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© Actex 2015 | Johnny Li | SoA Exam MFE / CAS Exam 3F

P-1 Preface

Preface

Thank you for choosing ACTEX. Since Exam MFE was introduced in May 2007, there have been quite a few changes to its syllabus and its learning objectives. To cope with these changes, ACTEX decided to launch a brand new study manual, which adopts a completely different pedagogical approach. The most significant difference is that this edition is fully self-contained, by which we mean that, with this manual, you do not even have to read the “required” text (Derivatives Markets by Robert L. McDonald). By reading this manual, you should be able to understand the concepts and techniques you need for the exam. Sufficient practice problems are also provided in this manual. As such, there is no need to go through the textbook’s end-of-chapter problems, which are either too trivial (simple substitutions) or too computationally intensive (Excel may be required). Note also that the textbook’s end-of-chapter problems are not at all similar (in difficulty and in format) to the questions released by the Society of Actuaries (SoA). We do not want to overwhelm students with verbose explanations. Whenever possible, concepts and techniques are demonstrated with examples and integrated into the practice problems. Another distinguishing feature of this manual is that it covers the exam materials in a different order than it occurs in Derivatives Markets. There are a few reasons for using an alternative ordering: 1. Some topics are repeated quite a few times in the textbook, making students difficult to fully

understand them. For example, “estimation of volatility” is discussed four times in Derivatives Markets (Sections 11.4, 12.5, 18.5, 24.2)! In sharp contrast, our study manual presents this topic fully in one single section (Module 3, Lesson 4.1).

2. The focus of the textbook is somewhat different from what SoA expects from the candidates.

According to the SoA, the purpose of the exam is to develop candidates’ knowledge of the theoretical basis. Nevertheless, the first half of the textbook is almost entirely devoted to applications. Therefore, we believe that reading the textbook or following the textbook’s ordering is not the best use of your precious time.

Perhaps you have passed some SoA exams by memorizing formulas. However, from the released exam questions, you can easily tell that it is difficult, if not impossible, to pass Exam MFE/3F simply by memorizing all formulas in the textbook. In this connection, in this writing, we help you really learn the materials. By having the reasoning skills, you will discover that there is not really much to memorize. To help you better prepare for the exam, we intentionally write the practice problems and the mock exams in a similar format as the released exam and sample questions. This will enable you to, for example, retrieve information more quickly in the real exam. Further, we have integrated

Preface


P-2

the sample and previous exam questions provided by the SoA into the study manual into our examples, our practice problems, and our mock exams. This seems to be a better way to learn how to solve those questions, and of course, you will need no extra time to review those questions. Our recommended procedure for use of this study manual is as follows: 1. Read the lessons in order. 2. Immediately after reading a lesson, complete the practice problems we provide for that

lesson. Make sure that you understand every single practice problem. 3. Immediately after completing a module, work on the “grand quiz” we provide for that

module. The “grand quizzes” allow you to try out some real exam problems before working on our mock exams.

3. After studying all 22 lessons, work on the mock exams. For your convenience, a grand formula sheet containing all important formulas involved in Exam MFE is appended to this study manual. If you find a possible error in this manual, please let us know at the “Customer Feedback” link on the ACTEX homepage (www.actexmadriver.com). Any confirmed errata will be posted on the ACTEX website under the “Errata & Updates” link.


P-3 Preface

A Note on Rounding and Using the Normal Table To achieve the desired accuracy, we recommend that you store values in intermediate steps in your calculator. If you prefer not to, please keep at least six decimal places. In this study guide, normal probability values and z-values are based on the standard normal distribution table, which is provided on page T0-3. When using the standard normal distribution table, do not interpolate.

• Use the nearest z-value in the table to find the probability. Example: Suppose that you are to find Pr(Z < 0.759), where Z denotes a standard normal random variable. Because the z-value in the table nearest to 0.759 is 0.76, your answer is Pr(Z < 0.76) = 0.7764.

• Use the nearest probability value in the table to find the z-value. Example: Suppose that you are to find z such that Pr(Z < z) = 0.7. Because the probability value in the table nearest to 0.7 is 0.6985, your answer is 0.52.

