Introduction to Financial Mathematics Options 62 8.1 European Option Deﬁnitions ... Normal...

Introduction to Financial Mathematics

Kyle Hambrook

December 15, 2016

Contents

1 Probability Theory: Random Variables 5

1.1 Sample Space and Random Variable . . . . . . . . . . . . . . . . . . . . . 5

1.2 Probability Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.3 Discrete Random Variables . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.4 Expectation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.5 Properties of Expectation . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.6 Law of the Unconscious Statistician (LOTUS) . . . . . . . . . . . . . . . . 10

1.7 Variance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2 Arbitrage 12

2.1 Assets and Portfolios . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.2 Arbitrage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.3 What is the Sample Space and the Probability Measure? . . . . . . . . . . . 13

2.4 Monotonicity and Replication . . . . . . . . . . . . . . . . . . . . . . . . 13

3 Compound Interest, Discounting, and Basic Assets 16

3.1 Interest Rates and Compounding . . . . . . . . . . . . . . . . . . . . . . . 16

3.2 Zero Coupon Bonds and Discounting . . . . . . . . . . . . . . . . . . . . . 19

3.3 Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3.4 Stocks and Bonds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3.5 Foreign Exchange Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

4 Forward Contracts 26

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4.1 Derivative Contracts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4.2 Forward Contracts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4.3 Value of Forward . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4.4 Value at Maturity and Payoff . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.5 Forward Price . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

4.6 Value of Forward and Forward Price . . . . . . . . . . . . . . . . . . . . . 29

4.7 Forward Price for Asset Paying No Income . . . . . . . . . . . . . . . . . 30

4.8 Forward Price for Asset Paying Known Income . . . . . . . . . . . . . . . 32

4.9 Forward Price for Stock Paying Known Dividend Yield . . . . . . . . . . . 34

4.10 Forward Price for Currency . . . . . . . . . . . . . . . . . . . . . . . . . . 35

5 Forward Rates and Libor 36

5.1 Forward Zero Coupon Bond Prices . . . . . . . . . . . . . . . . . . . . . . 36

5.2 Forward Interest Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

5.3 Libor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

5.4 Forward Rate Agreement Definition . . . . . . . . . . . . . . . . . . . . . 42

5.5 Forward Libor Rate and Value of Forward Rate Agreement . . . . . . . . . 43

5.6 Valuing Floating and Fixed Cashflows . . . . . . . . . . . . . . . . . . . . 46

6 Interest Rate Swaps 48

6.1 Swap Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

6.2 Value of Swap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

6.3 Forward Swap Rate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

6.4 Value of Swap in Terms of Forward Swap Rate . . . . . . . . . . . . . . . 52

6.5 Swaps as Difference Between Bonds . . . . . . . . . . . . . . . . . . . . . 52

6.6 Par or Spot-Starting Swaps . . . . . . . . . . . . . . . . . . . . . . . . . . 53

7 Futures Contracts 55

7.1 Futures Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

7.2 Futures Prices When Rates Are Constant: Result and Examples . . . . . . . 57

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7.3 Futures Prices When Rates Are Constant: Proof . . . . . . . . . . . . . . . 58

7.4 Futures Convexity Correction . . . . . . . . . . . . . . . . . . . . . . . . . 59

7.5 Optional Topic: Futures Convexity Correction and Correlation . . . . . . . 60

8 Options 62

8.1 European Option Definitions . . . . . . . . . . . . . . . . . . . . . . . . . 62

8.2 American Option Definitions . . . . . . . . . . . . . . . . . . . . . . . . . 63

8.3 More Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

8.4 Option Prices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

8.5 Put-Call Parity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

8.6 Call Prices for Assets Paying No Income . . . . . . . . . . . . . . . . . . . 67

8.7 Put Prices for Assets Paying No Income . . . . . . . . . . . . . . . . . . . 69

8.8 Call and Put Prices for Stocks Paying Known Dividend Yield . . . . . . . . 71

8.9 Call and Put Spreads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

8.10 Butterflies and Convexity of Option Price . . . . . . . . . . . . . . . . . . 73

8.11 Digital Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

9 Probability Theory: Conditioning 79

9.1 Conditional Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

9.2 Conditional Expectation . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

9.3 Condition Expectation as Random Variable . . . . . . . . . . . . . . . . . 81

9.4 Properties of Conditional Expectation . . . . . . . . . . . . . . . . . . . . 81

9.5 Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

10 The Fundamental Theorem of Asset Pricing and the Binomial Tree 84

10.1 Fundamental Theorem of Asset Pricing . . . . . . . . . . . . . . . . . . . 84

10.2 Binomial Tree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

10.3 Arbitrage-Free Binomial Tree . . . . . . . . . . . . . . . . . . . . . . . . 91

10.4 Arbitrage-Free Pricing on the Binomial Tree . . . . . . . . . . . . . . . . . 92

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11 Probability Theory: Normal Distribution and Central Limit Theorem 98

11.1 Normal Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

11.2 Standard Normal Distribution . . . . . . . . . . . . . . . . . . . . . . . . . 99

11.3 Central Limit Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

12 Continuous-Time Limit and Black-Scholes Formula 100

12.1 Black-Scholes Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

12.2 Binomial Tree to Black-Scholes . . . . . . . . . . . . . . . . . . . . . . . 101

12.3 Black-Scholes Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

12.4 Properties of Black-Scholes Formula . . . . . . . . . . . . . . . . . . . . . 105

12.5 The Greeks: Delta and Vega . . . . . . . . . . . . . . . . . . . . . . . . . 107

12.6 Volatility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

Chapter 1

Probability Theory: Random Variables

Financial mathematics is built on probability theory. Indeed, the following assumption isfundamental: The value of an asset at a future time is a random variable.

We will give a brief introduction to probability theory (without measure-theoretic subtletiesand with minimal set theory).

1.1 Sample Space and Random Variable

The set of all possible outcomes of an experiment is called the sample space and is denotedΩ. Individual outcomes are typically denoted by ω.

A random variable X is a function from the sample space Ω to the set of real numbers R.In other words, X assigns to each outcome ω ∈ Ω a real number X(ω). (The symbol “∈”means “in”).

Example 1.1.1. Experiment: Flip a fair coin twice.Sample space: Ω = HH,TT,HT, TH.An example of a random variable is X = number of heads.If the outcome is ω = HT , then X(ω) = 1.If the outcome is ω = TT , then X(ω) = 0.

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An event is a set of possible outcomes. In other words, an event is a subset of the samplespace.

Example 1.1.1 Continued. Experiment: Flip a fair coin twice.Sample space: Ω = HH,TT,HT, TH.The set HH,HT is an event. It is the event that the first flip is heads. 4

We often consider events that are defined in terms of random variables.

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Example 1.1.1 Continued. Experiment: Flip a fair coin twice.Sample space: Ω = HH,TT,HT, TH.X = number of heads.Here is an event:

X ≥ 1 = ω ∈ Ω : X(ω) ≥ 1= set of all outcomes ω in the sample space Ω such that X(ω) ≥ 1

= HH,HT, TH

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1.2 Probability Measure

Fix a sample space Ω.

A probability measure P is a function that assigns to each event A a real number 0 ≤P (A) ≤ 1 (in such a way that the properties discussed below hold). The number P (A) iscalled the probability of A.

Interpretation: The probability P (A) encodes our knowledge or belief about how likelyevent A is. P (A) = 0 means the event cannot occur. P (A) = 1 means the event is certainto occur.

A probability measure P must satisfy the following properties for all random variables Xand all real numbers a ≤ b.

• P (X ≤ a) + P (X > a) = 1.

• P (X ≤ a) + P (a < X ≤ b) = P (X ≤ b)

• limx→a+

P (X ≤ x) = P (X ≤ a)

From these properties, many others can be deduced. For example,

P (a ≤ X ≤ b) = 1− P (X < a or X > b).

Example 1.2.1. Experiment: Roll two fair six-sided dice.The possible outcomes ω are pairs (i, j), where i is the number shown on the first die andj is the number shown on the second die.The sample space is

Ω = (1, 1), (1, 2), (1, 3), . . . , (6, 6) .

The dice are fair, so all outcomes are equally likely: P (ω) = 136

for all ω ∈ Ω.

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X1 = number on first dieX2 = number on second dieY = sum of the dice

For the outcome ω = (2, 5),

X1(ω) = X1((2, 5)) = 2

X2(ω) = X2((2, 5)) = 5

Y (ω) = Y ((2, 5)) = 2 + 5 = 7

Sum of the dice is at least 11Event: Y ≥ 11 = (5, 6), (6, 5), (6, 6)Probability: P (Y ≥ 11) = 3

36= 1

12

Sum of the dice is less than 11Event: Y < 11Probability: P (Y < 11) = 1− P (Y ≥ 11) = 1− 1

12= 11

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Both dice show same numberEvent: X1 = X2 = (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)Probability: P (X1 = X2) = 6

36= 1

64

If the sample space is infinite, it may not be possible to define the probability measure Pjust by defining P (ω) for every outcome ω.

Example 1.2.2. Experiment: Pick a point uniformly at random from the interval [0, 1].Sample space: Ω = [0, 1].The probability that the point belongs to a given subinterval [a, b] in [0, 1] is proportional tothe length of the interval:P ([a, b]) = b− a for 0 ≤ a ≤ b ≤ 1In particular, P (c) = P ([c, c]) = 0 for any 0 ≤ c ≤ 1. 4

Multiple probability measures can be defined on the same sample space.

Example 1.2.3. Sample space: Ω = 1, 2, 3, 4, 5, 6The probability measure P on Ω satisfying P (ω) = 1

6for all ω ∈ Ω could represent an

experiment where a fair six-sided die is rolled.The probability measure P∗ on Ω satisfying P∗(ω) = 1

7for all ω ∈ 1, 2, 3, 4, 5 and

P∗(6) = 27

could represent an experiment where an unfair six-sided die is rolled. 4

Exercise 1.2.1. Flip a fair coin 5 times. Let X = totals number of heads.(a) Write down three possible outcomes ω from the sample space Ω.(b) How many outcomes are in the sample space?

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(c) Compute P (ω) for each outcome ω you wrote down in part (a).(d) Compute X(ω) for each outcome ω you wrote down in part (a).(e) Find P (X ≤ 3). Hint: Find P (X > 3) first.

Exercise 1.2.2. Consider a coin where the probability of heads is 0 < p < 1. Do notassume p = 1/2. Flip it until the first tails occurs. Let X = number of flips needed to seethe first tails.(a) How many outcomes are in the sample space?(b) Find P (X = 1), P (X = 2), and P (X = 3).(c) Write down a formula for P (X = k), where k is a positive integer.

1.3 Discrete Random Variables

Fix a sample space Ω and probability measure P .

Let X be a random variable. The range of X is

R(X) = X(ω) : ω ∈ Ω = the set of all possible outputs X(ω)

X is called a discrete random variable if R(X) is a countable set.

A countable set is a set whose elements can be listed. For example, −1, 0, 1, 1, 1/2, . . . , 1/10,N = 1, 2, 3, . . ., and Z = . . . ,−2,−1, 0, 1, 2, . . . are countable sets. R is an uncount-able set.

Example 1.3.1. Experiment: Flip a fair coin until the first tails occurs.X = number of heads in the first two flips.Y = total number of heads.R(X) = 0, 1, 2 and R(Y ) = 0, 1, 2, . . ..X and Y are discrete. 4

1.4 Expectation


If X is a discrete random variable, the expectation of X is

E(X) =∑

k∈R(X)

kP (X = k).

It is a weighted average of the possible outputs ofX , with the weights being the probabilityof each output.

Terminology: expectation = expected value = average value = mean = first moment

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Example 1.4.1. Experiment: Roll a fair six-sided die.X = the number shownR(X) = 1, 2, 3, 4, 5, 6E(X) =

∑k∈R(X)

kP (X = k) = (1)(1/6) + (2)(1/6) + ...+ (5)(1/6) + (6)(1/6) = 21/6 =

3.5 4

Example 1.4.2. Experiment: Roll a six-sided die where the probability of a 6 is 27

and theprobability of each other number is 1 7. X = the number shownR(X) = 1, 2, 3, 4, 5, 6E(X) =

∑k∈R(X)

kP (X = k) = (1)(1/7)+(2)(1/7)+ ...+(5)(1/7)+(6)(2/7) = 54/14 =

3.85714285714 4

1.5 Properties of Expectation


Let X and Y be random variables and let c ∈ R. Then

• E(c) = c,

• E(cX) = cE(X),

• E(X + Y ) = E(X) + E(Y ).

Example 1.5.1. Roll two fair six-sided dice. Let Z be the sum of the dice. Find EZ.Define Xi = number on i-th die. Then Z = X1 +X2, EXi = 3.5, and

EZ = EX1 + EX2 = 3.5 + 3.5 = 7.

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Exercise 1.5.1. Prove E(cX) = cE(X) for every discrete random variable X and constantc.

Exercise 1.5.2. Consider a class of 50 students. For each student, a fair six-sided die willbe rolled to determine the student’s final grade. If the die shows 6, the grade is 90. If thedie shows any other number, the grade is 40. Let Xi be the grade of the i-th student.(a) Let Xi be the grade of the i-th student. Find E(Xi).(b) Let Z be the class average. Write down the formula for Z in terms of X1, . . . , X50.(c) If only 7 students roll a 6, what is the class average?(d) What is the expectation of the class average?

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Exercise 1.5.3. Consider a coin where the probability of heads is p. Flip the coin n times.Xi = 1 if i-th flip is heads, 0 if i-th flip is tails. Define Y =

∑ni=1Xi.

(a) Express the event that there are exactly k heads in terms of Y .(b) Find E(Y ).

1.6 Law of the Unconscious Statistician (LOTUS)

Let X be a discrete random variable and let g : R→ R be a function. Then

E(g(X)) =∑

k∈R(X)

g(k)P (X = k)

Notice g(X) is also a random variable. Using LOTUS we can calculate the expectation ofg(X) using P (X = k) (which we presume is well-understood) instead of using P (g(X) =g(k)) (which may be more complicated to use).

Example 1.6.1. For any random variableX and constantK, we define the random variableI X > K by

I X > K (ω) =

1 if X(ω) > K0 otherwise

Let X be the number shown after rolling a fair six-sided die. Show that EI X > 3 =P (X > 3).Note R(X) = 1, 2, 3, 4, 5, 6. We use LOTUS with

g(k) =

1 if k > 30 otherwise.

EI X > 3 = Eg(X) =∑

k∈R(X)

g(k)P (X = k) =∑

k∈R(X)

I k > 3P (X = k)

= I 1 > 3P (X = 1) + I 2 > 3P (X = 2) + I 3 > 3P (X = 3)

+ I 4 > 3P (X = 4) + I5 > 3P (X = 5) + I 6 > 3P (X = 6)

= (0)P (X = 1) + (0)P (X = 2) + (0)P (X = 3)

+ (1)P (X = 4) + (1)P (X = 5) + (1)P (X = 6)

= P (X > 3).

Let X be any discrete random variable. Show EI X > K = P (X > K).

EI X > K =∑

k∈R(X)

I k > KP (X = k) =∑

k∈R(X)k>K

P (X = k) = P (X > K).

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1.7 Variance

Exercise 1.7.1. The variance of a random variable X is

Var(X) = E((X − E(X))2).

It is the average squared-distance between X and its average E(X).(a) Use the properties of expectation to prove Var(X) = E(X2)− (E(X))2.(b) LetX be the number shown after rolling of a fair six-sided die. Find E(X2) and Var(X).

Chapter 2

Arbitrage

2.1 Assets and Portfolios

An asset is a valuable thing that can be owned. Examples are shares (stocks), bonds, cash(domestic or foreign currency), real estate, and resource rights. (Don’t worry if you don’tknow what these are yet.)

We assume we can hold any amount of an asset, including negative amounts. A holding of−1 asset is a debt of 1 asset. For example, if you have no apples and you owe Johnny 3apples, you have −3 apples.

The value of an asset is the price at which it can be sold.

A portfolio is a collection of assets.

The value of a portfolio is the sum of the values of the assets in the portfolio. V A(T )denotes the value of a portfolio A at time T .

Our mathematical model is based on the following assumption: If t is the current time andT > t, then V A(t) is a constant and V A(T ) is a random variable.

2.2 Arbitrage

A portfolio A is called an arbitrage portfolio if the following conditions hold:

• At current time t,V A(t) ≤ 0.

• At some future time T > t,

V A(T ) ≥ 0 with probability one and V A(T ) > 0 with positive probability.

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In symbols, the second condition is: P (V A(T ) ≥ 0) = 1 and P (V A(T ) > 0) > 0 for someT > t.

An arbitrage portfolio represents “free lunch” or “getting something for nothing.” It isperhaps more accurate (but less snappy) to say an arbitrage portfolio represents “free lunchwithout risk” or “getting something for nothing without risk.”

No-arbitrage. There are no arbitrage portfolios. 4

Informally, ”there is no such thing as a free lunch.” Again, it is more accurate to say ”thereis no such thing as a free lunch without risk”.

We will always assume no-arbitrage unless otherwise indicated.

2.3 What is the Sample Space and the Probability Mea-sure?

We have assumed V A(T ) is a random variable and referred to probabilities involving it, butwe haven’t explicitly defined a sample space or probability measure. There is an advancedtheorem in probability theory (called the Kolmogorov existence theorem) that says we don’thave to. It says (essentially) that for any collection of random variables we encounter, if wespecify (know) the probabilities associated with the random variables (like P (a ≤ X ≤ b)),then there exists a sample space where the variables are defined and a probability measurethat gives the desired (correct) probabilities.

If it helps, we can have in mind a finite sample space Ω = ω1, . . . , ωn and a probabilitymeasure P with P (ωi) > 0 for all i. Then A is an arbitrage portfolio if

• At current time t, V A(t) ≤ 0.

• At some future time T > t, V A(T )(ωi) ≥ 0 for all i and V A(T )(ωj) > 0 for some j.

We can think of the sample space as all possible states of the universe.

2.4 Monotonicity and Replication

The monotonicity theorem and replication theorem are important consequences of the no-arbitrage assumption. We will use them frequently.

Monotonicity Theorem. Let A and B be portfolios and let T > t, where t is the currenttime.

(a) If V A(T ) ≥ V B(T ) with probability one, then V A(t) ≥ V B(t).

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(b) If V A(T ) ≥ V B(T ) with probability one, and V A(T ) > V B(T ) with positive proba-bility, then V A(t) > V B(t). 4

We prove (b) first because it is easier.

Proof of (b). Assume (1) V A(T ) ≥ V B(T ) with probability one and (2) V A(T ) > V B(T )with positive probability. We want to show V A(t) > V B(t). So we assume (3) V A(t) ≤V B(t), and we show this leads to a contradiction.

Consider the portfolio C consisting of A minus B. For example, if A consists of 5 sharesof AAPL and B consists of 3 shares of GOOGL, then C consists of 5 shares of AAPL and−3 shares of GOOGL. Then (by our assumptions (3),(1),(2) [in that order]) we have

• V C(t) = V A(t)− V B(t) ≤ 0

• V C(T ) = V A(T )− V B(T ) ≥ 0 with probability one

• V C(T ) = V A(T )− V B(T ) > 0 with positive probability

Therefore C is an arbitrage portfolio. This contradicts the no-arbitrage assumption.

Proof of (a). Assume V A(T ) ≥ V B(T ) with probability one. We want to show V A(t) ≥V B(t). So we assume V A(t) < V B(t) and show this leads to a contradiction.

Define ε = V B(t) − V A(t). Consider the portfolio C consisting of A minus B plus ε ofcash. Then

• V C(t) = V A(t)− V B(t) + ε = 0,

• V C(T ) = V A(T )− V B(T ) + ε ≥ ε > 0 with probability one.

Therefore C is an arbitrage portfolio. This contradicts the no-arbitrage assumption.

Remark. The ε cash in the previous proof is simply held. We discuss investing cash in thenext chapter.

Replication Theorem. Let A and B be portfolios and let T > t, where t is the currenttime. If V A(T ) = V B(T ) with probability one, then V A(t) = V B(t). 4

Proof. V A(T ) = V B(T ) means V A(T ) ≥ V B(T ) and V A(T ) ≤ V B(T ). The mono-tonicity theorem implies V A(t) ≥ V B(t) and V A(t) ≤ V B(t), which means V A(t) =V B(t).

