stochastic calculus for finance - michigan state...

Stochastic Calculus for Finance

Albert Cohen

Actuarial Sciences ProgramDepartment of Mathematics

Department of Statistics and ProbabilityA336 Wells Hall

Michigan State UniversityEast Lansing MI

[email protected]@stt.msu.edu

Albert Cohen (MSU) STT 888: Stochastic Calculus for Finance MSU 2016 1 / 249

Course Information

Syllabus to be posted on class page in first week of classes

Homework assignments will posted there as well

Page can be found athttps://www.stt.msu.edu/Academics/ClassPages/

Course Information

Many examples within these slides are used with kind permission ofProf. Dmitry Kramkov, Dept. of Mathematics, Carnegie MellonUniversity.

Course Textbook: Stochastic Calculus for Finance (Springer Finance.)

Some examples here will be similar to those practice questionspublicly released by the SOA. Please note the SOA owns thecopyright to these questions.

This book will be our reference, and some questions for assignmentswill be chosen from it. Copyright for all questions used from this bookbelongs to Springer.

From time to time, we will also follow the format of Marcel Finan’s ADiscussion of Financial Economics in Actuarial Models: A Preparationfor the Actuarial Exam MFE/3F. Some proofs from there will bereferenced as well. Please find these notes here

What are financial securities?

Traded Securities - price given by market.

For example:

StocksCommodities

Non-Traded Securities - price remains to be computed.

Is this always true?

We will focus on pricing non-traded securities.

For example:

StocksCommodities

For example:

StocksCommodities

For example:

StocksCommodities

How does one fairly price non-traded securities?

By eliminating all unfair prices

Unfair prices arise from Arbitrage Strategies

Start with zero capitalEnd with non-zero wealth

We will search for arbitrage-free strategies to replicate the payoff of anon-traded security

This replication is at the heart of the engineering of financial products

More Questions

Existence - Does such a fair price always exist?

If not, what is needed of our financial model to guarantee at least onearbitrage-free price?

Uniqueness - are there conditions where exactly one arbitrage-freeprice exists?

And What About...

Does the replicating strategy and price computed reflect uncertaintyin the market?

Mathematically, if P is a probabilty measure attached to a series ofprice movements in underlying asset, is P used in computing theprice?

And What About...

Does the replicating strategy and price computed reflect uncertaintyin the market?

Mathematically, if P is a probabilty measure attached to a series ofprice movements in underlying asset, is P used in computing theprice?

Notation

Forward Contract:

A financial instrument whose initial value is zero, and whose finalvalue is derived from another asset. Namely, the difference of thefinal asset price and forward price:

V (0) = 0,V (T ) = S(T )− F (1)

Value at end of term can be negative - buyer accepts this in exchangefor no premium up front

Notation

Forward Contract:

V (0) = 0,V (T ) = S(T )− F (1)

Notation

Forward Contract:

V (0) = 0,V (T ) = S(T )− F (1)

Notation

Interest Rate:

The rate r at which money grows. Also used to discount the valuetoday of one unit of currency one unit of time from the present

V (0) =1

1 + r,V (1) = 1 (2)

Notation

Interest Rate:

The rate r at which money grows. Also used to discount the valuetoday of one unit of currency one unit of time from the present

V (0) =1

1 + r,V (1) = 1 (2)

An Example of Replication

Forward Exchange Rate: There are two currencies, foreign anddomestic:

SBA = 4 is the spot exchange rate - one unit of B is worth SB

A of Atoday (time 0)

rA = 0.1 is the domestic borrow/lend rate

rB = 0.2 is the foreign borrow/lend rate

Compute the forward exchange rate FBA . This is the value of one unit

of B in terms of A at time 1.

An Example of Replication: Solution

At time 1, we deliver 1 unit of B in exchange for FBA units of domestic

currency A.

This is a forward contract - we pay nothing up front to achieve this.

Initially borrow some amount foreign currency B, in foreign market togrow to one unit of B at time 1. This is achieved by the initial

amountSBA

1+rB(valued in domestic currency)

Invest the amountFBA

1+rAin domestic market (valued in domestic

currency)

currency A.

amountSBA

currency)

currency A.

amountSBA

currency)

currency A.

amountSBA

currency)

This results in the initial value

V (0) =FBA

1 + rA−

1 + rB(3)

Since the initial value is 0, this means

FBA = SB

1 + rA

1 + rB= 3.667 (4)

This results in the initial value

V (0) =FBA

1 + rA−

1 + rB(3)

Since the initial value is 0, this means

FBA = SB

1 + rA

1 + rB= 3.667 (4)

Outline1 Discrete Multiperiod Model

ArbitrageRisk Neutral ProbabilityAmerican OptionsExotic OptionsValuation via Simulation

2 Continuous Model-Ito CalculusBrownian MotionBSMExamplesOptions on FuturesPath Dependent Options

3 Advanced TopicsHeat EquationGeneral Solution of Heat EquationApplication to B-S-M PDE

4 Continuous Model-ProbabilityExpected ValuesApplication of Option GreeksAlbert Cohen (MSU) STT 888: Stochastic Calculus for Finance MSU 2016 13 / 249

Discrete Probability Space

Let us define an event as a point ω in the set of all possible outcomes Ω.This includes the events ”The stock doubled in price over two tradingperiods” or ”the average stock price over ten years was 10 dollars”.

In our initial case, we will consider the simple binary spaceΩ = H,T for a one-period asset evolution. So, given an initialvalue S0, we have the final value S1(ω), with

S1(H) = uS0,S1(T ) = dS0 (5)

with d < 1 < u. Hence, a stock increases or decreases in price,according to the flip of a coin.

Let P be the probability measure associated with these events:

P[H] = p = 1− P[T ] (6)

S1(H) = uS0,S1(T ) = dS0 (5)

P[H] = p = 1− P[T ] (6)

S1(H) = uS0,S1(T ) = dS0 (5)

P[H] = p = 1− P[T ] (6)

S1(H) = uS0,S1(T ) = dS0 (5)

P[H] = p = 1− P[T ] (6)

S1(H) = uS0,S1(T ) = dS0 (5)

P[H] = p = 1− P[T ] (6)

Arbitrage

Assume that S0(1 + r) > uS0

Where is the risk involved with investing in the asset S ?

Assume that S0(1 + r) < dS0

Why would anyone hold a bank account (zero-coupon bond)?

Lemma Arbitrage free ⇒ d < 1 + r < u

Arbitrage

Derivative Pricing

Let S1(ω) be the price of an underlying asset at time 1. Define thefollowing instruments:

Zero-Coupon Bond : V B0 = 1

1+r ,VB1 (ω) = 1

Forward Contract : V F0 = 0,V F

1 = S1(ω)− F

Call Option : V C1 (ω) = max(S1(ω)− K , 0)

Put Option : V P1 (ω) = max(K − S1(ω), 0)

In both the Call and Put option, K is known as the Strike.Once again, a Forward Contract is a deal that is locked in at time 0 forinitial price 0, but requires at time 1 the buyer to purchase the asset forprice F .

What is the value V0 of the above put and call options?

Derivative Pricing

1+r ,VB1 (ω) = 1

1 = S1(ω)− F

Derivative Pricing

1+r ,VB1 (ω) = 1

1 = S1(ω)− F

In both the Call and Put option, K is known as the Strike.

Once again, a Forward Contract is a deal that is locked in at time 0 forinitial price 0, but requires at time 1 the buyer to purchase the asset forprice F .

Derivative Pricing

1+r ,VB1 (ω) = 1

1 = S1(ω)− F

Derivative Pricing

1+r ,VB1 (ω) = 1

1 = S1(ω)− F

Put-Call Parity

Can we replicate a forward contract using zero coupon bonds and put andcall options?

Yes: The final value of a replicating strategy X has value

V C1 − V P

1 + (K − F ) = S1 − F = X1(ω) (7)

This is achieved (replicated) by

Purchasing one call option

Selling one put option

Purchasing K − F zero coupon bonds with value 1 at maturity.

all at time 0.Since this strategy must have zero initial value, we obtain

V C0 − V P

0 =F − K

1 + r(8)

Question: How would this change in a multi-period model?

Put-Call Parity

Can we replicate a forward contract using zero coupon bonds and put andcall options?Yes: The final value of a replicating strategy X has value

V C1 − V P

1 + (K − F ) = S1 − F = X1(ω) (7)

V C0 − V P

0 =F − K

1 + r(8)

Put-Call Parity

V C1 − V P

1 + (K − F ) = S1 − F = X1(ω) (7)

V C0 − V P

0 =F − K

1 + r(8)

Put-Call Parity

V C1 − V P

1 + (K − F ) = S1 − F = X1(ω) (7)

all at time 0.

Since this strategy must have zero initial value, we obtain

V C0 − V P

0 =F − K

1 + r(8)

Put-Call Parity

V C1 − V P

1 + (K − F ) = S1 − F = X1(ω) (7)

V C0 − V P

0 =F − K

1 + r(8)

Put-Call Parity

V C1 − V P

1 + (K − F ) = S1 − F = X1(ω) (7)

V C0 − V P

0 =F − K

1 + r(8)

General Derivative Pricing -One period model

If we begin with some initial capital X0, then we end with X1(ω). To pricea derivative, we need to match

X1(ω) = V1(ω) ∀ ω ∈ Ω (9)

to have X0 = V0, the price of the derivative we seek.

A strategy by the pair (X0,∆0) wherein

X0 is the initial capital

∆0 is the initial number of shares (units of underlying asset.)

What does the sign of ∆0 indicate?

Replicating Strategy

Initial holding in bond (bank account) is X0 −∆0S0

Value of portfolio at maturity is

X1(ω) = (X0 −∆0S0)(1 + r) + ∆0S1(ω) (10)

Pathwise, we compute

V1(H) = (X0 −∆0S0)(1 + r) + ∆0uS0

V1(T ) = (X0 −∆0S0)(1 + r) + ∆0dS0

Algebra yields

∆0 =V1(H)− V1(T )

(u − d)S0(11)

X1(ω) = (X0 −∆0S0)(1 + r) + ∆0S1(ω) (10)

V1(H) = (X0 −∆0S0)(1 + r) + ∆0uS0

V1(T ) = (X0 −∆0S0)(1 + r) + ∆0dS0

Algebra yields

∆0 =V1(H)− V1(T )

(u − d)S0(11)

X1(ω) = (X0 −∆0S0)(1 + r) + ∆0S1(ω) (10)

V1(H) = (X0 −∆0S0)(1 + r) + ∆0uS0

V1(T ) = (X0 −∆0S0)(1 + r) + ∆0dS0

Algebra yields

∆0 =V1(H)− V1(T )

(u − d)S0(11)

X1(ω) = (X0 −∆0S0)(1 + r) + ∆0S1(ω) (10)

V1(H) = (X0 −∆0S0)(1 + r) + ∆0uS0

V1(T ) = (X0 −∆0S0)(1 + r) + ∆0dS0

Algebra yields

∆0 =V1(H)− V1(T )

(u − d)S0(11)

Risk Neutral Probability

Let us assume the existence of a pair (p, q) of positive numbers, and usethese to multiply our pricing equation(s):

pV1(H) = p(X0 −∆0S0)(1 + r) + p∆0uS0

qV1(T ) = q(X0 −∆0S0)(1 + r) + q∆0dS0

Addition yields

X0(1 + r) + ∆0S0(pu + qd − (1 + r)) = pV1(H) + qV1(T ) (12)

Risk Neutral Probability

Let us assume the existence of a pair (p, q) of positive numbers, and usethese to multiply our pricing equation(s):

pV1(H) = p(X0 −∆0S0)(1 + r) + p∆0uS0

qV1(T ) = q(X0 −∆0S0)(1 + r) + q∆0dS0

Addition yields

X0(1 + r) + ∆0S0(pu + qd − (1 + r)) = pV1(H) + qV1(T ) (12)

If we constrain

0 = pu + qd − (1 + r)

1 = p + q

0 ≤ p

0 ≤ q

then we have a risk neutral probability P where

V0 = X0 =1

1 + rE[V1] =

pV1(H) + qV1(T )

p = P[X1(ω) = H] =1 + r − d

u − d

q = P[X1(ω) = T ] =u − (1 + r)

u − d

If we constrain

0 = pu + qd − (1 + r)

1 = p + q

0 ≤ p

0 ≤ q

then we have a risk neutral probability P where

V0 = X0 =1

1 + rE[V1] =

pV1(H) + qV1(T )

p = P[X1(ω) = H] =1 + r − d

u − d

q = P[X1(ω) = T ] =u − (1 + r)

u − d

Example: Pricing a forward contract

Consider the case of a stock with

S0 = 400

u = 1.25

d = 0.75

r = 0.05

Then the forward price is computed via

1 + rE[S1 − F ]⇒ F = E[S1] (13)

Example: Pricing a forward contract

Consider the case of a stock with

S0 = 400

u = 1.25

d = 0.75

r = 0.05

Then the forward price is computed via

1 + rE[S1 − F ]⇒ F = E[S1] (13)

This leads to the explicit price

F = puS0 + qdS0

= (p)(1.25)(400) + (1− p)(0.75)(400)

= 500p + 300− 300p = 300 + 200p

= 300 + 200 · 1 + 0.05− 0.75

1.25− 0.75= 300 + 200 · 3

Homework Question: What is the price of a call option in the caseabove,with strike K = 375?

This leads to the explicit price

F = puS0 + qdS0

= (p)(1.25)(400) + (1− p)(0.75)(400)

= 500p + 300− 300p = 300 + 200p

= 300 + 200 · 1 + 0.05− 0.75

1.25− 0.75= 300 + 200 · 3

Homework Question: What is the price of a call option in the caseabove,with strike K = 375?

General one period risk neutral measure

We define a finite set of outcomes Ω ≡ ω1, ω2, ..., ωn and anysubcollection of outcomes A ∈ F1 := 2Ω an event.

Furthermore, we define a probability measure P, not necessarily thephysical measure P to be risk neutral if

P[ω] > 0 ∀ ω ∈ ΩX0 = 1

1+r E[X1]

for all strategies X .

The measure is indifferent to investing in a zero-coupon bond, or arisky asset X

The same initial capital X0 in both cases produces the same”‘average”’ return after one period.

Not the physical measure attached by observation, experts, etc..

In fact, physical measure has no impact on pricing

Example: Risk Neutral measure for trinomial case

Assume that Ω = ω1, ω2, ω3 with

S1(ω1) = uS0

S1(ω2) = S0

S1(ω3) = dS0

Given a payoff V1(ω) to replicate, are we assured that a replicatingstrategy exists?

Assume that Ω = ω1, ω2, ω3 with

S1(ω1) = uS0

S1(ω2) = S0

S1(ω3) = dS0

Given a payoff V1(ω) to replicate, are we assured that a replicatingstrategy exists?

Homework:

Try our first example with

(S0, u, d , r) = (400, 1.25.0.75, 0.05)

V digital1 (ω) = 1S1(ω)>450(ω).

Now, assume you are observe the price on the market to be

V digital0 =

1 + rE[V digital

1 ] = 0.25. (14)

Use this extra information to price a call option with strike K = 420.

Homework:

Try our first example with

(S0, u, d , r) = (400, 1.25.0.75, 0.05)

V digital1 (ω) = 1S1(ω)>450(ω).

Now, assume you are observe the price on the market to be

V digital0 =

1 + rE[V digital

1 ] = 0.25. (14)

Use this extra information to price a call option with strike K = 420.

Solution: Risk Neutral measure for trinomial case

The above scenario is reduced to finding the risk-neutral measure(p1, p2, p3). This can be done by finding the rref of the matrix M:

1 1 1 1500 400 300 420

1 0 0 0.25(1.05)

which results in

rref (M) =

1 0 0 0.26250 1 0 0.6750 0 1 0.0625

. (16)

It follows that (p1, p2, p3) = (0.2625, 0.675, 0.0625), and so

V C0 =

1.05E[(S1 − 420)+ | S0 = 400]

=0.2625

1.05× (500− 420) = 20.

Could we perhaps find a set of digital options as a basis setV d1

1 (ω),V d21 (ω),V d3

1 (ω)

= 1A1(ω), 1A2(ω), 1A3(ω) (18)

with A1,A2,A3 ∈ F1 to span all possible payoffs at time 1?

How about (A1,A2,A3) = (ω1 , ω2 , ω3) ?

It follows that (p1, p2, p3) = (0.2625, 0.675, 0.0625), and so

V C0 =

1.05E[(S1 − 420)+ | S0 = 400]

=0.2625

1.05× (500− 420) = 20.

Could we perhaps find a set of digital options as a basis setV d1

1 (ω),V d21 (ω),V d3

1 (ω)

= 1A1(ω), 1A2(ω), 1A3(ω) (18)

with A1,A2,A3 ∈ F1 to span all possible payoffs at time 1?

How about (A1,A2,A3) = (ω1 , ω2 , ω3) ?

Exchange one stock for another

Assume now an economy with two stocks, X and Y . Assume that

(X0,Y0, r) = (100, 100, 0.01) (19)

(X1(ω),Y1(ω)) =

(110, 105) : ω = ω1

(100, 100) : ω = ω2

(80, 95) : ω = ω3.

Consider two contracts, V and W , with payoffs

V1(ω) = max Y1(ω)− X1(ω), 0W1(ω) = Y1(ω)− X1(ω).

Price V0 and W0.

Exchange one stock for another

In this case, our matrix M is such that

1 1 1 1110 100 80 101105 100 95 101

which results in

rref (M) =

1 0 0 310

0 1 0 610

0 0 1 110

. (22)

It follows that

W0 =E[Y1]− E[X1]

1.01= Y0 − X0 = 0

1.01· (15p3) =

1.01= 1.49.

Existence of Risk Neutral measure

Let P be a probability measure on a finite space Ω. The following areequivalent:

P is a risk neutral measure

For all traded securities S i , S i0 = 1

1+r E[S i

]Proof: Homework (Hint: One direction is much easier than others. Also,strategies are linear in the underlying asset.)

