over small periods of time, changes in the value of the process
are normally distributed with a variance that is proportional to the length of the time period.
4
5 See Theorem 40.2 in [3].
Diﬀerential forms of Brownian motions are called stochastic diﬀerential equations 58 BROWNIAN MOTION 395 Additional Properties of Z (t)
We next discuss two important properties of Brownian motion that prove to be extremely important.
We continue to use the binomial approximation to the Brownian process. We ﬁrst introduce the
following deﬁnition: Partition an interval [a, b] into n equal subintervals. The quadratic variation
of a stochastic process {Z (t)}a≤t≤b is deﬁned to be
n [Z (ti ) − Z (ti−1 )]2 lim n→∞ i=1 if the limit exists (under convergence in probability). In the case of a standard Brownian motion
as deﬁned above we have
n n n→∞ √
Yih h [Z (ih) − Z ((i − 1)h)]2 = lim lim n→∞ i=1 i=1
n
2
Yih h = T < ∞. = lim n→∞ 2 i=1 An important implication of the fact that the quadratic variation of a Brownian process is ﬁnite is
that higherorder variations are all zero.
Example 58.6
Let Z be a standard Brownian motion. Show that limn→∞ n
i=1 [Z (ih) Solution.
We have
n n
4 [Z (ih) − Z ((i − 1)h)] √
Yih h = i=1 i=1
n
4
Yih h2 =
i=1
n h2 ≤
i=1 T2
=2
n
Hence,
n [Z (ih) − Z ((i − 1)h)]4 = 0 lim n→∞ i=1 4 − Z ((i − 1)h)]4 = 0. 396 BROWNIAN MOTION The total variation of the standard Brownian process is
n Z (ti ) − Z (ti−1 ) = lim lim n→∞ n i=1 n→∞ i=1
n = lim n→∞ √
Yih  h
√ √
h = T lim n→∞ i=1 √
= T lim n→∞ n n i=1 1
√
n √
n=∞ i=1 This means that the Brownian path moves up and down very rapidly in the interval [0, T ]. That is,
the path will cross its starting point an inﬁnite number of times in the interval [0, T ]. 58 BROWNIAN MOTION 397 Practice Problems
Problem 58.1
Given that Z follows a Brownian motion. Find the variance of Z (17) − Z (5).
Problem 58.2
Let Z be a Brownian motion. Suppose that 0 ≤ s1 < t1 ≤ s2 < t2 . Given that E (Z (t1 ) − Z (s1 )) = 4
and E (Z (t2 ) − Z (s2 )) = 5. Find E [(Z (t1 ) − Z (s1 ))(Z (t2 ) − Z (s2 ))].
Problem 58.3
Let Z represent a Brownian motion. Show that for any constant α we have limt→0 Z (t + α) = Z (α).
Problem 58.4
Let Z represent a standard Brownian motion. Show that {Z (t + α) − Z (α)}t≥0 is also a standard
Brownian motion for a ﬁxed α > 0.
Problem 58.5
Let {Z (t)}0≤t≤1 represent a standard Brownian motion. Show that {Z (1) − Z (1 − t)}0≤t≤1 is also
a standard Brownian motion.
Problem 58.6
1
Let {Z (t)}t≥0 represent a standard Brownian motion. Show that {s− 2 Z (st)}t≥0 is a standard
Brownian motion, where s > 0 is ﬁxed.
Problem 58.7 ‡
Let Z be a standard Brownian motion. Show that limn→∞ n
i=1 [Z (ih) − Z ((i − 1)h)]3 = 0. 398 BROWNIAN MOTION 59 Arithmetic Brownian Motion
With a standard Brownian motion, dZ (t) has an mean of 0 and variance of 1 per unit time. We
can generalize this to allow nonzero mean and an arbitrary variance. Deﬁne X (t) by
√
X (t + h) − X (t) = αh + σY (t + h) h.
√
where Y (t) is a random draw from a binomial distribution. We call αh the drift term and σ h
the noise term. Let T > 0. Partition [0, T ] into n subintervals each of length h = T . Then
n
n X (T ) − X (0) = α
i=1 T
+ σY (ih)
n
n √
=αT + σ T
i=1 √
By the Central Limit Theorem T
variance T. Hence, we can write n
Y (ih)
√
i=1
n T
n Y (ih)
√
n approaches a normal distribution with mean 0 and X (T ) − X (0) = αT + σZ (t)
where Z is the standard Brownian motion. The stochastic diﬀerential form of this expression is
dX (t) = αdt + σdZ (t). (59.1) A stochastic process {X (t)}t≥0 that satisﬁes (59.1) is called an arithmetic Brownian motion.
Note that E (X (t) − X (0)) = αt and Var(X (T ) − X (0)) = Var(αt + σZ (t)) = σ 2 . We call α the
instantaneous mean per unit time or the drift factor and σ 2 the instantaneous variance
per unit time. Note that X (t) − X (0) is normally distributed with expected mean of αt and
variance σ 2 t. Hence, X (t) is normally distributed with mean X (0) + αt and variance α2 t.
Example 59.1
Let {X (t)}t≥0 be an arithmetic Brownian motion with drift factor α and volatility σ. Show that
√
X (t) = X (a) + α(t − a) + σ t − aW (t) where W is a standard normal random variable.
Solution.
√
Using (59.1) we can write X (t) − X (a) = α(t − a) + σdZ (t) = α(t − a) + σ t − aW (t). Thus, X (t)
is normally distributed with mean X (a) + α(t − a) and variance σ 2 (t − a) 59 ARITHMETIC BROWNIAN MOTION 399 Example 59.2
{X (t)}t≥0 follows an arithmetic Brownian motion such that X (30) = 2. The drift factor of this
Brownian motion is 0.435, and the volatility is 0.75. What is the probability that X (34) < 0?
Solution.
The mean of the normal distribution√ (34) is X (30) + α(34 − 30) = 2 + 0.435 × 4 = 3.74. The
X
√
standard deviation is σ t − a = 0.75 34 − 30 = 1.5. Thus,
P r(X (34) < 0) = P r Z < 0 − 3.74
1.5 = N (Z < −2.49) = 0.006387 Example 59.3
A particular arithmetic Brownian motion is as follows:
dX (t) = 0.4dt + 0.8dZ (t).
(a) What is the drift factor of this Brownian motion?
(b) What is the instantaneous variance per unit time?
Solution.
(a) The drift factor is the instantaneous mean per unit time α = 0.4.
(b) The instantaneous variance per unit time is σ 2 = 0.82 = 0.64
The following observations about equation (59.1) are worth mentioning:
• X (t) is normally distributed.
• The random term dZ (t) is multiplied by a scale factor that enables us to change the variance.
• The αdt term introduces a nonrandom drift into the process. Adding αdt has the eﬀect of
adding α per unit time to X (0).
Example 59.4
Let {X (t)}t≥0 be an arithmetic Brownian motion starting from 0 with α = 0.2 and σ 2 = 0.125.
Calculate the probability that X (2) is between 0.1 and 0.5.
Solution.
We have that X (2) is normally distributed with mean αT = 0.4 and variance σ 2 T = 0.125×2 = 0.25.
Thus,
P r(0.1 < X (2) < 0.5) = N 0.5 − 0.4
0.5 −N 0.1 − 0.4
0.5 = 0.57926 − 0.274253 = 0.305007 400 BROWNIAN MOTION The OrnesteinUhlenbeck Process
When modeling commodity prices, one notices that these prices exhibit a tendency to revert to the
mean. That is, if a value departs from the mean in either direction, in the long run, it will tend to
revert back to the mean. Thus, meanreversion model has more economic logic than the arithmetic
Brownian process described above. We can incorporate mean reversion by modifying the drift term
in (59.1) and obtain:
dX (t) = λ(α − X (t))dt + σdZ (t)
(59.2)
where α is the longrun equilibrium level (or the longrun mean value which X (t) tends to revert);
σ is the volatility factor, λ is the speed of reversion or the reversion factor, and Z (t) is the standard
Brownian motion. Equation (59.2) is known as the OrnsteinUhlenbeck process.
Example 59.5 ‡
Solve equation (59.2).
Solution.
We introduce the change of variables Y (t) = X (t) − α. In this case, we obtain the diﬀerential form
dY (t) = −λY (t)dt + σdZ (t)
or
dY (t) + λY (t)dt = σdZ (t).
This can be written as
d[eλt Y (t)] = eλt σdZ (t).
Hence, integrating from 0 to t we obtain
t eλs dZ (s) eλt Y (t) − Y (0) = σ
0 or t Y (t) = e−λt Y (0) + σ e−λ(t−s) dZ (s).
0 Writing the answer in terms of X we ﬁnd
t X (t) = X (0)e−λt + α(1 − e−λt ) + σ e−λ(t−s) dZ (s)
0 59 ARITHMETIC BROWNIAN MOTION 401 Practice Problems
Problem 59.1
Let {X (t)}t≥0 be an arithmetic Brownian motion with X (0) = 0. Show that the random variable
X (t)−αt
is normally distributed. Find the mean and the variance.
σ
Problem 59.2
Let {X (t)}t≥0 be an arithmetic Brownian motion with X (0) = 0. Show that the random variable
X (t)−αt
√
has the standard normal distribution.
σt
Problem 59.3
Let {X (t)}t≥0 be an arithmetic Brownian motion with X (0) = 0, α = 0.1 and σ 2 = 0.125. Calculate
the probability that X (2) is between 0.1 and 0.3.
Problem 59.4
Consider the OrnsteinUhlenbeck process:
dZ (t)
dX (t)
= −0.4dt + 0.20
.
X (t)
X (t)
Determine the parameters α, λ, and σ.
Problem 59.5
Consider the OrnsteinUhlenbeck process:
dX (t) = 0.048dt + 0.30dZ (t) − 0.40X (t)dt.
Determine the parameters α, λ, and σ.
Problem 59.6
Solve the equation
dX (t) = 0.4[0.12 − X (t)]dt + 0.30dZ (t)
subject to X (0) = 0.24.
Problem 59.7
The price of Stock R follows a mean reversion process, where the instantaneous mean per unit
time is 54, and the volatility factor is 0.46. The reversion factor for this process is 3.5. At some
time t, you know that dt = 0.24, dZ (t) = 0.94, and the stock price is $46 per share. What is the
instantaneous change in the stock price, i.e., dX (t)? 402 BROWNIAN MOTION Problem 59.8
The price of Stock S follows a mean reversion process, but the mean is currently unknown. You do
know that at time t, dt = 0.125, dZ (t) = 0.3125, and dX (t), the instantaneous change in the stock
price, is 0.634. Also, X (t), the stock price, is 613, the volatility is 0.53, and the reversion factor for
this process is 0.1531. What is the instantaneous mean per unit time for this process?
Problem 59.9
The price of Stock Q follows an OrnsteinUhlenbeck process. The reversion factor for this process
is 0.0906, and the volatility factor is 0.63. At some time t, dt = 1, dZ (t) = 0.9356, and dX (t), the
instantaneous change in the stock price, is 0.0215. What is the price of Stock Q at time t?
Problem 59.10
{X (t)}t≥0 follows an arithmetic Brownian motion with a drift factor of 0.35 and a volatility of 0.43.
Given that X (4) = 2.
(a) Find the mean of the normal distribution X (13).
(b) Find the standard deviation of the normal distribution X (13).
(c) What is the probability that X (13) > 9?
Problem 59.11
{X (t)}t≥0 follows an arithmetic Brownian motion such that X (45) = 41. The drift factor of this
Brownian motion is 0.153, and the volatility is 0.98. What is the probability that X (61) < 50? 60 GEOMETRIC BROWNIAN MOTION 403 60 Geometric Brownian Motion
The arithmetic Brownian motion has many drawbacks, namely,
• X (t) can be negative and therefore the arithmetic Brownian process is a poor model for stock
prices.
• The mean and variance of changes in dollar terms are independent of the level of the stock price.
In practice, if a stock price doubles, we would expect both the dollar expected return and the dollar
standard deviation of returns to approximately double.
These drawbacks and others can be eliminated with geometric Brownian motion which we discuss
next.
When the drift factor and the volatility in the arithmetic Brownian motion are functions of X (t),
then the stochastic diﬀerential form
dX (t) = α(X (t))dt + σ (X (t))dZ (t)
is called an Itˆ process. In particular, if α(X (t)) = αX (t) and σ (X (t)) = σX (t) then the previous
o
equation becomes
dX (t) = αX (t)dt + σX (t)dZ (t)
or
dX (t)
= αdt + σdZ (t).
X (t) (60.1) The process in the previous equation is known as geometric Brownian motion. Note that
equation (60.1) says that the the percentage change in the asset value is normally distributed with
instantaneous mean α and instantaneous variance σ 2 .
Example 60.1
A given geometric Brownian motion can be expressed as follows:
dX (t)
= 0.215dt + 0.342dZ (t).
X (t)
(a) What is the instantaneous mean of the percentage change in the asset value?
(b) What is the instantaneous variance of the percentage change in the asset value?
Solution.
(a) The instantaneous mean of the percentage change in the asset value is α = 0.215.
(b) The instantaneous variance of the percentage change in the asset value is σ 2 = 0.3422 = 404 BROWNIAN MOTION 0.116964
For an arbitrary initial value X (0) equation (60.1) has the analytic solution1
X (t) = X (0)e(α−0.5σ 2 )t+σ √ tZ where Z is a normal random variable with parameters 0 and 1. Alternatively, we can write
X (t) = X (0)e(α−0.5σ 2 )t+σZ (t) where Z is a normal random variable with parameters 0 and t. Note that X (t) is an exponential and
therefore is not normal. However, X (t) is lognormally distributed with mean E (X (t)) = X (0)eαt
2
(See Example 48.2) and variance Var(X (t)) = e2αt X (0)2 (eσ t − 1). See Section 47. It follows that
ln (X (t)) is normally distributed with mean ln (X (0)) + (α − 0.5σ 2 )t and variance σ 2 t. That is,
ln X (t) = ln X (0) + (α − 0.5σ 2 )t + σZ (t).
It follows that when X follows a geometric Brownian motion its logarithm follows an arithmetic
Brownian motion2 , that is,
d[ln X (t)] = (α − 0.5σ 2 )dt + σdZ (t).
Example 60.2
The current price of a stock is 100. The stock price follows a geometric Brownian motion with drift
rate of 10% per year and variance rate of 9% per year. Calculate the probability that two years
from now the price of the stock will exceed 200.
Solution.
We want
P r(S (2) > 200) = P r ln
But ln S (2)
S (0)
2 S (2)
S (0) > ln 2 . is normally distributed with mean (α − 0.5σ 2 )t = (0.10 − 0.5(0.09) × 2 = 0.11 and variance σ t = 0.09 × 2 = 0.18. Therefore,
ln 2 − 0.11
√
0.18 P r(S (2) > 200) = P r Z > = 1 − N (1.37) = 1 − 0.915 = 0.085 Example 60.3
1
Given a geometric Brownian motion with drift factor 0.10. For h = 365 , suppose that the ratio of
the noise term to the drift term is found to be 22.926. Find the standard deviation σ.
Solution.
We are given that
1
2 √
σh
αh = 22.926 or σ
√
0.10 1
365 = 22.926. Solving for σ we ﬁnd σ = 0.12 The correctness of the solution can be veriﬁed using Itˆ’s lemma to be discussed in Section 64.
o
This follows from Itˆ’s lemma.
o 60 GEOMETRIC BROWNIAN MOTION 405 Practice Problems
Problem 60.1
Stock A follows a geometric Brownian motion where the drift factor is 0.93 and the variance factor
is 0.55. At some particular time t, it is known that dt = 0.035, and dZ (t) = 0.43. At time t, the
stock trades for $2354 per share. What is the instantaneous change in the price of stock A?
Problem 60.2
Stock B follows a geometric Brownian motion where the drift factor is 0.0234 and the variance
factor is 0.953. At some particular time t, it is known that dt = 0.531, dZ (t) = 0.136, and dX (t)−
the instantaneous change in the price of the stock− is 0.245. What is the stock price at time t?
Problem 60.3
Stock C follows a geometric Brownian motion where the variance factor is 0.35. At some particular
time t, it is known that dt = 0.0143, dZ (t) = 0.0154, and dX (t)− the instantaneous change in the
price of the stock− is 0.353153. The price of Stock C at time t is $31.23. What is the drift factor
for this Brownian motion?
Problem 60.4
1
Given a geometric Brownian motion with drift factor 0.09. For h = 365 , suppose that the ratio of
the noise term to the drift term is found to be 13.428. Find the standard deviation σ.
Problem 60.5
{X (t)}t≥0 follows a geometric Brownian motion with a drift factor of 0.35 and a volatility of 0.43.
We know that X (0) = 2.
(a) What is the mean of the normal distribution ln X (13) ?
X (0)
(b) What is the standard deviation of the normal distribution ln
(c) Find P r(X (13) > 9). X (13)
X (0) ? Problem 60.6 ‡
At time t = 0, Jane invests the amount of W (0) in a mutual fund. The mutual fund employs a
proportional investment strategy: There is a ﬁxed real number φ, such that, at every point of time,
100φ% of the funds assets are invested in a nondividend paying stock and 100(1 − φ)% in a riskfree
asset.
You are given:
(i) The continuously compounded rate of return on the riskfree asset is r.
(ii) The price of the stock, S (t), follows a geometric Brownian motion,
dS (t)
= αdt + σdZ (t), t ≥ 0
S (t) 406 BROWNIAN MOTION where {Z (t)} is a standard Brownian motion.
Let W (t) denote the Janet’s fund value at time t, t ≥ 0. Show that
S (t)
W (t) = W (0)
S (0) φ
2 e(1−φ)(r+0.5φσ )t . Problem 60.7 ‡
Consider the BlackScholes framework. You are given the following three statements on variances,
conditional on knowing S (t), the stock price at time t. Determine the ones that are true.
(i) Var[ln S (t + h)S (t)] = σ 2 h
S
(ii) Var dS ((tt)) S (t) = σ 2 dt
(iii) Var[S (t + dt)S (t)] = S 2 σ 2 dt. 61 ITO PROCESS MULTIPLICATION RULES 407 61 Ito Process Multiplication Rules
In many problems involving Itˆ processes and geometric Brownian motion, you will encounter
o
situations where you need to multiply dZ by itself, dt by itself, or dZ by dt. Rules for these products
are known as multiplication rules. But ﬁrst we deﬁne the following: For α > 1 we will deﬁne
(dt)α = 0. This, makes senses since dt represents a very small number.
Example 61.1
Show that (dt)2 = 0.
Solution.
This follows from our deﬁnition of (dt)α = 0 with α = 2
Example 61.2
(a) Show that E [dZ × dt] = 0.
(b) Show that E [(dZ × dt)2 ] = 0.
(c) Show that Var[dZ × dt] = 0.
(d) Show that dZ × dt = 0.
Solution.
(a) We have E [dZ × dt] = dtE [Z (t)] = 0 since dt is a constant.
(b) We have E [(dZ × dt)2 ] = (dt)2 E [(dZ )2 ] = (dt)2 Var(dZ ) = (dt)2 (dt) = (dt)3 = 0.
(c) We have Var[dZ × dt] = E [(dZ × dt)2 ] − (E [dZ × dt])2 = 0 − 0 = 0.
(d) This follows from (a) and (c)
Example 61.3
(a) Show that E [(dZ )2 ] = dt.
(b) Show that Var[(dZ )2 ] = 0.
(c) Show that (dZ )2 = dt.
Solution.
(a) We know that E [dZ ] = 0 and Var(dZ ) = dt. Thus, E [(dZ )2 ] = E [(dZ )2 ] − [E (dZ )]2 = Var(dZ ) =
dt.
(b) We have
Var[(dZ )2 ] =E [(dZ )4 ] − E [(dZ )2 ]
=E [Y 4 (t)(dt)2 ] − (dt)2
=(dt)2 − (dt)2 = 0. 2 (c) It follows from (b) that (dZ )2 is a constant. Thus, using (a) we ﬁnd that (dZ )2 = dt 408 BROWNIAN MOTION Example 61.4
Let Z and Z be two standard Brownian motions. Show that dZ × dZ = ρdt where ρ = E [Y (t)Y (t)].
ρ is also known as the correlation of the underlying assets driven by the diﬀerent Brownian motions.
Solution.
√
√
We know that dZ (t) = Y (t) dt and dZ (t) = Y (t) dt. Thus, E (dZ × dZ ) = dtE [Y (t)Y (t)] = ρdt.
We also have,
2 Var(dZ × dZ ) =E [(dZ × dZ )2 ] − (E [dZ × dZ ])
notag =(dt)2 E [(Y Y )2 ] − ρ2 (dt)2 = 0. (61.1) Hence, dZ × dZ must be a constant. Since E (dZ × dZ ) = ρdt we conclude that dZ × dZ = ρdt .
Summarizing, the multiplication rules are
(dt)2 =0
dZ × dt =0
(dZ )2 =dt
dZ × dZ =ρdt.
Example 61.5
A geometric Brownian motion is described by the stochastic form
dX (t) = 0.07X (t)dt + 0.03X (t)dZ (t).
Calculate (dX (t))2 .
Solution.
We have
(dX (t))2 =[0.07X (t)dt + 0.03X (t)dZ (t)]2
=(0.07X (t))2 (dt)2 + 2(0.07)(0.03)dtdZ (t) + (0.03X (t))2 (dZ (t))2
=0 + 0 + 0.0009X 2 (t)dt
Example 61.6
Let X be a Brownian motion characterized by dX (t) = (α − δ − 0.5σ 2 )dt + σdZ (t). For > 0, split
the interval [0, T ] into n equal subintervals each of length h = T . Evaluate
n
n [X (ih) − X ((i − 1)h)]2 . lim n→∞ i=1 61 ITO PROCESS MULTIPLICATION RULES 409 Solution.
We have (See Problem 61.5)
n
n→∞ T [dX (t)]2 [X (ih) − X ((i − 1)h)]2 = lim 0 i=1 T [(α − δ − 0.5σ 2 )2 (dt)2 + 2σ (α − δ − 0.5σ 2 )dtdZ (t) + σ 2 (dZ (t))2 ] =
0
T σ 2 dt = σ 2 T =
0 410 BROWNIAN MOTION Practice Problems
Problem 61.1
Let X be an arithmetic Brownian motion with drift factor α and volatility σ. Find formulas for
(dX )2 and dX × dt.
Problem 61.2
Find the simplest possible equivalent expression to (34dZ + 45dt)(42dZ + 3dt).
Problem 61.3
Given that the assets have a correlation ρ = 0.365. Find the simplest possible equivalent expression
to (95dZ + 424dZ )(2dZ + 241dt).
Problem 61.4
Let Z be a standard Brownian motion. Show that the stochastic integral T
[dZ (t)]4
0 is equal to 0. Problem 61.5
Let Z be a standard Brownian motion and T > 0. Partition the interval [0, T ] into n equal subintervals each of length h = T . For any positive integer k we deﬁne
n
n
n→∞ T [dZ (t)]k . [Z (ih) − Z ((i − 1)h)]k = lim 0 i=1 Using the multiplication rules, simplify
n i=1 n→∞ n
2 [Z (ih) − Z ((i − 1)h)] + lim X = lim n→∞ n [Z (ih) − Z ((i − 1)h)]3 . [Z (ih) − Z ((i − 1)h)] + lim
i=1 n→∞ i=1 Problem 61.6
Find the mean and the variance of the random variable of the previous problems.
Problem 61.7
Given the following Brownian motion
dX (t) = (0.05 − X (t))dt + 0.10dZ (t).
Find (dX (t))2 . 61 ITO PROCESS MULTIPLICATION RULES 411 Problem 61.8 ‡
Consider the BlackScholes framework. Let S (t) be the stock price at time t, t ≥ 0. Deﬁne X (t) =
ln [S (t)].
You are given the following three statements concerning X (t).
(i) {X (t), t ≥ 0} is an arithmetic Brownian motion.
(ii) Var[X (t + h) − X (t)] = σ 2 h, t ≥ 0, h > 0.
(iii) limn→∞ n=1 [X (ih) − X ((i − 1)h)]2 = σ 2 T.
i
Which of these statements are true?
Problem 61.9 ‡
Deﬁne
(i) W (t) = t2 .
(ii) X (t) = [t], where [t] is the greatest integer part of t; for example, [3.14] = 3, [9.99] = 9, and
[4] = 4.
(iii) Y (t) = 2t + 0.9Z (t), where {Z (t) : t ≥ 0} is a standard Brownian motion.
Let VT2 (U ) denote the quadratic variation of a process U over the time interval [0, T ].
Rank the quadratic variations of W, X and Y over the time interval [0, 2.4]. 412 BROWNIAN MOTION 62 Sharpe ratios of Assets that Follow Geometric Brownian
Motions
We have ﬁrst encountered the concept of Sharpe ratio in Section 32: In ﬁnance, the risk premium
on an asset is deﬁned to be the excess return of the asset over the riskfree rate. Thus, if r denotes
the riskfree rate and α the expected return on the asset, then the risk premium on the asset is the
diﬀerence
Risk premium = α − r.
The Sharpe ratio of an asset is the risk premium of a stock divided by its volatility, or, standard
deviation. It is given by
α−r
.
Sharpe ratio = φ =
σ
The Sharpe ratio is used to characterize how well the return of an asset compensates the investor
for the risk taken.
Consider two nondividend paying stocks that are perfectly correlated. That is, their stock prices
are driven by the same dZ (t). In this case, we can write
dS1 = α1 S1 dt + σ1 S1 dZ
dS2 = α2 S2 dt + σ2 S2 dZ
It is important to remember that a stock price is assumed to follow a geometric Brownian motion.
We claim that φ1 = φ2 . Suppose not. Without loss of generality, we can assume φ1 > φ2 . Buy σ11S1
1
1
shares of asset 1 (at a cost of σ1 ) and sell σ21S2 shares of asset 2 (at a cost of σ2 ). Invest the cost
1
1
− σ1 in the riskfree bond, which has the rate of return rdt. The return on this portfolio (i.e. the
σ2
arbitrage proﬁt) after a short time dt is
dS1
dS2
−
+
σ1 S1 σ2 S2 1
1
−
σ2 σ1 α1 S1 dt + σ1 S1 dZ α2 S2 dt + σ2 S2 dZ
−
+
σ1 S1
σ2 S2
=(φ1 − φ2 )dt. rdt = 1
1
−
σ2 σ1 rdt Thus, we have constructed a zeroinvestment portfolio with a positive riskfree return. Therefore,
to preclude arbitrage, the two stocks must have equal Sharpe ratio.
Example 62.1
A nondividend paying stock follows a geometric Brownian motion given by
dS (t) = 0.10Sdt + 0.12Sdt.
Find the riskfree interest rate r is the Sharpe ratio of the stock is 0.10. 62 SHARPE RATIOS OF ASSETS THAT FOLLOW GEOMETRIC BROWNIAN MOTIONS413
Solution.
We are given α = 0.10, σ = 0.12, and φ = 0.10. Hence, 0.10 =
r = 8.8% 0.10−r
.
0.12 Solving this equation we ﬁnd Example 62.2
Nondividendpaying stocks S1 and S2 are perfectly correlated and follow these geometric Brownian
motions:
dS1
= Adt + 0.884dZ
S1
dS2
= 0.3567dt + 0.643dZ.
S2
The annual continuously compounded riskfree interest rate is 0.1. What is A?
Solution.
