FINANCIAL OPTION ANALYSIS HANDOUTS

FINANCIAL OPTION ANALYSIS

HANDOUTS

FAIR PRICING

There is a market for an object called “S”. The prevailing price “today” is S0 = 100. At this price the

object “S” can be bought or sold by anyone for any number. A person buying the object “S” today must pay 100 now, and a person selling the object “S” will receive 100 now. It is common knowledge that the prevailing price “tomorrow” S1 will either be S1(u) = 125 if the market for “S” is “up” or S1(d) = 80 if the

market for “S” is “down”. A person who purchases a unit of “S” today will receive either 125 or 80 tomorrow (depending on the outcome), and the person who sells a unit of “S” today is obligated to “buy back” a unit of “S” at the new prevailing market price (125 or 80).

There is also a market for an object called “C”. It may be bought or sold by anyone for any number. A person who buys the object “C” today must pay $C0 now, and a person who sells the object “C” will

receive $C0 now. A person who buys “C” today has the option but not the obligation to buy 1 unit of the

object “S” tomorrow at today’s market price of 100. A person who sells “C” today has the obligation of supplying a unit of “S” tomorrow for the price of 100 if a holder of a “C” wishes to exercise his option.

There is also a market for dollars, the so-called money market M. A person who buys m dollars today receives m dollars today and owes (1.10)m tomorrow; that is, the person is taking out a loan. A person who sells m dollars today must give m dollars today (to the person buying), and will receive (1.10)m the next period; that is, when a person sells dollars today, they are acting like a banker in that they are loaning money to the person buying.

OPTION EXAMPLE 1

In each period the stock price may go up by a factor of u = 1.25 or down by a factor of d = 0.80. The stock price at time 0 is 100. Risk-free rate each period is a constant 10%. Price a one-period call option with strike price 100. Determine the self-financed replicating portfolio.

125 25 2/3 100 15.15 0.555S – 40.40M 1/3 80 0 NOTES 1. 0.5555 = ΔC/ΔS = [25 – 0] ÷ [125 – 80]

2. The value of the call option at time 0 = (1/1.11)[ 2/3*25 + 1/3*0] = 15.15 3. The money market amount = 15.15 – 0.5555[100] = -40.40

4. State Equations:

h*S1(u) + m*R = C1(u) 125h + 1.1m = 25

OPTION EXAMPLE 2

In each period the stock price may go up by a factor of u = 1.25 or down by a factor of d = 0.80. The stock price at time 0 is 100. Risk-free rate each period is a constant 10%. Price a one-period put option with strike price 100. Determine the self-financed replicating portfolio.

125 0 2/3 100 6.06 -0.444S + 50.50M 1/3 80 20 NOTES 5. -0.444 = ΔC/ΔS = [0 – 20] ÷ [125 – 80]

6. The value of the put option at time 0 = (1/1.11)[ 2/3*0 + 1/3*20] = 6.06 7. The money market amount = 6.06 – -0.444*[100] = 50.50

8. State Equations:

h*S1(u) + m*R = P1(u) 125h + 1.1m = 0

OPTION EXAMPLE 3

In each period the stock price may go up by a factor of u = 1.25 or down by a factor of d = 0.80. The stock price at time 0 is 100. Risk-free rate each period is a constant 10%. Price a 2-period call option with strike price 100. Determine the self-financed replicating portfolio.

156.25 56.25 2/3 125 34.09 2/3 1.00S – 90.90M 100 1/3 100 20.666 0 0.7575S – 55.10M 2/3 1/3 80 0 1/3 64 0 NOTES 9. a. 56.25 = max(156.25 – 100, 0). b. 0 = max(100 – 100, 0) c. 0 = max(64 – 100, 0) d. 34.09 = (1/1.1)[2/3*56.25 + 1/3*0] e. 1.00 = ΔC/ΔS = [56.25 – 0] ÷ [156.25 – 100] f. -90.90 = 34.09 – 1.00*125 g. 20.666 = (1/1.1)[2/3*34.09 + 1/3*0] ΔC/ΔS = [34.09 – 0] ÷ [125 – 80]

OPTION EXAMPLE 4

In each period the stock price may go up by a factor of u = 1.25 or down by a factor of d = 0.80. The stock price at time 0 is 100. Risk-free rate each period is a constant 10%. Price a 2-period put option with strike price 100. Determine the self-financed replicating portfolio.