Exercise 2.4.1. This exercise is to practice “proof by contradiction.” Prove:(a) Let a ≥ 0. If a ≤ ε for every ε > 0, then a = 0.(b)√

2 is irrational.

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Exercise 2.4.2. We showed that no-arbitrage implies the monotonicity theorem. In thisexercise we do the reverse. Show that the monotonicity theorem implies no-arbitrage. Hint:If A is an arbitrage portfolio, apply the monotonicity theorem to A and an empty portfolioB to deduce a contradiction.

Chapter 3

Compound Interest, Discounting, andBasic Assets

3.1 Interest Rates and Compounding

If we invest (lend, deposit) N dollars at interest rate r compounded annually, then:

• After one year the value of the investment is N(1 + r) dollars

• After two years: N(1 + r)2 dollars

• After T years: N(1 + r)T dollars

N is called the notional or principle. Unless we are dealing with multiple currencies, wewill usually omit writing “dollar” or currency symbols like $.

If we owe a debt of N at interest rate r compounded annually, then after T years we oweN(1+r)T . In other words, after T years we have−N(1+r)T . A debt ofN is like investing−N .

If we invest N at interest rate r compounded m times per year, then:

• After 1/m years the value of the investment is N(1 + r/m).

• After 2/m years: N(1 + r/m)2.

• After T years (i.e., mT/m years): N(1 + r/m)mT .

m is called the compounding frequency. The time T is measured in years and can be anynon-negative real number.

Remember that limm→∞

(1 + r/m)mT = erT .

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If we invest N at interest rate r compounded continuously, then:

• After T years we have NerT .

Example 3.1.1. If we invest 500 at rate 3% = 0.03 with compounding frequency 4 (quar-terly compounding), the value after 3 years is

500(1 + 0.03/4)4·3 = 546.9034 . . .

4

Example 3.1.2. If we invest 500 at rate 3% = 0.03 with daily compounding (assuming 365days per year), the value after 3 years is

500(1 + 0.03/365)365·3 = 547.0851 . . .

4

Example 3.1.3. If we invest 500 at rate 3% = 0.03 with continuous compounding, thevalue after 3 years is

500e0.03·3 = 547.0871 . . .

4

We will always deal with per-year interest rates. This means time is measured in units ofyears. The only exception is the next example, where we consider per-month interest rates.

Example 3.1.4. If we borrow 500 at rate 1.2% = 0.012 per month with monthly com-pounding, then after T months we owe

500(1 + 0.012)T .

Notice T is measured in months.

If we borrow 500 at rate 1.2% = 0.012 per month with compounding twice a month, thenafter T months we owe

500(1 + 0.012/2)2T .

If we borrow 500 at rate 1.2% = 0.012 per month with daily compounding (assuming eachmonth has 30 days), then after T months we owe

500(1 + 0.012/30)30T .

4

Result 3.1.5. If the interest rate with compounding frequency m is rm and the interest ratewith continuous compounding is r, then

(a)(

1 +rmm

)mT= erT for all T > 0

18

(b) r = m ln(

1 +rmm

)(c) rm = m

(er/m − 1

)4

Idea of Proof. If (a),(b),(c) do not hold, then we can build an arbitrage portfolio.

Proof. If we have any one of (a),(b),(c), then we get the other two by algebra. So we onlyprove (a).

Assume(

1 +rmm

)mT> erT for some T > 0. Fix t < T and consider the portfolio

A: At time t, invest 1 at rate rm with compounding frequencym and borrow 1 at rate r withcontinuous compounding.

Then V A(t) = 0 and V A(T ) =(

1 +rmm

)mT− erT > 0 with probability one.

So A is an arbitrage portfolio. This contradicts the no-arbitrage assumption.

If(

1 +rmm

)mT< erT for some T > 0, then a similar arguments also leads to a contradic-

tion.

The only remaining possibility is that(

1 +rmm

)mT= erT for all times T > 0.

A proof based on the monotonicity theorem is also possible.

Proof. Assume(

1 +rmm

)mT> erT for some T > 0. Fix t < T and consider portfolios A

and B.A: At time t, invest 1 at rate rm with compounding frequency m.B: At time t, invest 1 with at rate r with continuous compounding.Then V A(t) = V B(t) = 1 and

V A(T ) =(

1 +rmm

)mT> erT = V B(T ) with probability one.

This contradicts the monotonicity theorem. Note also the portfolio C = A − B is anarbitrage portfolio.

If(

1 +rmm

)mT< erT for some T > 0, then a similar arguments also leads to a contradic-

tion.

The only remaining possibility is that(

1 +rmm

)mT= erT for all times T > 0.

The money market account, Mt, is the value at time t of an investment of 1 dollar at time0. If the continuously compounded interest rate over period 0 to t is the constant r, thenMt = ert.

19

Remark. Interest rates are typically not constant in time. They depend on the period overwhich we lend/borrow. We will consider this generalization later.

Remark. We will always assume we can lend and borrow at non-negative interest rates.It is possible to consider negative interest rates, but we will not do so for simplicity. Aswe move through the course, the reader should think about how results would change ifinterest rates were negative.

Exercise 3.1.1. Show that if the interest rate with compounding frequency m1 is r1 and theinterest rate with compounding frequency m2 is r2, then

(a)(

1 +r1

m1

)m1T

=

(1 +

r2

m2

)m2T

for all times T

(b) r1 = m1

[(1 +

r2

m2

)m2/m1

− 1

]Exercise 3.1.2. If the six-month compounded interest rate is 4.3% = 0.043, find the con-tinuously compounded rate r and the annually compounded rate rA. Hint: The six-monthcompounded interest rate is the interest rate with compounding twice per year.

3.2 Zero Coupon Bonds and Discounting

A zero coupon bond (ZCB) with maturity T is an asset that pays 1 at time T (and nothingelse).

What is the value of a ZCB at time t ≤ T ? In other words, what is the value today of thepromise of a dollar tomorrow?

Let Z(t, T ) be the value at time t of a ZCB with maturity T , where t ≤ T . By definition,Z(T, T ) = 1.

Result 3.2.1. If the continuously compounded interest rate from time t to time T has con-stant value r, then

Z(t, T ) = e−r(T−t).

4

Proof. Consider two portfolios.A: At time t, a ZCB with maturity TB: At time t, investment of N = e−r(T−t) with continuously compounded interest rate r.Then V A(T ) = 1 and V B(T ) = Ner(T−t) = e−r(T−t)er(T−t) = 1.Therefore V A(T ) = V B(T ) with probability one.By replication, V A(t) = V B(t).In other words, Z(t, T ) = e−r(T−t).

20

Z(t, T ) is also called a discount factor. It depends on the interest rate and compoundingfrequency over the period from t to T .

Determining the value of an asset at time t based on its value at some future time T > t iscalled discounting. The value at time t is called the present value.

Example 3.2.2. Consider an asset that pays 500 and matures 3 years from now. If thecontinuously compounded interest rate is 2.1%, what is its present value?

If the present time is t, the asset is is equivalent to 500 ZCBs with maturity T = 3 + t, andthe present value is

500Z(t, T ) = 500e−r(T−t) = 500e−0.021(3).

4

We can define interest rates in terms of Z(t, T ). The number rA such that

Z(t, T ) = (1 + rA)−(T−t)

is called the annually compounded zero rate for the period t to T . The number r suchthat

Z(t, T ) = e−r(T−t)

is called the continuous zero rate for the period t to T .

The result above says that if the continuous interest rate is constant over the period t to T ,then it equals the continuous zero rate for the period t to T .

Exercise 3.2.1. Consider an asset that pays N at maturity 3 years from now. Suppose theannually compounded interest rate is 3% and the present value is 300. Find N .

Exercise 3.2.2. Show that if the interest rate with compounding frequency m from time tto time T has constant value r, then

Z(t, T ) = (1 + r/m)−m(T−t).

3.3 Annuities

An annuity is a series of fixed payments C at times T1, . . . , Tn. It is equivalent to thefollowing collection of ZCBs:

• C ZCBs with maturity T1

• C ZCBs with maturity T2

• ...

21

• C ZCBs with maturity Tn

Its value at time t ≤ T1 is

Vt = C

n∑i=1

Z(t, Ti).

Its value at time T1 < t ≤ T2 is

Vt = C

n∑i=2

Z(t, Ti).

because the 1st payment has already been made.

Example 3.3.1. Consider an annuity starting at time 0 that pays 1 each year for M years.Assume the annually compounded zero rate is rA for all maturities T = 1, . . . ,M . Thismeans that

Z(0, T ) = (1 + rA)−T for T ∈ 1, . . . ,MThe value of the annuity at time t = 0 is

V0 =M∑T=1

Z(0, T ) =M∑T=1

1

(1 + rA)T.

We simplify this geometric sum by a standard trick. Observe that

1

1 + rAV0 − V0 =

M+1∑T=2

1

(1 + rA)T−

M∑T=1

1

(1 + rA)T=

1

(1 + rA)M+1− 1

1 + rA

and solve for V to obtain

V0 =1− (1 + rA)−M

rA.

4

By arguing as in Example 3.3.1 we get:

Result 3.3.2. Consider an annuity starting at time t that pays C each year for M years.Assume the annually compounded zero rate is rA for all maturities T = t + 1, . . . , t + M .The value at its starting time t is

Vt = C · 1− (1 + rA)−M

rA.

4Example 3.3.3. In the US, a $100 million Powerball lottery jackpot is typically structuredas an annuity paying $4 million per year for 25 years. With an annually compoundedinterest rate of 3%, the value of the jackpot at time t = 0 is

4 · 106

25∑T=1

Z(0, T ) = 4 · 106

25∑T=1

1

(1 + 0.03)T=

1− (1 + 0.03)−25

0.03≈ 69.65 million

4

22

Example 3.3.4. A loan of 1000 is to be paid back in 5 equal installments due yearly.Interest of 15% of the balance is applied each year, before the installment is paid. This typeof loan is called an amoritized loan.

Find the amount C of each installment.

For the lender, the loan is equivalent to an annuity.Assume it starts at t = 0. So it pays C at times T = 1, 2, 3, 4, 5.The 15% yearly interest on the balance is equivalent to a 15% annually compounded interestrate.By Result 3.3.2, the value at time 0 is

V0 = C1− (1 + (0.15))−5

0.15

On the other hand, we know V0 = 1000.Therefore

C = 1000 · 0.15

1− (1 + (0.15))−5≈ 298.32.

4

Example 3.3.5. Consider an amortized loan with initial value V due inM years with equalannual installments C and annually compounded interest rate r. As in Example 3.3.4, eachinstallment is

C = Vr

1− (1 + r)−M.

The balance after the 1st installment is

B1 = V (1 + r)− C.

The balance after the 2nd installment is

B2 = (V (1 + r)− C)(1 + r)− C = V (1 + r)2 − C(1 + r)− C.

The balance after the k-th installment is

Bk = V (1 + r)k − Ck−1∑i=0

(1 + r)i.

We can rewrite this expression by substituting

C = Vr

1− (1 + r)−Mand

k−1∑i=0

(1 + r)i =1− (1 + r)k

1− (1 + r)

and doing some algebra. We find the balance after the k-th installment is

Bk = V(1 + r)M − (1 + r)k

(1 + r)M − 1.

23

B0 = V is the initial balance.

The interest at the 1st installment is

B0r = V r.

The interest at the 2nd installment is

B1r = (V (1 + r)− C)r

The interest at the k-th installment is

Bk−1r.

The amount of the initial loan repaid in the k-th installment (that is, the amount of the k-thinstallment that does not go towards interest) is

C −Bk−1r

4

Exercise 3.3.1. Prove Result 3.3.2.

Exercise 3.3.2. Consider the loan in Example 3.3.4.(a) What is the amount of interest included in each installment?(b) How much of the initial loan is repaid in each installment?(c) What is the outstanding balance after each installment is paid?

Exercise 3.3.3. Consider an annuity starting at time 0 that pays 1 each year for M years.Assume the annually compounded zero rate is rA for all maturities T = 1, . . . ,M . InExample 3.3.1, we found that the value of this annuity at its starting time 0 is

V0 =M∑i=1

Z(0, i) =1− (1 + rA)−M

rA.

Find the value of the annuity at time t′, where 0 < t′ < 1.

Exercise 3.3.4. Consider an annuity starting at time t that pays 1 each year for M years.Assume the annually compounded zero rate is rA for all maturities T = t + 1, . . . , t + M .According to Result 3.3.2, the value of this annuity at its starting time t is

Vt =M∑i=1

Z(t, t+ i) =1− (1 + rA)−M

rA.

Find the value of the annuity at time t′, where t < t′ < t+ 1.

24

3.4 Stocks and Bonds

A stock or share is an asset giving ownership of a fraction of a company. The price ofa stock at time T is denoted by ST . If t is the current time, then the known price St iscalled the spot price, and ST is a random variable for T > t. A stock may sometimes paya dividend, which is a cash payment usually expressed as a percentage of the stock price.

A fixed rate bond with notional N , coupon c, start date T0, maturity Tn, and term length αis an asset that pays N at time Tn and coupon payments αNc at times Ti for i = 1, . . . , n,where Ti+1 = Ti + α.

We define floating rate bonds later.

As is often done in finance, we will use the word security as a synonym for asset.

Result 3.4.1. Consider a fixed rate bond with coupon c, notional N , maturity M yearsfrom now, and annual coupon payments. Assume the annually compounded interest ratehas constant value rA. The value of the bond at present time t is

Vt = cN1− (1 + rA)−M

rA+N(1 + rA)−M .

4

Proof. The bond is equivalent to an annuity paying cN each year forM years plusN ZCBswith maturity M . Use Result 3.3.2 and Exercise 3.2.2.

Exercise 3.4.1. (a) Consider an annuity that pays 1 every quarter for M years. In otherwords, the payment times are T = t+ 1

4, t+ 2

4, . . . , t+ 4M

4. Show that the value at present

time t is

Vt =1− (1 + r4/4)−4M

r4/4,

assuming the quarterly compounded interest rate has constant value r4.(b) Consider a fixed rate bond with notional N and coupon c that starts now, matures Myears from now, and has quarterly coupon payments. Show that the value at present time tis

Vt =cN

4· 1− (1 + r4/4)−4M

r4/4+N(1 + r4/4)−4M ,

assuming the quarterly compounded interest rate has constant value r4.

Exercise 3.4.2. (a) Consider an annuity that pays 1 every quarter for M years. In otherwords, the payment times are T = t+ 1

4, t+ 2

4, . . . , t+ 4M

4. Show that the value at present

time t is

Vt =1− (1 + r8/8)−8M

(1 + r8/8)2 − 1,

assuming the interest rate with compounding 8 times per year has constant value r8.(b) Consider a fixed rate bond with notional N and coupon c that starts now, matures M

25

years from now, and has quarterly coupon payments. Show that the value at present time tis

Vt =cN

4· 1− (1 + r8/8)−8M

(1 + r8/8)2 − 1+N(1 + r8/8)−8M ,

assuming the interest rate with compounding 8 times per year has constant value r8.

3.5 Foreign Exchange Rates

Example 3.5.1. The current euro (EUR) to US dollar (USD) exchange rate is

0.89 EUR/USD.

Therefore the USD to EUR exchange rate is

1

0.89 EUR/USD≈ 1.12 USD/EUR.

Then

150 USD = (150 USD)

(0.89

EURUSD

)= 150(0.89) EUR = 133.50 EUR.

4

Chapter 4

Forward Contracts

4.1 Derivative Contracts

A derivative contract or derivative is a financial contract between two entities whosevalue is a function of (derives from) the value of another variable.

The variable could be the price of an asset, an interest rate, or even the weather.

The two entities in the contract are called counterparties.

4.2 Forward Contracts

Our first derivative is the forward contract.

In a forward contract or forward, two counterparties agree to trade a specific asset (likea stock) at a certain future time T and a certain price K.

One counterparty agrees to buy the asset at time T and price K, and the other counterpartyagrees to sell the asset at time T and price K.

We say the buyer is long the forward contract, and the seller is short the forward contract.

K is the called the delivery price. T is called the maturity or delivery date.

4.3 Value of Forward

Fix an asset.

Consider a forward on the asset with delivery price K and maturity T .

26

27

K and T are fixed at the time the forward contract is agreed to, but the value of the forwardcontract may change over time.

VK(t, T ) denotes the value of the forward to the long counterparty at time t ≤ T .

Then −VK(t, T ) is the value of the forward to the short counterparty at time t ≤ T .

To take at time t the long position in a forward contract with maturity T and delivery priceK, we must pay VK(t, T ) at time t to the counterparty party taking the short position. Notethat we must also pay K at time T to buy the asset.

You can think of VK(t, T ) as the amount the long counterparty must pay upfront (time t)to convince the short counterparty to agree to the forward contract. The long counterpartymust still pay K at time T to buy the asset.

Note that if VK(t, T ) is negative, it is actually the short counterparty that pays upfront.Paying a negative amount means receiving.

Here is one more way to understand the value of a forward contract. Suppose two coun-terparties have agreed (at some time in the past) to a forward contract with maturity T anddelivery price K. At current time t ≤ T , we want to buy we buy the long position from thelong counterparty, so that we become the long counterparty. To do so, we need to pay thecurrent long counterparty VK(t, T ) at time t. Note that we will still need to pay pay K attime T to buy the asset itself.

4.4 Value at Maturity and Payoff

Fix an asset. Let St be its price at time t. Consider a forward on the asset with deliveryprice K and maturity T

At time T , we know the counterparty long the forward must pay K to buy the asset whosevalue is ST . Therefore

VK(T, T ) = ST −K.

This is the value at maturity long the forward. It is also called the payoff or payout longthe forward.

The function g(x) = x − K is called the forward payoff function. Then the payoff(payout, value at maturity) long the forward can be written as

VK(T, T ) = g(ST ) = ST −K.

The payoff (payout, value at maturity) short the forward is

−g(ST ) = −VK(T, T ) = K − ST .

In later sections, we will study VK(t, T ) for general t < T .

28

Example 4.4.1. Consider a forward with delivery price 100 and maturity T = 2 years. Ifthe underlying asset has price 95.10 at maturity, find value of the forward to the long partyat maturity.

We have K = 100, T = 2, ST = 95.10. Therefore VK(T, T ) = ST −K = 95.10− 100 =−4.90. 4

Example 4.4.2. Consider a forward with delivery price 100 and maturity T . Supposeunderlying asset has price

ST =

110 with probability 0.690 with probability 0.4.

Find the expected payoff long the forward.

The payoff at maturity long the forward is

g(ST ) = ST−100 =

110− 100 with probability 0.690− 100 with probability 0.4

=

10 with probability 0.6−10 with probability 0.4

The expected payoff at maturity long the forward is

E(g(ST )) =∑

k∈R(g(ST ))

kP (g(ST ) = k) = (10)(0.6) + (−10)(0.4) = 2

4

Exercise 4.4.1. Consider a forward with delivery price 200 and maturity T . Suppose theunderlying asset has price

ST =

150 with probability 0.3200 with probability 0.5250 with probability 0.2

.

(a) Find the payoff long the forward.(b) Find the expected value at maturity long the forward.(c) Find the expected value of the forward to the short counterparty at maturity.

4.5 Forward Price

Pre-emptive clarification:

forward price 6= price of forward = value of forward = VK(t, T )

The terminology is unfortunate, but we are stuck with it.

Fix an asset. Fix times t ≤ T .

29

The forward price of the asset is the number F (t, T ) such that the forward contract withmaturity T and delivery price K = F (t, T ) has value VK(t, T ) = 0 at time t.

We will describe the forward price in two more ways.

Consider a forward contract on the asset with maturity T . Consider all possible deliveryprices for this forward contract. The special delivery price K such that VK(t, T ) = 0 iscalled the forward price of the asset and is denoted F (t, T ).

Consider a forward contract on the asset with maturity T . Suppose the two counterpartiesare negotiating the delivery price K before they agree to the contract. Remember that atfuture time T the long counterparty will pay K to the short counterparty in exchange forthe asset. If K is too small, the long counterparty must pay VK(t, T ) at current time tto convince the short counterparty to agree to the contract. If K is too large, the shortcounterparty must pay −VK(t, T ) (VK(t, T ) < 0 in this situation) at current time t toconvince the long counterparty to agree to the contract. The delivery price K where eachcounterparty pays zero (VK(t, T ) = 0) at t to convince the other is called the forwardprice and is denoted F (t, T ). A better name for the forward price may be the “fair deliveryprice.”

