Solution to Derivatives Markets : for Exam FM
Yufeng Guo
June 24, 2007
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c Yufeng Guo
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ii
Contents
Intro duction
1
vii
Intro duction to derivatives
1
2
Intro duction to forwards and options
3
Insurance, collars, and other strategies
29
4
Intro duction to risk management
79
5
Financial forwards and futures
129
8
Swaps
141
iii
7
CONTENTS
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CONTENTS
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iv
Preface
This is Guo’s solution to Derivatives Markets (2nd edition ISBN 0-321-28030X) for Exam FM. Unlike the official solution manual published by AddisonWesley, this solution manual provides solutions to both the even-numbered and
odd-numbered problems for the chapters that are on the Exam FM syllabus.
Problems that are out of the scope of the FM syllabus are excluded.
Please report any errors to yufeng_guo@m
yufeng_guo@msn.com
sn.com.
This book is the excl
exclusiv
usivee proper
property
ty of Yufeng Guo. Redistrib
Redistribution
ution of this
book in any form is prohibited.
v
PREFACE
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PREFACE
c Yufeng Guo
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vi
Introduction
Recommendations on using this solution manual:
1. Obviously
Obviously,, you’ll need to buy Derivativ
Derivatives
es Markets (2nd edition) to see the
problems.
2. Make sure you do
download
wnload the textbook errata from http://www.ke
http://www.kellogg.
llogg.
northwestern.edu/faculty/mc
northwestern.e
du/faculty/mcdonald/htm/ty
donald/htm/typos2e_01.html
pos2e_01.html
vii
CHAPTER
CHAPTE
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INTRODUCTIO
DUCTION
N
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viii
Chapter 1
Introduction to derivatives
Problem 1.1.
Derivatives on weather are not as farfetched as it might appear. Visit http:
//www.cme.com/trading/ and you’ll find more than a dozen weather derivativ
derivatives
es
currently traded at CME such as "CME U.S. Monthly Weather Heating Degree
Day Futures" and "CME U.S. Monthly Weather Cooling Degree Day Futures."
a. Soft drink
drink sales greatly
greatly depend on wea
weather.
ther. Generally
Generally,, warm weather
weather
boosts soft drink sales and cold weather
weather reduc
reduces
es sales. A soft drink producer
can use weather futures contracts to reduce the revenue swing caused by weather
and smooth its earnings. Shareholders of a company generally want the earnings
to be steady. They don’t want the management to use weather as an excuse for
poor earnings or wild fluctuations of earnings.
b. The recreation
recreational
al skiing industry
industry grea
greatly
tly dependents
dependents on weat
weather.
her. A ski
resort can lose money due to warm temperatures, bitterly cold temperatures, no
snow,
sno
w, too little
little snow, or too much snow. A resort can use weather
weather derivativ
derivatives
es
to reduce its revenue risk.
c. Extremely hot or cold weather will result in greater demand for electricity.
An electric utility company faces the risk that it may have to buy electricity at
a higher spot price.
d. Fewer
ewer people will visit an amu
amusemen
sementt park under extrem
extremee wea
weather
ther conditions. An amusement park can use weather derivatives to manage its revenue
risk.
How can we buy or sell weather? No one can accurately predict weather. No
one can deliver
deliver weather.
weather. For people
p eople to trade on wea
weather
ther deriv
derivativ
atives,
es, weather
indexes
index
es need to be invent
invented
ed and agree
agreed
d upon. Once we have weat
weather
her indexes
indexes,,
we can link the payoff of a wea
weather
ther deriv
derivativ
ativee to a weather
weather index. For more
information on weather derivatives, visit:
• http://hometown.aol.com/gml
http://hometown.aol.com/gml1000/wrms.htm
1000/wrms.htm
• http://www.investopedia.com
http://www.investopedia.com
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DUCTION TO DERIVATIVES
Problem 1.2.
• Anyone (such as speculators and investors) who wants to earn a pro fit can
enter weather futures. If you can better predict a weather index than does
the market maker, you can enter weather futures and make a pro fit. Of
course, it’s hard to predict a weather index and hence loss may occur.
• Two companies with opposite risks may enter weather futures as counter
parties.
partie
s. For exa
exampl
mple,
e, a sof
softt drink
drink compan
company
y and a ski
ski-re
-resor
sortt operato
operatorr
havee opposite hedging need
hav
needss and can enter a futures contract.
contract. The soft
drink company can have a positive payo ff if the weather is too cold and
a negative payoff if warm.
warm. This way,
way, when the weather
weather is too cold, the
soft drink company can use the gain from the weather futures to o ffset its
loss in sales. Since the soft drink company makes good money when the
weather is warm, it doesn’t mind a negative payoff when the weather is
cold.. On the other hand, the ski resort can have a negativ
cold
negativee payoff if the
weather is too cold and a positive payoff if too warm. The ski resort can
use the gain from the futures to o ffset its loss in sales.
Problem 1.3.
a.
100 × 41.
41.05 + 20 = 4125
b.
100 × 40.
40.95 − 20 = 4075
c.
For each
each stoc
stock,
k, you
you bu
buy
y at $41.
41.05 and sell it an instant later for $40
$ 40..95.
95.
The total loss due to the ask-bid spread: 100(41
100(41..05 − 40.
40.95) = 10.
10. In addition,
you pay $20 twice. Your total transaction cost is 10
1000 (41.
(41.05 − 40.
40.95)
95) + 2 (2
(20)
0) =
50
Problem 1.4.
a.
b.
100 × 41.
41.05 + 100 × 41.
41.05 × 0.003 = 4117.
4117. 315
100 × 40.
40.95 − 100 × 40.
40.95 × 0.003 = 4082.
4082. 715
c.
For each
each stoc
stock,
k, you
you bu
buy
y at $41.
41.05 and sell it an instant later for $40
$ 40..95.
95.
The total loss due to the ask-bid spread: 100(41
100(41..05 − 40.
40.95) = 10.
10. In addition,
your pay commission 100 × 41.
41.05 × 0.003 + 100 × 40.
40.95 × 0.003 = 24.
24. 6. Your
total transaction cost is 10 + 24.
24. 6 = 34.
34. 6
Problem 1.5.
The market maker buys a security for $100 and sells it for $100
$100..12.
12. If the
the
market maker buys 100 securities and immediately sells them, his pro fit is
100(100..12 − 100) = 12
100(100
Problem 1.6.
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N TO DERIV
DERIVA
ATIVES
Your sales proceeds: 300(30
300(30..19) − 300(30
300(30..19)
19) (0.
(0.005) = 9011.
9011. 715
Your cost of buying 300 shares from the market to close your short position
is:
300 (29.
(29.87) + 300
300 (29.
(29.87)
87) (0.
(0.005) = 9005.
9005. 805
Your profit: 9011
9011.. 715 − 9005
9005.. 805 = 5.
5. 91
Problem 1.7.
a. Consider
Consider the bid-ask
bid-ask spread but ignore commissi
commission
on and interest.
interest.
Your sales proceeds: 400(25
400(25..12) = 10048
Your cost of buying back: 40
400
0 (23
(23..06) = 9224
Your profit: 10048 − 9224 = 824
b. If the bid-ask spread and 0.3% commission
Your sales proceeds: 400(25
400(25..12) − 400(25
400(25..12)(0
12)(0..003) = 10017.
10017. 856
Your cost of buying back: 40
400
0 (23
(23..06) + 400(23.
400(23.06)
06) (0
(0..003) = 9251.
9251. 672
Your profit: 10017
10017.. 856 − 9251
9251.. 672 = 766.
766. 184
Profit drops by: 824 766
766.. 184 = 57.
57. 816
−
c. Your sales proceeds stay in your
your marg
margin
in account,
account, serving
serving as a collater
collateral.
al.
Since you earn zero interest on the collateral, your lost interest is
If ignore commission: 1004
10048
8 (0.
(0.03) = 301.
301. 44
If consider commission: 10017
10017.. 856
856 (0.
(0.03) = 300.
300. 54
Problem 1.8.
By signing the agreement, you allow your broker to act as a bank, who lends
your stocks to someone else and possibly earns interest on the lent stocks.
Short sellers typically leave the short sale proceeds on deposit with the broker,, along with additional
ker
additional capital called
called a haircut.
haircut. The short sale proceeds
proceeds and
the haircut serve as a collateral. The short seller earns interest on this collateral.
This interest is called the short rebate in the stock market.
The rebate rate is often
often equal to the preva
prevailin
iling
g mark
market
et interest
interest rate.
rate. HowHowever, if a stock is scarce, the broker will pay far less than the prevailing interest
rate, in which case the broker earns the di fference between the short rate and
the prevailing interest rate.
This arrangement makes short selling easy. Also short selling can be used to
hedge financ
nancial
ial risks, which
which is good for the economy
economy. By the wa
way
y, you are not
hurt in any way by allowing your broker to lend your shares to short sellers.
Problem 1.9.
According to http://www.
http://www.investorwords
investorwords.com
.com, the ex-dividend date was
created to allow all pending transactions to be completed before the record date.
If an investor does not own the stock before the ex-dividend date, he or she will
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CHAPTER
CHAPTE
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INTRODUCTION
DUCTION TO DERIVATIVES
be ineligible for the dividend payout. Further, for all pending transactions that
have not been completed by the ex-dividend date, the exchanges automatically
reduce
red
uce the pric
pricee of the stock
stock by the amoun
amountt of the divide
dividend.
nd. This
This is done
because a dividend payout automatically reduces the value of the company (it
comes from the company’s cash reserves), and the investor would have to absorb
that reduction in value (because neither the buyer nor the seller are eligible for
the dividend).
If you borrow stock to make a short sale, you’ll need to pay the lender the
dividend
divi
dend distrib
distributed
uted while you maintain
maintain your short position.
position. According
According to the
IRS, you can deduct these payments on your tax return only if you hold the
short sale open for a minimum period (such as 46 days) and you itemize your
deductions.
In a perfect market, if a stock pays $5 dividend, after the ex-dividend date,
the stock
stock price
price will be reduce
reduced
d by $5. Then
Then you
you could
could buy back
back stocks
stocks from
the market
market at a reduc
reduced
ed price to close your short positi
position.
on. So you don’t need
to worry whether the dividend is $3 or $5.
However, in the real world, a big increase in the dividend is a sign that
a compan
company
y is doing bette
betterr tha
than
n expe
expecte
cted.
d. If a compan
company
y pa
pays
ys a $5 dividen
dividend
d
instead of the expected $3 dividend, the company’s stock price may go up after
the announcement that more dividend will be paid. If the stock price goes up,
you have
have to buy back stocks
stocks at a higher price to close your short position.
position. So
an unexpected increase
increase of the dividend may hurt you.
In addition, if a higher dividend is distributed, you need to pay the lender
the dividend while you maintain your short position. This requires you to have
more capital on hand.
In the real world, as a short seller, you need to watch out for unexpected
increases of dividend payout.
Problem 1.10.
http://www.investopedia.c
http://www.i
nvestopedia.com/articles/0
om/articles/01/082201.asp
1/082201.asp offers a good ex-
planation of short interest:
Short Interest
Short interest is the total number of shares of a particular stock that have
be
been
en sold short by investors
investors but have not yet been
been covered
covered or closed
closed out. This
can be expressed as a number or as a percentage.
When expresse
expressed
d as a pe
perc
rcenta
entage,
ge, short interest
interest is the numb
number
er of shorte
shorted
shares
shar
es divi
divide
ded
d by the numb
number
er of shares outstandin
outstanding.
g. For example,
example, a stock
stock with
1.5 million shares sold short and 10 million shares outstanding has a short interest of 15% (1.5 million/10 million = 15%).
Most stock exchanges track the short interest in each stock and issue reports
at month’s end. These reports
reports are great
great because
because by showing
showing what short sellers
are doing, they allow investors to gauge overall market sentiment surrounding
a pa
partic
rticular
ular stock.
stock. Or alternati
alternatively,
vely, most exchanges
exchanges provide
provide an online tool to
calculate short interest for a particular security.
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CHAPTER
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INTRODUCTION
N TO DERIV
DERIVA
ATIVES
Reading Short Interest
A lar
large
ge increa
increase
se or de
decr
crea
ease
se in a stock’s
stock’s short
short intere
interest
st fr
from
om the previo
previous
us
month
mon
th can be a very
very telling
telling in
indic
dicato
atorr of invest
investor
or senti
sentimen
ment.
t. Let’s
et’s say that
Microsoft’s (MSFT) short interest increased by 10% in one month. This means
that there was a 10% increase in the amount of people who believe the stock will
de
decr
crea
ease.
se. Such a significant shift provides good cause for us to find out more.
We would need to check the current research and any recent news reports to
see what is happening with the company and why more investors are selling its
stock.
A high short-int
short-inter
erest
est stock
stock should
should be approache
approached
d for buying
buying with extreme
extreme
caution but not necessarily avoided at all costs. Short sellers (like all investors)
aren’t perfect and have been known to be wrong from time to time.
In fact, many contrarian investors use short interest as a tool to determine
the direc
direction
tion of the market. The rationa
rationale
le is that if everyone
everyone is selling, then
the stock
stock is alr
alreeady
ady at its low and
and can only
only mov
movee up
up.. Thus,
Thus, contr
contrari
arians
ans feel
feel
that a high short-interest ratio (which we will discuss below) is bullish - because
eventually there will be significant upward pressure on the stock’s price as short
sellers cover
cover their
their short positi
positions
ons (i.e. buy back the stocks
stocks they borrowe
borrowed
d to
return to the lender).
The more likely that investors can speculate on the stock, the higher the
demand for the stock and the higher the short interest.
A broker
broker can short sell more than his existin
existing
g inv
invent
entory
ory.. For example, if a
broker has 500 shares of IBM stocks, he can short sell 600 shares of IBM stocks
as long as he knows where to find the additional
additional 100 shares of IBM stocks. If
all the brokers simultaneously lend out more than what they have in their stock
inventories, then the number of stocks sold short might exceed the total number
of the stocks outstanding.
NASDAQ short interest is available by issue for a rolling twelve months
and is based
based on a mid
mid-mo
-mont
nth
h settle
settlemen
mentt date.
date. For more
more inf
inform
ormati
ation,
on, vis
visit
it
http://www.nasdaqtrader.co
http://www.na
sdaqtrader.com/asp/short_i
m/asp/short_interest.asp
nterest.asp.
Problem 1.11.
You go to a bank. The bank uses its custome
customers’
rs’ deposit
depositss and lends
lends you an
asset worth
worth $100. Then 90 days later you buy back the asset at $102 from the
open market
market (i.e. you come up with $102 from whatev
whatever
er sources)
sources) and return
return
$102 to the bank. Now your short position is closed.
Problem 1.12.
We need to borrow an asset called money from a bank (the asset owner) to
pay for a new house.
house. The asset owne
ownerr faces credit risk (the risk that we may
not be able to repay the loan). To protect itself, the bank needs collateral.
The house is collateral. If we don’t pay back our loan, the bank can foreclose
the house.
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INTRODUCTION
DUCTION TO DERIVATIVES
To protect against the credit risk, the bank requires a haircut (i.e. requires
that the collateral
collateral is great
greater
er than the loan). Typical
Typically
ly,, a bank lends only 80%
of the purchase price of the house, requiring the borrower to pay a 20% down
payment.
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Chapter 2
Introduction to forwards
and options
Problem 2.1.
Long a stock=Own a stock (or buy a stock).
If you own a stock, your payoff at any time is the stock’s market price because
you can sell it any time at the market price. Let S represent the stock price at
T = 1.
1.
Your payoff at T = 1 is S .
Your profit at T = 1 is:
Payoff - FV(initial investment)= S − 50(1
50(1..1) = S − 55.
55.
You can see that the pro fit is zero when the stock price S = 55.
55. Alternatively,
set S − 55 = 0 → S = 55.
55.
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Payoff=S
INTR
INTRODUCTIO
ODUCTION
N TO F
FOR
ORW
WARDS AND OPTIONS
OPTIONS
Profit=
t=S
S − 55
80
60
Payoff
40
20
0
10
20
30
40
50
60
70
80
Stock Price
-20
Profit
-40
Payoff and profit: Long one stock
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INTR
INTRODUCTIO
ODUCTION
N TO F
FOR
ORW
WARDS AND OPTION
OPTIONS
S
Problem 2.2.
Short a stock=Short
stock=Short sell a stock. If you short sell a stock
stock,, your
your payoff at
any time after
after the short sale is the negativ
negativee of the stock’s
stock’s market price.
price. This
is because to close your short position you’ll need to buy the stock back at the
market
mark
et price and return it to the broker. Your payoff at T = 1 is − S . Your
profit at T = 1 is: Payoff - FV(initial investment)= −S + 50(1.
50(1.1) = 55 − S
You can see that the pro fit is zero when the stock price S = 55.
55. Alternatively,
set 55 − S = 0 → S = 55.
55.
Payoff or Profit
50
Profit
40
30
20
10
0
10
20
30
-10
40
50
60
Stock Price
-20
-30
Payoff
-40
-50
-60
Payoff and profit: Short one stock
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CHAPTER
CHAPTE
R 2.
INTR
INTRODUCTIO
ODUCTION
N TO F
FOR
ORW
WARDS AND OPTIONS
OPTIONS
Problem 2.3.
The opposite of a purchased call is written call (or sold call).
The opposite of a purchased put is written put (or sold put).
The main idea of this problem is:
= a purchased put
• The opposite of a purchased call 6
• The opposite of a purchased put 6
= a purchased call
Problem 2.4.
a. Long forward means being a buyer in a forward contract.
Payoff of a buyer in a forward at T is
Payoff = S T − F = ST − 50
ST Payoff = ST − 50
40
45
50
55
60
−−105
0
5
10
b. Payoff of a long call (i.e. owning a call) at expiration T is:
max (0
(0,, ST − 50)
Payoff = max
max (0,
(0, ST − K ) = max
ST
40
45
50
55
60
Payoff = ma
max
x (0,
(0, ST
0
0
0
5
10
− 50)
c. A call option is a privilege. You exercise a call and buy the stock only if
your payoff is positive.
In contrast,
contrast, a forward is an obli
obligatio
gation.
n. You need to buy the stock even if
your payoff is negative.
A privilege is better than an obligation.
Consequently, a long call is more expensive than a long forward on the same
underlying stock with the same time to expiration.
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CHAPTER
CHAPTE
R 2.
INTR
INTRODUCTIO
ODUCTION
N TO F
FOR
ORW
WARDS AND OPTION
OPTIONS
S
Problem 2.5.
a. Short forward = Enter into a forward as a seller
Payoff of a seller in a forward at T is
Payoff = F − ST = 50 − ST
ST Payoff = 50 − ST
40
10
45
5
50
0
55
−5
60
−10
b. Payoff of a long put (i.e. owning a put) at expiration T is:
Payoff = max
max (0,
(0, K − ST ) = max(0,
max(0, 50 − ST )
ST
Payoff = max
max (0,
(0, 50 − ST )
40
45
50
55
60
10
5
0
0
0
c. A put option is a privi
privilege
lege.. You exerci
exercise
se a put and sell the stock only if
your payoff is positive.
In contrast,
contrast, a forw
forward
ard is an oblig
obligation
ation.. You need to sell the stock
stock even if
your payoff is negative.
A privilege is better than an obligation.
Consequently, a long put is more expensive than a short forward on the same
underlying stock with the same time to expiration.
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CHAPTER
CHAPTE
R 2.
INTR
INTRODUCTIO
ODUCTION
N TO F
FOR
ORW
WARDS AND OPTIONS
OPTIONS
Problem 2.6.
91(1 + r ) = 100
→ r = 0.0989
The effective annual interest rate is 9.89%.
If you buy the bond at t = 0, your payoff at t = 1 is 100
Your profit at t = 1 is 100 − 91 (1 + 0.0989) = 0 regardless of the stock price
at t = 1.
If you buy a bond, you just earn the risk
risk-fre
-freee intere
interest
st rate. Beyond
Beyond this,
your profit is zero.
Payoff
101.0
100.5
100.0
99.5
99.0
0
10
20
30
40
50
60
Stock Price
Payoff: Long a bond
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CHAPTER
CHAPTE
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Profit
INTR
INTRODUCTIO
ODUCTION
N TO F
FOR
ORW
WARDS AND OPTION
OPTIONS
S
1.0
0.8
0.6
0.4
0.2
0.0
10
20
30
40
50
60
70
-0.2
80
90
100
Stock Price
-0.4
-0.6
-0.8
-1.0
Profit of longing a bond is zero.
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CHAPTER
CHAPTE
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INTR
INTRODUCTIO
ODUCTION
N TO F
FOR
ORW
WARDS AND OPTIONS
OPTIONS
If you sell the bond at t = 0, your payoff at t = 1 is −100 (you need to pay
the bond holder 100).
Your profit at t = 1 is 91 (1 + 0.0989) − 100 = 0 regardless of the stock price
at tIf=you
1 sell a bond, you just earn the risk-free interest rate. Beyond this, your
profit is zero.
0
Payoff
10
20
30
40
Stock Price
50
60
-99.0
-99.5
-100.0
-100.5
-101.0
Payoff: Shorting a bond
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CHAPTER
CHAPTE
R 2.
Profit
INTR
INTRODUCTIO
ODUCTION
N TO F
FOR
ORW
WARDS AND OPTION
OPTIONS
S
1.0
0.8
0.6
0.4
0.2
0.0
10
20
30
40
50
60
70
-0.2
80
90
100
Stock Price
-0.4
-0.6
-0.8
-1.0
Profit of shorting a bond is zero.
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CHAPTER
CHAPTE
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INTR
INTRODUCTIO
ODUCTION
N TO F
FOR
ORW
WARDS AND OPTIONS
OPTIONS
Problem 2.7.
a. It costs nothi
nothing
ng for one to ent
enter
er a forward
forward contract.
contract. Hence
Hence the payo
payoff of
a forward is equal to the profit.
Suppose we long a forward (i.e. we are the buyer in the forward). Our payoff
and profit at expiration is:
ST − F = S T − 55
Payoff (Profit)
40
30
20
10
0
10
20
30
40
50
60
70
80
90
100
Stock Price
-10
-20
-30
-40
-50
Payoff (and profit) of a long forward
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16
CHAPTER
CHAPTE
R 2.
INTR
INTRODUCTIO
ODUCTION
N TO F
FOR
ORW
WARDS AND OPTION
OPTIONS
S
Suppose we short a forward (i.e. we are the seller in the forward), our payo ff
and profit at expiration is:
F − ST = 55 − ST
Payoff (Profit)
50
40
30
20
10
0
10
20
30
40
50
60
-10
70
80
90
100
Stock Price
-20
-30
-40
Payoff (and profit) of a short forward
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17
CHAPTER
CHAPTE
R 2.
INTR
INTRODUCTIO
ODUCTION
N TO F
FOR
ORW
WARDS AND OPTIONS
OPTIONS
b. If the stock doesn’t pay dividend,
dividend, buying
buying a stock outrigh
outrightt at t = 0 and
getting a stock at T = 1 through a forward are identical. There’s no benefit to
owning a stock early.
c. Suppose the stock pays dividend before the forward expiration date T = 1.
1.
Please note that if you own a stock prior to the dividend date, you will receive
the dividend.
dividend. In contr
contrast,
ast, if you are a buyer in a forward contract
contract,, at T = 1,
you’ll get a stock but you won’t receive any dividend.
• If the stock is expected to pay dividend, then the stock price is expected
to drop after the dividend is paid. The forward price agreed upon at t = 0
already considers that a dividend is paid during (0
(0,, T ); the dividend will
reduce the forward rate. There’s no advantage to buying a stock outright
over buying a stock through a forward. Otherwise, there will be arbitrage
opportunities.