156.25 0 2/3 125 0 2/3 100 1/3 100 3.306 0 -0.242S + 27.55M 2/3 1/3 80 10.90 -1.00S + 90.90M 1/3 64 36 NOTES 11. a. 0 = max(100 – 156.25, 0). b. 0 = max(100 – 100, 0) c. 36 = max(100 – 64, 0) d. 10.90 = (1/1.1)[2/3*0 + 1/3*36] e. -1.00 = ΔC/ΔS = [0 – 36] ÷ [100 – 64] f. 90.90 = 10.90 – -1.00*80 g 3.306 = (1/1.1)[2/3*0 + 1/3*10.90] h 0.2424 = ΔC/ΔS = [0 – 10.90] ÷ [125 – 80] i 27.55 = 3.306 – -0.2424*100

OPTION EXAMPLE 5

In each period the stock price may go up by a factor of u = 1.25 or down by a factor of d = 0.80. The stock price at time 0 is 100. Risk-free rate each period is a constant 10%. Price a 2-period call option with strike price 80. Determine the self-financed replicating portfolio.

156.25 76.25 2/3 125 52.27 2/3 1.00S – 72.72M 100 1/3 100 35.36 20 0.8922S – 53.87M 2/3 1/3 80 12.12 0.5556S – 32.32M 1/3 64 0 NOTES 14. a. 76.25 = max(156.25 – 80, 0). b. 20 = max(100 – 80, 0) c. 0 = max(64 – 80, 0) d. 52.27 = (1/1.1)[2/3*76.25 + 1/3*20] e. 1.00 = ΔC/ΔS = [76.25 – 20] ÷ [156.25 – 100] f. -72.72 = 52.27 – 1.00*125 g. 12.12 = (1/1.1)[2/3*20 + 1/3*0] ΔC/ΔS = [20 – 0] ÷ [100 – 64]

OPTION EXAMPLE 6

In each period the stock price may go up by a factor of u = 1.25 or down by a factor of d = 0.80. The stock price at time 0 is 100. Risk-free rate each period is a constant 10%. Price a 2-period put option with strike price 80. Determine the self-financed replicating portfolio.

156.25 0 2/3 125 0 2/3 100 1/3 100 1.469 0 -0.107S + 12.24M 2/3 1/3 80 4.84 -0.444S + 40.40M 1/3 64 16 NOTES 16. a. 0 = max(80 – 156.25, 0). b. 0 = max(80 – 100, 0) c. 16 = max(80 – 64, 0) d. 4.84 = (1/1.1)[2/3*0 + 1/3*16] e. -0.444 = ΔC/ΔS = [0 – 16] ÷ [100 – 64] f. 40.40 = 4.84 – -0.444*80 g. 1.469 = (1/1.1)[2/3*0 + 1/3*4.84] h. -0.107 = ΔC/ΔS = [0 – 4.84] ÷ [125 – 80] i. 12.24 = 1.469 – -0.1077*100

CHOOSER OPTION

Consider a non-dividend paying stock whose initial stock price is 62 and which has a

log-volatility of σ = 0.20. The interest rate r = 2.5% continuously compounded.

Consider a 5-month option with a strike price of 60 in which after exactly 3 months the

purchaser may declare this option to be a (European) call or put option.

QUESTIONS:

1. Determine the value of u and d for the binomial lattice.

The value for U = exp{σ(Δt)

} = exp{0.20(1/12)

} = 1.05943.

Note that D = 1/U = 0.94390.

2. Determine the values for the binomial lattice for 5 1-month periods.

0 1 2

3

4 5

62.00 65.68 69.59

73.72

78.11 82.75

58.52 62.00

65.68

69.59 73.72

55.24

58.52

62.00 65.68

Stock Price

52.14

55.24

58.52

49.21

52.14

46.45

3. Determine the appropriate risk-free rate.

The interest rate per month R = exp(0.025*1/12) = 1.00209.

4. Determine the risk-neutral probability q of going “UP”.

The value for q satisfies q(US

) + (1-q)(DS

) = RS

, which implies that

5. Determine the values for the call option and put option along the lattice.

0 1 2

3

4 5

4.6686 7.0172 10.1423

13.9743

18.2315 22.7488

2.3054 3.876

6.2971

9.7137

13.7248

0.7216

1.4359

2.8571 5.6849

Call Option

0

0

0

0

0

0

0 1 2

3

4 5

2.0469 0.8344 0.1797

0

0

0

3.2858 1.5022

0.3627

0

0

5.1091

2.6646

0.7322

0

Put Option

7.6107

4.6364

1.4782

10.6604

7.8602

13.5462

6. Find the value of this Chooser Option.

Compute the terminal value of the Chooser Option at t = 3 as the maximum of the call

and put options at t = 3. From there we work backwards in the usual manner.

0 1 2

3

4 5

6.0483 7.3187

10.1423

13.9743

4.7847 4.4847

6.2971

5.1091

2.6646

Table 1: A stochastic volatility, random interest rate model

t = 0 t = 1 t = 2

S0 = 4, r0 = 25% S1(U ) = 8, r1(U ) = 25% S2(U U ) = 12

S1(D) = 2, r1(D) = 50% S2(U D) = S2(DU ) = 8

S2(DD) = 2

Option Analysis with Stochastic Interest Rates

In this problem we consider a two-period, stochastic volatility, random interest rate model. The stock prices and interest rates are provided in Table 1.