We know that α2 − r
α1 − r
=
,
σ1
σ2
where α1 = A, r = 0.1, σ1 = 0.884, α2 = 0.3567, and σ2 = 0.643. Thus,
A − 0.1
0.3567 − 0.1
=
.
0.884
0.643
Solving this equation we ﬁnd A = 0.453
Example 62.3
Nondividendpaying stocks S1 and S2 follow these geometric Brownian motions:
dS1
= 0.08dt + 0.23dZ
S1
dS2
= 0.07dt + 0.25dZ.
S2
The annual continuously compounded riskfree interest rate is 0.04. Demonstrate an arbitrage
opportunity.
Solution.
The Sharpe ratio of the ﬁrst stock is φ1 = 0.08.−0.04 = 0.174. For the second stock, we have φ2 =
0 23
0.07−0.04
1
= 0.12. Since φ1 > φ2 , an arbitrage opportunity occurs by buying 0.23S1 shares of stock 1,
0.25
1
selling 0.25S2 of stock 2, and borrowing 0.1 − 0.1 . The arbitrage proﬁt is
23
25
(φ1 − φ2 )dt = (0.174 − 0.12)dt = 0.054dt 414 BROWNIAN MOTION Example 62.4 ‡
Consider two nondividendpaying assets X and Y. There is a single source of uncertainty which
is captured by a standard Brownian motion {Z (t)}. The prices of the assets satisfy the stochastic
diﬀerential equations
dX (t)
= 0.07dt + 0.12dZ (t)
X (t)
and dY (t)
= Adt + BdZ (t),
Y (t) where A and B are constants. You are also given:
(i) d[ln Y (t)] = µdt + 0.085dZ (t);
(ii) The continuously compounded riskfree interest rate is 4%.
Determine A.
Solution.
Since Y follows a geometric Brownian motion, we have d[ln Y (t)] = (A − 0.5B 2 )dt + BdZ (t) =
µdt + 0.085dZ (t). Hence, A − 0.5B 2 = µ and B = 0.085. Since X and Y have the same underlying
source of risk Z , they have equal Sharpe ratios. That is,
αY − r
αX − r
=
.
σX
σY
or A − 0.04
0.07 − 0.04
=
.
0.12
0.085
Solving this equation for A we ﬁnd A = 0.06125
Sharpe Ratio Under Capital Asset Pricing Model
In CAPM model, the riskpremium of an asset is deﬁned to be
αS − r = βSM (αM − r)
with
βSM = ρSM σS
σM where ρSM is the correlation of asset with the market. It follows from this deﬁnition that
φ= αS − r
= ρSM φM .
σS 62 SHARPE RATIOS OF ASSETS THAT FOLLOW GEOMETRIC BROWNIAN MOTIONS415
Example 62.5
A nondividendpaying Stock A has annual price volatility of 0.20. Its correlation with the market is
0.75. A nondividendpaying Stock B has a market correlation equals to 0.4 and Sharpe ratio equals
0.0625. Find the riskpremium of Stock A.
Solution.
We are given that ρBM = 0.4 and φB = 0.0625 so that 0.0625 = 0.4 ×
0.15625. Thus, the Sharpe ratio of Stock A is αM − r
.
σM φA = ρAM φM = 0.75 × 0.15625 = 0.1171875
and the riskpremium of Stock A is
αA − r = σA × φA = 0.20 × 0.1171875 = 0.0234375 Hence, ρM = αM −r
σM = 416 BROWNIAN MOTION Practice Problems
Problem 62.1
A nondividend paying stock follows a geometric Brownian motion given by
dS (t) = 0.08Sdt + 0.12Sdt.
Find the riskfree interest rate r if the Sharpe ratio of the stock is 0.31.
Problem 62.2
A nondividendpaying stock follows a geometric Brownian motion such that dS = 0.5dt + 0.354dZ.
S
The annual continuously compounded riskfree interest rate is 0.034. What is the Sharpe ratio of
the stock?
Problem 62.3
Nondividendpaying stocks S1 and S2 are perfectly correlated and follow these geometric Brownian
motions:
dS1
= 0.453dt + 0.884dZ
S1
dS2
= 0.3567dt + AdZ.
S2
The annual continuously compounded riskfree interest rate is 0.1. What is A?
Problem 62.4
Nondividendpaying stocks S1 and S2 are perfectly correlated and follow these geometric Brownian
motions:
dS1
= 0.152dt + 0.251dZ
S1
dS2
= 0.2dt + 0.351dZ.
S2
What is the annual continuously compounded riskfree interest rate?
Problem 62.5
Nondividendpaying stocks S1 and S2 follow these geometric Brownian motions:
dS1
= 0.07dt + 0.12dZ
S1
dS2
= 0.05dt + 0.11dZ.
S2
The annual continuously compounded riskfree interest rate is 0.03. Demonstrate an arbitrage
opportunity. 62 SHARPE RATIOS OF ASSETS THAT FOLLOW GEOMETRIC BROWNIAN MOTIONS417
Problem 62.6
A nondividendpaying Stock A has annual price volatility of 0.28 and Sharpe ratio equals to 0.15.
The beta of the stock with the market is 0.7. Find the market riskpremium.
Problem 62.7
A nondividendpaying Stock A has annual price volatility of 0.28 and Sharpe ratio equals to 0.15.
The beta of the stock with the market is 0.7. A nondividendpaying stock has price volatility of
0.555 and beta 1.3. Find the Sharpe ratio of Stock B.
Problem 62.8 ‡
Consider an arbitragefree securities market model, in which the riskfree interest rate is constant.
There are two nondividendpaying stocks whose price processes are
S1 (t) = S1 (0)e0.1t+0.2Z (t)
S2 (t) = S2 (0)e0.125t+0.3Z (t)
where Z (t) is a standard Brownian motion and t ≥ 0.
Determine the continuously compounded riskfree interest rate.
Problem 62.9 ‡
Consider two nondividendpaying assets X and Y, whose prices are driven by the same Brownian
motion Z. You are given that the assets X and Y satisfy the stochastic diﬀerential equations:
dX (t)
= 0.07dt + 0.12dZ (t)
X (t)
and dY (t)
= Gdt + HdZ (t),
Y (t) where G and H are constants.
You are also given:
(i) d[ln Y (t)] = 0.06dt + σdZ (t);
(ii) The continuously compounded riskfree interest rate is 4%.
(iii) σ < 0.25.
Determine G.
Problem 62.10 ‡
The prices of two nondividendpaying stocks are governed by the following stochastic diﬀerential
equations:
dS1 (t)
= 0.06dt + 0.02dZ (t)
S1 (t) 418 BROWNIAN MOTION
dS2 (t)
= 0.03dt + kdZ (t)
S2 (t) where Z (t) is a standard Brownian motion and k is a constant.
The current stock prices are S1 (0) = 100 and S2 (0) = 50. The continuously compounded riskfree
interest rate is 4%.
You now want to construct a zeroinvestment, riskfree portfolio with the two stocks and riskfree
bonds.
If there is exactly one share of Stock 1 in the portfolio, determine the number of shares of Stock 2
that you are now to buy. (A negative number means shorting Stock 2.) 63 THE RISKNEUTRAL MEASURE AND GIRSANOV’S THEOREM 419 63 The RiskNeutral Measure and Girsanov’s Theorem
Consider a stock that pays dividend at the compounded yield δ. The true price process that is
observed in the world is given by
dS (t)
= (α − δ )dt + σdZ (t)
S (t) (63.1) where dZ (t) is the unexpected portion of the stock return and Z (t) is a martingale under the true
probability distribution.
In riskneutral pricing, we need a riskneutral version of the above process. The random part of
˜
the process will involve a Brownian motion Z that is martingale under a transformed probability
distribution, let’s call it Q. This transformed probability distribution is referred to as the riskneutral measure.
˜
˜
To ﬁnd Z and the associated Q, we let Z (t) = Z (t)+ α−r t. A result, known as Girsanov’s theorem,
σ
˜
asserts the existence of a unique riskneutral measure Q under which Z (t) is a standard Brownian
˜ is martingale under Q. Diﬀerentiating Z and rearranging yields
˜
motion and Z
α−r
˜
dt.
dZ (t) = dZ (t) −
σ
Put this back in equation (63.1) we obtain
dS (t)
˜
= (r − δ )dt + σdZ (t).
S (t)
We refer to this equation as the riskneutral price process. Notice that the volatility is the same
for the true price process and the riskneutral price process.
Example 63.1
The true price process of a stock that pays dividends at the continuously compounded yields δ is
dS (t)
= 0.09dt + σdZ (t).
S (t)
The corresponding riskneutral price process is
dS (t)
˜
= 0.05dt + σdZ (t).
S (t)
The continuously compounded riskfree interest rate is 0.08. Find δ and α. 420 BROWNIAN MOTION Solution.
We know that r − δ = 0.05 or 0.08 − δ = 0.05. Solving for δ, we ﬁnd δ = 0.03. On the other hand,
we know that α − δ = 0.09 or α − 0.03 = 0.09. Thus, α = 0.12
Example 63.2
The riskneutral price process of a dividend paying stock is given by
dS (t)
= 0.05dt + 0.12dZ (t).
S (t)
It is known that the riskpremium of the stock is 0.14. Find the drift factor of the true price process
model.
Solution.
We have α − r − (α − δ ) = 0.05 and α − r = 0.14. Thus, the drift factor is α − δ = 0.14 − 0.05 = 0.09
Example 63.3
The riskneutral price process for a nondividendpaying stock is
dS (t)
˜
= 0.07dt + 0.30dZ (t).
S (t)
The expected rate of return on the stock is 0.13. Find the Sharpe ratio of the stock.
Solution.
We have r − δ = r − 0 = 0.07 so that r = 0.07. The price volatility of the stock is σ = 0.30. Thus,
the Sharpe ratio of the stock is
α−r
0.13 − 0.07
=
= 0.20
σ
0.30 63 THE RISKNEUTRAL MEASURE AND GIRSANOV’S THEOREM 421 Practice Problems
Problem 63.1
Consider a stock that pays dividends. The volatility of the stock price is 0.30. The Sharpe ratio of
the stock is 0.12. Find the true price model.
Problem 63.2
The riskneutral price process of a dividend paying stock is given by
dS (t)
= 0.05dt + 0.12dZ (t).
S (t)
It is known that the drift factor in the true price process is 0.09. Find the riskpremium of the
stock.
Problem 63.3
The riskneutral price model of a dividend paying stock is given by
dS (t)
= 0.05dt + 0.12dZ (t).
S (t)
The continuously compounded yield on the stock is 0.03. Find the continuously compounded riskfree interest rate.
Problem 63.4
The true price process of a stock that pays dividends at the continuously compounded yields δ is
dS (t)
= 0.13dt + σdZ (t).
S (t)
The corresponding riskneutral price process is
dS (t)
˜
= 0.08dt + σdZ (t).
S (t)
The expected rate of return on the stock is 0.15. Find δ and r.
Problem 63.5
The riskneutral price process of dividend paying stock is
dS (t)
˜
= 0.04dt + 0.12dZ (t).
S (t)
The continuously compounded yield is 0.03. It is known that the expected rate of return on the
stock is twice the riskfree interest rate. Find α. 422 BROWNIAN MOTION Problem 63.6 ‡
Assume the BlackScholes framework.
You are given:
(i) S (t) is the price of a stock at time t.
(ii) The stock pays dividends continuously at a rate proportional to its price. The dividend yield is
1%.
(iii) The stockprice process is given by
dS (t)
= 0.05dt + 0.25dZ (t)
S (t)
where {Z (t)} is a standard Brownian motion under the true probability measure.
(iv) Under the riskneutral probability measure, the mean of Z (0.5) is −0.03.
Calculate the continuously compounded riskfree interest rate. ˆ
64 SINGLE VARIATE ITO’S LEMMA 423 64 Single Variate Itˆ’s Lemma
o
In this section we discuss Itˆ’s Lemma which is essential in the derivation of BlackScholes Equation.
o
Lemma 64.1 (Itˆ’s)
o
Assume that the process S satisﬁes the following stochastic diﬀerential equation
dS (t) = µ(S, t)dt + σ (S, t)dZ (t)
where Z is the standard Brownian motion. Then for any twice continuously diﬀerentiable function
f (S, t), the change in f is given by
df (S, t) = ∂f
∂f
12
∂ 2f
+ µ(S, t)
+ σ (S, t) 2
∂t
∂S 2
∂S dt + σ (S, t) ∂f
dZ (t).
∂S Proof.
Using Taylor expansion we ﬁnd
df (S, t) =f (S + dS, t + dt) − f (S, t)
∂f
1 ∂ 2f
∂f
dt +
(dS )2
= dS +
2
∂S
∂t
2 ∂S
3
1 ∂ 2f
∂ 2f
+
(dt)2 +
dSdt + terms in(dt) 2 and higher
2
2 ∂t
∂S∂t
Now, using the multiplication rules we have:
(dt)α = 0, α > 1.
dt × dZ = 0
(dZ )2 = dt
(dS )2 = σ 2 (S, t)dt
dS × dt = 0.
Hence,
∂f
1 ∂ 2f 2
∂f
(µ(S, t)dt + σ (S, t)dZ (t)) +
dt +
σ (S, t)dt
∂S
∂t
2 ∂S 2
∂f
1
∂ 2f
∂f
∂f
=
+ µ(S, t)
+ σ 2 (S, t) 2 dt + σ (S, t) dZ (t)
∂t
∂S 2
∂S
∂S df (S, t) = 424 BROWNIAN MOTION Example 64.1
Suppose that S (t) follows a geometric Brownian motion:
dS (t) = αS (t)dt + σS (t)dZ (t).
Find d[sin S (t)].
Solution.
Let f (S, t) = sin S (t). Then
∂f
=0
∂t
∂f
= cos S (t)
∂S
∂ 2f
= − sin S (t).
∂S 2
Using Itˆ’s lemma we can write
o
∂f
1 ∂ 2f
∂f
dt +
(dS (t))2 +
dS (t)
2
∂t
2 ∂S
∂S
1
= − sin S (t)(dS )2 + cos S (t)dS (t)
3
1
= − sin S (t)σ 2 S 2 (t)dt + cos S (t)[αS (t)dt + σS (t)dZ (t)]
3
1
= αS (t) cos S (t) − σ 2 S 2 (t) sin S (t) dt + σS (t) cos S (t)dZ (t)
2 d[sin S (t)] = Example 64.2
Suppose that S (t) follows a geometric Brownian motion:
dS (t) = (α − δ )S (t)dt + σS (t)dZ (t).
Let f be a twice diﬀerentiable function of S and t. Find df (S, t).
Solution.
We have µ(S, t) = (α − δ )S (t) and σ (S, t) = σS (t). Hence,
df (S, t) = ∂f
1
∂ 2f
∂f
+ (α − δ )S (t)
+ σ 2 S 2 (t) 2
∂t
∂S 2
∂S dt + σS (t) ∂f
dZ (t)
∂S ˆ
64 SINGLE VARIATE ITO’S LEMMA 425 Example 64.3
Suppose that S (t) satisﬁes the geometric Brownian motion
dS = (α − δ )S (t)dt + σS (t)dZ (t).
Show that
d[ln S (t)] = (α − δ − 0.5σ 2 )dt + σdZ (t).
Solution.
Let f (S, t) = ln S (t). Then
∂f
=0
∂t
∂f 1
=
∂S S
∂ 2f
1
=− 2
2
∂S
S
Now, the result follows from Itˆ lemma. Using stochastic integration we ﬁnd
o
S (t) = S (0)e(α−δ−0.5σ 2 )t+σZ (t) Example 64.4 ‡
The price of a stock is a function of time. You are given:
2
• The expression for a lognormal stock price is S (t) = S (0)e(α−δ−0.5σ )t+σZ (t) .
• The stock price is a function of the Brownian process Z.
• Itˆ’s lemma is used to characterize the behavior of the stock as a function of Z (t).
o
Find an expression for dS.
Solution.
We let f (Z, t) = S (t). Thus,
∂f ∂S (t)
=
= (α − δ − 0.5σ 2 )S (t)
∂t
∂t
∂f
=σS (t)
∂Z
∂ 2f
=σ 2 Z (t).
∂Z 2 426 BROWNIAN MOTION Using Itˆ’s lemma we can write
o
∂f
1 ∂ 2f
∂f
dt +
dZ (t)
(dZ (t))2 +
2
∂t
2 ∂Z
∂Z
1
=(α − δ − 0.5σ 2 )S (t)dt + σ 2 S (t)dt + σS (t)dZ (t)
2
=(α − δ )S (t)dt + σS (t)dZ (t) dS (t) = ˆ
64 SINGLE VARIATE ITO’S LEMMA 427 Practice Problems
Problem 64.1
Assume that S (t) follows an arithmetic Brownian motion: dS (t) = αdt + σdZ (t). Use Itˆ’s Lemma
o
2
to ﬁnd d(2S (t)).
Problem 64.2
Assume that S (t) follows a meanreverting process: dS (t) = λ(α − S (t))dt + σdZ (t). Use Itˆ’s
o
Lemma to ﬁnd d(2S 2 (t)).
Problem 64.3
Assume that S (t) follows a geometric Brownian process:
ﬁnd d(2S 2 (t)). dS (t)
S (t) = αdt + σdZ (t). Use Itˆ’s Lemma to
o Problem 64.4
Assume that S (t) follows an arithmetic Brownian motion: dS (t) = αdt + σdZ (t). Use Itˆ’s Lemma
o
5
to ﬁnd d(35S (t) + 2t).
Problem 64.5 ‡
You are given the following information:
(i) S (t) is the value of one British pound in U.S. dollars at time t.
(ii) dS (t) = 0.1S (t)dt + 0.4S (t)dZ (t).
(iii) The continuously compounded riskfree interest rate in the U.S. is r = 0.08.
(iv) The continuously compounded riskfree interest rate in Great Britain is r∗ = 0.10.
∗
(v) G(t) = S (t)e(r−r )(T −t) is the forward price in U.S. dollars per British pound, and T is the
maturity time of the currency forward contract.
Based on Itˆ’s Lemma, ﬁnd an expression for dG(t).
o
Problem 64.6 ‡
X (t) is an OrnsteinUhlenbeck process deﬁned by
dX (t) = 2[4 − X (t)]dt + 8dZ (t),
where Z (t) is a standard Brownian motion.
Let
Y (t) = 1
.
X (t) You are given that
dY (t) = α(Y (t))dt + β (Y (t))dZ (t)
for some functions α(y ) and β (y ).
Determine α(0.5). 428 BROWNIAN MOTION Problem 64.7 ‡
Let {Z (t)} be a standard Brownian motion. You are given:
(i) U (t) = 2Z (t) − 2
(ii) V (t) = [Z (t)]2 − t
(iii)
t W (t) = t2 Z (t) − 2 sZ (s)ds.
0 Which of the processes deﬁned above has / have zero drift?
Problem 64.8 ‡
The stochastic process {R(t)} is given by
t es−t R(t) = R(0)e−t + 0.05(1 − e−t ) + 0.1 R(s)dZ (s). 0 where {Z (t)} is a standard Brownian motion. Deﬁne X (t) = [R(t)]2 .
Find dX (t).
Problem 64.9 ‡
The price of a stock is governed by the stochastic diﬀerential equation:
dS (t)
= 0.03dt + 0.2dZ (t)
S (t)
where {Z (t)} is a standard Brownian motion. Consider the geometric average
1 G = [S (1)S (2)S (3)] 3 .
Find the variance of ln [G].
Problem 64.10 ‡
Let x(t) be the dollar/euro exchange rate at time t. That is, at time t, e1 = $x(t).
Let the constant r be the dollardenominated continuously compounded riskfree interest rate. Let
the constant re be the eurodenominated continuously compounded riskfree interest rate.
You are given
dx(t)
= (r − re )dt + σdZ (t),
x(t)
where {Z (t)} is a standard Brownian motion and σ is a constant.
1
Let y (t) be the euro/dollar exchange rate at time t. Thus, y (t) = x(t) . Show that
dy (t)
= (re − r + σ 2 )dt − σdZ (t).
y (t) 65 VALUING A CLAIM ON S A 429 65 Valuing a Claim on S a
In this section, we want to compute the price of a claim whose payoﬀ depends on the stock price
raised to some power.
Suppose that a stock with an expected instantaneous return of α, dividend yield of δ, and instantaneous velocity σ follows a geometric Brownian motion given by
dS (t) = (α − δ )S (t)dt + σS (t)dZ (t).
Consider a claim with payoﬀ S (T )a at time T. Let’s examine the process followed by S a . Let
f (S, t) = S (t)a . Then
∂f
=0
∂t
∂f
=aS (t)a−1
∂S
∂ 2f
=a(a − 1)S (t)a−2 .
∂S 2
Using Itˆ’s lemma we can write
o
∂f
1 ∂ 2f
∂f
dt +
(dS (t))2 +
dS (t)
∂t
2 ∂S 2
∂S
1
= a(a − 1)S (t)a−2 (dS (t))2 + aS (t)a−1 dS (t)
2
1
= a(a − 1)S (t)a−2 [(α − δ )S (t)dt + σS (t)dZ (t)]2 + aS (t)a−1 [(α − δ )S (t)dt + σS (t)dZ (t)]
2
1
= a(a − 1)S (t)a−2 σ 2 dt + aS (t)a−1 [(α − δ )S (t)dt + σS (t)dZ (t)]
2
1
=[aS (t)a (α − δ ) + a(a − 1)S (t)a σ 2 ]dt + aS (t)a σdZ (t)
2 d[S (t)a ] = Hence,
d[S (t)a ]
1
= [a(α − δ ) + a(a − 1)σ 2 ]dt + aσdZ (t).
a
S (t)
2
1
Thus, S a follows a geometric Brownian motion with drift factor a(α − δ ) + 2 a(a − 1)σ 2 and risk
aσdZ (t). Now we expect the claim to be perfectly correlated with the stock price S which means
that the stock and the claim have equal Sharpe ratio. But the Sharpe ratio of the stock is (α − r)/σ.
Since the volatility of the claim is aσ we expect the claim riskpremium to be a(α − r). Hence, the
expected return of the claim is a(α − r) + r.
The following result gives us the forward and prepaid forward prices for the claim. 430 BROWNIAN MOTION Theorem 65.1
The prepaid forward price of a claim paying S (T )a at time T is
p
F0,T [S (T )a ] = e−rT S (0)a e[a(r−δ)+0.5a(a−1)σ 2 ]T . The forward price for this claim is
F0,T [S (T )a ] = S (0)a e[a(r−δ)+0.5a(a−1)σ 2 ]T . Proof.
The riskneutral price process of the claim is given by
d[S (t)a ]
1
˜
= [a(r − δ ) + a(a − 1)σ 2 ]dt + aσdZ (t).
a
S (t)
2
The expected value of the claim at time T under the riskneutral measure, which we denote by E ∗
is
2
E [S (T )a ] = S (0)a e[a(r−δ)+0.5a(a−1)σ ]T .
By Example 15.2, the expected price under the riskneutral measure is just the forward price. That
is, the forward price of the claim is
F0,T [S (T )a ] = E [S (T )a ] = S (0)a e[a(r−δ)+0.5a(a−1)σ 2 ]T . Discounting this expression at the riskfree rate gives us the prepaid forward price
p
F0,T [S (T )a ] = e−rT S (0)a e[a(r−δ)+0.5a(a−1)σ 2 ]T Example 65.1
Examine the cases of a = 0, 1, 2.
Solution.
• If a = 0 the claim does not depend on the stock price so it is a bond. With a = 0 the prepaid
forward price is
2
p
F0,T [S (T )0 ] = e−rT S (0)0 e[0(r−δ)+0.50(0−1)σ ]T = e−rT
which is the price of a bond that pays $1 at time T.
• If a = 1 then the prepaid forward price is
p
F0,T [S (T )] = e−rT S (0)e[(r−δ)+0.5(1−1)σ 2 ]T = S (0)e−δT which is the prepaid forward price on a stock.
• If a = 2 then the prepaid forward price is
p
F0,T [S (T )2 ] = e−rT S (0)2 e[2(r−δ)+0.5(2)(2−1)σ 2 ]T = S (0)2 e(r−2δ+σ 2 )T 65 VALUING A CLAIM ON S A 431 Example 65.2
Assume the BlackScholes framework. For t ≥ 0, let S (t) be the time−t price of a nondividendpaying stock. You are given:
(i) S (0) = 0.5
(ii) The stock price process is
dS (t)
= 0.05dt + 0.2dZ (t)
S (t)
where Z (t) is a standard Brownian motion.
(iii) E [S (1)a ] = 1.4, where a is a negative constant.
(iv) The continuously compounded riskfree interest rate is 3%.
Determine a.
Solution.
Since
S (T ) = S (0)e(α−δ−0.5σ 2 )T +σZ (T ) we can write
S (T )a = S (0)a ea(α−δ−0.5σ 2 )T +aσZ (T ) Thus,
E [S (T )a ] = S (0)a e[a(α−δ)+0.5a(a−1)σ 2 ]T . From (iii), we have
2 1.4 = 0.5a ea(0.05−0)+0.5a(a−1)0.2 .
Taking the natural logarithm of both sides to obtain
ln 1.4 = a ln 0.5 + 0.05a + 0.5a(a − 1)(0.04).
This leads to the quadratic equation
0.02a2 + (ln(0.5) + 0.03)a − ln 1.4 = 0.
Solving this equation for a we ﬁnd a = −0.49985 or a = 33.65721. Since a < 0 we discard the
positive value 432 BROWNIAN MOTION Practice Problems
Problem 65.1
Assume that the BlackScholes framework holds. Let S (t) be the price of a nondividendpaying
stock at time t ≥ 0. The stock’s volatility is 18%, and the expected return on the stock is 8%. Find
an expression for the instantaneous change d[S (t)a ], where a = 0.5.
Problem 65.2
Assume that the BlackScholes framework holds. Let S (t) be the price of a stock at time t ≥ 0. The
stock’s volatility is 20%, the expected return on the stock is 9%, and the continuous compounded
yield is 2%. Find an expression for the instantaneous change d[S (t)a ], where a = 0.5.
Problem 65.3
Assume that the BlackScholes framework holds. Let S (t) be the price of a stock at time t ≥ 0. The
stock’s volatility is 20%, the expected return on the stock is 9%, and the continuous compounded
yield is 2%. Suppose that S (1) = 60. Find E [d[S (1)0.5 ]].
Problem 65.4 ‡
Assume the BlackScholes framework. For t ≥ 0, let S (t) be the time−t price of a nondividendpaying stock. You are given:
(i) S (0) = 0.5
(ii) The stock price process is
dS (t)
= 0.05dt + 0.2dZ (t)
S (t)
where Z (t) is a standard Brownian motion.
(iii) E [S (1)a ] = 1.4, where a is a negative constant.
(iv) The continuously compounded riskfree interest rate is 3%.
Consider a contingent claim that pays S (1)a at time 1. Calculate the time0 price of the contingent
claim.
Problem 65.5 ‡
Assume that the BlackScholes framework holds. Let S (t) be the price of a nondividendpaying
stock at time t ≥ 0. The stock’s volatility is 20%, and the continuously compounded riskfree
interest rate is 4%.
You are interested in contingent claims with payoﬀ being the stock price raised to some power. For
0 ≤ t < T, consider the equation
P
Ft,T [S (T )x ] = S (T )x
where the lefthand side is the prepaid forward price at time t of a contingent claim that pays S (T )x
at time T. A solution for the equation is x = 1.