1+r E[S i

Proof: Homework (Hint: One direction is much easier than others. Also,strategies are linear in the underlying asset.)

1+r E[S i

Complete Markets

A market is complete if it is arbitrage free and every non-traded asset canbe replicated.

Fundamental Theorem of Asset Pricing 1: A market is arbitrage freeiff there exists a risk neutral measure

Fundamental Theorem of Asset Pricing 2: A market is complete iffthere exists exactly one risk neutral measure

Proof(s): We will go over these in detail later!

Complete Markets

Optimal Investment for a Strictly Risk Averse Investor

Assume a complete market, with a unique risk-neutral measure P.

Characterize an investor by her pair (x ,U) of initial capital x ∈ X andutility function U : X → R+.

Assume U ′(x) > 0.

Assume U ′′(x) < 0.

Define the Radon-Nikodym derivative of P to P as the randomvariable

Z (ω) :=P(ω)

P(ω). (24)

Note that Z is used to map expectations under P to expectationsunder P: For any random variable X , it follows that

E[X ] = E[ZX ]. (25)

A strictly risk-averse investor now wishes to maximize her expected utilityof a portfolio at time 1, given initial capital at time 0:

u(x) := maxX1∈Ax

E[U(X1)]

Ax := all portfolio values at time 1 with initial capital x .(26)

Theorem

Define X1 via the relationship

U ′(

):= λZ (27)

where λ sets X1 as a strategy with an average return of r under P:

E[X1] = x(1 + r). (28)

Then X1 is the optimal strategy.

Proof.

Assume X1 to be an arbitrary strategy with initial capital x . Then for

f (y) := E[U(yX1 + (1− y)X1)] (29)

it follows that

f ′(0) = E[U ′(X1)

(X1 − X1

X1 − X1

)]= λE

[(X1 − X1

f ′′(y) = E[

U ′′(yX1 + (1− y)X1)(

X1 − X1

)2]< 0

and so f attains its maximum at y = 0. We conclude thatE[U(X1)] < E[U(X1)] for any admissible strategy X1.

Optimal Investment: Example

Assume an investor and economy defined by

U(x) = ln (x)

(S0, u, d , p, q, r) = (400, 1.25.0.75, 0.5, 0.5, 0.05).(31)

It follows that

(p, q) =

)(Z (H),Z (T )) =

Since U ′(x) = 1x , we have

X1(ω) =1

Z (ω)

x = X0 =1

1 + rE[X1] =

(p · 1

p+ q · 1

1 + r.

Combining the previous results, we see that

X1(ω) =x(1 + r)

Z (ω)

u(x) = p ln X1(H) + (1− p) ln X1(T )

= p ln

(x(1 + r)

)+ (1− p) ln

(x(1 + r)

Z (T )

((1 + r)

Z (H)pZ (T )1−p x

)= ln (1.0717x) > ln (1.05x).

In terms of her actual strategy, we see that

π0 :=∆0S0

X1(H)− X1(T )

S1(H)− S1(T )=

u − d

Z (H)− 1

Z (T )

u − d

p− 1− p

1− p

6− 5

)= −0.875.

Therefore, the optimal strategy is to sell a stock portfolio worth 87.5% ofher initial wealth x and invest the proceeds into a safe bank account.

In fact, since π0 = 1+ru−d

(pp −

1−p1−p

), we see that qualitatively, her optimal

strategy involves

> 0 : p > p= 0 : p = p< 0 : p < p.

This links with her strategy via

1 + r1(ω) :=X1(ω)

X0= (1− π0)(1 + r) + π0

S1(ω)

S0(36)

and so for our specific case where (r , u, d , π0) = (0.05, 1.25, 0.75,−0.875),we have

1 + r1(H) = (1− π0)(1 + r) + π0u = 0.875

1 + r1(T ) = (1− π0)(1 + r) + π0d = 1.3125.(37)

Optimal Investment: U(x) =√x

Consider now the same set-up as before, only that the utility functionchanges to U(x) =

√x .

It follows that

U ′(X1) =1

1√X1

⇒ X1 =1

Solving for λ returns

x(1 + r) = E[X1]

= E[Z X1]

4λ2E[

Optimal Investment: U(x) =√x

Combining the results above, we see that

X1 =x(1 + r)

]⇒ u(x) = E

[√X1

[√x(1 + r)

x(1 + r)

Question: Is it true for all (p, p) ∈ (0, 1)× (0, 1) that√E[

(1− p)2

1− p≥ 1? (41)

Optimal Betting at the Omega Horse Track!

Imagine our investor with U(x) = ln (x) visits a horse track.

There are three horses: ω1, ω2 and ω3.

She can bet on any of the horses to come in 1st .

The payoff is 1 per whole bet made.

She observes the price of each bet with payoff 1 right before the raceto be

(B10 ,B

30 ) = (0.5, 0.3, 0.2). (42)

Symbolically,B i

1(ω) = 1ωi(ω). (43)

Our investor feels the physical probabilities of each horse winning is

(p1, p2, p3) = (0.6, 0.35, 0.05). (44)

How should she bet if the race is about to start?

In this setting, we can assume r = 0.

This directly implies that

(p1, p2, p3) = (0.5, 0.3, 0.2). (45)

Our Radon-Nikodym derivative of P to P is now

(Z (ω1),Z (ω2),Z (ω3)) =

). (46)

Her optimal strategy X1 reflects her betting strategy, and satisfies

X1(ω) =X0

Z (ω)

(X1(ω1)

X1(ω2)

X3(ω1)

So, per dollar of wealth, she buys 65 of a bet for Horse 1 to win, 7

6 ofa bet for Horse 2 to win, and 1

4 of a bet for Horse 3 to win.

The total price (per dollar of wealth) is thus

5· 0.5 +

6· 0.3 +

4· 0.2 = 0.6 + 0.35 + 0.05 = 1. (48)

Dividends

What about dividends? How do they affect the risk neutral pricing ofexchange and non-exchange traded assets? What if they are paid atdiscrete times? Continuously paid?

Recall that if dividends are paid continuously at rate δ, then 1 share attime 0 will accumulate to eδT shares upon reinvestment of dividends intothe stock until time T .

It follows that to deliver one share of stock S with initial price S0 at timeT , only e−δT shares are needed. Correspondingly,

Fprepaid = e−δTS0

F = erT e−δTS0 = e(r−δ)TS0.(49)

Dividends

Binomial Option Pricing w/ cts Dividends and Interest

Over a period of length h, interest increases the value of a bond by afactor erh and dividends the value of a stock by a factor of eδh.

Once again, we compute pathwise,

V1(H) = (X0 −∆0S0)erh + ∆0eδhuS0

V1(T ) = (X0 −∆0S0)erh + ∆0eδhdS0

and this results in the modified quantities

∆0 = e−δhV1(H)− V1(T )

(u − d)S0

p =e(r−δ)h − d

u − d

q =u − e(r−δ)h

u − d

Binomial Option Pricing w/ cts Dividends and Interest

Over a period of length h, interest increases the value of a bond by afactor erh and dividends the value of a stock by a factor of eδh.

Once again, we compute pathwise,

V1(H) = (X0 −∆0S0)erh + ∆0eδhuS0

V1(T ) = (X0 −∆0S0)erh + ∆0eδhdS0

and this results in the modified quantities

∆0 = e−δhV1(H)− V1(T )

(u − d)S0

p =e(r−δ)h − d

u − d

q =u − e(r−δ)h

u − dAlbert Cohen (MSU) STT 888: Stochastic Calculus for Finance MSU 2016 48 / 249

Binomial Models w/ cts Dividends and Interest

For σ, the annualized standard deviation of continuously compoundedstock return, the following models hold:

Futures - Cox (1979)

u = eσ√h

d = e−σ√h.

General Stock Model

u = e(r−δ)h+σ√h

d = e(r−δ)h−σ√h.

Currencies with rf the foreign interest rate, which acts as a dividend:

u = e(r−rf )h+σ√h

d = e(r−rf )h−σ√h.

1- and 2-period pricing

Consider the case r = 0.10, δ = 0.05, h = 0.01, σ = 0.1,S0 = 10 = K .

Now price two digital options, using the

1 General Stock Model

2 Futures-Cox Model

with respective payoffs

V K1 (ω) := 1S1≥K(ω)

V K2 (ω) := 1S2≥K(ω).

Consider the case r = 0.10, δ = 0.05, h = 0.01, σ = 0.1,S0 = 10 = K .

Now price two digital options, using the

1 General Stock Model

2 Futures-Cox Model

with respective payoffs

V K1 (ω) := 1S1≥K(ω)

V K2 (ω) := 1S2≥K(ω).

Calibration Exercise

Assume table below of realized gains & losses over a ten-period cycle.

Use the adjusted values (r , δ, h,S0,K ) = (0.02, 0, 0.10, 10, 10).

Calculate binary options from last slide using these assumptions.

Period Return

1 S1S0

= 1.05

2 S2S1

= 1.02

3 S3S2

= 0.98

4 S4S3

= 1.01

5 S5S4

= 1.02

6 S6S5

= 0.99

7 S7S6

= 1.03

8 S8S7

= 1.05

9 S9S8

= 0.96

10 S10S9

= 0.97

Calibration Exercise: Linear Approximation

We would like to compute σ for the logarithm of returns ln(

SiSi−1

Assume the returns per period are all independent.

Q: Can we use a linear (simple) return model instead of a compoundreturn model as an approximation?

If so, then for our observed simple return rate values:

Calculate the sample variance σ2∗.

Estimate that σ ≈ σ∗√h

Note that if SiSi−1

= 1 + γ for γ 1, then

Si−1

)≈ γ =

Si − Si−1

Si−1. (50)

Approximation: Convert our previous table, using simple interest.

Over small time periods h, define linear return values for i th period:

Xih :=Si − Si−1

Si−1. (51)

In other words, for simple rate of return Xi for period i :

Si = Si−1 · (1 + Xih). (52)

Our returns table now looks like

Period Return

1 S1−S0S0

= 0.05

2 S2−S1S1

= 0.02

3 S3−S2S2

= −0.02

4 S4−S3S3

= 0.01

5 S5−S4S4

= 0.02

6 S6−S5S5

= −0.01

7 S7−S6S6

= 0.03

8 S8−S7S7

= 0.05

9 S9−S8S8

= −0.04

10 S10−S9S9

= −0.03

sample standard deviation σ∗ = 0.0319estimated return deviation σ ≈ 0.1001

We estimate, therefore, that under the Futures-Cox model

(u, d) = (e0.0319, e−0.0319) = (1.0324, 0.9686)

p =e0.002 − e−0.0319

e0.0319 − e−0.0319= 0.5234.

For the one-period digital option:

V0 = e−rhE0[1S1≥10] = e−0.002 · p = 0.5224. (54)

For the two-period digital option:

V0 = e−2rhE0[1S2≥10] = e−0.004 ·[p2 + 2pq

]= 0.7698. (55)

V0 = e−rhE0[1S1≥10] = e−0.002 · p = 0.5224. (54)

V0 = e−2rhE0[1S2≥10] = e−0.004 ·[p2 + 2pq

]= 0.7698. (55)

Calibration Exercise: No Approximation

Without the linear approximation, we can directly estimate

σY√

h = 0.03172

(u, d) = (e0.03172, e−0.03172) = (1.0322, 0.9688)

p =e0.002 − e−0.03172

e0.03172 − e−0.03172= 0.5246.

V0 = e−rhE0[1S1≥10] = e−0.002 · p = 0.5236. (57)

V0 = e−2rhE0[1S2≥10] = e−0.004 ·[p2 + 2pq

]= 0.7721. (58)

Calibration Exercise: No Approximation

Without the linear approximation, we can directly estimate

σY√

h = 0.03172

(u, d) = (e0.03172, e−0.03172) = (1.0322, 0.9688)

p =e0.002 − e−0.03172

e0.03172 − e−0.03172= 0.5246.

V0 = e−rhE0[1S1≥10] = e−0.002 · p = 0.5236. (57)

V0 = e−2rhE0[1S2≥10] = e−0.004 ·[p2 + 2pq

]= 0.7721. (58)

We can solve for 2-period problems

on a case-by-case basis, or

by developing a general theory for multi-period asset pricing.

In the latter method, we need a general framework to carry out ourcomputations

Risk Neutral Pricing Formula

Assume now that we have the ”regular assumptions” on our coin flipspace, and that at time N we are asked to deliver a path dependentderivative value VN . Then for times 0 ≤ n ≤ N, the value of thisderivative is computed via

Vn = e−rhEn [Vn+1] (59)

and so

X0 = E0 [XN ]

Xn :=Vn

Vn = e−rhEn [Vn+1] (59)

and so

X0 = E0 [XN ]

Xn :=Vn

Vn = e−rhEn [Vn+1] (59)

and so

X0 = E0 [XN ]

Xn :=Vn

Vn = e−rhEn [Vn+1] (59)

and so

X0 = E0 [XN ]

Xn :=Vn

Computational Complexity

Consider the case

p = q =1

S0 = 4, u =4

3, d =

but now with term n = 3.

There are 23 = 8 paths to consider.

However, there are 3 + 1 = 4 unique final values of S3 to consider.

In the general term N, there would be 2N paths to generate SN , butonly N + 1 distinct values.

At any node n units of time into the asset’s evolution, there are n + 1distinct values.

Computational Complexity

At each value s for Sn, we know that Sn+1 = 43 s or Sn+1 = 3

Using multi-period risk-neutral pricing, we can generate forvn(s) := Vn(Sn(ω1, ..., ωn)) on the node (event) Sn(ω1, ..., ωn) = s:

vn(s) = e−rh[pvn+1

+ qvn+1

4s)]. (62)

An Example:

Assume r , δ, and h are such that

2= q, e−rh =

S0 = 4, u = 2, d =1

2V3 := max 10− S3, 0 .

It follows thatv3(32) = 0

v3(8) = 2

v3(2) = 8

v3(0.50) = 9.50.

Compute V0.

Markov Processes

If we use the above approach for a more exotic option, say a lookbackoption that pays the maximum over the term of a stock, then we findthis approach lacking.

There is not enough information in the tree or the distinct values forS3 as stated. We need more.

Consider our general multi-period binomial model under P.

Definition We say that a process X is adapted if it depends only on thesequence of flips ω := (ω1, ..., ωn)

Definition We say that an adapted process X is Markov if for every0 ≤ n ≤ N − 1 and every function f (x) there exists another function g(x)such that

En [f (Xn+1)] = g(Xn). (65)

Markov Processes

This notion of Markovity is essential to our state-dependent pricingalgorithm.

Indeed, our stock process evolves from time n to time n + 1, usingonly the information in Sn.

We can in fact say that for every f (s) there exists a g(s) such that

g(s) = En [f (Sn+1) | Sn = s] . (66)

In fact, that g depends on f :

g(s) = e−rh[pf(4

+ qf(3

4s)]. (67)

So, for any f (s) := VN(s), we can work our recursive algorithmbackwards to find the gn(s) := Vn(s) for all 0 ≤ n ≤ N − 1

Markov Processes

Some more thoughts on Markovity:

Consider the example of a Lookback Option.

Here, the payoff is dependent on the realized maximumMn := max0≤i≤nSi of the asset.

Mn is not Markov by itself, but the two-factor process (Mn, Sn) is.Why?

Let’s generate the tree!

Homework Can you think of any other processes that are not Markov?

Call Options on Zero-Coupon Bonds

Assume an economy where

One period is one year

The one year short term interest rate from time n to time n + 1 is rn.

The rate evolves via a stochastic process:

r0 = 0.02

rn+1 = Xrn

P[X = 2k ] =1

3for k ∈ −1, 0, 1 .

Consider now a zero-coupon bond that matures in 3−years withcommon face and redemption value F = 100.

Also consider a call option on this bond that expires in 2−years withstrike K = 97.

Denote Bn and Cn as the bond and call option values, respectively.

Note that we iterate backwards from the values

B3(r) = 100

C2(r) = max B2(r)− 97, 0 .(69)

Compute (B0,C0).

Our general recursive formula is

Bn(r) =1

1 + rE[Bn+1(rn+1) | rn = r ]

Cn(r) =1

1 + rE[Cn+1(rn+1) | rn = r ].

Iterating backwards, we see that at t = 2,

B2(r) =1

1∑k=−1

B3(2k r). (71)

At time t = 2, we have that

r2 ∈ 0.08, 0.04, 0.02, 0.01, 0.005 . (72)

Our associated Bond and Call Option values at time 2:

r2 B2 C2

0.08 92.59 00.04 96.15 00.02 98.04 1.040.01 99.01 2.01

0.005 99.50 2.50

r1 B1 C1

0.04 91.92 0.330.02 95.82 1.000.01 97.87 1.83

r0 B0 C0

0.02 93.34 1.03

Question: What if the delivery time of the option is changed to 3?Symbolically, what if

C3(r) = max B3(r)− 97, 0? (73)

Capital Structure Model

As an analyst for an investments firm, you are tasked with advisingwhether a company’s stock and/or bonds are over/under-priced.

You receive a quarterly report from this company on it’s return onassets, and have compiled a table for the last ten quarters below.

Today, just after the last quarter’s report was issued, you see that inbillions of USD, the value of the company’s assets is 10.

There are presently one billions shares of this company that are beingtraded.

The company does not pay any dividends.

Six months from now, the company is required to pay off a billionzero-coupon bonds. Each bond has a face value of 9.5.

Assume Miller-Modigliani holds with At = Bt + St , where the assetsof a company equal the sum of its share and bond price.

Presently, the market values are (B0, S0, r) = (9, 1, 0.02).

The Merton model for corporate bond pricing asserts that atredemption time T ,

Bt = e−r(T−t)E [min AT ,F]St = e−r(T−t)E [max AT − F , 0] .

With all of this information, your job now is to issue a Buy or Sell onthe stock and the bond issued by this company.

Table of return on assets for Company X, with h = 0.25.