Consider the European option whose final payoffs are V2= max{S2− 7, 0}. Determine the

Table 2: Solution to the stochastic volatility, random interest rate model

t = 0 t = 1 t = 2

1.00¯4 = _1.251 [0.5(2.4) + 0.5(0.¯1)] 2.4 = _1.251 [0.5(5) + 0.5(1)] 5 0.¯1 = _1.51 [(1/6)(1) + (5/6)(0)] 1 0

The risk-neutral q changes along the tree since the risk-free rate is stochastic. Otherwise, all the calculations are the same, since at each node the replicating portfolio and corresponding risk-neutral discounted expectation ideas still apply. The solution is provided in Table 2.

Asian Option

Consider a non-dividend paying stock S whose price process follows a binomial lattice with U = 2 and D = 0.5. R = 1.25 and S0= 4. Define

Yt:= t

X

k=0

Sk, t = 0, 1, 2, 3

to be the sum of the stock prices between times zero and t.

1. Consider a (European) Asian call option that expires at time three and has a strike price K = 4; that is, its payoff at time three is

maxnY3 4 − 4, 0

o

.

This is like a European call option, except the payoff of the option is based on the average stock price rather than the final stock price. Let Vt(s, y) denote the price of this option

at time n if St= s and Yt= y. In particular,

V3(s, y) = max

ny

4 − 4, 0

o

.

(a) Develop an algorithm for computing Vt recursively. In particular, write a formula

for Vt in terms of Vt+1.

(b) Apply the algorithm developed in (a) to compute V0(4, 4), the price of the Asian

option at time zero.

(c) Provide a formula for δt(s, y), the number of shares of stock that should be held by

the replicating portfolio at time t if St= s and Yt= y.

2. What is the value of a (European) Asian put option that expires at time three and has a strike price K = 4; that is, its payoff at time three is

maxn4 −Y3 4 , 0

o

1. Risk-neutral q = 0.5. (a) Vt(s, y) = (1/1.25)  0.5Vt+1(us, y + us) + 0.5Vt+1(ds, y + ds)  . (b) V3(32, 60) = 11; V3(8, 36) = 5; V3(8, 24) = 2; V3(2, 18) = 0.5; V3(8, 18) = 0.5; V3(2, 12) = V3(2, 9) = V3(0.5, 7.5) = 0. V2(16, 28) = (1/1.25)[0.5(11) + 0.5(5)] = 6.40. V2(4, 16) = (1/1.25)[0.5(2) + 0.5(0.5)] = 1.0. V2(4, 10) = (1/1.25)[0.5(0.5) + 0.5(0)] = 0.20. V2(1, 7) = (1/1.25)[0.5(0) + 0.5(0)] = 0. V1(8, 12) = (1/1.25)[0.5(6.4) + 0.5(1.0)] = 2.96. V1(2, 6) = (1/1.25)[0.5(0.2) + 0.5(0)] = 0.08. V0(4, 4) = (1/1.25)[0.5(2.96) + 0.5(0.08)] = 1.216.

Remark. The value of this Asian option, 1.216, equals the discounted expectation of the final payoffs using the risk-free rate and the risk-neutral probability, i.e.,

1 (1.25)3

11 + 5 + 2 + 0.5 + 0.5 + 0 + 0 + 0

8 .

Simulation is an especially useful computational approach for valuing path-dependent, European-style derivative securities, since their value can be obtained as a (discounted) sample average of the value along a “sample path.”

(c)

δt(s, y) =

Vt+1(us, y + us) − Vt+1(ds, y + ds)

No Arbitrage Bounds

Consider a family of call options on a non-dividend paying stock, each option being identical except for its strike price. The value of the call with strike price K is denoted by C(K). Prove the following two general relations using arbitrage arguments:

1. If K2 > K1, then K2− K1 ≥ C(K1) − C(K2). 2. If K3 > K2 > K1, then C(K2) ≤ K₃− K₂ K3− K1  C(K1) + K₂− K₁ K3− K1  C(K3).

Hint: For both parts find a portfolio that is guaranteed to have no negative but sometimes positive final payoffs. If there is to be no arbitrage, such a portfolio must cost something to acquire today. In each part plot the final payoffs for each portfolio, otherwise known as the payoff diagram.

Solution: For both parts below one finds a portfolio that is guaranteed to have no negative but sometimes positive final payoffs. If there is to be no arbitrage, such a portfolio must cost something to acquire today. In each case below it will be instructive to plot the final payoffs for each portfolio, otherwise known as the payoff diagram.