Let St be the price of the asset at time T . Since VK(T, T ) = ST −K and since F (T, T ) isthe value of K that makes VK(T, T ) = 0, we have

F (T, T ) = ST .

In later sections, we will study F (t, T ) for general t < T .

4.6 Value of Forward and Forward Price

For the rest of the chapter, the continuous zero rate for period t to T is denoted by r.

The next result connects the value of a forward on an asset, the delivery price of the forward,and the forward price of the asset.

Result 4.6.1. For any asset,

VK(t, T ) = (F (t, T )−K)Z(t, T ) = (F (t, T )−K)e−r(T−t).

4

Proof. Consider two portfolios at time t.A: one long forward with maturity T and delivery priceK, one short forward with maturityT and delivery price F (t, T )B: (F (t, T )−K) ZCBs with maturity TWe haveV A(T ) = (ST −K) + (F (t, T )− ST ) = F (t, T )−K

30

V B(T ) = F (t, T )−KSo V A(T ) = V B(T ) with probability one.By replication, V A(t) = V B(t).ButV A(t) = VK(t, T ) + 0 = VK(t, T )V B(t) = (F (t, T )−K)Z(t, T )Therefore

VK(t, T ) = (F (t, T )−K)Z(t, T ).

Example 4.6.2. Assume VK(t, T ) < (F (t, T )−K)Z(t, T ). Find an arbitrage portfolio.A is “A from the proof” minus VK(t, T ) cash (or VK(t, T )/Z(t, T ) ZCBs with maturity T )V A(t) = 0.

V A(T ) = (ST −K) + (F (t, T )− ST )− VK(t, T )

Z(t, T )> 0 with probability one. 4

Exercise 4.6.1. Assume VK(t, T ) > (F (t, T )−K)Z(t, T ). Find an arbitrage portfolio.

4.7 Forward Price for Asset Paying No Income

Result 4.7.1. Suppose an asset pays no income (no dividends, no coupons, no payment atmaturity, no rent, etc). Then

F (t, T ) =St

Z(t, T )= Ste

r(T−t),

where St is the price of asset at time t. 4

Proof. Consider two portfolios.A: At time t, one unit of the asset.B: At time t, one long forward contract with maturity T and delivery priceK plusK ZCBs.

Then V A(T ) = ST and V B(T ) = (ST −K) +K = ST .So V A(T ) = V B(T ).By replication, V A(t) = V B(t).This means

St = VK(t, T ) +KZ(t, T ). (4.7.1)

The forward price F (t, T ) is the value of K such that VK(t, T ) = 0.Setting K = F (t, T ) and VK(t, T ) = 0 in (4.7.1) leads to

F (t, T ) =St

Z(t, T ).

31

Example 4.7.2. The current price of a certain stock paying no income is 20. Assumethe continuous zero rate will be 2.1% for the next 6 months. Find the forward price withmaturity in 6 months.

T − t = 6 months = 0.5 years.

F (t, T ) =St

Z(t, T )=

20

e−(0.021)(0.5)

4

Example 4.7.3. (a) Find a formula for the value at time t of a forward contract on a stockpaying no income if the delivery price is K and maturity is T .(b) The current price of a certain stock paying no income is 20. Assume the continuouszero rate will be 2.1% for the next 6 months. Use the formula from part (a) to find the valueof a forward contract on the stock if the delivery price is 25 and maturity is in 6 months.

(a) Combining Result 4.6.1 and Result 4.7.1, we get

VK(t, T ) = (F (t, T )−K)Z(t, T ) =

(St

Z(t, T )−K

)Z(t, T ) = St−KZ(t, T ) = St−Ke−r(T−t)

where r is the continuous zero rate for period t to T .(b) T − t = 0.5 years, St = 20, K = 25, r = 0.021

VK(t, T ) = 20− 25e−(0.021)(0.5)

4

If the price of an asset does not match the no-arbitrage price (i.e., the price implied by theno-arbitrage assumption), then we can construct an arbitrage portfolio.

Example 4.7.4. At current time t, a certain stock paying no income has price 45, theforward price with maturity T on the stock is 48, and the price of a zero coupon bond withmaturity T is 0.95. Determine whether there is an arbitrage opportunity. If so, construct anarbitrage portfolio.

Given St = 45, F (t, T ) = 48, Z(t, T ) = 0.95. Notice

StZ(t, T )

=45

0.95≈ 47.37.

So F (t, T ) >St

Z(t, T ). This violates the result above deduced from the no-arbitrage as-

sumption. So there must be arbitrage portfolio.

At current time t, portfolio A is

• −St cash (i.e., St cash debt)

32

• 1 stock

• 1 short forward contract with maturity T and delivery price equal to the forward priceF (t, T )

V A(t) = −St + St + VF (t,T )(t, T ) = 0.

At time T , we must sell 1 stock at price F (t, T ), and our debt has grown to Ster(T−t) =St/Z(t, T ), where r is the continuous zero rate for period t to T .

At time T , portfolio A is:

• −Ster(T−t) cash

• F (t, T ) cash

The value of A is

V A(T ) = F (t, T )− Ster(T−t) = F (t, T )− StZ(t, T )

> 0.

with probability one.

Therefore A is an arbitrage portfolio. 4

Exercise 4.7.1. At current time t, a certain stock has price 45, the forward price withmaturity T on the stock is 40, and the price of a zero coupon bond with maturity T is 0.95.Construct an arbitrage portfolio if possible. Explain the transactions needed to realize thearbitrage. Verify the portfolio you construct is an arbitrage portfolio.

4.8 Forward Price for Asset Paying Known Income

Result 4.8.1. Suppose an asset an pays a known amount of income (for example, dividends,coupons, payment at maturity, rent) during the life of the forward contract. Then

F (t, T ) =St − ItZ(t, T )

= (St − It)er(T−t),

where St is the price of asset at time t, and It is the present value of the income at time t.4

Notice the forward price for an asset paying known income is lower than it would be if theasset paid no income. Until maturity of the forward, the short counterparty holds the assetand collects any income it pays, while the long counterparty gets no income. Thus, for anasset paying income, there is an advantage to buying it spot (i.e., buying it immediately attime t) rather than buying it forward. So the forward price is lower to compensate.

33

Proof. A: At time t, one unit of the asset plus −It cashB: At time t, one long forward contract with maturity T and delivery priceK plusK ZCBs.

We haveV A(T ) = ST + Ite

r(T−t) − Iter(T−t) = ST ,V B(T ) = ST −K +K = ST .So V A(T ) = V B(T ) with probability 1By replication, V A(t) = V B(t).Therefore

St − It = VK(t, T ) +KZ(t, T ). (4.8.1)

The forward price F (t, T ) is the value of K such that VK(t, T ) = 0.Setting K = F (t, T ) and VK(t, T ) = 0 in (4.8.1) leads to

F (t, T ) =St − ItZ(t, T )

.

By the definition of continuous zero rate, Z(t, T ) = e−r(T−t).

Example 4.8.2. Assume the continuously compounded interest rate is a constant r. Con-sider an asset that pays d at times T1 ≤ T2 ≤ · · · ≤ Tn. Find the forward price F (t, T )when t ≤ T1 and Tn ≤ T .It = present value of the income = present value of the stream of payments.

The payment of d at Ti is equivalent to d ZCBs with maturity Ti. Therefore

It = dn∑i=1

Z(t, Ti) = dn∑i=1

e−r(Ti−t)

Hence

F (t, T ) =St − ItZ(t, T )

=St − d

∑ni=1 e

−r(Ti−t)

e−r(T−t).

4

Exercise 4.8.1. The income may be negative if the asset has carrying costs, such as insur-ance or storage costs. Suppose you have a single gold bar (400 troy ounces) in a storage unitin Mountain View, California. Rent for the storage unit is 100 per month (with paymentsstarting immediately, not at the end of the month). The current price of gold is 1325 pertroy ounce. Find the current forward price for the gold bar if maturity is 1 year from now,assuming the continuously compounded interest rate has constant value 3%. Hint: Fromthe point of view of the rental company, the rent is a sequence of ZCBs with maturitiesafter 0 months, 1 month, 2 months,. . ., 11 months.

34

4.9 Forward Price for Stock Paying Known Dividend Yield

Result 4.9.1. Suppose a stock pays dividends as a percentage q of the stock price on acontinuously compounded per-year basis. q is called the dividend yield. Suppose thedividends are automatically reinvested in the stock. Then

F (t, T ) =Ste−q(T−t)

Z(t, T )= Ste

(r−q)(T−t)

4

Dividends are often assumed to be paid continuously, rather than at discrete times. Thisassumption is appropriate for a portfolio of many stocks paying dividends at many differ-ent times throughout the year. If you wish, replace the word “stock” in Result 4.9.1 by“portfolio of many stocks.”

Proof. Consider two portfolios at time t.A: N = e−q(T−t) units of stockB: 1 forward with delivery price K and maturity T , K ZCBs with maturity TAt time T , the number of units of stock in A is Neq(T−t) = e−q(T−t)eq(T−t) = 1. ThereforeV A(T ) = STV B(T ) = (ST −K) +K = STSince V A(T ) = V B(T ) with probability one, the replication theorem gives V A(t) =V B(t).We have V A(t) = e−q(T−t)StV B(t) = VK(t, T ) +KZ(t, T )Therefore

e−q(T−t)St = VK(t, T ) +KZ(t, T ).

Setting K = F (t, T ) gives VK(t, T ) = 0 and leads to

F (t, T ) =Ste−q(T−t)

Z(t, T ).

Exercise 4.9.1. Suppose a stock pays dividends m times per year at equally spaced timeswith annual yield q. So each dividend payment is equal to q/m of the stock price and thepayments are made at times t + 1/m, t + 2/m, . . . , where t is the current time. Supposethe dividends are automatically reinvested in the stock.(a) If you have 1 unit of stock at time t, how many will you have 1/m years later when thefirst dividend is paid?(b) If T − t is an integer multiple of 1/m, show that the forward price for the stock is

F (t, T ) =St(1 + q/m)−m(T−t)

Z(t, T ). (4.9.1)

35

(c) Compute the limit as m→∞.(d) Suppose m = 1 and T − t = 0.5 (so T − t is not an integer multiple of m). Show that if(4.9.1) holds, then you can build an arbitrage portfolio. Verify the portfolio is an arbitrageportfolio.

4.10 Forward Price for Currency

Result 4.10.1. Suppose Xt is the price at time t in USD of one unit of foreign currency.(For example, 1 CAD = 0.77 USD.) Let r$ be the zero rate for USD. Let rf be the zero ratefor foreign currency, both constant and compounded continuously. Then the forward pricefor one unit of foreign currency is

F (t, T ) = Xte(r$−rf )(T−t).

4

Exercise 4.10.1. Prove Result 4.10.1. Hint: In the proof of Result 4.9.1, replace the stockby foreign currency. The foreign currency is analogous to a stock paying known dividendyield. The foreign interest rate corresponds to the dividend yield.

Chapter 5

Forward Rates and Libor

5.1 Forward Zero Coupon Bond Prices

Fix T1 ≤ T2. Consider a forward contract with maturity T1 where the underlying asset isa ZCB with maturity T2. The (spot) price of the ZCB is St = Z(t, T2) for t ≤ T2. Theforward price of the ZCB is denoted F (t, T1, T2) for t ≤ T1.

Result 5.1.1.F (t, T1, T2) =

Z(t, T2)

Z(t, T1)

4

Proof. Consider two portfolios.A: 1 ZCB with maturity T2

B: 1 long forward contract with delivery price K and maturity T1 on a ZCB with maturityT2; K ZCBs with maturity T1

We have V A(T1) = Z(T1, T2) and V B(T1) = (Z(T1, T2)−K) +K = Z(T1, T2).Therefore V A(T1) = V B(T2) with probability one.By the replication theorem,

V A(t) = V B(t).

We have V A(t) = Z(t, T2) and V B(t) = VK(t, T1) +KZ(t, T1)Therefore

Z(t, T2) = VK(t, T1) +KZ(t, T1) (5.1.1)

The forward price F (t, T1, T2) is the value of K such that VK(t, T ) = 0.Setting K = F (t, T1, T2) and VK(t, T ) = 0 in (5.1.1) leads to

F (t, T1, T2) =Z(t, T2)

Z(t, T1).

36

37

5.2 Forward Interest Rates

Fix times t ≤ T1 ≤ T2. The forward rate at current time t for period T1 to T2 is the interestrate agreed at t at which one can borrow or lend money from T1 to T2.

f12 = forward rate at current time t for period T1 to T2

r1 = zero rate for period t to T1.

r2 = zero rate for period t to T2.

Figure 5.1: Forward and zero rates

Figure 5.1 depicts two strategies we can follow.

Strategy 1. We agree at time t to lend/borrow for period t to T2 at rate r2.

Strategy 2. We agree at time t to lend/borrow for period t to T1 at rate r1, and to lend/borrowfor period T1 to T2 at rate f12.

The next result says the two strategies are equivalent.

Result 5.2.1. Assume all rates are for continuous compounding. Then

er1(T1−t)ef12(T2−T1) = er2(T2−t) (5.2.1)

which (by rearranging) means

f12 =r2(T2 − t)− r1(T1 − t)

T2 − T1

.

4

Proof. Consider a portfolio A where we borrow N according to Strategy 1 and lend Naccording to Strategy 2.

In other words, suppose A starts empty at time t. We borrow N at t until T2 at rate r2. Sowe have N cash in hand and a debt of N for period t to T2 at rate r2. The debt will becomeNer2(T2−t) by time T2. We lend theN cash at t until T1 at rate r1 to obtainNer(T1−t) at timeT1. Then we lend the Ner1(T1−t) from T1 to T2 at rate f12 to obtain Ner1(T1−t)ef12(T2−T1) attime T2.

38

Therefore

V A(t) = N −N = 0 and

V A(T2) = Ner1(T1−t)ef12(T2−T1) −Ner2(T2−t).

If (5.2.1) is not true, then either

V A(T2) > 0 or V A(T2) < 0.

In the first case, A is an arbitrage portfolio. In the second case,−A is an arbitrage portfoliobecause V −A(t) = −V A(t) = 0 and V −A(T2) = −V A(T2) > 0. Either way we get anarbitrage portfolio. This contradicts the no-arbitrage assumption. Therefore (5.2.1) mustbe true.

Example 5.2.2. One year from now, your business plan requires a loan of 100, 000 topurchase new equipment. You plan to repay the loan one year after that. You want toarrange the interest rate of the loan today, rather than gamble on the interest rate one yearfrom now. The current time is t = 0. The interest rate for period 0 to 1 is 8%. The interestrate for period 0 to 2 is 9%. Assuming (as usual) no-arbitrage, what must the interest rateon the loan be?

Assume all rates are for continuous compounding.

We have t = 0, T1 = 1, T2 = 2, r1 = 0.08, r2 = 0.09. Want f12.

f12 =r2(T2 − t)− r1(T1 − t)

T2 − T1

=0.09(2− 0)− 0.08(1− 0)

2− 1= 0.1 = 10%

4

Result 5.2.3. Assume all rates are for annual compounding. Then

(1 + r1)T1−t(1 + f12)T2−T1 = (1 + r2)T2−t, (5.2.2)

which we can solve for f12, r1, r2, T1 − t, T2 − t, or T2 − T1. 4

Result 5.2.4. With annual compounding the forward price F (t, T1, T2) of a ZCB with ma-turity T2 is

F (t, T1, T2) = (1 + f12)−(T2−T1) =Z(t, T2)

Z(t, T1).

With continuous compounding the forward price F (t, T1, T2) of a ZCB with maturity T2 is

F (t, T1, T2) = e−f12(T2−T1) =Z(t, T2)

Z(t, T1).

4

39

Proof. Assume all rates are for annual compounding. Then

Z(t, Ti) = (1 + ri)−(Ti−t).

Substituting in (5.2.2) gives

(Z(t, T1))−1(1 + f12)T2−T1 = (Z(t, T2))−1.

Rearranging gives

(1 + f12)−(T2−T1) =Z(t, T2)

Z(t, T1).

But remember the forward price F (t, T1, T2) of a ZCB with maturity T2 is also

F (t, T1, T2) =Z(t, T2)

Z(t, T1).

This proves the result for annual compounding. A similar argument works for any com-pounding frequency.

Example 5.2.5. Assume t ≤ T1 ≤ T2, where t = current time. What can you say aboutinterest rates between T1 and T2 if(a) Z(t, T1) = Z(t, T2).(b) Z(t, T1) > 0 and Z(t, T2) = 0.

For annual compounding:

F (t, T1, T2) =Z(t, T2)

Z(t, T1)= (1 + f12)−(T2−T1).

(a) We have

(1 + f12)−(T2−T1) =Z(t, T2)

Z(t, T1)= 1,

so the forward rate f12 between T1 and T2 must be 0.(b) We have

(1 + f12)−(T2−T1) =Z(t, T2)

Z(t, T1)= 0,

so the forward rate f12 between T1 and T2 must be∞.

The same is true for any compounding frequency.

For continuous compounding:

F (t, T1, T2) =Z(t, T2)

Z(t, T1)= e−f12(T2−T1).

(i) We have

e−f12(T2−T1) =Z(t, T2)

Z(t, T1)= 1,

40

so the forward rate f12 between T1 and T2 must be 0.(ii) We have

e−f12(T2−T1) =Z(t, T2)

Z(t, T1)= 0,

so the forward rate f12 between T1 and T2 must be∞. 4

Remark. Our interest rate and forward rate notation is not perfect. For example, f12

with annual compounding is not f12 with continuous compounding. The notation hidesinformation. It should indicate the current time, the start and end dates of the interestperiod, and the compounding frequency. A notation like

ft,T1,T2,m

would be good, but it is tedious to write. Without introducing a tedious notation, the solu-tion is simply to be very careful to clearly state and understand the hidden parameters ofthe rates you work with.

For the next example, we need some definitions.

The zero rate for the period starting now and ending M years later is called the M -yearzero rate. It is the zero rate for period t to T = t+M .

The forward rate for the period starting M years from now and and ending N years later isthe M -year forward N -year rate. It is denoted fMN . It is also called the MyNy rate. Itis the forward rate at current time t for period T1 = t+M to T2 = T1 +N = t+M +N .

Remark. Notice that the notation for the one-year forward two-year rate is f12. Thisconflicts with the notation f12 used earlier for the forward rate for period T1 to T2. Thisis unfortunate, but both notations are used by the textbook. If you are careful, you will beable to keep things straight.

Example 5.2.6. Assume all rates are annually compounded. The one-year and two-yearzero rates are 1% and 2%, respectively.

(a) What is the one-year forward one-year rate?

Given:r1 = 0.01 = zero rate for period starting now and ending 1 year later,r2 = 0.02 = zero rate for period starting now and ending 2 years later.Want: f11 = 1y1y forward rate = one-year forward one-year rate = forward rate for theperiod starting one year from now and ending one year later.

We can assume current time t = 0. Then T1 = 1, T2 = 2.Givenr1 = 0.01 = zero rate for period 0 to 1r2 = 0.02 = zero rate for period 0 to 2Wantf11 = forward rate for period 1 to 2

41

Draw a picture like Figure 5.1.

Then(1 + r1)T1−t(1 + f11)T2−T1 = (1 + r2)T2−t

i.e.,(1 + 0.01)1−0(1 + f11)2−1 = (1 + 0.02)2−0.

Solving for f11 gives

f11 =

((1 + r2)T2−t

(1 + r1)T1−t

)1/(T2−T1)

− 1 =

((1 + 0.02)2−0

(1 + 0.01)1−0

)1/(2−1)

− 1

= 0.0300990099 . . . = 3.00990099 . . .%

(b) If the two-year one-year forward rate is 3%, what is the three-year zero rate?

In class, I will draw a picture like Figure 5.1.