Period Return on Assets

1 A1−A0A0

= 0.05

2 A2−A1A1

= 0.02

3 A3−A2A2

= −0.02

4 A4−A3A3

= 0.01

5 A5−A4A4

= 0.02

6 A6−A5A5

= −0.01

7 A7−A6A6

= 0.03

8 A8−A7A7

= 0.05

9 A9−A8A8

= −0.04

10 A10−A9A9

= −0.03

sample standard deviation σ∗ = 0.0319estimated return deviation σ ≈ 0.0638

We scale all of our calculation in terms of billions ($, shares, bonds).

Using the Futures- Cox model, we have

(u, d) = (e0.0319, e−0.0319) = (1.0324, 0.9686)

p =e0.005 − e−0.0319

e0.0319 − e−0.0319= 0.5706.

Using this model, the only time the payoff of the bond is less than theface is on the path ω = TT .

The price of the bond and stock are thus modeled to be

B0 = e−0.02·(2·0.25)[p2 · 9.5 + 2pq · 9.5 + q2 · 9.38

]= 9.38 > 9.00

S0 = 10− 9.38 = 0.62 < 1.00.

It follows that,according to our model, one should Buy the bond as itis underpriced and one should Sell the stock as it is overpriced.

The Interview Process

Consider the following scenario:

After graduating, you go on the job market, and have 4 possible jobinterviews with 4 different companies.

So sure of your prospects that you know that each company will makean offer, with an identically, independently distributed probabilityattached to the 4 possible salary offers:

P [Salary Offer=50, 000] = 0.1

P [Salary Offer=100, 000] = 0.2.

Questions:

How should you interview?

Specifically, when should you accept an offer and cancel theremaining interviews?

How does your strategy change if you can interview as many times asyou like, but the distribution of offers remains the same as above?

Questions:

The Interview Process: Strategy

Some more thoughts...

At any time the student will know only one offer, which she can eitheraccept or reject.

Of course, if the student rejects the first three offers, than she has toaccept the last one.

So, compute the maximal expected salary for the student after thegraduation and the corresponding optimal strategy.

The Interview Process: Optimal Strategy

The solution process Xk4k=1 follows an Optimal Stopping Strategy:

Xk(s) = max

s, E[Xk+1 | kth offer = s

]. (78)

At time 4, the value of this game is X4(s) = s, with s being the salaryoffered.

At time 3, the conditional expected value of this game is

E[X4 | 3rd offer = s

]= E[X4]

= 0.1× 50, 000 + 0.3× 70, 000

+ 0.4× 80, 000 + 0.2× 100, 000

= 78, 000.

Hence, one should accept an offer of 80, 000 or 100, 000, and rejectthe other two.

This strategy leads to a valuation:

X3(50, 000) = 78, 000

X3(70, 000) = 78, 000

X3(80, 000) = 80, 000

X3(100, 000) = 100, 000.

At time 2, similar reasoning using E2[X3] leads to the valuation

X2(50, 000) = 83, 200

X2(70, 000) = 83, 200

X2(80, 000) = 83, 200

X2(100, 000) = 100, 000.

At time 1,X1(50, 000) = 86, 560

X1(70, 000) = 86, 560

X1(80, 000) = 86, 560

X1(100, 000) = 100, 000.

Finally, at time 0, the value of this optimal strategy is

E0[X1] = E[X1] = 0.8× 86, 560 + 0.2× 100, 000 = 89, 248. (83)

So, the optimal strategy is, for the first two interviews, accept only anoffer of 100, 000. If after the third interview, and offer of 80, 000 or100, 000 is made, then accept. Otherwise continue to the last interviewwhere you should accept whatever is offered.

Review

Let’s review the basic contracts we can write:

Forward Contract Initial Value is 0, because both buyer and sellermay have to pay a balance at maturity.

(European) Put/Call Option Initial Value is > 0, because both onlyseller must pay balance at maturity.

(European) ”Exotic” Option Initial Value is > 0, because both onlyseller must pay balance at maturity.

During the term of the contract, can the value of the contract ever fallbelow the intrinsic value of the payoff? Symbolically, does it ever occurthat

vn(s) < g(s) (84)

where g(s) is of the form of g(S) := max S − K , 0, in the case of a Calloption, for example.

Review

vn(s) < g(s) (84)

Review

vn(s) < g(s) (84)

Review

vn(s) < g(s) (84)

Review

vn(s) < g(s) (84)

Review

vn(s) < g(s) (84)

Review

vn(s) < g(s) (84)

Early Exercise

If σ = 0, and so uncertainty vanishes, then an investor would seek toexercise early if

rK > δS . (85)

If σ > 0, then the situation involves deeper analysis.

Whether solving a free boundary problem or analyzing a binomialtree, it is likely that a computer will be involved in helping theinvestor to determine the optimal exercise time.

For Freedom! (we must charge extra...)

What happens if we write a contract that allows the purchaser to exercisethe contract whenever she feels it to be in her advantage? By allowing thisextra freedom, we must

Charge more than we would for a European contract that is exercisedonly at the term N.

Hedge our replicating strategy X differently, to allow for thepossibility of early exercise.

American Options

In the end, the option v is valued after the nth value of the stockSn(ω) = s is revealed via the recursive formula along each path ω:

vn(Sn(ω)) = max

g(Sn(ω)), e−rhE[v(Sn+1(ω)) | Sn(ω)

]τ∗(ω) = inf k ∈ 0, 1, ..,N | vk(Sk(ω)) = g(Sk(ω)) .

Here, τ∗ is the optimal exercise time.

In the Binomial case, we reduce to

vn(s) = max

g(s), e−rh [pvn+1(us) + qvn+1(ds)]

τ∗ = inf k | vk(s) = g(s) .(87)

American Options

Some examples:

”American Bond:” g(s) = 1

”American Digital Option:” g(s) = 16≤s≤10

”American Square Option:” g(s) = s2.

Does an investor exercise any of these options early? Consider again thesetting

2= q, e−rh =

S0 = 4, u = 2, d =1

American Square Options

Consider the American Square Option. We know via Jensen’s Inequalitythat

e−rhE[g(Sn+1) | Sn

]= e−rhE

[S2n+1 | Sn

]≥ e−rh

(E[Sn+1 | Sn

]2)= e−rh

(erhSn

= erhS2n > S2

n = g(Sn).

American Square Options

It follows that vN(s) = s2 and

vN−1(s) = max

g(s), e−rhE[vN(SN) | SN−1 = s]

s2, e−rhE[S2N | SN−1 = s]

= e−rhE[S2

N | SN−1 = s]

= e−rhE[vN(SN) | SN−1 = s].

In words, with one period to go, don’t exercise yet!!

The American and European option values coincide. Keep going.

How about with two periods left before expiration?

American Options

Let’s return to the previous European Put example, where

2= q, e−rh =

S0 = 4, u = 2, d =1

2V3 := max 10− S3, 0 .

It follows that S3(ω) ∈

12 , 2, 8, 32

Use this to compute v3(s) and the American Put recursively.

Matching Interest Rates to Market Conditions

Consider again a series of coin flips (ω1, ..., ωn) where at time n, theinterest rate from n to n + 1 is modeled via

rn = rn(ω1, ..., ωn) (92)

and a stochastic volatility σ at time n via

σn =1

(rn(ω1, ..., ωn−1, ωn = H)

rn(ω1, ..., ωn−1, ωn = T )

). (93)

Keep in mind that we will build a recombining binomial tree for thismodel. So, for example,

r2(H,T ) = r2(T ,H). (94)

Futhermore, we can define the yield rate y(t,T , r(t)) for a zero-couponbond B(t,T , r(t)) via

B(t,T , r(t)) =1

(1 + y(t,T , r(t)))T−t(95)

and the corresponding yield rate volatility

σn =1

(y(1, n, r1(H))

y(1, n, r1(T ))

). (96)

We can in fact see that

y(1, 2, r1(H)) = r1(H)

y(1, 2, r1(T )) = r1(T ).(97)

But, for example, it is not clear how to obtain

(y(1, 3, r1(H)), y(1, 3, r1(T ))) . (98)

Furthermore, how can we match to market conditions and update ourestimates for interest rates rn?

One way forward is the Black Derman Toy Model 1

Consider a market with the following observations:

Maturity Yield to Maturity y(0,T , r) Yield Volatility σT1 0.100 N.A.2 0.110 0.1903 0.120 0.1804 0.125 0.1505 0.130 0.140

(For Daily US Treasury Real Yield Curve Rates click here )

1Black, Fischer, Emanuel Derman, and William Toy. ”A one-factor modelof interest rates and its application to treasury bond options.” FinancialAnalysts Journal 46.1 (1990): 33-39.

In this setting, we match market conditions to a model where at eachtime, an interest rate moves up or down with a ”risk-neutral”probability of 1

It follows that we have from time t = 0 to t = 1, with an initial rater0 = r ,

1 + y(0, 1, r)=

⇒ y(0, 1, r) = r .

From time t = 0 to t = 2, again with our initial rate r0 = r ,

Connecting our observed two-year yield with yearly interest ratesreturns

(1 + y(0, 2, r))2=

1 + r1(H)+

1 + r1(T )

)⇒ 1

1.112=

1 + r1(H)+

1 + r1(T )

Also, connecting our one-year yields with yearly interest rates leads to

σ1 =1

(r1(H)

r1(T )

(y(1, 2, r1(H))

y(1, 2, r1(T ))

)= σ2 = 0.190.

Solution leads to the pair

(r1(H), r1(T )) = (0.1432, 0.0979). (102)

From time t = 0 to t = 3, again with our initial rate r0 = r , we try toestimate the matching (r2(H,H), r2(H,T ), r2(T ,T )).

Connecting our observed two-year yield with yearly interest ratesreturns

(1 + y(0, 3, r))3=

1 + r1(H)

1 + r2(H,H)

1 + r1(H)

1 + r2(H,T )

1 + r1(T )

1 + r2(T ,H)

1 + r1(T )

1 + r2(T ,T )

We can now substitute our values(r , r1(H), r1(T ), y(0, 3, r)) = (0.10, 0.1432, 0.0979, 0.120).

We also know that σ2 6= σ2(ω1, ω2) and so

σ2 =1

(r2(H,H)

r2(H,T )

)σ2 =

(r2(T ,H)

r2(T ,T )

)⇒ r2(H,T ) = r2(T ,H) =

√r2(H,H)r2(T ,T ).

Finally, we have one more matching condition via

0.180 = σ3 =1

(y(1, 3, r1(H))

y(1, 3, r1(T ))

)(105)

(1 + y(1, 3, r1(H))2=

1 + r1(H)

1 + r2(H,H)+

1 + r2(H,T )

(1 + y(1, 3, r1(T ))2=

1 + r1(T )

1 + r2(T ,H)+

1 + r2(T ,T )

(106)Now solve for (r2(H,H), r2(H,T ), r2(T ,T ))!!HW1: Price a bond that matures in two years, with the aboveobservations for term structure, and with F = 100 and coupon rate 10%.HW2: What about r3, r4, r5? Can we compute them?

Pricing a Two-year Coupon Bond Using Market Conditions

Consider the previous observations for term structure, and a bond withF = 100 and coupon rate 10%. In this setting, we have

Time Interest Rate Value

0 r0 0.1001 r1(H) 0.14321 r1(T ) 0.09792 r2(H,H) 0.19422 r2(H,T ) 0.13772 r2(T ,H) 0.13772 r2(T ,T ) 0.0976

We can decompose the two-year coupon bond into two componentzero-coupon bonds.

The first is B(1), which has face 10, maturity T = 1, and initial priceB(1)(0,T , r) at t = 0.

The first is B(2), which has face 110, maturity T = 2, and initial priceB(2)(0,T , r) at t = 0.

The total coupon bond price is thus

B(0, 2, r) = B(1)(0, 1, r) + B(2)(0, 2, r). (107)

Our component bonds thus have prices

Time Interest Rate B1(t,T , r)

0 r0 = 0.10 B(1)(0, 1, r0) = 9.09

1 r1(H) = 0.1432 B(1)(1, 1, r1(H)) = 10

1 r1(T ) = 0.0979 B(1)(1, 1, r1(T )) = 10

Time Interest Rate B2(t,T , r)

0 r0 = 0.10 B(2)(0, 2, r0) = 89.28

1 r1(H) = 0.1432 B(2)(1, 2, r1(H)) = 96.22

1 r1(T ) = 0.0979 B(2)(1, 2, r1(T )) = 100.19

2 r2(H,H) = 0.1942 B(2)(2, 2, r2(H,H)) = 110

2 r2(H,T ) = 0.1377 B(2)(2, 2, r2(H,T )) = 110

2 r2(T ,H) = 0.1377 B(2)(2, 2, r2(T ,H)) = 110

2 r2(T ,T ) = 0.0976 B(2)(2, 2, r2(T ,T )) = 110

Finally, we have our coupon bond with price

Time Interest Rate B(t,T , r)

0 r0 = 0.10 B(0, 2, r0) = 98.371 r1(H) = 0.1432 B(1, 2, r1(H)) = 106.221 r1(T ) = 0.0979 B(1, 2, r1(T )) = 110.192 r2(H,H) = 0.1942 B(2, 2, r2(H,H)) = 1102 r2(H,T ) = 0.1377 B(2, 2, r2(H,T )) = 1102 r2(T ,H) = 0.1377 B(2, 2, r2(T ,H)) = 1102 r2(T ,T ) = 0.0976 B(2, 2, r2(T ,T )) = 110

Note: Note that with our coupons, the yield yc(0, 2, r) = 0.1095, which isobtained by solving

98.37 =10

1 + yc(0, 2, r)+

(1 + yc(0, 2, r))2. (108)

Pricing a 1-year Call Option on our 2-year Coupon Bond

Consider the previous term structure, and a European Call Option on thetwo year bond with K = 97 and expiry of 1 year.

Compute the initial Call price C0(r).

Compute the initial number of bonds to hold to replicate this option.

Pricing a 1-year Call Option on our 2-year Coupon Bond

In this case, we look at the price of the bond minus the accrued interest:

Time Interest Rate B(t,T , r)

0 r0 = 0.10 B(0, 2, r0) = 98.371 r1(H) = 0.1432 B(1, 2, r1(H)) = 96.221 r1(T ) = 0.0979 B(1, 2, r1(T )) = 100.192 r2(H,H) = 0.1942 B(2, 2, r2(H,H)) = 1002 r2(H,T ) = 0.1377 B(2, 2, r2(H,T )) = 1002 r2(T ,H) = 0.1377 B(2, 2, r2(T ,H)) = 1002 r2(T ,T ) = 0.0976 B(2, 2, r2(T ,T )) = 100

Pricing and hedging is accomplished via

C0(0.10) =1

2· 0 +

)= 1.45

∆0(0.10) =0− 3.19

96.22− 100.19= 0.804.

Asian Options

In times of high volatility or frequent trading, a company may want toprotect against large price movements over an entire time period, using anaverage. For example, if a company is looking at foreign exchange marketsor markets that may be subject to stock pinning due to large actors.

As an input, the average of an asset is used as an input against a strike,instead of the spot price.

There are two possibilities for the input in the discrete case: h = TN and

Arithmetic Average: IA(T ) := 1N

∑Nk=1 Skh .

Geometric Average: IG (T ) :=(

ΠNk=1Skh

HW: Is there an ordering for IA, IG that is independent of T ?

Asian Options

∑Nk=1 Skh .

ΠNk=1Skh

Asian Options

∑Nk=1 Skh .

ΠNk=1Skh

Asian Options

∑Nk=1 Skh .

ΠNk=1Skh

Asian Options: An Example:

Notice that these are path-dependent options, unlike the put and calloptions that we have studied until now. Assume r , δ, and h are such that

S0 = 4, u = 2, d =1

2, e−rh =

10g(I ) = max I − 2.5, 0 .

Consider an arithmetic average with N = 2. Then

v2(HH) = max

8 + 16

2− 2.5, 0

v2(HT ) = max

2− 2.5, 0

v2(TH) = max

2− 2.5, 0

v2(TT ) = max

2− 2.5, 0

Compute v0, assuming a European structure. How about an Americanstructure?

v2(HH) = max

8 + 16

2− 2.5, 0

v2(HT ) = max

2− 2.5, 0

v2(TH) = max

2− 2.5, 0

v2(TT ) = max

2− 2.5, 0

Compute v0, assuming a European structure.

How about an Americanstructure?

v2(HH) = max

8 + 16

2− 2.5, 0

v2(HT ) = max

2− 2.5, 0

v2(TH) = max

2− 2.5, 0

v2(TT ) = max

2− 2.5, 0

Compute v0, assuming a European structure. How about an Americanstructure?

Lognormality

Analyzing returns, we assume:

A probability space(

Ω,F ,P).

Our asset St(ω) has an associated return over any period (t, t + u)defined as

rt,u(ω) := ln

(St+u(ω)

St(ω)

). (112)

Lognormality

Partition the interval [t,T ] into n intervals of length h = T−tn , then:

The return over the entire period can be taken as the sum of thereturns over each interval:

rt,T−t(ω) = ln

(ST (ω)

St(ω)

n∑k=1

rtk ,h(ω)

tk = t + kh.

We model the returns as being independent and possessing a binomialdistribution.

Employing the Central Limit Theorem, it can be shown that asn→∞, this distribution approaches normality.

Binomial Tree and Discrete Dividends

Another issue encountered in elementary credit and investment theoryis the case of different compounding and deposit periods.

This also occurs in the financial setting where a dividend is not paidcontinuously, but rather at specific times.

It follows that the dividend can be modeled as delivered in the middleof a binomial period, at time τ(ω) < T .

This view is due to Schroder and can be summarized as viewing theinherent value of St(ω) as the sum of a prepaid forward PF and thepresent value of the upcoming dividend payment D:

PFt(ω) = St(ω)− De−r(τ(ω)−t)

u = erh+σ√h

d = erh−σ√h.

Now, the random process that we model as having up and downmoves is PF instead of S .