1. Verify that the portfolio −C(K1) + C(K2) + (K2− K1) has no negative but sometimes

positive final payoffs. Hence, its cost today must be nonnegative, which establishes the result.

2. Consider the portfolio mC(K1) − C(K2) + nC(K3) with m, n > 0 and m < 1. This

portfolio’s final payoffs are zero if ST ≤ K1 and will be positive on the interval K1 ≤

ST ≤ K2. Since m < 1 the payoffs decline on the interval K2 ≤ ST ≤ K3. If we set m so

that

m(K3− K1) = (K3− K2),

then the payoffs will remain nonnegative in the interval K2 ≤ ST ≤ K3. In particular,

the payoff when ST = K3 will be zero. If we further set n so that m + n = 1, the final

payoffs will be zero on the interval ST ≥ K3. We conclude that the portfolio

K₃− K₂ K3− K1  C(K1) − C(K2) + K₂− K₁ K3− K1  C(K3)

has no negative but sometimes positive final payoffs. Hence, its cost today must be nonnegative, which establishes the result.

FINANCIAL OPTION ANALYSIS PROBLEMS

A. We consider a single period binomial lattice with S0 = 50, u = 1.20 (d = 1/1.20) and R = 1.04.

1. a. What is the objective probability of an upward movement in the stock price, p, if the market’s required expected percentage return on the stock, rS, is 8%? [p = 0.6727]

b. Suppose the objective probability of an upward movement in the stock price p is 0.70. What is the expected percentage return, rS , on the stock? [rS = 9%]

2. A derivative security C has final payoffs given by C1 = (max [S1 – 50, 0])2, where S1 is the final stock

price. Assume an objective probability of an upward movement in the stock price p = 0.70. a. Determine the no-arbitrage value C0 for C. [54.1958]

b. Determine the market’s expected percentage return on C (using objective probabilities). [29.16%] c. Determine the replicating portfolio hS + mM. [5.4545S - 218.5315M]

d. Determine the portfolio weights, wS and wM, on the stock and the money market in the replicating

portfolio. [wS = 5.032, wM = -4.032]

e. Determine the portfolio’s expected percentage return using the portfolio weights and the expected percentage returns on the stock and bond. [29.16%]

B. A non-dividend paying stock has an initial price = 100. Its price path is modeled as a binomial lattice with U = 1.3 and D = 1/1.3. Period length = 1 year. The risk-free rate is 6% per year.

1. Determine the value of a 2 year European put option with strike price K = 100.

2. An American put permits the holder to exercise at any date t, and receive the intrinsic value

max (0, K – St). Hence, at each time t, the holder can take one of two possible actions: no exercise, as in a

European option, or exercise. Determine the value of a 2 year American put option with strike price K = 100.

C. Smith knows that the value of a six month European call option with strike price K = 24 on a non-dividend paying stock is 6.80. The current value of the stock is 26. However, he wishes to price a

European put option on the same stock with the same strike price and maturity. He knows that the risk-free rate is 2.50% per year, but he is stuck because he does not know the log-volatility σ upon which he would calculate the value of U. He comes to you for help. What say you?

D. Do the following markets exhibit arbitrage? If so, demonstrate. If not, state why not. 1. Securities S1 S2 price at time 0 5 10 up state at time 1 20 60 down state at time 1 10 30 2.

Securities

S1 S2 S3 price at time 0 60 115 1

up state at time 1 90 100 1.05 down state at time 1 40 150 1.05 3.

Securities

S1 S2 S3 S4 S5 S6 S7 S8 S9 Price at time 0 20 1 100 100 100 100 200 200 305 Up state at time 1 28 1.05 100 110 140 50 260 160 385 Down state at time 1 14 1.05 110 100 70 160 160 260 245

E. The evolution of the stock price over 2 periods is shown in the figure below. Let S2 denote the

(random) value of the stock price at t = 2. The appropriate risk-adjusted rate of return (cost of capital) is 20% per period. The risk-free rate is 4% per period. In this problem we shall consider pricing a European “square root” derivative security with strike price 140 that pays off ( | S2 – 140 | )1/2 at time t =2.

220

150

115 120

90

80

1. Determine the objective probabilities along the lattice.

2. Determine the final period payoffs for the “square root” option. (Enter them in the figure.)

3. Determine the Decision-Tree value of the “square root” option by (1) first computing the expectation of the final period payoffs using the objective probabilities, and then (2) discounting using the risk-adjusted rate of return.

4. Fill out the above figure by placing the correct value of the “square root” option as the 2nd _{entry, and}

recording the self-financed replicating portfolio as the 3rd _{entry. Determine the risk-neutral probabilities}

along the lattice and mark them with an asterisk.