We can assume current time t = 0.f21 = 0.03 = 2y1y forward rate = two-year forward one-year rate = forward rate for theperiod starting two years from now and ending one year later = forward rate at t = 0 forT2 = 2 to T3 = 3r2 = 0.02 = zero rate for 0 to 2r3 = zero rate for 0 to 3 = ???Then

(1 + r2)T2−t(1 + f21)T3−T2 = (1 + r3)T3−t

i.e.,(1 + 0.02)2−0(1 + 0.03)3−2 = (1 + r3)3−0.

Solving for r3 givesr3 = 0.02332249903 . . . .

4

5.3 Libor

If the simple interest rate for period t to t+α is r, then a deposit of N at time t with growto N(1 + αr) at time t+ α.

The simple interest rate at which banks borrow or lend money to each other is called libor orlibor rate. Libor stands for London InterBank Offered Rate. Most interest rate derivativesare libor derivatives.

The libor at current time t for period t to t+ α is denoted Lt[t, t+ α].

Banks can deposit (or borrow) N at time t and receive (or pay back) N(1 + αLt[t, t + α])at time t+ α.

42

The libor LT [T, T + α] for a future date T > t is a random variable. It is sometimes calledthe libor fix at T .

For α = 1, the libor Lt[t, t+α] for is called the 1-year libor or twelve-month libor or 12mL.

For α = 0.25, the libor Lt[t, t + α] for α = 0.25 is called the quarter-year libor or three-month libor or 3mL.

Example 5.3.1. The current six-month libor is 3.46%. If a bank lends 2, 000, 000 at thisrate, how much will they receive at term?

N(1 + αLt[t, t+ α]) = 2, 000, 000(1 + (0.5)(0.0346)) = 2, 034, 600

4

In practice, libor (London InterBank Offered Rate) is the interest rate at which a bank offersto lend money and libid (London InterBank Bid Rate) is the interest rate at which a bankis willing to borrow money. For large transactions the two rates are very close together. Tokeep things simple, we assume libor = libid.

5.4 Forward Rate Agreement Definition

A forward rate agreement FRA is a contract to exchange two cashflows. It has threeparameters: T = maturity; K = delivery price or fixed rate; α = term length or daycountfraction.

The buyer (long counterparty) of the FRA agrees at t ≤ T to pay αK and receive αLT [T, T+α] at time T + α. The seller (short counterparty) agrees to do the opposite.

The value or payout at time T + a of the FRA for the buyer is

αLT [T, T + α]− αK.

αLT [T, T + α] is the amount of simple interest that would be accrued on notional 1 atfloating (unknown, random) rate LT [T, T + α] from T to T + α.

αK is the amount of simple interest that would be accrued on notional 1 at fixed (known,non-random) rate K from T to T + α.

The FRA is a new type of derivative contract. It is similar to a forward contract, butαLT [T, T + α] is not the price of an asset that is being traded.

43

5.5 Forward Libor Rate and Value of Forward Rate Agree-ment

Consider a FRA with maturity T , fixed rate (delivery price) K, and term length α. Thevalue of the FRA at current time t ≤ T is denoted VK(t, T ).

The forward libor rate Lt[T, T + α] is the special number such that the value at currenttime t of a FRA with maturity T , term length α, and delivery price K = Lt[T, T + α] isVK(t, T ) = 0.

Compare: The forward price F (t, T ) of an asset is the special number such that the valueat t of the forward contract on the asset with maturity T and delivery price K = F (t, T ) isVK(t, T ) = 0.

Make sure you understand the difference betweenLt[t, t+α], LT [T, T+α], andLt[T, T + α]:

• Lt[t, t + α] = libor rate at current time t for period t to t + α. Banks can deposit (orborrow) N at time t and receive (or pay back) N(1 + αLt[t, t+ α]) at time t+ α.

• LT [T, T + α] = libor rate at future time T for period T to T + α. Banks can deposit(or borrow) N at time T and receive (or pay back) N(1 + αLt[T, T + α]) at timeT + α.

• Lt[T, T + α] = forward libor rate at t for period T to T + α. It is the number suchthat the value at t of the FRA with maturity T , term length α, and delivery priceK = Lt[T, T + α], and is VK(t, T ) = 0.

• Lt[t, t+ α] and Lt[T, T + α] are known (non-random) at time t, while LT [T, T + α]is a random variable.

Result 5.5.1. The value at current time t of a FRA with maturity T , delivery price K, andterm length α is

VK(t, T ) = Z(t, T )− Z(t, T + α)− αKZ(t, T + α).

The forward libor rate at time t for maturity T and term length α is

Lt[T, T + α] =Z(t, T )− Z(t, T + α)

αZ(t, T + α).

4

Proof. Our strategy is to find a portfolio that replicates the FRA.

A: 1 long FRA with maturity T , delivery price K, and term length α.

B: At time t, 1 ZCB with maturity T and −(1 + αK) ZCBs with maturity T + α. At timeT , put the 1 from the ZCB maturing at T in a deposit with libor rate LT [T, T + α].

44

At time T + α:

V A(T + α) = α(LT [T, T + α]−K)

V B(T + α) = −(1 + αK) + (1 + αLT [T, T + α]) = α(LT [T, T + α]−K)

Since V A(T+α) = V B(T+α) with probability one, the replication theorem gives V A(t) =V B(t).

At time t:

V A(t) = value of the FRA = VK(t, T ).

V B(T ) = Z(t, T )− (1 + αK)Z(t, T + α).

Therefore:VK(t, T ) = Z(t, T )− Z(t, T + α)− αKZ(t, T + α).

By the definition of the forward libor rate Lt[T, T + α], setting K = Lt[T, T + α] givesVK(t, T ) = 0 and so

0 = Z(t, T )− Z(t, T + α)− αLt[T, T + α]Z(t, T + α).

Rearranging gives

Lt[T, T + α] =Z(t, T )− Z(t, T + α)

αZ(t, T + α).

Rearranging the formula for the forward libor rate in Result 5.5.1 in another way gives thediscounting formula

Result 5.5.2.Z(t, T + α) = Z(t, T )

1

1 + αLt[T, T + α].

4

Figure 5.2: Discounting by ZCBs and the forward libor rate

Example 5.5.3. Suppose the one-year and two-year continuous zero rates are 1% and 2%,respectively. What is the one-year forward one-year libor rate?

Assume current time t = 0.

45

one-year continuous zero = continuous zero rate for period 0 to 1 = 1%

two-year continuous zero = continuous zero rate for period 0 to 2 = 2%

The m-year forward n-year libor rate is the forward libor rate for the period starting myears from now and ending n years after that.

Want: one-year forward one-year libor rate = Lt[T, T + α]with t = 0, T = 1, α = 1, T + α = 2.

Lt[T, T + α] =Z(t, T )− Z(t, T + α)

αZ(t, T + α)=Z(0, 1)− Z(0, 2)

Z(0, 2)

=e−(0.01)(1) − e−(0.02)(2)

e−(0.02)(2)= 0.0304545 . . .

4

Example 5.5.4. A bank plans to borrow 1 at future time T until T + α.

(a)Show how the bank can at current time t lock in the interest cost for the period T toT + α by combining a FRA trade today with a libor loan at T . What is the total interestcost?

(b) What is the interest cost if instead the bank plans to borrow $5 million for a term lengthα = 2 years and the current forward libor rate for period T to T + 2 is 0.03?

At current time t: Go long one FRA with maturity T , term length α, and fixed rate (deliveryprice) K equal to the forward libor rate Lt[T, T + α]. There is no cost to do this.

At time T : Borrow 1 at the (random) libor rate LT [T, T + α] until T + α.

Portfolio at T : 1 cash, libor debt, long FRA at forward libor rate.

At time T + α:

libor debt⇒ payback 1 + αLT [T, T + α]

FRA⇒ receive αLT [T, T + α] and pay αK = αLt[T, T + α]

Value of Portfolio at T + α:

1− (1 + αLT [T, T + α]) + (αLT [T, T + α]− αLt[T, T + α]) = −αLt[T, T + α]

The total interest cost is αLt[T, T + α].

(b) If the bank borrows N , the above argument gives total interest cost

NαLt[T, T + α] = (5 · 106)(2)(0.03) = 300, 000.

4

46

Remark. In Example 5.5.4, we saw how to arrange at time t to borrow N at time T andpay back N(1 + αLt[T, T + α]) at time T + α. Thus Example 5.5.4 gives an importantcharacterization of the forward libor rate Lt[T, T + α]. Specifically, Lt[T, T + α] is thelibor rate agreed to at time t at which one can borrow (or lend) money from T to T + α.Thus the forward libor rate is (unsurprisingly) a kind of forward rate (see Section 5.2).

5.6 Valuing Floating and Fixed Cashflows

Result 5.6.1. The value at t ≤ T of an agreement to receive the fixed (i.e., known, non-random) payment αK at T + α is

αKZ(t, T + α).

4

Result 5.6.2. The value at t ≤ T of an agreement to receive the floating (i.e., unknown,random variable) payment αLT [T, T + α] at T + α is

Z(t, T )− Z(t, T + α) = αLt[T, T + α]Z(t, T + α).

4

Proof. This agreement is a FRA without the αK payment. In other words, the agreement isa FRA with K = 0. So we set K = 0 in the formula for the value of a FRA (first equationin Result 5.5.1) to get

Z(t, T )− Z(t, T + α).

To see that this equalsαLt[T, T + α]Z(t, T + α),

rearrange the second equation in Result 5.5.1.

Remark. Notice the value is non-random. It does not even depend on the distribution ofthe random variable LT [T, T + α].

Remark. Receiving an unknown libor interest payment from T to T + α on a unit of cashhas the same value as receiving the unit of cash at T and then paying the unit back at T +α.This is intuitively clear, since one can invest the unit of cash in a libor deposit at T andreceive the libor interest payment at T + α.

Here is an example of a floating rate annuity.

Example 5.6.3. Let T0 ≤ T1 ≤ · · · ≤ Tn be a sequence of times with Ti+1 = Ti + αfor a constant α > 0. The sequence of payments αLTi [Ti, Ti + α] at times Ti+1 for i =0, 1, · · · , n− 1 is called a floating leg of libor payments.

47

Its value at time t ≤ T0 is

Vt =n−1∑i=0

(value at t of receiving αLTi [Ti, Ti + α] at Ti+1 = Ti + α)

=n−1∑i=0

(Z(t, Ti)− Z(t, Ti+1))

= Z(t, T0)− Z(t, Tn).

Suppose the annuity is spot-starting, which means T0 = t. Then

Z(t, T0) = Z(t, t) = 1.

Moreover,

T0 = t, T1 = T0 + α = t+ α, T2 = T1 + α = t+ 2α, . . . , Tn = t+ nα.

SoZ(t, Tn) = Z(t, t+ nα)→ 0 as n→∞.

ThereforeVt → 1 as n→∞.

4

Example 5.6.4. Recall: If a fixed rate annuity pays 1 at times Ti = t + i for i = 1, . . . , n(i.e., each year for n years) and the annual zero rates for all maturities are r, then value atcurrent time t of the annuity is

Vt =n∑i=1

Z(t, Ti) =n∑i=1

(1 + r)−(Ti−t) =n∑i=1

(1 + r)−i =1− (1 + r)−n

r.

by the geometric sum trick.

NoteVt →

1

ras n→∞.

4

Chapter 6

Interest Rate Swaps

Interest rate swaps are the most widely traded and most liquid of all over-the-counterderivative contracts.

6.1 Swap Definition

A swap is an agreement between two counterparties to exchange a sequence of cash flows.

Parameters: K = fixed rate or delivery price; T0 = start date; Tn = maturity; T1, . . . , Tn =payment dates.

We assume Ti+1 = Ti + α for every i with fixed α > 0.

So knowing n, T0, and Tn is enough to determine α and every Ti: Tn = T0 + nα andTi = T0 + iα.

One counterparty (called “buyer”, “payer”, or “long”) agrees at t to pay αK and receiveαLTi [Ti, Ti+α] at Ti+α for each i = 0, . . . , n−1. The other counterparty (called “seller”,“receiver”, or “short”) does the opposite.

Therefore a swap is a sequence of forward rate agreements (FRAs).

The floating leg of the swap consists of the payments αLTi [Ti, Ti+α] at times Ti+1 = Ti+αfor i = 0, . . . , n− 1.

The fixed leg of the swap consists of the payments αK at times Ti+1 = Ti + α for i =0, . . . , n− 1.

The buyer pays the fixed leg to and receives the floating leg from the seller. The seller doesthe opposite.

αLTi [Ti, Ti + α] is the amount of simple interest that would be accrued on notional 1 atfloating (unknown, random) rate LTi [Ti, Ti + α] from Ti to Ti + α and paid at Ti + α.

48

49

αK is the amount of simple interest that would be accrued on notional 1 at fixed (known,non-random) rate K from Ti to Ti + α and paid at Ti + α.

Figure 6.1: A swap

Remark. This is a swap with notional 1. To treat a swap with notional N , either considerN swaps of notional 1 or make the payments NαLTi [Ti, Ti + α] and NαK.

Remark. There are more general types of swaps. For example, α can be different for eachperiod, so that Ti+1 = Ti + αi. As another example, the payment times for the floating andfixed legs may differ. This will be explored in the exercises.

6.2 Value of Swap

Consider a swap from T0 to Tn with fixed rate K.

For t ≤ T0,

V SWK (t) = value of the swap at t

V FL(t) = value of the floating leg at t

V FXDK (t) = value of the fixed leg t

They are related by

Result 6.2.1.

V SWK (t) = V FL(t)− V FXD

K (t).

4

Note that V SWK (t) is the value to the buyer. The value to the seller is the −V SW

K (t).

The fixed leg is a sequence of ZCBs with maturities at T1, . . . , Tn, so

Result 6.2.2.

V FXDK (t) =

n∑i=1

αKZ(t, Ti) = KPt[T0, Tn],

50

where we define

Pt[T0, Tn] =n∑i=1

αZ(t, Ti).

4

We call Pt[T0, Tn] the pv01 of the swap.

The floating leg is a sequence of libor payments at T1, . . . , Tn, so by Result 5.6.2 we have

Result 6.2.3.

V FL(t) =n∑i=1

(value at t of payment αLTi−1[Ti−1, Ti] at Ti = Ti−1 + α)

=n∑i=1

αLt[Ti−1, Ti]Z(t, Ti) =n∑i=1

[Z(t, Ti−1)− Z(t, Ti)]

= Z(t, T0)− Z(t, Tn)

4

This says the value of receiving a stream of libor interest payments on an investment of 1 isequal to the value of receiving one dollar at the beginning of the stream and paying it backat the end. We simply take the dollar and repeatedly invest in a sequence of libor deposits.

By combining Results 6.2.1, 6.2.2, and 6.2.3, we can write V SWK (t) as a linear combination

of ZCB prices.

Result 6.2.4.

V SWK (t) = Z(t, T0)− Z(t, Tn)− αK

n∑i=1

Z(t, Ti).

4

Example 6.2.5. Consider a swap starting now with fixed rate 3%, quarterly payment fre-quency, and ending in 2 years. Suppose the quarterly compounded zero rates for all pay-ment times are 2%. Find the present value of(a) the floating leg(b) the swap

(a) Given t = T0, Tn − t = 2, α = 0.25, r4 = 0.02.

V FL(t) = Z(t, T0)− Z(t, Tn) = 1− (1 + r4/4)−4(Tn−t)

= 1− (1 + 0.02/4)−4(2) = 0.039114 . . .

(b) Since V SWK (t) = V FL(t)− V FXD

K (t), we just need V FXDK (t).

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Have K = 0.03. Have Ti − t = Ti − T0 = iα = i0.25. Using Tn = T0 + nα (or drawing apicture), we get n = 8. Then

V FXDK (t) = αK

n∑i=1

Z(t, Ti) = αK

n∑i=1

(1 + r4/4)−4(Ti−t) = αK

n∑i=1

(1 + r4/4)−i

The sumn∑i=1

(1 + r4/4)−i =8∑i=1

(1 + 0.02/4)−i

only has 8 terms, so we could just compute it directly. Or we can use the formula

n∑i=1

(1 + r4/4)−i =1− (1 + r4/4)−n

r4/4.

Then

V FXDK (t) = αK

1− (1 + r4/4)−n

r4/4= (0.25)(0.03)

1− (1 + 0.02/4)−8

0.02/4= 0.05867219 . . .

Therefore


K (t)

= 1− (1 + 0.02/4)−4(2) − (0.25)(0.03)1− (1 + 0.02/4)−8

0.02/4

= −0.0195573 . . .

4

6.3 Forward Swap Rate

The forward swap rate yt[T0, Tn] is the special number such that the value at t of the swapfrom T0 to Tn with fixed rate K = yt[T0, Tn] has value V SW

K (t) = 0.

Result 6.3.1. The forward swap rate at t ≤ T0 for a swap from T0 to Tn is

yt[T0, Tn] =Z(t, T0)− Z(t, Tn)

Pt[T0, Tn]=

∑ni=1 αLt[Ti−1, Ti]Z(t, Ti)∑n

i=1 αZ(t, Ti). (6.3.1)

4

Proof. We use Results 6.2.1, 6.2.2, 6.2.3. Setting K = yt[T0, Tn] gives

52

V SWK (t) = 0 ⇔ V FXD

K (t) = V FL(t) (6.3.2)⇔ KPt[T0, Tn] = Z(t, T0)− Z(t, Tn) (6.3.3)

⇔ K

n∑i=1

αZ(t, Ti) =n∑i=1

αLt[Ti−1, Ti]Z(t, Ti) (6.3.4)

Solving (6.3.3) for K gives the first equality in (6.3.1).

Solving (6.3.4) for K gives the second equality in (6.3.1).

6.4 Value of Swap in Terms of Forward Swap Rate

Result 6.4.1. The value at t ≤ T0 of a swap from T0 to Tn with fixed rate K is

V SWK (t) = (yt[T0, Tn]−K)Pt[T0, Tn].

4

Remark. Compare to the value of a forward contract:

VK(t, T ) = (F (t, T )−K)Z(t, T ).

Proof. By Results 6.2.1, 6.2.2, and 6.2.3, we have


K (t) = Z(t, T0)− Z(t, Tn)−KPt[T0, Tn].

By Result 6.3.1, we have

Z(t, T0)− Z(t, Tn) = yt[T0, Tn]Pt[T0, Tn].

6.5 Swaps as Difference Between Bonds

A fixed rate bond with notional N , coupon c, start date T0, maturity Tn, and term length αis an asset that pays N at time Tn and coupon payments αNc at times Ti for i = 1, . . . , n,where Ti+1 = Ti + α.

If N = 1, the price at t of the fixed rate bond is denoted BFXDc (t).

A floating rate bond with notional N , start date T0, maturity Tn, and term length α isan asset that pays N at time Tn and coupon payments αNLTi−1

[Ti−1, Ti] at times Ti fori = 1, . . . , n, where Ti+1 = Ti + α.

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If N = 1, the price at t of the floating rate bond is denoted BFL(t).

Consider a swap from T0 to Tn with fixed rate K.

Result 6.5.1. For t ≤ T0,

V SWK (t) = BFL(t)−BFXD

K (t)

4

Proof. The fixed rate bond with notional 1 and coupon K equals the fixed leg of the swapplus a payment of 1 at Tn.

The floating rate bond with notional 1 equals the floating leg of the swap plus a payment of1 at Tn.

Therefore

BFL(t)−BFXDK (t) = (V FL(t) + Z(t, Tn))− (V FXD

K (t) + Z(t, Tn))

= V FL(t)− V FXDK (t)

= V SWK (t).

6.6 Par or Spot-Starting Swaps

When t = T0, we call yT0 [T0, Tn] a par swap rate or spot-starting swap rate.

Given par swap rates yT0 [T0, Tk] for every Tk, we can recover the ZCB price Z(T0, Tk) forevery Tk by using Result 6.3.1:

yt[T0, Tk] =Z(t, T0)− Z(t, Tk)

Pt[T0, Tk]

This process is known as bootstrapping and is used frequently in practice. It will be ex-plored in the exercises.

Consider the fixed rate bond with notional 1, coupon c, start date T0, maturity Tn, and termlength α. It pays αNc at times T1, . . . , Tn (where Ti+1 = Ti +α) and pays 1 at time Tn. Itsvalue at time T0 is BFXD

c (T0).

Result 6.6.1.