Determine another x that solves the equation. 65 VALUING A CLAIM ON S A 433 Problem 65.6 ‡
Assume the BlackScholes framework. Consider a derivative security of a stock.
You are given:
(i) The continuously compounded riskfree interest rate is 0.04.
(ii) The volatility of the stock is σ.
(iii) The stock does not pay dividends.
(iv) The derivative security also does not pay dividends.
(v) S (t) denotes the timet price of the stock.
k
(iv) The timet price of the derivative security is [S (t)]− σ2 , where k is a positive constant.
Find k.
Problem 65.7 ‡
Assume the BlackScholes framework.
Let S (t) be the timet price of a stock that pays dividends continuously at a rate proportional to
its price.
You are given:
(i)
dS (t)
˜
= µdt + 0.4dZ (t),
S (t)
˜
where {Z (t)} is a standard Brownian motion under the riskneutral probability measure;
(ii) for 0 ≤ t ≤ T, the timet forward price for a forward contract that delivers the square of the
stock price at time T is
Ft,T (S 2 ) = S 2 (t)e0.18(T −t) .
Calculate µ.
Problem 65.8 ‡
Assume the BlackScholes framework. For a stock that pays dividends continuously at a rate
proportional to its price, you are given:
(i) The current stock price is 5.
(ii) The stock’s volatility is 0.2. (iii) The continuously compounded expected rate of stockprice
appreciation is 5%.
Consider a 2year arithmetic average strike option. The strike price is
1
A(2) = [S (1) + S (2)].
2
Calculate Var[A(2)]. 434 BROWNIAN MOTION The BlackScholes Partial Diﬀerential
Equation
In the derivation of the option pricing, Black and Scholes assumed that the stock price follows
a geometric Brownian motion and used Itˆ’s lemma to describe the behavior of the option price.
o
Their analysis yields a partial diﬀerential equation, which the correct option pricing formula must
satisfy. In this chapter, we derive and examine this partial diﬀerential equation. 435 436 THE BLACKSCHOLES PARTIAL DIFFERENTIAL EQUATION 66 Diﬀerential Equations for Riskless Assets
In this section we will describe the pricing of a bond or a stock under certainty, i.e., the owner of
the asset receives the riskfree return. It turns out that the pricing formula is the solution of a
boundary value diﬀerential equation. That is, the diﬀerential equation describing the change in the
price together with a boundary condition determine the price of either the bond or the stock at any
point in time.
Let S (t) the value of a riskfree zerocoupon bond that pays $1 at time T. Then over a small time
interval h, the price of the bond changes by rhS (t). That is,
S (t + h) − S (t) = rhS (t)
or S (t + h) − S (t)
= rS (t).
h
Letting h → 0 we obtain the diﬀerential equation
dS (t)
= rS (t).
dt
Example 66.1
Solve the diﬀerential equation
dS (t)
= rS (t).
dt
subject to the condition S (T ) = $1.
Solution.
Separating the variables we ﬁnd
dS (t)
= rdt.
S (t)
Integrate both sides from t to T to obtain
T
t dS
=r
S T dt.
t Thus,
ln (S (T )/S (t)) = r(T − t)
or
S (t) = S (T )e−r(T −t) . 66 DIFFERENTIAL EQUATIONS FOR RISKLESS ASSETS 437 Using the boundary condition, S (T ) = 1 we ﬁnd that the price of the bond is given by the
formula
S (t) = e−r(T −t) .
This solution conﬁrms what we already know: The price of a risk free zero coupon bond at time t
is the discount factor back from the time when the bond matures to time t
Now, let S (t) be the value of a stock that pays continuous dividend at the rate of δ. Keep in
mind that we are pricing the stock under certainty so that the random nature of future stock prices
is ignored and the stock receives only the riskfree rate. Then for a small time period h, the current
price of the stock is equal to the dividends paid plus the future stock price, discounted back for
interest:
S (t) = [δhS (t) + S (t + h)](1 + rh)−1 .
This can be rewritten in the form
S (t + h) − S (t) + δhS (t) = rhS (t).
This equation says that the stock return is the change in the stock price plus dividend paid. The
previous equation can be rewritten in the form
S (t + h) − S (t)
= (r − δ )S (t).
h
Letting h → 0 we ﬁnd
dS (t)
= (r δ )S (t).
dt
Using the method of separation of variables, the general solution to this equation is
S (t) = Ae−(r−δ)(T −t) .
If S0 is the stock price at time t = 0 then we have the initial boundary condition S (0) = S0 . In
this case, A = S0 e(r−δ)T . Hence,
S (t) = S0 e(r−δ)t .
The price of a risk free stock at time t is the initial price accumulated at a rate of r − δ. The price
of a risk free stock at time t is its forward price.
Now, if the stock pays continuous dividends in the amount D(t), then we have the equation
S (t) = [S (t + h) + hD(t + h)](1 + rh)−1 438 THE BLACKSCHOLES PARTIAL DIFFERENTIAL EQUATION which can be rewritten as
S (t + h) − S (t) + hD(t + h) = rhS (t)
or S (t + h) − S (t)
+ D(t + h) = rS (t).
h
Letting h → 0 we ﬁnd the diﬀerential equation
dS (t)
+ D(t) = rS (t).
dt
˜
If we inpose the terminal boundary condition S (T ) = S, then by solving the intial value problem
by the method of integrating factor1 we ﬁnd the solution
T ˜
D(s)e−r(s−t) ds + Se−r(T −t) . S (t) =
t Example 66.2
A stock pays continuous dividends in the amount of D(t) = $8. The continuously compounded
riskfree interest rate is 0.10. Suppose that S (3) = $40. Determine the price of the stock two years
from now. Assume that the future random nature of the stock price is ignored.
Solution.
We have
3 8e−0.10(s−2) ds + 40e−0.10(3−2) S (2) =
2 3 e−0.10s ds + 40e−0.10 =8e0.20
2
0.20 =8e 1 See Practice Problems 66.166.3. −10e−0.10s 3
2 + 40e−0.10 = $43.81 66 DIFFERENTIAL EQUATIONS FOR RISKLESS ASSETS Practice Problems
Problem 66.1
Consider the diﬀerential equation
dS (t)
− rS (t) = −D(t).
dt
Show that this equation is equivalent to
d −rt
e S (t) = −e−rt D(t).
dt
Problem 66.2
Find the general solution of the diﬀerential equation
dS (t)
− rS (t) = −D(t).
dt
Problem 66.3
Find the solution to the boundary value problem
dS (t)
− rS (t) = −D(t), 0 ≤ t ≤ T
dt
˜
with S (T ) = S.
Problem 66.4
The price S (t) of a bond that matures in T ≥ 4 years satisﬁes the diﬀerential equation
dS (t)
= 0.05S (t)
dt
subject to the boundary condition S (T ) = $100. Find T if S (2) = $70.50.
Problem 66.5
The price S (t) of a bond that matures in 3 years satisﬁes the diﬀerential equation
dS (t)
= rS (t)
dt
subject to the boundary condition S (3) = $100. Find r if S (1) = $81.87.
Problem 66.6
Find the diﬀerential equation satisﬁed by the expression
T ˜
δS (s)e−0.10(s−t) ds + Se−0.10(T −t) . S (t) ==
t 439 440 THE BLACKSCHOLES PARTIAL DIFFERENTIAL EQUATION 67 Derivation of the BlackScholes PDE
There are several assumptions involved in the derivation of the BlackScholes equation. It is important to understand these properly so as to see the limitations of the theory. These assumptions
are summarised below :
1. The eﬃcient market hypothesis is assumed to be satisﬁed. In other words, the markets are
assumed to be liquid, have pricecontinuity, be fair and provide all players with equal access to
available information. This implies that zero transaction costs are assumed in the BlackScholes
analysis.
2. It is assumed that the underlying asset is perfectly divisible and that short selling with full use
of proceeds is possible.
3. Constant riskfree interest rates are assumed. In other words, we assume that there exists a
riskfree security which returns $1 at time T when e−r(T −t) is invested at time t.
4. As we will see later, the BlackScholes analysis requires continuous trading. This is, of course,
not possible in practice as the more frequently one trades, the larger the transaction costs.
5. The prinicple of no arbitrage is assumed to hold.
6. The price of the underlying asset is assumed to follow a geometric Brownian process of the form
dS (t) = αS (t)dt + σS (t)dZ (t) where Z (t) is the risk term or the stochastic term.
It should be obvious that none of these principles can be perfectly satisﬁed. Transaction costs exist
in all markets, all securities come in discrete units, short selling with full use of proceeds is very
rare, interest rates vary with time and we will later see that there is evidence that the price of most
stocks do not precisely follow a geometric Brownian process.
Let V = V [S, t] be the value of a call or a put option written on an underlying asset with value
S (t) at time t. Then according to Itˆ’s lemma, V changes over the inﬁnitesimal time interval dt
o
according to
1
∂ 2V
∂V
∂V
∂V
+ σ 2 S 2 2 + αS
dt + σS
dZ (t).
dV =
∂t
2
∂S
∂S
∂S
Let us now assume that we have one option with value V and ∆ shares of the underlying asset
where ∆ as yet undetermined, with ∆ > 0 for shares held long and ∆ < 0 for shares held short.
The value of this portfolio at any time t is
π = V [S, t] + ∆S.
Over the time dt, the gain in the value of the portfolio is
dπ = dV + ∆dS.
or
dπ = ∂V
1
∂ 2V
∂V
+ σ 2 S 2 2 + αS
∂t
2
∂S
∂S dt + σS ∂V
dZ (t) + ∆(αSdt + σSdZ (t)).
∂S 67 DERIVATION OF THE BLACKSCHOLES PDE 441 If we choose ∆ = − ∂V then the stochastic terms cancel so that the gain is deterministic which
∂S
means the gain cannot be more or less than the gain in the value of the portfolio were it invested
at the riskfree interest rate r. That is, we must have
dπ = rπdt = r V − ∂V
S dt.
∂S Now, equating the two expressions of dπ we ﬁnd
∂ 2V
∂V
1
∂V
+ σ 2 S 2 2 + rS
− rV = 0.
∂t
2
∂S
∂S
This is the famous BlackScholes equation for the value of an option.
For a dividend paying asset the corresponding BlackScholes formula is given by
1
∂V
∂ 2V
∂V
+ σ 2 S 2 2 + (r − δ )S
− rV = 0.
∂t
2
∂S
∂S
The BlackScholes equation applies to any derivative asset. What distinguishes the diﬀerent assets
are the boundary conditions. For a zerocoupon bond that matures at time T and pays $1 at
maturity, the boundary condition is that it must be worth $1 at time T. For a prepaid forward
contract on a share of stock, the boundary condition is that the prepaid forward contract is worth a
share at maturity. For a European call option the boundary condition is max{0, S (T ) − K } whereas
for a European put the boundary condition is max{0, K − S (T )}. A derivation of the solution of
the BlackScholes for European options was discussed in Section 50.
Binary Options
A binary option is a type of option where the payoﬀ is either some ﬁxed amount of some asset or
nothing at all. The two main types of binary options are the cashornothing binary option and
the assetornothing binary option.1
Example 67.1
A cashornothing binary option pays some ﬁxed amount of cash if the option expires inthemoney.
Suppose you buy a cashornothing option on a stock with strike price $100 and payoﬀ of $350.
What is the payoﬀ of the option if the stock price at expiration is $200? What if the stock price is
$90?
Solution.
If the stock price at expiration is $200 then the payoﬀ of the option is $350. If its stock is trading
below $100, nothing is received
1 These options are also called allornothing options or digital options. 442 THE BLACKSCHOLES PARTIAL DIFFERENTIAL EQUATION Example 67.2
A cashornothing call option pays out one unit of cash if the spot is above the strike at maturity.
Its BlackSholes value now is given by
V [S (t), t] = e−r(T −t) N (d2 )
where
d2 = ln (S (t)/K ) + (r − δ − 0.5σ 2 )(T − t)
√
.
σ T −t Show that V satisﬁes the BlackScholes partial diﬀerential equation.
Solution.
For this problem, we refer the reader to Section 29 and Section 30 where expressions for
and ∂N (d2 ) were established. We have
∂S
Vt =rV + e−r(T −t) 1 S − d2 (r−δ)(T −t)
1
√
e 2e
2π K ln (S/K )
2σ (T − t) 3
2 − ∂N (d1 ) ∂N (d2 ) ∂N (d1 )
, ∂t , ∂S ,
∂t r − δ − 0.5σ 2 r − δ − 0.5σ 2
√
√
+
σ T −t
2σ T − t d2
1
1 e(r−δ)(T −t)
1
VS =e−r(T −t) √ e− 2 ·
·√
K σ T −t
2π
d2
1 e(r−δ)(T −t)
1
−r(T −t) 1
√ e− 2 ·
·
VSS = − d1 e
K Sσ 2 (T − t)
2π Thus,
1
Vt + σ 2 S 2 VSS + (r − δ )SVS − rV = 0
2
Example 67.3
Suppose that V = ert ln S is a solution to BlackScholes equation. Given that r = 0.08 and σ = 0.30,
ﬁnd δ.
Solution.
We have
V
Vt
VS
VSS =ert ln S
=rert ln S
=ert S −1
= − ert S −2 67 DERIVATION OF THE BLACKSCHOLES PDE
Substituting into the diﬀerential equation we ﬁnd
1
rert ln S + (0.30)2 S 2 (−ert S −2 ) + (0.08 − δ )S (ert S −1 ) − rert ln S = 0.
2
Reducing this expression we ﬁnd
−0.5(0.3)2 ert + (0.08 − δ )ert = 0
or
−0.5(0.30)2 + 0.08 − δ = 0.
Solving this equation we ﬁnd δ = 3.5% 443 444 THE BLACKSCHOLES PARTIAL DIFFERENTIAL EQUATION Practice Problems
Problem 67.1
The price of a zerocoupon bond that pays $1 at time T is given by V (t, T ) = e−r(T −t) . Show that
V satisﬁes the BlackScholes diﬀerential equation with boundary condition V (T, T ) = $1.
Problem 67.2
The price of a prepaid forward contract on a share of stock is given by V [S (t), t] = S (t)e−δ(T −t) . Show
that V satisﬁes the BlackScholes diﬀerential equation with boundary condition V [S (T ), T ] = S (T ).
Problem 67.3
2r
The underlying asset pays no dividends. Show that the value function V [S (t), t] = ert S 1− σ2 satisﬁes
the BlackScholes partial diﬀerential equation.
Problem 67.4
Suppose you were interested in buying binary call options for common shares of ABC company
with a strike price of $50 per share and a speciﬁed binary payoﬀ of $500. What would you receive
if the stock is trading above $50 when the expiration date is reached? What if the stock is trading
below $50 per share at the expiration date.
Problem 67.5
A cashornothing put option pays out one unit of cash if the spot is below the strike at maturity.
Its BlackSholes value now is given by
V [S (t), t] = e−r(T −t) N (−d2 ) = e−r(T −t) (1 − N (d2 ))
where ln (S (t)/K ) + (r − δ − 0.5σ 2 )(T − t)
√
.
σ T −t
Show that V satisﬁes the BlackScholes partial diﬀerential equation.
d2 = Problem 67.6
An assetornothing call option pays out one unit of asset if the spot is above the strike at maturity.
Its BlackSholes value now is given by
V [S (t), t] = Se−δ(T −t) N (d1 )
where ln (S (t)/K ) + (r − δ + 0.5σ 2 )(T − t)
√
d1 =
.
σ T −t
Show that V satisﬁes the BlackScholes partial diﬀerential equation. 67 DERIVATION OF THE BLACKSCHOLES PDE 445 Problem 67.7
An assetornothing put option pays out one unit of asset if the spot is below the strike at maturity.
Its BlackSholes value now is given by
V [S (t), t] = Se−δ(T −t) N (−d1 ) = Se−δ(T −t) (1 − N (d1 ))
where ln (S (t)/K ) + (r − δ + 0.5σ 2 )(T − t)
√
.
σ T −t
Show that V satisﬁes the BlackScholes partial diﬀerential equation.
d1 = Problem 67.8
When a derivative claim makes a payout D(t) at time t, the BlackScholes equation takes the form
1
Vt + σ 2 S 2 VSS + (α − δ )SVS + D(t) − rV = 0.
2
You are given the following: r = 0.05, δ = 0.02, D(t) = 0.04V (t), V (t) = e0.01t ln S. Determine the
value of σ.
Problem 67.9 ‡
Assume the BlackScholes framework. Consider a stock and a derivative security on the stock.
You are given:
(i) The continuously compounded riskfree interest rate, r, is 5.5%.
(ii) The timet price of the stock is S (t).
(iii) The timet price of the derivative security is ert ln [S (t)].
(iv) The stock’s volatility is 30%.
(v) The stock pays dividends continuously at a rate proportional to its price.
(vi) The derivative security does not pay dividends.
Calculate δ, the dividend yield on the stock. 446 THE BLACKSCHOLES PARTIAL DIFFERENTIAL EQUATION 68 The BlackScholes PDE and Equilibrium Returns
In this section we derive the BlackScholes partial diﬀerential equation based on the condition that
the expected return on the option must be equal the equilibrium expected return on the underlying
asset.
Let V = V [S (t), t] denote the value of an option with a stock as underlying asset. Using Itˆ’s lemma
o
we can write
1 ∂ 2V
∂V
∂V
dt +
dS.
dV =
(dS )2 +
∂t
2 ∂S 2
∂S
If the stock price follows a geometric Brownian motion then
dS = (α − δ )Sdt + σSdZ (t).
Thus, using the mutliplication rules, we can write
dV = 122
σ S VSS + (α − δ )SVS + Vt dt + SVS σdZ (t).
2 The term involving dZ is the random element or the unexpected return on the option. In contrast,
the term involving dt is deterministic or the expected return on the option.
We deﬁne the instantaneous expected return on the option by the expression
αoption = 0.5σ 2 S 2 VSS + (α − δ )SVS + Vt
.
V The unexpected return on a stock is given by σSdZ (t). Likewise, we deﬁne the unexpected return
on an option by σoption V dZ (t). But from the equation above, the unexpected return is SVS σdZ (t).
Hence, σoption V dZ (t) = SVS σdZ (t) which implies
σoption = SVS
σ.
V Since elasticity is deﬁned as the percentage change of option price divided by the percentage change
in stock price we ﬁnd the option’s elasticity to be
Ω= ∂V
V
∂S
S = ∂V
∂S
V
S = Therefore,
σoption = Ωσ. SVS
.
V 68 THE BLACKSCHOLES PDE AND EQUILIBRIUM RETURNS 447 Since the option value and stock value are driven by the same standard Brownian motion, they
must have the same Sharpe ratios. That is,
α−r
αoption − r
=
σ
σoption
and from this we ﬁnd the riskpremium on the option
αoption − r = Ω(α − r).
The quantity Ω(α − r) + r is referred to as the equilibrium expected return on the stock. It
follows that the expected return on the option is equal to the equilibrium expected return on the
stock. From this fact we have 122
σ S VSS
2 αoption − r =Ω(α − r)
SVS
122
σ S VSS + (α − δ )SVS + Vt − r =
(α − r)
2
V
SVS
+ (α − δ )SVS + Vt − r −
(α − r) =0
V
1
Vt + σ 2 S 2 VSS + (r − δ )SVS − rV =0
2 which is the BlackScholes equation.
Example 68.1
The following information are given regarding an option on a stock: r = 0.08, δ = 0.02, σ =
0.30, S (t) = 50, Vt = −3.730, VS = 0.743, and VSS = 0.020. Find the value of the option.
Solution.
Using the BlackScholes formula, we ﬁnd
0.5(0.30)2 (50)2 (0.020) + (0.08 − 0.02)(50)(0.743) − 3.730 = 0.08V.
Solving this equation for V we ﬁnd V = 9.3625
Example 68.2
The following information are given regarding an option on a stock: r = 0.08, δ = 0.02, σ =
0.30, S (t) = 50, Vt = −3.730, VS = 0.743, and VSS = 0.020. Find the instantaneous expected return
on the option if the stock riskpremium is 0.13. 448 THE BLACKSCHOLES PARTIAL DIFFERENTIAL EQUATION Solution.
We have
αoption = 0.5(0.30)2 (50)2 (0.020) + 0.13(50)(0.743) − 3.730
0.5σ 2 S 2 VSS + (α − δ )SVS + Vt
=
= 0.3578
V
9.3625 Example 68.3
Consider a European put option on a nondividendpaying stock with exercise date T, T > 0. Let
S (t) be the price of one share of the stock at time t, t ≥ 0. For 0 ≤ t ≤ T, let P (s, t) be the price
of one unit of the call option at time t, if the stock price is s at that time. You are given:
(i) dS((tt)) = 0.11dt + σdZ (t), where σ is a positive constant and {Z (t)} is a Brownian motion.
S
S ,t
(ii) dP((S ((tt)),t)) = γ (S (t), t)dt + σP (S (t), t)dZ (t), 0 ≤ t ≤ T.
P
(iii) P (S (0), 0) = 7.
(iv) At time t = 0, the cost of shares required to deltahedge one unit of the put option is 18.25.
(v) The continuously compounded riskfree interest rate is 8%.
Determine γ (S (0), 0). Solution.
We have
γoption − r = SVS
(α − r )
V which, for this problem, translates to
γ (S (t), t) − 0.08 = S (t)∆
(0.11 − 0.08).
P (S (t), t) We are given that
S (0)∆
−18.25
=
= −2.61.
P (S (0), 0)
7
Hence,
γ (S (0), 0) = −2.61(0.11 − 0.08) + 0.08 = 0.0017
Remark 68.1
The BlackScholes equation in the previous section was derived by assuming no arbitrage; this
involved hedging the option. The equilibrium and noarbitrage prices are the same. The equilibrium
pricing approach shows that the BlackScholes equation does not depend on the ability to hedge
the option. 68 THE BLACKSCHOLES PDE AND EQUILIBRIUM RETURNS 449 Practice Problems
Problem 68.1
The following information are given regarding an option on a stock: r = 0.08, σ = 0.30, S (t) =
50, Vt = 3.730, VS = 0.743, VSS = 0.020, and V = 9.3625. Find the continuously compounded
yield δ.
Problem 68.2
The following information are given regarding an option on a stock: δ = 0.02, σ = 0.30, S (t) =
50, Vt = −3.730, VS = 0.743, VSS = 0.020, and V = 9.3625. Find the continuously compounded
riskfree rate r.
Problem 68.3
The following information are given regarding an option on a stock: r = 0.08, δ = 0.02, σ =
0.30, S (t) = 50, Vt = −3.730, VS = 0.743, and VSS = 0.020. Find the option’s elasticity.
Problem 68.4
The following information are given regarding an option on a stock: r = 0.08, δ = 0.02, σ =
0.30, S (t) = 50, Vt = −3.730, VS = 0.743, and VSS = 0.020. Find the option’s riskpremium if the
stock’s riskpremium is 0.13.
Problem 68.5
The following information are given regarding an option on a stock: r = 0.08, δ = 0.02, σ =
0.30, S (t) = 5, Vt = −0.405, VS = 0.624, VSS = 0.245. Find the option’s elasticity.
Problem 68.6 ‡
Consider a European call option on a nondividendpaying stock with exercise date T, T > 0. Let
S (t) be the price of one share of the stock at time t, t ≥ 0. For 0 ≤ t ≤ T, let C (s, t) be the price
of one unit of the call option at time t, if the stock price is s at that time. You are given:
(i) dS((tt)) = 0.1dt + σdZ (t), where σ is a positive constant and {Z (t)} is a Brownian motion.
S
S ,t
(ii) dC((S ((tt)),t)) = γ (S (t), t)dt + σC (S (t), t)dZ (t), 0 ≤ t ≤ T.
C
(iii) C (S (0), 0) = 6.
(iv) At time t = 0, the cost of shares required to deltahedge one unit of the call option is 9.
(v) The continuously compounded riskfree interest rate is 4%.
Determine γ (S (0), 0). 450 THE BLACKSCHOLES PARTIAL DIFFERENTIAL EQUATION 69 The BlackScholes Equation and the Risk Neutral Pricing
Since α, the expected return on the stock, does not appear in the BlackScholes equation, the
actual expected return on a stock is irrelevant for pricing an option on the stock. The pricing is
only dependent on the riskfree rate and so it was observed that the BlackScholes pricing model is
consistent with any possible world where there is no arbitrage. Thus, it makes sense when valuing
an option to select the world with the easiest option valuation. The easiet valuation environment
is the riskneutral environment.
As disucssed in Section 63, the true stock price process that is observed in the world follows a
Brownian motion given by
dS (t)
= (α − δ )dt + σdZ (t).
S (t)
The riskneutral price process is given by
dS (t)
˜
= (r − δ )dt + σdZ (t)
S (t)
˜
where Z is a standard Brownian process (the riskneutral process). Notice that the volatility is the
same for the true price process and the riskneutral price process.
Since,
1
dV = (Vt + σ 2 S 2 VSS + (α − δ )SVS )dt + SσVS dZ (t),
2
the actual expected change per unit of time in the option price is given by
E (dV )
1
= Vt + σ 2 S 2 VSS + (α − δ )SVS .
dt
2
Likewise, we deﬁne the expected change per unit of time in the option value with respect to the
riskneutral distribution to be
1
E ∗ (dV )
= Vt + σ 2 S 2 VSS + (r − δ )SVS
dt
2
where E ∗ is the expectation with respect to the riskneutral probability distribution. Hence, the
BlackScholes equation can be rewritten as
E ∗ (dV )
= rV.
dt (69.1) In other words, in the riskneutral environment, the expected increase in the value of the option is
at the risk free rate. 69 THE BLACKSCHOLES EQUATION AND THE RISK NEUTRAL PRICING 451 Example 69.1
Given the following information about a European option: r = 0.08, δ = 0.02, σ = 0.3, Vt =
−4.567, VS = 0.123, VSS = 0.231, S = 25. Find the expected change of the option value with respect
to the riskneutral distribution.
Solution.
Using the BlackScholes equation we have
1
rV = −4.567 + (0.3)2 (25)2 (0.231) + (0.08 − 0.02)(25)(0.123) = 2.11.
2
Thus, E ∗ (dV )
= rV = 2.11
dt
In their paper, Black and Scholes established that the solution to equation (69.1) is just the expected
value of the derivative payoﬀ under the riskneutral probability distribution discounted at the riskfree rate. For example, in the case of a European call option on the stock with boundary condition
C (T ) = max{S (T ) − K }, the price of the call is given by
∞ C = e−r(T −t) [S (T ) − K ]f ∗ (S (T ))dS (T )
K ∗ where f (S (T )) equals the riskneutral probability density function for the stock price at time T,
given the observed stock price is S (t) at time t < T and the integral is the expected value of the
the payoﬀ on the call in the risk neutral environment.
Example 69.2
Find the premium of a European put in the riskneutral environment that satisﬁes equation (69.1).
Solution.
Since the payoﬀ of the put is K − S (T ) for S (T ) < K, the expected payoﬀ in the riskneutral
environment is:
K [K − S (T )]f ∗ (S (T ))dS (T ).
0 Discounting back to time t at the riskfree rate, the put premium is:
K P = e−r(T −t) [K − S (T )]f ∗ (S (T ))dS (T )
0 Remark 69.1
The above result shows that the discounted riskneutral expectations are prices of derivatives. This
is also true for riskneutral probabilities. See pp. 6923 of [1]. 452 THE BLACKSCHOLES PARTIAL DIFFERENTIAL EQUATION Practice Problems
Problem 69.1
Given the following information about a European option: r = 0.08, δ = 0.02, σ = 0.3, Vt =
−0.405, VS = 0.624, VSS = 0.245, S = 5. Find the expected change per unit time of the option value
with respect to the riskneutral distribution.