5. Suppose the current market value of the “square root” option is the Decision-Tree value. Conceptually explain how you would use your answer to (d) to obtain a risk-free profit from the incorrect pricing. Be specific about your dynamic trading strategy. (No further calculations are required.)

Sample Worksheet 1: 220 150 115 120 90 80 Sample Worksheet 2: 220 150 115 120 90

/** FOA_VanillaOptionAnalysis.hava */ import Lattice;

import FOA_ReplicatingPortfolio;

/* PROBLEM CLASSIFICATION */

token EUROPEAN, AMERICAN, CALL, PUT; /** */ type = AMERICAN; kind = PUT; strike = 60; r = 0.10; p0 = 62; maturity = 5/12; sigma = 0.20; dt = 1/12; dividendRate = 0.00; /** */ /* PROBLEM SPECIFICATION */ numPeriods = round(maturity/dt);

initialState = State(numPeriods, p0, false); private SOLUTION = sdp_Result(initialState); /** */

/* STATE DEFINITION */

// n = date, measured in periods // p = asset price on this date // eState = TRUE (if exercised) struct State(n, p, eState);

/* ACTION SET DEFINITION */

// Actions = exercise or continue (depending on option type) token CONT, EX;

function sdp_ActionSet(state)= if (state.n == 0) {collect(EX)}

else if (type == AMERICAN) {(CONT, EX)} else {collect(CONT)};

/* NEXT STATE DEFINITION */

function sdp_NextState(state, action, event) = if (action == CONT)

{State(state.n-1, (1-dividendRate*dt)*event.outcome, false)} else

{State(state.n, state.p, true)};

/* DISCOUNT DEFINITION */ discount = exp(-r*dt);

private sdp_Discount(state, action, event) = discount;

/* TERMINAL CONDITION DEFINITION */

function sdp_TerminalCondition(state) = (state.eState == true);

/* SAMPLE SPACE GENERATION */

private lattice = BinomialLattice(sigma, r, dt); function sdp_GenerateSampleSpace(state, action) = BL_GenerateSampleSpace(state, action, lattice); table latticeData = BL_LatticeData(lattice); /*

private lattice = UpDownLattice(U, D, r, dt); function sdp_GenerateSampleSpace(state, action) = UD_GenerateSampleSpace(state, action, lattice); table latticeData = UD_LatticeData(lattice); */

/* PROBLEM REPORTING */

function sdp_Information(state, action) = rp_ReplicatingPortfolio(state, action);

FOA_VanillaOptionAnalysis.hava type AMERICAN kind PUT strike 60 r 0.1 p0 62 maturity 0.416667 sigma 0.2 dt 0.0833333 dividendRate 0.0 numPeriods 5

initialState State(5, 62, false)

SDP_AO sdp_Result state value

n p eState policy information

action value hRatio cash assetW

0 46.4538 true IGNORE 0.0 0 46.4538 false EX 13.5462 0 52.1398 true IGNORE 0.0 0 52.1398 false EX 7.8602 0 52.1398 true IGNORE 0.0 0 52.1398 false EX 7.8602 0 58.5218 true IGNORE 0.0 0 58.5218 false EX 1.4782 0 58.5218 true IGNORE 0.0 0 58.5218 false EX 1.4782 0 58.5218 true IGNORE 0.0 0 58.5218 false EX 1.4782 0 65.6849 true IGNORE 0.0 0 65.6849 false EX 0.0 0 65.6849 true IGNORE 0.0 0 65.6849 false EX 0.0 0 73.7248 true IGNORE 0.0

1 62.0 false CONT 0.6479 -0.2064 13.4449 -19.7463 1 69.5889 true IGNORE 0.0 1 69.5889 false CONT 0.0 0.0 0.0 1 78.1066 true IGNORE 0.0 1 78.1066 false CONT 0.0 0.0 0.0 2 52.1398 true IGNORE 0.0 2 52.1398 false EX 7.8602 2 58.5218 true IGNORE 0.0 2 58.5218 false CONT 2.4456 -0.6084 38.0503 -14.5585 2 58.5218 true IGNORE 0.0 2 58.5218 false CONT 2.4456 -0.6084 38.0503 -14.5585 2 65.6849 true IGNORE 0.0 2 65.6849 false CONT 0.284 -0.0854 5.89357 -19.7463 2 73.7248 true IGNORE 0.0 2 73.7248 false CONT 0.0 0.0 0.0 3 55.2387 true IGNORE 0.0 3 55.2387 false CONT 4.7987 -0.8484 51.6632 -9.76623 3 62.0 true IGNORE 0.0 3 62.0 false CONT 1.2291 -0.3018 19.9408 -15.2221 3 69.5889 true IGNORE 0.0 3 69.5889 false CONT 0.1245 -0.0353 2.58089 -19.7463 4 58.5218 true IGNORE 0.0 4 58.5218 false CONT 2.7836 -0.5279 33.6771 -11.0991 4 65.6849 true IGNORE 0.0 4 65.6849 false CONT 0.6076 -0.1456 10.1715 -15.7351 5 62 true IGNORE 0.0 5 62 false CONT 1.5564 -0.3038 20.392 -12.1018