Back to the Continuous Time Case

Consider the case of a security whose binomial evolution is modeled as anup or down movement at the end of each day. Over the period of oneyear, this amounts to a tree with depth 365. If the tree is not recombining,then this amounts to 2365 branches. Clearly, this is too large to evaluatereasonably, and so an alternative is sought.

Whatever the alternative, the concept of replication must hold. This isthe reasoning behind the famous Black-Scholes-Merton PDE approach.

Back to the Continuous Time Case

Consider the case of a security whose binomial evolution is modeled as anup or down movement at the end of each day. Over the period of oneyear, this amounts to a tree with depth 365. If the tree is not recombining,then this amounts to 2365 branches. Clearly, this is too large to evaluatereasonably, and so an alternative is sought.

Whatever the alternative, the concept of replication must hold. This isthe reasoning behind the famous Black-Scholes-Merton PDE approach.

Monte Carlo Techniques

Our model for asset evolution is

⇒ St = S0e(α−δ− 12σ2)t+σ

Z ∼ N(0, 1).(115)

Consider now the possibility of simulating the stock evolution bysimulating the random variable Z , or in fact an i.i.d. sequence

For a European option with time expiry T , we can simulate the expirytime payoff mulitple times:

S (i),T)

S (i))

S0e(α−δ− 12σ2)T+σ

√TZ (i)

). (116)

⇒ St = S0e(α−δ− 12σ2)t+σ

Z ∼ N(0, 1).(115)

S (i),T)

S (i))

√TZ (i)

). (116)

⇒ St = S0e(α−δ− 12σ2)t+σ

Z ∼ N(0, 1).(115)

S (i),T)

S (i))

√TZ (i)

). (116)

If we sample uniformly from our simulated values

S (i),T)n

can appeal to a sampling-convergence theorem with the appoximation

V (S , 0) = e−rT1

n∑i=1

S (i),T). (117)

The challenge now is to simulate our lognormally distributed assetevolution.

One can simulate the value ST directly by one random variable Z , or amultiple of them to simulate the path of the evolution until T .

The latter method is necessary for Asian options.

S (i),T)n

V (S , 0) = e−rT1

n∑i=1

S (i),T). (117)

S (i),T)n

V (S , 0) = e−rT1

n∑i=1

S (i),T). (117)

S (i),T)n

V (S , 0) = e−rT1

n∑i=1

S (i),T). (117)

There are multiple ways to simulate Z . One way is to find a randomnumber U taken from a uniform distribution U[0, 1].

It follows that one can now map U → Z via inversion of the Nornal cdf N:

Z = N−1(U). (118)

It can be shown that in a sample, the standard deviation of the sampleaverage σsample is related to the standard deviation of an individual drawvia

σsample =σdraw√

n. (119)

If σdraw = σ, then we can see that we must increase our sample size by22k if we wish to cut our σsample by a factor of 2k .

Z = N−1(U). (118)

σsample =σdraw√

n. (119)

Z = N−1(U). (118)

σsample =σdraw√

n. (119)

Brownian Motion

Consider a probability space (Ω,F ,P) and a process (Wt ,Ft) that lives onit, where Ft represents all the information about Wu0≤u≤t . Assumethat our pair satisfies, for s, t ≥ 0 and Aini=1 ⊆ F

P[W0 = 0] = 1

P[Wt ∈ dx ] = 1√2πt

e−x2

P[limt→s Wt = Ws ] = 1

P[Wt+s −Ws ∈ A | Fs ] = P[Wt ∈ A] for all A ∈ FP[∩ni=1

Wti −Wti−1 ∈ Ai

] = Πn

i=1P[Wti −Wti−1 ∈ Ai ]

What’s so hard about that? Take a Z ∼ N(0, 1) and define

Xt =√

tZ (120)

Brownian Motion

P[W0 = 0] = 1

P[Wt ∈ dx ] = 1√2πt

e−x2

] = Πn

Xt =√

tZ (120)

Brownian Motion

P[W0 = 0] = 1

P[Wt ∈ dx ] = 1√2πt

e−x2

] = Πn

Xt =√

tZ (120)

Brownian Motion

P[W0 = 0] = 1

P[Wt ∈ dx ] = 1√2πt

e−x2

] = Πn

Xt =√

tZ (120)

Quadratic Variation

Clearly, the last quantity, also known as independent increments, is whatmakes Brownian motion truly special. We can use this property to defineother, related properties. The first is the notion of quadratic variation.Simply put,

Wt+∆t −Wt ∼W∆t (121)

and so, for an i.i.d. N(0, 1) sequence Zini=1

n∑j=1

(Wtj+1 −Wtj

)2 ∼n∑

(Wtj+1−tj

∼n∑

(√tj+1 − tjZj

Quadratic Variation

Assuming we have an even partition of [0,T ], with tj+1 − tj = Tn , then

n∑j=1

(Wtj+1 −Wtj

)2 ∼ T

n∑j=1

(Zj)2 = T · χ

n→ T . (123)

To see why, recall that, for γ = 1√1−2t

Mχ2n(t) =

(E[et·Z

E[et·Z2] =

1√2π

∫ ∞−∞

etx2e−

=1√2π

∫ ∞−∞

e− x2

2γ2 dx

1− 2t

⇒ Mχ2nn

(t) =1√

(1− 2 tn )n→ et = E[et·1].

Quadratic Variation

n∑j=1

(Wtj+1 −Wtj

)2 ∼ T

n∑j=1

(Zj)2 = T · χ

n→ T . (123)

Mχ2n(t) =

(E[et·Z

E[et·Z2] =

1√2π

∫ ∞−∞

etx2e−

2 dx =1√2π

∫ ∞−∞

e− x2

2γ2 dx

1− 2t

⇒ Mχ2nn

(t) =1√

(1− 2 tn )n→ et = E[et·1].

Quadratic Variation

n∑j=1

(Wtj+1 −Wtj

)2 ∼ T

n∑j=1

(Zj)2 = T · χ

n→ T . (123)

Mχ2n(t) =

(E[et·Z

E[et·Z2] =

1√2π

∫ ∞−∞

etx2e−

2 dx =1√2π

∫ ∞−∞

e− x2

2γ2 dx

1− 2t

⇒ Mχ2nn

(t) =1√

(1− 2 tn )n→ et = E[et·1].

Quadratic Variation

n∑j=1

(Wtj+1 −Wtj

)2 ∼ T

n∑j=1

(Zj)2 = T · χ

n→ T . (123)

Mχ2n(t) =

(E[et·Z

E[et·Z2] =

1√2π

∫ ∞−∞

etx2e−

2 dx =1√2π

∫ ∞−∞

e− x2

2γ2 dx

1− 2t

⇒ Mχ2nn

(t) =1√

(1− 2 tn )n→ et = E[et·1].

k th order Variation, where k > 2

n∑j=1

(Wtj+1 −Wtj

)k ∼ (T

n∑j=1

(Wtj+1 −Wtj

)k ∝ Tk2 · n

Quadratic Variation

We can use this to define integration against Brownian Motion. This isdefined as∫ T

0f (Wt , t)dWt = lim

n→∞

n∑i=1

f (Wjh, jh) (Wjh+h −Wjh) (126)

In general, we can define the stochastic differential equation

dXt = µ(Xt , t)dt + σ(Xt , t)dWt (127)

and the accompanying Ito Equation for a new, stochastic calculus basedon the relationship

dWt · dWt = dt. (128)

Black-Scholes-Merton Analysis

Independent of the model for evolution of underlying asset, pricingmust be arbitrage-free

Achieve this via replication of derivative up to end of term of contract

Mathematical tools we require are now more involved than linearalgebra; namely Ito Calculus and Partial Differential Equations

To enable this analysis, assume the following for the model

∃ (Ω,F ,P) for which Wt is a standard Brownian motion.

dSt = St (µdt + σdWt), a Geometric Brownian Motion models theasset evolution

For a function f (x , t) ∈ C 2,1(R2 × R

)and Yt ≡ f (Wt , t), Ito

Calculus gives us:

dYt =(ft(Wt , t) + 1

2 fxx(Wt , t))

dt + fx(Wt , t)dWt

Calculus gives us:

2 fxx(Wt , t))

dt + fx(Wt , t)dWt

Calculus gives us:

2 fxx(Wt , t))

dt + fx(Wt , t)dWt

Calculus gives us:

2 fxx(Wt , t))

dt + fx(Wt , t)dWt

Calculus gives us:

2 fxx(Wt , t))

dt + fx(Wt , t)dWt

Calculus gives us:

2 fxx(Wt , t))

dt + fx(Wt , t)dWt

Calculus gives us:

2 fxx(Wt , t))

dt + fx(Wt , t)dWt

Calculus gives us:

2 fxx(Wt , t))

dt + fx(Wt , t)dWt

Calculus gives us:

2 fxx(Wt , t))

dt + fx(Wt , t)dWt

In general, for Zt = g(Xt , t), with dXt = µ(Xt , t)dt + σ(Xt , t)dWt , wehave

dZt = gt(Xt , t)dt +1

2gxx(Xt , t)dXtdXt + gxdXt

(gt(Xt , t) + µ(Xt , t)gx(Xt , t) +

2σ(Xt , t)2gxx(Xt , t)

+ σ(Xt , t)gx(Xt , t)dWt

Using the above assumptions, we arrive at the conclusion that we mustconstruct a portfolio Xt that matches the value Vt of the derivative wewish to price at all times t ≤ T , where T is the term of the contract.

In general, for Zt = g(Xt , t), with dXt = µ(Xt , t)dt + σ(Xt , t)dWt , wehave

dZt = gt(Xt , t)dt +1

2gxx(Xt , t)dXtdXt + gxdXt

(gt(Xt , t) + µ(Xt , t)gx(Xt , t) +

2σ(Xt , t)2gxx(Xt , t)

+ σ(Xt , t)gx(Xt , t)dWt

Using the above assumptions, we arrive at the conclusion that we mustconstruct a portfolio Xt that matches the value Vt of the derivative wewish to price at all times t ≤ T , where T is the term of the contract.

We match

dXt = dVt

XT = VT = G (ST )(130)

where G (S) is the payoff of the contract at time T if the value of theunderlying asset ST = S .

We match

dXt = dVt

XT = VT = G (ST )(130)

We match

dXt = dVt

XT = VT = G (ST )(130)

To achieve this, we recognize that Xt = Vt ⇒ dXt = dVt and by theassumption V = V (St , t), Ito provides:

dXt = r (Xt −∆tSt) dt + ∆tdSt = r (Vt −∆tSt) dt + ∆tdSt

∂t(St , t) +

2σ2S2

∂S2(St , t)

∂S(St , t)dSt .

This is due to the fact that dSt = µStdt + σStdWt ⇒ (dSt)2 = σ2S2

To achieve this, we recognize that Xt = Vt ⇒ dXt = dVt and by theassumption V = V (St , t), Ito provides:

∂t(St , t) +

2σ2S2

∂S2(St , t)

∂S(St , t)dSt .

This is due to the fact that dSt = µStdt + σStdWt ⇒ (dSt)2 = σ2S2

Black Scholes PDE

Matching terms implies

∆t =∂V

∂S(S , t)

∂t(S , t) +

2σ2S2∂

∂S2(S , t) = r

(V (S , t)− S

∂S(S , t)

)V (S ,T ) = G (S)

This is the famous B-S-M PDE formulation for European option pricing,with payoff G (S). The question now - how do we solve it?!

Note:For dSt = µStdt + σStdWt , we have as the solution

St = Sue(µ− 12σ2)(t−u)+σ(Wt−Wu)

∼ Sue(µ− 12σ2)(t−u)+σWt−u

∼ Sue(µ− 12σ2)(t−u)+σ

√t−uZ .

Black Scholes PDE

Matching terms implies

∆t =∂V

∂S(S , t)

∂t(S , t) +

2σ2S2∂

∂S2(S , t) = r

(V (S , t)− S

∂S(S , t)

)V (S ,T ) = G (S)

This is the famous B-S-M PDE formulation for European option pricing,with payoff G (S). The question now - how do we solve it?!

Note:For dSt = µStdt + σStdWt , we have as the solution

St = Sue(µ− 12σ2)(t−u)+σ(Wt−Wu)

∼ Sue(µ− 12σ2)(t−u)+σWt−u

∼ Sue(µ− 12σ2)(t−u)+σ

√t−uZ .

Our First Arbitrage Opportunity

Consider two perfectly correlated assets X ,Y whose evolution is modeledby

dXt = µ1Xtdt + σ1XtdWt

dYt = µ2Ytdt + σ2YtdWt

σ1 6= σ2

Assume wlog that 1σ1> 1

σ2. Design a portfolio that consists of

going long 1σ1Xt

units of X

short 1σ2Yt

units of Y ,

borrowing 1σ1− 1

σ2at continuous rate r .

Our First Arbitrage Opportunity

Borrowing the amount at time t means that the total net output at time tis nothing, but the evolution of our portfolio is

σ1XtdXt −

σ2YtdYt −

σ1− 1

)rdt =

(µ1 − r

σ1− µ2 − r

)dt (135)

Unless the respective Sharpe Ratios are equivalent, ie unless

µ1 − r

σ1=µ2 − r

σ2(136)

one could make a deterministic profit with zero upfront capital.

Spring 2009 Q10

Consider two perfectly correlated, non-dividend paying assets X ,Y whoseevolution is modeled by

dXt = 0.08Xtdt + 0.2XtdWt

dYt = 0.0925Ytdt − 0.25YtdWt(137)

An investor wishes to synthesize a risk-free asset by allocating 1000between X and Y . How much should she initially invest in the X ?

Spring 2009 Q10

If she goes long 5Xt

units of X for every 4Yt

units of Y she goes long, thenher portfolio has a deterministic growth rate. Symbolically,

dPt = ∆Xt dXt + ∆Y

Xt(0.08Xtdt + 0.2XtdWt) +

Yt(0.0925Ytdt − 0.25YtdWt)

= 0.77dt(138)

So, for every 9 units she spends initially, she has 5 invested in X . It followsthat if she spends 1000 initially, 5

9 · 1000 is invested in X to obtain a riskfree asset.

Spring 2009 Q10

= 0.77dt(138)

Spring 2009 Q10

= 0.77dt(138)

Spring 2009 Q18

Consider two perfectly correlated, non-dividend paying assets X ,Y whoseevolution is modeled by

Xt = X0e0.1t+0.2Wt

Yt = Y0e0.125t+0.3Wt(139)

with a constant risk-free rate r for all t ≥ 0.

If you are constrained to a non-arbitrage market, solve for r .

Spring 2009 Q18

Recall that for constant µ, σ,

dSt = µStdt + σStdWt

⇒ St = S0e(µ− 12σ2)t+σWt .

For stock X , σ1 = 0.2 and µ1 = 0.1 + 12 · 0.2

2 = 0.12.For stock Y , σ2 = 0.3 and µ2 = 0.125 + 1

2 · 0.32 = 0.17.

Since the two assets are perfectly correlated, their Sharpe ratios are equal:

µ1 − r

σ1=µ2 − r

⇒ r = 0.02.(141)

Spring 2009 Q18

Recall that for constant µ, σ,

dSt = µStdt + σStdWt

⇒ St = S0e(µ− 12σ2)t+σWt .

For stock X , σ1 = 0.2 and µ1 = 0.1 + 12 · 0.2

2 = 0.12.For stock Y , σ2 = 0.3 and µ2 = 0.125 + 1

2 · 0.32 = 0.17.

Since the two assets are perfectly correlated, their Sharpe ratios are equal:

µ1 − r

σ1=µ2 − r

⇒ r = 0.02.(141)

Practice Aug 2010 Q13

Let Wt be a Brownian motion and define

Xt = 2Wt − 2

Yt = W 2t − t

Zt = t2Wt − 2

0sWsds

Which of these has zero drift?

Consider the SDE

dXt = λ (α− Xt) dt + σdWt (143)

where λ, α, σ > 0 and X0 are known. Solve for Xt .

Answer:Using integrating factor eλt , we compute

eλtXt

)= αλeλtdt + σeλtdWt

⇒ eλtXt − X0 = α(eλt − 1) + σ

0eλsdWs

⇒ Xt = X0e−λt + α(1− e−λt) + σe−λt∫ t

0eλsdWs .

Bonus: Can we compute limt→∞ Xt in some meaningful way?

Consider the SDE

where λ, α, σ > 0 and X0 are known. Solve for Xt .Answer:Using integrating factor eλt , we compute

eλtXt

⇒ eλtXt − X0 = α(eλt − 1) + σ

0eλsdWs

0eλsdWs .

Consider the SDE

where λ, α, σ > 0 and X0 are known. Solve for Xt .Answer:Using integrating factor eλt , we compute

eλtXt

⇒ eλtXt − X0 = α(eλt − 1) + σ

0eλsdWs

0eλsdWs .

Practice: Brownian Bridge

Consider the SDE

dXt = − Xt

1− tdt + dWt

X0 = 0.(145)

Compute Xt .

Answer:Using integrating factor 1

1−t , we compute

1− t

1− tdWt

⇒ Xt

1− t− X0 =

1− sdWs

⇒ Xt = (1− t)

1− sdWs .

Bonus: Can we compute limt→1 Xt in some meaningful way?

Consider the SDE

dXt = − Xt

1− tdt + dWt

X0 = 0.(145)

Compute Xt .Answer:Using integrating factor 1

1−t , we compute

1− t

1− tdWt

⇒ Xt

1− t− X0 =

1− sdWs

⇒ Xt = (1− t)

1− sdWs .

Consider the SDE

dXt = − Xt

1− tdt + dWt

X0 = 0.(145)

Compute Xt .Answer:Using integrating factor 1

1−t , we compute

1− t

1− tdWt

⇒ Xt

1− t− X0 =

1− sdWs

⇒ Xt = (1− t)

1− sdWs .