Problem 69.2
Given the following information about a European option: α = 0.05, δ = 0.01, σ = 0.15, Vt =
−0.105, VS = 0.624, VSS = 0.245, S = 5. Find the expected change per unit time of the option value
with respect to the actual probability distribution.
Problem 69.3
The expected change per unit time of a European option with respect to the riskneutral distribution
is found to be 2.11. Assuming a riskfree interest rate of 8%, ﬁnd the value of the option.
Problem 69.4
Given the following information about a European option: r = 0.08, δ = 0.02, σ = 0.3, VS =
0.123, VSS = 0.231, S = 25. Find the value of Vt if the expected change of the option value with
respect to the riskneutral distribution is 2.11.
Problem 69.5
Find the riskfree interest rate r if the expected change of an option value with respect to the
riskneutral distribution is 1.72 and with option value of 18.1006.
Problem 69.6
Suppose that the instantaneous expected return on an option is 0.3578 and the option value is
9.3625. Find the actual expected change in the option price. Binary Options
Binary options, also known as digital options, were ﬁrst introduced in Section 67. In this chapter,
we discuss a family of binary options known as allornothing options. They include two types
of options: cashornothing and assetornothing. 453 454 BINARY OPTIONS 70 CashorNothing Options
A cashornothing call option with strike K and expiration T is an option that pays its owner
$b if S (T ) > K and zero otherwise. In the BlackScholes framework, P r(S (T ) > K ) = N (d2 ) (See
Section 49) where
ln (S (t)/K ) + (r − δ − 0.5σ 2 )(T − t)
√
d2 =
.
σ T −t
From Remark 69.1, the price of this call is just the discounted riskneutral probability. That is,
CashCall = be−r(T −t) N (d2 ).
Example 70.1
Find the price of a cashornothing put.
Solution.
A cashornothing put option with strike K and expiration T is an option that pays its owner $b
if S (T ) < K and zero otherwise. In the BlackScholes framework, P r(S (T ) < K ) = N (−d2 ) =
1 − N (d2 ) (See Section 49). The price of this put is the discounted riskneutral probability. That
is,
CashPut = be−r(T −t) N (−d2 ).
Remark 70.1
Note that the pricing formula of a cashornothing options satisfy the BlackScholes equation. See
Example 67.2. Also, note that if b = K we obtain the second term of the BlackScholes formula.
Example 70.2
Find an expression for CashCall + CashPut.
Solution.
We have
CashCall + CashPut = be−r(T −t) N (d2 ) + be−r(T −t) (1 − N (d2 )) = be−r(T −t)
We next examine the delta of cashornothing options. For a cashornothing call option we have
∆CashCall =
∂
∂N (d2 ) ∂d2
(be−r(T −t) N (d2 )) = be−r(T −t)
∂S
∂d2 ∂S
d2
2 e− 2
1
√
=be−r(T −t) · √ ·
.
2π Sσ T − t 70 CASHORNOTHING OPTIONS 455 Example 70.3
An option will pay its owner $10 three months from now if the stock price at that time 3 is greater
than $40. You are given: S0 = 40, σ = 30%, r = 8%, and δ = 2%. Find the delta of this option.
Solution.
0.0252 e− 2
1
√
∆CashCall = 10e−0.08×0.25 √
·
= 0.6515
2π
40(0.30) 0.25
The delta for a cashornothing put is
∆CashPut = ∂ (1 − N (d2 )
∂
(be−r(T −t) N (−d2 )) = be−r(T −t)
∂S
∂S
d2
2 e− 2
1
√
= − be−r(T −t) · √ ·
.
2π Sσ T − t
Example 70.4
An option will pay its owner $10 three months from now if the stock price at that time 3 is less
than $40. You are given: S0 = 40, σ = 30%, r = 8%, and δ = 2%. Find the delta of this option.
Solution.
− 0.025
2
−0.08×0.25 e ∆CashPut = −10e 2 1
√
√
·
= −0.6515
2π
40(0.30) 0.25 DeltaHeding of CashorNothing Options
Recall the formula for delta in the case of a cashornothing call option:
d2
2 ∆CashCall
Since e− 2
1
√
= be−r(T −t) · √ ·
.
2π Sσ T − t √
ln (S/K )
(r − δ − 0.5σ 2 ) T − t
√
+
d2 =
Sσ
Sσ T − t
we see that as t approaches T , d2 → 0 for ST = K and therefore ∆CashCall approaches inﬁnity.
For ST = K, we have d2 → ±∞ so that ∆CashCall approaches 0. Thus, if a cashornothing option
is getting very close to expiration, and the current stock price is close to K, then a very small
movement in the stock price will have a very large eﬀect on the payoﬀ of the option. The delta for
an atthemoney cashornothing option gets very big as it approaches expiration. Thus, it is very
hard to hedge cashornothing options. That is why cashornothing options are rarely traded. 456 BINARY OPTIONS Example 70.5
An option will pay its owner $40 three months from now if the stock price at that time 3 is greater
than $40. You are given: S0 = 40, σ = 30%, r = 8%, K = $40 and δ = 0%. Find payoﬀ diagram of
this cashornothing call as function of ST . Compare this diagram with an ordinary call option.
Solution.
The diagrams are shown in Figure 70.1 Figure 70.1
Remark 70.2
An ordinary put or call is easier to hedge because the payoﬀ is continuous−there is no discrete jump
at the strike price as the option approaches expiration.
Example 70.6
A collectondelivery call (COD) costs zero initialy, with the payoﬀ at expiration being 0 if
S < K ,and S − K − P if S ≥ K. The problem in valuing the option is to determine P, the amount
the option holder pays if the option is inthemoney at expiration. The premium P is determined
once for all when the option is created. Find a formula for P.
Solution.
The payoﬀ on the collectondelivery call is that from an ordinary call with with strike price K, 70 CASHORNOTHING OPTIONS 457 minus that from a cashornothing call with with strike K that pays P.
Since we pay nothing initially for the collectondelivery call, the initial value of the portfolio must
be zero. Thus, the premium on the cashornothing option is equal to the premium of the ordinary
European call at time t = 0. This implies the equation
P e−r(T −t) N (d2 ) = Se−δ(T −t) N (d1 ) − Ke−r(T −t) N (d2 ).
Solving this equation for P we ﬁnd
P= Se−δ(T −t) N (d1 ) − Ke−r(T −t) N (d2 )
e−r(T −t) N (d2 ) 458 BINARY OPTIONS Practice Problems
Problem 70.1
An option will pay its owner $10 three years from now if the stock price at that time is greater than
$40. You are given: S0 = 40, σ = 30%, r = 8%, and δ = 2%. What is the premium for this option?
Problem 70.2
An option will pay its owner $10 three years from now if the stock price at that time 3 is less than
$40. You are given: S0 = 40, σ = 30%, r = 8%, and δ = 2%. What is the premium for this option?
Problem 70.3
Option A pays $10 three years from now if the stock price at that time 3 is greater than $40. Option
B pays $40 three years from now if the stock price at that time 3 is less than $40. You are given:
S0 = 40, σ = 30%, r = 8%, and δ = 0%. What is the value of the two options?
Problem 70.4
The premium for a cashornothing call that pays one is 0.5129. The premium for a similar cashornothing put is 0.4673. If the time until expiration is three months, what is r?
Problem 70.5
The premium for a cashornothing call that pays one is 0.5129. The premium for a similar cashornothing put is 0.4673. The continuouslycompounded riskfree interest rate is 0.08. If the time
until expiration is T years, what is T ?
Problem 70.6
An option will pay its owner $40 three months from now if the stock price at that time 3 is less
than $40. You are given: S0 = 40, σ = 30%, r = 8%, K = $40 and δ = 0%. Find payoﬀ diagram of
this cashornothing put as function of ST . Compare this diagram with an ordinary put option.
Problem 70.7
A collectondelivery call (COD) costs zero initialy, with the payoﬀ at expiration being 0 if
S < 90 ,and S − K − P if S ≥ 90. You are also given the following: r = 0.05, δ = 0.01, σ = 0.4, S =
$70T − t = 0.5. Find P.
Problem 70.8
A cashornothing call that will pay $1 is atthemoney, with one day to expiration. You are given:
σ = 20%, r = 6%, δ = 0%.
A market maker has written 1000 of these calls. He buys shares of stock in order to delta hedge his
position. How much does he pay for these shares of stock? 70 CASHORNOTHING OPTIONS 459 Problem 70.9 ‡
Assume the BlackScholes framework. You are given:
(i) S (t) is the price of a nondividendpaying stock at time t.
(ii) S (0) = 10
(iii) The stock’s volatility is 20%.
(iv) The continuously compounded riskfree interest rate is 2%.
At time t = 0, you write a oneyear European option that pays 100 if [S (1)]2 is greater than 100
and pays nothing otherwise.
You deltahedge your commitment.
Calculate the number of shares of the stock for your hedging program at time t = 0.
Problem 70.10 ‡
Assume the BlackScholes framework. For a European put option and a European gap call option
on a stock, you are given:
(i) The expiry date for both options is T.
(ii) The put option has a strike price of 40.
(iii) The gap call option has strike price 45 and payment trigger 40.
(iv) The time0 gamma of the put option is 0.07.
(v) The time0 gamma of the gap call option is 0.08.
Consider a European cashornothing call option that pays 1000 at time T if the stock price at that
time is higher than 40.
Find the time0 gamma of the cashornothing call option. Hint: Consider a replicating portfolio
of the cashornothing option. 460 BINARY OPTIONS 71 AssetorNothing Options
An assetornothing call option with strike K and expiration T is an option that gives its owner
one unit of a share of the stock if S (T ) > K and zero otherwise. In the BlackScholes framework,
the value of this option in riskneutral environment is1
AssetCall = e−r(T −t) E [St St > K ]P r(St > K ) = S0 e−r(T −t) e(r−δ)(T −t) N (d1 )
N (d2 ) = S0 e−δ(T −t) N (d1 )
N (d2 ) where ln (S/K ) + (r − δ + 0.5σ 2 )(T − t)
√
.
σ T −t
Note that the price given above is the ﬁrst term of the BlackScholes formula for a European option.
d1 = Example 71.1
An option will give its owner a share of stock three months from now if the stock price at that time
is greater than $40. You are given: S0 = $40, σ = 0.30, r = 0.08, δ = 0. What is the premium for
this option?
Solution.
We have
d1 = ln (40/40) + (0.08 − 0 + 0.5(0.3)2 )(0.25)
ln (S/K ) + (r − δ + 0.5σ 2 )(T − t)
√
√
=
= 0.208.
σ T −t
0.30 0.25 Thus, N (d1 ) = 0.582386 and
AssetCall = S0 e−δ(T −t) N (d1 ) = 40e−0×0.25 (0.582386) = $23.29
Example 71.2
Find the price of an assetornothing put.
Solution.
An assetornothing put option with strike K and expiration T is an option that gives its owner to
receive one unit of a share of stock if S (T ) < K and zero otherwise. In the BlackScholes framework,
the time 0 value of this option is
AssetPut = e−r(T −t) E [St St < K ]P r(St < K ) = S0 e−r(T −t) e(r−δ)(T −t) N (−d1 )
N (−d2 ) = S0 e−δ(T −t) N (−d1 )
N (−d2 ) which is the second term in the BlackScholes formula for a European put
1 See Section 50. 71 ASSETORNOTHING OPTIONS 461 Example 71.3
An option will gives its owner to receive a share of stock three years from now if the stock price at
that time is less than $40. You are given S0 = 40, σ = 0.03, r = 0.08, δ = 0. What is the premium
for this option? Solution.
We have
d1 = ln (S/K ) + (r − δ + 0.5σ 2 )(T − t)
ln (40/40) + (0.08 − 0 + 0.5(0.3)2 )(0.25)
√
√
=
= 0.208.
σ T −t
0.30 0.25 Thus, N (d1 ) = 0.582386 and
AssetPut = S0 e−δ(T −t) N (−d1 ) = 40e−0×0.25 (1 − 0.582386) = $16.70
In monetary value, the owner of a European call gets ST − K, if ST > K. Therefore, a European
call is equivalent to the diﬀerence between an assetornothing call and a cashornothing call that
pays K. That is,
Premium of a European call = Premium of an assetornothing call − Premium of a
cashornothing call that pays K
or
BSCall = Se−δ(T −t) N (d1 ) − Ke−r(T −t) N (d2 )
which is the BlackScholes formula for a European call. Likewise, the owner of a European put
gets K − ST , if ST < K. Therefore, a European put is equivalent to the diﬀerence between a
cashornothing put that pays K and an assetornothing put. That is,
Premium of a European put = Premium of a cashornothing put that pays K − Premium of an
assetornothing put
or
BSPut = Ke−r(T −t) N (−d2 ) − Se−δ(T −t) N (−d1 ) 462 BINARY OPTIONS which is the BlackScholes formula for a European put.
Next, let’s examine the delta of an assetornothing call. We have
∆AssetCall = ∂ AssetCall
∂S ∂N (d1 )
∂S
∂N (d1 ) ∂d1
=e−δ(T −t) N (d1 ) + Se−δ(T −t)
∂d1 ∂S =e−δ(T −t) N (d1 ) + Se−δ(T −t) d2
1 e− 2
1
√
=e−δ(T −t) N (d1 ) + Se−δ(T −t) · √ ·
2π Sσ T − t
d2
1 1
e− 2
=e−δ(T −t) N (d1 ) + e−δ(T −t) · √ · √
2π σ T − t
Example 71.4
An option will give its owner a share of stock three months from now if the stock price at that
time is greater than $40. You are given: S0 = $40, σ = 0.30, r = 0.08, δ = 0. Find the delta of this
option.
Solution.
We have
d1 = ln (40/40) + (0.08 − 0 + 0.5(0.3)2 )(0.25)
ln (S/K ) + (r − δ + 0.5σ 2 )(T − t)
√
√
=
= 0.208.
σ T −t
0.30 0.25 Thus, N (d1 ) = 0.582386 and
d2
1 ∆AssetCall =e −δ (T −t) −δ (T −t) N (d1 ) + e 0.2082
2 e−
=0.582386 + √ 2π · e− 2
1
·√ · √
2π σ T − t 1
√
0.30 0.25 =3.185
Example 71.5
Express the premium of a gap call option with payoﬀ ST − K1 if ST > K2 in terms of assetornothing
call option and cashornothing call. 71 ASSETORNOTHING OPTIONS 463 Solution.
The owner of a gap call gets ST − K1 , if ST > K2 . Therefore, a gap call is equivalent to the diﬀerence
between an assetornothing call with K = K2 , and a cashornothing call with K = K2 that pays
K1 .
Therefore, as discussed previously, the premium of a gap call is
AssetCall − K1 CashCall = Se−δ(T −t) N (d1 ) − K1 e−r(T −t) N (d2 )
where d1 and d2 are calculated using K = K2
Remark 71.1
For the same reason as with cashornothing options, the delta for an atthemoney assetornothing
option gets very big as it approaches expiration. Thus, it is very hard to hedge assetornothing
options. That is why assetornothing options are rarely traded. 464 BINARY OPTIONS Practice Problems
Problem 71.1
An option will give its owner a share of stock three years from now if the stock price at that time
is greater than $100. You are given: S0 = $70, σ = 0.25, r = 0.06, δ = 0.01. What is the premium
for this option?
Problem 71.2
An option will give its owner a share of stock three years from now if the stock price at that time
is less than $100. You are given: S0 = $70, σ = 0.25, r = 0.06, δ = 0.01. What is the premium for
this option?
Problem 71.3
Find an expression for AssetCall + AssetPut.
Problem 71.4
An option will give its owner a share of stock three years from now if the stock price at that time
is greater than $100. You are given: S0 = $70, σ = 0.25, r = 0.06, δ = 0.01. What is the delta of
this option?
Problem 71.5
Find the delta of an assetornothing put.
Problem 71.6
An option will give its owner a share of stock three years from now if the stock price at that time
is less than $100. You are given: S0 = $70, σ = 0.25, r = 0.06, δ = 0.01. What is the delta of this
option?
Problem 71.7
Express the premium of a gap put option with payoﬀ K1 − ST if ST < K2 in terms of assetornothing
put option and cashornothing put.
Problem 71.8
The premium for a cashornothing call is 0.5129. The premium for a similar assetornothing call
is 23.30. If K = 40, what is the premium for a similar European Call.
Problem 71.9
The premium for a European call is 2.7848. The premium for a similar cashornothing call is
0.5129. The premium for a similar assetornothing call is 23.30. Determine the strike price K. 71 ASSETORNOTHING OPTIONS 465 Problem 71.10
The premium for a assetornothing call is 47.85. The premium for a similar assetornothing put
is 40.15. The stock pays no dividends. What is the current stock price?
Problem 71.11 ‡
Your company has just written one million units of a oneyear European assetornothing put option
on an equity index fund.
The equity index fund is currently trading at 1000. It pays dividends continuously at a rate
proportional to its price; the dividend yield is 2%. It has a volatility of 20%.
The option’s payoﬀ will be made only if the equity index fund is down by more than 40% at the
end of one year. The continuously compounded riskfree interest rate is 2.5%
Using the BlackScholes model, determine the price of the assetornothing put options. 466 BINARY OPTIONS 72 Supershares
In this section, we discuss a type of binary options known as supershares. In an article published
in 1976, Nils Hakansson proposed a ﬁnancial intermediary that would hold an underlying portfolio
and issue claims called supershares against this portfolio to investors. A supershare is a security,
which on its expiration date entitles its owner to a given dollar value proportion of the assets of
the underlying portfolio, provided the value of those assets on that date lies between a lower value
KL and an upper value KU . Otherwise, the supershare expires worthless. That is, the payoﬀ of a
supershare is: ST < KL
0
ST
KL ≤ ST ≤ KU
Payoﬀ = KL
0
ST > KU
1
Consider a portfolio that consists of K1 units of long assetornothing call options with strike K1 and
1
units of short assetornothing call options with strike price K2 . Then the payoﬀ at expiration of
K1
this portfolio is exactly the payoﬀ of the supershare mentioned above with KL = K1 and KU = K2
with K1 < K2 . Hence, Premium of the supershare = 1
K1 (AssetCall(K1 ) − AssetCall(K2 )) That is,
SS =
where S −δ(T −t)
e
[N (d1 ) − N (d1 )]
K1 d1 = ln (S/K1 ) + (r − δ + 0.5σ 2 )(T − t)
√
σ T −t d1 = ln (S/K2 ) + (r − δ + 0.5σ 2 )(T − t)
√
σ T −t and Example 72.1
Find the premium of a supershare for the following data: S = 100, r = 0.1, δ = 0.05, σ = 0.2, K1 =
100, K2 = 105, and T − t = 0.5 years.
Solution.
We have
d1 = ln (S/K1 ) + (r − δ + 0.5σ 2 )(T − t)
ln (100/100) + (0.1 − 0.05 + 0.5(0.2)2 )(0.5)
√
√
=
= 0.247487
σ T −t
0.2 0.5 72 SUPERSHARES 467 and
d1 = ln (S/K2 ) + (r − δ + 0.5σ 2 )(T − t)
ln (100/105) + (0.1 − 0.05 + 0.5(0.2)2 )(0.5)
√
√
=
= −0.097511.
σ T −t
0.2 0.5 Thus, N (0.247487) = 0.597734 and N (−0.097511) = 0.461160. The premium of the supershare is
SS = 100 −0.05×0.5
S −δ(T −t)
e
[N (d1 ) − N (d1 )] =
e
[0.597734 − 0.461160] = $0.1332
K1
100 468 BINARY OPTIONS Practice Problems
Problem 72.1
Consider a supershare based on a portfolio of nondividend paying stocks with a lower strike of 350
and an upper strike of 450. The value of the portfolio on November 1, 2008 is 400. The riskfree
rate is 4.5% and the volatility is 18%. Using this data, calculate the values of d1 and N (d1 ) for
T − t = 0.25.
Problem 72.2
Consider a supershare based on a portfolio of nondividend paying stocks with a lower strike of 350
and an upper strike of 450. The value of the portfolio on November 1, 2008 is 400. The riskfree
rate is 4.5% and the volatility is 18%. Using this data, calculate the values of d1 and N (d1 ) for
T − t = 0.25.
Problem 72.3
Consider a supershare based on a portfolio of nondividend paying stocks with a lower strike of 350
and an upper strike of 450. The value of the portfolio on November 1, 2008 is 400. The riskfree
rate is 4.5% and the volatility is 18%. Using this data, calculate the price of the supershare option
on February 1, 2009.
Problem 72.4
Given the following information regarding a supershare: S = 100, K1 = 95, K2 = 110, r = 0.1, δ =
0.05, σ = 0.2, and T − t = 0.5. Calculate d1 and N (d1 ).
Problem 72.5
Given the following information regarding a supershare: S = 100, K1 = 95, K2 = 110, r = 0.1, δ =
0.05, σ = 0.2, and T − t = 0.5. Calculate d1 and N (d1 ).
Problem 72.6
Given the following information regarding a supershare: S = 100, K1 = 95, K2 = 110, r = 0.1, δ =
0.05, σ = 0.2, and T − t = 0.5. Find the premium of this supershare.
Problem 72.7
Using the previous problem and the results of Example 72.1, what can you conclude about the value
of supershares and the value K2 − K1 ? Interest Rates Models
In this chapter we examine pricing models for derivatives with underlying assets either bonds
or interest rates. That is, pricing models for bond options and interest rate options. As with
derivatives on stocks, prices of interest rates or bonds derivatives are characterized by a partial
diﬀerential equation that is essentially the same as the BlackScholes equation. 469 470 INTEREST RATES MODELS 73 Bond Pricing Model with Arbitrage Opportunity
In this section, we examine a bond pricing model that gives rise to arbitrage opportunities. We
examine the hedging of one bond with other bonds. The type of hedging that we consider in this
section is referred to as durationhedging.
By analogy to modeling stocks, we will assume that the timet bond price P (r, t, T ) follows an Itˆ
o
process
dP
= α(r, t)dt − q (r, t)dZ (t).
P (73.1) Also, we will assume that the shortterm interest rate follows the Itˆ process
o
dr = a(r)dt + σ (r)dZ (t). (73.2) We ﬁrst examine a bond pricing model based on the assumption that the yield curve is ﬂat.1 We
will show that this model gives rise to arbitrage opportunities.
Under the ﬂat yield curve assumption, the price of zerocoupon bonds is given by
P (r, t, T ) = e−r(T −t) .
The purchase of a T2 −year zerocoupon bond (T2 < T1 ) can be deltahedged by purchasing N
T1 −year zerocoupon bonds and lending W = −P (r, t, T2 ) − N × P (r, t, T1 ) at the shortterm
interest rate r. The number W can be positive or negative. If W > 0 then we lend. If W < 0 then
we borrow. Also, N can be positive or negative. If N < 0 (i.e., t < T2 < T1 ) then we sell rather
than buy the T2 −year bond.
Now, let I (t) be the value of the deltahedged position at time t. Then we have
I (t) = N × P (r, t, T1 ) + P (r, t, T2 ) + W = 0.
Since W is invested in shortterm (zeroduration) bonds, at the next instant we must have
dW (t) = rW (t)dt.
1 In ﬁnance, the yield curve is the relation between the interest rate (or cost of borrowing) and the time to
maturity of the debt for a given borrower. A ﬂat yield curve is a yield curve where the interest rate is the same
for all maturities. 73 BOND PRICING MODEL WITH ARBITRAGE OPPORTUNITY 471 Also, using Itˆ’s lemma we have
o
dI (t) =N dP (r, t, T1 ) + dP (r, t, T2 ) + dW
1
=N Pr (r, t, T1 )dr + Prr (r, t, T1 )(dr)2 + Pt (r, t, T1 )dt
2
1
+ Pr (r, t, T2 )dr + Prr (r, t, T2 )(dr)2 + Pt (r, t, T2 )dt + dW (t)
2
1
=N −(T1 − t)P (r, t, T1 )dr + (T1 − t)2 σ 2 P (r, t, T1 )dt + rP (r, t, T1 )dt
2
1
+ −(T2 − t)P (r, t, T2 )dr + (T2 − t)2 σ 2 P (r, t, T2 )dt + rP (r, t, T2 )dt + rW (t)dt
2
We select N in such a way to eliminate the eﬀect of interest rate change, dr, on the value of the
portfolio. Thus, we set
(T2 − t)P (r, t, T2 )
, t = T1 , t < T2 .
N =−
(T1 − t)P (r, t, T1 )
T2 −t
Note that the ratio T1 −t is the ratio of the duration of the bond maturing at T2 to the duration of
the bond maturing at T1 . With this choice of N, we have dI (t) =N 1
(T1 − t)2 σ 2 P (r, t, T1 )dt + rP (r, t, T1 )dt
2 1
(T2 − t)2 σ 2 P (r, t, T2 )dt + rP (r, t, T2 )dt + rW (t)dt
2
1
=[N P (r, t, T1 ) + P (r, t, T2 ) + W (t)]rdt + σ 2 N (T1 − t)2 P (r, t, T1 ) + (T2 − t)2 P (r, t, T2 ) dt
2
(T2 − t)P (r, t, T2 )
12
× (T1 − t)2 P (r, t, T1 )dt + (T2 − t)2 P (r, t, T2 )dt
=σ −
2
(T1 − t)P (r, t, T1 )
1
= σ 2 (T2 − T1 )(T2 − t)P (r, t, T2 )dt = 0.
2
+ Thus, this gives rise to an arbitrage since the deltahedged portfolio is riskfree and has zero investment.
With the above pricing model, we encounter two diﬃculties with pricing bonds:
• A casual speciﬁed model may give rise to arbitrage opportunities.
• In general, hedging a bond portfolio based on duration does not result in a perfect hedge.
Example 73.1
Suppose the yield curve is ﬂat at 8%. Consider 3−year and 6−year zerocoupon bonds. You buy 472 INTEREST RATES MODELS one 3−year bond and sell an appropriate quantity N of the 6−year bond to durationhedge the
position. Determine N.
Solution.
We are given that t = 0, T1 = 6, T2 = 3, and r = 0.08. Thus, P (0.08, 0, 3) = e−0.08(3−0) = $0.78663
and P (0.08, 0, 6) = e−0.08(6−0) = $0.61878. The number of 6−year bonds that must be sold to
durationhedge the position is
N =− (T2 − t)P (r, t, T2 )
(3 − 0) × 0.78663
=−
= −0.63562
(T1 − t)P (r, t, T1 )
(6 − 0) × 0.61878 Example 73.2
Suppose the yield curve is ﬂat at 8%. Consider 3−year and 6−year zerocoupon bonds. You buy
one 3−year bond and sell an appropriate quantity N of the 6−year bond to durationhedge the
position. What is the total cost of the durationhedge strategy? How much will you owe in one
day?
Solution.
The total cost of the durationhedge strategy is
W (0) = −P (r, t, T2 ) − N × P (r, t, T1 ) = −0.78663 + 0.63562 × 0.61878 = −$0.39332.
Since W < 0, we have to borrow $0.39332 at the shortterm rate of 8% to ﬁnance the position.