SDP_AO sdp_ExpValue state action value

n p eState

0 46.4538 false EX 13.5462

0 52.1398 false EX 7.8602

1 62.0 false CONT 0.6479 1 62.0 false EX 0.0 1 69.5889 false CONT 0.0 1 69.5889 false EX 0.0 1 78.1066 false CONT 0.0 1 78.1066 false EX 0.0 2 52.1398 false CONT 7.3623 2 52.1398 false EX 7.8602 2 58.5218 false CONT 2.4456 2 58.5218 false EX 1.4782 2 58.5218 false CONT 2.4456 2 58.5218 false EX 1.4782 2 65.6849 false CONT 0.284 2 65.6849 false EX 0.0 2 73.7248 false CONT 0.0 2 73.7248 false EX 0.0 3 55.2387 false CONT 4.7987 3 55.2387 false EX 4.7613 3 62.0 false CONT 1.2291 3 62.0 false EX 0.0 3 69.5889 false CONT 0.1245 3 69.5889 false EX 0.0 4 58.5218 false CONT 2.7836 4 58.5218 false EX 1.4782 4 65.6849 false CONT 0.6076 4 65.6849 false EX 0.0 5 62 false CONT 1.5564 5 62 false EX 0.0

SDP_AO sdp_ExpPresentValue state action value

n p eState 0 46.4538 false EX 0.0 0 52.1398 false EX 0.0 0 52.1398 false EX 0.0 0 58.5218 false EX 0.0 0 58.5218 false EX 0.0

1 62.0 false EX 0.0 1 62.0 false CONT 0.647942 1 62.0 false EX 0.0 1 69.5889 false CONT 0.0 1 69.5889 false EX 0.0 1 78.1066 false CONT 0.0 1 78.1066 false EX 0.0 2 52.1398 false CONT 7.36225 2 52.1398 false EX 0.0 2 58.5218 false CONT 2.44556 2 58.5218 false EX 0.0 2 58.5218 false CONT 2.44556 2 58.5218 false EX 0.0 2 65.6849 false CONT 0.283995 2 65.6849 false EX 0.0 2 73.7248 false CONT 0.0 2 73.7248 false EX 0.0 3 55.2387 false CONT 4.7987 3 55.2387 false EX 0.0 3 62.0 false CONT 1.22914 3 62.0 false EX 0.0 3 69.5889 false CONT 0.124486 3 69.5889 false EX 0.0 4 58.5218 false CONT 2.78357 4 58.5218 false EX 0.0 4 65.6849 false CONT 0.607648 4 65.6849 false EX 0.0 5 62 false CONT 1.55637 5 62 false EX 0.0

sdp_CashFlow state action value

n p eState

1 62.0 false CONT 0 1 62.0 false EX 0 1 62.0 false CONT 0 1 62.0 false EX 0 1 69.5889 false CONT 0 1 69.5889 false EX 0 1 78.1066 false CONT 0 1 78.1066 false EX 0 2 52.1398 false CONT 0 2 52.1398 false EX 7.86016 2 58.5218 false CONT 0 2 58.5218 false EX 1.4782 2 58.5218 false CONT 0 2 58.5218 false EX 1.4782 2 65.6849 false CONT 0 2 65.6849 false EX 0 2 73.7248 false CONT 0 2 73.7248 false EX 0 3 55.2387 false CONT 0 3 55.2387 false EX 4.76127 3 62.0 false CONT 0 3 62.0 false EX 0 3 69.5889 false CONT 0 3 69.5889 false EX 0 4 58.5218 false CONT 0 4 58.5218 false EX 1.4782 4 65.6849 false CONT 0 4 65.6849 false EX 0 5 62 false CONT 0 5 62 false EX 0 discount 0.991701 latticeData UP 1.05943 DOWN 0.9439 RETURN 1.00837

FINANICAL OPTION ANALYSIS SAMPLE PROBLEMS

1. The price of asset S at time t = 0 is S0 = 20. The asset’s value at time t = 1 is S1(u) = 28 in the up state

is S1(d) = 14 in the down state. The risk-free rate is 5%. The objective probability of the up state is 0.60.

A derivative security V has final payoffs at time t = 1 of V1(u) = 126 in the up state and V1(d) = 84 in the

down state.

a. Determine the expected return, rS, on the asset S.

b. Determine the no-arbitrage value for V. c. Determine the replicating portfolio for V.

d. Determine the portfolio weights of S and M in the replicating portfolio for V. e. Determine the expected return, rV, on the derivative security V.