• If the stock is not expected to pay dividend but actually pays dividend
(a surprise dividend), then the forward price F agreed upon at t = 0 was
set without knowing
knowing the surprise divide
dividend.
nd. So F is the forward price on
a non-dividend paying stock. Since dividend reduces the value of a stock,
F is higher than the forward price on an otherwise identical but dividendpa
payin
yingg stock
stock.. If yo
you
u own a stock
stock at t = 0, you’ll
you’ll receive
receive the windfall
windfall
divide
div
idend.
nd. If you buy a stoc
stock
k thr
throug
ough
h a for
forwa
ward,
rd, you’ll
you’ll pay F , which
which
is higher than the forward price on an otherwise identical but dividendpaying stock. Hence owning a stock outright is more beneficial than buying
a stock through a forward.
Problem 2.8.
r =risk
= risk free interest rate
Under the no-arbitrage principle, you get the same pro fit whether you buy
a stock outright or through a forward.
Profit at T = 1 if you buy a stock at t = 0 is: S T 50 (1 + r)
− ST − 53
Profit at T = 1 if you buy a stock through a forward:
→ ST − 50 (1 + r) = ST − 53
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50 ((11 + r) = 53
°
c Yufeng Guo
r = 0.06
18
CHAPTER
CHAPTE
R 2.
INTR
INTRODUCTIO
ODUCTION
N TO F
FOR
ORW
WARDS AND OPTION
OPTIONS
S
Problem 2.9.
a. Price of an index forward contract expiring in one year is:
F Index = 1000
1000 (1.
(1.1) = 1100
To see why: If the seller borrows 1000 at t = 0, buys an index, and holds it
for one year, then he’ll have one stock to deliver at T = 1. The seller’s
seller’s cost is
10000 (1
100
(1..1) = 1100.
1100. To avoid arbitrage, the forward price must be 1100
1100..
Profit at T = 1 of owning a forward on an index:
ST − F Index = ST − 1100
If you buy an index at t = 0, your profit at T = 1 is ST
ST − 1100
− 100
10000 (1
(1..1) =
So you get the same profit whether you buy the index outright or buy the
index
inde
x through a forward.
forward. This should mak
makee sense. If ownin
owning
g a stock outrigh
outrightt
and buying it through a forward have di fferent profits, arbitrage opportunities
exist.
b. The forward
forward price
price 1200 is greater than the fair forward price 1100
1100.. No
rational
rati
onal person will want to ent
enter
er such an unfai
unfairr forw
forward
ard contract.
contract. Thus the
seller
seller need
needss to pay the buy
buyer
er an up-fr
up-fron
ontt pre
premi
mium
um to incite
incite the buyer.
buyer. The
1200 − 1100
buyer in the forward needs to receive
= 90.
90. 91 at t = 0 to make
1.1
the forward
forward contrac
contractt fair
fair.. Of course, the buy
buyer
er needs to pay the forward
forward price
price
1200 at T = 1.
1.
c. No
Now
w the forward
forward price
price 100
1000
0 is low
lower
er tha
than
n the fair forw
forward
ard price
price 1100.
1100.
You can imagine thousands of bargain hunters are waiting in line to enter this
forward contract. If you want to enter the forward contract, you have to pay the
1100 − 1000
seller a premium in the amount of
= 90.
90. 91 at t = 0. In addition,
addition,
1.1
you’ll need to pay the forward price 1000 at T = 11..
Problem 2.10.
95.68
a. Profit=
t=ma
max
x (0,
(0, ST − 1000) − 95.
Set profit to zero:
95.68 = 0
max
ma
x (0
(0,, ST − 1000) − 95.
→ ST − 1000 − 95.
95.68 = 0
ST = 1000 + 95.
95.68 = 1095.
1095. 68
b. The profit of a long forward (i.e. being a buyer in a forward): S T
95.68
1020
0 = max (0,
(0, ST − 1000) − 95.
ST − 102
If S T > 1000
1000,, ther
there’s
e’s no solution
solution
If S T ≤ 1000
1000::
ST − 1020 = 0 − 95.
95.68
→ ST = 1020 − 95.
95.68 = 924.
924. 32
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− 1020
19
CHAPTER
CHAPTE
R 2.
INTR
INTRODUCTIO
ODUCTION
N TO F
FOR
ORW
WARDS AND OPTIONS
OPTIONS
Problem 2.11.
a. Profit of a long (i.e. owning) put is max(0
max(0,, 1000 − ST ) − 75.
75.68
max
ma
x (0,
(0, 1000 − ST ) − 75.
75.68 = 0
75.68 = 0
ST = 924.
924. 32
1000 − ST − 75.
b. Profit of a short forward (i.e. being a seller in a forward) is 1020 − ST
max (0,
max
(0, 1000 − ST ) − 75.
75.68 = 1020 − ST
If 1000 ≥ ST
1000 − S − 75.
75.68 = 1020 − S
If 1000 ≤ ST
−75.
75.68 = 1020 − ST
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no solution
ST = 1020 + 75.
75.68 = 1095.
1095. 68
°
c Yufeng Guo
20
CHAPTER
CHAPTE
R 2.
INTR
INTRODUCTIO
ODUCTION
N TO F
FOR
ORW
WARDS AND OPTION
OPTIONS
S
Problem 2.12.
Table 2.4 is:
Position
Long
Long for
orw
ward
ard (bu
(buyer in forw
rwar
ard)
d)
Short forward (seller in forward)
Long call (own a call)
Short call (sell a call)
Long put (own a put)
Sho
Short put
put (sel
sell a put
put)
Maximum Loss
-For
-Forw
ward
ard pric
pricee
Unlimited
-FV (premium)
Unlimited
-FV(premium)
PV
PV((prem
premiium)um)-S
Str
trik
ikee Pr
Priice
Maximum Gain
Unl
nliimite
mited
d
Forward Price
Unlimited
FV(premium)
Strike Price - FV(premium)
FV(p
FV(prremi
emium)
um)
• If you are a buyer in a forward, the worst that can happen to you is ST = 0
(i.e. stock price
price at T is zero). If this happens, you stil
stilll have to pay the
forward price F at T to buy the stock which is worth zero. You’ll lose F .
Your best case is S T = ∞, where you have an unlimited gain.
• If you are a seller in a forward, the worst case is that S T =
unlimited
unlim
ited loss.
loss. Your best case is that ST
worthless asset for the forward price F .
; you’ll incur
∞ you sell a
= 0, in which case
• If you buy a call, your worst case is ST < K , where K is the strike price. If
this happens, you just let the call expire worth
worthless
less.. You’ll
ou’ll lose the future
future
value of your premium (if you didn’t buy the call and deposit your money
in a bank account, you could earn the future value of your deposit). Your
best case is that S T = ∞, where you’ll have an unlimited gain.
• If you sell a call, your worst case is S T =
∞, in which case you’ll incur an
unlimited loss. Your best case is ST < K , in which case the call expires
unlimited
worthless; the call holder wastes his premium and your profit is the future
value of the premium you received from the buyer.
≥ K , in which case you’ll
let your put expire worthless and you’ll lose the future value of the put
premium.
premi
um. Your best case is ST = 0, in which case you sell a worthless
stock for K . Your profit is K − F V (premium
premium)).
• If you buy a put, your worst case is that ST
• If you sell a put, your worst case is S T = 0, in which case the put holder
sells you a worthless stock for K ; your profit is F V (premium
premium)) − K . Your
best case is ST ≥ K , where the written put expires worthless and your
profit is F V (premium
premium)).
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21
CHAPTER
CHAPTE
R 2.
INTR
INTRODUCTIO
ODUCTION
N TO F
FOR
ORW
WARDS AND OPTIONS
OPTIONS
Problem 2.13.
Let S represent the stock price at the option expiration date.
I’ll draw a separate diagram for the payo ff and a separate diagram for the
profit.
a. Suppose you long a call (i.e. buy a call).
½
if S < 35
S − 35 if S ≥ 35
Your profit at expiration= Payoff - FV (premium)
= max
max (0,
(0, S − 35) − 9.12(1
12(1..08) = max
max (0
(0,, S − 35) − 9. 8496
0
if S < 35
−9. 8496 if S < 35
=
849 6 =
− 9. 849
S − 35 if S ≥ 35
S − 44.
44. 8496 if 35 ≤ S
(i) Payoff at expiration is max(0
max(0,, S − 35) =
½
0
½
60
50
Payoff
40
30
Profit
20
10
0
10
20
30
40
50
60
70
80
90
100
Stock Price
-10
Payoff and Profit: Long a 35 strike call
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22
CHAPTER
CHAPTE
R 2.
INTR
INTRODUCTIO
ODUCTION
N TO F
FOR
ORW
WARDS AND OPTION
OPTIONS
S
½
if S < 40
S − 40 if S ≥ 40
Your profit at expiration= Payoff - FV (premium)
(ii) Payoff at expiration is max(0
max(0,, S − 40) =
0
− 6. 7176
= max(0,
max(00, S − if40) S−<
6.22(1
22(1.
max (0,
(0, S −
40 .08) = max
−640)
. 7176
if S < 40
=
−
6. 717
717 6 =
S − 40 if S ≥ 40
S − 46.
46. 7176 if S ≥ 40
½
½
60
50
Payoff
40
30
Profit
20
10
0
10
20
30
40
50
60
70
80
90
100
Stock price at expiration
Payoff and Profit: Long a 40 strike call
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23
CHAPTER
CHAPTE
R 2.
INTR
INTRODUCTIO
ODUCTION
N TO F
FOR
ORW
WARDS AND OPTIONS
OPTIONS
½
0
if S < 45
S − 45 if S ≥ 45
Your profit at expiration= Payoff - FV (premium)
(iii) Payoff at expiration is max(0
max(0,, S − 45) =
− 4. 4064
= max
max (0,
(0
4.08(1
08(1.
max (0
(0,, S −
0, S − if45) S−<
45 .08) = max
−445)
. 4064
if S < 45
=
−
4. 406
406 4 =
S − 45 if S ≥ 45
S − 49.
49. 4064 if S ≥ 45
½
½
50
Payoff
40
30
Profit
20
10
0
10
20
30
40
50
60
70
80
90
100
Stock price at expiration
Payoff and Profit: Long a 45 strike call
b. The pa
payo
yoff of a long call is ma
max
x (0
(0,, S − K ). As K increases, the payoff
gets worse
worse and the option becomes less val
valuable
uable.. Everythi
Everything
ng else equal, the
higher the strike price, the lower the call premium.
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CHAPTER
CHAPTE
R 2.
INTR
INTRODUCTIO
ODUCTION
N TO F
FOR
ORW
WARDS AND OPTION
OPTIONS
S
Problem 2.14.
Suppose we own a put (i.e. long put).
35
a. Payoff at expiration is ma
max
x (0,
(0, 35 − S ) =
S
if S 35
if S ≤
> 35
0−
½
Your profit at expiration = Payoff - FV (premium)
= max(0,
max(0, 35 − S ) − 1.53(1
53(1..08) = max
max (0,
(0, 35 − S ) − 1. 6524
35 − S if S ≤ 35
33.
33. 3476 − S if S ≤ 35
=
− 1. 652
652 4 =
0
if S > 35
−1. 6524 if S > 35
½
½
30
20
10
0
10
20
30
40
50
60
70
80
90
100
Stock price at expiration
Payoff and profit: Long a 35 strike put
The blue line is the payo ff. The black line is the pro fit.
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25
CHAPTER
CHAPTE
R 2.
INTR
INTRODUCTIO
ODUCTION
N TO F
FOR
ORW
WARDS AND OPTIONS
OPTIONS
½
40 − S
0
Your profit at expiration = Payoff - FV (premium)
b. Payoff at expiration is max(0
max(0,, 40 − S ) =
if S ≤ 40
if S > 40
= max
ma40
x (0,
(0−, 40
3.26(1
26(1.
max (0
(0,, 40
S ) −−3.S5208
36.
36.−4792
if S ≤ 40
S −ifS ) S−≤
40 .08) = max
− 3. 520
520 8 =
=
−3. 5208 if S > 40
0
if S > 40
½
½
40
30
20
10
0
10
20
30
40
50
60
70
80
90
100
Stock price at expiration
Payoff and profit: Long a 40 strike put
The blue line is the payoff. The black line is the profit.
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26
CHAPTER
CHAPTE
R 2.
INTR
INTRODUCTIO
ODUCTION
N TO F
FOR
ORW
WARDS AND OPTION
OPTIONS
S
½
45 − S
0
Your profit at expiration = Payoff - FV (premium)
c. Payoff at expiration is ma
max
x (0,
(0, 45 − S ) =
if S ≤ 45
if S > 45
= max(0,
max(0
5.75(1
75(1.
max (0,
(0
, 45
38.
38
. 79−−SS) − if6. 21S ≤ 45
45 −, 45
S −ifS ) S−≤
45 .08) = max
−
6. 21 =
=
−6. 21 if S > 45
0
if S > 45
½
½
40
30
20
10
0
10
20
30
40
50
60
70
80
90
100
Stock price at expiration
Payoff and profit: Long a 45 strike put
The blue line is the payo ff. The black line is the pro fit.
As the strik
strikee pri
price
ce inc
increa
reases
ses,, the pa
payo
yoff of a put goes up and the more
valuab
valuable
le a put is. Everythi
Everything
ng else equal, the highe
higherr the strike
strike price, the more
expensive a put is.
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CHAPTER
CHAPTE
R 2.
INTR
INTRODUCTIO
ODUCTION
N TO F
FOR
ORW
WARDS AND OPTIONS
OPTIONS
Problem 2.15.
If you borrow money from a bank to buy a $1000 S&R index, your borrowing
cost is known at the time of borrowing.
borrowing. Suppose the annual
annual effective risk free
interest rate is r.
r . If you borrow $1000 at t = 0, then at T you just pay the bank
10000 (1 + r)T . You can slee
100
sleep
p wel
welll knowing that your borrowing
borrowing cost is fixed in
advance.
In contrast, if you short-sell n number of IBM stocks and use the short sale
proceeds to buy a $1000 S&R index, you own the brokerage firm n number of
IBM stocks.
stocks. If you wan
wantt to close your short position
position at time T , you need to
buy n stocks at T . The cost of n stocks at T is nS T , where S T is the price of
IBM stocks per share at T . Since S T is not known in advance, if you use short
selling to fi nance your purchase of a $1000 S&R index, your borrowing cost nS T
cannot
cann
ot be known
known in adv
advance.
ance. This bring
bringss additional
additional risk to your position.
position. As
such, you can’t determine your profit.
Problem 2.16.
Skip this problem.
problem. SOA
SOA is unlikel
unlikely
y to ask you to design a spreadshee
spreadsheett on
the exam.
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28
Chapter 3
Insurance, collars, and
other strategies
Problem 3.1.
The put premium is 74.
74 .201
201.. At t = 0, you
• spend 1000 to buy an S&R index
74 .201 to buy a 1000-strike put
• spend 74.
980..39
• borrow 980
74.201)
• take out (1000 + 74.
980..39 = 93.
93. 811 out of your own pocket.
− 980
So your total borrowing is 980
980..39 + 93.
93. 811 = 1074.
1074. 20.
20.
The future value is 1074
1074.. 20(1
20(1..02) = 1095.
1095. 68
S&R
S&
R in
inde
dex
x
S&
S&R
R Pu
Putt
900
950
1000
1050
1100
1150
1200
100
50
0
0
0
0
0
Payoff
-(Cost+I
-(Cos
t+Int
ntere
erest)
st)
1000
1000
1000
1050
1100
1150
1200
−1095
1095.. 68
−1095
1095.. 68
−1095
1095.. 68
1095.. 68
−1095
1095.. 68
−1095
1095.. 68
−1095
1095.. 68
−1095
29
Profit
1000 − 1095
1095.. 68 = −95.
95. 68
1000 − 1095
1095.. 68 = −95.
95. 68
1000 − 1095
1095.. 68 = −95.
95. 68
1050 − 1095
1095.. 68 = −45.
45. 68
1100 − 1095
1095.. 68 = 4.
4. 32
1150 − 1095
1095.. 68 = 54.
54. 32
1200 − 1095
1095.. 68 = 104.
104. 32
CHAPTER
CHAPTE
R 3. INSURAN
INSURANCE,
CE, COLLARS,
COLLARS, AND OTHER STRA
STRATEGIES
TEGIES
Payoff. The payoff of owning an index is S , where S is the price of the index
at the put expirati
expiration
on
The payoff of owning a put is ma
max
x (0
(0,, 1000 − S ) at expiration.
Total payoff:
S +m
+max
ax (0,
(0, 1000 − S ) = S +
½
Profit is:
1000 if S ≤ 1000
S
if S > 1000
½
Payoff (Profit)
1000 − S
0
− 1095
1095.. 68 =
if S ≤ 1000
=
if S > 1000
½
1000 if S ≤ 1000
S
if S > 1000
½−
95.
95. 68
if S ≤ 1000
S − 1095
1095.. 68 if S > 1000
2000
Payoff
1500
1000
Profit
500
0
20
200
0
40
400
0
60
600
0
80
800
0 10
1000
00 1
120
200
0 1400
1400 1
160
600
0 1800
1800 2
200
000
0
Stock price at expiration
Payoff and Profit: index + put
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30
CHAPTER
CHAPTE
R 3. INSURANCE,
INSURANCE, COLLARS,
COLLARS, AND OTHER STRA
STRATEGIES
TEGIES
Problem 3.2.
At t = 0 you
• short sell one S&R index, receiving $1000
$ 1000
• sell a 1000-strike put, receiving $74
$ 74..201
74.201 = 1074.
1074. 201 in a savings
savings accoun
account.
t. This grows into
into
• deposit 1000 + 74.
1074.. 201(1
1074
201(1..02) = 1095.
1095. 68 at T = 1
The payoff of the index sold short is −S
The payoff of a sold put: − max(0
max(0,, 1000 − S )
The total payoff at expiration is:
½
max
x (0,
(0, 1000 − S ) = −S −
−S −ma
1000 − S
0
if S ≤ 1000
=
if S > 1000
The profit at expiration is:
≤ 1000
1095.. 68 =
1000
if S
1000 + 1095
S
if
S>
½ −−
S&R
S&
R in
inde
dex
x
−900
−950
−1000
−1050
−1100
−1150
−1200
S&
S&R
R Pu
Putt
−100
−50
0
0
0
0
0
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Payoff
−1000
−1000
−1000
−1050
−1100
−1150
−1200
½
95.
95. 68. 68 − S
1095.
1095
-(Cost+
-(Cos
t+In
Inter
terest
est))
1095. 68
1095. 68
1095. 68
1095. 68
1095. 68
1095. 68
−1095
1095.. 68
°
c Yufeng Guo
½−
1000 if S ≤ 1000
−S if S > 1000
if S
S≤
1000
if
> 1000
Profit
1095. 68 = 95.
95. 68
−1000 + 1095.
1095. 68 = 95.
95. 68
−1000 + 1095.
−1000 + 1095.
1095. 68 = 95.
95. 68
−1050 + 1095.
1095. 68 = 45.
45. 68
−1100 + 1095.
1095. 68 = −4. 32
−1150 + 1095.
1095. 68 = −54.
54. 32
−1200 + 1095.
1095. 68 = −104
104.. 32
31
CHAPTER
CHAPTE
R 3. INSURAN
INSURANCE,
CE, COLLARS,
COLLARS, AND OTHER STRA
STRATEGIES
TEGIES
Payoff=
½−
1000 if S ≤ 1000
−S if S > 1000
½−
1000 if S ≤ 1000
1095. 68
−S if S > 1000 + 1095.
Profit=
200
400
600
800
1000
1200
1400
Index Price
1600
1800
2000
0
Profit
-500
-1000
-1500
Payoff
-2000
short index +sell put
You can verify that the profit diagram above matches the textbook Figure
3.5 (d).
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32
CHAPTER
CHAPTE
R 3. INSURANCE,
INSURANCE, COLLARS,
COLLARS, AND OTHER STRA
STRATEGIES
TEGIES
Problem 3.3.
Option 1: Buy S&R index for 1000 and buy a 950
950-strike
-strike put
Option 2: Invest 931
931..37 in a zero-coupon bond and buy a 950
950-strike
-strike call.
Verify that Option 1 and 2 have the same payo ff and the same profit.
Option 1:
If you own an index, your payoff at any time is the spot price of the index
S . The payoff of owning a 950-strike put is ma
max
x (0,
(0, 950 − S ). Your total payoff
at the put expiration is
½
S +m
+max
ax (0
(0,, 950 − S ) = S +
950 − S
0
if S ≤ 950
=
if S > 950
½
950 if S ≤ 950
S if S > 950
To calculate the profit, we need to know the initial inves
investmen
tment.
t. At t = 0,
we spend 1000 to buy an index and 51.
51 .777 to buy the 950-strike put. The total
investment is 1000 + 51.
51.777 = 1051.
1051. 777
777.. The future value of the investment is
1051.. 777
1051
777 (1
(1..02) = 1072.
1072. 81
So the profit is:
950 if S ≤ 950
S if S > 950
½
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− 1072
1072.. 81 =
½
122.. 81
−122
S − 1072
1072.. 81
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if S > 950
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Payoff=
½
½
950 if S ≤ 950
S if S > 950
Profit=
950 if S ≤ 950
S if S > 950
1072.. 81
− 1072
2000
Payoff
1500
1000
500
Profit
0
200
400
600
800
1000
1200
1400
1600
1800
2000
Index
index + put
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Option 2:
Payoff of the zero-coupon bond at T = 00..5 year is: 931.
931.37(1
37(1..02) = 950
Payoff of owning a 950-strike call: max(0
max(0,, S − 950)
Total payoff:
½
950+max
950+
max (0,
(0, S − 950) = 950+
0
if S ≤ 950
=
S − 950 if S > 950
½
950 if S ≤ 950
S if S > 950
To calculate the profit, we need to know the initial inv
investm
estment
ent.. We spend
931..37 to buy a bond and 120
931
120..405 to buy a 950-strike call. The future value of
the investment is (931
(931..37 + 120.
120.405)1
405)1..02 = 1072.
1072. 81.
81. The profit is:
½
950 if S ≤ 950
S if S > 950
− 1072
1072.. 81 =
½
122.. 81
if S ≤ 950
−122
S − 1072
1072.. 81 if S > 950
Option 1 and 2 have the same payoff and the same profit. But
But why?
why? It’s
It’s
because the put-call parity:
C ((K,
K, T ) + P V (K ) = P (K,
( K, T ) + S0
Option 1 consists of buying S&R index and a 950
950-strike
-strike put
Option 2 consists of investing P V (K ) = 950 1.02−1 = 931.
931. 37 and buying
a 950-st
950-strik
rikee call.
call. Due to the putput-cal
calll par
parit
ity
y, Option
Option 1 and 2 ha
have
ve the same
payoff and the same profit.
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Problem 3.4.
Option 1: Short sell S&R index for 1000 and buy a 950
950-strike
-strike call
Option 2: Borrow 931
931..37 and buy a 950
950-strike
-strike put
Verify that Option 1 and 2 have the same payo ff and the same profit.
Option 1: At t = 0.5, your payoff from the short sale of an index is − S ,
Option
where S is the index price at T = 0.5. At T = 0.5, your payoff from owning a
0
if S < 950
call is max(0
max(0,, S − 950) =
.
S − 950 if S ≥ 950
½
Your total payoff is
−S + 0S − 950 ifif SS <≥ 950
=
950
½
½−
S
−950
if S < 950
if S ≥ 950
Please note that when calculating the payo ff, we’ll ignore the sales price of
the index $1
$1, 000 and the call purchase price 120
120.. 405
405.. These two numbers affect
your profit, but they don’t affect your payoff. Your payoff is the same no matter
whether
you sold
index
for $1
the 950-strike
call your
for $10
or $120
$120.
. 41.
41or
. $1000, and no matter whether you buy
Next, let’s find the profit at T = 0.5. At t = 0, you sell an index for 1000
1000..
Of 1000 you get, you spend 120
120.. 405 ≈ 120.
120. 41 to buy a 950-st
950-strik
rikee cal
call.
l. You
have 1000 − 120
120.. 41 = 879.
879. 59 left
left.. This will grow
grow into 879.
879. 59 × 1.02 = 897.
897. 18
at T = 0.