Consider the non-dividend stock evolution

dSt = αStdt + σStdWt (147)

in a market that also provides a continuously compounded rate r onriskless asset. At time 0, Jane invests in a fund that employs aproportional investment strategy where at every point in time, 100kpercent is invested in S and 100(1− k) percent in the riskless asset. Solvefor Jane’s wealth Xt .

Assuming there are no transaction costs,

St+ (1− k)rdt

= k · (αdt + σdWt) + (1− k)rdt

= (kα + (1− k)r) dt + kσdWt

Xt = X0e(kα+(1−k)r− 12

(kσ)2)t+kσWt

Assuming there are no transaction costs,

St+ (1− k)rdt

= k · (αdt + σdWt) + (1− k)rdt

= (kα + (1− k)r) dt + kσdWt

Xt = X0e(kα+(1−k)r− 12

(kσ)2)t+kσWt

Consider the stock evolution

dSt = 0.03Stdt + 0.2StdWt (149)

If G = (S1 · S2 · S3)13 , calculate Var (ln (G ))

Bonds, Forwards,Barriers

In any market, we have two basic assets that we use to price derivativeswith - a stock and a bond. Let’s see how these carry through to the BSMframework

If G (S) = 1 then at time T , one unit of bond (or currency) ispresented. What is the price at time 0?

If G (S) = Sa then at time T , one unit of stock (or asset) ispresented. What is the price at time 0?

If S0 = 50, and we agree to pay 1 unit of currency if the stock everhits 100, otherwise we pay nothing, what is the value of this financialinstrument at time 0?

By the BSM technology, we must find a solution to a PDE that satisfiesall conditions imposed by the definition of the contract we wish to price.For the first two cases above, there are only final time conditions.

For the case of presenting a unit of currency at time T , it is a fairassumption that the value V (S , t) should be independent of S , as thepayoff G (S) = 1 is.

Our resulting PDE is in fact an ODE for V (t):

V ′(t) = rV (t)

V (T ) = 1(150)

The solution of course is V (t) = e−r(T−t), which is the discountedvalue of one unit of currence T − t units of time in the future.

The hedging strategy is ∆ = ∂V∂S = 0, which means no stocks are held

to replicate this, and all the money taken at time t is invested inbuying e−r(T−t) units of bond.

V ′(t) = rV (t)

V (T ) = 1(150)

V ′(t) = rV (t)

V (T ) = 1(150)

Consider the case V (S ,T ) = G (S) = Sa. Then if we assume the Ansatz

V (S , t) = SaV (t), (151)

substitution into the BSM PDE returns

V ′(t) +1

2σ2a(a− 1)V (t) = r(1− a)V (t)

V (T ) = 1.(152)

The solutiion of this ODE with a final time condition is

V (t) = exp([

r(1− a) +1

2σ2a(1− a)

](t − T )

−[r(1−a)+ 1

2σ2a(1−a)

](T−t)

Another way to see this is our solution is to again recall that under the riskneutral measure P, from u to time t, we have the evolution

St = Su exp([

r − 1

2σ2](t − u) + σWt−u

⇒ G (ST ) = SaT = Sa

t exp(

a[r − 1

2σ2](T − t) + aσWT−t

)⇒ V (S , t) = e−r(T−t)E[G (ST ) | St = S ]

= Sae−r(T−t)ea

[r− 1

](T−t)

· E[eaσWT−t ]

= Sae−r(T−t)ea

[r− 1

](T−t)

· e12a2σ2(T−t)

[−r+ar− 1

2σ2a+ 1

2a2σ2

](T−t)

[−r(1−a)− 1

2σ2a(1−a)

](T−t)

St = Su exp([

r − 1

2σ2](t − u) + σWt−u

)⇒ G (ST ) = Sa

T = Sat exp

(a[r − 1

⇒ V (S , t) = e−r(T−t)E[G (ST ) | St = S ]

= Sae−r(T−t)ea

[r− 1

](T−t)

· E[eaσWT−t ]

= Sae−r(T−t)ea

[r− 1

](T−t)

· e12a2σ2(T−t)

[−r+ar− 1

2σ2a+ 1

2a2σ2

](T−t)

[−r(1−a)− 1

2σ2a(1−a)

](T−t)

St = Su exp([

r − 1

2σ2](t − u) + σWt−u

)⇒ G (ST ) = Sa

T = Sat exp

(a[r − 1

)⇒ V (S , t) = e−r(T−t)E[G (ST ) | St = S ]

= Sae−r(T−t)ea

[r− 1

](T−t)

· E[eaσWT−t ]

= Sae−r(T−t)ea

[r− 1

](T−t)

· e12a2σ2(T−t)

[−r+ar− 1

2σ2a+ 1

2a2σ2

](T−t)

[−r(1−a)− 1

2σ2a(1−a)

](T−t)

In the last case, time is not explicitly given as a condition, meaning thatthe option is perpetual - only events to do with the the underlying stockvalue S will factor into the price of the contract. Hence, we have again asimpler ODE for V (S), but now with boundary conditions:

2σ2S2V ′′(S) = r

(V (S)− SV ′(S)

)V (0) = 0

V (100) = 1

General solution of this ODE is

V (S) = c1S + c2S−2rσ2 . (156)

Solution of this ODE with our boundary conditions leads to the closedform V (S) = 1

100 S , and so the specific price asked for is V (50) = 0.50

2σ2S2V ′′(S) = r

(V (S)− SV ′(S)

)V (0) = 0

V (100) = 1

V (S) = c1S + c2S−2rσ2 . (156)

2σ2S2V ′′(S) = r

(V (S)− SV ′(S)

)V (0) = 0

V (100) = 1

V (S) = c1S + c2S−2rσ2 . (156)

Some Notes..

In our pricing examples, and in the BSM PDE, it turns out that theexpected return µ of the stock doesn’t factor directly into our calculations.Nevertheless, we note that

[µdt + σdWt

]= µdt

[rdt + σdWt

]= rdt

Multivariable BSM PDE

Following in the same fashion as the above arguments, we can show for

dSi (t)

Si (t)= µidt +

m∑j=1

σijdW(j)t

αij =m∑

σikσjk

qi = the dividend rate of the i th asset

n∑i ,j=1

αijSiSj∂2V

∂Si∂Sj+

n∑i=1

(r − qi )Si∂V

∂Si= rV

V (S1,S2, ..,Sn,T ) = G (S1, S2, ..,Sn)

Forward Delivered in a Foreign Currency

Consider an asset with value S1 in yen and an exchange rate of S2 dollarsper yen. If the final payoff a contract is [S1(T )]a, then the risk-neutralprice of a contract that delivers this value in dollars is V (S1,S2, t), withpde

2∑i ,j=1

αijSiSj∂2V

∂Si∂Sj+

2∑i=1

(r − qi )Si∂V

∂Si= rV

V (S1,S2,T ) = Sa1 S2.

Once again, we assume the Ansatz V (S1, S2, t) = Sa1 S2V (t) to obtain the

ODE with final condition:

V ′(t) + γV (t) = 0

V (T ) = 1,(161)

where γ := 12 (α11a(a− 1) + a(α12 + α21)) + a(r − q1) + (r − q2)− r .

It follows that

V (S1,S2, t) = Sa1 S2eγ(T−t). (162)

Note that for a = 1, our solution is

V (S1,S2, t) = S1S2er(T−t). (163)

Once again, we assume the Ansatz V (S1, S2, t) = Sa1 S2V (t) to obtain the

ODE with final condition:

V ′(t) + γV (t) = 0

V (T ) = 1,(161)

where γ := 12 (α11a(a− 1) + a(α12 + α21)) + a(r − q1) + (r − q2)− r .

It follows that

V (S1,S2, t) = Sa1 S2eγ(T−t). (162)

Note that for a = 1, our solution is

V (S1,S2, t) = S1S2er(T−t). (163)

Options on Futures

It may be that it would be to an investor’s advantage to hold an optioncontract on an asset associated with the underlying (ie commodity, stock,etc...). One that is more liquid and involves lower transaction costsanyway than the one involving the actual asset.

One way to enable this liquidity is to write the contract on the associatedFuture or Forward contract. Recall that

F = Ser(T−t) (164)

and so

∂S= er(T−t) ∂

∂F∂2

∂S2=

(er(T−t) ∂

)= e2r(T−t) ∂

∂F 2

∂t− rF

Options on Futures

F = Ser(T−t) (164)

and so

∂S= er(T−t) ∂

∂F∂2

∂S2=

(er(T−t) ∂

)= e2r(T−t) ∂

∂F 2

∂t− rF

Options on Futures

F = Ser(T−t) (164)

and so

∂S= er(T−t) ∂

∂F∂2

∂S2=

(er(T−t) ∂

)= e2r(T−t) ∂

∂F 2

∂t− rF

Options on Futures

F = Ser(T−t) (164)

and so

∂S= er(T−t) ∂

∂F∂2

∂S2=

(er(T−t) ∂

)= e2r(T−t) ∂

∂F 2

∂t− rF

Options on Futures

F = Ser(T−t) (164)

and so

∂S= er(T−t) ∂

∂F∂2

∂S2=

(er(T−t) ∂

)= e2r(T−t) ∂

∂F 2

∂t− rF

Options on Futures

The resulting B-S-M equation is

2σ2F 2VFF − rV = 0

V (F ,T ) = G (F )(166)

Question: How do we solve for V (F , t) ? Certainly, it is much closer toHeat equation than the original form of the B-S-M PDE.

Options on Futures

2σ2F 2VFF − rV = 0

V (F ,T ) = G (F )(166)

Options on Futures

2σ2F 2VFF − rV = 0

V (F ,T ) = G (F )(166)

Path Dependent Options: Types and Technology

So far, we have mainly described the methodology for fairly pricing whatare known as Vanilla options,namely Puts, Calls and Forwards. These areonly time-dependent, in that the only event that the price is contingent onis the payoff at the end of the term, T .

A more exotic option would provide extra security by enabling thepurchaser to receive a payment contingent on an event related to theunderlying asset the option is written on. This event is most often the firsttime a stock price hits a certain theshold. However, the event could berelated to the average or running maximum of the stock price over a giventime window.

Barrier Options

One of the initial examples of the B-S-M equation we were able to pricewas a perpetual barrier option. In that case, if a stock hit an upper barrierin its price, one dollar was paid to the option purchaser, otherwise theoption continues on.

In most cases, a time limit is put on such an option. What happens if wewish to put an extra barrier condition on an otherwise plain vanillaEuropean Call option?

Recall: European Barrier Options

Consider now the case where an option is activated or deactivated,depending on whether or not the asset has reached a certain price level(barrier) before the option expires.

Hence, we have another path dependent option with the need for anothertracking variable. In this case, we can add the maximum or minimum ofthe asset over its path. They are knocked in or knocked out of action ifthe barrier is reached before expiry, depending on the contract design.

Recall: European Barrier Options

Consider now the case where an option is activated or deactivated,depending on whether or not the asset has reached a certain price level(barrier) before the option expires.

Hence, we have another path dependent option with the need for anothertracking variable. In this case, we can add the maximum or minimum ofthe asset over its path. They are knocked in or knocked out of action ifthe barrier is reached before expiry, depending on the contract design.

Barrier Options: PDE Valuation

Consider a Down-and-Out Call Option:

∂t(S , t) +

2σ2S2∂

∂S2(S , t) + r

(S∂V

∂S(S , t)− V (S , t)

V (S ,T ) = (S − K )+

V (S , t) = 0 if S ≤ X

Barrier Options

How do we solve such a PDE? Certainly, without this condition at X , wehave a solution of our plain Vanilla Call C (S , t). And, if X → 0, whatvalue should we should recover? Perhaps a modification of the solutionC (S , t) is a good place to start. With this in mind, we assume

V (S , t) = C (S , t)− B(S , t) (168)

where B(X , t) = C (X , t) and B(S , t) satisfies the same B-S-M PDE asthe European Call. This is reasonable assumption, as the B-S-M equationis linear.

Barrier Options

Our next assumption is that we can rescale our function C (S , t) and stillhave it solve the B-S-M equation for Call options. Namely, try

B(S , t) =

)αC (SβX γ , t)

) (169)

Substitution into B-S-M equation leads us to α = 1− r2σ2 , β = −1, γ = 2

Hence,

V (S , t) = C (S , t)−(

)1− r2σ2

)(170)

General Barrier Options

Imagine the case now where we have an option that pays G (S) atexpiration time T , but option is knocked out if St /∈ [L,U] for any t < T .In this case, our PDE is

∂t(S , t) +

2S2σ2∂

∂S2(S , t) + r

(S∂V

∂S(S , t)− V (S , t)

V (S ,T ) = G (S)

V (S , t) = 0 if S ≥ U

V (S , t) = 0 if S ≤ L

where G (S) is a general payoff function.

Barrier Options

Once again, we can attempt a solution V (S , t) = A(t)B(S) and so weobtain:

A′(t)B(S) +1

2σ2A(t)S2B ′′(S) + r

(A(t)SB ′(S)− A(t)B(S)

B(U) = 0

B(L) = 0

A(T )B(S) = G (S)

Modifications need to be made here if G (S) does not solve thecorresponding ODE for B(S).

Barrier Options

First, assume

2S2B ′′(S) + rSB ′(S)

B(S)= λ

A′(t)− rA(t)

A(t)= −λ

B(U) = 0

B(L) = 0

A(T ) = 1

We now have two separate ODE’s, one of which is easily solved:

A(t) = e−(r−λ)(T−t). (174)

But the question remains: what is λ?

Barrier Options:Eigenvalue Approach

Again, we try the Ansatz B(S) = Sn. Solution of the ODE requires

2σ2n(n − 1) + rn = λ. (175)

However, neither boundary condition is satisfied. We need a moresophisticated approach, that of eigenfunction expansions.

The solution of the boundary value problem

2S2B ′′(S) + rSB ′(S)

B(S)= λ

B(U) = 0 = B(L) = 0

can be shown to exist only for specific λ, namely a countable set of them:

(r − σ2

)Bn(S) =

σ√ln U

S−νσ sin

(πn ln X

λn = r +ν2

2+σ2π2n2

2 ln2 UL

Each Bn(S) and corresponding e(r−λn)(t−T ) solve their respecitveODE’s with boundary conditions.

Hence, any linear combination∑∞

n=0 cne(r−λn)(t−T )Bn(S) shouldsolve the PDE. The only condition left is to solve the final timecondition

V (S ,T ) = G (S) (178)

We can do this by taking the right sequence cn∞n=0 such that

G (S) = V (S ,T ) =∞∑n=0

cnBn(S) (179)

For an explanation of how to do this, proof of the form of thesolutions Bn(S), and even more on Barrier options see PricingOptions on Scalar Diffusions: An Eigenfunction ExpansionApproach by Dmitry Davydov and Vadim Linetsky

European Barrier Options

There are three types of barrier options:

Knock-out options: Options are deactivated when the asset pricereaches the barrier before option maturity. These are non-zero tobegin with.

down-and-out when the asset price has to decrease to reach thebarrier.up-and-out when the price has to increase to reach the barrier.

Knock-in options: Options are activated when asset price reaches thebarrier before option maturity. These are only activated if barrier isreached before option expiry.

down-and-in when the asset price has to decrease to reach the barrier.up-and-in when the price has to increase to reach the barrier.

Rebate options: Guaranteed level payment made to holder if the assetprice reaches the barrier before expiry. Payment can be made at thetime the barrier is reached or deferred to expiry time. Up Rebate ifbarrier is above the current spot price, Down Rebate if barrier isbelow the current spot price.

Parity for Barrier Options

It is straightforward to see that

V knock-in0 + V knock-out

0 = V0 (180)

To see this, we have the explicit formulation usingτB := inf t <∞ : St = B:

V knock-out at B0 = e−rT E0

[G (ST )1τB>T

]V knock-in at B

0 = e−rT E0

[G (ST )1τB≤T

]V0 = e−rT E0 [G (ST )]

= e−rT E0

[G (ST )

(1τB>T + 1τB≤T

)]= V knock-out at B + V knock-in at B .

Barrier Options: Example

Consider the following continuous economy with a non-dividend payingstock S , where S0 = K = 100, σ = 0.1 and r = 0.05. For a one-yearoption, we have the standard calculations

d1 =ln S0

K + (r − δ + 0.5σ2)T

T= 0.55

d2 = d1 − σ√

T = 0.55− 0.1 = 0.45

N(d1) = P[Z ≤ d1] = 0.7088

N(d2) = P[Z ≤ d2] = 0.6736

⇒ V C0 = 100 · 0.7088− 100 · e−0.05 · 0.6736 = 6.805

⇒ V C ,knock-in at 990 = V C

0 = 6.805

⇒ V C ,knock-out at 990 = V C

0 − V C ,knock-in at 990 = 0.

d1 =ln S0

K + (r − δ + 0.5σ2)T

T= 0.55

d2 = d1 − σ√

T = 0.55− 0.1 = 0.45

N(d1) = P[Z ≤ d1] = 0.7088

N(d2) = P[Z ≤ d2] = 0.6736

⇒ V C0 = 100 · 0.7088− 100 · e−0.05 · 0.6736 = 6.805

0 = 6.805

0 − V C ,knock-in at 990 = 0.

d1 =ln S0

K + (r − δ + 0.5σ2)T

T= 0.55

d2 = d1 − σ√

T = 0.55− 0.1 = 0.45

N(d1) = P[Z ≤ d1] = 0.7088

N(d2) = P[Z ≤ d2] = 0.6736

⇒ V C0 = 100 · 0.7088− 100 · e−0.05 · 0.6736 = 6.805

0 = 6.805

0 − V C ,knock-in at 990 = 0.

d1 =ln S0

K + (r − δ + 0.5σ2)T

T= 0.55

d2 = d1 − σ√

T = 0.55− 0.1 = 0.45

N(d1) = P[Z ≤ d1] = 0.7088

N(d2) = P[Z ≤ d2] = 0.6736

⇒ V C0 = 100 · 0.7088− 100 · e−0.05 · 0.6736 = 6.805

0 = 6.805

0 − V C ,knock-in at 990 = 0.