2. A non-dividend paying stock has an initial price S0 = 100. It’s price path is modeled as a binomial

lattice with u = 1.25 and d = 0.80. Period length is 1 year. The risk-free rate is 10% per year. a. Determine the value of a 2 year American put option with strike price K = 90.

b. Determine the value of a 2 year European put option with strike price K = 90. c. Use Put-Call Parity to value a 2 year European call option with strike price K = 90.

3. Does the following market exhibit arbitrage? If so, provide a concrete example of arbitrage. If not, state why not.

Securities

S1 S2 S3 S4 S5 S6 S7 S8 S9 Price at time 0 20 1 100 100 100 100 200 200 305 Up state at time 1 28 1.05 100 110 140 50 260 160 385 Down state at time 1 14 1.05 110 100 70 160 160 260 245 4. The price process of a non-dividend paying stock over the next 2 years is shown in the following table:

t = 0 t = 1 t = 2

100 130 171

80 102

57 The risk-free rate is 5% per year. Define Y2 = S0 + S1 + S2 denote the sum of the stock prices between

times zero and 2. A European Asian put option that expires at time t = 2 and has strike price K = 104 has its payoff at time t = 2 equal to max{104 - Y2 /3, 0}. (This is like a European put option, except that the

payoff of this option is based on the average stock price rather than the final stock price.) Determine the no-arbitrage value of this path-dependent option.

5. The evolution of the stock price over 2 periods is shown in Figure 1 below. Let S2 denote the (random)

value of the stock price at t = 2. The appropriate risk-adjusted rate of return (cost of capital) is 20% per period. The risk-free rate is 4% per period. In this problem we wish to price a European derivative security that pays off ( | S2 – 64 | + | S2 – 114 | + | S2 – 181.50 | ) at time t =2.

a. For each box in Figure 1 place the correct value of the derivative security as the 2nd entry and record the self-financed replicating portfolio as the 3rd entry.

Figure 1. 181.50 130 100 114 80 64

b. Determine the value at time t = 0 of this derivative security according to DTA.

c. Suppose the current market value of the derivative security is 150. Exactly explain how you could guarantee a risk-free profit from the incorrect pricing. Be specific with numbers.

6. Consider dividend-price data for a complete, no-arbitrage market with the following three securities: Security 1 Security 2 Security 3 Payoff Vector V

Price at t = 0 100 80 1.0 ?

7. a. Does the following market exhibit arbitrage? If so, provide a concrete example of arbitrage. If not, state why not.

Securities S1 S2 S3 S4 S5 Price at time 0 60 8 60 40 22 State 1 125 0 110 0 22 State 2 80 0 110 0 11 State 3 0 28 0 105 21 State 4 0 14 0 105 42

b. Determine the risk-free rate.

8. The evolution of the stock price over 2 periods is shown in Figure 1 below. Let S2 denote the (random)

value of the stock price at t = 2. The appropriate risk-adjusted rate of return (cost of capital) is 20% per period. The risk-free rate is 5% per period. In this problem we wish to price a European derivative security that pays off | S2 – 119 |1/2 at time t =2.

a. Determine the no-arbitrage value of this derivative security. b. Determine the replicating portfolio at time 0.

Figure 1. 189 160 100 119 70 49

FINANCIAL OPTION ANALYSIS SAMPLE PROBLEM SOLUTIONS

1. a. rS = [0.6(28) + 0.4(14)]/20 - 1 = 0.12.

b. Since [0.5(28) + 0.5(14)]/1.05 = 20, q = 0.5. Thus, V0 = [0.5(126) + 0.5(84)]/1.05 = 100.

c. (126-84)/(28-14) = 3. So h = 3. Thus, replicating portfolio is 3S + 40M. d. 60/100 is invested in S, so wS = 0.6 and wM = 0.4. e. rV = [0.6(126) + 0.4(84)]/100 - 1 = 0.092. 2. a. t = 0 t = 1 t = 2 100 125 [0] 156.25 [0] 3.03 = 1/3(10)/1.1 80* [10] 100 [0] since 10 > 1/3(26)/1.1 64 [26] b. (1/9)(26)/(1.1)2 = 2.3875. c. 90/(1.1)2 = S + P – C = 100 + 2.3875 – C. C = 28.0073.

3. First four securities show that q = 0.5.

To be consistent, the value S9 = [0.5(385) + 0.5(245)]/1.05 = 300, which is less than 305. The portfolio 10S1 + 100S2 replicates S9 and costs 300.

So sell S9 for 305 and buy this replicating portfolio for 300, and pocket the difference of 5. 4. There are four possible paths.