After one day, we owe the lender 0.39332e0.08/365 = $0.3934
Example 73.3
Suppose the yield curve is ﬂat at 8%. Consider 3−year and 6−year zerocoupon bonds. You buy
one 3−year bond and sell an appropriate quantity N of the 6−year bond to durationhedge the
position. Suppose that the yield curve can move up to 8.25% or down to 7.75% over the course of
one day. Do you make or lose money on the hedge?
Solution.
The hedged position has an initial value of $0. After one day, the shortterm rate increases to 8.25%
and the new value of the position is
1 1 e−0.0825(3− 365 ) − 0.63562e−0.0825(6− 365 ) − 0.39332e 365 = −$0.00002255.
0.08 The hedged position has an initial value of $0. After one day, the shortterm rate decreases to
7.75% and the new value of the position is
1 1 e−0.0775(3− 365 ) − 0.63562e−0.0775(6− 365 ) − 0.39332e 365 = −$0.00002819.
Thus, there is a lose of about $0.00003 0.08 73 BOND PRICING MODEL WITH ARBITRAGE OPPORTUNITY 473 Practice Problems
Problem 73.1
Suppose the yield curve is ﬂat at 8%. Consider 2−year and 7−year zerocoupon bonds. You buy
one 2−year bond and sell an appropriate quantity N of the 7−year bond to durationhedge the
position. Determine the time0 prices of the two bonds.
Problem 73.2
Suppose the yield curve is ﬂat at 8%. Consider 2−year and 7−year zerocoupon bonds. You buy one
2−year bond and sell an appropriate quantity N of the 7−year bond to durationhedge the position.
Determine the quantity of the 7−year bonds to be purchased to durationhedge the position.
Problem 73.3
Suppose the yield curve is ﬂat at 8%. Consider 2−year and 7−year zerocoupon bonds. You buy
one 2−year bond and sell an appropriate quantity N of the 7−year bond to durationhedge the
position. Find the total cost of ﬁnancing this position.
Problem 73.4
Suppose the yield curve is ﬂat at 8%. Consider 2−year and 7−year zerocoupon bonds. You buy
one 2−year bond and sell an appropriate quantity N of the 7−year bond to durationhedge the
position. Suppose that the yield curve can move up to 8.5% or down to 7.5% over the course of one
day. Do you make or lose money on the hedge?
Problem 73.5
Suppose the yield curve is ﬂat at 6%. Consider 4−year 5%coupon bond and an 8−year 7%coupon
bond. All coupons are annual. You buy one 4−year bond and purchase an appropriate quantity N
of the 8−year bond to durationhedge the position. Determine the time0 prices of the two bonds.
Problem 73.6
Suppose the yield curve is ﬂat at 6%. Consider 4−year 5%coupon bond and an 8−year 7%coupon
bond. All coupons are annual. You buy one 4−year bond and purchase an appropriate quantity
N of the 8−year bond to durationhedge the position. Determine the (modiﬁed) durations of each
bond. Hint: The modiﬁed duration is given by D = ∂P
∂t P . Problem 73.7
Suppose the yield curve is ﬂat at 6%. Consider 4−year 5%coupon bond and an 8−year 7%coupon
bond. All coupons are annual. You buy one 4−year bond and purchase an appropriate quantity N
of the 8−year bond to durationhedge the position. Determine the number of 8−year bonds to be
purchased for this position. 474 INTEREST RATES MODELS Problem 73.8
Suppose the yield curve is ﬂat at 6%. Consider 4−year 5%coupon bond and an 8−year 7%coupon
bond. All coupons are annual. You buy one 4−year bond and purchase an appropriate quantity N
of the 8−year bond to durationhedge the position. What is the total cost of the durationhedge
strategy? How much will you owe in one day?
Problem 73.9
Suppose the yield curve is ﬂat at 6%. Consider 4−year 5%coupon bond and an 8−year 7%coupon
bond. All coupons are annual. You buy one 4−year bond and purchase an appropriate quantity N
of the 8−year bond to durationhedge the position. Suppose that the yield curve can move up to
6.25% over the course of one day. Do you make or lose money on the hedge?
Problem 73.10
Suppose the yield curve is ﬂat at 6%. Consider 4−year 5%coupon bond and an 8−year 7%coupon
bond. All coupons are annual. You buy one 4−year bond and purchase an appropriate quantity N
of the 8−year bond to durationhedge the position. Suppose that the yield curve can move down
to 5.75% over the course of one day. Do you make or lose money on the hedge? 74 A BLACKSCHOLES ANALOGUE FOR PRICING ZEROCOUPON BONDS 475 74 A BlackScholes Analogue for Pricing ZeroCoupon Bonds
In this section, we continue studying the bondhedging problem with a general pricing model
P (r, t, T ) rather than the speciﬁc pricing model of the previous section.
At the next instant, the change in the bond’s price is
1
dP (r, t, T ) =Pr (r, t, T )dr + Prr (r, t, T )(dr)2 + Pt (r, t, T )dt
2
1
= a(r)Pr (r, t, T ) + Prr (r, t, T )σ (r)2 + Pt (r, t, T ) dt + Pr (r, t, T )σ (r)dZ (t)
2
Deﬁne
1
1
a(r)Pr (r, t, T ) + Prr (r, t, T )σ (r)2 + Pt (r, t, T )
P (r, t, T )
2
1
q (r, t, T ) = −
Pr (r, t, T )σ (r)
P (r, t, T ) α(r, t, T ) = In this case, we can describe the change in the price of the bond by the equation
dP (r, t, T )
= α(r, t, T )dt − q (r, t, T )dZ (t).
P (r, t, T )
Thus, the instantaneous rate of return on a bond maturing at time T has a mean α(r, t, T ) and
standard deviation q (r, t, T ). That is, the expected return on the bond over the next instant is
α(r, t, T ).1 Note that Pr (r, t, T ) < 0 so that q (r, t, T ) > 0.2
Now we consider again the deltahedged portofolio that consists of buying a T2 −year zerocoupon
bond, buying N T1 −year zerocoupon bonds, and ﬁnancing the strategy by lending W = −P (r, t, T2 )−
N × P (r, t, T1 ). Let I (t) denote the value of the portfolio at time t. Then
I (t) = N × P (r, t, T1 ) + P (r, t, T2 ) + W = 0
and
dI (t) =N [α(r, t, T1 )dt − q (r, t, T1 )dZ (t)]P (r, t, T1 )
+[α(r, t, T2 )dt − q (r, t, T2 )dZ (t)]P (r, t, T2 ) + rW (t)dt
1 We will assume that α(r, t, T ) > r since a bond is expected to earn more than the riskfree rate due to the fact
that owning a bond is riskier than lending at the riskfree rate.
2
Bond prices decrease when interest rates increase because the ﬁxed interest and principal payments stated in the
bond will become less attractive to investors. 476 INTEREST RATES MODELS For the shortterm interest rate to be the only source of uncertainty, that is, for the bond price to
be driven only by the shortterm interest rate, we choose N such that the dZ terms in the previous
equation are eliminated. This occurs when
N =− P (r, t, T2 )q (r, t, T2 )
Pr (r, t, T2 )
=−
.
P (r, t, T1 )q (r, t, T1 )
Pr (r, t, T1 ) In this case, we have
dI (t) =[N α(r, t, T1 )P (r, t, T1 ) + α(r, t, T2 )P (r, t, T2 ) + rW ]dt
P (r, t, T2 )q (r, t, T2 )
+ α(r, t, T2 )P (r, t, T2 ) − rP (r, t, T2 )
=[−α(r, t, T1 ) ×
q (r, t, T1 )
P (r, t, T2 )q (r, t, T2 )
+rdt
q (r, t, T1 )
P (r, t, T2 )
dt
= {q (r, t, T2 )[r − α(r, t, T1 )] + q (r, t, T1 )[α(r, t, T2 ) − r]} ×
q (r, t, T1 )
Since the cost of the portfolio is zero, and its return in not random, to preclude arbitrage we must
have dI (t) = 0. Thus, we obtain
α(r, t, T2 ) − r
α(r, t, T1 ) − r
=
.
q (r, t, T1 )
q (r, t, T2 )
This equation says that the two bonds have the same Sharpe ratio. This is consistent with what
we proved about prices of two assets driven by the same random term dZ (t). It follows that all
zerocoupon bonds have the same Sharpe ratio regardless of maturity. Denoting the Sharpe ratio
for a zerocoupon bond by φ(r, t) we have
φ(r, t) = α(r, t, T ) − r
.
q (r, t, T ) Substituting the expressions for α(r, t, T ) and q (r, t, T ) into the Sharpe ratio formula for a zerocoupon bond that matures at time T , we have
φ(r, t) =
φ(r, t) = α(r, t, T ) − r
q (r, t, T )
1
1
[a(r)Pr (r, t, T ) + 2 σ (r)2 Prr (r, t, T ) + Pt (r, t, T )] − r
P (r,t,T )
1
− P (r,t,T σ (r)Pr (r, t, T ) 1
−φ(r, t)σ (r)Pr (r, t, T ) =a(r)Pr (r, t, T ) + σ (r)2 Prr (r, t, T ) + Pt (r, t, T ) − rP (r, t, T )
2
1
rP (r, t, T ) = σ (r)2 Prr (r, t, T ) + [a(r) + σ (r)φ(r, t)]Pr (r, t, T ) + Pt (r, t, T )
2 (74.1) 74 A BLACKSCHOLES ANALOGUE FOR PRICING ZEROCOUPON BONDS 477 When the shortterm interest rate is the only source of uncertainty, equation (74.1) must be satisﬁed
by any zerocoupon bond. Equation (74.1) is analogous to the BlackScholes equation for Stocks
(See Section 67).
As with stocks, we deﬁne the Greeks of a bond as follows:
∂P
∂r
∂ 2P
Γ= 2
∂r
∂P
Θ=
∂t ∆= Example 74.1
Find the Greeks of the bond price model P (r, t, T ) = e−r(T −t) .
Solution.
We have
∂P
= −(T − t)e−r(T −t)
∂r
∂ 2P
Γ = 2 = (T − t)2 e−r(T −t)
∂r
∂P
= re−r(T −t)
Θ=
∂t
It follows that the ∆ΓΘ approximation of the change in price of a zerocoupon bond is given by
∆= 1
1
dP = Pr dr + Prr (dr)2 + Pt dt = ∆dr + Γ(dr)2 + Θdt.
2
2
Example 74.2 ‡
For t ≤ T, let P (r, t, T ) be the price at time t of a zerocoupon bond that pays $1 at time T, if the
shortrate at time t is r.
You are given:
(i) P (r, t, T ) = A(t, T )exp[−B (t, T )r] for some functions A(t, T ) and B (t, T ).
(ii) B (0, 3) = 2.
Based on P (0.05, 0, 3), you use the deltagamma approximation to estimate P (0.03, 0, 3), and denote
0,
Est
the value as PEst (0.03, 0, 3). Find PP (0(0.03,,3)3 .
.05,0
Solution.
By the deltagammatheta approximation we have
1
P (r(t + dt), t + dt, T ) − P (r(t), t, T ) ≈ [r(t + dt) − r(t)]Pr + [r(t + dt) − r(t)]2 Prr + Pt dt.
2 478 INTEREST RATES MODELS The question is asking us to use only the deltagamma approximation so that we will neglect the
term Pt dt. Also, we let r(t + dt) = 0.05, r(t) = 0.03, and t + dt ≈ t. Moreover, we have
P (r, t, T ) =A(t, T )e−B (t,T )r(t)
Pr (r(t), t, T ) = − B (t, T )P (r(t), t, T )
Prr (r(t), t, T ) =[B (t, T )]2 P (r(t), t, T )
Thus,
P (0.05, 0, 3) − PEst (0.03, 0, 3) = − (0.05 − 0.03)B (0, 3)PEst (0.03, 0, 3)
1
+ (0.05 − 0.03)2 [B (0, 3)]2 PEst (0.03, 0, 3)
2
which implies
PEst (0.03, 0, 3)[1 − 0.02 × 2 + 0.0002 × 22 ] = P (0.05, 0, 3).
Thus,
PEst (0.03, 0, 3)
1
=
= 1.0408
P (0.05, 0, 3)
1 − 0.02 × 2 + 0.0002 × 4
The riskpremium of a zerocoupon bond that matures at time T is the expected return minus
the riskfree return which is equal to the Sharpe ratio times the bond’s volatility
Riskpremium of bond = α(r, t, T ) − r = φ(r, t)q (r, t, T ).
Now consider an asset that has a percentage price increase of dr where
dr = a(r)dt + σ (r)dZ (t).
Since the percentage change in price of a zerocoupon bond and the percentage price increase of the
asset are driven by the same standard Brownian motion, they must have the same Sharpe ratio:
− a(r) − r
α(r, t, T ) − r
=
= φ(r, t).
σ (r)
q (r, t, T ) where the asset Sharpe ratio is deﬁned by3
−
3 a(r) − r
.
σ (r ) A bond decreases in value as the shortterm rate increases whereas the asset value increases as the shortterm
rate increases. Since the riskpremium of the bond is positive, the risk premium of the asset must be negative.
Otherwise, it would be possible to combine the asset with the zerocoupon bond to create a riskfree portfolio that
earned more than the riskfree of return. 74 A BLACKSCHOLES ANALOGUE FOR PRICING ZEROCOUPON BONDS 479 In this case, the riskpremium of the asset is deﬁned by
a(r) − r = −σ (r)φ(r, t).
Example 74.3
An asset has a percentage price increse of dr where
dr = 0.25(0.10 − r)dt + 0.01dZ.
Find the Riskpremium of the asset and the Sharpe ratio.
Solution.
We have a(r) = 0.25(0.10 − r) and σ (r) = 0.01. Thus, the riskpremium of the asset is a(r) − r =
0.025 − 1.25r. The Sharpe ratio is
− a(r) − r
0.025 − 1.25r
1.25r − 0.025
=−
=
= 125r − 2.5
σ (r)
0.01
0.01 Example 74.4
The change in a zerobond price is −$0.008651. Given the following information: Pr = −1.70126, Prr =
1
4.85536, dt = 365 , and dr = 0.0052342. Estimate ∆.
Solution.
We have
1
dP =Pr dr + Prr (dr)2 + Pt dt
2
1
−0.008651 = − 1.70126(0.0052342) + (4.85536)(0.0052342)2 + Pt
2 1
365 . Thus,
1
∆ = 365 −0.008651 + 1.70126(0.0052342) − (4.85536)(0.0052342)2
2 ≈ 0.06834 480 INTEREST RATES MODELS Practice Problems
Problem 74.1
The price of a zerocoupon bond that matures at time T is given by the expression P (r, t, T ) =
A(t, T )e−B (t,T )r where A(t, T ) and B (t, T ) are diﬀerentiable functions of t. Find ∆, Γ and Θ.
Problem 74.2
An asset has a percentage price increse of dr where
dr = 0.34(0.09 − r)dt + 0.26dZ.
Find the Riskpremium of the asset and the Sharpe ratio.
Problem 74.3
The following information are given: Pr = −1.70126, Prr = 4.85536, Pt = 0.0685, dt =
dr = 0.0052342. Estimate the change in the price P using the ∆ΓΘ approximation. 1
,
365 and Problem 74.4
∂
Show that q (r, t, T ) = −σ (r) ∂r ln [P (r, t, T )].
Problem 74.5
The price of a zerocoupon bond that matures to $1 at time T follows an Itˆ process
o
dP (r, t, T )
= α(r, t, T )dt − q (r, t, T )dZ (t).
P (r, t, T )
Find q (r, t, T ) if P (r, t, T ) = A(t, T )e−B (t,T )r .
Problem 74.6
Find α(r, t, T ) in the previous exercise if the the Sharpe ratio of the zerocoupon bond is a constant
φ. 75 ZEROCOUPON BOND PRICING: RISKNEUTRAL PROCESS 481 75 ZeroCoupon Bond Pricing: RiskNeutral Process
We next consider the riskneutral process for the shortterm interest rate. For that purpose, we
ﬁrst notice the following
dP (r, t, T )
=α(r, t, T )dt − q (r, t, T )dZ (t)
P (r, t, T )
=rdt − q (r, t, T )dZ (t) + [α(r, t, T ) − r]dt
α(r, t, T ) − r
=rdt − q (r, t, T ) dZ (t) −
dt
q (r, t, T )
=rdt − q (r, t, T )[dZ (t) − φ(r, t)dt]
Using Girsanov’s theorem, the transformation
˜
dZ (t) = dZ (t) − φ(r, t)dt
transforms the standard Brownian process, Z (t), to a new standard Brownian process that is martingale under the riskneutral probability measure. In this case, the process for the bond price can
be written as
dP (r, t, T )
˜
= rdt − q (r, t, T )dZ (t).
P (r, t, T )
This shows that the expected return on the bond is r. It follows that the Sharpe ratio of a zerocoupon bond under the riskneutral distribution is 0.
The Itˆ process for r using this new Brownian motion is
o
dr =a(r)dt + σ (r)dZ (t)
˜
=a(r)dt + σ (r)[dZ (t) + φ(r, t)]
˜
=[a(r) + σ (r)φ(r, t)]dt + σ (r)dZ (t).
It has been proven that the solution to equation (74.1) subject to the boundary condition P (r, T, T ) =
$1 is given by
P (r(t), t, T ) = E ∗ (e−R(t,T ) )
where E ∗ is the expectation based on the riskneutral distribution and R(t, T ) is the cumulative
interest rate given by
T R(t, T ) = r(s)ds.
t 482 INTEREST RATES MODELS
∗ Now by Jensen’s inequality1 , E ∗ (e−R(t,T ) ) = e−E (R(t,T )) . That is, it is not correct to say that the
price of a zerocoupon bond is found by discounting at the expected interest rate.
In summary, an approach to modeling zerocoupon bond prices is exactly the same procedure used
to price options of stocks:
• We begin with a model that describes the interest rate and then use equation (74.1) a partial
diﬀerential equation that describes the bond price.
• Next, using the PDE together with the boundary conditions we can determine the price of the
bond.
Example 75.1
The realistic process for the shortterm interest rate is given by
dr = 0.3(0.12 − r)dt + 0.03dZ
and the riskneutral process is given by
˜
dr = 0.3(0.125 − r)dt + 0.03dZ.
Determine the Sharpe ratio φ(r, t).
Solution.
We have that
0.3(0.125 − r) = 0.3(0.12 − r) + 0.03φ(r, t).
Solving this equation we ﬁnd
φ(r, t) = 0.05
DeltaGamma Approximations for Bonds
Using Itˆ lemma, the change in the price of a zerocoupon bond under the riskneutral probability
o
measure is given by
1
dP =Pr (r, t, T )dr + Prr (r, t, T )(dr)2 + Pt (r, t, T )dt
2
˜
=Pr (r, t, T ) [a(r) + σ (r)φ(r, t)]dt + σ (r)dZ
2
1
˜
+ Prr (r, t, T ) [a(r) + σ (r)φ(r, t)]dt + σ (r)dZ + Pt (r, t, T )dt
2
1
˜
=
σ (r)2 Prr (r, t, T ) + [a(r) + σ (r)φ(r, t)]Pr (r, t, T ) + Pt (r, t, T ) dt + σ (r)Pr dZ
2
1 See Section 84. 75 ZEROCOUPON BOND PRICING: RISKNEUTRAL PROCESS 483 The expected change of price per unit time is
1
E ∗ (dP )
= σ (r)2 Prr (r, t, T ) + [a(r) + σ (r)φ(r, t)]Pr (r, t, T ) + Pt (r, t, T ).
dt
2
This and equation (74.1) imply
E ∗ (dP )
= rP.
dt
This says that under the riskneutral probability distribution measure, a bond has an expected
return that is equal to the riskfree rate of return. The approximation is exact if the interest rate
moves one standard deviation.
Example 75.2
Find the diﬀerential of U (t) = σ t a(s−t)
e
0 r(s)dZ (s). Solution.
We have
t dU (t) =d σ e−at eas r(s)dZ (s) 0
t
−at =d[σe t e as −at r(s)dZ (s) + σe eas d r(s)dZ (s) 0 0
t = − aσe−at dt eas r(s)dZ (s) + σe−at eat eas r(s)dZ (s) + σ r(t)dZ (t) 0
t = − aσe−at dt r(t)dZ (t) 0 Example 75.3
Suppose that the shortterm interest rate is given by
t
−at r(t) = r(0)e + b(1 − e −at ea(s−t) )+σ
0 Find an expression for dr(t). r(s)dZ (s). 484 INTEREST RATES MODELS Solution.
We have
t
−at dr(t) =[−r(0) + b]ae −at dt − aσe eas r(s)dZ (s) + σ eas dt r(s)dZ (s) dt + σ r(t)dZ (t) 0
t = − a r(0)e−at − be−at + σe−at r(t)dZ (t) 0
t = − a r(0)e−at + b(1 − e−at ) + σe−at eas
0 = − ar(t)dt + abdt + σ
=a(b − r(t))dt + σ r(t)dZ (t) r(t)dZ (t) r(s)dZ (s) dt + abdt + σ r(t)dZ (t) 75 ZEROCOUPON BOND PRICING: RISKNEUTRAL PROCESS 485 Practice Problems
Problem 75.1
The realistic process for the shortterm interest rate is given by
˜
dr = 0.3(0.12 − r)dt + 0.03dZ.
For t ≤ T, let P (r, t, T ) be the price at time t of a zerocoupon bond that pays $1 at time T, if the
shortrate at time t is r. The Sharpe ratio of this bond is 0.05. Find the riskneutral process of the
shortterm interest rate.
Problem 75.2
The riskneutral process of a zerocoupon bond is given by
dP (0.08, 0, 10)
˜
= 0.08dt − 0.095dZ.
P (0.08, 0, 10)
Find the Sharpe ratio φ∗ (0.08, 0) of a zerocoupon bond under the riskneutral distribution.
Problem 75.3
˜
Suppose that the Sharpe ratio of a zerocoupon bond is φ(r, t) = 0.05. Let Z (t) be the transformation
˜
deﬁned in this section. Find Z (4) if Z (4) = −1.4766.
Problem 75.4
Suppose that r(t) = (br(0) − σ )e−at with r(0) =
satisﬁed by r. σ
,
b−1 Problem 75.5
Suppose that r(t) = (br(0) − σ )e−at with r(0) = b > 1. Find the stochastic diﬀerential equation σ
,
b−1 b > 1. Find an expression for R(t, T ). Problem 75.6
The riskneutral process for a zerocoupon bond is given by
dP (r, t, T )
˜
= 0.08dt − 0.095dZ.
P (r, t, T )
Determine E ∗ (dP )
.
P Problem 75.7
The riskneutral process for a zerocoupon bond is given by
dP (r, t, T )
˜
= rdt − 0.0453dZ.
P (r, t, T )
Determine r if E ∗ (dP )
P = 0.07dt. 486 INTEREST RATES MODELS 76 The RendlemanBartter ShortTerm Model
In this and the next two sections, we discuss several bond pricing models based on equation (74.1),
in which all bond prices are driven by the shortterm interest rate r with
dr = a(r)dt + σ (r)dZ.
In this section, we look at the RendlemanBartter model. The RendlemanBartter model in
ﬁnance is a short rate model describing the evolution of interest rates. It is a type of “one factor
model” as it describes interest rate movements as driven by only one source of market risk.
We ﬁrst start by looking at shortterm interest rate model that follows the arithmetic Brownian
motion
dr = adt + σdZ.
This says that the shortrate is normally distributed with mean r(0) + at and variance σ 2 t. There
are several problems with this model, namely:
• The shortrate can assume negative values which is not realistic in practice.
• The process is not meanreverting because the drift term a is constant. For example, if a > 0,
the shortrate will drift up over time forever. In practice, shortrate exhibits meanreversion1 .
• The volatility σ of the shortrate is constant regardless of the rate. In practice, the shortrate is
more volatile if rates are high.
To overcome the above mentioned obstacles, the RendlemanBartter model replaces the arithmetic
motion by a standard geometric Brownian motion, that is, the instantaneous interest rate follows
a geometric Brownian motion:
dr = ardt + σrdZ
where the drift parameter, a, represents a constant expected instantaneous rate of change in the
interest rate, while the standard deviation parameter, σ, determines the volatility of the interest
rate.
The process for r is of the same type as that assumed for a stock price in Section 48. That is, the
natural logarithm of the rate follows an arithmetic Brownian motion2 :
ln r(t)
r(0) = (a − 0.5σ 2 )t + σZ (t) and solving for r(t) we ﬁnd
r(t) = r(0)e(a−0.5σ
1 2 )t+σZ (t) . When rates are high, the economy tends to slow down and borrowers require less funds. As a result rates decline.
When rates are low, there tends to be a high demand for funds on the part of the borrowers and rates tend to rise
2
See Problems 76.3  76.4 76 THE RENDLEMANBARTTER SHORTTERM MODEL 487 This shows that the rate can never be negative. Also, the volatility increases with the shortterm
rate. This follows from the fact that the variance of the shortterm rate over a small increment of
time is
Var(r(t + dt)r(t)) = r2 σ 2 dt.
Unlike stock prices, the only disadvantage that was observed with this model is that it does not
incorporate meanreversion.
Example 76.1
The RendlemanBartter onefactor interest rate model with the shortterm rate is given by the
process
dr = 0.001rdt + 0.01rdZ (t).
Suppose that the relevant Sharpe ratio is φ(r) = 0.88. Find the process satisﬁed by r under the
riskneutral probability measure.
Solution.
We have
˜
˜
dr = [a(r) + φ(r)σ (r)]dt + σ (r)dZ (t) = (0.001r + 0.88 × 0.01r)dt + 0.01dZ (t)
Example 76.2
The RendlemanBartter onefactor interest rate model with the shortterm rate is given by the
process
dr = 0.001rdt + 0.01rdZ (t).
Determine Var(r(t + h)r(t)).
Solution.
We have Var(r(t + h)r(t)) = r2 σ 2 h = 0.012 r2 h 488 INTEREST RATES MODELS Practice Problems
Problem 76.1
Consider the RendlemanBartter model dr(t) = ar(t)dt + σr(t)dZ (t). Find an expression of r(t) in
integral form.
Problem 76.2
Consider the RendlemanBartter model dr(t) = ar(t)dt + σr(t)dZ (t). Show that
d[ln r(t)] = (a − 0.5σ 2 )dt + σdZ (t).
Hint: Apply Itˆ’s lemma to the function f (r, t) = ln [r(t)].
o
Problem 76.3
Consider the RendlemanBartter model dr(t) = ar(t)dt + σr(t)dZ (t). Show that
ln [r(t)] = ln [r(0)] + (a − 0.5σ 2 )t + σZ (t).
Problem 76.4
Consider the RendlemanBartter model dr(t) = ar(t)dt + σr(t)dZ (t). Show that
r(t) = r(0)e(a−0.5σ 2 )t+σZ (t) . Problem 76.5
Consider the RendlemanBartter model dr(t) = ar(t)dt + σr(t)dZ (t). Show that for t > s we have
r(t) = r(s)e(a−0.5σ 2 )(t−s)+σ (Z (t)−Z (s)) . Problem 76.6
Consider the RendlemanBartter model dr(t) = ar(t)dt + σr(t)dZ (t). Determine a formula for
E [r(t)r(s)], t > s. See Section 47.
Problem 76.7
Consider the RendlemanBartter model dr(t) = ar(t)dt + σr(t)dZ (t). Determine a formula for
Var[r(t)r(s)], t > s. See Section 47. 77 THE VASICEK SHORTTERM MODEL 489 77 The Vasicek ShortTerm Model
The Vasicek model for the shortterm interest rate is given by the process
dr = a(b − r)dt + σdZ (t).