Gap Options

In keeping with the extra freedom allowed in option design of the previoussections, consider a payoff

G (S) = (S − K1)1S>K2 = (S − K2)1S>K2 + (K2 − K1)1S>K2 (183)

where K1 is the familiar strike price and K2 is the trigger price thatactivates the payoff. It can be shown that

V C = Se−δTN(d1)− K1e−rTN(d2)

V P = K1e−rTN(−d2)− Se−δTN(−d1)

d1 =ln S

K2+ (r − δ + 0.5σ2)T

d2 = d1 − σ√

N(d1) = P[Z ≤ d1]

N(d2) = P[Z ≤ d2].

Exchange Options

Consider now the case where one can exchange S for K , where K is nownon-constant and risky. Then

G (ST ,KT ) = max ST − KT , 0 (185)

and so

V C (S ,K , t) = e−r(T−t)E[G (ST ,KT ) | St = S ,Kt = K ]

= Se−δS (T−t)N(d1)− Ke−δK (T−t)N(d2)

d1 =ln Se−δS (T−t)

Ke−δK (T−t) + 0.5σ2(T − t)

T − t

d2 = d1 − σ√

σ =√σ2S + σ2

K − 2ρσSσK .

Asian Options:Options on Averages

Another useful example of a path dependent option that is commonlyemployed is the Asian option. As an example, an airline may only buy afinite supply of jet fuel every year, but use it continuously during the year.To guard itself against large price movements, the company may choose touse financial instruments with a fixed purchase date to avoid having tocontinuously hedge against the commodity price movements. We assumethat the asset evolves according to

dSt = µ(St , t)dt + σ(St , t)dWt (187)

on some fixed probability space (Ω,F ,P)

A common instrument used is a forward contract. However, this fixes theprice once the contract is signed. Another choice companies have is topurchase an Asian option that pays, at time T ,

(S − 1

0Sudu, 0

)(188)

To be able to price this option at time t, two pieces of information arerequired:

St − the stock price

0Sudu − the running integral of the stock price

(S − 1

0Sudu, 0

)(188)

(S − 1

0Sudu, 0

)(188)

(S − 1

0Sudu, 0

)(188)

Furthermore, we also require the differential form dIt :

dIt = Stdt. (190)

Furthermore, we also require the differential form dIt :

dIt = Stdt. (190)

With the triple of information (St , It , t), we can again construct aB-S-M portfolio replication argument.

On the one hand, we have the self-financing portfolio we construct,Xt using the underlying asset St and the riskless bank account:

dXt = r (Xt −∆St) dt + ∆tdSt . (191)

With the triple of information (St , It , t), we can again construct aB-S-M portfolio replication argument.

On the one hand, we have the self-financing portfolio we construct,Xt using the underlying asset St and the riskless bank account:

dXt = r (Xt −∆St) dt + ∆tdSt . (191)

And on the other hand, we have the price of the contract Vt ≡ V (St , It , t)that satisfies Ito’s equation

2σ(St , t)2∂

∂SdSt +

∂IdIt

2σ(St , t)2∂

∂S2+ St

∂SdSt .

And on the other hand, we have the price of the contract Vt ≡ V (St , It , t)that satisfies Ito’s equation

2σ(St , t)2∂

∂SdSt +

∂IdIt

2σ(St , t)2∂

∂S2+ St

∂SdSt .

Matching terms once again, with the constraint that Vt = Xt , we obtain

2σ(S , t)2∂

∂S2+ S

∂I+ r

(S∂V

∂S− V

V (S , I ,T ) = max

(S − I

)∆ =

Matching terms once again, with the constraint that Vt = Xt , we obtain

2σ(S , t)2∂

∂S2+ S

∂I+ r

(S∂V

∂S− V

V (S , I ,T ) = max

(S − I

)∆ =

Asian Options:Similarity Solutions

This Asian option acts as a Call option contract, in that the strike Kis replaced by an average value over the path of the asset evolution.

But, the dimension of the PDE has increased by 1. This makes thesolution more difficult to obtain.

Still, we have the tools used in the solution of the heat equation andBarrier options, namely Similarity solutions.

We assume that

V (S , I , t) = S · H(R, t)

R ≡ I

V (S , I ,T ) = max

(S − I

)= S ·max

(1− I

)= S ·max

(1− R

)= S · H(R,T )

This Asian option acts as a Call option contract, in that the strike Kis replaced by an average value over the path of the asset evolution.

But, the dimension of the PDE has increased by 1. This makes thesolution more difficult to obtain.

Still, we have the tools used in the solution of the heat equation andBarrier options, namely Similarity solutions.

We assume that

V (S , I , t) = S · H(R, t)

R ≡ I

V (S , I ,T ) = max

(S − I

)= S ·max

(1− I

)= S ·max

(1− R

)= S · H(R,T )

Substitution into the B-S-M equation derived above returns

2σ2R2∂

∂R2+ (1− rR)

∂R= 0

H(R,T ) = max

(1− R

) (195)

Substitution into the B-S-M equation derived above returns

2σ2R2∂

∂R2+ (1− rR)

∂R= 0

H(R,T ) = max

(1− R

) (195)

For boundary conditions similar to those imposed on the European call, wesee that if R →∞, then it must be that S → 0, and so the value of theoption should be worthless:

H(∞, t) = 0 (196)

If R → 0, then assuming that all derivatives of H with respect to R arefinite, it is reasonable to return to the BSM equation to obtain

∂t(0, t) +

∂R(0, t) = 0 (197)

Use tools similiar to those used in explicit solution of original EuropeanCall option C (S , t) to solve this PDE.

H(∞, t) = 0 (196)

∂t(0, t) +

∂R(0, t) = 0 (197)

H(∞, t) = 0 (196)

∂t(0, t) +

∂R(0, t) = 0 (197)

H(∞, t) = 0 (196)

∂t(0, t) +

∂R(0, t) = 0 (197)

Lookback Options

Another exotic, path-dependent option that is studied is the Lookbackoption. This contract agrees to pay the investor the value

max (JT − ST , 0) (198)

where Jt ≡ max0≤u≤tSu

Such an option is valuable, as it helps to reduce the regret of an investorwho wishes to sell his asset S at its highest value over a window of time[0,T ].

Of course, only the actual asset S is traded in, not its maximum Jt at timet. It follows that our replicating portfolio Xt is constructed using S andthe bank account only.

Lookback Options

max (JT − ST , 0) (198)

Lookback Options

max (JT − ST , 0) (198)

Lookback Options

max (JT − ST , 0) (198)

Lookback Options

max (JT − ST , 0) (198)

Lookback Options

Once again, our B-S-M replication argument returns

2σ(St , t)2∂

∂SdSt +

∂JdJt

2σ(St , t)2∂

∂SdSt +

∂J1St=JtdSt

Lookback Options

2σ(St , t)2∂

∂SdSt +

∂JdJt

2σ(St , t)2∂

∂SdSt +

∂J1St=JtdSt

Lookback Options

2σ(St , t)2∂

∂SdSt +

∂JdJt

2σ(St , t)2∂

∂SdSt +

∂J1St=JtdSt

Lookback Options

Again, we can choose how we wish to hedge. It can be shown that∂V∂J (S ,S) = 0. This implies that ∆t = ∂V

∂S and we return our favouritePDE with an extra condition:

2σ(S , t)2∂

∂S2+ r

(S∂V

∂S− V

V (S , J,T ) = max (J − S , 0)

V (0, J, t) = Je−r(T−t)

∂J(S ,S) = 0

∆ =∂V

Lookback Options: Similarity Solutions

To enable a reduction in dimension, we again return to the notion of asimilarity solution. In the case of the Lookback option, we have

V (S , J, t) = J ·W (ζ, t)

ζ ≡ S

JV (S , I ,T ) = max (J − S , 0)

= J ·max

(1− S

)= J ·max (1− ζ, 0)

= JW (ζ,T )

Lookback Options: Similarity Solutions

The resulting PDE is now

2σ2ζ2Wζζ + rζWζ − rW = 0

W (0, t) = e−r(T−t)

W (ζ,T ) = max (1− ζ, 0)

Wζ(1, t) = W (1, t)

Estimating Historical Volatility

Given an interval [0,T ] one can look at the returns over a partition of theinterval into n equal subintervals:

rt,h := ln(St+h

)(203)

and so for h = Tn ,

σ2 :=1

n − 1·

n∑k=1

(r(k−1)h,h)2 (204)

Note that our estimator σ2 is stochastic.

Estimating Historical Volatility

Given an interval [0,T ] and an assumed model

St= αdt + σdWt (205)

one can look at the returns over a partition of the interval into n equalsubintervals for h = T

rt,h = ln(St+h

)∼ N

((α− 1

2σ2)· h, σ2h)

n∑k=1

r(k−1)h,h

σ2 =1

n − 1

n∑k=1

(r(k−1)h,h − r)2 =

∑nk=1(r(k−1)h,h)2

n − 1− nr 2

n − 1.

It follows that n · r is our estimator for α and n · σ2 for σ2 over the period[0,T ]. See Q.51 in the SOA MFE practice for a nice worked example.

What happens if σ 6= constant?

Returning to the continuous time setting, there are two models thataccount for a non-cosntant volatility.

Constant Elasticity of Variance:

St= αdt + σS

β−22 dWt

Heston Model (Stochastic Volatility):

St= αdt +

√σtdWt

dσt = k · (σ(t)− σt)dt + σHdWt

dWt · dWt = ρdt.

Both lead to more mathematically complex solution techniques. However,empirical data suggests that they calibrate significantly better withobserved option data.

St= αdt + σS

β−22 dWt

St= αdt +

√σtdWt

dWt · dWt = ρdt.

St= αdt + σS

β−22 dWt

St= αdt +

√σtdWt

dWt · dWt = ρdt.

Interest Rate Derivatives

Consider a zero-coupon, non-defaultable bond that pays V (T ,T ) = 1 atexpiry date T . The risk-free rate is a function of time r(t).

The corresponding boundary value problem is (holding T fixed)

V= r(t)dt

V (T ,T ) = 1.(208)

The solution is

V (t,T ) = e−∫ Tt r(s)ds . (209)

Consider a zero-coupon, non-defaultable bond that pays V (T ,T ) = 1 atexpiry date T . The risk-free rate is a function of time r(t).

The corresponding boundary value problem is (holding T fixed)

V= r(t)dt

V (T ,T ) = 1.(208)

The solution is

V (t,T ) = e−∫ Tt r(s)ds . (209)

Consider again a zero-coupon, non-defaultable bond that paysV (T ,T ) = 1 at expiry date T .

Now, the risk-free rate is a random variable indexed by time rt(ω) on a

probability space(

Ω,F ,P)

It follows that, if we have a related risk neutral measure P, then for astochastic process rt(ω), we expect the (martingale) solution

V (t,T ) = E[e−

∫ Tt rs(ω)ds

]. (210)

probability space(

Ω,F ,P)

V (t,T ) = E[e−

∫ Tt rs(ω)ds

]. (210)

probability space(

Ω,F ,P)

V (t,T ) = E[e−

∫ Tt rs(ω)ds

]. (210)

Short Rate Model

For the evolution of interest rates r on(

Ω,F ,P)

, we can assume the

general model

drt = µ(rt , t)dt + σ(rt , t)dWt

Vt= α(rt , t)dt + σV (rt , t)dWt

Short Rate Model

Consider now two bonds, one with expiry date T1 and the other withexpiry date T2. By the pricing formula

V (t,T1) = E[e−

∫ T1t rs(ω)ds

]V (t,T2) = E

∫ T2t rs(ω)ds

] (212)

where both are driven by the same Brownian motion W . It follows that,∀t, the Sharpe Ratios for bonds must be equal for any expiry date:

α(r , t)− r

σV (r , t)= λ(r , t) = Market Price of Risk . (213)

Market Price of Risk

Recall from our initial discussion of equity models that under constant(µ, σ, r), the rate of return on a share of equity is

St= µdt + σdWt

= rdt + σdWt .

⇒ dWt = dWt +µ− r

= dWt + λdt

In words, the market price of risk, or Sharpe ratio, is the drift correctionthat allows us to transition from a Brownian motion under the physicalmeasure P to a Brownian motion under the risk-neutral measure P.

Recall from our initial discussion of equity models that under constant(µ, σ, r), the rate of return on a share of equity is

St= µdt + σdWt

= rdt + σdWt .

⇒ dWt = dWt +µ− r

= dWt + λdt

In words, the market price of risk, or Sharpe ratio, is the drift correctionthat allows us to transition from a Brownian motion under the physicalmeasure P to a Brownian motion under the risk-neutral measure P.

It follows that we can apply the same method/reasoning to interest ratemodeling:

= [µ(rt , t)− σ(rt , t)λ(rt , t)]dt + σ(rt , t)dWt .

dWt = dWt + λ(rt , t)dt.

It follows that we can apply the same method/reasoning to interest ratemodeling:

= [µ(rt , t)− σ(rt , t)λ(rt , t)]dt + σ(rt , t)dWt .

dWt = dWt + λ(rt , t)dt.

Once we are in the risk-neutral framework, then for

V (r , t,T ) = E[e−

∫ Tt rs(ω)ds | rt = r

]drt = [µ(rt , t)− σ(rt , t)λ(rt , t)]dt + σ(rt , t)dWt

it follows by Ito’s Lemma and the equivalence of all Sharpe ratios that Valso satisfies

0 =∂V

∂t+ [µ(r , t)− σ(r , t)λ(r , t)]

2σ2(r , t)

∂r 2− rV

1 = V (r ,T ,T ).

Question: How do we determine the market price of risk λ?

Once we are in the risk-neutral framework, then for

V (r , t,T ) = E[e−

∫ Tt rs(ω)ds | rt = r

]drt = [µ(rt , t)− σ(rt , t)λ(rt , t)]dt + σ(rt , t)dWt

it follows by Ito’s Lemma and the equivalence of all Sharpe ratios that Valso satisfies

0 =∂V

∂t+ [µ(r , t)− σ(r , t)λ(r , t)]

2σ2(r , t)

∂r 2− rV

1 = V (r ,T ,T ).

Question: How do we determine the market price of risk λ?

Affine Models

As mentioned before, it is not clear what λ should be from just the shortrate evolution rt .

There are models, however, that make assumptions on the form of λ thatallow for easy computation of V (r , t,T ).

Assume the Ansatz

V (r , t,T ) = eA(t,T )+B(t,T )r . (218)

HW: What are the corresponding boundary value problems for A and B?

Affine Models

Two important examples are

Vasicek with Constant λ :

drt = [γ(µ− r)]dt + σdWt

= [γ(µ− r)− σλ]dt + σdWt

CIR with λ(r , t) =λ√

drt = [γ(µ− r)]dt + σ√

= [γ(µ− r)− λr ]dt + σ√

rtdWt .

HW: Solve the corresponding boundary value problems for A and B forboth the Vasicek and CIR models.

Affine Models

= [γ(µ− r)− λr ]dt + σ√

rtdWt .

Affine Models

= [γ(µ− r)− λr ]dt + σ√

rtdWt .

Capital Structure

A company that has shareholders can be valued by the equation

At = Et + Bt

At := Value of Company’s Assets at time t.

Et := Value of Company’s Equity at time t.

Bt := Value of Company’s Debt at time t.

Capital Structure

If we assume that the equity does not pay dividends, and that theoutstanding debt is of the form of a zero-coupon bond with face B,payable at time T , then

ET = max

AT − B, 0

BT = min

AT , B

= AT + min

B − AT , 0

= AT −max

AT − B, 0

= AT − ET

Here, default is implicitly assumed to coincide with the event

AT < B

.This results in a turnover of the company’s assets to bondholders if assetsare worth less than the total value of bond outstanding.

Capital Structure

It follows that, as BT = AT on the set

AT < B

B0 = e−rT E0[BT ]

= e−rT E0[min

AT , B

= e−rT E0[AT | AT < B] · P0[AT < B] + e−rT B · P0[AT > B]

= e−rT

[∫ B

0A · P0[AT ∈ dA] + B

∫ ∞B

P0[AT ∈ dA]

] (222)

Capital Structure

Furthermore, we define

E[Recovery Rate] =E[BT | AT < B]

E[Loss Given Default] =B − E[BT | AT < B]

= 1− E[Recovery Rate]

Credit Spread

We use the above to formulate

Credit Spread :=1

)− r

1− P0[AT < B] · E0[Loss Given Default]

)≈ 1

T· P0[AT < B] · E0[Loss Given Default].

Merton Default Model

Recall that under the BSM framework

Pt [Default] = N

B+ r − δ − 1

2σ2(T − t)

T − t

Et [AT | Default] = Ate(r−δ)(T−t)

(− ln At

B+r−δ+ 1

2σ2(T−t)

σ√T−t

(− ln At

B+r−δ− 1

2σ2(T−t)

σ√T−t

) (225)

Completing the analysis, under the assumption δ = 0,

B0 = e−rT[E0[AT | Default] · P0[Default] + B · P0[No Default]

]= A0N

B+ r + 1

)+ Be−rT · N

(ln A0

B+ r − 1

)= A0N(−d1) + Be−rTN(d2)

(226)This implies that

Credit SpreadMerton =1

Tln( B

)− r =

(e−rT B

= − 1

[N(d2) +

Be−rT· N(−d1)

] (227)

Q: What happens to the credit spread as T → 0 ?