0 .5 At T = 0.
0 .5, your profit is 897
897.. 18 plus the payoff:
Profit =897
897.. 18 +
½
−S
−950
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½
897.
897. 18 − S
−52.
52. 82
if S < 950
if S ≥ 950
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½−
Payoff=
S
if S < 950
−950 if S ≥ 950
½−
S
if S < 950
−950 if S ≥ 950
Profit =897
=897.. 18+
800
Profit
600
400
200
0
200
400
600
800
1000
1200
1400
1600
-200
1800
2000
Index Price
-400
-600
Payoff
-800
Payoff and Profit: Short index + Long call
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Option 2 payoff. At t = 0.5, you need to pay the lender 931
931..37 × 1.02 = 950,
950,
so, a payoff of −950 (you’ll write the lender a check of 950
950).
). At T = 0.
0 .5, the pay950 − S if S < 950
off from buying a 950-strike put is max(0
max(0,, 950 S ) =
.
−
≥
0
if S 950
½
Your total payoff at T = 00..5 is:
−950 +
½
950 − S
0
if S < 950
=
if S ≥ 950
½−
S
if S < 950
−950 if S ≥ 950 .
This is the same as the payoff in Option 1.
Option 2 profit. There are two ways to calculate the profit.
Method 1. The total
Method
total profit is the sum of profit earned from borrowing 931
931..37
and the profit ear
earned
ned by buyin
buying
g a 950
950-st
-strik
rikee put.
put. The profit from borrowing
931..37 is zero; you borrow 931
931
931..37 at t = 0. This grows into 931
931..37 × 1.02 = 950
at T = 0.5 in your savings
savings acco
account
unt.. Then at T = 0.5, you take out 950 from
your savings account and pay the lender. Now your savings account is zero. So
the profit earned from borrowing 931
931..37 is zero.
Next, let’s calculate the profit from buying
buying the put. The put premium
premium is
$51.78. So your profit earned from buying the put option is
−51.
51.78 × 1.02 + max
max (0,
(0, 950 − S ) = −52.
52. 82 + max (0
(0,, 950 − S )
= −52.
52. 82 +
½
950 − S
0
if S < 950
=
if S ≥ 950
½
−
½−
897.
897. 18 − S
52.
52. 82
if S < 950
if S ≥ 950
if S < 950
. We just
if S ≥ 950
need to deduct the future value of the initial investme
investment.
nt. At t = 0, you receive
931..37 from the lender and pay 51.
931
51.78 to buy the
the put. So your
your total
total cash is
931..37 − 51.
931
51.78 = 879.
879. 59,
59, which grows into 879.
879. 59 × 1.02 = 897.
897. 18 at t = 0.5.
Hence, your profit is:
Method
Meth
od 2. We already know the payoff is
½−
S
if S < 950
897. 18 =
−950 if S ≥ 950 + 897.
½
S
−950
897.
897. 18 − S
−52.
52. 82
if S < 950
if S ≥ 950
No matter whether you use Method 1 or Method 2, the Option 2 profit is
the same as the Option 1 pro fit.
You might wonder why Option 1 and Option 2 have the same payo ff and the
same profit. The parity formula is
Call (K, T ) − P ut (K, T ) = P V (F0,T
− K ) = P V (F0,T ) − P V (K )
Rearranging this equation, we get:
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Call (K, T ) + P V (K ) = P ut (K, T ) + P V (F0,T )
Since P V (F0,T ) = S 0 , now we have:
Call (K, T ) +
| {z }
own a call
P V (K )
= P ut (K, T ) +
own PV of strike price
own a put
| {z }
| {z }
S0
|{z}
own one index
The above equation can also be read as;
P V (K )
= P ut (K, T ) +
Call (K, T ) +
| {z }
buy a call
| {z }
invest
inve
st PV of strike price
| {z }
buy a put
Rearranging the above formula, we get:
Call (K, T ) +
−S0 = P ut (K, T ) +
| {z } |{z}
own a call
sell one index
| {z }
own a put
S0
|{z}
buy one index
P V (K )
|− {z }
borrow PV of strike price
The above equation can also be read as:
Call (K, T ) +
−S0
= P ut (K, T ) +
−P V (K )
sell one index
| {z }
borrow PV of strike price
| {z } |{z}
buy a call
buy a put
| {z }
According to the parity equation, Option 1 and Option 2 are identical portfolios and should have the same payo ff and the same profit. In this
this problem,
problem,
Option 1 consists of shorting an S&R index and buying a 950-str
950-strike
ike call.
call. Op−
1
tion 2 consists of borrowing P V (K ) = 950 1.02
= 931.
931. 37 and buying a
950-strike
950
-strike put. As a result, Option 1 and 2 have the same payo ff and the same
profit.
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Problem 3.5.
Option 1: Short sell index for 1000 and buy 1050-strike call
Option
2: Borrow
buy
1050-strike
Verify that
Option1029.41
1 and 2and
have
thea same
payo ff put.
and the same profit.
Option 1:
Payoff:
0
if S < 1050
−S if S < 1050
−S +m
+max
ax (0,
(0, S − 1050) = −S +
=
S − 1050 if S ≥ 1050
−1050 if S ≥ 1050
½
½
Profit:
Your receive 1000 from the short sale and spend 71.802 to buy the 1050-strike
call.
The future value is: (1000 − 71.
71.802)
802) 1.02 = 946.
946. 76
So the profit is:
−S if S < 1050 + 946
−1050 if S ≥ 1050 946.. 76 =
½
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946. 76 − S
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103.. 24
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½−
Payoff=
S
if S < 1050
−1050 if S ≥ 1050
½−
Profit=
S
−1050
if S < 1050
+ 946
946.. 76
if S ≥ 1050
800
Profit
600
400
200
0
200
400
600
800
1000
1200
1400
1600
-200
1800
2000
Index Price
-400
-600
-800
Payoff
-1000
Payoff and Profit: Short index + Long call
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Option 2:
Payoff:
If you borrow 1029
1029..41,
41, you’ll need to pay 1029
1029..41(1
41(1..02) = 1050 at T = 0.
0 .5
So the payoff of borrowing 1029
1029..41 is 1050
1050..
Payoff of the purchased put is ma
max
x (0
(0,, 1050 − S )
Total payoff is:
1050 − S
= −1050 +
0
½
if S < 1050
=
if S ≥ 1050
½−
S
−1050
if S < 1050
if S ≥ 1050
Initially, you receive 1029
1029..41 from a bank and spend 101.
101.214 to buy a 950strik
strikee put
put.. So your net rece
receipt
ipt at t = 0 is 1029
1029..41 − 101
101..214 = 928.
928. 196
196.. Its
future value is 928
928.. 196
196 (1.
(1.02) = 946.
946. 76.
76. Your profit is:
½−
S
if S < 1050
946.. 76 =
−1050 if S ≥ 1050 + 946
½
946.
946. 76 − S
−103
103.. 24
if S < 1050
if S ≥ 1050
Option 1 and 2 have the same payoff and the same profit. This is because
the put-call parity:
Call (K, T ) +
−S0 = P ut (K, T ) +
−P V (K )
| {z } |{z}
buy a call
sell one index
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| {z }
buy a put
| {z }
borrow PV of strike price
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Problem 3.6.
(a) buy an index for 1000
(b)
buythat
a 950-strike
call,
sell the
a 950-strike
put,
andthe
lend
931.pro
931
.37 fit.
Verify
(a) and (b)
have
same payo
same
ff and
(a)’s payoff is S . Profit is S
− 100
1000
0 (1.
(1.02) = S − 1020
(b)’s payoff:
buy a 950-strike call
sell a 950-strike put
lend 931.37
Total
Payoff
max(0, S − 950)
max(0,
max
x (0,
(0, 950 − S )
− ma
931.
931.37(1
37(1..02) = 950
Initial receipt
−120
120..405
51.
51.777
931..37
−931
120..405 + 51.
51.777 − 931
931..37 =
−120
Total payoff:
max
ma
x (0
(0,, S − 950) − max(0
max(0,, 950 − S ) + 950
=
=
0
S
½ −
½−
if S < 950
950 if S ≥ 950
−
½
950 − S
0
if S < 950 + 950
if S ≥ 950
(950 − S ) + 950 if S < 950
=S
S − 950 + 950
if S ≥ 950
Total profit:
S − 100
1000
0 (1.
(1.02) = S − 1020
(a) and (b) have the same payoff and the same profit. Why?
Call (K, T ) +
S0
| {z } |−{z}
buy a call
→
S0
buy one index
|{z}
sell one index
=
= P ut (K, T ) +
| {z }
buy a put
Call (K, T ) +
buy a call
P V (K )
|− {z }
borrow PV of strike price
−P ut (K, T ) +
sell a put
| {z } | {z }
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P V (K )
lend PV of strike price
| {z }
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(a) and (b) have the following common payoff and profit.
Payoff=S
2000
Profit=
t=S
S − 1020
Payoff
1000
0
200
400
600
800
1000
1200
1400
1600
1800
2000
Index Price
Profit
-1000
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Problem 3.7.
(a) short index for 1000
(b)
sellthat
1050-strike
buy athe
1050-strike
andthe
borrow
Verify
(a) and call,
(b) have
same payoput,
same 1029.41
profit.
ff and
(a) Payoff is −S . Profit is −S + 1000
1000 (1.
(1.02) = 1020 − S
(b) Payoff:
max
ma
x (0
(0,, 1050 − S ) − max(0
max(0,, S − 1050) − 1029
1029..41(1
41(1..02)
=
=
½
½
1050 − S
0
if S < 1050
if S ≥ 1050
−
½
0
if S < 1050
S − 1050 if S ≥ 1050
(1050 − S ) − 1050 if S < 1050
− (S − 1050) − 1050 if S ≥ 1050 =
½−
S
−S
− 1050
if S < 1050
= −S
if S ≥ 1050
We need to calculate the initial investment of (b).
At t = 0, we
71 .802 from selling a 1050
1050-strike
-strike call
• Receive 71.
101..214 to buy a 1050
1050-strike
-strike put
• Pay 101
1029..41 from a lender
• Receive 1029
Our net receipt is 71.
71 .802 − 101
101..214 + 1029.
1029.41 = 1000.
1000.
The future value is 100
1000
0 (1.
(1.02) = 1020
So the profit at T = 0.
0 .5 is
−S + 1020 = 1020 − S.
You see that (a) and (b) have the same payo ff and the same profit. Why?
Why?
From the put-call parity, we have:
−S0 = P ut (K, T ) +
−P V (K )
Call (K, T ) +
buy a call
sell one index
buy a put
borrow PV of strike price
| {z } |{z} | {z } | {z }
−| {z }
→ |−{z}
| {z } |− {z }
S0
=
sell one index
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buy a put
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borrow PV of strike price
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Payoff=
−S .
Profit= 1020 − S
1000
Profit
0
200
400
-1000
600
800
1000
1200
1400
1600
1800
2000
Index Price
Payoff
-2000
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Problem 3.8.
Put-call parity:
Call (K, T ) +
buy a call
P V (K )
inv
invest
est PV of strike price
| {z }
| {z }
109.2 + P V (K ) = 60.
109.
60.18 + 1000
P V (K ) = 60.
60.18 + 1000 − 109
109..2 = 950.
950. 98
P V (K ) =
K
1.02
= P ut (K, T )+
S0
buy a put
buy one index
| {z } |{z}
K = 950.
950. 98(1
98(1..02) = 970
Problem 3.9.
Buy a call (put) a lower strike + Sell an otherwise identical call (put) with
a higher strike=
strike=Bull call (put) spread
Option 1: buy 950-strike call and sell 1000-strike call
Option 2: buy 950-strike put and sell 1000-strike put.
Verify that option 1 and 2 have the same profit.
Option 1:
Payoff=ma
max
x (0,
(0, S − 950) − max(0
max(0,, S − 1000)
0
if S < 950
0
if S < 1000
=
−
S − 950 if S ≥ 950
S − 1000 if S ≥ 1000
½
½
⎧⎨ 0
⎧⎨ 0
if S < 950
if S < 950
950 if 1000 > S ≥ 950 −
> S ≥ 950
=
⎩ SS −− 950
⎩ 0S − 1000 ifif 1000
if S ≥ 1000
S ≥ 1000
⎧⎨ 0
⎧⎨ 0
if S < 950
if S < 950
if 1000 > S ≥ 950 =
S − 950 if 1000 > S ≥ 950
=
⎩ SS −− 950
⎩ 50
950 − (S − 1000) if S ≥ 1000
if S ≥ 1000
Initial cost:
• Spend 120
120..405 to buy 950-strike call
• Sell 1000-strike call receiving 93.
93 .809
Total initial investment: 120
120..405 − 93.
93.809 = 26.
26. 596
The future value is 26.
26 . 596(1
596(1..02) = 27.
27. 12792
Profit is:
0
if S < 950
=
S − 950 if 1000 > S ≥ 950
50
if S ≥ 1000
⎧⎨
⎩
⎧⎨ −27.
27.13
− 977
977.. 13
=
⎩ S22.
22. 87
⎧⎨ −27.
27.13
27.13 =
S − 950 − 27.
27.13
−27.
⎩ 50 − 27.
27.13
if S < 950
if 1000 > S ≥ 950
if S ≥ 1000
if S < 950
if 1000 > S ≥ 950
if S ≥ 1000
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Payoff=
⎧⎨ 0
if S < 950
S − 950 if 1000 > S ≥ 950
50
if S ≥ 1000
⎩
⎧⎨ 0
Pro t=
S − 950
⎩ 50
if S < 950
if 1000 > S ≥ 950
if S ≥ 1000
fi
27.13
− 27.
50
Payoff
40
30
20
Profit
10
0
200
400
600
800
1000
1200
1400
1600
1800
2000
Index Price
-10
-20
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Option 2: buy 950-strike put and sell 1000-strike put.
The payoff:
max
ma
x (0
(0,, 950 S ) max(0
max(0,, 1000 S )
=
½
− if − S < 950
950 − S
0
if S ≥ 950
−S
− − 1000
0
½
if S < 1000
if S ≥ 1000
⎧⎨ 950 − S if S < 950
⎧⎨ 1000 − S if S < 950
=
if 1000 > S ≥ 950 −
> S ≥ 950
⎩ 00
⎩ 01000 − S ifif 1000
if S ≥ 1000
S ≥ 1000
⎧⎨ (950 − S) − (1000 − S) if S < 950
⎧⎨ −50
if S < 950
if 1000 > S ≥ 950 =
> S ≥ 950
=
⎩ −0 (1000 − S)
⎩ S0 − 1000 ifif 1000
S ≥ 1000
S ≥ 1000
Initial cost:
• Buy 950-strike put. Pay 51.
51 .777
• Sell 1000-strike put. Receive 74.
74 .201
Net receipt: 74.
74 .201 − 51.
51.777 = 22.
22. 424
Future value: 22.
22. 424(1
424(1..02) = 22.
22. 87248
The profit is:
−50
if S < 950
S − 1000 if 1000 > S ≥ 950 + 22.
22. 87
0
if S ≥ 1000
⎧⎨
⎩
⎧⎨ −50 + 22.
22. 87
=
S − 1000 + 22.
22. 87
⎩ 22.
22. 87
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if S < 950
if 1000 > S ≥ 950 =
if S ≥ 1000
°
c Yufeng Guo
⎧⎨ −27.
27. 13
− 977
977.. 13
⎩ S22.
22. 87
if S < 950
if 1000 > S ≥ 950
if S ≥ 1000
49
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Payoff=
⎧⎨ −50
if S < 950
S − 1000 if 1000 > S ≥ 950
0
if S ≥ 1000
⎩
⎧⎨ −50
Pro t= S − 1000
⎩0
if S < 950
if 1000 > S ≥ 950 + 22.
22. 87
if S ≥ 1000
fi
Profit
20
10
0
200
400
600
800
1000
1200
1400
1600
1800
2000
Index Price
-10
-20
-30
-40
Payoff
-50
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TEGIES
The payoff of the first option is $50 greater than the payoff of the second
option. Howe
However,
ver, at t = 0, we pay 26.
26. 596 to set up option 1; we pay −22.
22. 424
(i.e.. we receive
(i.e
receive 22.
22. 424
424)) to set up option
option 2. It cos
costs
ts us 26.
26. 596 ( 22.
22. 424) =
− value
49.
49. 02 more initially
initially to set up opti
option
on 1 than optio
option
n 2. The future
v−alue of this
initial set up cost is 49.
49. 02(1
02(1..02) = 50.
50. As a res
result
ult,, option
option 1 and 2 have
have the
same profit at T = 00..5.
Thiss should
Thi
should make
make sense
sense in a world
world of no arb
arbitr
itrage
age.. Consid
Consider
er two portportfolios A and B. If for any stock price Payoff (A
(A) = Payoff ((B
B ) + c,
c , then
InitialCost (A) = InitialCost (B ) + P V (c) to avoid arbitrage.
Profit (A) = P ay
ayoff
off (A
(A) − F V [InitialCost (A)]
= P ay
ayoff
off (B
(B ) + c − F V [InitialCost (B )] − c
= P ay
ayoff
off (B
(B ) − F V [InitialCost (B )] = P rofit
rofit (B )
Similarly, if InitialCost (A) = I nitialCost
nitialCost (B ) + P V (c)
→ Payoff (A
(A) = P ayo
ayoff
ff (B
(B ) + c
→ Profit (A) = P rofit
rofit (B )
Finally, let’s see why option 1 is always $50 higher than option 2 in terms
of the payoff and the initial set up cost. The put-call parity is:
P V (K )
= P ut (K, T ) +
S0
Call (K, T ) +
| {z }
buy a call
| {z }
invest
inve
st PV of strike price
| {z }
buy a put
|{z}
buy one index
The timing of the put-call parity is at t = 0. The above equation means
+
P V (K )
Call (K, T )
| {z }
| {z }
| {z }
|{z}
Cost of buying a call at t=0
Cost of investing PV of strike price at t=0
=
+
P ut (K, T )
Cost of buying a put at t=0
S0
Cost of buying an index at t=0
If we are interested in the payoff at expiration date T , then the put-call
parity is:
= P ut (K, T ) +
K
S
Call (K, T ) +
Payo ff a call at T
strikee price at T
strik
Payo ff of a put at T
Index price at T
| {z }
|{z}
| {z }
|{z}
Now we set up the initial cost parity for two strike prices K 1 < K2
Call (K1 , T )
+ P V (K1 )
| {z }
| {z }
| {z }
| {z }
cost of buying a call
Call (K2 , T )
cost of buying a call
⎡
→ ⎢⎣
=
P ut (K1 , T ) +
=
| {z }
| {z }
invest PV of K1
+ P V (K2 )
invest PV of K2
Call (K1 , T )
| {z }
cost of buying a call
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−
cost of buying one index at t=0
cost of buying a put
P ut (K2 , T ) +
| {z }
cost of buying a call
S0
cost of buying one index at t=0
cost of buying a put
Call (K2 , T )
S0
|{z}
|{z}
⎤
⎥⎦ + [P
[P V (K ) − P V (K )]
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2
51
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⎡
⎢
=
⎣
⎡
→ ⎢⎣
P ut (K1 , T )
P ut (K2 , T )
−
cost of buying a put
cost of buying a put
| {z }
| {z }
|
⎡
⎢
=⎣
Call (K1 , T )
cost of buying a call
| {z }
cost of buying a call
{z
Call spread
P ut (K1 , T )
| {z }
P ut (K2 , T )
−
| {z }
cost of buying a put
|
Call (K2 , T )
−
| {z }
⎤
⎥
⎦
⎤
⎥⎦
cost of buying a put
{z
⎤
⎥⎦ + [P
[P V (K ) − P V (K )]
Put spread
⎡
→ ⎢⎣
|
⎡
⎢
=⎣
+
Call (K1 , T )
cost of buying a call
| {z }
{z
| {z }
+
cost of buying a put
|
2
1
}
⎤
|−Call{z(K , T}) ⎥⎦
2
cost of selling a call
}
Call spread
P ut (K1 , T )
}
⎤
|−P ut{z(K , T}) ⎥⎦ + [P[P V (K ) − P V (K )]
2
2
1
cost of selling a put
{z
}
Put spread
So the initial cost of setting up a call bull spread always exceeds the initial
set up cost of a bull put spread by a uniform amount P V (K2 ) − P V (K1 ).
In this problem, P V (K2 ) − P V (K1 ) = (1000 − 950)1
950)1..02−1 = 49.
49 . 02
Set up the payoff parity at T :
=
K1
Call (K1 , T ) +
| {z }
Payo ff a call at T
|{z}
strike price at T
Call (K2 , T ) +
| {z }
Payo ff a call at T
⎡
→ ⎢⎣
|
strike price at T
Call (K1 , T )
| {z }
−
Payo ff a long call at T
P ut (K2 , T )
| {z }
Payo ff of a put at T
Call (K2 , T )
| {z }
Payo ff a long call at T
{z
Call bull payo ff at T
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S
|{z}
Index price at T
Payo ff of a put at T
=
K1
|{z}
P ut (K1 , T )
| {z }
+
S
|{z}
Index price at T
⎤
⎥⎦
}
52
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⎡
⎢
=
|⎣
P ut (K2 , T )
Payo ff of a long put at T
|⎡
|
Payo ff of a long put at T
Put spread payo ff at T
Call (K1 , T )
| {z }
+
−Call (K2, T )
| {z }
Payo ff short call at T
Payo ff a long call at T
⎢⎣
2
| {z } {z | {z }
⎡
→ ⎢⎣
=
P ut (K1 , T )
−
{z
Call spread payoff at T
P ut (K2 , T )
| {z }
Payo ff of a long put at T
+
⎤
⎥+K −K
⎦}
⎤
⎥⎦
}
−P ut (K1, T )
| {z }
Payo ff of a short put at T
{z
1
Put spread payo ff at T
⎤
⎥⎦ + K − K
2
1
}
In this problem, K 2 − K1 = 1000 − 950 = 50
So the payoff of a call bull spread at T = 00..5 always exceeds the payo ff of a
put bull spread by a uniform amount 50.
50 .
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TEGIES
Problem 3.10.
Buy a call (put) a higher strike + Sell an otherwise identical call (put) with
lowerr strike
lowe
strike=
=Bear call (put) spread.
In this problem, K 1 = 1050,
1050, K 2 = 950 (a bear spread)
P V (K2 ) − P V (K1 ) = (950 − 105
1050)
0) 1.02−1 = ( −100)1
100)1..02−1 = −98.
98. 04
⎡
→ ⎢⎣
=
|⎡
⎢⎣
⎤
−| Call{z(K , T }) ⎥⎦
+
Call (K1 , T )
| {z }
cost of buying a call
2
cost of selling a call
{z
}⎤
Call spread
P ut (K1 , T )
+
[P V (K2 ) − P V (K1 )]
−P ut (K2, T ) ⎥⎦ + [P
| {z } | {z }
|
{z
}
cost of buying a put
cost of selling a put
Put spread
=
⎡⎢⎣
P ut (K1 , T )
−P ut (K2, T ) ⎤⎥⎦ − 98.
98. 04
+
| {z } | {z }
{z
}
|
cost of buying a put
cost of selling a put
Put spread
⎡
→ ⎢⎣
=
|⎡
⎢⎣
Call (K1 , T )
| {z }
+
Payo ff a long call at T
−Call (K2, T )
| {z }
Payo ff short call at T
{z
Call spread payo ff at T
P ut (K2 , T )
+
| {z }
Payo ff of a long put at T
⎤
⎥⎦
}
P ut (K1 , T )
|− {z }
Payo ff of a short put at T
⎤
⎥⎦ + K − K
2
1
Put spread payo ff at T
=
⎤}
|⎡⎢
{z
−| {z } ⎥⎦ −
⎣ | {z }
{z
}
|
P ut (K2 , T )
Payo ff of a long put at T
+
P ut (K1 , T )
100
Payo ff of a short put at T
Put spread payo ff at T
For any index price at expiration, the payoff of the call bear spread is always
100 less than the payoff of the put bear spread. Consequently, as we have seen,
to avoid arbitrage, the initial set-up cost of the call bear spread is less than
the initial set-up cost of the put bear spread by the amount of present value of
the 100
100.. The call bear spread and the put bear spread have the same pro fit at
expiration.