BFXDc (T0) = 1 if and only if c = yT0 [T0, Tn] = par swap rate

4

54

Because of this result, the par swap rate yT0 [T0, Tn] is sometimes called the coupon rate.

In other words, this result says we can invest 1 at time T0, receive 1 back at time Tn, andreceive fixed payments of αyT0 [T0, Tn] at times T1, . . . , Tn in between. And this isn’t trueif yT0 [T0, Tn] is replaced by any other coupon c.

Proof. We know

By definition of the par swap rate yT0 [T0, Tn], we have

V SWc (T0) = 0 if and only if c = yT0 [T0, Tn].

ButV SWc (T0) = BFL(T0)−BFXD

c (T0)

and

BFL(T0) = V FL(T0) +Z(T0, Tn) = Z(T0, T0)−Z(T0, Tn) +Z(T0, Tn) = Z(T0, T0) = 1.

Therefore1−BFXD

c (T0) = 0 if and only if c = yT0 [T0, Tn].

Chapter 7

Futures Contracts

7.1 Futures Definition

Fix an asset. Let St be its price at time t.

Fix times t ≤ T .

A forward contract (or forward) on the asset with maturity T and delivery price K is anagreement to trade the asset. At maturity T , the long counterparty receives (pays if nega-tive) value

ST −K = F (T, T )−K.There are no payments in between. If K equals the forward price F (t, T ), the contract hasno cost to enter (the value is zero) at time t.

We have previously always considered physically-settled forwards, where at T the longcounterparty receives the asset (value ST ) and pays K cash. In a cash-settled forward, thelong counterparty receives ST cash and pays K cash

A futures contract (or future) on the asset with maturity T and delivery price K is anagreement to trade the asset. Unlike a forward contract, there are payments every dayuntil the maturity date T . We describe the payments below. If K equals the futures priceΦ(t, T ), the contract has no cost to enter (the value is zero) at time t.

Let ∆ = 1365

. Since we measure time in years, day 1 is time t+ ∆ and day i is time t+ i∆.We define day n to be maturity T , so that T = t+ n∆.

On day 0, the long counterparty receives (pays if negative) the amount

Φ(t, T )−K.

On day 1, the long counterparty receives (pays if negative) the amount

Φ(t+ ∆, T )−K.

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56

On each day 1 < i < n, the long counterparty receives (pays if negative) the amount

Φ(t+ i∆, T )− Φ(t+ (i− 1)∆, T )

On day n, the long counterparty receives (pays if negative) the amount

Φ(T, T )− Φ(t+ (n− 1)∆, T )

The payment amount on each day is called the mark-to-market change (or variationmargin).

For the short counterparty, the payments are, of course, the negatives of these amounts.

Usually K = Φ(t, T ) (i.e., the delivery price is the futures price), so that there is no mark-to-market change on day zero.

Φ(t + i∆, T ) is the future price on day i. It costs nothing to enter a futures contract att+ i∆, if the delivery price is Φ(t+ i∆, T ) and maturity is T .

At current time t, Φ(t+ i∆, T ) is unknown (random) for i ≥ 1.

The future price at maturity T is defined to be Φ(T, T ) = ST .

The mark-to-market payments are invested (or borrowed) when they are paid, so they ac-crue interest.

Over the life of the contract, the long counterparty receives mark-to-market payments thattotal

ST −K = Φ(T, T )−K

= Φ(t+n∆, T )−Φ(t+(n−1)∆, T )+· · ·+Φ(t+i∆, T )−Φ(t+(i−1)∆, T )+· · ·+Φ(t+∆, T )−K

However, each payment is made at a different time and accrues interest, so the value of thepayments at T will not in general equal ST −K.

Remark. We have described the payments for cash-settlement, where the final paymentis Φ(T, T ) = ST cash minus Φ(t + (n − 1)∆, T ) cash. For physical-settlement, the finalpayment is the asset (value Φ(T, T ) = ST ) minus Φ(t+ (n− 1)∆, T ) cash.

Remark. In practice, at t each counterparty in a futures contract makes a deposit called aninitial margin or performance bond in an account at an exchange. Each day, the exchangetransfers the appropriate variation margin amount from one counterparties account to theother. The initial margin needs to be large enough to cover several days of likely variationmargin transfers. A typical initial margin is 5% to 15% of the value St −K. If the accountbalance of a counterparty drops below a certain level, called the maintenance margin,the exchange may issue a margin call to that counterparty. A margin call is a demand toreplenish the account. If the account is not replenished, then the futures contract is closedand the counterparties are paid the balance of their accounts. Note that the accounts accrueinterest. In other words, the initial margin and variation margin accrue interest.

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7.2 Futures Prices When Rates Are Constant: Result andExamples

Result 7.2.1. If interest rates are constant, then

Φ(t, T ) = F (t, T )

for all t ≤ T . 4

Before we give the proof, here are some examples.

Example 7.2.2. If interest rates are constant, then

Φ(t+ i∆, T ) = F (t+ i∆, T )

for all t ≤ T , i = 0, 1, . . . , n. 4

Example 7.2.3. If interest rates are constant and if the underlying asset is a stock payingno income, then

Φ(t, T ) = Ster(T−t).

4

Example 7.2.4. If interest rates are constant and if the underlying asset is a stock that paysdividends at continuous rate q, then

Φ(t, T ) = Ste(r−q)(T−t).

4

Example 7.2.5. The table below is for a future contract maturing on day 5 with deliveryprice equal to the futures price. The underlying asset is a stock paying no income, and theconstant continuously compounded interest rate 8%. The MTM column lists market-to-market payment The interest column lists the interest on the mark-to-market payment thatwill be accumulated over the life of the contract. All values are rounded off.

As an exercise, do the calculations to reproduce the last three columns in this table.

day St Φ(t, T ) MTM interest0 1000 1001.10 0 01 1020 1020.89 19.80 0.0172 980 980.64 -40.25 -0.0263 1020 1020.45 39.80 0.0174 1050 1050.23 29.78 0.0075 1030 1030 -20.23 0

sum: 28.90 0.015

Note that value of the corresponding forward contract at maturity is

ST − F (t, T ) = 1030− 1001.10 = 28.90.

4

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7.3 Futures Prices When Rates Are Constant: Proof

Proof of Result 7.2.1. For simplicity assume t = 0. Then T = n∆. Let the constantcontinuously compounded interest rate be r.

Note that we can make trades in a portfolio, as long as they have no cost. For example, wecan’t just add or subtract cash to a portfolio.

Consider the portfolio having the following strategy.

At time 0, go long e−r(n−1)∆ futures contracts with maturity T at deliver price equal to thefutures price Φ(0, T ). By the definition of the futures prices, we can do this at no cost.

At time ∆, increase position to e−r(n−2)∆ futures at futures price Φ(∆, T ). That is, we golong on e−r(n−2)∆− e−r(n−1)∆ futures with maturity T at delivery price equal to the futuresprice Φ(∆, T ). Again, by the definition of the futures prices, we can do this at no cost.

At time i∆ (for i = 2, . . . , n − 2), increase position to e−r(n−i−1)∆ futures at futures priceΦ(i∆, T ).

At time (n− 1)∆, increase position to 1 futures contract at futures price Φ((n− 1)∆, T ).

With this strategy we receive the following amounts.

At time ∆, we receive mark-to-market gain/loss

(Φ(∆, T )− Φ(0, T ))e−r(n−1)∆.

This will be invested at rate r. So by time T = n∆ (after time T − ∆ = (n − 1)∆ haspassed) it becomes

(Φ(∆, T )− Φ(0, T ))e−r(n−1)∆ · er(n−1)∆ = Φ(∆, T )− Φ(0, T ).

At time (i+ 1)∆, we receive mark-to-market gain/loss

(Φ((i+ 1)∆, T )− Φ(i∆, T ))e−r(n−i−1)∆.

This will be invested at rate r. So by time T = n∆ (after time T − (i+1)∆ = (n− i−1)∆has passed) it becomes

(Φ((i+ 1)∆, T )− Φ(i∆, T ))e−r(n−i−1)∆ · er(n−i−1)∆ = Φ((i+ 1)∆, T )− Φ(i∆, T ).

Therefore the value at time T = n∆ of the portfolio is

n−1∑i=0

Φ((i+ 1)∆, T )− Φ(i∆, T ) = Φ(n∆, T )− Φ(0, T ) = ST − Φ(0, T ).

Let A be the above portfolio plus Φ(0, T ) ZCBs maturing at T (or e−rT cash). Then

V A(T ) = ST .

59

Let B be the portfolio consisting at time 0 of one long forward contract on the asset withmaturity T and delivery price equal to the forward price F (0, T ) plus F (0, T ) ZCBs ma-turing at T (or e−rT cash). Then also

V B(T ) = ST .

By replication,V A(t) = V B(t),

which meansΦ(0, T )Z(0, T ) = F (0, T )Z(0, T ).

Divide by Z(0, T ) to conclude.

Result 7.3.1. Suppose that at the present time t we know what the interest rate betweenany two days will be. IThen

Φ(t, T ) = F (t, T )

for any given t ≤ T . 4

Exercise 7.3.1. Prove Result 7.3.1. You may assume t = 0 for simplicity. Use the notationrij for the continuous interest rate between day i and day j.

7.4 Futures Convexity Correction

In general, Φ(t, T ) 6= F (t, T ). The difference Φ(t, T ) − F (t, T ) is called the futuresconvexity correction.

Result 7.4.1 (General Features of Futures Convexity Correction).

(a) If the asset price ST increases (on average) as interest rates increase, then Φ(t, T ) −F (t, T ) > 0. If the asset price ST decreases (on average) as interest rates increase, thenΦ(t, T )− F (t, T ) < 0.

(b) The size of the convexity correction |Φ(t, T ) − F (t, T )| typically increases as T − tincreases, and it approaches zero as t approaches T .

4

A more precise version of (a) is given in the next section, which is optional.

Here is an intuitive explanation of (a).

Suppose asset price increases as interest rates increase. In both a long forward contract anda long futures contract, the total payment over the life of the contract is

ST −K.

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In the forward contract, the payment is made only at the end. In the futures contract, thetotal payment is the sum of mark-to-market payments

Φ(t+ i∆, T )− Φ(t+ (i− 1)∆, T )

over the life of the contract. Thus we receive gains and pay losses early in a future contract.If the interest rate at the time of a mark-to-market payment is relatively high, then the assetprice is high, so the mark-to-market payment is positive (gain), and we can invest it at arelatively high rate. If the interest rate at the time of a mark-to-market payment is relativelylow, then the asset price is low, so the mark-to-market payment is negative (loss), but wecan borrow it at a relatively low rate. Thus we expect the interest gains to exceed the intereston the losses over the life of the futures contract. Therefore we would prefer to hold thelong future rather than the long forward, hence Φ(t, T )− F (t, T ) > 0.

Example 7.4.2. Suppose T < T0. Is the future price Φ(t, T ) of a ZCB with maturity T0

likely to be higher or lower than the forward price F (t, T )?

We know the ZCB price is Z(T, T0) = e−r(T0−T ) where r is the continuous zero ratefor period T to T0 available at T . So the asset price increases as interest rates decrease.Therefore Φ(t, T ) < F (t, T ). 4

7.5 Optional Topic: Futures Convexity Correction and Cor-relation

The covariance of random variables X and Y is

Cov(X, Y ) = E[(X − E(X))(Y − E(Y ))].

If (X − E(X))(Y − E(Y )) > 0, then X and Y are either both above or both below theirmeans E(X) and E(Y ).

If Cov(X, Y ) > 0, then X and Y are, on average, on the same side of their means (bothabove or both below). If Cov(X, Y ) > 0, we say X and Y are positively correlated.

If (X − E(X))(Y − E(Y )) < 0, then X and Y are on opposite sides of their means E(X)and E(Y ).

If Cov(X, Y ) < 0, then X and Y are, on average, on opposite sides of their means. IfCov(X, Y ) < 0, we say X and Y are negatively correlated.

If Cov(X, Y ) = 0, we say X and Y are uncorrelated.

Example 7.5.1. If X and Y are the height and weight of a randomly selected person, thenX and Y are positively correlated. Height and weight tend to move together.

IfX is the snowfall and Y is the temperature in Rochester on a randomly selected day, thenX and Y are negatively correlated. On average, there is more snowfall on days with lowtemperature than on days with high temperature.

61

If X and Y are the numbers shown on two consecutive rolls of a die, then X and Y areuncorrelated. The number shown on the first roll tells us nothing about the second roll. 4

The value of the money market account at time t is denoted Mt. It is the value at time tof 1 invested at time 0. Thus M0 = 1, and if r is the constant continuously compoundedinterest rate, then Mt = ert.

Result 7.5.2. If the asset price ST and the money market account MT are positively corre-lated, then Φ(t, T )− F (t, T ) > 0.

If the asset price ST and the money market account MT are negatively correlated, thenΦ(t, T )− F (t, T ) < 0.

If the asset price ST and the money market account MT are uncorrelated, then Φ(t, T ) =F (t, T ). 4

The proof is beyond our scope.

Chapter 8

Options

8.1 European Option Definitions

Fix an asset. Let St be its price at time t.

A European call option on the asset with strike (or exercise price) K and maturity (orexercise date) T is the right—but not the obligation—to buy the asset for K at time T .Using the right to trade the asset is called exercising the option.

In contrast, a long forward contract is the obligation to buy the asset for K at time T .

We would only choose to pay K for an asset worth ST if ST ≥ K. Thus the payout orvalue at maturity of a European call option is

ST −K if ST ≥ K0 if ST ≤ K

= max ST −K, 0 = (ST −K)+.

We can also write the payout as g(ST ), where g(x) = (x−K)+ = max x−K, 0 is thepayout function.

Example 8.1.1. Suppose the strike is K = 100 for an call option with maturity T . IfST = 110, then we receive value ST −K = 110− 100 = 10 by exercising the option. Onthe other hand, if ST = 90, then ST − K = −10, so we let the option expire and receive0. 4

A European put option is the right to sell the asset for K at time T .

The payout or value at maturity of a European put option isK − ST if ST ≤ K

0 if ST ≥ K= max K − ST , 0 = (K − ST )+.

Notice that the payout of a put is not the negative of the payout of a call.

62

63

Suppose we buy a European call option at time t. The price we pay is denoted CK(t, T ).The party who sold us the call is said to be short on the call (or the writer of the call).Since we are holding the call, we are said to be long on the call (or the owner of the call).As the long counterparty, we have the right to buy the asset from the short counterparty forK at T . If this right is exercised, the short counterparty has the obligation to immediatelysell the asset to us. Note that

short 1 call 6= long 1 put.

The payout at maturity when short a call is

−max ST −K, 0

which is the negative of the payout when long the call. However, don’t forget that the shortcounterparty was paid CK(t, T ) at t by the long counterparty.

A European straddle option is a European call plus a European put. It has payout

|ST −K| =ST −K if ST ≥ KK − ST if ST ≤ K.

8.2 American Option Definitions

A European option allows exercise only at maturity T .

An American option allows exercise at any time t ≤ T . More precisely, an American calloption on an asset with strike (or exercise price) K and maturity (or exercise date) T isthe right to buy the asset for K at any time t ≤ T . For an American put option, replace“buy” with “sell.”

There are many other, less common types of options. We won’t study them.

Note that an option is automatically exercised at maturity if the payout is positive.

8.3 More Definitions

At time t ≤ T , a call option with strike K is said to be:

in-the-money (ITM) if St > K (the call option will have positive value at maturity if theasset price remains unchanged)

out-of-the-money (OTM) if St < K (the call option will be worthless if the asset priceremains unchanged)

at-the-money (ATM) if St = K

64

in-the-money-forward (ITMF) if F (t, T ) > K

out-of-the-money-forward (OTMF) if F (t, T ) < K

at-the-money-forward (ATMF) if F (t, T ) = K

As usual, F (t, T ) is the forward price on the asset at t for maturity T .

For a put option, we reverse these inequalities. For example, a put option is ITM ifK > St.

The intrinsic value at t of a call option is max St −K, 0. The intrinsic value at t ofa put option is max K − St, 0. Thus an option is in-the-money if its intrinsic value ispositive.

The intrinsic value is the payout if we had to immediately exercise the option or let it expire.

The intrinsic value is not the value/price of the option.

8.4 Option Prices

We begin our option valuation

Fix an asset. Let St be its price at time t. At this point, we make no assumptions about thenature of the asset.

For t ≤ T :

CK(t, T ) = price/value at time t of a European call with strike K and maturity T .

PK(t, T ) = price/value at time t of a European put with strike K and maturity T .

CK(t, T ) = price/value at time t of an American call with strike K and maturity T .

PK(t, T ) = price/value at time t of an American put with strike K and maturity T .

Clearly,CK(T, T ) = CK(T, T ) = max ST −K, 0 .

andPK(T, T ) = PK(T, T ) = max K − ST , 0 .

Result 8.4.1.

CK(t, T ) ≥ CK(t, T ) ≥ 0 and PK(t, T ) ≥ PK(t, T ) ≥ 0.

4

Proof. It is intuitively clear that

CK(t, T ) ≥ CK(t, T ) and PK(t, T ) ≥ PK(t, T )

65

because an American option gives the same rights as a European option, and more. (Asan exercise, prove these inequalities using the no-arbitrage principle or the monotonicitytheorem.)

In light of the assertion above, we just need to prove.

CK(t, T ) ≥ 0 and PK(t, T ) ≥ 0.

These inequalities should also be intuitively clear because an option can never lose moneyfor its owner. For completeness, we prove the first inequality anyway. We apply the mono-tonicity theorem to the portfolios

A = 1 call and B = empty.

We see thatV A(T ) = CK(T, T ) = max ST −K, 0 ≥ 0 = V B(T ).

So by monotonicity,V A(t) ≥ V B(t)

that is,CK(t, T ) ≥ 0.

The proof of the second inequality is similar.

8.5 Put-Call Parity

Fix an asset. Let St be its price at t. Again, we make no assumptions about the nature ofthe asset.

Recall that the value of a forward is related to the forward price and delivery price by

VK(t, T ) = (F (t, T )−K)Z(t, T ).

The next result relates the value of a forward to the price of a European call and the priceof a European put.

Result 8.5.1 (Put-Call Parity).

VK(t, T ) = CK(t, T )− PK(t, T )

4

Proof. Consider two portfolios.

A: long 1 forward with delivery price K and maturity T .

66

B: long 1 European call and short 1 European put (i.e., +1 call and −1 put), both withstrike K and maturity T .

We have

V A(T ) = ST −K

V B(T ) =

(ST −K)− 0 if ST ≥ K0− (K − ST ) if ST ≤ K

= ST −K.

By the replication theorem,V A(t) = V B(t)

which meansVK(t, T ) = CK(t, T )− PK(t, T ).

In words, put-call parity says:

VK(t, T ) = CK(t, T )− PK(t, T )

long 1 forward equals long 1 call and short 1 put

CK(t, T ) = VK(t, T ) + PK(t, T )

long 1 call equals long 1 forward and long 1 put

PK(t, T ) = −VK(t, T ) + CK(t, T )

long 1 put equals short 1 forward and long 1 call

Here “equals” means “has the same value as.”

Remember the options are European.

There is a version of put-call parity for American options, but it involves inequalities ratherthan an equality, and it depends on the income paid by the stock. We will not study it here.

Result 8.5.2.CK(t, T ) = PK(t, T )

if and only if K = F (t, T ) if and only if K = F (t, T ) if and only if the call and put areboth at-the-money-forward (ATMF) at t. 4

Proof. Use put-call parity and the definition of forward price.

67

Example 8.5.3. Suppose that a stock paying no income is trading at price 30 per share.European puts on the stock with strike 35 and exercise date in six months are trading atprice 10. The six-month libor rate is 4%. What is the price of a European call with thesame strike and exercise date?

Put-call parity says VK(t, T ) = CK(t, T )− PK(t, T ), so

CK(t, T ) = VK(t, T ) + PK(t, T ).

Since the stock pays no income, we have

VK(t, T ) = (F (t, T )−K)Z(t, T ) = St −KZ(t, T ) = St −K(1 + (T − t)Lt[t, T ])−1

Therefore

CK(t, T ) = St −K(1 + (T − t)Lt[t, T ])−1 + PK(t, T ) = 30− 35(1 + (0.5)(0.04))−1 + 10 = 5.68627 . . . .