Completing the analysis, under the assumption δ = 0,

B0 = e−rT[E0[AT | Default] · P0[Default] + B · P0[No Default]

]= A0N

B+ r + 1

)+ Be−rT · N

(ln A0

B+ r − 1

)= A0N(−d1) + Be−rTN(d2)

(226)This implies that

Credit SpreadMerton =1

Tln( B

)− r =

(e−rT B

= − 1

[N(d2) +

Be−rT· N(−d1)

] (227)

Q: What happens to the credit spread as T → 0 ?Albert Cohen (MSU) STT 888: Stochastic Calculus for Finance MSU 2016 196 / 249

Calibration

Recall that under the BSM framework

Et [AT | Default] = Ate(α−δ)(T−t)

(− ln At

B+α−δ+ 1

2σ2(T−t)

σ√T−t

(− ln At

B+α−δ− 1

2σ2(T−t)

σ√T−t

Distance to Default :=E[ln (AT )]− B

T − t

=ln (At) + α + 1

2σ2(T − t)− ln (B)

T − t

Calibration

Under this framework, we have also derived an important result forcalibration, where recovery upon default is priced via risk neutral measure,but may only be observed under a physical measure.

The two are related via

Et [AT | Default] = γ(t) · Et [AT | Default]

γ := e(r−α)(T−t)

(− ln At

B+r−δ+ 1

2σ2(T−t)

σ√T−t

(− ln At

B+r−δ− 1

2σ2(T−t)

σ√T−t

(− ln At

B+α−δ− 1

2σ2(T−t)

σ√T−t

(− ln At

B+α−δ+ 1

2σ2(T−t)

σ√T−t

)(229)

Note that γ(t) = 1 for all 0 ≤ t ≤ T if α = r .

Calibration

Under this framework, we have also derived an important result forcalibration, where recovery upon default is priced via risk neutral measure,but may only be observed under a physical measure.

The two are related via

Et [AT | Default] = γ(t) · Et [AT | Default]

γ := e(r−α)(T−t)

(− ln At

B+r−δ+ 1

2σ2(T−t)

σ√T−t

(− ln At

B+r−δ− 1

2σ2(T−t)

σ√T−t

(− ln At

B+α−δ− 1

2σ2(T−t)

σ√T−t

(− ln At

B+α−δ+ 1

2σ2(T−t)

σ√T−t

)(229)

Note that γ(t) = 1 for all 0 ≤ t ≤ T if α = r .

Solving the BSM PDE:

Under a change of variables, solving the BSM PDE is equivalent to solvingthe Heat Equation

uτ = uxx

u(x , 0) = u0(x)(230)

ProofWe show this for a European Call - namely G (S) = (S − K )+ for somestrike K . Let us further write V (S , t) = C (S , t) to denote that we arelooking at a Call option. Our PDE is now

2σ2S2CSS + r (SCS − C ) = 0

C (0, t) = 0

limS→∞

C (S , t)

C (S ,T ) = (S − K )+

uτ = uxx

u(x , 0) = u0(x)(230)

2σ2S2CSS + r (SCS − C ) = 0

C (0, t) = 0

limS→∞

C (S , t)

C (S ,T ) = (S − K )+

uτ = uxx

u(x , 0) = u0(x)(230)

2σ2S2CSS + r (SCS − C ) = 0

C (0, t) = 0

limS→∞

C (S , t)

C (S ,T ) = (S − K )+

uτ = uxx

u(x , 0) = u0(x)(230)

2σ2S2CSS + r (SCS − C ) = 0

C (0, t) = 0

limS→∞

C (S , t)

C (S ,T ) = (S − K )+Albert Cohen (MSU) STT 888: Stochastic Calculus for Finance MSU 2016 199 / 249

Assume the change of variables

S = Kex

t = T − 1

2σ2τ

C (S , t) = Kv(x , τ)

and so

vτ = vxx +

σ2− 1

)vx −

Kv(x , 0) = max(Kex − K , 0)⇒ v(x , 0) = max(ex − 1, 0)

Furthermore, assume v(x , τ) = eαx+βτu(x , τ).

S = Kex

t = T − 1

2σ2τ

C (S , t) = Kv(x , τ)

and so

vτ = vxx +

σ2− 1

)vx −

S = Kex

t = T − 1

2σ2τ

C (S , t) = Kv(x , τ)

and so

vτ = vxx +

σ2− 1

)vx −

S = Kex

t = T − 1

2σ2τ

C (S , t) = Kv(x , τ)

and so

vτ = vxx +

σ2− 1

)vx −

S = Kex

t = T − 1

2σ2τ

C (S , t) = Kv(x , τ)

and so

vτ = vxx +

σ2− 1

)vx −

Thus means that if we choose

α = −1

σ2− 1

)β = −1

σ2+ 1

uτ = uxx , −∞ < x <∞u(x , 0) = e−αxmax(ex − 1, 0)

QEDAlbert Cohen (MSU) STT 888: Stochastic Calculus for Finance MSU 2016 201 / 249

Question - How do we solve for u(x , τ)?

Answer - depends!

Try solving the following basic problem on the half real line:

uτ = uxx , x ≥ 0

u(0, τ) = 1

with the basic guess that we can reduce to an ordinary differentialequation via the transformation

ζ :=x√τ

and the solution u(x , τ) = U(ζ)

Answer - depends!

uτ = uxx , x ≥ 0

u(0, τ) = 1

ζ :=x√τ

Answer - depends!

uτ = uxx , x ≥ 0

u(0, τ) = 1

ζ :=x√τ

Answer - depends!

uτ = uxx , x ≥ 0

u(0, τ) = 1

ζ :=x√τ

General Solution of Heat Equation

The resulting ODE is

U ′′(ζ) = −1

2ζU ′(ζ)

U(0) = 1

U(∞) = 0

which has solution

U(ζ) = C

∫ ζ

4 ds + D (234)

which, upon satisfaction of the boundary conditions at 0 and ∞ leads to

U(ζ) =1√π

∫ ∞ζ

e−s2

4 ds ⇒ u(x , τ) =1√π

∫ ∞x√τ

e−s2

4 ds (235)

The resulting ODE is

U ′′(ζ) = −1

2ζU ′(ζ)

U(0) = 1

U(∞) = 0

which has solution

U(ζ) = C

∫ ζ

4 ds + D (234)

which, upon satisfaction of the boundary conditions at 0 and ∞ leads to

U(ζ) =1√π

∫ ∞ζ

e−s2

4 ds ⇒ u(x , τ) =1√π

∫ ∞x√τ

e−s2

4 ds (235)

Again for the basic equation

uτ = uxx ∀x ∈ (−∞,∞) (236)

we look for solutions of the form

u(x , τ) =1√τ

ζ :=x√τ

which leads to the ODE of the form

[U ′(ζ) +

2ζU(ζ)

]= 0 (237)

uτ = uxx ∀x ∈ (−∞,∞) (236)

u(x , τ) =1√τ

ζ :=x√τ

[U ′(ζ) +

2ζU(ζ)

]= 0 (237)

uτ = uxx ∀x ∈ (−∞,∞) (236)

u(x , τ) =1√τ

ζ :=x√τ

[U ′(ζ) +

2ζU(ζ)

]= 0 (237)

uτ = uxx ∀x ∈ (−∞,∞) (236)

u(x , τ) =1√τ

ζ :=x√τ

[U ′(ζ) +

2ζU(ζ)

]= 0 (237)

uτ = uxx ∀x ∈ (−∞,∞) (236)

u(x , τ) =1√τ

ζ :=x√τ

[U ′(ζ) +

2ζU(ζ)

]= 0 (237)

uτ = uxx ∀x ∈ (−∞,∞) (236)

u(x , τ) =1√τ

ζ :=x√τ

[U ′(ζ) +

2ζU(ζ)

]= 0 (237)

from which we take the fundamental solution

U(ζ) =1

2√π

e−ζ2

u(x , τ) =1

2√πτ

e−x2

4τ =: Γ(x , τ)

Now, consider the family of functions δε(x), parametrized by ε > 0, with

δε(x) :=1

2ε1[−ε,ε](x) (239)

It follows that

limε→0

∫ ∞−∞

δε(x)φ(x)dx = φ(0) (240)

for any reasonable function φ(x)

U(ζ) =1

2√π

e−ζ2

u(x , τ) =1

2√πτ

e−x2

4τ =: Γ(x , τ)

δε(x) :=1

2ε1[−ε,ε](x) (239)

It follows that

limε→0

∫ ∞−∞

δε(x)φ(x)dx = φ(0) (240)

U(ζ) =1

2√π

e−ζ2

u(x , τ) =1

2√πτ

e−x2

4τ =: Γ(x , τ)

δε(x) :=1

2ε1[−ε,ε](x) (239)

It follows that

limε→0

∫ ∞−∞

δε(x)φ(x)dx = φ(0) (240)

U(ζ) =1

2√π

e−ζ2

u(x , τ) =1

2√πτ

e−x2

4τ =: Γ(x , τ)

δε(x) :=1

2ε1[−ε,ε](x) (239)

It follows that

limε→0

∫ ∞−∞

δε(x)φ(x)dx = φ(0) (240)

U(ζ) =1

2√π

e−ζ2

u(x , τ) =1

2√πτ

e−x2

4τ =: Γ(x , τ)

δε(x) :=1

2ε1[−ε,ε](x) (239)

It follows that

limε→0

∫ ∞−∞

δε(x)φ(x)dx = φ(0) (240)

In general,

limε→0

∫ ∞−∞

δε(x − x0)φ(x)dx = φ(x0) (241)

We call the (weak) limit of δε(x) the Dirac delta δ(x), after the physicist.

So, for any function u0(x), we have the relation

limε→0

∫ ∞−∞

δε(y − x)u0(y)dx = u0(x) (242)

It also follows that our fundamental solution Γ(x , τ)⇒ δ(x) as τ → 0.Since u(x , 0)” = ”δ(x) and Γ(x − s, τ) = Γ(s − x , τ) for fixed s, we have

∂τ[u0(s)Γ(s − x , τ)] =

∂x2[u0(s)Γ(s − x , τ)]

u0(s)Γ(s − x , 0) = u0(s)δ(s − x)

limε→0

∫ ∞−∞

δε(y − x)u0(y)dx = u0(x) (242)

∂τ[u0(s)Γ(s − x , τ)] =

∂x2[u0(s)Γ(s − x , τ)]

u0(s)Γ(s − x , 0) = u0(s)δ(s − x)

limε→0

∫ ∞−∞

δε(y − x)u0(y)dx = u0(x) (242)

It also follows that our fundamental solution Γ(x , τ)⇒ δ(x) as τ → 0.

Since u(x , 0)” = ”δ(x) and Γ(x − s, τ) = Γ(s − x , τ) for fixed s, we have

∂τ[u0(s)Γ(s − x , τ)] =

∂x2[u0(s)Γ(s − x , τ)]

u0(s)Γ(s − x , 0) = u0(s)δ(s − x)

limε→0

∫ ∞−∞

δε(y − x)u0(y)dx = u0(x) (242)

∂τ[u0(s)Γ(s − x , τ)] =

∂x2[u0(s)Γ(s − x , τ)]

u0(s)Γ(s − x , 0) = u0(s)δ(s − x)

limε→0

∫ ∞−∞

δε(y − x)u0(y)dx = u0(x) (242)

∂τ[u0(s)Γ(s − x , τ)] =

∂x2[u0(s)Γ(s − x , τ)]

u0(s)Γ(s − x , 0) = u0(s)δ(s − x)

u0(x) =

∫ ∞−∞

u0(s)δ(s − x)ds (244)

and u0(x)u(s − x , τ) is a solution of the PDE for all s, we have that

v(x , τ) =

∫ ∞−∞

u0(s)u(s − x , τ)ds

∫ ∞−∞

u0(s)1

2√πτ

e−(s−x)2

4τ ds

is a solution of

vτ = vxx

v(x , 0) = u0(x)(246)

u0(x) =

∫ ∞−∞

u0(s)δ(s − x)ds (244)

v(x , τ) =

∫ ∞−∞

u0(s)1

2√πτ

e−(s−x)2

4τ ds

is a solution of

vτ = vxx

v(x , 0) = u0(x)(246)

u0(x) =

∫ ∞−∞

u0(s)δ(s − x)ds (244)

v(x , τ) =

∫ ∞−∞

u0(s)1

2√πτ

e−(s−x)2

4τ ds

is a solution of

vτ = vxx

v(x , 0) = u0(x)(246)

The focus we have is on solving a specific PDE, namely the European Call.The previous slides have shown us that the solution can be derived fromthe fundamental solution of the Heat equation. After making thenecessary substitutions, and by using the notation

d1 :=ln S

K +(r + 1

(T − t)

T − t

d2 :=ln S

K +(r − 1

(T − t)

T − t

N(x) :=1√2π

−∞e−

General Solution of BSM PDE revisited

Then we have the closed form solution

C (S , t) = SN(d1)− Ke−r(T−t)N(d2) (248)

HW1: Can we use any theorems/ideas to use this work to find thecorresponding value of a European Put option without resorting to solvingthe associated PDE?

HW2: Prove that ∆C = ∂C∂S = N(d1)

HW3: Find the value of an option that pays 1 if the stock value is aboveK at time T , and 0 otherwise. Hint - use the above analysis!

C (S , t) = SN(d1)− Ke−r(T−t)N(d2) (248)

Black Scholes Pricing using Underlying Asset

In the next section, we will derive the following solutions to theBlack-Scholes PDE:

V C (S , t) = e−r(T−t)E [(ST − K )+ | St = S ]

= Se−δ(T−t)N(d1)− Ke−r(T−t)N(d2)

V P(S , t) = e−r(T−t)E [(K − ST )+ | St = S ]

= Ke−r(T−t)N(−d2)− Se−δ(T−t)N(−d1)

d1 =ln(

)+ (r − δ + 1

2σ2)(T − t)

T − t

d2 = d1 − σ√

T − t

N(x) =1√2π

−∞e−

2 dz .

Notice that V C (S , t)− V P(S , t) = Se−δ(T−t) − Ke−r(T−t).Question: What underlying model of stock evolution leads to this value?How can we support such a probability measure?

Normal Random Variables

We say that X is a Normal Random Variable with parameters µ, σ2 if

fX (x) =1√2πσ

e−(x−µ)2

2σ2 for −∞ < x <∞ (250)

Furthermore, we say that Z is a Standard Normal Random Variable if it isNormal with parameters µ = 0, σ = 1. Consider

z =x − µσ∫ ∞

−∞fX (x)dx =

∫ ∞−∞

1√2πσ

e−(x−µ)2

2σ2 dx

∫ ∞−∞

1√2π

e−(z)2

∫ ∞−∞

fZ (z)dz

We user the notation X ∼ N(µ, σ2) and Z ∼ N(0, 1). By ourtransformation above, it can be seen that if Z ∼ N(0, 1), thenX = µ+ σ · Z ∼ N(µ, σ2). We can see this via

Φ(z) := FZ (z) = P[Z ≤ z ]

= P[X − µσ

≤ x − µσ

= P[X ≤ x ] =: FX (x)

(x − µσ

dxFZ (z)

dxFX (x) = fX (x)

fZ (z) =1√2π

e−(z)2

Φ(z) := FZ (z) = P[Z ≤ z ]

= P[X − µσ

≤ x − µσ

= P[X ≤ x ] =: FX (x)

(x − µσ

dxFZ (z)

dxFX (x) = fX (x)

fZ (z) =1√2π

e−(z)2

Φ(z) := FZ (z) = P[Z ≤ z ]

= P[X − µσ

≤ x − µσ

= P[X ≤ x ] =: FX (x)

(x − µσ

dxFZ (z)

dxFX (x) = fX (x)

fZ (z) =1√2π

e−(z)2

Convolutions and Sums of Independent Random Variables

fX+Y (a) =d

(∫ ∞−∞

FX (a− y)fY (y)dy

∫ ∞−∞

daFX (a− y)fY (y)dy by cty of fX (Leibniz)

∫ ∞−∞

fX (a− y)fY (y)dy

Sums of Normal Random Variables

For a sequence cini=1 of real numbers, we have for a sequence ofcorrelated normal random variables Xini=1, where the Xi ∼ N(µiσi ) withcovariance ρij that

n∑i=1

ciXi ∼ N

n∑i=1

n∑j=1

cicjρijσiσj

). (254)

Lognormal Random Variables

We say that Y ∼ LN(µ, σ) is Lognormal if ln(Y ) ∼ N(µ, σ2).

As sums of normal random variables remain normal, products of lognormalrandom variables remain lognormal.

Recall that the moment-generating function ofX ∼ N(µ, σ2) ∼ µ+ σN(0, 1) is

MX (t) = E[etX ] = eµt+ 12σ2t2

If Y = eµ+σZ , then, it can be seen that

E[Y n] = E[enX ] = eµn+ 12σ2n2

fY (y) =1

2πyexp

(− (ln(y)− µ)2

)(257)

Lognormal Random Variables

We say that Y ∼ LN(µ, σ) is Lognormal if ln(Y ) ∼ N(µ, σ2).

As sums of normal random variables remain normal, products of lognormalrandom variables remain lognormal.

Recall that the moment-generating function ofX ∼ N(µ, σ2) ∼ µ+ σN(0, 1) is

MX (t) = E[etX ] = eµt+ 12σ2t2

If Y = eµ+σZ , then, it can be seen that

E[Y n] = E[enX ] = eµn+ 12σ2n2

fY (y) =1

2πyexp

(− (ln(y)− µ)2

)(257)

Stock Evolution and Lognormal Random Variables

One application of lognormal distributions is their use in modeling theevolution of asset prices S . If we assume a physical measure P with α theexpected return on the stock under the physical measure, then

((α− δ − 1

2σ2)t, σ2t

)⇒ St = S0e(α−δ− 1

2σ2)t+σ

We can use the previous facts to show

E[St ] = S0e(α−δ)t

P[St > K ] = N

(ln S0

K + (α− δ − 0.5σ2)t

Note that under the risk-neutral measure P, we exchange α with r , therisk-free rate:

E[St ] = S0e(r−δ)t

P[St > K ] = N(d2).(260)

We can use the previous facts to show

E[St ] = S0e(α−δ)t

P[St > K ] = N

(ln S0

K + (α− δ − 0.5σ2)t

Note that under the risk-neutral measure P, we exchange α with r , therisk-free rate:

E[St ] = S0e(r−δ)t

P[St > K ] = N(d2).(260)

The next challenge is to construct a process St that possesses the aboveproperties as well as continuity of paths.