Thus, the interest rate follows an OrnsteinUhlenbeck process (See equation (59.2)). The term
a(b − r)dt induces meanreversion. The parameter b is the level to which the short term revert. If
r > b, the shortterm rate is expected to decrease toward b. If r < b, the shortterm rate is expected
to increase toward b. The parameter a is a positive number that reﬂects the speed of the reversion
to b. The higher a is, the faster the reversion is.
The Vasicek model is meanreverting, but it exhibits the two problems of the arithmetic model
discussed at the beginning of Section 76, namely:
• Interest rates in this model can become negative.
• The volatility of the shortrate is the same whether the rate is high or low.
When solving for the zerocoupon bond price, it is assumed that the Sharpe ratio for interest
rate risk is some constant φ. In this case, we have the following diﬀerential equation
12
σ Prr + [a(b − r) + σφ]Pr + Pt − rP = 0.
2
For a = 0, the solution to this diﬀerential equation subject to the condition P (r, T, T ) = $1 is given
by
P (r, t, T ) = A(t, T )e−B (t,T )r(r)
where
A(t, T ) =er(B (t,T )+t−T )− B 2 (t,T )σ 2
4a 1 − e−a(T −t)
a
σφ 0.5σ 2
r =b +
−
a
a2
where r is the yield to maturity on an inﬁnitely lived bond, that is the value the yield will approach
as T goes to inﬁnity.
For a = 0, we have
B (t, T ) = B (t, T ) =T − t
2 A(t, T ) =e0.5σ(T −t) φ+
r =undeﬁned σ 2 (T −t)3
6 490 INTEREST RATES MODELS Example 77.1
Show that A(0, T − t) = A(t, T ) and B (0, T − t) = B (t, T ).
Solution.
For a = 0, we have
B (0, T − t) = 1 − e−a(T −t)
1 − e−a(T −t−0)
=
= B (t, T )
a
a and
A(0, T − t) = erB (0,T −t)+0−(T −t))− B 2 (0,T −t)σ 2
4a = er(B (t,T )+t−T )− B 2 (t,T )σ 2
4a = A(t, T ). Likewise, we can establish the results for the case a = 0
Example 77.2
A Vasicek model for a shortrate is given by
dr = 0.2(0.1 − r)dt + 0.02dZ (t).
The Sharpe ratio for interest rate risk is φ = 0. Determine the yield to maturity on an inﬁnitely
lived bond.
Solution.
We are given: a = 0.2, b = 0.1, σ = 0.02, and φ = 0. The yield to maturity on an inﬁnitely lived
bond is
σφ 0.5σ 2
0.02 0.5 × 0.022
r =b+
−
= 0.1 + 0 ×
−
= 0.095
a
a2
0.2
0.22
Example 77.3
A Vasicek model for a shortrate is given by
dr = 0.2(0.1 − r)dt + 0.02dZ (t).
The Sharpe ratio for interest rate risk is φ = 0. Determine the value of B (0, 10).
Solution.
We use the formula B (t, T ) =
φ = 0. Thus, 1−e−a(T −t)
.
a We are given: a = 0.2, b = 0.1, σ = 0.02, t = 0, T = 10 and 1 − e−0.2(10−0)
= 4.323
0.2
Now, the price P (r, t, T ) of a zerocoupon bond in the Vasicek model follows an Itˆ process
o
B (1, 2) = dP (r, t, T )
= α(r, t, T ) − q (r, t, T ).
P (r, t, T ) 77 THE VASICEK SHORTTERM MODEL 491 Example 77.4
Show that q (r, t, T ) = σB (t, T ).
Solution.
We have
Pr (r, t, T )
∂P (r, t, T )
= −σ
P (r, t, T )
∂r
= − σ × −B (t, T ) = σB (t, T ) q (r, t, T ) = − σ Example 77.5
Show that α(r, t, T ) = r + B (t, T )σφ.
Solution.
Since the Sharpe ratio for interest rate risk in the Vasicek model is given by a constant φ, we can
write
α(r, t, T ) − r
α(r, t, T ) − r
=
.
φ=
q (r, t, T )
σB (t, T )
From this, it follows that
α(r, t, T ) = r + B (t, T )σφ
Example 77.6 ‡
For a Vasicek bond model you are given the following information:
• a = 0.15
• b = 0.1
• r = 0.05
• σ = 0.05
Calculate the expected change in the interest rate, expressed as an annual rate.
Solution.
The given Vasicek model is described by the process
dr = 0.15(0.10 − r)dt + 0.05dZ (t).
The expected change in the interest rate is
E (dr) = 0.15(0.10 − r)dt = (0.015 − 0.15r)dt.
Since r = 0.05 we obtain
E (dr) = (0.015 − 0.15 × 0.05)dt = 0.0075dt.
The expected change in the interest rate, expressed as an annual rate, is then
E (dr)
= 0.0075
dt 492 INTEREST RATES MODELS Practice Problems
Problem 77.1
A Vasicek model for a shortrate is given by
dr = 0.34(0.09 − r)dt + 0.26dZ (t).
The Sharpe ratio for interest rate risk is φ = 0.88. Determine the yield to maturity on an inﬁnitely
lived bond.
Problem 77.2
A Vasicek model for a shortrate is given by
dr = 0.34(0.09 − r)dt + 0.26dZ (t).
The Sharpe ratio for interest rate risk is φ = 0.88. Determine the value of B (2, 3).
Problem 77.3
A Vasicek model for a shortrate is given by
dr = 0.2(0.1 − r)dt + 0.02dZ (t).
The Sharpe ratio for interest rate risk is φ = 0. Determine the value of A(0, 10).
Problem 77.4
A Vasicek model for a shortrate is given by
dr = 0.2(0.1 − r)dt + 0.02dZ (t).
The Sharpe ratio for interest rate risk is φ = 0. Determine the price of a zerocoupon bond with
par value of $100 and that matures in 10 years if the riskfree annual interest rate is 8%.
Problem 77.5 ‡
You are using the Vasicek onefactor interestrate model with the shortrate process calibrated as
dr(t) = 0.6[b − r(t)]dt + σdZ (t).
For t ≤ T, let P (r, t, T ) be the price at time t of a zerocoupon bond that pays $1 at time T, if the
shortrate at time t is r. The price of each zerocoupon bond in the Vasicek model follows an Itˆ
o
process,
dP (r, t, T )
= α(r, t, T )dt − q (r, t, T )dZ (t), t ≤ T.
P (r, t, T )
You are given that α(0.04, 0, 2) = 0.04139761. Find α(0.05, 1, 4). 77 THE VASICEK SHORTTERM MODEL 493 Problem 77.6 ‡
You are given:
(i) The true stochastic process of the shortrate is given by
dr(t) = [0.09 − 0.5r(t)]dt + 0.3dZ (t),
where {Z (t)} is a standard Brownian motion under the true probability measure.
(ii) The riskneutral process of the shortrate is given by
˜
dr(t) = [0.15 − 0.5r(t)]dt + σ (r(t))dZ (t),
˜
where {Z (t)}is a standard Brownian motion under the riskneutral probability measure.
(iii) g (r, t) denotes the price of an interestrate derivative at time t, if the shortrate at that time is
r. The interestrate derivative does not pay any dividend or interest.
(iv) g (r(t), t) satisﬁes
dg (r(t), t) = µ(r(t), g (r(t), t))dt − 0.4g (r(t), t)dZ (t).
Determine µ(r, g ).
Problem 77.7 ‡
You are given:
(i) The true stochastic process of the shortrate is given by
dr(t) = [0.008 − 0.1r(t)]dt + 0.05dZ (t),
where {Z (t)} is a standard Brownian motion under the true probability measure.
(ii) The riskneutral process of the shortrate is given by
˜
dr(t) = [0.013 − 0.1r(t)]dt + 0.05dZ (t),
˜
where {Z (t)}is a standard Brownian motion under the riskneutral probability measure.
(iii) For t ≤ T, let P (r, t, T ) be the price at time t of a zerocoupon bond that pays $1 at time T, if
the shortrate at time t is r. The price of each zerocoupon bond follows an Itˆ process,
o
dP (r, t, T )
= α(r, t, T )dt − q (r, t, T )dZ (t), t ≤ T.
P (r, t, T )
Determine α(0.04, 2, 5). 494 INTEREST RATES MODELS Problem 77.8 ‡
Let P (r, t, T ) denote the price at time t of $1 to be paid with certainty at time T, t ≤ T, if the
short rate at time t is equal to r. For a Vasicek model you are given: P (0.04, 0, 2) =0.9445
P (0.05, 1, 3) =0.9321
P (r∗ , 2, 4) =0.8960
Calculate r∗ . 78 THE COXINGERSOLLROSS SHORTTERM MODEL 495 78 The CoxIngersollRoss ShortTerm Model
The CoxIngersollRoss (CIR) bond pricing model assumes that the risk natural process for the
short interest rate is given by
√
dr = a(b − r)dt + σ rdZ.
Note that the variance of the shortterm rate over a small increment of time is
Var(r(t + dt)r(t)) = Var(dr) = σ 2 rdt.
This says that the standard deviation of the shortterm rate is proportional to the square root of r.
Thus, as r increases the volatility increases. Like the Vasicek model, the drift term a(b − r) induces
mean reversion. Also, it should be noted that the interest rate in this model can never become
negative. For example, √ r = 0 in the CIR model, the drift factor a(b − r) becomes ab > 0, while
if
the volatility factor is σ 0 = 0, so the interest rate will increase.
The Sharpe ratio in the CIR model takes the form
√
r
φ(r, t) = φ
σ
and the bond price satisﬁes the diﬀerential equation
∂P
∂P
1 2 ∂ 2P
σ r 2 + [a(b − r) + rφ]
+
− rP = 0.
2
∂r
∂r
∂t
The solution to this equation that satisﬁes the condition P (r, T, T ) = $1 is given by
P (r, t, T ) = A(t, T )e−B (t,T )r(t)
where
2γe(a+φ+γ )(T −t)/2
A(t, T ) =
(a + φ + γ )(eγ (T −t) − 1) + 2γ
B (t, T ) = 2ab
σ2 2(eγ (T −t) − 1)
(a + φ + γ )(eγ (T −t) − 1) + 2γ γ = (a + φ)2 + 2σ 2
As in the Vasicek model, the terms A(t, T ) and B (t, T ) are independent of the shortterm rate. The
yield to maturity of a long lived bond in the CIR model is given by
r = lim T →∞ − ln [P (r, t, T )]
2ab
=
.
T −t
a+φ+γ 496 INTEREST RATES MODELS From the process of the bond price
dP (r, t, T )
= α(r, t, T )dt − q (r, t, T )dZ (t)
P (r, t, T )
we ﬁnd
q (r, t, T ) = −σ (r) √
√
Pr (r, t, T )
1
=−
[−B (t, T )]P (r, t, T )σ r = B (t, T )σ r
P (r, t, T )
P (r, t, T ) and
α(r, t, T ) − r
φ(r, t) =
q (r, t, T )
√
r α(r, t, T ) − r
√
φ
=
σ
B (t, T )σ r
α(r, t, T ) =r + rφB (t, T )
Example 78.1
√
A CIR shortrate model is described by the process dr = 0.18(0.1 − r)dt + 0.17 rdZ. Given that
the Sharpe ratio is zero. Find φ.
Solution.
√
We have 0 = φ(r, t) = φ σr . This implies that φ = 0
Example 78.2
√
A CIR shortrate model is described by the process dr = 0.18(0.1 − r)dt + 0.17 rdZ. Given that
the Sharpe ratio is zero. Determine the value of γ.
Solution.
We have a = 0.18, b = 0.10, σ = 0.17, and φ = 0. Thus,
γ= (a + φ)2 + 2σ 2 = 0.18 + 0 + 2(0.17)2 = 0.30033 Example 78.3
√
A CIR shortrate model is described by the process dr = 0.18(0.1 − r)dt + 0.17 rdZ. Given that
the Sharpe ratio is zero. Determine the value of B (0, 5). 78 THE COXINGERSOLLROSS SHORTTERM MODEL 497 Solution.
We have a = 0.18, b = 0.10, σ = 0.17, φ = 0, γ = 0.30033, t = 0, and T = 5. Substituting these
values into the formula
2(eγ (T −t) − 1)
B (t, T ) =
(a + φ + γ )(eγ (T −t) − 1) + 2γ
we ﬁnd B (0, 5) = 3.06520
Example 78.4
√
A CIR shortrate model is described by the process dr = 0.18(0.1 − r)dt + 0.17 rdZ. Given that
the Sharpe ratio is zero. Determine the value of A(0, 5).
Solution.
We have a = 0.18, b = 0.10, σ = 0.17, φ = 0, γ = 0.30033, t = 0, and T = 5. Substituting these
values into the formula
(a+φ+γ )(T −t)/2 A(t, T ) = 2γe
(a + φ + γ )(eγ (T −t) − 1) + 2γ 2ab
σ2 we ﬁnd A(0, 5) = 0.84882
Example 78.5
√
A CIR shortrate model is described by the process dr = 0.18(0.1 − r)dt + 0.17 rdZ. Given that
the Sharpe ratio is zero. Determine the price of a zerocoupon bond that pays $ in ﬁve years if the
riskfree interest rate is assumed to be 8%. What is the yield on the bond?
Solution.
We have
P (0.08, 0, 5) = A(0, 5)e−B (0,5)(0.08) = 0.84882e−3.06520×0.08 = $0.66424.
The yield on the bond is
yield = − ln [P (0.08, 0, 5)0
= 0.08182
5 Example 78.6
What is the yield on a long lived zero √
coupon bond under the CIR shortrate model described by
the process dr = 0.18(0.1 − r)dt + 0.17 rdZ with zero Sharpe ratio? 498 INTEREST RATES MODELS Solution.
The yield is
2ab
2(0.18)(0.10)
=
= 0.075
a+σ+γ
0.18 + 0 + 0.30033
Vasicek Model Versus CIR Model
We next compare the various features of Vasicek and CIR models:
• In the Vasicek model, interest rates can be negative. In the CIR model, negative interest rates are
impossible. As one’s time horizon T increases, the likelihood of interest rates becoming negative in
the Vasicek model greatly increases as well.
• In the Vasicek model, the volatility of the shortterm interest rate is constant. In the CIR model,
the volatility of the shortterm interest rate increases as the shortterm interest rate increases.
• The shortterm interest rate in both models exhibit mean reversion.
• In both the Vasicek and the CIR model, the delta and gamma Greeks for a zerocoupon bond are
based on the change in the shortterm interest rate.
• With a relatively high volatility (lower panel of Figure 78.1), the CIR yields tend to be higher
than the Vasicek yields. This occur because the Vasicek yields can be negative.
• With a relatively low volatility (upper panel of Figure 78.1), the meanreversion eﬀect outweighs
the volatility eﬀect, and the Vasicek yields tend to be a bit higher than the CIR yields. Also, both
models produce upward sloping yield curves.
r= An Exogenously Prescribed Time Zero Yield Curve
An exogenous prescription of the timezero yield curve in a ﬁnancial model would mean that
you know−empirically, or from some source external to the model−the data regarding the yields
to maturity for many diﬀerent time horizons, where currently, t = 0. Then you would be able to
put that data into the model and get consistent results as well as the ability to predict yields to
maturity for other time horizons. A model based on m parameters can only be used consistently
with exogenously prescribed data consisting of m − 1 or fewer data points. To elborate, suppose
your model is described by the twoparameter equation x + y = 2. If we are just given x, we can
solve for y consistently with the model. If we are given both x and y, however, we will not always
be able to do so. For instance, if x = 3 and y = 4, there is no way for x + y to equal 5. If you have
even more externally prescribed data, the likelihood of the model working decreases even further.
The Vasicek and CIR have each exactly four inputs: a, b, σ, and r. A yield curve can consist of
many more than four yields which means that the two models don’t have enough inputs to fully
describe an existing yield curve. Another way to say this is that the time zero yield curve cannot
be exagenously prescribed using either the Vasicek or CIR model.
Example 78.7
For which of these exogenously prescribed data sets regarding yields to maturity for various time 78 THE COXINGERSOLLROSS SHORTTERM MODEL 499 horizons can the Vasicek and CIR models be used? More than one answer may be correct. Each
data has the form [t, T, r].
Set A: [0, 1, 0.22], [0, 2, 0.26], [0, 3, 0.28]
Set B: [0, 0.5, 0.04], [0, 5, 0.05]
Set C: [0, 1, 0.12], [0, 2, 0.09], [0, 3, 0.18], [0, 6, 0.21], [0, 7, 0.03]
Set D: [0, 1, 0.11], [0, 2, 31], [0, 7, 0.34], [0, 8, 0.54].
Solution.
The timezero yield curve for an interest rate model can only be interpreted consistently with the
model (in most cases) when the number of data points in the timezero yield curve is less than the
number of parameters in the model. The Vasicek and CIR models each have 4 parameters: a, b, σ,
and r. So only data sets with 3 points or fewer can have the models consistently applied to them.
Thus, only Sets A and B, with 3 and 2 data points respectively, can have the models applied to
them Figure 78.1 500 INTEREST RATES MODELS Practice Problems
Problem 78.1
State two shortcomings of the Vasicek model that the CIR does not have.
Problem 78.2
Which of the following statements are true? More than one answer may be correct.
(a) The Vasicek model incorporates mean reversion.
(b) The CIR model incorporates mean reversion.
(c) In the Vasicek model, interest rates can be negative.
(d) In the CIR model, interest rates can be negative.
(e) In the Vasicek model, volatility is a function of the interest rate.
(f) In the CIR model, volatility is a function of the interest rate.
Problem 78.3
Assume the CIR model holds. A particular interest rate follows this Brownian motion:
√
dr = 0.22(0.06 − r)dt + 0.443 rdZ.
At some particular time t, r = 0.11. Then, r suddenly becomes 0.02. What is the resulting change
in the volatility?
Problem 78.4
Assume that the CIR model holds. When a particular interest rate is 0, the Brownian motion it
follows is dr = 0.55dt.
You know that a + b = 1.7 and a > 1. What is the drift factor of the Brownian motion in this model
when r = 0.05?
Problem 78.5
Find the riskneutral process that corresponds to the CIR model
√
dr = a(b − r)dt + σ rdZ.
Problem 78.6
The realistic process of the CIR model is given by
√
dr = a(0.08 − r)dt + 0.04 rdZ
and the riskneutral process is given by
√
dr = 0.2(0.096 − r)dt + 0.04 rdZ.
Determine the values of a and φ. 78 THE COXINGERSOLLROSS SHORTTERM MODEL 501 Problem 78.7
The CIR shortrate process is given by
√
dr = 0.0192(0.08 − r)dt + 0.04 rdZ
Determine the value of φ(0.07, t).
Problem 78.8
√
A CIR shortrate model is described by the process dr = 0.18(0.1 − r)dt + 0.17 rdZ. Given that
the Sharpe ratio is zero. Find the delta and the gamma of a zerocoupon bond that pays $1 in ﬁve
years under this model if the riskfree interest rate is assumed to be 0.08.
Problem 78.9
√
A CIR shortrate model is described by the process dr = 0.18(0.1 − r)dt + 0.17 rdZ. Given that
the Sharpe ratio is zero. Find the theta of a zerocoupon bond that pays $1 in ﬁve years under this
model if the riskfree interest rate is assumed to be 0.08.
Problem 78.10 ‡
The CoxIngersollRoss (CIR) interestrate model has the shortrate process:
dr(t) = a[b − r(t)]dt + σ r(t)dZ (t), where {Z (t)} is a standard Brownian motion.
For t ≤ T, let P (r, t, T ) be the price at time t of a zerocoupon bond that pays $1 at time T, if the
shortrate at time t is r. The price of each zerocoupon bond in the Vasicek model follows an Itˆ
o
process,
dP (r, t, T )
= α(r, t, T )dt − q (r, t, T )dZ (t), t ≤ T.
P (r, t, T )
You are given that α(0.05, 7, 9) = 0.06. Find α(0.04, 11, 13).
Problem 78.11 ‡
The shortrate process {r(t)} in a CoxIngersollRoss model follows
dr(t) = [0.011 − 0.1r(t)]dt + 0.08 r(t)dZ (t),
where {Z (t)} is a standard Brownian motion under the true probability measure.
For t ≤ T, let P (r, t, T ) denote the price at time t of a zerocoupon bond that pays $1 at time T, if
the shortrate at time t is r.
You are given:
√
(i) The Sharpe ratio takes the form φ(r, t) = c r.
r,
(ii) limT →∞ ln [P (T 0,T )] = −0.1 for each r > 0.
Find the constant c. 502 INTEREST RATES MODELS 79 The Black Formula for Pricing Options on Bonds
In this section, we discuss the Black model, a variant of the BlackScholes pricing model, for pricing
zerocoupon bond options.
Consider a s−year zerocoupon bond. When this bond is purchased at time T then it will pay $1
at time T + s. Consider a forward contract signed at time t ≤ T that calls for time−T delivery of
the bond maturing at time T + s. Let Pt (T, T + s) be the forward price of this contract. That is,
Pt (T, T + s) is the time−t price agreed upon to be paid at time T. Let P (T, T + s) = PT (T, T + s)
be the (spot) price of the bond at time T. For t < T, we let Ft,T [P (T, T + s)] = Pt (T, T + s) be the
forward price at time t for an agreement to buy a bond at time T that pays $ at tim T + s. At time
t, if you want $1 at time T + s, there are two ways to achieve this:
1. You may buy a bond immediately maturing for $1 at time T + s. This would cost P (t, T + s).
2. You can enter into a forward agreement to buy a bond at time T maturing for $1 at time T + s.
At time T, you would pay Ft,T [P (T, T + s)].
Therefore, discounting Ft,T [P (T, T + s)] to time t, we must have
Ft,T [P (T, T + s)]P (t, T ) = P (t, T + s)
or
Ft,T [P (T, T + s)] = P (t, T + s)
P (t, T ) where P (t, T ) is the timet price of a bond that pays $1 at time T.
Example 79.1
The time0 price of a 2year zerocoupon bond with par value $1 is $0.8495. The time0 price of a
3year zerocoupon bond with par value $1 is $0.7722. At current time t = 0, what is the forward
price for a 1year bond in year 2?
Solution.
We are asked to ﬁnd P0 (T, T + s) = F0,T [P (T, T + s)] = P (0,T +s)
P (0,T ) with T = 2 and s = 1. That is, P (0, 3)
0.7722
=
= $0.90901
P (0, 2)
0.8495
Now, consider a European call option with strike K, expiration date T, and underlying asset the
forward price of a s−year zerocoupon bond. The payoﬀ of this option at time T is
Call option payoﬀ = max{0, P (T, T + s) − K }. 79 THE BLACK FORMULA FOR PRICING OPTIONS ON BONDS 503 Under the assumption that the bond forward price is lognormally distributed with constant volatility
σ, the Black formula for such a European call option is given by
C = P (0, T )[F N (d1 ) − KN (d2 )],
where
d1 =
d2 = ln (F/K ) + (σ 2 /2)T
√
σT √
ln (F/K ) − (σ 2 /2)T
√
= d1 − σ T
σT and where F = F0,T [P (T, T + s)].
The price of an otherwise equivalent European put option on the bond is
P = P (0, T )[KN (−d2 ) − F N (−d1 )].
Since P (0, T )F = P (0, T + s), the Black formula simply uses the time0 price of a zerocoupon bond
that matures at time T + s as the underlying asset.
Example 79.2
The time0 price of a 2year zerocoupon bond with par value $1 is $0.8495. The time0 price of a
3year zerocoupon bond with par value $1 is $0.7722. What is the price of a European call option
that expires in 2 years, giving you the right to pay $0.90 to buy a bond expiring in 1 year? The
1year forward price volatility is 0.105.
Solution.
We have
F=
d1 = P (0, 3)
= 0.90901
P (0, 2) ln (F/K ) + (σ 2 /2)T
ln (0.90901/0.90) + 0.5(0.105)2 (2)
√
√
=
= 0.14133
0.105 2
σT
√
√
d2 = d1 − σ T = 0.14133 − 0.105 2 = −0.00716
N (d1 ) = 0.556195, N (d2 ) = 0.497144. Thus,
C = 0.8495[0.90901 × 0.556195 − 0.90 × 0.497144] = $0.0494 504 INTEREST RATES MODELS Example 79.3
The time0 price of a 2year zerocoupon bond with par value $1 is $0.8495. The time0 price of a
3year zerocoupon bond with par value $1 is $0.7722. What is the price of a European put option
that expires in 2 years, giving you the right of selling a bond expiring in 1 year for the price of
$0.90? The 1year forward price volatility is 0.105.
Solution.
By the Black model, the price is
P = P (0, T )[KN (−d2 )−F N (−d1 )] = 0.8495[0.90(1−0.497144)−0.90901(1−0.556195)] = $0.04175 79 THE BLACK FORMULA FOR PRICING OPTIONS ON BONDS 505 Practice Problems
Problem 79.1
The time0 price of a 1year zerocoupon bond with par value $1 is $0.9259. The time0 price of a
2year zerocoupon bond with par value $1 is $0.8495. At current time t = 0, what is the forward
price of a 1year bond in year 1?
Problem 79.2
The time0 price of a 1year zerocoupon bond with par value $1 is $0.9259. The time0 price of a
2year zerocoupon bond with par value $1 is $0.8495. What is the price of a European call option
that expires in 1 year, giving you the right to pay $0.9009 to buy a bond expiring in 1 year? The
1year forward price volatility is 0.10.
Problem 79.3
The time0 price of a 1year zerocoupon bond with par value $1 is $0.9259. The time0 price of a
2year zerocoupon bond with par value $1 is $0.8495. What is the price of a European put option
that expires in 1 year, giving you the right of selling a bond expiring in 1 year for the price of
$0.9009? The 1year forward price volatility is 0.10.
Problem 79.4
The time0 price of a 2year zerocoupon bond with par value $1 is $0.8853. The time0 price of
a 5year zerocoupon bond with par value $1 is $0.6657. A European call option that expires in
2 years enables you to purchase a 3year bond at expiration at a purchase price of $0.7799. The
forward price of the bond is lognormally distributed and has a volatility of 0.33. Find the value of
F in the Black formula for the price of this call option.
Problem 79.5
The time0 price of a 2year zerocoupon bond with par value $1 is $0.8853. The time0 price of
a 5year zerocoupon bond with par value $1 is $0.6657. A European call option that expires in
2 years enables you to purchase a 3year bond at expiration at a purchase price of $0.7799. The
forward price of the bond is lognormally distributed and has a volatility of 0.33. Find the value of
d1 in the Black formula for the price of this call option.
Problem 79.6
The time0 price of a 2year zerocoupon bond with par value $1 is $0.8853. The time0 price of
a 5year zerocoupon bond with par value $1 is $0.6657. A European call option that expires in
2 years enables you to purchase a 3year bond at expiration at a purchase price of $0.7799. The
forward price of the bond is lognormally distributed and has a volatility of 0.33. Find the value of
d2 in the Black formula for the price of this call option. 506 INTEREST RATES MODELS Problem 79.7
The time0 price of a 2year zerocoupon bond with par value $1 is $0.8853. The time0 price of
a 5year zerocoupon bond with par value $1 is $0.6657. A European call option that expires in
2 years enables you to purchase a 3year bond at expiration at a purchase price of $0.7799. The
forward price of the bond is lognormally distributed and has a volatility of 0.33. Assume the Black
framework, ﬁnd the price of this call option.