4

8.6 Call Prices for Assets Paying No Income

Result 8.6.1. For an asset paying no income, the European call price satisfies

St −KZ(t, T ) ≤ CK(t, T ) ≤ St.

4

Proof. We haveCK(T, T ) = max ST −K, 0 ≤ ST

To prove the upper boundCK(t, T ) ≤ St,

apply the monotonicity theorem to the following portfolios.A: 1 call.B: 1 stock.

We haveST −K ≤ max ST −K, 0 = CK(T, T ).

Applying the monotonicity theorem to the portfoliosA: 1 stock, −K ZCBs with maturity TB: 1 callproves the lower bound

St −KZ(t, T ) ≤ CK(t, T ).

68

Result 8.6.2. For an asset paying no income, the price of an American or European call isat least the intrinsic value of the call: CK(t, T ) ≥ CK(t, T ) ≥ max St −K, 0. 4

Proof. We know CK(t, T ) ≥ CK(t, T ) and CK(t, T ) ≥ 0. So we just need to observe that

CK(t, T ) ≥ St −KZ(t, T ) ≥ St −K

by Result 8.6.1.

Result 8.6.3. For an asset paying no income, the price of an American call and the price ofa European call (with the same strike and maturity) are always equal, i.e.,

CK(t, T ) = CK(t, T )

for all t ≤ T . 4

Proof. We know CK(t, T ) ≥ CK(t, T ). To prove CK(t, T ) ≤ CK(t, T ), consider twoportfolios at t:A: +1 American callB: +1 European call

Case 1. The American call is not exercised before T . Then either both calls are exercisedat T (happens if ST > K) or both calls expire (happens if ST ≤ K). Therefore

V A(T ) = V B(T ) = max ST −K, 0 .

Case 2. The American option is exercised at time T0 with t ≤ T0 < T .

A at T0 consists of one unit of the asset and a debt of K cash (i.e., −K cash). (Why?Portfolio A has nothing besides the call option. When the call is exercised, A needs toborrow K cash to pay for the asset.

A at T consists of one unit of the asset and −Ker(T−T0) = −K/Z(T0, T ) cash, where r isthe continuous zero rate for period T0 to T .

Therefore

V A(T ) = ST −Ker(T−T0) ≤ ST −K ≤ max ST −K, 0 = V B(T )

Combining both cases, we have

V A(T ) ≤ V B(T )

with probability one. So, by the monotonicity theorem,

V A(t) ≤ V B(t),

which meansCK(t, T ) ≤ CK(t, T ).

69

Result 8.6.4. For an asset paying no income, an American call will never be exercisedbefore maturity if interest rates are positive. 4

Proof. Suppose t < T . By Result 8.6.1, we have

CK(t, T ) ≥ CK(t, T ) ≥ St −KZ(t, T ).

If the interest rates for period t to T are positive, then Z(t, T ) < 1, and so

CK(t, T ) > St −K.

We would receive value St−K if we exercise an American call at t. But we would receivea strictly larger value CK(t, T ) if we sell the call instead. So we would never choose toexercise at t < T .

8.7 Put Prices for Assets Paying No Income

Result 8.7.1. For an asset paying no income,

KZ(t, T )− St ≤ PK(t, T ) ≤ KZ(t, T ), (8.7.1)

K − St ≤ PK(t, T ) ≤ K (8.7.2)

4

Remember that American puts are always worth at least as much as European puts (withthe same strike and maturity). In contrast to calls, American puts are usually worth strictlymore than European puts, since by exercising early we receive the strike price K early andcan invest this amount.

Example 8.7.2. The current price of a stock paying no income is St = 10. The one-yearannually compounded interest rate is r = 16%.

Consider an American put expiring in one year with strike K = 80. Exercising now, wecan gain

ST −K = 80− 10 = 70,

which can be invested to become

70(1 + 0.16) = 81.20

after one year.

Consider a European put expiring in one year with strike K = 80. The payout at T is atmost

max 80− ST , 0 ≤ 80 < 81.20

after one year.

Since we can potentially make more money with the American put, its current price will bestrictly larger than the price of a European put. 4

70

Proof of (8.7.1). As with the European call bounds of Result 8.6.1, we can prove (8.7.1)directly by replication. Instead, to illustrate another method, we prove (8.7.1) using Euro-pean call bounds and put-call parity.

We prove the upper bound PK(t, T ) ≤ KZ(t, T ) in detail. According to Result 8.6.1,

CK(t, T ) ≤ St.

By put-call parity,CK(t, T ) = VK(t, T ) + PK(t, T )

HencePK(t, T ) ≤ St − VK(t, T ).

But, for an asset paying no income,

VK(t, T ) = (F (t, T )−K)Z(t, T ) = St −KZ(t, T ).

ThereforePK(t, T ) ≤ KZ(t, T ).

To prove the lower bound KZ(t, T )− St ≤ PK(t, T ), start with

0 ≤ CK(t, T )

and argue the same way.

Proof of (8.7.2). The lower bound is easy. If we exercise the American put at t, we receivevalue K − St, so the price PK(t, T ) must be at least this much.

To prove the upper bound PK(t, T ) ≤ K, we assume PK(t, T ) > K and deduce a contra-diction. Consider portfolio C with the following strategy.

At time t, the portfolio is empty. Write and sell 1 American put. The portfolio becomesshort 1 American put and PK(t, T ) cash. It is not necessary to invest the cash, we just holdit.

Case 1. The American put expires without being exercised. Then

V C(T ) = PK(t, T ) ≥ PK(t, T )−K > 0.

Case 2. The American is exercised at some time. At that moment, we must pay K to buy 1asset from the holder of the put. The portfolio is then PK(t, T )−K non-invested cash and1 asset. Nothing else happens until T , so

V C(T ) = PK(t, T )−K + ST > 0.

Combining both cases, we have V C(T ) > 0 with probability one and V C(t) = 0. Thiscontradicts the no-arbitrage assumption.

Exercise 8.7.1. Give a proof of the upper bound in Result 8.7.2 using the monotonicitytheorem.

71

8.8 Call and Put Prices for Stocks Paying Known Divi-dend Yield

Result 8.8.1. For a stock paying dividends at continuous yield q with automatic reinvest-ment,

Ste−q(T−t) −KZ(t, T ) ≤ CK(t, T ) ≤ Ste

−q(T−t)

KZ(t, T )− Ste−q(T−t) ≤ PK(t, T ) ≤ KZ(t, T )

St −K ≤ CK(t, T ) ≤ St

K − St ≤ PK(t, T ) ≤ K

4

Recall that for a stock paying no income, American and European calls have the samevalue. For a stock paying dividends, an American call is often worth more than a Europeancall. This is because, by exercising early, we may receive dividends from the stock whichwe would not have received otherwise.

Example 8.8.2. Consider a stock paying dividends at continuous yield q = 5% with auto-matic reinvestment. The current price of the stock is St = 10. The one-year continuouslycompounded interest rate is r = 5%.

Consider an American call expiring in one year with strike K = 20. Exercising now, weget the stock and a debt of K = 20. If the stock price in one year is ST = 30, then we willhave gained

ST eq(T−t) −Ker(T−t) = 30e0.05 − 20e0.05 ≈ 16.49

after one year.

A European call expiring in one year with strike K = 20 would payout only

ST −K = 30− 20 = 10

after one year. 4

Exercise 8.8.1. Prove Result 8.8.1.

8.9 Call and Put Spreads

From now on, we consider only European options.

Fix an asset. Let St be its price at t.

Let K1 < K2.

72

A (K1, K2) call spread is a portfolio consisting of long 1 call option with strike K1 andshort 1 call option with strike K2, both with maturity T .

Result 8.9.1. For a (K1, K2) call spread:

• The value at t ≤ T isCK1(t, T )− CK2(t, T ).

• The payout at maturity T is0 if ST ≤ K1 (neither exercised)ST −K1 if K1 ≤ ST ≤ K2 (only K1-call exercised)K2 −K1 if ST ≥ K2 (both exercised)

• The value at t satisfies

0 ≤ CK2(t, T )− CK1(t, T ) ≤ (K2 −K1)Z(t, T ).

4

Proof. Only the last point needs justification.

The payout at T is ≥ 0, so (by the monotonicity theorem) the value at t is also ≥ 0.

The payout at T is always ≤ K2 −K1, which is the value at T of a portfolio of K2 −K1

ZCBs with maturity T . So (by the monotonicity theorem) the value of the spread at t is≤ (K2 −K1)Z(t, T ).

Figure 8.1: Payout of (80, 110) call spread

73

A (K2, K1) put spread is a portfolio consisting of short 1 put option with strike K1 andlong 1 put option with strike K2, both with maturity T .

Result 8.9.2. For a (K2, K1) put spread:

• The value at t ≤ T isPK2(t, T )− PK1(t, T ).

• The payout at maturity T isK2 −K1 if ST ≤ K1 (both exercised)K2 − ST if K1 ≤ ST ≤ K2 (only K2-put exercised)0 if ST ≥ K2 (neither exercised)

• The value at t satisfies

0 ≤ PK2(t, T )− PK1(t, T ) ≤ (K2 −K1)Z(t, T )

4

The previous two results imply

Result 8.9.3. If K1 ≤ K2,

CK1(t, T ) ≥ CK2(t, T ) and PK1(t, T ) ≤ PK2(t, T )

In other words, CK(t, T ) is a decreasing function of the strike K, and PK(t, T ) is an in-creasing function of the strike K. 4

8.10 Butterflies and Convexity of Option Price


Let K1 < K∗ < K2. Then there is a unique λ ∈ (0, 1) such that

K∗ = λK1 + (1− λ)K2.

The portfolio consisting of

• +2λ calls with strike K1,

• −2 calls with strike K∗,

• +2(1− λ) calls with strike K2,

74

all with the same maturity, is called a (K1, K∗, K2) call butterfly.

If λ 6= 1/2, it is called an asymmetric call butterfly.

If λ = 1/2, it is called an symmetric call butterfly.

Since symmetric call butterflies are vastly more common in practice, if the shorter term callbutterfly is used without further qualification, we will always assume it means symmetriccall butterfly.

For a symmetric call butterfly, the portfolio is

• +1 call with strike K1,

• −2 calls with strike K∗ = 12(K1 +K2),

• +1 call with strike K2.

Result 8.10.1. For a (K1, K∗, K2) symmetric call butterfly, at maturity T the payout is

0 if ST ≤ K1

ST −K1 if K1 ≤ ST ≤ K∗

K2 − ST if K∗ ≤ ST ≤ K2

0 if ST ≥ K2

4

Figure 8.2: Payout of a (90, 105, 120) call butterfly.

75

Result 8.10.1 is a special case of

Result 8.10.2. For a (K1, K∗, K2) asymmetric or symmetric call butterfly:

• The value at t ≤ T is

2λCK1(t, T ) + 2(1− λ)CK2(t, T )− 2CK∗(t, T ).

• The payout at maturity T is0 if ST ≤ K1

2λ(ST −K1) if K1 ≤ ST ≤ K∗

2(1− λ)(K2 − ST ) if K∗ ≤ ST ≤ K2

0 if ST ≥ K2

• The payout at T is ≥ 0, so (by the monotonicity theorem) the value at t is also ≥ 0.

4

Proof. The first and third points are clear.

For the second point, the cases ST ≥ K1 and K1 ≤ ST ≤ K∗ are clear. If K∗ ≤ ST ≤ K2,the payout is

2λ(ST−K1)−2(ST−K∗) = 2λ(ST−K1)−2(ST−λK1−(1−λ)K2) = 2(1−λ)(K2−ST )

If ST ≥ K2, the payout is

2λ(ST −K1)− 2(ST −K∗) + 2(1− λ)(ST −K1) = 0

The third point of Result 8.10.2 implies

Result 8.10.3. If K1 < K∗ < K2, then

CK∗(t, T ) ≤ λCK1(t, T ) + (1− λ)CK2(t, T )

whereK∗ = λK2 + (1− λ)K2.

Geometrically, this means that the graph of CK(t, T ) versus K is always below the secantline between any two points on the graph.

In other words, CK(t, T ) is a concave up (convex) function of K. 4

76

Remember Result 8.9.3 says CK(t, T ) is a decreasing function of K.

In class, I will draw a graph of CK(t, T ) versus K.

In the exercises, we outline how to show that PK(t, T ) is a concave up (convex) functionof K. Remember Result 8.9.3 says PK(t, T ) is an increasing function of K.

Exercise 8.10.1. Draw the payout diagram for an asymmetric call butterfly with strikes(70, 80, 110).

Exercise 8.10.2. Consider an arbitrary (K1, K∗, K2) asymmetric or symmetric call butter-

fly. (a) Use put-call parity to express it as a portfolio of only puts.

(b) Show that the value at t is

2λPK1(t, T ) + 2(1− λ)PK2(t, T )− 2PK∗(t, T ).

(c) Show that PK(t, T ) is a concave up (convex) function of K by showing that

2PK∗(t, T ) ≤ 2λPK1(t, T ) + 2(1− λ)PK2(t, T ).

8.11 Digital Options


A digital call option with strike K and maturity T has payout at T1 if ST ≥ K0 if ST < K

A digital put option with strike K and maturity T has payout at T1 if ST ≤ K0 if ST > K

Digital options are also called binary options or all-or-nothing options.

Figure 8.3: Payout of a K-strike digital call

77

Result 8.11.1. The price/value at time t of a digital call option with strike K and maturityT is

limh→0+

CK−h(t, T )− CK(t, T )

h.

It is equal to

− d

dKCK(t, T )

if CK(t, T ) is differentiable with respect to K. 4Result 8.11.2. When h > 0 is small, a digital call option with strike K and maturity T isapproximately replicated by the portfolio

A(h): 1/h (K − h,K) call spreads (i.e., long 1/h calls with strike K − h, short 1/h callswith strike K) with maturity T

which at t has valueV A(h)(t) =

CK−h − CKh

.

4

Proof of Both Results. If ST ≥ K, then all the calls in A(h) are exercised and

V A(h)(T ) =1

h(ST − (K − h) +K − ST ) = 1

for all h > 0.

If ST < K, then ST < K − h for all small enough h > 0, so nothing in A(h) is exercisedand

V A(h)(T ) = 0.

Therefore

limh→0+

V A(h)(T ) =

1 if ST ≥ K0 if ST < K

= V B(T )

where portfolio B is a digital call with strike K and maturity T .

By a no-arbitrage argument (details omitted), we have that

limh→0+

V A(h)(t) = limh→0+


h= V B(t).

If CK(t, T ) is differentiable with respect to K, then

− d

dKCK(t, T ) = − lim

h→0

CK+h(t, T )− CK(t, T )

h

= − limh→0

CK+(−h)(t, T )− CK(t, T )

−h

= limh→0+


h

= value at t of a digital call with strike K and maturity T

78

Exercise 8.11.1. What is the price at time t of a digital put option with strike K? Prove it.

Chapter 9

Probability Theory: Conditioning

9.1 Conditional Probability


Let A and B be events (i.e., subsets of Ω). The conditional probability of A given B is

P (A|B) =P (A and B)

P (B).

It is undefined if P (B) = 0. Out of the probability assigned to the outcomes in B, P (A|B)is the fraction assigned to those outcomes that are in both A and B.

A B

Ω

Interpretation: P (A|B) is the probability that event A occurs assuming event B occurs.

Example 9.1.1. Experiment: Flip a fair coin three times.Sample space: Ω = HHH,HHT,HTH,THH,HTT,THT,TTH,TTT Probability Measure:P (ω) = 1

8for all ω ∈ Ω

X = number of headsY = 1 if 1st flip heads, 0 otherwiseZ = 1 if 2nd flip heads, 0 otherwise

79

80

Find P (X = 2), P (X = 2|Y = 1), and P (X = 2|Y = 1, Z = 0).Notation: Y = 1, Z = 0 = Y = 1 and Z = 0.

P (X = 2) = P (HHT,HTH,THH) =3

8

P (X = 2|Y = 1) =P (X = 2, Y = 1)

P (Y = 1).

Y = 1 = HHH,HHT,HTH,HTT and X = 2, Y = 1 = HHT,HTH

P (X = 2|Y = 1) =P (X = 2, Y = 1)

P (Y = 1)=

2/8

4/8=

1

2.

P (X = 2|Y = 1, Z = 0) =P (X = 2, Y = 1, Z = 0)

P (Y = 1, Z = 0)Y = 1, Z = 0 = HTH,HTT and X = 2, Y = 1, Z = 0 = HHT

P (X = 2|Y = 1, Z = 0) =P (X = 2, Y = 1, Z = 0)

P (Y = 1, Z = 0)=

1/8

2/8=

1

2. 4

9.2 Conditional Expectation

Let X and Y be discrete random variables.

Recall the expected value of X is

E(X) =∑

x∈R(X)

xP (X = x),

where R(X) is the range of X . It is a weighted average of the possible outputs of X , withthe weights being the probability of each output.

Terminology: expectation = expected value = average value = mean = first moment

The conditional expectation of X given Y = y is

E(X|Y = y) =∑

x∈R(X)

xP (X = x|Y = y).

Interpretation: E(X|Y = y) is the expected value of X assuming that Y = y occurs.

Example 9.2.1. Roll two fair six-sided dice.

X = sum of the dice, Y = number on first die, Z = number on second die.

E(X) = E(Y + Z) = E(Y ) + E(Z) = 3.5 + 3.5 = 7

E(X|Y = 6) = E(Y + Z|Y = 6) = E(Y |Y = 6) + E(Z|Y = 6) = 6 + 3.5 = 9.5.

81

4

9.3 Condition Expectation as Random Variable

Let X and Y be discrete random variables.

Recall that if g : R→ R is any function, then g(Y ) is a random variable. It is defined by

g(Y )(ω) = g(Y (ω))

for all ω ∈ Ω.

Example 9.3.1. If g(x) = (x−K)+ and ST is the price of an asset at a future time T , then

g(ST ) = (ST −K)+

is a random variable representing the payout of a call option. 4

If we define h by h(y) = E(X|Y = y), the conditional expectation of X given Y is therandom variable

E(X|Y ) = h(Y ).

In other words, if yω stands for the real number Y (ω), then

E(X|Y )(ω) = E(X|Y = yω)

for all ω ∈ Ω.

Remember E(X|Y = y) is a real number and E(X|Y ) is a random variable.

9.4 Properties of Conditional Expectation

Let X, Y, Z be random variables and let c ∈ R.

Linear Properties:

• E(c|Z = z) = c and E(c|Z) = c

• E(cX|Z = z) = cE(X|Z = z) and E(cX|Z) = cE(X|Z)

• E(X+Y |Z = z) = E(X|Z = z)+E(Y |Z = z) and E(X+Y |Z) = E(X|Z)+E(Y |Z)

82

When conditioning on Y = y, replace Y by y in the expectation:

E(g(X, Y )|Y = y) = E(g(X, y)|Y = y)

for any function g(x, y).

Conditioning with respect to Y means that Y , hence g(Y ), should be interpreted as knownor constant, so that g(Y ) can be moved outside the expectation:

E(Xg(Y )|Y ) = g(Y )E(X|Y )

for any function g(y).

We conclude with the tower law or the law of iterated expectations.

Basic form:E(E(X|Y )) = E(X) and E(E(X)|Y ) = E(X).

General Form:

E(E(X|Y, Z)|Z) = E(X|Z) and E(E(X|Z)|Y, Z) = E(X|Z).

9.5 Independence

Discrete random variables X and Y are called independent if

P (X = x, Y = y) = P (X = x)P (Y = y) for all x, y ∈ R.

Using the formulas

P (X = x|Y = y) =P (X = x, Y = y)

P (Y = y)and P (Y = y|X = x) =

P (X = x, Y = y)

P (X = x),

we have

Result 9.5.1. The following statements are equivalent:

• X and Y are independent

• P (X = x, Y = y) = P (X = x)P (Y = y) for all x, y ∈ R

• P (X = x|Y = y) = P (X = x) for all x, y ∈ R

• P (Y = y|X = x) = P (Y = y) for all x, y ∈ R 4

Intuitively, X and Y are independent if knowing one of them gives no information aboutthe other.

83

Example 9.5.2. Roll two fair six-sided dice. X = sum of the dice, Y = number on first die,Z = number on second die.