The average values along the uu, ud, du, and dd paths are, respectively, 133.66, 110.67, 94, and 79. The values of the Asian put option for these paths are, respectively, 0, 0, 10, and 25.

The q probability of path du is (0.5)(0.6) = 0.3, and the q probability of path dd is (0.5)(0.4) = 0.2. Therefore, the value of this option is [(0.3)(10) + (0.2)(25)]/(1.05)2 = 7.2562.

5. a. Figure 1. Objective probabilities are in parentheses.

181.50 185 0.314 (0.622) 130 133.36 0.48 (0.80) S+3.36M 100 114 132.85 0.686 (0.377) 117.5

b. [185(0.80*0.622) + 117.5(0.80*0.377 + 0.20*0.64) + 167.5(0.20*0.36)] / (1.2)2 = 107.43. c. Derivative security is overpriced, so you would want to sell it. Collect 150 and use 132.85 of it to purchase the replicating portfolio of –0.1848S + 151.33B. Invest the 17.15 in the bank or buy lunch. If the price goes up next period to 130, rebalance the portfolio to S + 3.36B, which you can afford to do since it will cost 133.36 and this is precisely what the portfolio –0.1848S + 151.33B equals when the price = 130. If the price goes down next period to 80, rebalance the portfolio to –S + 222.60, which you can afford to do by the same reasoning as before. Finally, when period 2 comes around, your updated portfolio will exactly match the final payoffs of the derivative security (185, 117.5 or 167.5) regardless of the final state and so you will be able to meet your obligations. (If you were forced to buy back the derivative security at time 1, you would have the exact money to do so.)

6. a. Since S2 does not payoff in either state 1 or 2 only S1 and S3 can be used to replicate the payoffs of the Vector V in these two states. (S3 is of course our old friend M.) We’re back to the “hS1 + M”: here, h = (420-460)/(200-240) = 1 and M = S3 = 200. Now use state 3 to pin down the number of units of S2 to hold: 176(S2) + 1.1(200) = 44

⇒

b. Let q = (q1, q2, q3) denote the risk-neutral probability vector. Obviously R = 1.1.

Discounted expectation using q and R applied to S2 implies that (1/1.1)176q3 = 80

⇒

Thus, q1 + q2 = 0.5. Discounted expectation using q and R applied to S1 implies that

(1/1.1)[200q1 + 240q2] = 100 q1 =q2 = 0.25. Risk-neutral valuation says that (1/R)Eq[V] =

(1/R) qTV = p for any vector V. Thus, the value of V is (1/1.1)[0.25(420) + 0.25(460) + 0.5(44)] = 220, which coincides with the answer in part (a) as it should.

⇒

7. a. We can use the first four assets to determine the state-prices. The equations are:

125y1 + 80y2 = 60. 110y1 + 110y2 = 60. These equations imply that y1 = 4/11, y2 = 2/11.

28y3 + 14y4 = 8. 105y3 + 105y4 = 40. These equations imply that y3 = 4/21, y4 = 4/21.

Now we can use these state-prices to determine the no-arbitrage value of asset 5 as: (4/11)22 + (2/11)11 + (4/21)21 + (4/21)42 = 22.

Since this IS the price of asset 5, this market does NOT exhibit arbitrage.

b. Recall that the reciprocal of the sum of the state-prices equals R = 1 + risk-free rate. Thus, R = 1/(4/11 + 2/11 + 4/21 + 4/21) = 231/214 = 1.0794. r = 7.94%.

Alternatively, one can easily combine assets 3 and 4 to obtain a constant payoff vector. For example, a purchase of 1 unit of asset 3 and 110/105 units of asset 4 yields a constant payoff of 110. The cost of this portfolio is 60(1) + 40(110/105) = 101.905. Thus, the total return on the risk-free security is 110/101.905 = 1.0794, same as above, as it should.

8. Figure 1. 189 11.83 0.7 160 7.888 0.389 0.169S-19.152M 0.3 100 119 7.185 0 0.00626S+6.559M 0.611 0.35 70 7.325 -0.169S+19.155M 0.65 49 11.83 9.

Stock price paths Lookback option value over time

4 8 16 uuu: 32 1.376 2.24 3.2 32-32 = 0 uud: 8 16-8 = 8 4 udu: 8 2.4 8-8 = 0 udd: 2 8-2 = 6 2 4 duu: 8 1.20 0.8 8-8 = 0 dud: 2 4-2 = 2 1 ddu: 2 2.2 4-2 = 2 ddd: 0.5 4-0.5 = 3.5 Risk-neutral probability = 0.5. 1.376 = [(0 + 8 + 0 + 6 + 0 + 2 + 2 + 3.5)/8]/(1.25)3. 2.24 = [(0 + 8 + 0 + 6)/4](1.25)2. 1.20 = [(0 + 2 + 2 + 3.5)/4](1.25)2.