Click here for a neat article relating actuarial reserving to option pricing!

Risk managers are also interested in Conditional Tail Expectations (CTE’s)of random variables:

CTEX (k) := E[X | X > k] =E[X 1X>k

]P[X > k]

. (261)

In our case,

E[St | St > K ] =

[S0e(α−δ− 1

2σ2)t+σ

√tZ1

S0e(α−δ− 1

2σ2)t+σ

√tZ>K

P[S0e(α−δ− 1

2σ2)t+σ

√tZ > K

= S0e(α−δ)t

+(α−δ+0.5σ2)t

σ√t

+(α−δ−0.5σ2)t

σ√t

⇒ E[St | St > K ] = S0e(r−δ)t N(d1)

N(d2)(262)

In fact, we can use this CTE framework to solve for the European Calloption price in the Black-Scholes framework, where P0[A] = P[A | S0 = S ]and

V C (S , 0) := e−rT E[(ST − K )+ | S0 = S

]= e−rT E0

[ST − K | ST > K

]· P0[ST > K ]

= e−rT E0

[ST | ST > K

]· P0[ST > K ]− Ke−rT P0[ST > K ]

= e−rTSe(r−δ)T N(d1)

N(d2)· N(d2)− Ke−rTN(d2)

= Se−δTN(d1)− Ke−rTN(d2).(263)

Black Scholes Pricing using Prepaid Forwards

In order to apply the previous formulae to a myriad of underlying assets,we rewrite in terms of prepaid forwards:

d1(S , t) =ln(Se−δ(T−t)

Ke−r(T−t)

2σ2(T − t)

T − t

=ln(F St,T

2σ2(T − t)

T − t

d2 = d1 − σ√

T − t.

Black Scholes Analysis: Option Greeks

For any option price V (S , t), define its various sensitivities as follows:

∆ =∂V

Γ =∂∆

∂S=∂2V

ν =∂V

Θ =∂V

ρ =∂V

Ψ =∂V

∂δ.

These are known accordingly as the Option Greeks.

Straightforward partial differentiation leads to

∆C = e−δ(T−t)N(d1)

∆P = −e−δ(T−t)N(−d1)

ΓC = ΓP =e−δ(T−t)N ′(d1)

σS√

T − t

νC = νP = Se−δ(T−t)√

T − tN ′(d1)

as well as..

ρC = (T − t)Ke−r(T−t)N(d2)

ρP = −(T − t)Ke−r(T−t)N(−d2)

ΨC = −(T − t)Se−δ(T−t)N(d1)

ΨP = (T − t)Se−δ(T−t)N(−d1).

What do the signs of the Greeks tell us?

HW: Compute Θ for puts and calls.

as well as..

ρC = (T − t)Ke−r(T−t)N(d2)

ρP = −(T − t)Ke−r(T−t)N(−d2)

ΨC = −(T − t)Se−δ(T−t)N(d1)

ΨP = (T − t)Se−δ(T−t)N(−d1).

What do the signs of the Greeks tell us?

HW: Compute Θ for puts and calls.

Portfolio Sensitivity Analysis

For a portfolio of M options, each with weighting λi and∑M

i=1 λi = 1:

Greek(Portfolio) =M∑i=1

λiGreek(i thasset). (268)

If P(i) is the price of a portfolio of income streams: P(i) =∑n

k=1 Pk(i).

⇒ D = −(1 + i) · P ′(i)

P(i)= −(1 + i) ·

∑nk=1 P ′k(i)

= −(1 + i) ·n∑

P ′k(i)

P(i)= −(1 + i) ·

n∑k=1

P(i)·

P ′k(i)

Dk · qk

and so the portfolio duration is the weighted average of the individualdurations. What if the interest rate i is a random variable itself?

i=1 λi = 1:

k=1 Pk(i).

⇒ D = −(1 + i) · P ′(i)

P(i)= −(1 + i) ·

∑nk=1 P ′k(i)

= −(1 + i) ·n∑

P ′k(i)

P(i)= −(1 + i) ·

n∑k=1

P(i)·

P ′k(i)

Dk · qk

and so the portfolio duration is the weighted average of the individualdurations.

What if the interest rate i is a random variable itself?

i=1 λi = 1:

k=1 Pk(i).

⇒ D = −(1 + i) · P ′(i)

P(i)= −(1 + i) ·

∑nk=1 P ′k(i)

= −(1 + i) ·n∑

P ′k(i)

P(i)= −(1 + i) ·

n∑k=1

P(i)·

P ′k(i)

Dk · qk

and so the portfolio duration is the weighted average of the individualdurations. What if the interest rate i is a random variable itself?

Option Elasticity

Define

Ω(S , t) := limε→0

V (S+ε,t)−V (s,t)V (S,t)

S+ε−SS

V (S , t)limε→0

V (S + ε, t)− V (s, t)

S + ε− S

=∆ · S

V (S , t).

Consequently,

ΩC (S , t) =∆C · S

V C (S , t)=

Se−δ(T−t)

Se−δ(T−t) − Ke−r(T−t)N(d2)≥ 1

ΩP(S , t) =∆P · S

V P(S , t)≤ 0.

Option Elasticity

Theorem

The volatility of an option is the option elasticity times the volatility of thestock:

σoption = σstock× | Ω | . (272)

The proof comes from Finan: Consider the strategy of hedging a portfolioof shorting an option and purchasing ∆ = ∂V

∂S shares.The initial and final values of this portfolio are

Initally: V (S(t), t)−∆(S(t), t) · S(t)

Finally: V (S(T ),T )−∆(S(t), t) · S(T )

Option Elasticity

Proof.

If this portfolio is self-financing and arbitrage-free requirement, then

er(T−t)(

V (S(t), t)−∆(S(t), t) ·S(t))

= V (S(T ),T )−∆(S(t), t) ·S(T ).

(273)It follows that for κ := er(T−t),

V (S(T ),T )− V (S(t), t)

V (S(t), t)= κ− 1 +

[S(T )− S(t)

S(t)+ 1− κ

⇒ Var

[V (S(T ),T )− V (S(t), t)

V (S(t), t)

]= Ω2Var

[S(T )− S(t)

]⇒ σoption = σstock× | Ω | .

Option Elasticity

If γ is the expected rate of return on an option with value V , α is theexpected rate of return on the underlying stock, and r is of course the riskfree rate, then the following equation holds:

γ · V (S , t) = α ·∆(S , t) · S + r ·(

V (S , t)−∆(S , t) · S). (275)

In terms of elasticity Ω, this reduces to

Risk Premium(Option) := γ − r = (α− r)Ω. (276)

Furthermore, we have the Sharpe Ratio for an asset as the ratio of riskpremium to volatility:

Sharpe(Stock) =(α− r)

(α− r)Ω

σΩ= Sharpe(Call). (277)

HW Sharpe Ratio for a put? How about elasticity for a portfolio ofoptions? Now read about Calendar Spreads, Implied Volatility, andPerpetual American Options.

Option Elasticity

γ · V (S , t) = α ·∆(S , t) · S + r ·(

V (S , t)−∆(S , t) · S). (275)

(α− r)Ω

Option Elasticity

γ · V (S , t) = α ·∆(S , t) · S + r ·(

V (S , t)−∆(S , t) · S). (275)

(α− r)Ω

Example: Hedging

Under a standard framework, assume you write a 4− yr European Calloption a non-dividend paying stock with the following:

S0 = 10 = K

σ = 0.2

r = 0.02.

Calculate the initial number of shares of the stock for your hedgingprogram.

Example: Hedging

Recall

∆C = e−δ(T−t)N(d1)

d1 =ln(

)+ (r − δ + 1

2σ2)(T − t)

T − t

d2 = d1 − σ√

T − t.

It follows that

∆C = N(

= 0.6554. (280)

Example: Risk Analysis

Assume that an option is written on an asset S with the followinginformation:

The expected rate of return on the underlying asset is 0.10.

The expected rate of return on a riskless asset is 0.05.

The volatility on the underlying asset is 0.20.

V (S , t) = e−0.05(10−t)(

Compute Ω(S , t) and the Sharpe Ratio for this option.

By definition,

Ω(S , t) =∆ · S

V (S , t)=

S · ∂V (S,t)∂S

V (S , t)

dS (S2eS)

(S2eS)=

S · (2SeS + S2eS)

= 2 + S .

Furthermore, since Ω = 2 + S ≥ 2, we have

Sharpe =0.10− 0.05

0.20= 0.25. (282)

By definition,

Ω(S , t) =∆ · S

V (S , t)=

S · ∂V (S,t)∂S

V (S , t)

dS (S2eS)

(S2eS)=

S · (2SeS + S2eS)

= 2 + S .

Furthermore, since Ω = 2 + S ≥ 2, we have

Sharpe =0.10− 0.05

0.20= 0.25. (282)

Example: Black Scholes Pricing

Consider a portfolio of options on a non-dividend paying stock S thatconsists of a put and a call, both with strike K = 5 = S0. What is the Γfor this option as well as the option value at time 0 if the time toexpiration is T = 4, r = 0.02, σ = 0.2.

In this case,

V = V C + V P

Γ =∂2

(V C + V P

)= 2ΓC .

Consider a portfolio of options on a non-dividend paying stock S thatconsists of a put and a call, both with strike K = 5 = S0. What is the Γfor this option as well as the option value at time 0 if the time toexpiration is T = 4, r = 0.02, σ = 0.2.

In this case,

V = V C + V P

Γ =∂2

(V C + V P

)= 2ΓC .

Consequently, d1 = 0.4 and d2 = 0.4− 0.2√

4 = 0, and so

V (5, 0) = V C (5, 0) + V P(5, 0)

N(d1) + e−4rN(−d2)− e−4rN(d2)− N(−d1))

N(0.4) + e−4rN(0)− e−4rN(0)− N(−0.4))

= 1.5542

Γ(5, 0) =2N ′(0.4)

0.2 · 5 ·√

4= N ′(0.4) =

1√2π

e−0.5·(0.4)2= 0.4322.

Market Making

On a periodic basis, a Market Maker, services the option buyer byrebalancing the portfolio designed to replicate the payoff written into theoption contract.Define

Vi = Option Value i periods from inception

∆i = Delta required i periods from inception

∴ Pi = ∆iSi − Vi

Rebalancing at time i requires an extra (∆i+1 −∆i ) shares.

Market Making

On a periodic basis, a Market Maker, services the option buyer byrebalancing the portfolio designed to replicate the payoff written into theoption contract.Define

Vi = Option Value i periods from inception

∆i = Delta required i periods from inception

∴ Pi = ∆iSi − Vi = Cost of Strategy

Rebalancing at time i requires an extra (∆i+1 −∆i ) shares.

Market Making

Define

∂Si = Si+1 − Si

∂Pi = Pi+1 − Pi

∂∆i = ∆i+1 −∆i

∂Pi = Net Cash Flow = ∆i∂Si − ∂Vi − rPi

= ∆i∂Si − ∂Vi − r(

∆iSi − Vi

) (288)

Under what conditions is the Net Flow = 0?

Market Making

For a continuous rate r , we can see that if ∆ := ∂V∂S , Pt = ∆tSt − Vt ,

dV := V (St + dSt , t + dt)− V (St , t)

≈ Θdt + ∆ · dSt +1

2Γ · (dSt)

⇒ dPt = ∆tdSt − dVt − rPtdt

≈ ∆tdSt −(

Θdt + ∆ · dSt +1

2Γ · (dSt)

)− r (∆tSt − Vt) dt

≈ −

(Θdt + r(∆St − V (St , t))dt +

2Γ · [dSt ]

Market Making

If dt is small, but not infinitessimally small, then on a periodic basis giventhe evolution of St , the periodic jump in value from St → St + dSt may beknown exactly and correspond to a non-zero jump in Market Maker profitdPt .

If dSt · dSt = σ2S2t dt, then if we sample continuously and enforce a zero

net-flow, we retain the BSM PDE for all relevant (S , t):

∂t+ r(

∂S− V

2σ2S2∂

∂S2= 0

V (S ,T ) = G (S) for final time payoff G (S).

Market Making

If dt is small, but not infinitessimally small, then on a periodic basis giventhe evolution of St , the periodic jump in value from St → St + dSt may beknown exactly and correspond to a non-zero jump in Market Maker profitdPt .

If dSt · dSt = σ2S2t dt, then if we sample continuously and enforce a zero

net-flow, we retain the BSM PDE for all relevant (S , t):

∂t+ r(

∂S− V

2σ2S2∂

∂S2= 0

V (S ,T ) = G (S) for final time payoff G (S).

Note: Delta-Gamma Neutrality vs Bond Immunization

In an actuarial analysis of cashflow, a company may wish to immunizeits portfolio. This refers to the relationship between a non-zero valuefor the second derivative with respect to interest rate of the(deterministic) cashflow present value and the subsequent possibilityof a negative PV.

This is similar to the case of market maker with a non-zero Gamma.In the market makers cash flow, a move of dS in the stockcorresponds to a move 1

2 Γ(dS)2 in the portfolio value.

In order to protect against large swings in the stock causing non-lineareffects in the portfolio value, the market maker may choose to offsetpositions in her present holdings to maintain Gamma Neutrality or shewish to maintain Delta Neutrality, although this is only a linear effect.

Option Greeks and Analysis - Some Final Comments

It is important to note the similarities between Market Making andActuarial Reserving. In engineering the portfolio to replicate thepayoff written into the contract, the market maker requires capital.

The idea of Black Scholes Merton pricing is that the portfolio shouldbe self-financing.

One should consider how this compares with the capital required byinsurers to maintain solvency as well as the possibility of obtainingreinsurance.

Practice Solutions

Consider an economy where :

The current exchange rate is x0 = 0.011 $yen .

A four-year dollar-denominated European put option on yen with astrike price of 0.008$ sells for 0.0005$.

The continuously compounded risk-free interest rate on dollars is 3%.

The continuously compounded risk-free interest rate on yen is 1.5%.

Compute the price of a 4−year dollar-denominated European call optionon yens with a strike price of 0.008$.

ANSWER: By put call parity, and the Black Scholes formula, with theasset S as the exchange rate, and the foreign risk-free rate rf = δ,

V C (x0, 0) = V P(x0, 0) + x0e−rf T − Ke−rT

= 0.0005 + 0.011e−0.015·4 − 0.008e−0.03·4

= 0.003764.

Practice Solutions

Consider an economy where :

The current exchange rate is x0 = 0.011 $yen .

A four-year dollar-denominated European put option on yen with astrike price of 0.008$ sells for 0.0005$.

The continuously compounded risk-free interest rate on dollars is 3%.

The continuously compounded risk-free interest rate on yen is 1.5%.

Compute the price of a 4−year dollar-denominated European call optionon yens with a strike price of 0.008$.ANSWER: By put call parity, and the Black Scholes formula, with theasset S as the exchange rate, and the foreign risk-free rate rf = δ,

V C (x0, 0) = V P(x0, 0) + x0e−rf T − Ke−rT

= 0.0005 + 0.011e−0.015·4 − 0.008e−0.03·4

= 0.003764.

Practice Solutions

An investor purchases a 1−year, 50− strike European Call option on anon-dividend paying stock by borrowing at the risk-free rate r . Theinvestor paid V C (S0, 0) = 10. Six months later, the investor finds out thatthe Call option has increased in value by one: V C (S0.05, 0.5) = 11.Assuming (σ, r) = (0.2, 0.02). Should she close out her position after 6months?

ANSWER: Simply put, her profit if she closes out after 6 months is

11− 10e0.02 12 = 0.8995. (292)

So, yes, she should liquidate her position.

Practice Solutions

An investor purchases a 1−year, 50− strike European Call option on anon-dividend paying stock by borrowing at the risk-free rate r . Theinvestor paid V C (S0, 0) = 10. Six months later, the investor finds out thatthe Call option has increased in value by one: V C (S0.05, 0.5) = 11.Assuming (σ, r) = (0.2, 0.02). Should she close out her position after 6months?ANSWER: Simply put, her profit if she closes out after 6 months is

11− 10e0.02 12 = 0.8995. (292)

So, yes, she should liquidate her position.

Practice Solutions

Consider a 1−year at the money European Call option on a non-dividendpaying stock. If you are told that ∆C = 0.65, and the economy bears a1% rate, can you estimate the volatility σ?

ANSWER: By definition,

∆C = e−δTN(d1) = N( r + 1

(0.01 + 12σ

)= 0.65

⇒0.01 + 1

σ= 0.385

⇒ σ ∈ 0.0269, 0.7431 .

More information is needed to choose from the two roots computed above.

Practice Solutions

Consider a 1−year at the money European Call option on a non-dividendpaying stock. If you are told that ∆C = 0.65, and the economy bears a1% rate, can you estimate the volatility σ?ANSWER: By definition,

∆C = e−δTN(d1) = N( r + 1

(0.01 + 12σ

)= 0.65

⇒0.01 + 1

σ= 0.385

⇒ σ ∈ 0.0269, 0.7431 .

Practice Solutions

∆C = e−δTN(d1) = N( r + 1

(0.01 + 12σ

)= 0.65

⇒0.01 + 1

σ= 0.385

⇒ σ ∈ 0.0269, 0.7431 .

Practice Solutions

∆C = e−δTN(d1) = N( r + 1

(0.01 + 12σ

)= 0.65

⇒0.01 + 1

σ= 0.385

⇒ σ ∈ 0.0269, 0.7431 .

Review

Are you comfortable with...

The definition of the Black-Scholes pricing formulae for Europeanputs and calls.

What are the Greeks? Given a specific option, could you compute theGreeks?

What is the Option Elasticity? How is it useful? How about theSharpe ratio of an option? Can you compute the Elasticity andSharpe ration of a given option?

Review

And...

What is Delta Hedging?

If the Delta and Gamma values of an option are known, can youcalculate the change in option value given a small change in theunderlying asset value?

How does this correspond the Market Maker’s profit?

stochastic calculus for finance - michigan state...

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