Y and Z are independent. Indeed, it’s intuitively clear (and we can easily check) that

P (Y = y|Z = z) = P (Y = y) for all y, z ∈ R.

X and Y are not independent because (for example)

P (X = 2|Y = 6) = 0 6= 1

36= P (X = 2).

4

Random variables X1, . . . , Xn are independent if

P (X1 = x1, . . . , Xn = xn) = P (X1 = x1) · · · · · P (Xn = xn) for all x1, . . . , xn.

An infinite sequence of random variables X1, X2, . . . is independent if any finite sub-collection is independent.

If X1, X2 . . . , Xi, . . ., are independent, then we have, for example,

P (X1 = x1|X2 = x2) = P (X1 = x1)

P (X1 = x1|X2 = x2, . . . , Xn = xn) = P (X1 = x1)

P (X1 = x1|X2 + . . .+Xn = y) = P (X1 = x1).

Chapter 10

The Fundamental Theorem of AssetPricing and the Binomial Tree

In the previous chapter, we were only able to derive upper and lower bounds for optionprices. Finding exact formulas for option prices (and other derivatives) requires additionaltheory. We develop that theory in this and the following chapters.

10.1 Fundamental Theorem of Asset Pricing

Recall that asset prices at future times are random variables. These random variables aredefined on a sample space Ω. There is also a probability measure P defined on Ω.

Remember that Ω is the set of all possible outcomes or all possible states of the world. AndP is a function that assigns a number 0 ≤ P (A) ≤ 1 to every event (set of outcomes).

Recall that a portfolio A is an arbitrage portfolio if the following conditions are satisfied:

• At current time t,V A(t) ≤ 0.

• At some future time T ,

P (V A(T ) ≥ 0) = 1 and P (V A(T ) > 0) > 0.

Notice how the definition of an arbitrage portfolio depends on the probability measure P .

Consider an asset. We want to find its price/value Dt at current time t.

Original Method: Let A be the portfolio containing the asset at time t. If we can find aportfolio B whose value at t is known and which has the same value as A at some futuretime T , then the no-arbitrage principle implies (via the replication theorem) that A and Bhave the same value at t. Hence Dt = V B(t).

84

85

Problem: If the value DT of the asset at T is even slightly complicated, it may be difficultto find a suitable portfolioB. This difficulty occurs when the asset is a European call optionon some stock and DT = max ST −K, 0.

New Method???: Suppose the random variable DT has known distribution. This meansP (DT = k) is known for all k.

Without using the no-arbitrage principle, a natural approach to find Dt is: 1. Compute theexpected value of the asset at T :

E(DT ) =∑

k∈R(DT )

kP (DT = k),

2. Discount the expected value back to today:

Z(t, T )E(DT ),

3. Use the discounted expected value as the current asset price

Dt = Z(t, T )E(DT ). (10.1.1)

Problem I: The pricing formula 10.1.1 ignores any income or payout at times other thanT . For example, consider an asset that pays 1 at time (T + t)/2 and 1 at time T . The valueof the asset at T is DT = 1, so 10.1.1 would say Dt = Z(t, T ). But the correct price at isclearly Dt = Z(t, (T + t)/2) + Z(t, T ).

Solution I: Restrict to assets that pay nothing except possibly at T . Alternatively, considerportfolios rather than assets because the value of a portfolio accounts for things like divi-dends, exercises, and trades. For example, if A at t contains the asset paying 1 at (t+T )/2and 1 at T , then V A(T ) 6= DT . In fact, V A(T ) > 2 > 1 = DT .

Problem II: The pricing formula (10.1.1) can lead to an arbitrage portfolio, as the nextexample illustrates. (In fact, this example already appeared as Exercise 3 of HW8.)

Example 10.1.1. Consider a one-year European call option on a stock with strike 40. Sup-pose:

• Current stock price = St = 35

• There are three possible states of the world at maturity T : ω1, ω2, ω3

• P (ω1) = 1/2, P (ω2) = 1/3, P (ω3) = 1/6

• ST (ω1) = 50, ST (ω2) = 55, ST (ω3) = 30

• Z(t, T ) = 0.9.

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(a) Find the discounted present value of the expected payout: Z(t, T )E(ST −K)+.

Since

(ST −K)+ =

50− 40 = 10, with probability 1/2,

55− 40 = 15, with probability 1/3,

0 with probability 1/6,

we haveE(ST −K)+ = 10 · 1

2+ 15 · 1

3+ 0 · 1

6= 10.

Therefore Z(t, T )E(ST −K)+ = 0.9 · 10 = 9.

(b) Suppose we take the call price to be discounted present value of the expected payout:

CK(t, T ) = Z(t, T )E(ST −K, 0)+ = 9.

Does this represent an arbitrage opportunity? If so, build an arbitrage portfolio. Verify it isan arbitrage portfolio.

Yes, there is an arbitrage opportunity.

Start with portfolio C empty at t. Sell one K = 40 call for 9 cash, borrow additional 26cash, buy 1 stock for 35 cash.

C at t: short one K = 40 call, one stock, -26 cash.

V C(t) = −9 + 35− 26 = 0.

V C(T ) = ST−(ST−K)+− 26

Z(t, T )=

50− 10− 26/0.9 = 11.1111, with probability 1/2,

55− 15− 26/0.9 = 11.1111, with probability 1/3,

30− 0− 26/0.9 = 1.1111, with probability 1/6.

Therefore V C(T ) > 0 with probability one. So C is an arbitrage portfolio. 4

Solution II: The example above shows the pricing formula

Dt = Z(t, T )E(DT ) (10.1.2)

can lead to an arbitrage portfolio. However, by replacing the probability measure P byanother probability measure P ∗ (i.e., by adjusting how we assign probabilities), the pricingformula (10.1.2) can be made compatible with the no-arbitrage principle.

The fundamental theorem says that we can make the pricing formula (10.1.2) work if andonly if there are no arbitrage portfolios.

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Result 10.1.2 (Fundamental Theorem of Asset Pricing (T -Payout Assets)). There are noarbitrage portfolios if and only if there is a probability measure P ∗ satisfying

• P ∗ is equivalent to P ,

• for any times t ≤ T and for any T -payout asset, the asset prices Dt and DT arerelated by

Dt = Z(t, T )E∗(DT ). (10.1.3)

4

Definitions of Terms and Notation.

• Probability measures P and P ∗ are called equivalent if the events that have probabil-ity one according to P are exactly the events that have probability one according toP ∗. In other words, the arbitrage portfolios with respect to P are exactly the arbitrageportfolios with respect to P ∗.

• An asset is called a T -payout asset if it has no income or payout except possiblyat time T . Examples: Stock paying no income; European option with maturity T .Non-examples: Stock paying dividends; American option with maturity T .

• E∗ means expectation with respect to P ∗. For example,

E∗(X) =∑

k∈R(X)

kP ∗(X = x),

E∗(X|Y = y) =∑

k∈R(X)

kP ∗(X = x|Y = y).

• If P ∗ is a probability measure satisfying the two conditions of the theorem, then P ∗

is called a risk-neutral probability measure (with respect to Z(t, T ) and P ).

• The formula (10.1.3) is called risk-neutral pricing (with respect to Z(t, T )).

Remark. Technically, the formula

Dt = Z(t, T )E∗(DT )

should beE∗(DT |It).

It stands for all relevant information at t. For a derivative that depends on a stock price,

It = St = stock price at t.

For a derivative that depends on interest rates, It stands for the interest rates at t.

We can restate the fundamental theorem in terms of portfolios.

88

Result 10.1.3 (Fundamental Theorem of Asset Pricing (Portfolios)). There are no arbitrageportfolios if and only if there is a probability measure P ∗ satisfying

• P ∗ is equivalent to P ,

• for any times t ≤ T and for any portfolio A,

V A(t) = Z(t, T )E∗(V A(T )). (10.1.4)

4

We will prove only the easy half of the fundamental theorem.

Proof: Existence of P ∗ Implies No-Arbitrage. Assume there is a probability measure P ∗

satisfying the two conditions in the theorem.

Let A be any portfolio. We want to show A cannot be an arbitrage portfolio. We willassume A is an arbitrage portfolio and deduce a contradiction. So we have

V A(t) ≤ 0, P (V A(T ) ≥ 0) = 1, P (V A(T ) > 0) > 0.

Since P ∗ is equivalent to P , we must have

V A(t) ≤ 0, P ∗(V A(T ) ≥ 0) = 1, P ∗(V A(T ) > 0) > 0.

SinceZ(t, T )E∗(V A(T )) = V A(t) ≤ 0,

we must have

E∗(V A(T )) =∑

k∈R(V A(T ))

kP ∗(V A(T ) = k) ≤ 0. (10.1.5)

Since P ∗(V A(T ) ≥ 0) = 1, we must have

P ∗(V A(T ) = k) = 0 for all k < 0.

Therefore the sum in (10.1.5) has only non-negative terms. But since the sum is ≤ 0, allthe terms must be zero. That is,

kP ∗(V A(T ) = k) = 0 for all k.

ThereforeP ∗(V A(T ) = k) = 0 for all k 6= 0.

ThereforeP ∗(V A(T ) = 0) = 1.

This contradicts thatP ∗(V A(T ) > 0) > 0.

Remark. According to the fundamental theorem, if we can find P ∗, then we can find theprice of any asset. We will soon study the binomial tree model and the Black-Scholesmodel, which are natural models in which we can find P ∗.

89

10.2 Binomial Tree

Consider a stock paying no income with price ST at time T .

A binomial tree for the stock is a model of the stock price at discrete times.

For simplicity, we assume the times are 0, 1, 2, . . . , n, . . ., so the time step is ∆T = 1.

It has parameters r, u, d, and p, where r > 0, d < u, and 0 < p < 1.

r = constant annually compounded interest rate.

At time 0 the stock price S0 is known.

At time n, the stock price goes from Sn−1 up to Sn = (1 + u)Sn−1 with probability p, ordown to Sn = (1 + d)Sn−1 with probability (1− p). Whether the price goes up or down attime n is independent of whether the price goes up or down at any other time.

Figure 10.1: Branch of a Binomial Tree

Figure 10.2: One-Step Binomial Tree with d = −0.1 and u = 0.2

90

Figure 10.3: Two-Step Binomial Tree with d = −0.1 and u = 0.2

Here is an equivalent way to define how the price changes at each time:

Sn = ξnSn−1 for n = 1, 2, . . . , (10.2.1)

where ξ1, ξ2, . . ., is a sequence of independent random variables with

ξi =

(1 + u) with probability p(1 + d) with probability 1− p

By applying (10.2.1) repeatedly, we see

Sn = ξnSn−1 = ξnξn−1Sn−2 = . . . = ξn · · · ξ1S0.

The next example shows how the probability p of an “up” movement determines the prob-ability measure P .

Example 10.2.1. From time 0 to time n, the stock price changes n times.

IfSn = (1 + u)k(1 + d)n−kS0,

it means the stock price moves “up” exactly k times out of the n possible times.

If n = 5 and k = 3, a typical path with exactly k “up” movements is

up-down-up-up-down.

There are(nk

)possible paths with exactly k up movements: Out of the n changes, choose k

of them to be “up” and the rest “down.”

Each such path has probability pk(1− p)n−k.

Therefore

P (Sn = (1 + u)k(1 + d)n−kS0) =

(n

k

)pk(1− p)n−k

for 0 ≤ k ≤ n. 4

91

The previous example proves

Result 10.2.2.

P (Sn = (1+u)k(1+d)n−kS0) = P (exactly k “up” movements in period 0 to n) =

(n

k

)pk(1−p)n−k.

for 0 ≤ k ≤ n. 4

10.3 Arbitrage-Free Binomial Tree

Consider a binomial tree (parameters r,u,d,p) for a stock paying no income. The stockprice at T is ST .

Example 10.3.1. Suppose the binomial tree has no arbitrage portfolios. By the fundamentaltheorem, there is a risk-neutral probability measure P ∗. Therefore

S0 = Z(0, 1)E∗(S1).

Recall p is the probability of an “up” movement in the stock price with respect to P .

Let p∗ denote the probability of an “up” movement in the stock price with respect to P ∗.

IN CLASS, I WILL DRAW A PICTURE OF THE TREE HERE

We have

S0 = Z(t, T )E∗(S1)

= (1 + r)−1E∗(S1)

= (1 + r)−1 · (p∗(1 + u)S0 + (1− p∗)(1 + d)S0)

= (1 + r)−1 · (p∗(1 + u) + (1− p∗)(1 + d)) · S0.

Divide both sides by S0 to get

1 = (1 + r)−1(p∗(1 + u) + (1− p∗)(1 + d)).

Solving for p∗ gives

p∗ =r − du− d

.

4

The previous example proves

Result 10.3.2. If the binomial tree has no arbitrage portfolios, then

p∗ =r − du− d

,

where p∗ is the probability of an “up” movement in the stock price with respect to therisk-neutral probability measure P ∗. 4

92

10.4 Arbitrage-Free Pricing on the Binomial Tree

Consider a binomial tree (parameters r,u,d,p) for a stock paying no income. The stockprice at T is ST . Assume there are no arbitrage portfolios.

Example 10.4.1. Find the price at 0 of European call with maturity T = n and strike K.

We haveCK(T, T ) = max ST −K, 0 = max Sn −K, 0 = g(Sn).

andZ(0, T ) = Z(0, n) = (1 + r)−n.

By the fundamental theorem,

CK(0, T ) = Z(0, T )E∗(CK(T, T )) = (1 + r)−nE∗(g(Sn)).

By LOTUS,

CK(0, T ) = (1 + r)−n∑s∈R(S)

g(s)P ∗(Sn = s).

The possible value of Sn are

(1 + u)k(1 + d)n−kS0 for k = 0, 1, . . . , n.

Hence

CK(0, T ) = (1 + r)−nn∑k=0

g((1 + u)k(1 + d)n−kS0)P ∗(Sn = (1 + u)k(1 + d)n−kS0).

By Result 10.3.2,

p∗ =r − du− d

,

is the probability of an “up” movement in the stock price with respect to the risk-neutralprobability measure P ∗. Therefore, by Example 10.2.1,

P ∗(Sn = (1 + u)k(1 + d)n−kS0) =

(n

k

)p∗k(1− p∗)n−k for k = 0, 1, . . . , n.

Hence

CK(0, T ) = (1 + r)−nn∑k=0

g((1 + u)k(1 + d)n−kS0)

(n

k

)p∗k(1− p∗)n−k

Since g(x) = max x−K, 0 and p∗ =r − du− d

,

CK(0, T ) = (1+r)−nn∑k=0

max

(1 + u)k(1 + d)n−kS0 −K, 0(n

k

)(r − du− d

)k (u− ru− d

)n−kIt’s complicated, but it’s exact. 4

93

With minor changes in notation, the previous example proves

Result 10.4.2. The price at 0 of a derivative of the stock with payout g(ST ) at T = n (andno payout otherwise) is

D(0, T ) = Z(0, T )E∗(g(ST )) = (1+r)−nn∑k=0

g((1+u)k(1+d)n−kS0)

(n

k

)p∗k(1−p∗)n−k.

where p∗ =r − du− d

. 4

Let’s do a computation with this formula.

Example 10.4.3. The constant annually compounded interest rate is 10% At current time0, a stock paying no income has price 50. Suppose that at each time point, the stock pricecan go up by 30% or down by 20%. Find the price at 0 of a European put with strike 47and maturity 2.

4

94

Let’s do the example again with a different method.

Example 10.4.4. The constant annually compounded interest rate is 10% At current time0, a stock paying no income has price 50. Suppose that at each time point, the stock pricecan go up by 30% or down by 20%. Find the price at 0 of a European put with strike 47and maturity 2. 4

95

Let’s do the example again, but with an American put.

Example 10.4.5. The constant annually compounded interest rate is 10% At current time0, a stock paying no income has price 50. Suppose that at each time point, the stock pricecan go up by 30% or down by 20%. Find the price at 0 of a American put with strike 47and maturity 2.

97

4

Remark. The examples above show that an American put can have price strictly greaterthan a European put with the same strike and maturity.

Chapter 11

Probability Theory: NormalDistribution and Central Limit Theorem

11.1 Normal Distribution

Let X be a random variable. If there are constants µ ∈ R and σ > 0 such that

P (a ≤ X ≤ b) =

∫ b

a

1√2πσ2

e−(x−µ)2/2σ2

dx

for all real numbers a ≤ b, then we write

X ∼ N (µ, σ2)

and we say thatX has normal distribution (with respect to P ). Note thatX is non-discretewith R(X) = R.

Suppose X ∼ N (µ, σ2). Then:

(i) For any function g(x), the expectation of g(X) is

E(g(X)) =

∫ ∞−∞

1√2πσ2

e−(x−µ)2/2σ2

g(x)dx

(ii) The expectation of X isE(X) = µ.

(iii) The variance of X is

Var(X) = E((X − µ)2) = E(X2)− µ2 = σ2.

(iv) If A is a constant, thenA+X ∼ N (A+ µ, σ2).

98

99

11.2 Standard Normal Distribution

If X ∼ N(0, 1), we say that X has standard normal distribution and the function

Φ(t) = P (X ≤ t) =

∫ t

−∞

1√2πe−x

2/2dx

is called the standard normal cumulative distribution function or standard normal cdf.

11.3 Central Limit Theorem

The central limit theorem is the reason the normal distribution is so important. Basically,it says that any sum of many independent random effects is approximately normally dis-tributed.

Central Limit Theorem. Suppose that for each fixed value of n, we have a sequence of nrandom variables

X(n)1 , X

(n)2 , . . . , X(n)

n

that are independent and all have the same distribution. Suppose

limn→∞

E(n∑i=1

Xi(n)) = µ and limn→∞

Var(n∑i=1

Xi(n)) = σ2.

Thenn∑i=1

Xi(n)→ N (µ, σ2),

which is an abbreviation for

limn→∞

P (a ≤n∑i=1

Xi(n) ≤ b) =

∫ b

a

1√2πσ2

e−(x−µ)2/2σ2

dx.

Remark. This version of the central limit theorem may be different from versions you haveseen before. It is a central limit theorem for triangular arrays.

Chapter 12

Continuous-Time Limit andBlack-Scholes Formula

12.1 Black-Scholes Model

In the Black-Scholes model, we consider a stock paying no income and derivatives on thestock.

Let St and ST be the stock prices at current time t and future time T .

The Black-Scholes model makes two assumptions:

No-Arbitrage. There are no arbitrage portfolios.

Lognormal Stock Price. Under the risk-neutral probability measure P ∗,

lnST ∼ N(lnSt + (r − 1

2σ2)(T − t), σ2(T − t))

where r is the constant continuously compounded interest rate, and σ > 0 is a constantcalled the volatility of the stock price.

The Lognormal Stock Price assumption means that

P ∗(a ≤ lnST ≤ b) =

∫ b

a

1√2πσ2(T − t)

e−(x−ν)2/2σ2(T−t)dx

whereν = lnSt + (r − 1

2σ2)(T − t).

100

101

12.2 Binomial Tree to Black-Scholes

We will outline how the Black-Scholes model can be obtained as a limit of the binomialtree model.

Consider a stock paying no income. For simplicity, the current time is t = 0. Fix a futuretime T . Choose a large integer N and define

∆T =T

N.

Consider a binomial tree for the stock with step size ∆T . We will letN →∞, so ∆T → 0.

The times on the tree are 0,∆T, 2∆T, . . ..

Given the stock price at (i− 1)∆T is S(i−1)∆, the stock price at i∆T is

Si∆T = ξiS(i−1)∆T .

Here ξ1, ξ2, . . . are independent random variables with

ξi =

1 + u with P probability 1/21 + d with P probability 1/2

Under the risk-neutral probability measure P ∗, we have

ξi =

1 + u with P ∗ probability p∗

1 + d with P ∗ probability 1− p∗

By Exercise 4 of HW11, we know that

p∗ =r∆T − du− d

where

r = interest rate with compounding at every time point 0,∆T, 2∆T, . . .

(compounding frequency m = 1/∆T ) .

Reminder of proof: By the fundamental theorem,

S0 = Z(0,∆T )E∗(S∆T )

= (1 + r/m)−m∆TE∗(S∆T )

= (1 + r∆T )−1(p∗(1 + u)S0 + (1− p∗)(1 + d)S0),

then solve for p∗.