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Next, let’s draw the payoff and profit diagram for each spread.
Payoff of the call bear spread:
Payoff=ma
max
x (0,
(0, S − 1050) − ma
max
x (0,
(0, S − 950)
=
Buy 1050-strike call
Sell 950-strike call
Total
⎧⎨ 0
=
−S
⎩ 950
−100
S < 950
0
0
0
950 ≤ S < 105
0500
0
950 − S
950 − S
105
0500 ≤ S
S − 1050
950 − S
−100
if S < 950
if 950 ≤ S < 1050
if S ≥ 1050
The initial set-up cost of the call bear spread:
71 .802
• Buy 1050-strike call. Pay 71.
• Sell 950-strike call. Receive 120
120..405
Net receipt: 120
120..405 − 71.
71.802 = 48.
48. 603
Future value: 48.
48 . 603(1
603(1..02) = 49.
49. 57506
So the profit of the call bear spread at expiration is
⎧⎨ 0
=
−S
⎩ 950
−100
if S < 950
if 950 ≤ S < 1050 +49
+49.. 58 =
if S ≥ 1050
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⎧⎨ 49.
49. 58
999.. 58 − S
⎩ 999
50. 42
−50.
if S < 950
if 950 ≤ S < 1050
if S ≥ 1050
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Payoff=
⎧⎨ 0
if S < 950
if 950 ≤ S < 1050
if S 1050
950 − S
100
⎩−
⎧⎨ 0
Pro t=
−S
⎩ 950
−100
≥
if S < 950
if 950 ≤ S < 1050 + 49.
49. 58
if S ≥ 1050
fi
40
20
0
200
400
600
800
1000
1200
1400
1600
1800
2000
Index Price
-20
Profit
-40
-60
-80
Payoff
-100
Payoff and Profit: Call bear spread
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Payoff of the put bear spread:
Payoff=ma
max
x (0,
(0, 1050 − S ) − ma
max
x (0,
(0, 950 − S )
Buy 1050-strike put
=
Sell 950-strike put
Total
⎧⎨ 100
=
⎩ 01050 − S
Payoff
S < 950
1050 − S
S − 950
100
950 ≤ S < 10
1050
50
1050 − S
0
1050 − S
10
10550 ≤ S
0
0
0
if S < 950
if 950 ≤ S < 1050
if S ≥ 1050
100
90
80
70
60
50
40
30
20
10
0
0
200
400
600
800
1000
1200
1400
1600
1800
2000
Index Price
Payoff of the put bear spread
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The initial set-up cost:
• Buy 1050-strike put. Pay 101
101..214
• Sell 950-strike put. Receive 51.
51 .777
Net cost: 101
101..214 − 51.
51.777 = 49.
49. 437
Future value: 49.
49 . 437(1
437(1..02) = 50.
50. 42
The profit at expiration is:
100
if S < 950
1050 − S if 950 ≤ S < 1050
=
0
if S ≥ 1050
⎧⎨
⎩
⎧⎨ 49.
49. 58
999.. 58 − S
−50.
50. 42 =
⎩ 999
−50.
50. 42
if S < 950
if 950 ≤ S < 1050
if S ≥ 1050
We see that the call bear spread and the put bear spread have the same
profit.
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TEGIES
Problem 3.11.
Buy
S&R indexput
Buy 950-strike
Sell 1050-strike call
Total
FV (initial cost)
Initial Cost
Payoff
1000
51.
51.777
71.802
−71.
1000 + 51.
51.777 − 71.
71.802 = 979.
979. 975
979.
979. 975(1
975(1..02) = 999.
999. 5745
S
max (0
(0,, 950 − S )
max
(0,, S − 1050)
− (0
The net option premium is: 51.
51 .777 − 71.
71.802 = −20.
20. 025
025.. So we receiv
receivee 20.
20.
025 if we enter this collar.
Payoff
S < 950 950 ≤ S < 10
1050
50 10
10550 ≤ S
Buy S&R index
S
S
S
Buy 950-strike put 950 − S 0
0
Sell 1050-strike call 0
0
1050 − S
Total
950
S
1050
The payoff at expiration is:
950 if S < 950
S
if 950 ≤ S < 1050
1050 if S ≥ 1050
⎧⎨
⎩
The profit at expiration is:
950 if S < 950
S
if 950 ≤ S < 1050
1050 if S ≥ 1050
⎧⎨
⎩
⎧⎨ −49.
49. 57
S − 999
999.. 57
=
50. 43
⎩ 50.
⎧⎨ 950 − 999
999.. 57
S
−
999.
999
. 57
−999
999.. 57 =
⎩ 1050 − 999
999.. 57
if S < 950
if 950 ≤ S < 1050
if S ≥ 1050
if S < 950
if 950 ≤ S < 1050
if S 1050
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Profit=
Profit
50
⎧⎨ −49.
49. 57
if S < 950
S − 999
999.. 57 if 950 ≤ S < 1050
50.
50. 43
if S 1050
≥
⎩
40
30
20
10
0
200
400
600
800
1000
1200
-10
1400
1600
1800
2000
Index Price
-20
-30
-40
-50
Profit: long index, long 950-strike put, short 1050-strike call
The net option premium is −20.
20. 025
025.. So we receiv
receivee 20.
20 . 025 if we enter this
col
collar
lar.. To con
constr
struct
uct a zer
zero-c
o-cost
ost collar
collar and keep 950-str
950-strik
ikee put
put,, we need to
increase the strike price of the call such that the call premium is equal to the
put premium of 51.
51 .777
777..
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Problem 3.12.
Buy
S&R indexput
Buy 950-strike
Sell 1107-strike call
Total
FV (initial cost)
Initial Cost
Payoff
1000
51.
51.777
51.873
−51.
1000 + 51.
51.777 − 51.
51.873 = 999.
999. 904
999.
999. 904(1
904(1..02) = 1019.
1019. 90208
S
max (0
(0,, 950 − S )
(0,, S − 1050)
−max (0
The net option premium is: 51.
51 .777 − 51.
51.873 = −0.096 . So we receive 00..096
if we enter
enter this
this collar
collar.. Thi
Thiss is very close
close to a zero-c
zero-cost
ost collar,
collar, where the net
premium is zero.
Payoff
Buy S&R index
Buy 950-strike put
Sell 1050-strike call
Total
S < 950
S
950 S
0 −
950
The profit is:
950 if S < 950
S
if 950 ≤ S < 1107
1107 if S ≥ 1107
⎧⎨
⎩
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950 ≤ S < 11
1107
07
S
0
0
S
11
11007 ≤ S
S
0
1107 − S
1107
⎧⎨ −69.
69. 9
− 1019
1019.. 90
−1019
1019.. 90 =
⎩ S87.
87. 1
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if S < 950
if 950 ≤ S < 1107
if S ≥ 1107
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Profit=
Profit
⎧⎨ −69.
69. 9
if S < 950
S − 1019
1019.. 90 if 950 ≤ S < 1107
87.
87. 1
if S 1107
≥
⎩
80
60
40
20
0
200
400
600
800
1000
1200
1400
1600
1800
2000
Index Price
-20
-40
-60
Profit: long index, long 950-strike put, short 1107-strike call
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Problem 3.13.
a. 1050-strike S&R straddle
Straddle = buy a call and put with the same strike price and time to expiration.
Buy1050-strike call
Buy 1050-strike put
Total
FV (initial cost)
Initial Cost
71.
71.802
101
101..214
71.
71.802 + 101.
101.214 = 173.
173. 016
173
173.. 016(1
016(1..02) = 176.
176. 476
476 32
Payoff
max (0
(0,, S − 1050)
max (0
(0,, 1050 − S )
Payoff
Buy1050-strike call
Buy 1050-strike put
Total
S < 1050
0
1050 − S
1050 − S
The profit is:
1050 − S if S < 1050
S − 1050 if S ≥ 1050
½
Profit
S ≥ 1050
S − 1050
0
S − 1050
− 176
176.. 48 =
½
873
873.. 52 − S
if S < 1050
S − 1226
1226.. 48 if S ≥ 1050
900
800
700
600
500
400
300
200
100
0
200
400
600
800
1000 1200 1400 1600 1800 2000 2200
-100
Index Price
Profit: long 1050-strike call and long 1050-strike put
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b. written 950-strike S&R straddle
Initial revenue
short 950-strike call 120
120..405
short 950-strike put
Total
FV (initial cost)
Payoff
max (0
(0,, S
51.
51.777
120.
120.405 + 51.
51.777 = 172.
172. 182
172.
172. 182(1
182(1..02) = 175.
175. 62564
950)
−max (0
−− S)
(0,, 950
Payoff
S < 950
0
S − 950
S − 950
sell 950-strike call
sell 950-strike put
Total
S ≥ 950
950 − S
0
950 − S
The profit is:
S − 950 if S < 950
+ 175
175.. 66 =
950 − S if S ≥ 950
½
Profit
½
S − 774
774.. 34
1125. 66 − S
1125.
if S < 950
if S ≥ 950
100
0
20
200
0
40
400
0
60
600
0
800
10
1000
00 1200
1200 14
140
00 1600
1600 18
180
00 200
000
0 22
2200
00
-100
Index Price
-200
-300
-400
-500
-600
-700
-800
-900
-1000
Profit: short 950-strike call and short 950-strike put
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c. simultaneous purchase of 1050-straddle and sale of 950-straddle
Profit = Profit of purchase of 1050-straddle + Profit of Sale of 950-straddle
Profit
873.
873. 52 − S
if S < 1050
S − 774
774.. 34 if S < 950
+
S − 1226
1226.. 48 if S ≥ 1050
1125.. 66 − S if S ≥ 950
1125
½
½
⎧⎨ 873
873.. 52 − S
873.. 52 − S
⎩ S873
− 1226
1226.. 48
⎧⎨ S − 774
774.. 34
1125.. 66 − S
1125
⎩ 1125
1125.. 66 − S
if S < 950
if 950 ≤ S < 1050 +
if S ≥ 1050
⎧⎨ 873
873.. 52 − S + (S
(S − 774
774.. 34)
873.. 52 − S + (1125.
(1125. 66 − S )
=
⎩ 873
S − 1226
1226.. 48 + (1125.
(1125. 66 − S )
⎧⎨ 99.
99. 18
1999.. 18 − 2S
=
⎩ 1999
100.. 82
−100
Profit
if
if
if
if
if
if
if S < 950
if 950 ≤ S < 1050
if S ≥ 1050
S < 950
950 ≤ S < 1050
S ≥ 1050
S < 950
950 ≤ S < 1050
S ≥ 1050
100
80
60
40
20
0
950
1000
1050
1100
1150
1200
Index Price
-20
-40
-60
-80
-100
Profit: long 1050-strike straddle and short 950 straddle
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Problem 3.14.
The put-call parity is:
Call (K, T ) +
P V (K )
= P ut (K, T ) +
S0
buy a put
buy one index
| {z } | {z } | {z } |{z}
→ | {z }
| {z } | {z } |{z}
→ | {z }
| {z } | {z } |{z}
buy a call
invest
inve
st PV of strike price
P V (K1 )
= P ut (K1 , T ) +
S0
inv
invest
est PV of strike price
buy a put
buy one index
P V (K2 )
= P ut (K2 , T ) +
S0
inv
invest
est PV of strike price
buy a put
buy one index
Call (K1 , T ) +
buy a call
Call (K2 , T ) +
buy a call
⎤
⎤
⎡
⎡
⎥
⎥
⎢
(K , T ) − Call (K , T )⎦+P V (K − K ) = ⎣P ut (K , T ) − P ut (K , T )⎦
→ ⎢⎣Call
| {z } | {z }
| {z } | {z }
2
1
buy a call
2
1
2
1
buy a call
buy a put
buy a put
⎡
⎤
⎤
⎡
⎥
⎥
⎢
→ ⎢⎣Call
(K , T ) + −Call (K , T )⎦+P V (K − K ) = ⎣P ut (K , T ) + −P ut (K , T )⎦
| {z } | {z }
| {z } | {z }
1
2
buy a call
1
2
sell a call
2
1
buy a put
sell a put
⎡
⎤ ⎡
⎤
⎥ ⎢
⎥
(K , T ) + −Call (K , T )⎦−⎣P ut (K , T ) + −P ut (K , T )⎦ = P V (K − K )
→ ⎢⎣Call
| {z } | {z } | {z } | {z }
1
1
2
buy a call
sell a call
2
2
buy a put
1
sell a put
⎤ ⎡
⎤
⎡
⎥ ⎢
⎥
(K , T ) + −Call (K , T )⎦+⎣−P ut (K , T ) + P ut (K , T )⎦ = P V (K − K )
→ ⎢⎣Call
| {z } | {z } | {z } | {z }
1
2
1
buy a call
sell a call
sell a put
2
2
buy a put
The initial cost is P V (K2 − K1 ) at t = 0. The payoff at expiration T = 0.
0 .5
is K2 − K1 . The transaction is equivalent to investing P V (K2 − K1 ) in a savings
account at t = 0 and receiving K2 − K1 at T , regardless of the S&R price at
expirati
expi
ration.
on. So the trans
transacti
action
on doesn’t have
have any S&R price
price risk. We just earn
the risk free interest rate over the 6- month period.
In this problem, K 1 = 950 and K 2 = 1000
⎡
⎤ ⎡
⎤
⎢⎣Call (K , T ) + −Call (K , T )⎥⎦ + ⎢⎣−P ut (K , T ) + P ut (K , T )⎥⎦
| {z } | {z } | {z } | {z }
1
buy a call
1
2
sell a call
¡
sell a put
¢
2
buy a put
= P V (1000 − 950) = P V (50) = 50 1.02−1 = 49.
49. 02
So the total initial cost is 49.
49 .02.
02. The payoff is 49.
49 .02(1
02(1..02) = 50.
50. The profit
is 0.
0 . We earn a 2% risk-free interest rate over the 6-month period.
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Problem 3.15.
a. Buy a 950-strik
950-strikee call and sell two 10501050-strik
strikee call
callss
buy a 950-stri
950-strike
ke call
call
sell two 1050-strike calls
Total
FV (initial cost)
Initial cost
120
120..405
2(71..802) = −143
143.. 604
−2(71
120.
120.405 − 143
143.. 604 = −23.
23. 199
23. 199(1
199(1..02) = −23.
23. 66298
−23.
Payoff
max (0
(0,, S − 950)
2max(0,, S − 1050)
−2max(0
Payoff
buy a 950-strike call
sell two1050-strike calls
Total
S < 950
0
0
0
950 ≤ S < 1050
S − 950
0
S − 950
The profit is:
0
if S < 950
S 950 if 950 S < 1050 +23
+23.. 66 =
−
≤
−
≥
1150 S if S 1050
⎧
⎨⎩
⎧⎨ 23.
23. 66
926.. 34
=
− 926
⎩ S1173
1173.. 66 − S
S ≥ 1050
S − 950
−2 (S − 1050)
S − 950 − 2 (S − 1050) = 1150 − S
⎧ 23.
23. 66
950 + 23.
23. 66
⎨⎩ S1150
− − S + 23.
23 . 66
if S < 950
if 950 S < 1050
if S ≥ ≤
1050
if S < 950
if 950 ≤ S < 1050
if S ≥ 1050
Profit 120
100
80
60
40
20
0
850
900
950
1000
1050
1100
1150
1200
Index Price
-20
long 950-strike call and short two1050-strike calls
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b. Buy two 950-strike calls and sell three 1050-strike calls
Initial cost
Payoff
buy two 950-strike calls
2(120..405) = 240.
2(120
240. 81
2 max (0, S 950)
−− 1050)
−3max(0
sell three 1050-strike calls −3(71
3(71..802) = −215
215.. 406
3max(0,, S
Total
240.
240. 81 − 215
215.. 406 = 25.
25. 404
FV (initial cost)
25.
25. 404(1
404(1..02) = 25.
25. 91208
Payoff
buy two 950-strike calls
sell three 1050-strike calls
Total
S < 950
0
0
0
950 ≤ S < 1050
2 (S − 950)
0
2 (S − 950)
S ≥ 1050
2 (S − 950)
−3 (S − 1050)
2 (S − 950) − 3 (S − 1050) = 1250 − S
⎧⎨Pro0 t:
⎧⎨ −25.
if S < 950
25. 91
2 (S − 950) if 950 ≤ S < 1050 −25.
2 (S − 950) − 25.
25. 91
25. 91 =
⎩ 1250 − S if S ≥ 1050
⎩ 1250 − S − 25.
25. 91
⎧⎨ −25.
25. 91
if S < 950
=
2S − 1925
1925.. 91 if 950 ≤ S < 1050
⎩ 1224
1224.. 09 − S
if S ≥ 1050
fi
Profit
if S < 950
if 950 ≤ S < 1050
if S ≥ 1050
160
140
120
100
80
60
40
20
0
950
1000
1050
1100
-20
1150
1200
125
Index Price
long two 950-strike calls and short three1050-strike calls
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c. Buy n 950-strike calls and short m 1050-strike calls such that the initial
premium is zero.
n
71.
71.802
120..405 = 00..5963
120..405
120
405n
n = 71.
71.802
802m
m → m = 120
Problem 3.16.
A spread consists of buying one option at one strike price and selling an
otherwise identical option with a different strike price.
A bull spread consists of buying one option at one strike and selling an
otherwise identical option but at a higher strike price.
A bear spread
spread con
consis
sists
ts of buyi
buying
ng one option
option at one strik
strikee and sellin
sellingg an
otherwise identical option but at a lower strike price.
A bull spread and a bear spread will never have zero premium because the
two options don’t have the same premium.
A butterfly spread might have a zero net premium.
Problem 3.17.
According to http://www.da
http://www.daytradeteam.com
ytradeteam.com/dtt/butterfl
/dtt/butterfly-options-trading.
y-options-trading.
asp , a butterfly spread combines a bull and a bear spread. It uses three strike
prices. The lower
prices.
lower two strike prices
prices are used in the bull spread, and the highe
higherr
strikee price in the bear spread. Both puts and calls
strik
calls can be used.
used. A very large
large
profit is made if the stock is at or very near the middle strike price on expiration
day.
When you enter a butterfly spread, you are entering 3 options orders at once.
If the stock remains or moves into a de fined range, you profit, and if the stock
moves
mov
es out of the desired
desired range, you lose. The closer the stock is to the middle
middle
strike price on expiration day, the larger your profit.
For the strike price K 1 < K2 < K3
K2 = λK1 + (1 − λ) K3
1 -strike
So for
each and
K2 -strike
sold,
there needs
to bought.
be λ units
of K
options
bought
(1 − λ)option
units of
K 3 -strike
options
In this
problem,
K1 = 950,
950, K 2 = 1020,
1020, K 3 = 1050
1020 = 950λ
950λ + (1 − λ) 1050
1050
→ λ = 0.3
For every ten 1020
1020-strike
-strike calls written, there needs to be three 950
950-strike
-strike calls
purchased and seven 1050
1050-strike
-strike calls purchased (so we buy three 950 − 1020
bull spreads and seven 1020 − 1050 bear spreads).
sell ten 1020
1020-strike
-strike calls
three 950
950-strike
-strike calls
seven 1050
1050-strike
-strike calls
Total
FV (initial cost)
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Initial cost
10 (−84.
84.47) = −844
844.. 7
3(120..405) = 361.
3(120
361. 215
7(71..802) = 502.
7(71
502. 614
844.. 7 + 361.
361. 215 + 502.
502. 614 = 19.
19. 129
−844
19.
19. 129(1
129(1..02) = 19.
19. 51158
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Payoff
10max(0,, S − 1020)
−10max(0
3 max (0, S − 950)
7 max (0, S − 1050)
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Payoff
S < 950 950 ≤ S < 10
1020
20
950-strike
950
-strike calls
0
3 (S 950)
1020-strike
1020
-strike calls 0
0 −
1050-strike
1050
-strike calls 0
0
Total
0
3 (S − 950)
3 (S − 950) − 10 (S − 1020) = 7350 − 7S
3 (S − 950) − 10 (S − 102
1020)
0) + 7 (S − 1050) = 0
The payoff
⎧⎪ 0
⎨ 3 (S − 950)
=
⎪⎩ 7350 − 7S
0
if
if
if
if
10
10220 ≤ S < 10
1050
50
3 ( S 950)
−10 (−S − 1020)
0
7350 − 7S
105
0500 ≤ S
3 (S 950)
−10 (−S − 1020)
7 (S − 1050)
0
S < 950
950 ≤ S < 1020
1020 ≤ S < 1050
1050 ≤ S
A key point to remember is that for a butter fly spread K 1 < K2 < K3 , the
payoff is zero if S ≤ K1 or S ≥ K3 .
The profit is:
⎧⎪⎨ −19.
⎧⎪⎨ 0
51950) − 19.
if 950
S <≤
950
3 (19
S .−
19. 51
3 (S − 950) if
S < 1020
−
19.
19. 51 =
19. 51
⎪⎩ 7350 − 7S − 19.
⎪⎩ 7350 − 7S if 1020 ≤ S < 1050
−19.
19. 51
0
if 1050 ≤ S
⎧⎪ −19.
if S < 950
⎨ 3S19−. 512869
2869.. 51 if 950 ≤ S < 1020
=
7330.. 49 − 7S if 1020 ≤ S < 1050
⎪⎩ 7330
−19.
19. 51
if 1050 ≤ S
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if
if
if
if
70
S
<≤
950
950
S < 1020
1020 ≤ S < 1050
1050 ≤ S
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Payoff
⎧⎪ 0
⎨ 3 (S − 950)
=
7S
⎪⎩ 7350
−
0
⎧⎪ 0
⎨ 3 (S − 950)
Pro t=
⎪⎩ 7350 − 7S
0
fi
if
if
if
if
if
if
if
if
S < 950
950 ≤ S < 1020
1020 S < 1050
1050 ≤ S
S < 950
950 ≤ S < 1020
1020 ≤ S < 1050
1050 ≤ S
− 19.
19. 51
200
180
160
140
120
100
80
60
40
20
0
850
900
950
1000
1050
-20
1100
1150
1200
Stock Price
Butterfly spread K 1 = 950,
950, K 2 = 1020,
1020, K 3 = 1050
The black line is the profit line. The blue line is the payoff line.
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Problem 3.18.
The option price table is:
Stri
Strike
ke
35
40
45
Call
Call prem
premiu
ium
m
6.13
2.78
0.97
Pu
Putt prem
premiu
ium
m
0.44
1.99
5.08
Time to expiration is T = 91
91/
/365 ≈ 0.25
The annual effective rate is 8.
8 .33%
√
The quarterly effective rate is 1.0833 − 1 = 202%
4
a. Buy 35—stri
35—strike
ke call, sell two 40-strik
40-strikee calls, and buy 45-strik
45-strikee call. Let’s
Let’s
reproduce
repr
oduce the text
textbook
book Figure
Figure 3.14.
buy a 35-strike
35 -strike call
sell two 40-strike
40 -strike calls
buy a 45-strike
45 -strike call
Total
FV (initial cost)
Initial cost
6.13
2 (−2.78) = −5. 56
0.97
6.13 − 5. 56 + 0.
0.97 = 1.
1. 54
1. 54(1
54(1..0202) = 1.
1. 571108
Payoff
S < 35 35 ≤ S < 40
35-strike
35-strike call
0
S − 35
40-strike
40-strike calls 0
0
45-strike
45-strike call
0
0
Total
0
S − 35
S − 35 − 2 (S − 40) = 45 − S
S − 35 − 2 (S − 40) + S − 45 = 0
⎧⎪The0 payo
⎨ S − 35
⎪⎩ 45 − S
0
ff
is:
if
if
if
if
The profit is:
0
if
S − 35 if
45 − S if
0
if
⎧⎪
⎨
⎪⎩
40 ≤ S < 45
S − 35
−2 (S − 40)
0
45 − S
45 ≤ S
S − 35
−2 (S − 40)
S − 45
0
S < 35
35 ≤ S < 40
40 ≤ S < 45
45 ≤ S
S < 35
35 ≤ S < 40
40 ≤ S < 45
45 ≤ S
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⎧⎪ −1. 57
⎨ − 36.