Problem 79.8
The time0 price of a 2year zerocoupon bond with par value $1 is $0.8853. The time0 price of
a 5year zerocoupon bond with par value $1 is $0.6657. A European put option that expires in 2
years enables you to sell a 3year bond at expiration for the price of $0.7799. The forward price of
the bond is lognormally distributed and has a volatility of 0.33. Find the price of this put option. 80 THE BLACK FORMULA FOR PRICING FRA, CAPLETS, AND CAPS 507 80 The Black Formula for Pricing FRA, Caplets, and Caps
In this section we can extend the Black formula to price options on interest rates. More speciﬁcally,
we will use the Black model to price forward rate agreement, interest rate caplets and caps.
Forward Rate Agreement
A forward rate agreement (FRA) is an agreement to pay or receive the diﬀerence between the
prevailing rate at a certain time and a ﬁxed rate R, times some notional amount. For example,
suppose that ABC Inc. enters into an FRA with MBF in which Company ABC will receive a ﬁxed
rate of 4% for one year on a principal of $1 million in three years. In return, Company MBF will
receive the oneyear LIBOR rate, determined in three years’ time, on the principal amount. The
agreement will be settled in cash in three years.
If, after three years’ time, the LIBOR is at 5.5%, the settlement to the agreement will require that
ABC Inc. pay MBF. This is because the LIBOR is higher than the ﬁxed rate. Mathematically, $1
million at 5% generates $50,000 of interest for ABC Inc. while $1 million at 5.5% generates $55,000
in interest for MBF. Ignoring present values, the net diﬀerence between the two amounts is $5,000,
which is paid to Company MBF.
Now, if after three years’ time, the LIBOR is at 4.5%, the settlement to the agreement will require
that MBF pay ABC Inc. This is because the LIBOR is lower than the ﬁxed rate. The amount
$5,000 is paid to Company ABC Inc.
Consider a borrower who wishes to have a loan of $1 at time T that matures at time T + s. To hedge
against interest rate ﬂuctuation, the borrower enters into a forward rate agreement (FRA) that
allows the borrowers to lockin an interest rate, the forward interest rate. Thus, at maturity time
T + s of the FRA, the FRA payoﬀ is given by
Payoﬀ to FRA = RT (T, T + s) − R0 (T, T + s)
where RT (T, T + s) is the spot s−period from time T to time T + s (ot the time−T forward rate
from time T to time T + s) and R0 (T, T + s) is the time0 forward interest rate from time T
to time T + s. If RT (T, T + s) > R0 (T, T + s), the borrower pays the lender the diﬀerence. If
RT (T, T + s) < R0 (T, T + s), the lender pays the borrower the diﬀerence.
But1 we know that
P (t, T )
−1
Rt (T, T + s) =
P (t, T + s)
so that
R0 (T, T + s) =
1 See Page 218 of [1]. P (0, T )
−1
P (0, T + s) 508 INTEREST RATES MODELS and
RT (T, T + s) = P (T, T )
1
−1=
−1
P (T, T + s)
P (T, T + s) where P (T, T ) = $1 (since a bond paying $1 at expiration will have a value of $1 right when it
expires) so that the payoﬀ of the FRA at time T + s takes the form
Payoﬀ to FRA = 1
P (0, T )
−
.
P (T, T + s) P (0, T + s) We should point out here that R0 (T, T + s) is nonannualized: If you invest $1 at time T, you will
have [1 + R0 (T, T + s)] after s time periods elapse.
Example 80.1
The time0 price of a 6year zerocoupon bond that pays $1 at expiration is $0.6464. The time0 price of a 8year zerocoupon bond that pays $1 at expiration is $0.5461. What is the time0
nonannualized forward interest rate from time t = 6 years to time t = 8 years?
Solution.
We use the formula of R0 (T, T + s) with T = 6, s = 2, P (0, 6) = 0.6464, and P (0, 8) = 0.5461. Thus,
R0 (T, T + s) = 0.6462
P (0, T )
−1=
− 1 = 0.18367
P (0, T + s)
0.5461 Example 80.2
GS has entered into a forward rate agreement which expires in 4 years on a loan to be made in
two years and which expires in four years. A zerocoupon bond paying $1 in 4 years has a price of
$0.5523 today. A zerocoupon bond paying $1 in 2 years has a price of $0.8868 today. Now assume
that two years have passed and a zerocoupon bond paying $1 in yet another 2 years has a price of
$0.7769. What will be GS’ payoﬀ to the forward rate agreement at the end of 4 years?
Solution.
The payoﬀ of the FRA is given by the formula
Payoﬀ to FRA = P (0, T )
1
−
P (T, T + s) P (0, T + s) with T = s = 2, P (0, 2) = 0.8868, P (0, 4) = 0.5523, and P (2, 4) = 0.7769. Thus,
Payoﬀ to FRA = 1
0.8868
−
= −$0.3184821582,
0.7769 0.5523 80 THE BLACK FORMULA FOR PRICING FRA, CAPLETS, AND CAPS 509 which means that GS will lose money in this ﬁnancial transaction
Pricing Interest Rate Caplets
Consider a loan of $1 to be made at time T for a period of length s. One way to hedge against rate
ﬂactuation is by buying a caplet. By a caplet we mean a European call option with underlying
asset a forward rate at the caplet maturity. For a caplet with strike rate KR and maturity T + s,
the payoﬀ at maturity is given by
Payoﬀ of caplet = max{0, RT − KR } (80.1) where RT = RT (T, T + s) is the lending rate between time T and time T + s. The caplet permits
the borrower to pay the time−T market interest rate if it is below KR , but receive a payment for
the diﬀerence in rates if the rate is above KR .
Example 80.3
GS has bought a caplet which expires in 4 years. The caplet’s strike rate of 0.2334. A zerocoupon
bond paying $1 in 4 years has a price of $0.5523 today. A zerocoupon bond paying $1 in 2 years
has a price of $0.8868 today. Now assume that two years have passed and a zerocoupon bond
paying $1 in yet another 2 years has a price of $0.7769. What will be GS’ payoﬀ to the caplet at
time t = 4?
Solution.
We have: T = s = 2, P (0, 2) = 0.8868, P (0, 4) = 0.5523, P (2, 4) = 0.7769 and KR = 0.2334. Thus,
using the formula
1
−1
RT (T, T + s) =
P (T, T + s)
we ﬁnd
1
RT (2, 4) =
− 1 = 0.2871669456.
0.7769
The payoﬀ of the caplet at time T + s is given by
Payoﬀ of caplet = max{0, RT − KR }.
Hence, the ﬁnal answer is
Payoﬀ of caplet = max{0, 0.2871669456 − 0.2334} = $0.0537669456
The caplet can be settled at time T. In this case, the value of the caplet at time T is the discounted
value of (80.1):
1
max{0, RT − KR }.
1 + RT 510 INTEREST RATES MODELS Note that 1+1 T is the time−T price of a bond paying $1 at time T + s. The previous expression
R
can be written as
RT − KR
1
max{0, RT − KR } = max{0,
}
1 + RT
1 + RT
RT − KR
}
(1 + RT )(1 + KR )
1
1
=(1 + KR ) max 0,
−
1 + KR 1 + RT
=(1 + KR ) max{0, But the expression
max 0, 1
1
−
1 + KR 1 + RT is the payoﬀ of a European put option on a bond with strike 1+1 R . Thus, a caplet is equivalent to
K
(1 + KR ) European call options on a zerocoupon bond at time T paying $1 at time T + s and with
strike price 1+1 R and maturity T. The price of the caplet is then found by ﬁnding the price of the
K
put option which is found by the Black formula discussed in the previous section.
Example 80.4
The time0 price of a 2year zerocoupon bond with par value $1 is $0.8495. The time0 price of a
3year zerocoupon bond with par value $1 is $0.7722. The 1year forward price volatility is 0.105.
What is the price of an interest rate caplet with strike rate 11% (eﬀective annual rate) on a 1year
loan 2 years from now.
Solution.
We are given T = 2, s = 1 and KR = 0.11. We ﬁrst ﬁnd the price of a European put with strike
rate 1+1 R = 1.1 = 0.9009 and maturity in 2 years. We use the Black formula. We have
K
11
F=
d1 = 0.7722
P (0, 3)
=
= 0.90901
P (0, 2)
0.8495 ln (F/K ) + (σ 2 /2)T
ln (0.90901/0.9009) + 0.5(0.105)2 (2)
√
√
=
= 0.1346
0.105 2
σT
√
√
d2 = d1 − σ T = 0.1346 − 0.105 2 = −0.0139
N (d1 ) = 0.553536, N (d2 ) = 0.494455. Thus, the price of the put option is
P = P (0, T )[KN (−d2 )−F N (−d1 )] = 0.8495[0.9009(1−0.494455)−0.90901(1−0.553536)] = $0.04214. 80 THE BLACK FORMULA FOR PRICING FRA, CAPLETS, AND CAPS 511 The price of the caplet is
(1 + Kr )P = 1.11 × 0.04214 = 0.0468
Pricing Interest Rate Caps
An interest rate cap with strike price of KR that makes payments at time ti is a series of caplets
with common strike price of KR such that at time ti the owner of the cap receives a payment in the
amount
max{0, Rti (ti , ti+1 ) − KR }
whenever Rti (ti , ti+1 ) > KR . The value of the cap is the sum of the values of the caplets that make
up the cap.
Example 80.5
Suppose you own an interest rate cap with a strike price of 0.0345. The cap makes payments at
times t = 3, t = 4, and t = 5. The forward rates for these time periods are as follows: R2 (2, 3) =
0.04343, R3 (3, 4) = 0.09661, R4 (4, 5) = 0.01342, and R5 (5, 6) = 0.04361. What will be the sum of
the payments you receive from it?
Solution.
We have
Cap payment at time t=2 = max{0, 0.04343 − 0.0345} = 0.00893
Cap payment at time t=3 = max{0, 0.09661 − 0.0345} = 0.06211
Cap payment at time t=4 = max{0, 0.01342 − 0.0345} = 0
Thus, the total sum you will receive from this cap is
0.00893 + 0.06211 + 0 = 0.07104 512 INTEREST RATES MODELS Practice Problems
Problem 80.1
Calculate the forward interest rate for a forward rate agreement for a 1year loan initiated at year
3, the beginning of the fourth year.
Problem 80.2
GS has bought a caplet which expires in 4 years and has a strike rate of 0.2334. A zerocoupon bond
paying $1 in 4 years has a price of $0.5523 today. A zerocoupon bond paying $1 in 2 years has a
price of $0.8868 today. Now assume that two years have passed and a zerocoupon bond paying $1
in yet another 2 years has a price of $0.7769. What will be GS’ payoﬀ to the caplet at time t = 2?
Problem 80.3
A caplet with a certain strike rate has a payoﬀ of 0.6556 at time 10 years and a payoﬀ of 0.3434 in
3 years. The caplet is based on the forward rate R3 (3, 10). What is the strike price for this caplet?
Problem 80.4
The time0 price of a 2year zerocoupon bond with par value $1 is $0.8573. The time0 price of a
3year zerocoupon bond with par value $1 is $0.7722. The 1year forward price volatility is 0.105.
You want to borrow $1 for a period 1 year two years from now. To hedge against rate ﬂuctuation you
buy a caplet with strike rate 10.75%. This caplet is equivalent to a certain number N of European
put option with maturity of 2 years and strike price K. Determine N and K.
Problem 80.5
The time0 price of a 2year zerocoupon bond with par value $1 is $0.8573. The time0 price of a
3year zerocoupon bond with par value $1 is $0.7722. The 1year forward price volatility is 0.105.
You want to borrow $1 for a period 1 year two years from now. To hedge against rate ﬂuctuation you
buy a caplet with strike rate 10.75%. This caplet is equivalent to a certain number N of European
put option with maturity of 2 years and strike price K. The price of the put is found using the
Black formula. Determine the value of F.
Problem 80.6
The time0 price of a 2year zerocoupon bond with par value $1 is $0.8573. The time0 price of a
3year zerocoupon bond with par value $1 is $0.7722. The 1year forward price volatility is 0.105.
You want to borrow $1 for a period 1 year two years from now. To hedge against rate ﬂuctuation you
buy a caplet with strike rate 10.75%. This caplet is equivalent to a certain number N of European
put option with maturity of 2 years and strike price K. The price of the put is found using the
Black formula. Find d1 , d2 , N (d1 ), and N (d2 ). 80 THE BLACK FORMULA FOR PRICING FRA, CAPLETS, AND CAPS 513 Problem 80.7
The time0 price of a 2year zerocoupon bond with par value $1 is $0.8573. The time0 price of a
3year zerocoupon bond with par value $1 is $0.7722. The 1year forward price volatility is 0.105.
You want to borrow $1 for a period 1 year two years from now. To hedge against rate ﬂuctuation you
buy a caplet with strike rate 10.75%. This caplet is equivalent to a certain number N of European
put option with maturity of 2 years and strike price K. The price P of the put is found using the
Black formula. Find P.
Problem 80.8
The time0 price of a 2year zerocoupon bond with par value $1 is $0.8573. The time0 price of a
3year zerocoupon bond with par value $1 is $0.7722. The 1year forward price volatility is 0.105.
You want to borrow $1 for a period 1 year two years from now. To hedge against rate ﬂuctuation
you buy a caplet with strike rate 10.75%. Find the price of the caplet.
Problem 80.9
A 2year cap on $100 has a strike price of 14%. The forward rate for time 0 to time 1 is 10%, and
the forward rate from time 1 to time 2 is 15.17%. What is the payoﬀ of each caplet at time T = 1?
Problem 80.10
The time0 price of a 1year zerocoupon bond with par value $1 is $0.9259. The time0 price of a
2year zerocoupon bond with par value $1 is $0.8495. The 1year forward price volatility is 0.10.
You want to borrow $1 for a period 1 year one year from now. To hedge against rate ﬂuctuation
you buy a caplet with strike rate 11.5%(eﬀectice annual). Find the price of this caplet.
Problem 80.11
The time0 price of a 2year zerocoupon bond with par value $1 is $0.8495. The time0 price of a
3year zerocoupon bond with par value $1 is $0.7722. The 1year forward price volatility is 0.105.
You want to borrow $1 for a period 1 year two years from now. To hedge against rate ﬂuctuation
you buy a caplet with strike rate 11.5%(eﬀectice annual). Find the price of this caplet.
Problem 80.12
The time0 price of a 3year zerocoupon bond with par value $1 is $0.7722. The time0 price of a
4year zerocoupon bond with par value $1 is $0.7020. The 1year forward price volatility is 0.11.
You want to borrow $1 for a period 1 year three years from now. To hedge against rate ﬂuctuation
you buy a caplet with strike rate 11.5%(eﬀectice annual). Find the price of this caplet.
Problem 80.13
What is the price of a 3year interest rate cap with strike rate 11.5% (eﬀective annual)? 514 INTEREST RATES MODELS Binomial Models for Interest Rates
In this chapter we examine binomial interest rate models, in particular the BlackDermanToy
model. 515 516 BINOMIAL MODELS FOR INTEREST RATES 81 Binomial Model for Pricing of a ZeroCoupon Bond
In this section we discuss pricing a zerocoupon bond based on interest rates. The model is based
on a simple binomial tree that models interest rates ﬂuctuations.
The binomial interest rate tree is patterned after the binomial option pricing model, this model
assumes that interest rates follow a binomial process in which in each period the rate is either higher
or lower.
Assume a 1period, riskfree spot rate r follows a process in which in each period the rate is either
equal to a proportion u times its beginningoftheperiod rate or a proportion d times its beginningoftheperiod rate with u > d. After one period, there would be two possible oneperiod spot rate:
ru = ur0 and rd = dr0 . Assuming u and d are constants over diﬀerent periods, then after two
periods there would be three possible rates: ruu = u2 r0 , rud = udr0 , and rdd = d2 r0 . This creates a
twoperiod binomial interest rate tree.
Example 81.1
Construct a threeyear interest rate binomial tree assuming that the current 1year spot rate is 10%
and that each year rate moves up 1% or down 0.5%, each with riskneutral probability 0.5
Solution.
The upward parameter is u = 1.1 and the downward parameter is d = 0.95. The tree is shown in
Figure 81.1
Now, the pricing of a zerocoupon bond that pays $1 at maturity can be determined in much
the same way we determined option prices in a binomial stockprice tree. We illustrate the process
in the next example.
Example 81.2
Use Figure 81.1, to ﬁnd the price of 1year, 2year and 3year zerocoupon defaultfree1 bonds that
pay $1 at maturity.
Solution.
The price of a 1year bond is
P (0, 1) = e−0.10 = $0.9048.
The twoyear bond is priced by working backward along the tree. In the second period, the price
of the bond is $1. One year from today, the bond will have the price e−ru with probability p = 0.5
or e−rd with probability 1 − p = 0.5. Thus, the price of the bond is
P (0, 2) = e−0.10 [0.5e−0.11 + 0.5e−0.095 ] = $0.81671.
1 A bond that has no default risk. 81 BINOMIAL MODEL FOR PRICING OF A ZEROCOUPON BOND 517 Likewise, a threeyear bond is priced as
P (0, 3) =e−r0 [pe−ru (pe−ruu + (1 − p)e−rud )
+(1 − p)e−rd (pe−rdu + (1 − p)e−rdd )]
=e−0.10 [0.5e−0.11 [0.5e−0.121 + 0.5e−0.1045 ] + 0.5e−0.095 [0.5e−0.1045 + 0.5e−0.09025 ]]
=$0.7353 Figure 81.1
Using the interest rate binomial tree one can construct a binomial tree of the prices of a bond. We
illustrate this point in the next example. Example 81.3
Use Figure 81.1, to construct the tree of prices of a 3year zerocoupon bond that pays $1 at
maturity. Solution.
The time2 price of the bond at the node uu is e−0.121 = $0.8860. The time2 price of the bond at
the node ud is e−0.1045 = $0.9008.The time1 price of the bond at the node u is e−0.11 (0.8860 × 0.5 +
0.9008 × 0.5) = $0.8003. We obtain the following tree of the bond prices. 518 BINOMIAL MODELS FOR INTEREST RATES Yields in the Binomial Model
The yield of an n−year zerocoupon bond is given by the formula
yield = − ln P (0, n)
.
n Example 81.4
Find the yields of the 1year, 2year, and 3year zerocoupon bond based on the interest rate tree
of Example 81.1.
Solution.
The yield of the 1year bond is
yield = − ln (0.9048)
ln P (0, 1)
=−
= 0.10004%
1
3 The yield of the 2year bond is
yield = − ln P (0, 2)
ln (0.81671)
=−
= 0.1012%
2
3 The yield of the 3year bond is
yield = − ln P (0, 3)
ln (0.7353)
=−
= 0.1025%
3
3 Remark 81.1
Note that in the previous example, the yields increase with maturity. This is not the case when
the up and down move are symmetric with probability 0.5. In this case, the yields decrease with
maturity. See Problem 81.6. Moreover, these yields are lower than current interest rate, which
means that the interest rate binomial model underprice the bonds. 81 BINOMIAL MODEL FOR PRICING OF A ZEROCOUPON BOND 519 Binomial Interest Rate Model in Bond Option Pricing
The binomial tree used for pricing a zerocoupon bond can be used to price an option with underlying
asset a zerocoupon bond. The pricing process is exactly the same as pricing option for stock, just
that the price of the underlying asset is bond instead of stock. The only thing to be careful about
is that the calculation here assumes there is a nonrecombining tree (the binomial tree constructed
above are examples of recombining trees).
We illustrate the process of pricing a bond option using an interest rate binomial tree in the next
example.
Example 81.5
Given the following interest rate binomial tree. Consider a 2year European put option on a 1year zerocoupon bond with strike price $0.88. What
is the price of this option?
Solution.
The price of the bond at the end of year 2 is e−0.18 = $0.8353 at the node uu; e−0.10 = $0.9048 at the
nodes ud and du; e−0.02 = $0.9802. The put option pays oﬀ only if the price of the bond is less than
$0.88. This occurs only at the node uu. The put payoﬀ in this case is 0.88 − 0.8353 = $0.0447. Using
the interest rates along the tree, and accounting for the 0.25 riskneutral probability of raching that
one node, we obtain the put option price of
0.0447 × (0.5)2 e−0.14−0.10 = $0.0088 520 BINOMIAL MODELS FOR INTEREST RATES Practice Problems
Problem 81.1
Construct a twoyear interest rate binomial tree assuming that the current 1year spot rate is 10%
and that each year rate moves up or down 4%, each with riskneutral probability 0.5.
Problem 81.2
Use Problem 81.1, to ﬁnd the price of 1year zerocoupon defaultfree bonds that pay $1 at maturity.
Problem 81.3
Use Problem 81.1, to ﬁnd the price of 2year zerocoupon defaultfree bonds that pay $1 at maturity.
Problem 81.4
Use Problem 81.1, to ﬁnd the price of 3year zerocoupon defaultfree bonds that pay $1 at maturity.
Problem 81.5
Use Problem 81.1, to construct the tree of prices of a 3year zerocoupon bond that pays $1 at
maturity.
Problem 81.6
(a) Construct a twoperiod interest rate tree assuming that the current 1year rate is 10% and that
each year the 1year rate moves up or down 2%, with probability 0.5.
(b) Find the prices of 1, 2, and 3year zero coupon bonds that pays $1 at maturity using the above
tree.
(c) Find the yields of the bonds in (b).
Problem 81.7
Construct a twoyear interest rate binomial tree assuming that the current 1year spot rate is 12%
and that each year rate moves up or down 4%, each with riskneutral probability 0.5.
Problem 81.8
Using the tree of the previous exercise, consider a 2year European put option on a 1year zerocoupon bond with strike price $0.85. What is the price of this option?
Problem 81.9 ‡
A discretetime model is used to model both the price of a nondividendpaying stock and the shortterm (riskfree) interest rate. Each period is one year.
At time 0, the stock price is S0 = 100 and the eﬀective annual interest rate is r0 = 5%.
At time 1, there are only two states of the world, denoted by u and d. The stock prices are Su = 110
and Sd = 95. The eﬀective annual interest rates are ru = 6% and rd = 4%.
Let C (K ) be the price of a 2year Kstrike European call option on the stock. Let P (K ) be the
price of a 2year Kstrike European put option on the stock.
Determine P (108) − C (108). 82 THE BASICS OF THE BLACKDERMANTOY MODEL 521 82 The Basics of the BlackDermanToy Model
The interest rate models that we have examined so far are arbitragefree only in a world consistent
with their assumptions. In the real world, however, they will generate apparent arbitrage opportunities, in the sense that observed prices will not match the theoretical prices obtained through these
models. So the important question is whether a particular interest rate model ﬁts given realworld
data. Recall that the Vasicek and CIR models often do not match data exogenous to the models,
because the models are based on four parameters while realworld data sets may have more than
four members. The BlackDermanToy (BDT) model attempts to incorporate an element of calibration1 to allow applications to realworld data.
The basic idea of the BlackDermonToy model is to construct a recombining binomial tree of shortterm interest rates2 , with a ﬂexible enough structure to match the realworld data. The shortterm
rates are assumed to be lognormally distributed. It is also assumed that the riskneutral probability
of an up move in the interest rate is 0.50.
We next describe the BDT tree: For each period in the tree there are two parameters, Rih and
σi , where the index i represents the number of time periods that have elapsed since time 0, and h
is the length of each time period. The parameter Rih is the rate level parameter, and σi is the
volatility parameter. Both parameters are used to match the tree with observed data.
√
At each node of the tree the ratio of√the up move to the down move is e2σ h . For example, at the
node r0 , if rd = Rh then ru = Rh e2σ1 h . Hence, in a single time period, the ratio between adjacent
nodes is the same. A general form of the BDT tree is shown in Figure 82.1.
Example 82.1
You are given the following values of a BlackDermanToy binomial tree: rdd = 0.0925 and rud =
0.1366. Find ruu .
Solution.
We have ruu
rdu
=
rdu
rdd or
ruu
0.1366
=
.
0.1366
0.0925
Solving this equation we ﬁnd ruu = 0.2017
1
2 Calibration is the matching of a model to ﬁt observed data.
All interest rates in the BDT are annual eﬀective, not continously compounded rates. 522 BINOMIAL MODELS FOR INTEREST RATES Figure 82.1
We next show through a series of examples how the BlackDermanToy model matches the observed
data.
Example 82.2
The table below lists market information about bonds that we would like to match using the BDT
model. All rates are eﬀective annual rates.
Maturity
(years)
1
2
3
4 Yield to
Maturity
10%
11%
12%
12.5% Bond
Price ($)
0.90901
0.8116
0.7118
0.6243 Volatility in
Year 1
NA
10%
15%
14% The BDT tree, shown in Figure 82.2, depicts 1year eﬀective annual rates based on the above table
. 82 THE BASICS OF THE BLACKDERMANTOY MODEL 523 Figure 82.2
Construct the tree of 1year bond prices implied by the tree in Figure 82.2.
Solution.
The price at each node is
Figure 82.3 1
,
1+r where r is at that node. Thus, we obtain the tree of prices shown in Example 82.3
Verify that the 2 and 3year zerocoupon bonds priced using BDT match the observed one listed
in the table of Example 82.2.
Solution.
The time 0 of a 2year zerocoupon bond is the discounted expected price at time 0 given by
0.9091(0.5 × 0.8832 + 0.5 × 0.9023) = $0.8116.
The time 0 of a 3year zerocoupon bond is the discounted expected price at time 0 given by
0.9091[0.5 × 0.8832(0.5 × 0.8321 + 0.5 × 0.8798)
+0.5 × 0.9023(0.5 × 0.8798 + 0.5 × 0.9153)] = $0.7118 524 BINOMIAL MODELS FOR INTEREST RATES The time 0 of a 4year zerocoupon bond can be shown to be $0.6243. See Problem 82.5 Figure 82.3
Now, let the time−h price of a zerocoupon bond maturing at time T when the time−h short term
rate is r(h) be denoted by P [h, T, r(h)]. Then the annualized yield of the bond is given by
1 y [h, T, r(h)] = P [h, T, r(h)]− T −h − 1.
Example 82.4
Verify that the annual yields found through the BDT model match the observed ones in the table
of Example 82.2.
Solution.
The time0 price of a 1year zerocoupon bond found by the BDT model is $0.9091. Thus, the
time0 yield of this bond is
1
0.9091− 1−0 − 1 = 10%. 82 THE BASICS OF THE BLACKDERMANTOY MODEL 525 Likewise, the time0 of a 2year zerocoupon bond is
1 0.8116− 2−0 − 1 = 11%.
The time0 of a 3year zerocoupon bond is
1 0.7118− 3−0 − 1 = 12%.
The time0 of a 4year zerocoupon bond is
1 0.6243− 4−0 − 1 = 12.5%
We deﬁne the annualized yield volatility by
0.5
P (h, T, ru )
Yield volatility = √ × ln
.
P (h, T, rd )
h
Example 82.5
Verify that the volatilities implied by the BDT tree match the observed volatilities.
Solution.