In the limit ∆→ 0,

r → interest rate with continuous compounding.

102

The stock price at T is

ST = SN∆T = ξNS(N−1)∆T = ξNξN−1S(N−2)∆T ,

and soST = ξNξN−1 · · · ξ1S0.

To change multiplication into addition, we take logarithms:

lnST = lnS0 +N∑i=1

ln ξi.

Note that for any fixed value of N , the random variables

ln ξ1, ln ξ2, . . . , ln ξN

are independent and all have the same distribution. This suggests we try the central limittheorem.

We define the volatility σ > 0 by

σ2T = Var∗(

ln

(STS0

))= Var∗

(N∑i=1

ln ξi

).

We can show (details omitted) that

limN→∞

E∗(

N∑i=1

ln ξi

)=

(r − 1

2σ2

)T.

By the central limit theorem,

N∑i=1

ln ξi → N (

(r − 1

2σ2

)T, σ2T ).

Therefore

lnST = lnS0 +N∑i=1

ln ξi → N (lnS0 +

(r − 1

2σ2

)T, σ2T ).

Remark. Notice we didn’t assume anything about u and d.

Remark. The normal distribution of the logarithm of the stock price in the Black-Scholesmodel comes from the multiplicative nature of the Binomial Tree model.

103

12.3 Black-Scholes Formula

Assume the Black-Scholes model for a stock paying no income. We now find a formula forthe price at t of a European call option with strike K and maturity T .

By the fundamental theorem,

CK(t, T ) = Z(t, T )E∗((ST −K)+)

= Z(t, T )E∗((elnST −K)+)

= Z(t, T )E∗(g(lnST ))

whereg(x) = (ex −K)+.

Under the risk-neutral probability measure P ∗,

lnST ∼ N (ν, σ2(T − t)),

where

ν = lnSt +

(r − 1

2σ2

)(T − t).

Therefore

CK(t, T ) = Z(t, T )1√

2πσ2(T − t)

∫ ∞−∞

e−(x−ν)2/2σ2(T−t)g(x)dx

= Z(t, T )1√

2πσ2(T − t)

∫ ∞−∞

e−(x−ν)2/2σ2(T−t)(ex −K)+dx.

Notice that

(ex −K)+ =

ex −K if x ≥ lnK0 if x ≤ lnK

Therefore

Result 12.3.1 (Black-Scholes Formula).

CK(t, T ) = Z(t, T )1√

2πσ2(T − t)

∫ ∞lnK

e−(x−ν)2/2σ2(T−t)(ex −K)dx.

4

By making a clever change of variable, we can write the integral above in terms of thestandard normal cdf

Φ(t) =

∫ t

−∞

1√2πe−x

2/2dx.

Doing so, we obtain

104


CK(t, T ) = StΦ(d1)−KZ(t, T )Φ(d2)

where

d1 =ln(St/K) + (r + 1

2σ2)(T − t)

σ√T − t

and d2 = d1 − σ√T − t.

4

Using the forward price formula for a stock paying no income

F (t, T ) =St

Z(t, T ),

we can rewrite the previous result as


CK(t, T ) = Z(t, T )(F (t, T )Φ(d1)−KΦ(d2)).

4

Results 12.3.1, 12.3.2, and 12.3.3 are different forms of the Black-Scholes formula.

Example 12.3.4. Use the Black-Scholes formula to find the current price of a Europeancall on a stock paying no income with strike 40 and maturity 18 months from now. Assumethe current stock price is 50, the stock volatility is 15%, and the constant continuouslycompounded interest rate is 5%.

d1 =ln(St/K) + (r + 1

2σ2)(T − t)

σ√T − t

=ln(50/40) + (0.05 + 1

2(0.15)2)(1.5)

0.15√

1.5

= 1.7147438 . . .

d2 = d1 − σ√T − t

= (1.7147438 . . .)− (0.15)√

1.5

= 1.531032 . . .


= 50Φ(1.7147438 . . .)− 40e−0.05(1.5)Φ(1.531032 . . .)

= 13.06 . . .

4

105

12.4 Properties of Black-Scholes Formula

Consider a stock paying no income. The Black-Scholes formula for a European call is


where

d1 =ln(St/K) + (r + 1

2σ2)(T − t)

σ√T − t

and d2 = d1 − σ√T − t.

Recall the bounds on a European call we obtained in Chapter 8:

max St −KZ(t, T ), 0 ≤ CK(t, T ) ≤ St

These bounds are illustrated when we plot the price of a 1-year 80-strike call under theBlack-Scholes model for various values of the volatility σ.

0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 1500

10

20

30

40

50

60

70

80

90

100

110

120

130

140

150

St

Pric

e

Black-Scholes price of 1-year 100-strike call (r = 5%)

StCK(t, T ) (σ = 300%)CK(t, T ) (σ = 100%)CK(t, T ) (σ = 50%)CK(t, T ) (σ = 30%)CK(t, T ) (σ = 10%)

max St −KZ(t, T ), 0

As the plot indicates, the bounds are achieved in limiting cases:

106

Result 12.4.1.

(a)CK(t, T )→ max St −KZ(t, T ), 0 as σ → 0.

(b)CK(t, T )→ St as σ →∞.

4

Proof. We prove (a) and leave (b) as an exercise. We have

limσ→0

d1 = limσ→0

ln(St/K) + (r + 12σ2)(T − t)

σ√T − t

= limσ→0

ln(St/K) + r(T − t)σ√T − t

+1

2σ√T − t

=

+∞ if ln(St/K) + r(T − t) > 00 if ln(St/K) + r(T − t) = 0−∞ if ln(St/K) + r(T − t) < 0

=

+∞ if St > Ke−r(T−t)

0 if St = Ke−r(T−t)

−∞ if St < Ke−r(T−t)

Therefore

limσ→0

Φ(d1) = limσ→0

P (X ≤ d1) =

1 if St > Ke−r(T−t)12

if St = Ke−r(T−t)

0 if St < Ke−r(T−t).

where X ∼ N(0, 1). Since

limσ→0

d2 = limσ→0

(d1 − σ√T − t) = lim

σ→0d1,

we have

limσ→0

Φ(d2) = limσ→0

Φ(d1) =

1 if St > Ke−r(T−t)12

if St = Ke−r(T−t)


Thus

limσ→0

CK(t, T ) = limσ→0

(StΦ(d1)−Ke−r(T−t)Φ(d2))

=

St −Ke−r(T−t) if St > Ke−r(T−t)

0 if St = Ke−r(T−t)


= max St −KZ(t, T ), 0

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12.5 The Greeks: Delta and Vega

The delta of an option (or other derivative contract) is the partial derivative of its price withrespect to the underlying asset price.

Result 12.5.1. Assuming the Black-Scholes model for a stock paying no income, the deltaof a European call is

∂CK(t, T )

∂St= Φ(d1).

4

We outline the proof at the end of this section.

Since Φ(d1) = P (X ≤ d1), where X ∼ N(0, 1), the delta of a European call satisfies

0 ≤ ∂CK(t, T )

∂St≤ 1.

If the delta of a contract is positive, the price of the contract increases as the stock priceincreases, and the party long on the contract is said to have a long delta position.

If the delta of a contract is negative, the price of the contract decreases as the stock priceincreases, and the party long on the contract is said to have a short delta position.

Example 12.5.2. The owner (long counterparty) of a European call has a long delta posi-tion, and so will profit if the stock price increases.

The writer (short counterparty) of a European call has a short delta position. Indeed, thedelta is

∂

∂St(−CK(t, T )) = −Φ(d1).

4

108

Figure 12.1: Black-Scholes delta of 1-year 100-strike call (for different volatilities σ)

Example 12.5.3 (Delta Hedging). Let ∆(s) be the delta of a European call when the stockprice is St = s:

∆(s) =∂CK(t, T )

∂St

∣∣∣∣St=s

.

For example, assume K = 100, T − t = 1, r = 10%, and σ = 20%. Then

∆(90) =∂CK(t, T )

∂St

∣∣∣∣St=90

= Φ(d1)|St=90

= Φ

(ln(St/K) + (r + 1

2σ2)(T − t)

σ√T − t

)= Φ

(ln(90/100) + (0.10 + 1

2(0.20)2)(1)

(0.20)2√

1

)= Φ(0.365987 . . .) = 0.642813.

Similarly,∆(80) = 0.00496022.

Fix a number s. Consider the portfolio consisting of +1 European call and short ∆(s)stocks. The delta of the portfolio is

∂

∂St(CK(t, T )−∆(s)St) =

∂CK(t, T )

∂St−∆(s)

∂St∂St

=∂CK(t, T )

∂St−∆(s).

Therefore when the stock price is St = s, the delta is

∂

∂St(CK(t, T )−∆(s)St)

∣∣∣∣St=s

=∂CK(t, T )

∂St

∣∣∣∣St=s

−∆(s) = 0.

109

So, when St = s, the value of the portfolio (consisting of +1 European call and short ∆(s))stocks is unaffected by small changes in the stock price. The portfolio is said to be delta-hedged or delta-neutral at St = s because its delta is zero when St = s. It has no exposureto changes in the stock price around St = s.

Similarly, if D(t, T ) is the price at t of a derivative contract with maturity T on the stock,then the portfolio consisting of +1 contract and short ∂D(t,T )

∂St

∣∣∣St=s

stocks is delta-hedgedat St = s. 4

The vega of an option (or other derivative contract) is the partial derivative of its price withrespect to volatility σ.

Result 12.5.4. Assuming the Black-Scholes model for a stock paying no income, the vegaof a European call is

∂CK(t, T )

∂σ= St√T − t 1√

2πe−d

21/2.

4

If the vega of a contract is positive, the price of the contract increases as the volatilityincreases, and the party long on the contract is said to have a long vega position.

If the vega of a contract is negative, the price of the contract increases as the volatilitydecreases, and the party long on the contract is said to have a short vega position.

Figure 12.2: Black-Scholes vega of 1-year 100-strike call (for different volatilities)

The next result shows how the delta (and vega) of a call and a put are related.

110

Result 12.5.5. In the Black-Scholes model for a stock paying no income,

(a)∂CK(t, T )

∂St− ∂PK(t, T )

∂St= 1

(b)∂CK(t, T )

∂σ− ∂PK(t, T )

∂σ= 0

4

Proof. Recall that put-call parity states

CK(t, T )− PK(t, T ) = VK(t, T ) = St −KZ(t, T )

Differentiating with respect to St gives (a). Differentiating with respect to σ gives (a).

By combining Results 12.5.1, 12.5.4, and 12.5.5, we obtain

Result 12.5.6. In the Black-Scholes model for a stock paying no income,

∂PK(t, T )

∂St=∂CK(t, T )

∂St− 1 = Φ(d1)− 1

and∂PK(t, T )

∂σ=∂CK(t, T )

∂σ= St√T − t 1√

2πe−d

21/2

4

Example 12.5.7. Suppose we are short a (K2, K1) put spread, where 0 < K1 < K2.

Determine whether this position is long vega (vega ≥ 0), short vega (vega ≤ 0), has novolatility exposure (vega = 0), or the volatility exposure is indeterminate

(a) Recall that a (K2, K1) put spread is +1 K2 put and −1 K1 put.

Since we are short the spread, our position is:

−1 K2 put and +1 K1 put

We will need to indicate that d1 depends on K. So we write

d1(K) =lnSt − lnK + (r + 1

2σ2)(T − t)

σ√T − t

.

Since K1 < K2, we haved1(K1) > d1(K2),

and soe−(d1(K1))2/2 < e−(d1(K2))2/2.

111

Then

vega = −∂PK2(t, T )

∂σ+∂PK1(t, T )

∂σ

= −St√T − t 1√

2πe−(d1(K2))2/2 + St

√T − t 1√

2πe−(d1(K1))2/2

= St√T − t 1√

2π

(e−(d1(K1))2/2 − e−(d1(K2))2/2

)Therefore vega < 0. It is a short vega position. 4

Outline of Proof of Result 12.5.1. The Black-Scholes formula says

CK(t, T ) = StΦ(d1)−Ke−r(T−t)Φ(d2)

where

d1 =ln(St/K) + (r + 1

2σ2)(T − t)

σ√T − t

and d2 = d1 − σ√T − t.

Notice d1 and d2 depend on St. Then

∂CK(t, T )

∂St=

∂

∂St

(StΦ(d1)−Ke−r(T−t)Φ(d2)

).

By the product rule

∂CK(t, T )

∂St= Φ(d1) + St

∂

∂StΦ(d1)−Ke−r(T−t) ∂

∂StΦ(d2)

By the fundamental theorem of calculus and the chain rule,

∂

∂StΦ(d1) =

∂

∂St

∫ d1

−∞

1√2πe−x

2/2dx

=1√2πe−d

21/2

∂

∂Std1

=1√2πe−d

21/2

∂

∂St

(lnSt − lnK + (r + 1

2σ2)(T − t)

σ√T − t

)=

1√2πe−d

21/2

∂

∂St

lnSt

σ√T − t

=1√2πe−d

21/2

1

Stσ√T − t

By a similar calculation,

∂

∂StΦ(d2) =

1√2πe−d

22/2

1

Stσ√T − t

Exercise: Complete the proof.

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12.6 Volatility

For a stock paying no income, the Black-Scholes model assumes for any t ≤ T that

lnST ∼ N (lnSt + (r − 1

2σ2)(T − t), σ2(T − t)).

under the risk-neutral measure probability measure P ∗.

r is the continuously compounded interest rate, which is assumed to be constant.

σ is a positive constant called the volatility of the stock. Roughly, it measures how muchthe stock price can be expected to change over time. Large volatility means large changesin the stock price are likely. Small volatility means large changes in the stock price areunlikely.

113

Price History of Stock A

Day Stock Price0 100.001 100.112 100.503 100.314 100.805 100.796 100.797 102.438 103.509 102.8810 102.6611 102.8012 101.6013 101.9514 101.9015 101.7516 101.5617 101.4618 101.5419 101.4820 102.6921 102.0622 102.8923 102.8324 103.6725 103.5326 102.4627 104.8128 105.4929 107.1530 109.21

0 5 10 15 20 25 30

90

95

100

105

110

Day

Stoc

kPr

ice

114

Price History of Stock B

Day Stock Price0 100.001 99.992 99.433 100.054 97.375 94.676 93.657 92.458 95.089 93.8010 88.4311 89.3512 92.6713 96.5214 97.4115 94.2416 92.7117 91.4318 92.4219 92.8320 90.2521 93.2522 92.3523 96.5524 101.5925 103.0026 103.5727 103.2728 100.9729 101.3030 107.34

0 5 10 15 20 25 30

90

95

100

105

110

Day

Stoc

kPr

ice

115

Based on the graphs, it looks like stock B has higher volatility than stock A.

We now explain how to estimate the volatility of stocks A and B based on the data above.

Some Theory.

Let Si∆T be the stock price on day i, where ∆T = 1/365 years = 1 day. Assuming theBlack-Scholes model for the stock, we have

lnSi∆T ∼ N(

lnS(i−1)∆T + (r − 1

2σ2)(T − t), σ2(T − t)

),

DefineXi = ln(Si∆T/S(i−1)∆T ).

Then

Xi = ln(Si∆T/S(i−1)∆T ) = ln(Si∆T )− ln(S(i−1)∆T ) ∼ N(

(r − 1

2σ2)∆T, σ2∆T.

)Therefore

σ2∆T = Var∗ (Xi) .

Procedure to Estimate σ From Data.

We compute Xi = ln(Si∆T/S(i−1)∆T ) for day i = 1, . . . , n = 30.

We compute the sample variance of the data points X1, . . . , Xn:

s2n =

1

n− 1

n∑i=1

(Xi −X)2 where X =1

n

n∑i=1

Xi.

Thens2n ≈ Var∗ (Xi) = σ2∆T.

Therefore

σ ≈√

1

∆Ts2n

Remark. We can use a spreadsheet to do the calculations. In Excel and Google Sheets,VAR() computes the sample variance of a set of data points.

116

Example 12.6.1 (Estimating the Volatility). We use a spreadsheet to estimate the volatilityof stocks A and B according to the procedure described above. You can find a samplespreadsheet on the course website.

Stock ADay Stock Price ln

(Si∆T

S(i−1)∆T

)0 100.001 100.11 0.0010582 100.50 0.0039583 100.31 -0.0019684 100.80 0.0048845 100.79 -0.0000506 100.79 0.0000277 102.43 0.0161018 103.50 0.0103899 102.88 -0.00605310 102.66 -0.00205911 102.80 0.00134712 101.60 -0.01180413 101.95 0.00345414 101.90 -0.00048015 101.75 -0.00144416 101.56 -0.00189117 101.46 -0.00099518 101.54 0.00078819 101.48 -0.00058820 102.69 0.01186821 102.06 -0.00616322 102.89 0.00807723 102.83 -0.00054824 103.67 0.00811925 103.53 -0.00137226 102.46 -0.01032127 104.81 0.02259828 105.49 0.00649429 107.15 0.01558630 109.21 0.019046

Sample Variance = 0.000066628Volatility = σ ≈ 0.1559

Stock BDay Stock Price ln

(Si∆T

S(i−1)∆T

)0 100.001 99.99 -0.0000532 99.43 -0.0057053 100.05 0.0063004 97.37 -0.0271705 94.67 -0.0281096 93.65 -0.0108647 92.45 -0.0129108 95.08 0.0280779 93.80 -0.013594

10 88.43 -0.05889511 89.35 0.01031212 92.67 0.03643313 96.52 0.04080814 97.41 0.00913415 94.24 -0.03309616 92.71 -0.01633017 91.43 -0.01395118 92.42 0.01078119 92.83 0.00445020 90.25 -0.02825121 93.25 0.03277022 92.35 -0.00970723 96.55 0.04444724 101.59 0.05090725 103.00 0.01381626 103.57 0.00545627 103.27 -0.00287828 100.97 -0.02251829 101.30 0.00326530 107.34 0.057947

Sample Variance = 0.00074733Volatility = σ ≈ 0.5222

From the data, we estimate the volatility asStock A: σ = 15.59%Stock B: σ = 52.22% 4

117

Remark. In fact, the price data was generated withStock A: σ = 15%Stock B: σ = 50% Our estimates are not bad for only 30 data points.

Example 12.6.2 (Option Pricing With Estimated Volatility). Assume the constant contin-uously compounded interest rate is 10%. Use the volatility estimates

Stock A: σ = 15.59%Stock B: σ = 52.22%

to find call prices for two-year 100-strike European calls on stocks A and B. The currenttime is Day 30.

The Black-Scholes formula is

CK(t, T ) = StΦ(d1)−KZ(t, T )Φ(d2).

Stock A:

d1 =ln(St/K) + (r + 1

2σ2)(T − t)

σ√T − t

=ln(109.21/100) + (0.10 + 1

2(0.1559)2)(2)

(0.1559)√

2

= 1.41697 . . .

d2 = d1 − σ√T − t

= (1.41697 . . .)− (0.1559)√

2

= 1.19649 . . .

Z(t, T ) = e−0.10(2)

Therefore


= 109.21Φ(1.41697 . . .)− 100e−0.10(2)Φ(1.19649 . . .)

= 28.26 . . .

Stock B:

118

d1 =ln(St/K) + (r + 1

2σ2)(T − t)

σ√T − t

=ln(107.64/100) + (0.10 + 1

2(0.5222)2)(2)

0.5222√

2

= 0.739761 . . .

d2 = d1 − σ√T − t

= (0.739761 . . .)− (0.15)√

1.5

= 0.00125836 . . .

Z(t, T ) = e−0.10(2)

Therefore


= 107.64Φ(0.739761 . . .)− 100e−0.10(2)Φ(0.00125836 . . .)

= 41.93 . . .

4

Bibliography

[1] S. Blyth, An Introduction to Quantitative Finance, 1 ed., 2014.

[2] M. Capinski, T. Zastawniak, Mathematics for Finance: An Introduction to FinancialEngineering, 2 ed., 2011.

[3] D. J. Higham, An Introduction to Financial Option Evaluation: Mathematics,Stochastics and Computation, 1 ed., 2004.

Kyle HambrookDepartment of Mathematics, University of Rochester, Rochester, NY, 14627 [email protected]

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