36. 57
− 1. 57 = ⎪ S43.
. 43 − S
⎩ 43
−1. 57
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if
if
if
if
S < 35
35 ≤ S < 40
40 ≤ S < 45
45 ≤ S
72
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⎧⎪ 0
⎨ S − 35
=
S
⎪⎩ 45
0 −
Payoff
if
if
if
if
S < 35
35 ≤ S < 40
40 S < 45
45 ≤ S
⎧⎪ 0
⎨ S − 35
Pro t=
⎪⎩ 45 − S
0
if
if
if
if
S < 35
35 ≤ S < 40
40 ≤ S < 45
45 ≤ S
fi
− 1. 57
5
4
Black line is Payoff
3
Blue line is Profit
2
1
0
25
30
35
40
45
50
55
60
Stock Price
-1
Butterfly spread K 1 = 35,
35, K 2 = 40,
40, K 3 = 45
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b. Buy a 35—
35—str
strik
ikee put
put,, sell
sell two
two 4040-str
strik
ikee puts,
puts, and buy a 45-str
45-strik
ikee put.
put.
Let’s reprodu
Let’s
reproduce
ce the text
textbook
book Figur
Figuree 3.14.
Initial cost
buy a 35-strike
35 -strike put
0.44
sell two 40-strike
40 -strike puts 2 (−1.99) = −3. 98
buy a 45-strike
45 -strike put
5.08
Total
0.44 − 3. 98 + 5.
5.08 = 1.
1. 54
FV (initial cost)
1. 54(1
54(1..0202) = 1.
1. 571108
Payoff
35-strike
35-strike put
40-strike
40-strike puts
45-strike
45-strike put
Total
S < 35
35 − S
−2(40 − S )
45 − S
0
35 ≤ S < 40
0
−2(40 − S )
45 − S
S − 35
40 ≤ S < 45
0
0
45 − S
45 − S
45 ≤ S
0
0
0
0
35 − S − 2(40 − S ) + 45 − S = 0
−2(40 − S ) + 45 − S = S − 35
⎧⎪ 0
S < 35
⎨ S − 35 ifif 35
≤ S < 40
The payo =
⎪⎩ 45 − S if 40 ≤ S < 45
0
if 45 ≤ S
⎧⎪ 0
⎧⎪ −1. 57
if S < 35
⎨ S − 35 if 35 ≤ S < 40
⎨ S − 36.
36. 57
The pro t =
1. 57 =
−
43. 43 − S
⎪⎩ 45 − S if 40 ≤ S < 45
⎪⎩ 43.
0
if 45 ≤ S
−1. 57
ff
fi
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if
if
if
if
S < 35
35 ≤ S < 40
40 ≤ S < 45
45 ≤ S
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⎧⎪ 0
⎨ S − 35
=
⎪⎩ 045 − S
if
if
if
if
S < 35
35 ≤ S < 40
40 S < 45
45 ≤ S
⎧⎪ 0
⎨ S − 35
Pro t =
⎪⎩ 45 − S
0
if
if
if
if
S < 35
35 ≤ S < 40
40 ≤ S < 45
45 ≤ S
Payoff
fi
− 1. 57
5
4
3
2
1
0
25
30
35
40
45
50
55
60
Stock Price
-1
Butterfly spread K 1 = 35,
35, K 2 = 40,
40, K 3 = 45
The black line is the payoff; the blue line is the profit.
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c. Buy one stock, buy a 35 put, sell two 40 calls, and buy a 45 call.
The put-call parity is:
Call (K, T ) +
P V (K )
= P ut (K, T ) +
S0
| {z }
buy a call
| {z }
| {z }
invest
inve
st PV of strike price
buy a put
|{z}
buy one stock
→Buy stock + buy 35 put=
put=buy 35 call + PV(35)
PV(35)
Buy one stock, buy a 35 put, sell two 40 calls, and buy a 45 call is the same
as:
buy a 35 call , sell two 40 calls, and buy a 45 call, and deposit PV (35) in a
savings account.
We already know
know from Part a. that "buy a 35 call , sell two 40 calls, and
buy a 45 call" reproduces the textbook pro fit diagram
diagram Figure 3.14.
−1
Depositing PV(35)
PV(35) = 35 1.0202
= 34.
34. 31 won’t change the pro fit because
any deposit in a savings account has zero profit.
Hence the profit diagram of "Buy one stock, buy a 35 put, sell two 40 calls,
¡
¢
and buy a 45 call" is Figure 3.14.
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CHAPTER
CHAPTE
R 3. INSURANCE,
INSURANCE, COLLARS,
COLLARS, AND OTHER STRA
STRATEGIES
TEGIES
Problem 3.19.
a. The parity is:
Call (K, T ) − P ut (K, T ) = P V (F0,T − K )
We are told that Call (K, T ) − P ut (K, T ) = 0
→ P V (F0,T − K ) = 0 P V (F0,T ) = P V (K )
Since P V (F0,T ) = S 0
→ P V (K ) = S 0
b. Buying a call and selling
selling an other
otherwise
wise iden
identica
ticall put creates
creates a syntheti
syntheticc
long forward.
c. We buy the call at the ask price and sell the put at the bid price. So we
have to pay the dealer a little more than the fair price of the call when we buy
the call from the dealer; we’ll get less than the fair price of put when we sell
a put to the dealer. To ensure that the call prem
premium
ium equals
equals the put premium
premium
given there’s a bid-ask spread, we need to make the call less valuable and the
put more valuable.
valuable. To mak
makee the call less valuab
aluable
le and the put more valuable
valuable,,
we can increase the strike
strike price. In other words, if there’s no bid-ask spread,
then K = F 0,T . If there’s bid-ask spread, K > F 0,T .
d. A synthet
synthetic
ic short stock positi
p osition
on means "buy put and sell call."
call." To have
zero
zer
o net premium
premium after
after the bid-ask
bid-ask spread,
spread, we need to make
make the call more
valuab
valuable
le and the put less valuab
valuable.
le. To ach
achiev
ievee this, we can decrease the strike
strike
price.. In other words,
price
words, if there
there’s
’s no bid-ask
bid-ask spread, then K = F0,T . If there
there’s
’s
bid-ask spread, K < F 0,T .
e. Transaction fees is not really a wash because there’s a bid-ask spread. We
pay more if we buy an option and we get less if we sell an option.
Problem 3.20.
This problem is about building a spreadsheet. You won’t be asked to build
a spreadsheet in the exam. Skip this problem.
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CHAPTER
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R 3. INSURAN
INSURANCE,
CE, COLLARS,
COLLARS, AND OTHER STRA
STRATEGIES
TEGIES
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Chapter 4
Introduction to risk
management
Problem 4.1.
Let
• S = the price of copper per pound at T = 1
• P BH =Profit per pound of copper at T = 1 before hedging
• P AH =Profit per pound of copper at T = 1 after hedging
For each pound of copper produced, XYZ incurs $0.5 fixed cost and $0.4
variable cost.
P BH = S − (0.
(0.5 + 0.
0.4) = S − 0.9
XYZ sells a forward. The pro fit of the forward at T = 1 is:
F0,T − S = 1 − S
→ P AH = (S
( S − 0.9) + (F
(F0,T
We are told that F 0,T = 1
→ P AH = 1 − 0.9 = 00..1
− S ) = F0,T − 0.9
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CHAPTER
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R 4. INTR
INTRODUCTI
ODUCTION
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P BH = S
Profit
− 0.9
P AH = 0.
0 .1
ing
dg
e
eh
r
fo
Be
0.3
0.2
After hedging
0.1
0.0
0.7
0.8
0.9
1.0
1.1
1.2
Copper price
-0.1
-0.2
-0.3
Profit: before hedging and after hedging
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CHAPTER
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N TO RISK MANA
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GEMENT
Problem 4.2.
P AH = F 0,T
0.9
If F 0,T = 0.
0 .8 − → P AH = 0.
0 .8 − 0.9 = −0.1
If XYZ shuts down its production, its pro fit at T = 1 is −0.5 (it still has to
pay the fixed cost)
If XYZ contin
continues
ues its production,
production, its after
after hedging
hedging pro fit at T = 1 is −0.1
−0.1 > −0.5
→XYZ should continue its production
If F 0,T = 0.
0 .45
0 .45 − 0.9 = −0.45
→ P AH = 0.
If XYZ shuts down its production, its pro fit at T = 1 is −0.5 (it still has to
pay the fixed cost)
If XYZ contin
continues
ues its production,
production, its after
after hedging
hedging pro fit at T = 1 is −0.45
−0.45 > −0.5
→XYZ should continue its production
Problem 4.3.
The profit of a long K -strike put at T = 11::
K −S
max
ma
x (0
(0,, K − S )−F V (Premium
Premium)) =
0
½
→
P AH
=S
=
½
=
P BH
+
− 0.9 +
K
S
½
½
K−S
0
K −S
0
if S < K
if S ≥ K
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if S < K
if S ≥ K
if S < K
if S ≥ K
if S < K
if S ≥ K
Premium))
−F V (Premium
− F V (Premium
Premium))
− F V (Premium
Premium))
Premium))
− 0.9 − F V (Premium
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CHAPTER
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R 4. INTR
INTRODUCTI
ODUCTION
ON TO RISK MANAGEMENT
MANAGEMENT
K = 0.95
F V (Premium
Premium)) = 0.0178(1
0178(1..06) = 0.
0.02
AH
→P
Profit
=
½
0.95 if S < 0.
0.95
S
if S ≥ 0.95
− 0.92
0.28
0.26
0.24
0.22
0.20
0.18
0.16
0.14
0.12
0.10
0.08
0.06
0.04
0.6
0.7
0.8
0.9
1.0
1.1
1.2
Copper Price
Long 0.95-strike put
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N TO RISK MANA
MANAGEMENT
GEMENT
K=1
→ P AH
F V (Premium
Premium)) = 0.037
0376
6 (1.
(1.06) = 0.
0.04
1 if S < 1
0.06
if S < 1
=
0.9 − 0.04 =
−
S if S 1
S 0.94 if S 1
½
½
≥
Profit
−
≥
0.26
0.24
0.22
0.20
0.18
0.16
0.14
0.12
0.10
0.08
0.06
0.6
0.7
0.8
0.9
1.0
1.1
1.2
Copper Price
Long 1-strike put
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CHAPTER
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R 4. INTR
INTRODUCTI
ODUCTION
ON TO RISK MANAGEMENT
MANAGEMENT
K = 1.05
→P
AH
=
F V (Premium
Premium)) = 0.0665(1
0665(1..06) = 0.
0.07
½
1.05 if S < 1.
1.05
S
if S ≥ 1.05
−0.9 −0.07 =
½
0.08
if S < 11.. 05
S − 0.97 if S ≥ 1.05
Profit 0.22
0.20
0.18
0.16
0.14
0.12
0.10
0.08
0.6
0.7
0.8
0.9
1.0
1.1
1.2
Copper Price
Long 1.05-strike put
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CHAPTER
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N TO RISK MANA
MANAGEMENT
GEMENT
Problem 4.4.
The profit of a short K -strike call at T = 11::
max(0,, S − K )+
)+F
F V (Premium
Premium)) = −
− max(0
→P
AH
=S
=
=P
BH
½
−
½
− 0.9 −
½
S
K
0
S−K
0
S−K
if S < K
if S ≥ K
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½
0
S−K
if S < K +F V (Premium
Premium))
if S ≥ K
if S < K
+ F V (Premium
Premium))
if S ≥ K
if S < K
+ F V (Premium
Premium))
if S ≥ K
− 0.9 + F V (Premium
Premium))
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CHAPTER
CHAPTE
R 4. INTR
INTRODUCTI
ODUCTION
ON TO RISK MANAGEMENT
MANAGEMENT
K = 0.95
→ P AH =
Profit
F V (Premium
Premium)) = 0.0649(1
0649(1..06) = 0.
0.07
½
S
if S < 0.
0.95
0.95 if S ≥ 0.95
S
− 0.9 + 0.07 =
½
0.83 if S < 00..95
if 0.95 ≤ S
−0.12
0.1
0.0
0.7
0.8
0.9
1.0
1.1
1.2
Copper Price
-0.1
-0.2
Short 0.95-strike call
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N TO RISK MANA
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GEMENT
K=1
F V (Premium
Premium)) = 0.037
0376
6 (1.
(1.06) = 0.
0.04
→ P AH =
½
S
1
if S < 1
if S ≥ 1
S
− 0.9 + 0.0.04 =
½
0.86 if S < 1
if 1 ≤ S
−0.14
Profit
0.1
0.0
0.7
0.8
0.9
1.0
1.1
1.2
Copper Price
-0.1
-0.2
Short 1-strike call
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CHAPTER
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INTRODUCTI
ODUCTION
ON TO RISK MANAGEMENT
MANAGEMENT
K = 1.05
→
P AH
=
F V (Premium
Premium)) = 0.0194(1
0194(1..06) = 0.
0.02
½
S
if S < 1.
1.05
1.05 if S ≥ 1.05
S
−0.9+ 0.02 =
½
0.88 if S < 11.. 05
−0.17 if 1. 05 ≤ S
Profit
0.1
0.0
0.7
0.8
0.9
1.0
1.1
1.2
Copper Price
-0.1
-0.2
Short 1.05-strike call
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CHAPTER
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INTRODUCTION
N TO RISK MANA
MANAGEMENT
GEMENT
Problem 4.5.
P AH = P BH + P Collar
P BH = S − 0.9
Suppose XYZ buy a K 1 -strike put and sells a K 2 -strike call.
The profit of the collar is:
P Collar = P ay
ayoff
off − F V (Net Initial Premium)
Premium)
= [max(0
[max(0,, K1 − S ) − ma
max
x (0,
(0, S − K2 )] − F V (Put Premium)
Premium) + F V (Call
Premium))
Premium
a. Buy 0.95-strike put and sell 1-strike call
The payoff is max(0
max(0,, 0.95 − S ) − max(0
max(0,, S − 1)
S < 0.
0 .95 0.95 ≤ S < 1
long 0.95-strike put 0.95 − S 0
short 1-strike call
Total
00.95 − S
00
S
0
≥1
− SS
11 −
−F V (P ut P remiu
remium
m) + F V (Call Premium)
Premium)
= ( −0.0178 + 0.
0.037
0376)
6) 1.06 = 0.
0.02
⎧⎨ 0.95 − S if S < 0.0.95
P Collar =
⎩ 10 − S ifif S0.95≥ 1≤ S < 1 + 0.0.02
P AH = P BH + P Collar
= S − 0.9 +
⎧⎨ 0.95 − S
0
1−S
⎧⎨ 0.07 ⎩ if
=
⎩ S −0.120.88 ifif
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if S < 0.
0.95
if 0.95 ≤ S < 1 + 0.
0.02
if S ≥ 1
S < 0.
0 .95
0.95 ≤ S < 1
1≤S
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CHAPTER
CHAPTE
R 4. INTR
INTRODUCTI
ODUCTION
ON TO RISK MANAGEMENT
MANAGEMENT
P
AH
=
⎧⎨
0.07
if
S < 0.
0.95
S − 0.88 if 0.95 ≤ S < 1
0.12
if
1 S
≤
⎩
Profit
0.12
0.11
0.10
0.09
0.08
0.07
0.6
0.7
0.8
0.9
1.0
1.1
1.2
Copper Price
Long 0.95-strike put and short 1-strike call
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GEMENT
b. Buy 0.975-strike put and sell 1.025-strike call
The payoff is max(0
max(0,, 0.975 S ) max(0
max(0,, S 1.025)
−
−
S < 0.
0.975 0.975 ≤−S < 11..025
long 0.975-strike put 0.975 − S 0
short 1-strike call
0
0
Total
0.975 − S 0
−F V (P ut P remiu
remium
m) + F V (Call Premium)
Premium)
= ( −0.0265 + 0.
0.027
0274)
4) 1.06 = 0.
0.000954 = 0.
0.001
⎧⎨ 0.975 − S if S < 0.0.975
1.025
P Collar =
≤ S < 1.
⎩ 10.025 − S ifif S0.975
≥ 1.025
S ≥ 1.025
0
1.025 − S
1.025 − S
+ 0.
0.001
P AH = P BH + P Collar
⎧⎨ 0.975 − S if S < 0.0.975
1.025
= S − 0.9 +
≤ S < 1.
⎩ 10.025 − S ifif S0.975
≥ 1.025
⎧⎨ 0.076
if S < 0.
0.975
S − 0.899 if 0.975 ≤ S < 1.
1.025
=
⎩ 0.126
if 1. 025 ≤ S
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91
CHAPTER
CHAPTE
R 4. INTR
INTRODUCTI
ODUCTION
ON TO RISK MANAGEMENT
MANAGEMENT
P
AH
=
⎧⎨ 0.076
S − 0.899
0.126
if S < 0.
0.975
if 0.975 ≤ S < 11..025
if 1. 025 S
≤
⎩
Profit
0.12
0.11
0.10
0.09
0.08
0.6
0.7
0.8
0.9
1.0
1.1
1.2
Copper Price
Long 0.975-strike put and short 1.025-strike call
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N TO RISK MANA
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GEMENT
c. Buy 1.05-st
1.05-strik
rikee put and sell 1.05-s
1.05-strik
trikee call
The payoff is max(0
max(0,, 1.05 S ) max(0
max(0,, S 1.05)
−
−
−
S < 1.
1.05 S ≥ 1.05
long 1.05-strike put
1.05 − S 0
short 1.05-strike call 0
1.05 − S
Total
1.05 − S 1.05 − S
−F V (P ut P remiu
remium
m) + F V (Call Premium)
Premium)
= ( −0.0665 + 0.
0.019
0194)
4) 1.06 = −0.05
P Collar = (1.
(1 .05 − S ) − 0.05 = 1 − S
AH
BH
P
=P
+ P Collar = S − 0.9 + 1 − S = 0.1
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P AH = 0.
0 .1
Profit
1.0
0.8
0.6
0.4
0.2
0.0
0.7
0.8
0.9
-0.2
1.0
1.1
1.2
Copper Price
-0.4
-0.6
-0.8
Long 1.05-strike put and short 1.05-strike call
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GEMENT
Problem 4.6.
a. Sell one 1.025-strik
1.025-strikee put and buy two 0.975-s
0.975-strik
trikee puts
The payoff is 2max(0
2max(0,, 0.975 − S ) − ma
max
x (0,
(0, 1.025 − S )
S < 0.
0.975
0.975 ≤ S < 11..025
long two 0.975-strike puts 2 (0.
(0.975 − S ) 0
short 1.025-strike put
S − 1.025
S − 1.025
Total
0.925 − S
S − 1.025
S
0
0
0
≥ 1.025
Initial net premium paid=2
paid=2 (0.
(0.0265) − 0.0509 = 0.
0.0021
Future value: 0.0021(1
0021(1..06) = 0.
0.0022
P
Paylater
AH
P
⎧⎨ 0.925 − S
=
⎩ 0S − 1.025
BH
=P
if S < 0.
0.975
if 0.975 ≤ S < 1.
1 .025
if S ≥ 1.025
− 0.0022
P aylater
+P
⎧⎨ 0.925 − S if S < 0.0.975
≤ S < 11..025 − 0.0022
= S − 0.9 +
⎩ S0 − 1.025 ifif 0S.975
≥ 1.025
⎧⎨ 0.0228
if S < 0.
0.975
=
2S − 1. 9272 if 0.975 ≤ S < 1.
⎩ S − 0.9022 if 1. 025 ≤ S 1.025
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P AH
Profit
⎧⎨ 0.0228
2S − 1. 9272
=
S
⎩ − 0.9022
if S < 0.
0 .975
if 0.975 ≤ S < 11..025
if 1. 025 S
≤
0.3
0.2
0.1
Hedged Profit
0.0
0.7
0.8
0.9
1.0
1.1
1.2
Copper Price
-0.1
-0.2
Unhedged Profit
-0.3
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N TO RISK MANA
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GEMENT
b. Sell two 1.034-stri
1.034-strike
ke puts and buy three 1-strike
1-strike puts
ff
The payo is 3max(0
3max(0,, 1 − S ) −
2max(0,
, 1.034 − S )
S 2max(0
<1
long three 1-strike puts
3 (1 − S )
short two1.034-strike puts 2 (S − 1.034)
Total
3 (1 − S ) + 2 (S − 1.034) = 0.
0.932
−S
1 ≤ S < 11..034
0
2 (S − 1.034)
2 (S − 1.034)
Initial premium paid: 3 (0.
(0.0376) − 2 (0.
(0.0563) = 0.
0.0002
Future value: 0.
0 .0002(1
0002(1..06) = 0.
0.000212
Profit:
⎧⎨ 0.932 − S
⎩ 2 (S −01.034)
if
S <1
if 1 ≤ S < 1.
1 .034
if
1.034 ≤ S
− 0.0002
P AH = P BH + P P aylater
⎧⎨ 0.932 − S if S < 1
S < 1.
1.034 − 0.0002
= S − 0.9 +
⎩ 2 (S −01.034) ifif 1 ≤1.034
≤S
⎧⎨ 0.0318
if S < 1
. 9682 if 1 ≤ S < 1.
1 .034
=
⎩ 3SS−−02.9002
if 1. 034 ≤ S
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S
0
0
0
≥ 1.034
CHAPTER
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R 4. INTR
INTRODUCTI
ODUCTION
ON TO RISK MANAGEMENT
MANAGEMENT
P AH
Profit
⎧⎨ 0.0318
3S − 2. 9682
=
⎩ S − 0.9002
if S < 1
if 1 ≤ S < 11..034
if 1. 034 S
≤
0.3
0.2
0.1
Hedged Profit
0.0
0.7
0.8
0.9
1.0
1.1
1.2
Copper Price
-0.1
-0.2
Unhedged Profit
-0.3
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GEMENT
Problem 4.7.
Telco buys copper wires from Wirco. Telco’s purchase price per unit of wire
is $5 plus the price of copper. Telco collects $6.2 per unit of wire used.
Telco’s unhedged profit is:
P BH = 6.
6 .2 − (5 + S ) = 1. 2 − S
If Telco buys a copper forward, the pro fit from the forward at T = 1 is
S − F0,T
The profit after hedging is
P AH = P BH + S − F0,T = 1.
1 . 2 − S + S − F0,T = 1.
1 . 2 − F0,T
F0,T = 1
1 . 2 − 1 = 0.2
→ P AH = 1.
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P AH = 1.
1 . 2 − F0,T
Profit
P AH = 1.
1 . 2 − 1 = 0.2
0.6
Un
hed
ged
P
0.5
rof
it
0.4
0.3
0.2
Hedged Profit
0.1
0.0
0.6
0.7
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Problem 4.8.
P BH = 1.
1. 2 S
Telco buys a−K -strike call. The profit from the long call is:
max
ma
x (0
(0,, S − K ) − F V (Premium
Premium))
→ P AH = 1.
1 . 2 − S + max
max (0,
(0, S − K ) − F V (Premium
Premium))
0
if S < K
= 1.
1. 2 − S +
− F V (Premium
Premium))
S − K if S ≥ K
½
→
P AH
= 1.
1 . 2 − F V (Premium
Premium)) −
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S
K
if S < K
if S ≥ K
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a. K = 0.95
P AH
F V (Premium
Premium)) = 0.064
06499 (1
(1..06) = 0.
0.069
S
if S < 00..95
1. 131 − S
= 1.
1 . 2 0.069
=
→
−
Profit
−
½
0.95 if S
≥ 0.95
½
0.181
if S < 00..95
if S
≥ 0.95
0.5
0.4
0.3
0.2
0.6
0.7
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0.9
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b. K = 1
P AH
P V (Premium
Premium)) = 0.037
0376
6 (1.