We will need to compute the implied bond yields in year 1 and then compute the volatility. For
−1
a 2year bond (1year bond in year 1) will be worth either $0.8832 with yield 0.8832− 2−1 − 1 or
1
$0.9023 with yield 0.9023− 2−1 − 1. Thus, the yield volatility in year 1 is
0.5 × ln 0.8832−1 − 1
0.9023−1 − 1 = 10%. For a 3year zerocoupon bond (2year bond in year 1) the price of the bond is either
0.8832(0.5 × 0.8321 + 0.5 × 0.8798) = $0.7560
1 with yield 0.7560− 3−1 − 1 or price
0.9023(0.5 × 0.8798 + 0.5 × 0.9153) = $0.8099
1 with yield 0.8099− 3−1 − 1. Thus, the yield volatility in year 1 is
1 0.5 × ln 0.7560− 2 − 1
1 0.8099− 2 − 1 = 15%. Both yield volatilities match the one listed in the table of Example 82.2. We ﬁnd the year 1 volatility
yield of a 4year zerocoupon bond in Problem 82.6 526 BINOMIAL MODELS FOR INTEREST RATES Practice Problems
Problem 82.1
In a BDT tree, you are given the following: ruu = 0.2017 and rdd = 0.0925. Find rud .
Problem 82.2
If we are given some number of nodes for a particular time period while the others are left blank,
it may be possible to ﬁll in the rest of the nodes, as the ratio between every two adjacent nodes in
a single time period is the same. From this one can show that for a time i we have
run di−n = rdi e2nσi √ h . where n is the number of times the interest rate has gone up and i − n is the number of times the
interest rate has gone down.
In a BDT model, you are given that rddddd = 0.0586, and σ5 = 0.21. Each time period in the
binomial tree is 2 years. Find ruuudd .
Problem 82.3
In a BDT model, you are given that ruudddd = 0.1185, and σ6 = 0.43. Each time period in the
binomial tree is 1 year. Find ruuuuuu .
Problem 82.4
You are given the following incomplete BlackDermanToy interest rate tree model for the eﬀective
annual interest rates: 82 THE BASICS OF THE BLACKDERMANTOY MODEL 527 Fill in the three missing interest rates in the model.
Problem 82.5
Verify that the time0 price of a 4year zerocoupon bond implied by the BDT model (Example
82.2) is $0.6243.
Problem 82.6
Verify that the annual yield volatility in year 1 of the 4year zerocoupon bond implied by the BDT
model is 14%.
Problem 82.7
The following is a BlackDermanToy binomial tree for eﬀective annual interest rates. Determine the value of rud .
Problem 82.8
Using the BDT tree of the previous problem, Construct the tree of bond prices implied by the BDT
model.
Problem 82.9 ‡
Using the tree of Problem 82.7, compute the volatility yield in year 1 of a 3year zerocoupon bond.
Problem 82.10 ‡
You are given the following market data for zerocoupon bonds with a maturity payoﬀ of $100. 528 BINOMIAL MODELS FOR INTEREST RATES
Maturity (years) Bond Price ($)
1
94.34
2
88.50 Volatility in Year 1
N/A
10% A 2period BlackDerman=Toy interest tree is calibrated using the data from above: (a) Calculate r0 and σ1 , the volatility in year 1 of the 2year zerocoupon bond.
(b) Calculate rd , the eﬀective annual rate in year 1 in the “down” state.
Problem 82.11 ‡
You are to use a BlackDermanToy model to determine F0,2 [P (2, 3)], the forward price for time2
delivery of a zerocoupon bond that pays 1 at time 3. In the BlackDermanToy model, each period
is one year. The following eﬀective annual interest rates are given: rd
ru
rdd
ruu
Determine 1000 × F0,2 [P (2, 3)]. =30%
=60%
=20%
=80% 83 BLACKDERMANTOY PRICING OF CAPLETS AND CAPS 529 83 BlackDermanToy Pricing of Caplets and Caps
In this section, we use the BlackDermanToy tree to price caplets and caps. We ﬁrst discuss the
pricing of a caplet.
Although a caplet makes a payment at the end of each year, it is valued at the beginning of the
year. Thus, the value of a T −year caplet with strike rate KR at time T − 1 is the discounted value
of the year−T payoﬀ. It is given by the formula
Notional × max{0, RT −1 − KR }
1 + RT −1 where RT −1 = rx is a rate in year T − 1 and x is some product of u’s and d’s (up and down
movements in the interest rates).
Example 83.1
The BDT interest rate tree is given in Figure 83.1. Figure 83.1
Consider a year4 caplet with strike rate of 10.5%. What are the year3 caplet values? 530 BINOMIAL MODELS FOR INTEREST RATES Solution.
A year4 caplet at year 3 has the following four values:
max{0, ruuu − KR }
1 + ruuu
max{0, ruud − KR }
Notional ×
1 + ruud
max{0, rudd − KR }
Notional ×
1 + rudd
max{0, rddd − KR }
Notional ×
1 + rddd Notional × 0.168 − 0.105
= 5.394
1.168
0.136 − 0.105
=100 ×
= 2.729
1.136
0.111 − 0.105
=100 ×
= 0.450
1.11
=100 × =0 Now, remember that the riskneutral probability on an up movement in the shortterm interest rate
is always 0.5 in the BDT model. Thus, if Px is the expected value of a caplet at some node in a
BDT binomial tree, where x is some product of u’s and d’s (up and down movements in the interest
rates), then we can obtain Px using the following formula:
Px = (1 + rx )−1 [0.5Pxu + 0.5Pxd ].
Example 83.2
Using the information of the previous exercise, ﬁnd the time−2 value of the caplet.
Solution.
At time−2 there are three nodes. The values of the caplet at these nodes are
(1 + 0.172)−1 [0.5 × 5.394 + 0.5 × 2.729] =3.4654
(1 + 0.135)−1 [0.5 × 2.729 + 0.5 × 0.450] =1.4004
(1 + 0.106)−1 [0.5 × 0.450 + 0.5 × 0] =0.2034
Example 83.3
Find the time−1 and time−0 values of the caplet of Example 83.1
Solution.
At year 1, there are two nodes. The values of the caplet at these nodes are
(1 + 0.126)−1 [0.5 × 3.4654 + 0.5 × 1.4004] =2.1607
(1 + 0.093)−1 [0.5 × 1.4004 + 0.5 × 0.2034] =0.7337. 83 BLACKDERMANTOY PRICING OF CAPLETS AND CAPS 531 The time−0 value of the caplet is
(1 + 0.09)−1 [0.5 × 2.1607 + 0.5 × 0.7337] = $1.3277
Next, we discuss the pricing of a cap. Recall that an interest rate cap pays the diﬀerence between
the realized interest rate in a period and the interest cap rate, if the diﬀerence is positive. Let us
illustrate how a cap works. Suppose you borrow $1,000,000 for 5 years at a ﬂoating rate which is
determined once a year. The initial rate is 6%, and the cap is 7%. The ﬂoating rate for years 25
are 5.5%, 8%, 9.5%, 6.5%. The amount of interest without the cap and with the cap are shown in
the table below.
End of Interest payment Interest payment
Year without a cap($)
with a cap($)
1
60,000
60,000
2
55,000
55,000
3
80,000
70,000
4
95,000
70,000
5
65,000
65,000
Example 83.4
Find the price of a 4year cap with strike rate 10.5% using the BDT model given in Figure 83.1 for
the notional amount of $100.
Solution.
The value of the cap is the sum of the values of the year3 cap payments, year2 cap payments, and
year1 cap payments. The cap payment in year3 has the following three values:
0.168 − 0.105
max{0, ruuu − KR }
=100 ×
= 4.5377
1 + ruuu
1.168
0.136 − 0.105
max{0, ruud − KR }
=100 ×
= 2.7289
Notional ×
1 + ruud
1.136
max{0, rudd − KR }
0.11 − 0.105
Notional ×
=100 ×
= 0.4505
1 + rudd
1.11
max{0, rddd − KR }
Notional ×
=0
1 + rddd
The node at the bottom is 0 because 8.9% is smaller than 10.5%.
The year3 cap payment in year2 has the following three possible values:
Notional × (1 + 0.172)−1 [0.5 × 4.5377 + 0.5 × 2.7289] =3.1000
(1 + 0.135)−1 [0.5 × 2.7289 + 0.5 × 0.4505] =1.4006
(1 + 0.106)−1 [0.5 × 0.4505 + 0.5 × 0] =0.2037. 532 BINOMIAL MODELS FOR INTEREST RATES The year3 cap payment in year1 has the two possible values
(1 + 0.126)−1 [0.5 × 3.1000 + 0.5 × 1.4006] =1.9985
(1 + 0.093)−1 [0.5 × 1.4006 + 0.5 × 0.2037] =0.7339
Thus, the year3 cap payment value is
(1 + 0.09)−1 [0.5 × 1.9985 + 0.5 × 0.7339] = 1.2534.
We next ﬁnd the value of the year2 cap payment. The cap payment in year 2 has the following
three possible values
0.172 − 0.105
max{0, ruu − KR }
=100 ×
= 5.7167
1 + ruu
1.172
max{0, rud − KR }
0.135 − 0.105
Notional ×
=100 ×
= 2.6432
1 + rud
1.135
max{0, rdd − KR }
0.106 − 0.105
Notional ×
=100 ×
= 0.0904
1 + rdd
1.106 Notional × The year2 cap payment in year1 has the two possible values
(1 + 0.126)−1 [0.5 × 5.7167 + 0.5 × 2.6432] =3.7122
(1 + 0.093)−1 [0.5 × 2.6432 + 0.5 × 0.0904] =1.2505
The year2 cap payment has the value
(1 + 0.09)−1 [0.5 × 3.7122 + 0.5 × 1.2505] = 2.2765.
The cap payment in year1 has the following two possible values
0.126 − 0.105
max{0, ru − KR }
=100 ×
= 1.8650
1 + ru
1.126
max{0, rd − KR }
Notional ×
=0
1 + rd Notional × The year1 cap payment has the value
(1 + 0.09)−1 [0.5 × 1.8650 + 0.5 × 0] = 0.8555
Finally, the 4year cap has the value
1.2534 + 2.2765 + 0.8555 = $4.3854 83 BLACKDERMANTOY PRICING OF CAPLETS AND CAPS 533 Practice Problems
Problem 83.1
In a particular BlackDermanToy binomial tree, you are given ruuu = 0.1, ruuuu = 0.11, and ruuud =
0.09. A particular caplet has a strike rate of 0.08 and a notional amount of $1. What is the expected
value of a caplet at node uuu of the BDT binomial tree? That is, ﬁnd Puuu .
Problem 83.2
The price of a caplet today is $3. The caplet has a strike rate of 0.06 and a notional amount of
$100. The oneyear shortterm interest rate today is 0.05. In one year, the oneyear shortterm
interest rate will be either 0.1 or Q. Find Q using a oneperiod BDT binomial tree.
Problem 83.3
You are given the following shortterm interest rates in a BlackDermanToy binomial tree: r0 =
0.04, rd = 0.02, ru = 0.11, rdd = 0.037, rdu = 0.074, ruu = 0.148. A 2year caplet (which actually pays
at time t = 3 years) has a strike rate of 0.07. The caplet has a notional amount of $100. What is
the current value of this caplet?
Problem 83.4
You are given the following shortterm interest rates in a BlackDermanToy binomial tree: r0 =
0.05, rd = 0.04, ru = 0.09, rdd = 0.06, ruu = 0.18, rddd = 0.04, ruud = 0.16. A 3year caplet (which
actually pays at time t = 4 years) has a strike rate of 0.07. The caplet has a notional amount of
$100. What is the current value of this caplet?
Problem 83.5
Find the price of a 4year cap with strike rate 12% using the BDT model given in Figure 83.1 for
the notional amount of $100. 534 BINOMIAL MODELS FOR INTEREST RATES Supplement
This supplement includes a discussion of Jensen’s inequality and its applications. It also includes a
discussion of utility theory and its relation to riskneutral pricing of a stock. 535 536 SUPPLEMENT 84 Jensen’s Inequality
In this section, we discuss Jensen’s inequality as applied to probability theory. We ﬁrst start with
the following deﬁnition: A function f is said to be convex on the open interval (a, b) if and only if
for any u, v in (a, b) and any 0 ≤ λ ≤ 1, we have
f (λu + (1 − λ)v ) ≤ λf (u) + (1 − λ)f (v ).
Graphically, this is saying that for any point on the line segment connecting u and v , the value of f at
that point is smaller than the value of the corresponding point on the line line segement connecting
(u, f (u)) and (v, f (v )). Thus, the arc from f (u) to f (v ) is below the line segment connecting f (u)
and f (v ). See Figure 84.1. Figure 84.1
It is easy to see that a diﬀerentiable function f is convex if and only if the slope of the tangent line
at a point is less than or equal to the slope of any secant line connecting the point to another point
on the graph of f.
The following result gives a characterization of convex functions.
Proposition 84.1
A twice diﬀerentiable function f is convex in (a, b) if and only if f (x) ≥ 0 in (a, b).
Proof.
Suppose that f is convex in (a, b). Fix a < v < b and let a < u < b. Then
f (u) ≤ f (v ) − f (u)
v−u 84 JENSEN’S INEQUALITY 537 and
f (v ) ≥ f (u) − f (v )
.
u−v Thus,
f (u) − f (v )
f (v ) − f (u) f (u) − f (v )
≥
−
= 0.
u−v
v−u
u−v
Hence,
f (u) − f (v )
≥ 0.
u→v
u−v f (v ) = lim Now, suppose that f (x) ≥ 0 for all x in (a, b). This means that f is increasing in (a, b). Let
a < u < b. Deﬁne the function g (v ) = f (v ) − f (u)(v − u). Then g (v ) = f (v ) − f (u) ≥ 0 if v ≥ u
and g (v ) ≤ 0 if v ≤ u. Thus, g is increasing to the right of u and decreasing to the left of u. Hence,
if v < u we have
0 ≤ g (u) − g (v ) = f (u) − f (v ) + f (u)(v − u)
which can be written as
f (u) ≤ f (v ) − f (u)
.
v−u For v > u we have
0 ≤ g (v ) − g (u) = f (v ) − f (u) − f (u)(v − u)
which can written as
f (u) ≤ f (v ) − f (u)
.
v−u It follows that f is convex
Example 84.1
Show that the exponential function f (x) = ex is convex on the interval (−∞, ∞).
Solution.
Since f (x) ≥ 0 for all real number x, the exponential function is convex by the previous proposition
Proposition 84.2 (Jensen)
If f is a convex function in (a, b) and X is a random variable with values in (a, b) then
E [f (X )] ≥ f [E (X )]. 538 SUPPLEMENT Proof.
The equation of the tangent line to the graph of f at E (X ) is given by the equation
y = f [E (X )] + f [E (X )](x − E (X )).
Since the function is convex, the graph of f (x) is at or above the tangent line. Therefore,
f (x) ≥ f [E (X )] + f [E (X )](x − E (X )).
Taking expected values of both sides:
E [f (X )] ≥ f [E (X )] + f [E (X )](E (X ) − E (X )) = f [E (X )]
Example 84.2
Let X be a binomial random variable that assumes the two values 1 and 2 with equal probability.
Verify that E [eX ] ≤ eE (X ) .
Solution.
We have E (X ) = 0.5 × 1 + 0.5 × 2 = 1.5 so that eE (X ) = e1.5 = 4.482. On the other hand,
E [eX ] = 0.5e + 0.5e2 = 5.054 > eE (X )
Example 84.3
Assume that the price of a stock at expiration is a binomial random variable and is equally likely
to be 35 or 45. Verify that Jensen’s inequality holds for the payoﬀ on an 40 strike call.
Solution.
Let f (x) = max{0, x − K }. Then f is a convex function. We have
E (X ) = 0.5 × 35 + 0.5 × 45 = 40
so that
f [E (X )] = max{0, 40 − 40} = 0.
On the other hand,
E [f (X )] = 0.5 × f (45) + 0.5 × f (35) = 0.5 × 5 + 0.5 × 0 = 2.5.
Hence, E [f (X )] ≥ f [E (X )]
We say that a function f is concave on the interval (a, b) if and only if for any u, v in (a, b)
and any 0 ≤ λ ≤ 1, we have
f (λu + (1 − λ)v ) ≥ λf (u) + (1 − λ)f (v ). 84 JENSEN’S INEQUALITY 539 Proposition 84.3
f is concave in (a, b) if and only if −f is convex in (a, b).
Proof. Let u, v be in (a, b) and 0 ≤ λ ≤ 1. Then f is concave if and only if f (λu + (1 − λ)v ) ≥
λf (u) + (1 − λ)f (v ) which is equivalent to (−f )f (λu + (1 − λ)v ) ≤ λ(−f )(u) + (1 − λ)(−f )(v ).
That is, if and only if −f is convex
Remark 84.1
It follows from Propositions and that f is concave in (a, b) if and only if f (x) ≤ 0 in (a, b).
Example 84.4
Show that the function f (x) = ln x is concave in (0, ∞).
Solution.
1
This follows from the above proposition since f (x) = − x2 < 0 540 SUPPLEMENT Practice Problems
Problem 84.1
Show that if f is convex on (a, b) then
f x+y
2 ≤ f (x) + f (y )
2 for all x, y in (a, b).
Problem 84.2
Show that the function f (x) = − ln x is convex for x > 0.
Problem 84.3
Show that if f is a concave function in (a, b) and X is a random variable with values in (a, b) then
E [f (X )] ≥ f [E (X )].
Problem 84.4
Show that for any random variable X, we have E [X 2 ] ≥ [E (X )]2 .
Problem 84.5
Assume that the price of a stock at expiration is equally likely to be 50, 100, or 150. Verify that
Jensen’s inequality holds for the payoﬀ on an 80 strike call.
Problem 84.6
Assume that the price of a stock at expiration is equally likely to be 60, 70, or 110. Verify that
Jensen’s inequality holds for the payoﬀ on an 80 strike put. 85 UTILITY THEORY AND RISKNEUTRAL PRICING 541 85 Utility Theory and RiskNeutral Pricing
In economics, the wellbeing of investors is not measured in dollars, but in utility. Utility is
a measure of satisfaction. An investor is assumed to have an utility function that translates an
amount of wealth to his/her amount of satisfaction. It is clearly an increasing function since an
investor is more satisﬁed with more wealth than less wealth. However, it was observed by economists
that the marginal utility of each extra dollar of wealth decreases as ones wealth increases. That is,
if the wealth keeps increasing, at one point, an additional dollar of wealth will make the investor
less happy than the previous dollars. Therefore, we can say that the utility function increases at a
decreasing rate which means in terms of calculus that the second derivative of the utility function is
negative. According to the previous section, we can say that a utility function is a concave function.
If we denote that utility function by U, by Jensen’s inequality we can write
E [U (X )] ≤ U [E (X )].
In other words, the expected utility associated with a set of possible outcomes is less than the utility
of getting the expected value of these outcomes for sure.
Example 85.1
Does the function U (x) = x
,x
x+100 ≥ 0 represent a utility function? Solution.
A utility function must have a positive ﬁrst derivative and a negative second derivative. We have
U (x) = 100
>0
(x + 100)2 and
U ( x) = − 200
< 0.
(x + 100)3 Hence, U can be a utility function
Investors are riskaverse which means that an investor will prefer a safer investment to a riskier
investment that has the same expected return. The next example illustrates this point.
Example 85.2
x
Suppose that your utility function is given by the formula U (x) = x+100 where x is your wealth.
You have the option of owning either a stock A with predicted future prices of $11 with probabiltiy
of 50% and $13 with probabiltiy 50% or a riskier stock B with predicted future prices of $10 with
probability of 80% and $20 with probability of 20%.
(a) Find the expected value of each stock.
(b) Evaluate the utility of each stock. 542 SUPPLEMENT Solution.
(a) The expected value of stock A is 0.5 × 11 + 0.5 × 13 = $12. The expected value of stock B is
0.8 × 10 + 0.2 × 20 = $12.
(b) The utilty value of owning stock A is
0.5 × 13
11
+ 0 .5 ×
= 0.1071.
11 + 100
13 + 100 The utilty value of owning stock B is
0.8 × 10
20
+ 0.2 ×
= 0.106.
10 + 100
20 + 100 This shows that the utility declines with the riskier investment. In order for the riskier stock to be
attractive for an investor, it must have higher expected return than the safer stock
In a fair game where gains and losses have the same expected return, a riskaverse investor will
avoid such a bet. We illustrate this point in the example below.
Example 85.3
x
Given that an investor’s with current wealth of 35 has a utility function given by U (x) = 1 − e− 100 ,
where x denotes the total wealth. The investor consider making a bet depending on the outcome
of ﬂipping a coin. The investor will receive 5 is the coin comes up heads, oherwise he will lose 5.
(a) What is the investor’s utility if he/she does not make the bet?
(b) What is the investor’s expected utility from making the bet?
Solution.
35
(a) The investor’s utility for not making the bet is 1 − e− 100 = 0.295.
(b) The investor’s utility for making the bet is
40 30 0.5(1 − e− 100 ) + 0.5(1 − e− 100 ) = 0.294
Now recall that the BlackScholes pricing model of a stock assumes a riskneutral environment. We
will next discuss why riskneutral pricing works using utility theory.
We will assume a world with only two assets: A riskfree bond and a risky stock.
Suppose that the economy in one period (for simplicity of the argument, we assume a period of one
year) will have exactly two states: A high state with probabiltiy p and a low state with probabiltiy
1 − p..
Let UH be value today of $1 received in the future in the high state of the economy. Let UL be value
today of $1 received in future in the low state of the economy. Since utility function is convex, the
marginal utility of a dollar is more when there is less wealth. Thus, UH < UL . 85 UTILITY THEORY AND RISKNEUTRAL PRICING 543 Example 85.4 (Valuing the riskfree bond)
Assume that the high and low states are equally likely, and UH = 0.90 and UL = 0.98. What would
one pay for a zero coupon bond today that pays $1 in one year? What is the riskfree rate of return?
Solution.
The bond will pay $1 in each state. The riskfree bond price is
0.5 × 0.90 + 0.5 × 0.98 = $0.94.
The riskfree rate of return is 1
− 1 = 6.4%
0.94
In general1 , the price of a zerocoupon bond is given by
r= pUH + (1 − p)UL
and the riskfree rate of return on a zerocoupon bond is given by
r= 1
− 1.
pUH + (1 − p)UL Thus, the price of a zerocoupon bond is just 1
.
1+r Now, let CH be the price of the stock in the high state and CL the price in the low state. The price
of the risky stock is given by
Price of stock = pUH CH + (1 − p)UL CL .
The expected future value on the stock is
pCH + (1 − p)CL .
Therefore, the expected return on the stock is given by
α= pCH + (1 − p)CL
− 1.
pUH CH + (1 − p)UL CL Example 85.5 (Valuing bond)
Assume that the high and low states are equally likely, and UH = 0.90 and UL = 0.98. The cash
ﬂow of the nondividend paying stock is $100 in the low state and $200 in the high state.
(a) What is the price of the stock?
(b) What is expected future value of the stock?
(c) What is the expected return on the stock?
1 See page 371 of [1] 544 SUPPLEMENT Solution.
(a) The price of the stock is
Price of stock = pUH CH + (1 − p)UL CL = 0.5 × 0.90 × 200 + 0.5 × 0.98 × 100 = 139.
(b) The expected future value on the stock is
pCH + (1 − p)CL = 0.5 × 200 + 0.5 × 100 = 150.
(c) The expected return on the stock is given by
α= pCH + (1 − p)CL
150
−1=
− 1 = 7.9%
pUH CH + (1 − p)UL CL
139 The current price of the asset can be rewritten as follows
pUH CH + (1 − p)UL CL =[pUH CH + (1 − p)UL CL ] × 1+r
1+r =[p(1 + r)UH CH + (1 − p)(1 + r)UL CL ] ×
=[pu CH + (1 − pu )CL ] × 1
1+r 1
1+r where the highstate riskneutral probability is
pu = pUH
pUH + (1 − p)UL and the lowstate riskneutralprobability is
1 − pu = (1 − p)UL
.
pUH + (1 − p)UL Thus, by taking the expected future value of the stock using the riskneutral probabilities and
discount back at the riskfree rate, we get the correct price for the stock.
Example 85.6
Assume that the high and low states are equally likely, and UH = .90 and UL = .98. In the high
state of the economy a nondividend paying stock will be worth 200, while in the low state it will
be worth 100. Use the riskneutral approach to ﬁnd the price of the stock. 85 UTILITY THEORY AND RISKNEUTRAL PRICING 545 Solution.
The riskneutral probability in the high state is
pu = 0.5 × 0.90
pUH
=
= 47.87%
pUH + (1 − p)UL
0.5 × 0.90 + 0.5 × 0.98 and the riskneutral probability in the low state is
1 − pu = 1 − 0.4787 = 52.13%.
On the other hand,
1
= pUH + (1 − p)UL = 0.5 × 0.90 + 0.5 × 0.98 = 0.94.
1+r
Thus, the price of the stock is
(1 + r)−1 [pu CH + (1 − pu )CL ] = 0.94[0.4787 × 200 + 0.5213 × 100] = $139
which matches the price found using real probabilties in Example 85.5
This riskneutral pricing worked because the same actual probabilities and the same utility weights,
UL and UH , applied to both the riskfree bond and the risky asset. This will be true for traded
assets. Thus, riskneutral pricing works for traded assets and derivatives on traded assets. Riskneutral pricing will not work if the asset can not be either traded, hedged, or have its cashﬂows
duplicated by a traded assets.2 2 See page 374 of [1]. 546 SUPPLEMENT Practice Problems
Problem 85.1
Verify that the function U (x) = 1 − e−αx , α > 0, is a valid utility function?
Problem 85.2
You are an art collector with wealth consisting of $40,000. There is a painting that you would like
to consider buying. The painting is for sale for $10,000. There is an annual probability of 10% of
total loss of the painting, and no partial loss of the painting is possible. What will be the maximum
premium that you have to pay to insure the painting using the utility function U (x) = 1 − e−0.0001x ?
For the remaining problems, you are given the following information:
• In one year the economy will be in one of two states, a high state or a low state.
• The value today of $1 received one year from now in the high state is UH = 0.87.
• The value today of $1 received one year from now in the low state is UL = 0.98.
• The probability of the high state is 0.52.
• The probability of the low state is 0.48.
• XYZ stock pays no dividends.
• The value of XYZ stock one year from now in the high state is CH = $180.
• The value of XYZ stock one year from now in the low state is CL = $30.
Problem 85.3
What is the price of a riskfree zerocoupon bond?
Problem 85.4
What is the riskfree rate?
Problem 85.5
What is pu , the riskneutral probability of being in the high state?
Problem 85.6
What is the price of XYZ Stock?
Problem 85.7
What is expected return on the stock?
Problem 85.8
What is the price of 1year European call option on XYZ stock with strike $130 under real probabilities?
Problem 85.9
What is the price of 1year European call option on XYZ stock with strike $130 under riskneutral
probabilities? Answer Key
The answer key to the book can be requested directly from the author through email: mﬁ[email protected] 547 548 ANSWER KEY Bibliography
[1] R.L. McDonald , Derivatives Markets, 2nd Edition (2006), Pearson.
[2] M.B. Finan , A Basic Course in the Theory of Interest and Derivatives Markets: A Preparation
for Exam FM/2, (2008).
[3] M.B. Finan , A Probability Course for the Actuaries: A Preparation for Exam P/1, (2007).
[4] SOA/CAS, Exam MFE Sample Questions.
[5] G. Stolyarov, The Actuary’s Free Study Guide for Exam 3F /MFE. 549 ...
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This note was uploaded on 04/09/2012 for the course MATH 172a taught by Professor Kong,l during the Fall '08 term at UCLA.
 Fall '08
 Kong,L
 Math, The Land

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