(1.06) = 0.
0.039 856
S if S < 1
1. 16 − S if S < 1
= 1.
1 . 2 0.04
=
→
−
Profit
−
½
1
if S
≥1
½
0.16
if 1 ≤ S
0.5
0.4
0.3
0.2
0.6
0.7
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0.9
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c. K = 1.05
P AH
→
F V (Premium
Premium)) = 0.019
01944 (1
(1..06) = 0.
0.020564
S
if S < 11..05
1. 18 − S if S < 11.. 05
= 1.
1 . 2 0.02
=
− − 1.05 if S ≥ 1.05
0.13
if 1. 05 ≤ S
½
Profit
½
0.5
0.4
0.3
0.2
0.6
0.7
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Problem 4.9.
P BH = 1.
1. 2 S
Telco sells a−put option with strike price K.
The profit from the written put is:
max
x (0
(0,, K − S ) + F V (Premium
Premium))
− ma
→ P AH = 1.
1 . 2 − S − ma
max
x (0,
(0, K − S ) + F V (Premium
Premium))
K − S if S < K
= 1.
1. 2 − S −
+ F V (Premium
Premium))
0
if S ≥ K
−K if S < K + 11.. 2 + F V (Premium
=
Premium))
−S if K ≤ S
½
½
→P
AH
=−
½
K
S
if S < K
+ 11.. 2 + F V (Premium
Premium))
if K ≤ S
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a. K = 0.95
P AH =
→
−
Profit
½
F V (Premium
Premium)) = 0.017
01788 (1
(1..06) = 0.
0.018868
0.95 if S < 0.
0 .95
0.269
+1.
+1. 2+0
2+0..019 =
S
if 0.95 ≤ S
1. 219 − S
½
if S < 00..95
if 0.95 ≤ S
0.2
0.1
0.0
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
Copper Price
-0.1
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b. K = 1
P AH =
→
−
Profit
F V (Premium
Premium)) = 0.037
0376
6 (1.
(1.06) = 0.
0.039856
1 if S < 1
0.24
+ 11.. 2 + 0.
0.04 =
S if 1 ≤ S
1. 24 − S
½
½
if S < 1
if 1 ≤ S
0.2
0.1
0.0
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
Copper Price
-0.1
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c. K = 1.05
P AH =
→
Profit
−
½
F V (Premium
Premium)) = 0.066
06655 (1
(1..06) = 0
0..07049
1.05 if S < 1.
1 .05
0.22
+1.
+1. 2+0
2+0..07 =
S
if 1.05 ≤ S
1. 27 − S
½
if S < 11.. 05
if 1. 05 ≤ S
0.2
0.1
0.0
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
Copper Price
-0.1
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Problem 4.10.
Suppose Telco sells a K 1 -strike put and buys a K 2 -strike call.
The profit of the collar is:
P Collar = P ay
ayoff
off − F V (Net Initial Premium)
Premium)
max (0,
(0, S − K2 )] + F V (Put Premium)
Premium) − F V (Call
= [− ma
max
x (0,
(0, K1 − S ) + max
Premium))
Premium
a. Sell 0.95-strik
0.95-strikee put and buy 1-strik
1-strikee call
The payoff is − max(0
max(0,, 0.95 − S )+ max(0,
max(0, S − 1)
S < 0.
0.95
0.95 ≤ S < 1 S ≥ 1
short 0.95-strike put − (0.
(0.95 − S ) 0
0
long 1-strike call
0
0
− (1 − S )
Total
(0.95 − S ) 0
− (0.
− (1 − S )
F V (P ut P remi
remium
um))−F V (Call P remium
remium)) = (0.
(0.0178 − 0.037
0376)
6) 1.06 =
Collar
P
⎧ S − 0.95
⎨
=⎩ 0
S −1
−0.02
if S < 0.
0.95
if 0.95 ≤ S < 1
if S ≥ 1
− 0.02
P AH = P BH + P Collar
⎧⎨ S − 0.95
= 1.
1 . 2−S + 0
⎩ S −1
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0.95
if 0.95 ≤ S < 1
if S ≥ 1
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⎧⎨ 0.23
−0.02 = ⎩ 1. 18 − S
0.18
if S < 00..95
if 0.95 ≤ S < 1
if 1 ≤ S
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P AH
Profit
⎧⎨ 0.23
1. 18 − S
=
⎩ 0.18
if S < 0.
0.95
if 0.95 ≤ S < 1
if 1 ≤ S
0.23
0.22
0.21
0.20
0.19
0.18
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
Copper Price
Short 0.95-strike put and long 1-strike call
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b. Sell 0.975-strike put and buy 1.025-strike call
The payoff is max(0
max(0,, 0.975 − S ) − max(0
max(0,, S − 1.025)
short 0.975-strike put
long 1-strike call
Total
S
0975
.975− S )
− <(0.
(0.0.
0
− (0.
(0.975 − S )
00.975 ≤ S < 11..025
0
0
S
0
≥ 1.025
− (1
(1..025 − S )
− (1
(1..025 − S )
F V (P ut P remi
remium
um))−F V (Call P remium
remium)) = (0.
(0.0265 − 0.027
0274)
4) 1.06 =
S − 0.975 if S < 0.
0.975
0
if 0.975 ≤ S < 1.
1.025 − 0.001
P Collar =
S − 1.025 if S ≥ 1.025
⎧⎨
⎩
−0.001
P AH = P BH + P Collar
⎧⎨ S − 0.975 if S < 0.0.975
0
if 0.975 S < 1.
1 .025
= 1.
1. 2 S +
0.001
⎩
−
−
S − 1.025 if S ≥ 1.≤
025
⎧⎨ 0.224
if S < 0.
0 .975
=
S + 1.
1. 199 if 0.975 ≤ S < 1.
1 .025
⎩ 0−.174
if 1. 025 ≤ S
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P AH
Profit
⎧⎨ 0.224
−S + 1.
1 . 199
=
⎩ 0.174
if S < 0.
0 .975
if 0.975 ≤ S < 11..025
if 1. 025 ≤ S
0.22
0.21
0.20
0.19
0.18
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
Copper Price
Short 0.95-strike put and long 1-strike call
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c. Sell 0.95-s
0.95-strik
trikee put and buy 0.95-st
0.95-strik
rikee call
The payoff is − max(0
max(0,, 0.95 − S ) + max
max (0,
(0, S − 0.95)
short 0.95-strike put
long 0.95-strike call
Total
S < 0.
0.95
− (0.
(0.95 − S )
0
S − 0.95
S ≥ 0.95
0
S − 0.95
S − 0.95
F V (P ut P remi
remium
um))−F V (Call P remium
remium)) = (0.
(0.0178 − 0.064
0649)
9) 1.06 =
P Collar = (S
( S − 0.95) − 0.05 = S − 1
P AH = P BH + P Collar = 1.
1 . 2 − S + S − 1 = 0.2
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P AH = 0.
0 .2
Profit
1.2
1.0
0.8
0.6
0.4
0.2
0.0
0.7
0.8
0.9
-0.2
1.0
1.1
1.2
Copper Price
-0.4
-0.6
-0.8
Short 0.95-strike put and long 0.95-strike call
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Problem 4.11.
a. sell 0.
0 .975
975-strike
-strike call and buy two 1.
1 .034
034-strike
-strike calls
The payoff is 2max(0
2max(0,, S − 1.034) − ma
max
x (0,
(0, S − 0.975)
S < 0.
0.975 0.975 ≤ S < 11..034
short 0.975-strike call 0
− (S − 0.975)
long 1.034-strike calls 0
0
Total
0
− (S − 0.975)
− (S − 0.975
975)) + 2 (S − 1.034) = S − 1. 093
⎧⎨ 0
=
−S
⎩ 0S.975
− 1. 093
The payoff
S
≥ 1.034
− (S − 0.975)
2 (S − 1.034)
S − 1. 093
if S < 0.
0.975
if 0.975 ≤ S < 1.
1.034
if S ≥ 1.034
Initial net premium paid=2
paid=2 (0.
(0.0243) − 0.05 = −0.0014
Future value: − 0.0014(1
0014(1..06) = −0.001 484 ≈ −0.001
P Paylater
⎧⎨ 0
=
−S
⎩ S0.975
− 1. 093
⎧⎨ 0.001
−S
=
⎩ S0.976
− 1. 092
if S < 0.
0.975
if 0.975 ≤ S < 1.
1 .034 + 0.
0.001
if S ≥ 1.034
if S < 0.
0.975
if 0.975 ≤ S < 1.
1.034
if 1. 034 ≤ S
P AH = P BH + P P aylater
0.001
if S < 0.
0.975
= 1.
1. 2 − S +
0.976 − S if 0.975 ≤ S < 11..034
S − 1. 092 if 1. 034 ≤ S
⎧⎨
⎩
1. 201 S
⎨
⎧
=
176 − 2S
⎩ 02..108
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if S < 0.
0.975
if 0.975 ≤ S < 1.
1.034
if 1. 034 ≤ S
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P AH
Profit
⎧⎨ 1. 201 − S
2. 176 − 2S
=
⎩ 0.108
if S < 0.
0.975
if 0.975 ≤ S < 11..034
if 1. 034 ≤ S
0.5
0.4
0.3
0.2
Hedged Profit
0.1
0.0
0.8
0.9
1.0
1.1
-0.1
1.3
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hed
ged
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b. sell two 1-stri
1 -strike
ke calls and buy three 1.
1 .034
034-strike
-strike calls
The payoff is −2max(0
2max(0,, S − 1) + 3 max
max (0,
(0, S − 1.034)
S < 1 1 ≤ S < 1.
1 .034
sell two 1-strike
1 -strike calls
0
−2 (S − 1)
buy three 1.
1 .034
034-strike
-strike calls 0
0
Total
0
−2 (S − 1)
−2 (S − 1) + 3 (S − 1.034) = S − 1. 102
⎧⎨ 0
=
⎩ −S 2−(S1.−1021)
The payoff
S ≥ 1.034
−2 (S − 1)
3 (S − 1.034)
S − 1. 102
if S < 1
if 1 ≤ S < 1.
1.034
if S ≥ 1.034
Initial net premium paid=3
paid=3 (0.
(0.0243) − 2 (0.
(0.0376) = −0.0023
Future value: − 0.0023(1
0023(1..06) = −0.002 438 ≈ −0.0024
P Paylater
0
⎨
⎧
=
⎩ S−2−(S1. −1021)
P AH = P BH + P P aylater
if S < 1
if 1 ≤ S < 1.
1.034 + 0.
0.0024
if S ≥ 1.034
⎧⎨ 0
if S < 1
< 1.
1 .034
= 1.
1. 2 − S +
⎩ S−2−(S1. −1021) ifif S1 ≤≥ S1.034
⎧⎨ 1. 2024 − S if S < 1
2024 − 3S if 1 ≤ S < 1.
1.034
=
⎩ 03..1004
if 1. 034 ≤ S
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0.0024
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P AH
Profit
⎧⎨ 1. 2024 − S
3. 2024 − 3S
=
⎩ 0.1004
0.5
if S < 1
if 1 ≤ S < 11..034
if 1. 034 ≤ S
0.4
0.3
0.2
Hedged Profit
0.1
0.0
0.8
0.9
1.0
1.1
1.2
1.3
1.4
Copper Price
-0.1
Unhe
dge
d
-0.2
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fit
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GEMENT
Problem 4.12.
Wirco’s total profit per unit of wire produced:
• Revenue. S + 5,
5, where S is the price of copper
• Copper cost is S
• Fixed cost is 3
• Variable cost 1.5
Profit before hedging:
hedging: P BH = S + 5 − (S + 3 + 1.
1.5) = 0.
0.5
The profit is fixed regardless
regardless of copper price
price..
If Wirco buys a copper forward, this will introduce the copper price risk (i.e.
Wirco’s profit will now depend on the copper price).
Wirco buys a copper forward. The pro fit from the forward is S − F0,T
The profit after hedging is
P AH = P BH + S − F0,T = 0.
0 .5 + S − F0,T
AH
If F 0,T = 1
P
= 0.
0 .5 + S 1 = S 0.5
fit depends on S . −
Now you the pro→
If S goes−down, the profit goes down.
Problem 4.13.
The unhedged profit is P BH = 0.
0 .5. This doesn’t depend on the copper price
at T = 1.
1 . However, if Wirco uses any derivatives (call, put, forward, etc.), this
will make the hedged profit as a function of the copper price at T = 1. Using
Using
any derivatives will make the hedged profit fluctuate with the copper price,
increasing the variability of the profit.
Problem 4.14.
The question "Did the firm earn $10 in pro fit (relative to accounting breakeven) or lose $30 in profit (relative to the profit that could be obtained by
hedging?" portrays derivatives a way to make profit. However, most companies
use derivatives to manage their risks, not to seek additional pro fit. If they have
idle money, they would rather invest in their core business than buy call or
put options
options to make
make money on stocks. This is all you need to know
know about this
question.
Problem 4.15.
Pre-tax operating income
Tax at 40%
After tax income
Price=$9
Price=$9
−$1
(0.4) = −0.4
−1 (0.
−1 − (−0.4) = −0.6
Because losses are fully tax deductible, we pay
us a check of 0.
0 .4)
Expected
Expect
ed profit is: 0.
0 .5 (−0.6 + 0.
0.72) = 0.
0.06
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Price=$11..20
Price=$11
$1.2
1.2 (0
(0..4) = 0.
0.48
1.2 − 0.48 = 0.
0.72
−0.4 tax (i.e.
(i.e.
IRS will send
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Problem 4.16.
a. Expected pre-tax profit:
Firm A:
0.5(1000 − 600) = 200
Firm B:
0.5 (300 + 100) = 200
b. Expected after-tax profit:
Firm A
Good State
Bad State
Pre-tax income
1000
−600
Tax at 40%
1000
100
0 (0.
(0.4) = 400
−600(0
600(0..4) = −240
After tax income 1000 − 400 = 600 −600 − (−240) = −360
Expected after-tax profit: 0.
0 .5(600 − 360) = 120
Firm B
Good State
Bad State
Pre-tax
income
300 .4) = 120
110000(0
Tax at 40%
300(0.
300(0
0(0..4) = 40
After tax income 300 − 120
120 = 180
180 100
100 − 40 = 60
Expected after-tax profit: 0.
0 .5 (180 + 60) = 120
Problem 4.17.
a. Expected pre-tax profit:
Firm A:
0.5(1000 − 600) = 200
Firm B:
0.5 (300 + 100) = 200
b. Expected after-tax profit:
Firm A
Good State
Bad State
−600
Pre-tax income
1000
Ta
Tax
x
1000
100
0 (0.
(0.4) = 400
0
After tax income 1000 − 400 = 600 −600
Expected after-tax profit: 0.
0 .5(600 − 600) = 0
Firm B
Good State
Bad State
Pre-tax income
300
100
Tax at 40%
300(0..4) = 120
300(0
100(0
0(0..4) = 40
After tax income 300 − 120
120 = 180
180 100
100 − 40 = 60
Expected after-tax profit: 0.
0 .5 (180 + 60) = 120
c. This question is vague. I’m not sure to whom A or B might pay. This is
what I guess the author wants us to answer:
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The expected cash flow Company A receives depends on the taw law.
120 if loss
loss iiss tax deduct
deductibl
iblee
E ProfitA =
0
if loss
loss is not tax deduct
deductibl
iblee
¡
¢ ½
Suppose A is unsure about the IRS’s
IRS’s tax policy (i.
(i.e.
e. not sure whether
whether the
loss is deductible or not next year). Then the present value of the di fference of
the tax law is
120//1.1 = 109.
120
109. 09.
09. So the effect of the tax law has a present value 109.09.
Compan
Com
pany
y B does
doesn’t
n’t have
have an
any
y loss.
loss. Its after
after tax profit doesn’t depend on
whether a loss is tax deductible. So the e ffect of the tax law has a present value
of zero.
Problem 4.18.
We are given:
r = δ = 4.879%
T =1
F0,T = S0 e−(r−δ)T = 420
→ S0 = 420
Call strike price K C = 440;
440; put strike price K P = 400
First, find the call and put premiums
premiums.. Instal
Installl the CD contai
contained
ned in the
textbook Derivatives Markets in your computer.
computer. Open the spreadsheet
spreadsheet titled
"optbasic2." Enter:
Inputs
Stock Price
420
Exercise Price
440
Volatility
5.500%
Risk-free interest rate
4.879%
Time to Expiration (years) 1
Dividend Yield
4.879%
You should get the call premium: C = 2.4944
Inputs
Stock Price
420
Exercise Price
400
Volatility
5.500%
Risk-free interest rate
4.879%
Time to Expiration (years) 1
Dividend Yield
4.879%
You should get the put premium: P = 22..2072
a. Buy 440-strike call and sell a 440-put
Let S represent the gold price at the option expiration date T = 1.
1.
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The payoff is:
Buy 440-strike call
Sell 400-strike put
Total
⎧⎨ S − 400
Payo =
⎩ 0S − 440
ff
S < 400
400 ≤ S < 440
−0 (400 − S )
S − 400
0
0
0
S
≥ 440
S − 440
0
S − 440
if S < 400
if 400 ≤ S < 440
if S ≥ 440
The initial cost of the collar is:
Premium = 2.4944 − 2.2072 = 0.
0.2872
0
.
04879(1)
F V (Premium
Premium)) = 0.2872
2872ee
= 0.
0 .30
So the profit from the collar is:
P Collar
S 400
⎨
⎧
=
⎩ S0 −− 440
if S < 400
if 400 ≤ S < 440
if S ≥ 440
− 0.30
The profit before hedging:
• Auric sells
sells eac
each
h widget fo
forr $800
• It has fixed cost: $340
• Input (gold) cost: S
Profit before hedgin
hedging:
g: P BH = 800 − (340 + S ) = 460 − S
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So Auric’s profit after hedging:
P AH = P BH + P Collar
⎧⎨ S − 400
= 460−S + 0
⎩ S − 440
Profit
if S < 400
if 400 ≤ S < 440
if S ≥ 440
⎧⎨ 59.
59. 7
459.. 7 − S
−0.30 = ⎩ 459
19.
19. 7
if S < 400
if 400 ≤ S < 440
if 440 ≤ S
60
50
40
30
20
360
360 370 38
380
0 390
390 400 41
410
0 42
420
0 43
430
0 440
440 450 46
460
0 470
470 48
480
0 490 50
500
0
Gold Price
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ODUCTION
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MANAGEMENT
b. We need to find
C=P
KC − KP = 30
Using the spreadsheet "optbasic2," after trial-and-error, we find that:
KC = 435.
435.52
KP = 405.
405.52
C = 3.4264
P = 33..4234
→C≈P
½
Let K C and K P represent the strike price for the call and the put.
Collar width 30 → K C − KP = 30
Let C and P represent the call and put premium calculated by the BlackScholes formula
When we buy the call, we pay C = C + 0.
0 .25
When we sell the put, we get P = P − 0.25
Zero collar→ C = P
C + 0.
0 .25 = P − 0.25
C = P − 0.5
So we need to find C and P such
C = P 0.5
KC − K−P = 30
0
0
0
0
½
This is all the concepts you need to know about this problem.
Using the spreadsheet "optbasic2," after trial-and-error, we find that:
KC = 436.
436.53
KP = 406.
406.53
C = 3.1938
P = 33..6938
Problem 4.19.
a. Sell 440440-stri
strike
ke call and buy two K -strike calls such the net premium is
zero.
The 440-strike call premium is: C 440 = 2.
2 .4944
440
We need to find K such that C
− 2C K = 0
K
→ 2.4944 − 2C
K
=0
C
= 2.
2 .4944
4944//2 = 11.. 2472
We know that K > 440 (otherwise C K ≥ C 440 )
Using the spreadsheet "optbasic2," after trial-and-error, we find that:
K = 448.
448.93
C K = 1.
1 .2469 ≈ 1. 2472
b. Profit before hedging: P BH = 800 − (340 + S ) = 460 − S
Zero cost collar →Profit = Payoff
The payoff is:
S < 440 440 ≤ S < 448
448..93 S ≥ 448
448..93
sell 440
440-strike
-strike call
0
− (S − 440)
− (S − 440)
buy two 448
448..93-strike
93-strike calls 0
0
2 (S − 448
448..93)
Total
0
440 − S
S − 457.
457. 86
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MANAGEMENT
GEMENT
− (S − 440
440)) + 2 (S − 448
448..93) = S − 457
457.. 86
P
Collar
⎧⎨ 0
S
=
⎩ S440−−457
457.. 86
if S < 440
if 440 ≤ S < 448
448..93
if S ≥ 448
448..93
So Auric’s profit after hedging:
P AH = P BH + P Collar
0
if S < 440
if 440 ≤ S < 448
448..93 =
= 460−S + 440 − S
S − 457
457.. 86 if S ≥ 448
448..93
⎧⎨
⎩
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⎧⎨ 460 − S
− 2S
⎩ 900
2. 14
if S < 440
if 440 ≤ S < 448
448..93
if 448.
448. 93 ≤ S
125
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MANAGEMENT
P
AH
Profit
⎧⎨ 460 − S
900 − 2S
=
⎩ 2. 14
if S < 440
if 440 ≤ S < 448
448..93
P BH = 460 − S
if 448
448.. 93 ≤ S
60
50
40
30
20
10
Hedged profit
0
410
420
430
440
450
460
470
-10
-20
480
490
500
Gold Price
Unhedged Profit
-30
-40
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GEMENT
Problem 4.20.
Ignore. Not on the FM syllabus.
Problem 4.21.
Ignore. Not on the FM syllabus.
Problem 4.22.
Ignore. Not on the FM syllabus.
Problem 4.23.
Ignore. Not on the FM syllabus.
Problem 4.24.
Ignore. Not on the FM syllabus.
Problem 4.25.
Ignore. Not on the FM syllabus.
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Chapter 5
Financial forwards and
futures
Problem 5.1.
Description
Sell asset outright
Sell asset through loan
Short prepaid forward
Short forward
Get paid at time
0
T
0
T
Deliver asset at time
0
0
T
T
payment
S0 at t = 0
S0 erT at T
S0 at t = 0
S0 erT at T
Problem 5.2.
a. F P = S e−δT P V (Div
Div))
0,T
−0.−
06(3/
06(3
/12)
06(6/
/12)
06(9/
/12)
06(12/
/12)
−0(1) 0
+ e−0.06(6
+ e−0.06(9
+ e−0.06(12
= 50e
50 e
−e
= 46.
46 . 1467
£
46 . 1467
1467ee0.06(1) = 49.
49 . 0003
b. F0,T = F0P,T erT = 46.
Problem 5.3.
a. F0P,T = S0 e−δT = 50e
50 e−0.08(1) = 46.
46 . 1558
46 . 1558
1558ee0.06(1) = 49.
49 . 0099
b. F0,T = F0P,T erT = 46.
Problem 5.4.
129
¤
CHAPTER 5. FINANCI
FINANCIAL
AL FORW
FORWARDS AND FUTURES
(0..05
05−
−0)0
0)0..5
a. F0,T = S0 e(r−δ)T = 35e
35 e(0
= 35.
35 . 8860
35.5 = 0.
b. 1 ln F 0,T = 1 ln 35.
0 .02837
T
S0
0.5
35
c. F0,T = S0 e(r−δ)T
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)0..5
35.
35.5 = 35e
35e(0.
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FINANCIAL FORW
FORWARDS AND FUTURES
Problem 5.5.
(0.05
05−
0)9/
/12
−0)9
a. F0,T = S0 e(r−δ)T = 1100e
1100e(0.
= 1142.
1142. 0332
b. As a buyer in a forward contract, we face the risk that the index may be
worth zero at T (i.e. ST = 0), yet we still have to pay F0,T to buy it. This
This is
how to hedge our risk:
Transactions
t=0 T
long a forward (i.e. be a buyer in forward) 0
ST − F0,T
short sell an index
S0
−ST
deposit S 0 in savings account
−S0 S0erT
Total
0
S0 erT − F0,T
For this problem:
Transactions
buy a forward
short sell an index
deposit S 0 in savings account
Total
t= 0
0
1100
1100
−110
0
T
ST 1142
1142.. 03
−ST−
(0..05)9
05)9/
/12
110
1100e(0
= 1142.
1142. 03
0
Afterr hedging,
Afte
hedging, our profit is zero.
c. As a sel
seller
ler in the forward
forward contra
contract,
ct, we face the risk that ST = ∞ . If
ST = ∞ and we don’t already have an index on hand for delivery at T , we have
to pay S T = ∞ and buy an index from the open market. We’ll be bankrupt.
This is how to hedge:
Transactions
t=0 T
sell a forward (i.e. be a seller in forward) 0
F0,T − ST
buy an index
−S0 ST
borrow S 0
S0
−S0erT
Total
For this problem:
Transactions
sell a forward (i.e. be a seller in forward)
buy an index
borrow S 0
Total
0
t=0
0
−1100
1100
0
F0,T
− S0 erT
T
1142. 03 − ST
ST
(0.05)9
05)9/
/12
−1100
1100ee(0.
= −1142
1142.. 03
0
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AL FORW
FORWARDS AND FUTURES
Problem 5.6.
(0.05
05−
015)9/
/12
−0.015)9
a. F0,T = S0 e(r−δ)T = 1100e
1100e(0.
= 1129.
1129. 26
b.
Transactions
long a forward (i.e. be a buyer in forward)
short sell e −δT index
deposit S 0 e−δT in savings account
Total
For this problem:
Transactions
buy a forward
short sell e −δT index
deposit S 0 e−δT in savings
Total
t= 0
0
S0 e−δT
−S0e−δT
0
t=0
0
015)9/
/12
1100ee(−0.015)9
1100
= 1087.
1087. 69
1087.. 69
1087
0−
c.
Transactions
short a forward (i.e. be a seller in forward)
buy e −δT index
borrow S 0 e−δT
Total
t=0
0
−S0e−δT
S0 e−δT
0
For this problem:
Transactions
t=0
short a forward 0
015)9/
/12
1100e(−0.015)9
= −1087
1087.. 69
buy e −δT index −1100e
−δT
borrow S 0 e
1087.. 69
1087
Total
0
T
ST − F0,T
−ST
S0 e(r−δ)T
S0 e(r−δ)T − F0,T
T
ST − 1129
1129.. 26
−ST
(0..05)9
05)9/
/12
1087. 69e
69e(0
= 1129.
1129. 26
0
T
F0,T − ST
ST
−S0e(r−δ)T
S0 e(r−δ)T − F0,T
T
1129.. 26 − ST
−1129
ST
(0.05)9
05)9/
/12
1087.. 69e
69e(0.
= −1129
1129.. 26
−1087
0
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Problem 5.7.
(0..05)0
05)0..5
F0,T = S0 erT = 1100e
1100e(0
= 1127.
1127. 85
a. The 6-month
6-month forward
forward price in the mark
market
et is 1135
1135,, which is greater than
the fair forward price 1127
1127.. 85.
85. So we have
have two identical
identical forwards
forwards (one in the
open market and one that can be synthetically built) selling at di fferent prices.
To arbitrage, always buy low and sell high.
Transactions
t=0
T = 0.
0 .5
sell expensive forward from market 0
1135 − ST
build cheap forward
buy an index −1100 ST
(0..05)0
05)0..5
borrow 110
1100 1100
100
−1100
1100ee(0
= −1127
1127.. 85
Total profit
0
1135 − 1127
1127.. 85 = 7.
7. 15
We didn’t pay anything at t = 0, but we have a pro fit 77.. 15 at T = 0.
0 .5.
b. The 6-month forward price in the market is 1115
1115,, which is cheaper than
the fair forward price 1127
1127.. 85.
85. So we have
have two identical
identical forwards
forwards (one in the
open market and one that can be synthetically built) selling at di fferent prices.
To arbitrage, always buy low and sell high.
Transactions
t= 0
T = 0.
0 .5
buy cheap forward from market 0
ST − 1115
build expensive forward for sale
short sell an index 1100
−ST
(0..05)0
05)0..5
lend 1100 −1100
1100 110
1100e(0
= 1127.
1127. 85
Total profit
0
1127. 85 − 1115 = 12.
12. 85
We didn’t pay anything at t = 0, but we have a pro fit 12.
12 . 85 at T = 0.
0 .5.
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CHAPTER 5. FINANCI
FINANCIAL
AL FORW
FORWARDS AND FUTURES
Problem 5.8.
(0.05
05−
02)0..5
−0.02)0
F0,T = S 0 e(r−δ)T = 1100e
1100e(0.
= 1116.
1116. 62
a. The 6-mon
6-month
th forward
forward price in the mark
market
et is 1120
1120,, which is greater than
the fair forward price 1116
1116.. 62.
62. So we have tw
two
o identical
identical forwards
forwards (one in the
open market and one that can be synthetically built) selling at different prices.
To arbitrage, always buy low and sell high.
Transactions
t=0
T = 0.
0 .5
sell expensive forward 0
1120 − ST
build cheap forward
02)0..5
buy e −δT index −1100
1100ee(−0.02)0
= −1089
1089.. 05 ST
−
δT
(
−
0
.
02)0.
02)0
.
5
(0..05)0
05)0..5 = −1116
borrow S 0 e
1100ee
1100
= 1089.
1089. 05
1089.. 05e
05e(0
1116.. 62
−1089
Total profit
0
1120 − 1116
1116.. 62 = 3.
3. 38
We didn’t pay anything at t = 0, but we have a profit 33.. 38 at T = 0.
0 .5.
b.
The 6-month forward price in the market is 1110
1110,, which is cheaper than the
fair forward price 1116
1116.. 62.
62. So we have two identical forwards (one in the open
market and one that can be synthetically built) selling at di fferen
erentt prices. To
arbitrage, always buy low and sell high.
Transactions
t=0
T = 0.
0 .5
buy cheap forward from market 0
ST − 1110
build expensive forward for sale
02)0..5
short sell e −δT index 1100
1100ee(−0.02)0
= 1089.
1089. 05 −ST
−δT
(0.05)0
05)0..5
lend S 0 e
−1089
1089.. 05
1089. 05e
05e(0.
= 1116.
1116. 62
Total profit
0
1116. 62 − 1110 = 6.
6. 62
We didn’t pay anything at t = 0, but we have a profit 66.. 62 at T = 0.
0 .5.
Problem 5.9.
This is a poorly designed problem, more amusing than useful for passing the
exam. Don’t waste any time on this. Skip.
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Problem 5.10.
(0.05
05−
)0..75
−δ )0
a. F0,T = S0 e(r−δ)T → 1129
1129..257 = 1100e
1100e(0.
1129..257
1129
(0..05−
05−δ )0.
)0.75
e(0
=
1100
1
1129..257
1129
0.05 − δ =
ln
= 3.
3 . 5%
0.75
1100
δ = 1.5%
b. If you believe that the true dividend yield is 0.
0 .5%
5%,, then the fair forward
price is:
(0.05
05−
−0.005)0
005)0..75
F0,T = 1100e
1100e(0.
= 1137.
1137. 759
The market forward price is 1129
1129..257
257,, which is cheaper than the fair price.
To arbitrage, buy low and sell high.
Transactions
t= 0
T = 0.
0 .5
1129..257
buy cheap forward from market 0
ST − 1129
build expensive forward
for sale
short sell e −δT index
lend S 0 e−δT
Total profit
005)0..75
1100e(−0.005)0
1100e
= 1095.
1095. 883
−1095
1095.. 883
0
−ST
(0..05)0
05)0..75 = 1137.
1095. 883
883ee(0
1137. 75 9
1137. 75 9 − 1129
1129..257 = 8.
8. 502
c. If you believe
believe that the true dividend yield
yield is 3%,
3%, then the fair forward
price is:
(0.05
05−
−0.03)0
03)0..75
F0,T = 1100e
1100e(0.
= 1116.
1116. 624
The market forward price is 1129
1129..257
257,, which is higher than the fair price.
To arbitrage, buy low and sell high.
Transactions
t=0
T = 0.
0 .5
sell expensive forward 0
1129.257 − ST
build cheap forward
03)0..75 = 1075
buy e −δT index
1100ee(−0.03)0
1100
1075.. 526 ST
δT
(0.05)0
05)0..75
−
−
−
borrow S 0 e
1075.. 526
1075
1075.. 526
526ee(0.
= −1116
1116.. 624
−1075
Total profit
0
1129.257 − 1116
1116.. 624 = 12.
12. 633
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CHAPTER 5. FINANCI
FINANCIAL
AL FORW
FORWARDS AND FUTURES
Problem 5.11.
a. One
One cont
contra
ract
ct is wor
worth
th 1200
1200 poin
points
ts.. Ea
Eacch poin
point is wort
worth
h $250
$250.. The
The
notional value of 4 S&P futures is:
4 × 1200 × 250 = $1,
$1, 200 , 000
b. The value of the initial margin: $1,
$1 , 200 , 000 × 0.1 = $120,
$120, 000
Problem 5.12.
a. Notional value of 10 S&P futures:
10 × 950 × 250 = $2,
$2, 375 , 000
The initial margin: $2375
$2375 000 × 0.1 = $237,
$237, 500
b. The maintenance margin: 237
237,, 500
500 (0
(0..8) = 190,
190, 000
At the end of Week 1, our initial margin grows to:
06(1/
/52)
237500ee0.06(1
237500
= 237774.
237774. 20
Suppose the futures price at the end of Week 1 is X . The future
futuress price at
t = 0 is 950
950.. After marking-to
marking-to-market,
-market, w
wee gain ((X
X − 950) points per contract.
The notional gain of the 4 futures after marking-to-market is:
(X − 950
950)) (10)
(10) (25
(250)
0)
After marking-to-market, our margin account balance is
237774.. 20 + (X
237774
(X − 950)
950)(10)
(10) (250) = 2500
2500X
X − 2137225
2137225.. 8
We get a margin call if
2500X
2500
X − 2137225
2137225.. 8 < 190000
→ X < 930
930.. 89032
For example, X = 930.
930. 89 will lead to a margin call.
Problem 5.13.
a.
Transactions
buy forward
lend S 0
Total
t0 = 0
−S0
−S0
b.
Transactions
buy forward
lend S 0 − P V (Div
Div))
Total
c.
Transactions
buy forward
lend S 0 e−δT
Total
T
ST − F0,T = S T
S0 erT
ST
t=0
0
Div))
−S0 + P V (Div
−S0 + P V (Div)
Div )
t=0
0
−S0e−δT
S0 e−δT
−
− S0erT
T
ST − F0,T = S T − S0 erT + F V (Div
Div))
rT
Div))
S0 e − F V (Div
ST
T
ST − F0,T = S T
S0 e(r−δ)T
ST
− S0e(r−δ)T
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FINANCIAL FORW
FORWARDS AND FUTURES
Problem 5.14.
If the forward price F 0,T is too low, this is how to make some free money.
1. Buy low.
low. At t = 0, enter a forw
forward
ard to buy one stock. Incur transacti
transaction
on
cost k
k..
2. Sell
Sell high.
high. At t = 0, sell one stock short and receive S 0b − k
3. The net cash flow after 1 and 2 is S0b
S0b − 2k er T at T
¢
¡
l
− 2k.
Len
Lend S0b
− 2k and receive
4. At T , pay F 0,T and receive one stock. Return the stock to the broker.
The net cash flow after 1 through 4 is zero. Your pro fit at T is:
S0b − 2k er T − F0,T
Arbitrage is possible if:
¡
¡
¢
¢
l
l
l
S0b − 2k er
T
− F0,T > 0
→ F0,T <
¡
S0b − 2k er
To avoid arbitrage, we need to have:
F0,T ≥ F − = S0b − 2k er T
To avoid arbitrage, we need to have:
S0b − 2k er T = F − ≤ F0,T ≤ F + = (S
( S0a + 22k
k) er
¡
¢
l
¡
¢
¢
T
l
b
T
Problem 5.15.
a. k = 0 and there’s no bid-ask spread (so S 0a = S0b = 800)
800)
So the non-arbitrage
non-arbitrage bound is:
(0..05)1
(0.055)1
800ee(0
800
= F − ≤ F0,T ≤ F + = 800e
800e(0.
→ 841
841.. 02 = F − ≤ F0,T ≤ F + = 845.
845. 23
Hence arbitrage is not profitable if 841
841.. 02 ≤ F0,T ≤ 845
845.. 23
800)
b. k = 1 and there’s no bid-ask spread (so S 0a = S0b = 800)
Please note that you can’t blindly copy the formula:
k) er T
( S0a + 22k
S0b − 2k er T = F − ≤ F0,T ≤ F + = (S
This is because the problem states that k is incurred for longing or shorting
a forward and that k is not incurr
incurred
ed for buyin
buying
g or selling an index. Giv
Given
en k is
incurred only once, the non-arbitrage bound is:
( S0a + k ) er T
S0b − k er T = F − ≤ F0,T ≤ F + = (S
(0.05)1 = F −
(800 − 1) e(0.
≤ F0,T ≤ F + = (800 + 1) e(0(0..055)1
→ 839
839.. 97 = F − ≤ F0,T ≤ F + = 846.
846. 29
¢
¡
¡
¢
b
l
b
l
c. Once again, you can’t blindly use the formula
k) er T
( S0a + 22k
S0b − 2k er T = F − ≤ F0,T ≤ F + = (S
The problem states that k 1 = 1 is incurred for longing or shorting a forward
and k2 = 2.4 is inc
incurr
urred
ed for buyi
buying
ng or selli
selling
ng an ind
index.
ex. The non-arb
non-arbitr
itrage
age
formula becomes:
¡
¢
l
b
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CHAPTER 5. FINANCI
FINANCIAL
AL FORW
FORWARDS AND FUTURES
b
l
( S0a + k1 + k2 ) er T
S0b k1 k2 er T = F − F0,T F + = (S
+ = (800 + 1 + 2.
(0.055)1
(0.05)1 = F≤
− ≤ F0≤
2.4) e(0.
(800−− 1 − 2.4) e(0.
,T ≤ F
→ 837
837.. 44 = F − ≤ F0,T ≤ F + = 848.
848. 82
¢
¡
d. Once again, you can’t blindly use the formula
S0b − 2k er T = F − ≤ F0,T ≤ F + = (S
( S0a + 2k
2k ) e r T
The problem states that k 1 = 1 is incurred for longing or shorting a forward;
k2 = 2.4 is incurred twice, for buying or selling an index, once at t = 0 and the
other at T . The non-arbitrage formula becomes:
S0b − k1 − k2 er T − k2 = F − ≤ F0,T ≤ F + = (S
( S0a + k1 + k2 ) er T + k2
(0.05)1
(0.055)1
(800 − 1 − 2.4) e(0.
− 2.4 = F − ≤ F0,T ≤ F + = (800 + 1 + 2.
2.4) e(0.
+
2.4
→ 837
837.. 44 − 2.4 = F − ≤ F0,T ≤ F + = 848.
848. 82 + 2.
2. 4
+
−
851. 22
835.. 04 = F ≤ F0,T ≤ F = 851.
→ 835
¡
¡
¢
l
b
¢
l
b
e. The non-arbitrage higher bound can be calculated as follows:
1. At t = 0 sell a forward contract. Incur cost k 1 = 1.
2. At t = 0 buy 1.003 ind
index.
ex. Thi
Thiss is wh
why
y we need
need to buy 1.00
1.003
3 index.
index. We
pay 0.3% of the index value
value to the brok
broker.
er. So if we buy one index, this
index becomes
becomes 1-0.3
1-0.3%=0.
%=0.997
997 index after the fee. To have one index,
index, we
1
1
need to have
=
≈ 1 + 00..3% = 1.1.003 (remember we need
0.997
1 − 0.3%
to deliver one index at T to the buyer in the forwar
forward).
d). To verify,
verify, if we
1
1
have
index, this will become
(1 − 0.3%) = 1 index after the
0.997
0.997
1
fee is deduc
deducted.
ted. Notice
Notice we use the Taylor
Taylor series
series
≈ 1 + x + x2 + ...
1−x
for a small x
3. At t = 0 borrow 1.003
003S
S0 + k1 = 1.003(800) + 1.
1. Repay
Repay this
this loan with
with
r T
(1.
(1.003
003S
S0 + k1 ) e
at T
b
4. At T deliver the index to the buyer and receive F 0,T . Pay the settlement
fee 0.
0 .3%
3%S
S0
Your initial cost for doing 1 through 4 is zero. Your pro fit at T is:
F0,T − (1.
(1.003
003S
S0 + k1 ) er T − 0.3%
3%S
S0
b
To avoid arbitrage, we need to have
F0,T − (1.
(1.003
003S
S0 + k1 ) er T − 0.3%
3%S
S0 ≤ 0
F0,T ≤ (1.
(1.003
003S
S0 + k1 ) er T + 0.
0.3%
3%S
S0
(0.055)1 + 0.
= (1.
(1 .003 × 800 + 1) e(0.
0.003 × 800 = 851.
851. 22
The lower bound price can also be calculated as follows:
b
b
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CHAPTER
CHAPTE
R 5. FINANCIAL
FINANCIAL FORW
FORWARDS AND FUTURES
1. Buy low
low.. At t = 0, enter a forward to buy 1.
1 .003 index (why 1.003 index
will be explained
explained later
later).
). Incu
Incurr transacti
transaction
on cost k 1 = 1.
2. Sell high.
high. At t = 0, sell 0.
0 .997 index short. Receive 00..997 (800)
(800) − 1. This is
why we need to short sell 0.997 index. If we short sell one index, the broker
charges us 0.3% of the index value and we’ll owe the broker 1+0.3%=1.003
index.. In order to ow
index
owee the brok
broker
er exactly one index
index,, we need to b
borro
orrow
w
1
≈ 1 − 0.003 = 0.0.997 index from the broker.
1.003
3. Lend
Lend 00..997 (800) − 1 and receive (0
(0..997 (800) − 1) er
l
T
at T
4. At T , pay 1.
1.003
003F
F0,T and receive 1.
1.003 index. Pay settlement fee 00..3%(1
3%(1..003)
003)..
After the settlement fee, we have (1.
(1.003
003)) (1 − 0.3%) ≈ 1 in
index
dex left. We
return this index to the broker.
The net initial cash flow after 1 through 4 is zero. Your pro fit at T is:
l
(0
(0..997 (800) − 1) er T − 1.003
003F
F0,T
To avoid arbitrage,
(0
(0..997 (800) − 1) er T − 1.003
003F
F0,T ≤ 0
(0.05)1
(0.
(0.997 × 800 − 1) e(0.
(0.997 (80
(800)
0) − 1) er T
= 834.
834. 94
=
→ F0,T ≥ (0.
1.003
1.003
The non-arbitrage bound is:
851.. 22
834.. 94 ≤ F0,T ≤ 851
→ 834
Makee sure you understan
Mak
understand
d part e, whi
which
ch pro
provid
vides
es a framew
framework
ork for finding the non-arbit
non-arbitrage
rage bound for complex proble
problems.
ms. Once you understand
understand this
framework, you don’t need to memorize non-arbitrage bound formulas.
l
l
Problem 5.16.
Not on the syllabus. Ignore.
Problem 5.17.
Not on the syllabus. Ignore.
Problem 5.18.
Not on the syllabus. Ignore.
Problem 5.19.
Problem 5.20.
1
1 91
× ×
= 11.. 7113%
100 4 90
a.
r91 = (100 − 93.
93.23) ×
b.
$10 (1 + 0.017113) = $10.
$10. 171
171 13 (million)
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CHAPTER 5. FINANCI
FINANCIAL
AL FORW
FORWARDS AND FUTURES
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Chapter 8
Swaps
Problem 8.1.
time t
annual interest during [0,
[0 , t]
fixed payment
floating payments
0
1
6%
R
22
2
6.5%
R
23
PV fixed payments=PV floating payments
22
1.06
+
23
1.0652
=
R
1.06
+
R
1.0652 ,
R = 22.
22. 4831
Problem 8.2.
a.
time t
annual interest during [0,
[0 , t]
fixed payment
floating payments
0
1
6%
R
20
2
6.5%
R
21
3
7%
R
2222
PV fixed payments=PV floating payments
20
1.06
+
21
1.0652
+
22
1.073
=
R
1.06
+
R
1.0652
+
R
,
1.073
R = 20.
20. 952
b. We are now standing at t = 1

P
(0,ti )f0 (ti ) .
recommend that initiall
initially
y you don’t memorize the complex formula R
P (0,ti )
Draw a cash fl ow diagram and set up the equation PV fi xed payments = PV fl oating payments.
Once you are familiar with the concept, you can use the memorized formula.
1I
=
141
CHAPTE
CHA
PTER
R 8. SW
SWAPS
APS
time t
annual interest during [1,
[1 , t]
fixed payment
payments
ts
floating paymen
1
2
6.5%
R
21
3
7%
R
22
PV fixed payments=PV floating payments
21
+ 1.22
= 1.R
+ 1.R
, R = 21.
21. 482
1.065
07
065
07
2
2
Problem 8.3.
The dealer pays fixed and gets floating. His risk is that oil’s spot price may
drop significan
cantly
tly.. For examp
example,
le, if the spot price at t = 2 is $18 per barrel
(as opposed to the expected $21 per barrel) and at t = 3 is $19 per barrel (as
opposed to the expected $22 per barrel), the dealer has overpaid the swap. This
is because the fixed swap rate R = 20.
20. 952 is calculated under the assumption
that the oil price is $21 per barrel at t = 2 and $22 per barrel at t = 3.
To hedge his risk, the dealer can enter 3 separate forward contracts, agreeing
at t = 0 to deliver oil to a buyer at $20 per barrel at t = 1, at $21 per barrel at
t = 2 , and at $22 per barrel at t = 3.
Next, let’s verify that the PV of the dealer’s locked-in net cash flow is zero.
time t
annual interest during [0,
[0 , t]
fixed payment
payments
ts
floating paymen
net cash flow
PV(net cash flows)=
ows)=
0.952
1.06
+
0
−0.048
1.0652
1
6%
20.
20. 952
20
20.
20. 952 − 20 = 0.
0.952
+
−1. 048
1.073
2
6.5%
20. 952
21
−0.048
3
7%
20. 952
22
−1. 048
=0
Problem 8.4.
The fixed payer overpaid 0.
0 .952 at t = 1. The implied interest rate in Year 2
1.065
(from t = 1 to t = 2) is 1.06 − 1 = 0.070024
070024.. So the overpayment 00..952 at t = 1
will grow into 0.
0 .952(1 + 0.
0.070024) = 1.
1. 0187 at t = 2. Then at t = 2, the fi xed
payer underpays 0.048 and his net overpayment is 1. 0187 − 0.048 = 0.
0.9707
9707..
1.07
The implied interest rate in Year 3 (from t = 2 to t = 3) is 1.065 − 1 = 0.0
80071.. So the fixed payer’s net overpayment 0.9707 at t = 2 will grow into
80071
0.970
970 7 (1 + 00..080
08007
07 1) = 1.
1. 0484
0484,, which exactly offsets his underpayment 11.. 048
at = 3. now the accumulative net payment after the 3rd payment is zero.
2
3
2
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CHAPTE
CHA
PTER
R 8. SW
SWAPS
APS
Problem 8.5.
5 basis points=5%%
points=5%% = 0.
0.5% = 0.
0.0005
a. immediately after the swap contract is signed the interest rate rises 0.5%
time t
original annual interest during [0,
[0 , t]
updated annual interest during [0,
[0 , t]
fixed payment
floating payments
20
1.065
0
1
6%
6.5%
R
20
2
6.5%
7%
7%
R
21
3
7%
7.5%
R
22
22
R
+ 1.21
+ 1.075
= 1.R
+ 1.R
+ 1.075
, R = 20.
20. 949 < 20.
20. 952
07
065
07
The fixed rate is worth 20.
20 . 949
949,, but the fixed payer pays 20.
20 . 952
952.. His loss is:
22
20
21
22
20
21
=
0
.
51063
+
+
−
+
+
1.075
1.065
1.07
1.07
1.06
1.065
2
3
3
2
2
3
2
3
¢
¡
b. immediately after the swap contract is signed the interest rate falls 0.5%
time t
0 1
2
3
original annual interest during [0,
[0 , t]
6%
6.5% 7%
updated annual interest during [0,
[0 , t]
5.5% 6%
6%
6.5%
R
R
R
fixed payment
20
21
22
floating payments
20
1.055
+
21
1.062
+
22
1.0653
=
R
1.055
+
R
1.062
+
R
1.0653
R = 20.
20. 955 > 20.
20. 952
The fixed rate is worth 20.
20 . 955
955,, but the fixed payer pays only 20.
20 . 952
The fixed payer’s gain is:
20
1.055
+
21
1.062
+
20
1.065
22
1.0653
−¡
+
21
1.072
+
22
1.0753
= 1. 0293
¢
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CHAPTE
CHA
PTER
R 8. SW
SWAPS
APS
Problem 8.6.
(1) calculate the per-barrel swap price for 4-quarter oil swap
time t (quarter)
0 1
2
3
4
R
R
R
R
fixed payment
payments
ts
21
21.1
20.
20.8
20.
20.5
floating paymen
−
1
−
2
−
3
discounting factor
1.015
1.015
1.015
1.015−4
P V of floating payments = P V of fixed payments
21 1.015−1 + 21.
21.1 1.015−2 + 20.
20.8 1.015−3 + 20.
20.5 1.015−4
−1
−2
−3
−4
= R 1.015 + 1.
1.015 + 1.
1.015 + 1.
1.015
R = 20.
20. 8533
¡¡ ¢
¡
¢
¡
¢¢
¡
(2) calculate the per-barrel swap price for 8-quarter oil swap
time t (quarter)
0 1
2
3
4
5
6
fi
xed payment
paymen
payments
ts
R
21
floating
R
21.1
R
20.
20.8
R
20.
20.5
R
20.
20.2
R
20
¢
7
8
R
19.9
R
19.
19.8
The discounting factor at t is 1.
1 .015−t (i.e. $1 at t is worth 11..015−t at t = 0)
P V of floating payments = P V of fixed payments
21 1.015−1 +21
+21..1 1.015−2 +20
+20..8 1.015−3 +20
+20..5 1.015−4 +20
+20..2 1.015−5 +
20 1.015−6 + 19.
19.9 1.015−7 + 19.
19.8 1.015−8
= R 1.015−1 + 1.
1.015−2 + 1.
1.015−3 + 1.
1.015−4 + 1.
1.015−5 + 1.
1.015−6 + 1.
1.015−7 + 1.
1.015−8
R = 20.
20. 4284
¡
¢
¡
¢
¡
¡ ¡¢ ¡ ¢ ¡
¢¢
¡
¢
¡
¢
(3) calculate the total cost of prepaid 4-quarter and 8-quarter swaps
cost of prepaid 4-quarter swap
21 1.015−1 + 21.
21.1 1.03−1 + 20.
20.8 1.045−1 + 20.
20.5 1.06−1 = 80.
80. 41902
cost of prepaid 8-quarter swap
21 1.015−1 +21
+21..1 1.015−2 +20
+20..8 1.015−3 +20
+20..5 1.015−4 +20
+20..2 1.015−5 +
152. 925604
19.8 1.015−8 = 152.
19.9 1.015−7 + 19.
20 1.015−6 + 19.
¡
¢
¡
¢
¡
¢
¡ ¡ ¢ ¢ ¡ ¡ ¢ ¢ ¡¡
¢¢
Total cost: 80.
80 . 419 02 + 152.
152. 9256 = 233.
233. 34462
¡
¡
¢
¢
¡
¢
¢
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144
CHAPTE
CHA
PTER
R 8. SW
SWAPS
APS
Problem 8.7.
The calculation
calculation is tedi
tedious.
ous. I’ll manually
manually solv
solvee the swap rates for the first 4
quarter but give you all the swaps.
Final result
quarter
forward
DiscFactor
R
1
21
0.9852
21
2
21.1
0.9701
21.05
3
20.
20.8
0.9546
20. 97
4
20.
20.5
0.9388
20. 85
I’ll manually solve for the first 4 swap rates.
quarter
1
2
forward price
21
21.1
R
R
fixed payment
2
zero coupon bond price
0.9852 0.9701
5
20.
20.2
0.9231
20.73
3
20.
20.8
R
0.9546
6
20
0.9075
20.61
7
19.9
0.8919
20.51
8
19.
19.8
0.8763
20.43
4
20.
20.5
R
0.9388
(1) if there’s only 1 swap occurring at t = 1 (quarter)
PV fixed=PV float
21(0..9852) = R (0.
21(0
(0.9852)
R = 21
(2) if there are two swaps occurring at t = 1 and t = 2
PV fixed=PV float
21(0..9852) + 21.
21(0
21.1 (0.
(0.9701) = R (0.
(0.9852 + 0.
0.9701)
21+21..1
21+21
= 21.
21 . 05
R = 21.
21. 0496 ≈
2
(3) if there are 3 swaps occurring at t = 1 ,2
, 2,3
PV fixed=PV float
21(0.
21(0
9852)
+ 21.
21
.21+21
1 (0.
(0.9701)
+=
2020.
20.
.8 .(0.
(0
.9546) = R (0
(0..9852 + 0.
0.9701 + 0.
0.9546)
21+21.
.1+20
1+20..8
20
96667
R
= .20.
20
. 9677
≈
3
(4)if there are 4 swaps occurring at t = 1 ,2
, 2,3,4
PV fixed=PV float
21(0..9852)+21
21(0
9852)+21..1 (0.
(0.9701)+20
9701)+20..8 (0.
(0.9546)+20
9546)+20..5 (0.
(0.9388) = R (0
(0..9852 + 0.
0.9701 + 0.
0.9546 + 0.
0.9388)
21+21..1+20
1+20..8+20
8+20..5
3
=
20
20.
.
85
R = 20.
20. 853 636 ≈ 21+21
4
2 Zero
coupon bond price is also the discounting factor.
3
you run
in the
exam, just take
oftenIfvery
closeout
to of
thetime
correct
answer.
R
fl
as the average oating payments. This is
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145
CHAPTE
CHA
PTER
R 8. SW
SWAPS
APS
By the way, please note that in the textbook Table 8.9, the gas swap prices
are not in line with the forward
forward price and discount
discounting
ing factor
factors.
s. This is because
the swap prices in Table 8.9 are stand-alone prices made up by the author of
Derivatives Markets so he can set up problems for you to solve:
ti (quarter) 1
2
3
4
5
6
7
8
R
2.25 2.4236 2.3503 2.2404 2.2326 2.2753 2.2583 2.2044
To av
avoid
oid conf
confusion
usion,, the author
author of Deri
Deriv
vatives
atives Markets
Markets should
should have
have used
multiple separate tables instead of combining separate tables into one.
Problem 8.8.
quarter
forward price
fixed payment
zero coupon bond price
1
2
3
20.
20.8
R
4
20.
20.5
R
5
20.
20.2
R
6
20
R
0.9546
0.9388
0.9231
0.9075
20..8+20
8+20..5+20
5+20..2+20
If you run out of time, then R = 20
= 20
20.. 375 ≈ 20.
20.38
4
The precise calculation is:
PV fixed=PV float
20.
20.8 (0.
(0.9546)+20
9546)+20..5 (0.
(0.9388)+20
9388)+20..2 (0.
(0.9231)
9231)+20
+20 (0
(0..9075) = R (0
(0..9546 + 0.
0.9388 + 0.
0.9231 + 0.
0.9075)
R = 20.
20. 38069 ≈ 20.
20.38
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CHAPTE
CHA
PTER
R 8. SW
SWAPS
APS
Problem 8.9.
If the problem didn’t give you R = 20.
20.43,
43, you can quickly estimate it as
(21 + 21.
21.1 + 20.
20.8 + 20.
20.5 + 20.
20.2 + 20 + 19.
19.9 + 19.
19.8) /8 = 20.
20. 4125
Back to the problem. Please note that this problem implicitly assumes that
the actual interest rates are equal to the expected interest rates implied in the
zero-coupon bonds. If the actual interest rates turn out to be di fferent than
the rates implied by the zero-coupon bonds, then you’ll need to know the actual
interest rates quarter-by-quarter to solve this problem. So for the sake of solving
this problem,
problem, we assume that the interest
interest rates implied
implied by the zero-coupon
zero-coupon
bonds are the actual interest
interest rates.
rates.
quarter
1
2
3
4
5
6
7
fwd price
21
21.1
20.
20.8
20.
20.5
20.
20.2
20
19.9
xed
pay
20.
20
.
4
3
2
0
.
4
3
2
0
.
4
3
2
0
.
4
3
2
0
.
4
3
2
0
.
4
3
2
0.43
fi
−0.57 −0.67 −0.37 −0.07 0.23 0.43 0.53
fixed−fwd
disct factor 0.9852 0.9701 0.9546 0.9388 0.9231 0.9075 0.8919
8
19.
19.8
20.43
0.63
0.8763
Loan balance at t = 0 is 0
Loan balance at t = 1 is −0.57.
57.
The implicit interest rate from t = 1 to t = 2 is solved by
0.9852
= 0.9701
9701.. So 1 + x = 00..9852
. The −0.57 loan will grow into
1+x
1+x
9701
0.9852
= −0.579 at t = 2.
0.9701
The loan balance at t = 2 is −0.579 − 0.67 = −1. 249
249..
−0.57 ×
= −1. 269 at t = 3.
−1. 249 will grow into −1. 249 × 00..9701
9546
The loan balance at t = 3 is −1. 269 − 0.37 = −1. 639
639,, which grows into
0.9546
−1. 639 × 0.9388 = −1. 667 at t = 4.
The loan balance at t = 4 is −1. 667 − 0.07 = −1. 737
−1. 737 grows into −1. 737 × 00..9388
9231 = −1. 767 at t = 5.
So the loan balance at t = 5 is −1. 767 + 0.
0.23 = −1. 537
−1. 537 grows into −1. 537 × 00..9231
9075 = −1. 563 at t = 6 .
The loan balance at t = 6 is −1. 563 + 0.
0.43 = −1. 133
= −1. 153 at t = 7.
−1. 133 grows into −1. 133 × 00..9075
8919
The loan balance at t = 7 is −1. 153 + 0.
0.53 = −0.623
= −0.634 at t = 8.
−0.623 grows into −0.623 × 00..8919
8763
So the loan balance at t = 8 is −0.634 + 0.
0.63 = −0.004
quarter
loan bal
0
0
1
2
3
4
≈
5
0
6
7
−0.57 −1. 249 −1. 639 −1. 737 −1. 537 −1. 133 −0.623
8
0
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CHAPTE
CHA
PTER
R 8. SW
SWAPS
APS
You can also work backward from t = 8 to t = 0. You know that the loan
loan
balance at t = 8 is zero; overall the fi xed payer and the fl oating payer each have
no gain and no loss if the expected yield curve turns out to the real yield curve.
quarter
fwd price
fixed pay
fixed−fwd
disct factor
1
21
20.
20.43
−0.57
0.9852
2
21.1
20.43
−0.67
0.9701
3
20.
20.8
20.43
−0.37
0.9546
4
20.
20.5
20.43
−0.07
0.9388
5
20.
20.2
20.43
0.23
0.9231
6
20
20.43
0.43
0.9075
7
19.9
20.43
0.53
0.8919
Since the loan balance at t = 8 is zero and the fixed payer overpays 00..6344
= −0.623
623..
at t = 8, the loan balance at t = 7 must be −0.634 ÷ 00..8919
8763
Similarly, the loan balance at t = 7 must be ( −0.623 − 0.53) ÷
133 .
0.9075
0.8919
= −1.
And the loan balance at t = 6 must be ( −1. 133 − 0.43) ÷ 00..9231
= −1. 537
537..
9075
So on and so forth.
This method is less intuitive. However, if the problem asks you to only find
the loan balance at t = 7, this backward method is lot faster than the forward
method.
4I
use 0.634 instead of 0.63 to show you that the backward method produces the same
correct answer as the forward method. If you use 0.63, you won’t be able to reproduce the
correct
answer
0.634). calculated by the forward method due to rounding (because 0.63 is rounded from
8
19.
19.8
20.43
0.63
0.8763
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CHAPTE
CHA
PTER
R 8. SW
SWAPS
APS
Problem 8.10.
The fl oating payer delivers 2 barrels at even numbered quarters and 1 barrel
at odd quarters.
The cash flow diagram is:
quarter
fwd price
fixed pay
disct factor
1
21
R
0.9852
2
21.1 (2
(2)
2R
0.9701
3
20.8
R
0.9546
4
20.
20.5 (2
(2)
2R
0.9388
5
20.2
R
0.9231
6
20 (2)
2R
0.9075
7
19.9
R
0.8919
8
19.
19.8(2)
2R
0.8763
7
19.9
R
0.8919
8
19.
19.8(2)
R
0.8763
PV float =PV fixed
21(0..9852) + 21.
21(0
21.1(2)(0
1(2)(0..9701) + 20.
20.8 (0.
(0.9546) + 20.
20.5(2)(0
5(2)(0..9388)
+20..2 (0
+20
(0..923
9231)
1) + 20(2) (0.
(0.9075) + 19.
19.9 (0.
(0.8919) + 19.
19.8(2)(0
8(2)(0..8763)
= R (0
(0..9852 + 2 × 0.9701 + 0.
0.9546 + 2 × 0.9388)
+R (0
(0..9231 + 2 × 0.9075 + 0.
0.8919 + 2 × 0.8763)
R = 20.
20. 40994
Please note that the cash flow diagram is not:
quarter
1
2
3
4
fwd price
21
21.1 (2
(2) 20.8
20.
20.5 (2
(2)
xed
pay
R
R
R
R
fi
disct factor 0.9852 0.9701
0.9546 0.9388
5
20.2
R
0.9231
6
20 (2)
R
0.9075
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CHAPTE
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R 8. SW
SWAPS
APS
Problem 8.11.
The key formula is the textbook equation 8.13:
R=
P
n
(0,, ti ) f0 (t
(ti )
i=1 P (0
n
(0,, ti )
i=1 P (0
P
From Table 8.9, we get:
ti (quarter) 1
2
R
2.25
2.4236
P (0
(0,, ti )
0.9852 0.9701
3
2.3503
0.9546
4
2.2404
0.9388
5
2.2326
0.9231
6
2.2753
0.9075
7
2.2583
0.8919
8
2.2044
0.8763
Notation:
• P (0
(0,, ti ) is the present value at t = 0 of $1 at the t i .
• R is the swap rate. For exampl
example,
e, for a 4-quarter
4-quarter swap,
swap, the swap rate is
2.2404
• f0 (t
(ti ) is the price of the forward contract signed at ti−1 and expiring at
ti
The 1-quarter swap rate is
P (0
(0,, t1 ) f0 (t
(t 1 )
R (1) =
P (0
(0,, ti )
→ f0 (t(t1) = R (1) = 2.2.2500
The 2-quarter swap rate is:
P (0
(0,, t1 ) f0 (t
(t1 ) + P (0
(0,, t2 ) f0 ((tt2 )
R (2) =
P (0
(0,, t1 ) + P (0
(0,, t2 )
0.9852(2
9852(2..25) + 0.
0.9701
9701ff0 ((tt2 )
→ 2.4236 =
f0 ((tt2 ) = 2.5999
0.9852 + 0.
0.9701
The 3-quarter swap rate is:
P (0
(0,, t1 ) f0 (t
(t1 ) + P (0
(0,, t2 ) f0 ((tt2 ) + P (0
(0,, t3 ) f0 ((tt3 )
R (3) =
P (0
(0,, t1 ) + P (0
(0,, t2 ) + P (0
(0,, t3 )
0.985
9852
2 (2.
(2.25) + 0.
0.970
9701
1 (2
(2..60) + 0.
0.9546
9546ff0 ((tt3 )
→ 0.
0 .9546 =
0.9852 + 0.
0.9701 + 0.
0.9546
→ f0 (t(t3) = 2.2002
So on and so forth. The result is:
ti
1
2
3
R
2.25
2.4236 2.3503
P (0
(0,, ti ) 0.9852 0.9701 0.9546
f0 (t
(ti )
2.2500 2.5999 2.2002
4
2.2404
0.9388
1.8998
5
2.2326
0.9231
2.2001
6
2.2753
0.9075
2.4998
7
2.2583
0.8919
2.1501
8
2.2044
0.8763
1.8002
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°
CHAPTE
CHA
PTER
R 8. SW
SWAPS
APS
Problem 8.12.
ti
R (8)
f0 ((tti )
P (0
(0,, ti )
i(ti−1 , ti )
1
2.2044
2.2500
0.9852
1.5022%
2
2.2044
2.5999
0.9701
1.5565%
3
2.2044
0.9546
0.9546
1.6237%
4
2.2044
0.9388
0.9388
1.6830%
ti
5
6
7
8
R (8)
2.2044
2.2044
2.2044
2.2044
f0 ((tti )
0.9231
0.9075
0.8919
0.8763
P (0
(0,, ti )
0.9231
0.9075
0.8919
0.8763
i(ti−1 , ti ) 1.7008% 1.7190% 1.7491% 1.7802%
First, let’s calculate the quarterly forward interest rate i(ti−1 , ti ), which is
the interest during[
during[ti−1 , ti ].
By the way, please note the di fference between f 0 ((tti ) and ii((ti−1 , ti ). f0 ((tti )
is the price of a forward contract and i(
i (ti−1 , ti ) is the forward interest rate.
The interest rate during the first quarter is i(t0 , t1 ). This
This is
is the
the effective
int
interest
erest per quarter
quarter from t = 0 to t = 0.25 (y
(year).
ear). Because P (0
(0,, t1 ) represents
the present value of $1 at t = 0.25
1
P (0
(0,, t1 ) =
1 + i(t0 , t1 )
1
1
0.9852 =
i(t0 , t1 ) =
− 1 = 1. 5022%
1 + i(t0 , t1 )
0.9852
The 2nd quarter
quarter interest
interest rate i(
i (t1 , t2 ) satisfies the following equation:
1
1
P (0
(0,, t1 )
P (0
(0,, t2 ) =
=
×
1 + i(t0 , t1 ) 1 + i(t1 , t2 )
1 + i(t1 , t2 )
0.9852
i(t1 , t2 ) = 1. 5566%
0.9701 =
1
2
1 + i(t , t )
Similarly,
1
1
1
P (0
(0,, t2 )
=
×
×
1 + i(t0 , t1 ) 1 + i(t1 , t2 ) 1 + i(t2 , t3 )
1 + i(t2 , t3 )
0.9701
→ 0.9546 = 1 + i(t , t ) i(t2, t3) = 1.6237%
2 3
P (0
(0,, t3 ) =
Keep doing
Keep
doing this,
this, you
you sho
should
uld be abl
ablee to calcul
calculate
ate all the forward
forward in
inter
terest
est
rates.
150
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151
CHAPTE
CHA
PTER
R 8. SW
SWAPS
APS
Next, let’s calculate the loan balance.
ti
1
2
R (8)
2.2044
2.2044
f0 (t
(ti )
2.2500
2.5999
f0 (t
(ti ) − R (8) 0.0456
0.3955
Loan balance 0.0456
0.4418
i(ti−1 , ti )
1.5022% 1.5565%
ti
R (8)
f0 (t
(ti )
f0 (t
(ti ) − R (8)
Loan balance
i(ti−1 , ti )
5
2.2044
0.9231
−0.0043
0.1451
1.7008%
6
2.2044
0.9075
0.2954
0.4430
1.7190%
3
2.2044
0.9546
−0.0042
0.4444
1.6237%
4
2.2044
0.9388
−0.3046
0.1470
1.6830%
7
2.2044
0.8919
−0.0543
0.3963
1.7491%
8
2.2044
0.8763
−0.4042
0.0009
−1.7802%
At t = 0, the
the loan
loan bal
balan
ance
ce is zero
zero.. A sw
swap
ap is a fa
fair
ir deal
deal and
and no money
money
changes hands.
At t = 1 (i.
(i.e.
e. the end of the first quarter), the floating payer lends 2.25 −
2.2044 = 0.
0.0456 to the fixed paye
payer.
r. Had th
thee floating payer signed a forward
contract at t = 0 agreeing to deliver the oil at t = 1, he would have received
2.25 at t = 1. How
Howeve
ever,
r, by ent
enterin
ering
g into
into an 8-quarter
8-quarter swap, the floating payer
receives only 2.2044 at the t = 1. So the
the floating payer lends 2.25 − 2.2044 =
0.0456 to the fixed payer.
The 0.
0 .0456 at t = 1 grows to 0.
0 .045
045 6 (1 + 0.
0.015022) = 0.
0.0463
The total loan balance at t = 2 is 0.
0 .0463 + 0.
0.3955 = 0.
0.4418
The loan balance 0.
0.4418 at t = 2 grows into 00..441
441 8 (1 + 0.
0.015565) = 0.
0.4487
at t = 3
The total loan balance at t = 3 is 0.
0 .4487 − 0.0042 = 0.
0.4444
I used Excel to do the calculation. Due to rounding, I got 00..4487 − 0.0042 =
0.4444 instead of 0.
0 .4445
4445..
So on and so forth. The final loan balance at t = 8 is:
0.396
39633 (1 + 0.
0.017491) − 0.4042 = −0.0009 ≈ 0
The loan balance at the end of the swap t = 8 should be zero. We didn’t get
zero due to rounding.
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CHAPTE
CHA
PTER
R 8. SW
SWAPS
APS
Problem 8.13.
ti
P (0
(0,, ti )
i(ti−1 , ti )
P (0
(0,, ti ) i(ti−1 , ti )
R=
P
6
i=2 P
2
0.9701
1.5565%
0.0151
(0
(0,, ti ) r (ti−1 , ti )
6
(0,, ti )
i=2 P (0
P
3
0.9546
1.6237%
0.0155
=
4
0.9388
1.6830%
0.0158
0.0777
= 11.. 655
655 3%
4.6941
Problem 8.14.
ti
1
2
3
4
Total
P (0
(0,, ti )
0.9852
0.9701
0.9546
0.9388
3.8487
i(ti−1 , ti )
1.5022%
1.5565%
1.6237%
1.6830%
P (0
(0,, ti ) i(ti−1 , ti )
0.0148
0.0151
0.0155
0.0158
0.0612
Calculate the 4-quarter swap rate.
R=
P
4
i=1 P
(0
(0,, ti ) r (ti−1 , ti )
4
(0,, ti )
i=1 P (0
P
=
0.0612
= 11.. 5901%
3.8487
Calculate the 8-quarter swap rate.
ti
P (0
(0,, ti ) i(ti−1 , ti ) P (0
(0,, ti ) i(ti−1 , ti )
1
0.9852
1.5022%
0.0148
2
3
4
5
6
7
8
Total
R=
0.9701
0.9546
0.9388
0.9231
0.9075
0.8919
0.8763
7.4475
P
8
i=1 P
1.5565%
1.6237%
1.6830%
1.7008%
1.7190%
1.7491%
1.7802%
(0
(0,, ti ) r (ti−1 , ti )
P
5
i=1 P
(0
(0,, ti )
0.0151
0.0155
0.0158
0.0157
0.0156
0.0156
0.0156
0.1237
=
Problem 8.15.
Not on the FM syllabus. Skip.
0.1237
= 11.. 6610%
7.4475
5
0.9231
1.7008%
0.0157
6
0.9075
1.7190%
0.0156
Total
4.6941
0.0777
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CHAPTE
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R 8. SW
SWAPS
APS
Problem 8.16.
Not on the FM syllabus. Skip.
Problem 8.17.
Not on the FM syllabus. Skip.
Problem 8.18.
Not on the FM syllabus. Skip.