Preface


P-4

Syllabus Reference Module 0 and Module 1

Our Study Manual The Required Text (Derivatives Markets, 3rd Edition) Module 0: Review

0.1 1.5, 5.2 (p.130 only) 0.2 5.1 − 5.3 0.3 2.2 − 2.3, 9.1 (up to formula (9.3)) 0.4 9.1 (p.270)

Module 1: Risk-Neutral Valuation in Discrete-time Lesson 1: Introduction to Binomial Trees

1.1.1 10.1 (up to the middle of p.297) 1.1.2 10.1 (from the middle of p.297 to the middle of p.298) 1.1.3 10.1 (from p.299 to the middle of p.300)

Lesson 2: Multiperiod Binomial Trees 1.2.1 10.3, 10.4 1.2.2 10.5

1.2.3 10.2 (p.303), 11.3 (from the middle of p.235 to the end of p.336)

Lesson 3: Options on Other Assets 1.3.1 10.5 (p.312) 1.3.2 10.5 (p.312 and 313), 9.1 (formula 9.4 only) 1.3.3 10.5 (from the middle of p.314 to the middle of p.315)

Lesson 4: Pricing with True Probabilities 1.4.1 11.2 (p.326 and 327) 1.4.2 11.2 (p.327 to the middle of p.330)

Lesson 5: State Prices 1.5.1 Appendix 11.B 1.5.2 Sample questions #27 1.5.3 Appendix 11.B


P-5 Preface

Module 2

Module 2: Risk-Neutral Valuation in Continuous-time Lesson 1: Brownian Motions

2.1.1 11.3 (up to the beginning of p.333) 2.1.2 20.2 (up to the bottom of p.604) 2.1.3 20.2 (from the middle of p.607 to the 4th line of p.610) 2.1.4 20.3 (up to the middle of p.608)

Lesson 2: Stochastic Calculus 2.2.1 Scattered in 20.2 and 20.3 2.2.2 20.4 (excluding multivariate Ito’s lemma) 2.2.3 Mainly scattered in 20.2 and 20.3, Example 20.1 2.2.4 20.2 (p.605 to p.607)

Lesson 3: Modeling Stock Price Dynamics 2.3.1 18.1, 18.2 2.3.2 20.1, 18.3 2.3.3 18.4 (up to the middle of p.559) 2.3.4 18.4 (from the bottom of p.559 to formula 18.30) 2.3.5 10.2 (up to the bottom of p.301)

Lesson 4: The Sharpe Ratio and the Black-Scholes Equation 2.4.1 12.1 (p.353) 2.4.2 20.5, 21.2 (p.630, but we generalize the approach here) 2.4.3 21.2 (from p.631 to the middle of p.635)

2.4.4 20.6 (except “Valuing a claim on SaQb”), 21.3 (except “The backward equation”)

Preface


P-6

Module 3

Module 3: The Black-Scholes Formula Lesson 1: Introduction to the Black-Scholes Formula

3.1.1 23.1 (up to the middle of p.686)

3.1.2 23.1 (p.686 “Ordinary options and gap options”), 12.1 (up to p.352)

3.1.3 12.2 Lesson 2: Greek Letters and Elasticity

3.2.1 12.3 (p.356 − 360 before “Rho”, p.361 “Greek measures for portfolio”), 13.4 (up to p.393)

3.2.2 12.3 (p.359, middle of p.360) 3.2.3 12.3 (p.362, to the end of the section 12.3) 3.2.4 Appendix 13.B

Lesson 3: Risk Management Techniques 3.3.1 13.2, 13.3 (up to the middle of p.387) 3.3.2 13.4 (p.394 to p.395) 3.3.3 13.3 (middle of p.387 to the end), 13.5 (p.397 to p.398) 3.3.4 13.5 (p.399 to the end of the section)

Lesson 4: Estimation of Volatilities and Expected Rates of Appreciation 3.4.1 12.5, 24.1, p.303, 24.2 (up to the middle of p.721), 18.5 3.4.2 18.6


P-7 Preface

Module 4

Module 4: Further Topics on Option Pricing Lesson 1: Exotic Options I

4.1.1 14.2 4.1.2 Exercise 14.20 4.1.3 14.3

4.1.4 14.4 (except “Options on dividend-paying stock” and Example 14.2)

Lesson 2: Exotic Options II 4.2.1 14.6 4.2.2 9.2 4.2.3 Exercise 14.21 4.2.4 14.5, 23.1 (p.686 “Ordinary options and gap options”)

Lesson 3: Simulation 4.3.1 19.2, 19.3 4.3.2 19.4 4.3.3 19.5

Lesson 4: General Properties of Options 4.4.1 9.3 (p.281 − 285 “Different strike prices”)

4.4.2 9.3 (p.276 − 277 “European versus American options” and “maximum and minimum option prices”)

4.4.3 9.3 (p.280 – 281 “Time to expiration”) Lesson 5: Early Exercise for American Options

4.5.1 9.3 (from the middle of p.277 to the middle of p.278), 11.1 4.5.2 9.3 (p.278 “Early exercise for puts”)

4.5.3 14.4 (p.419 “Options on dividend-paying stock” and Example 14.2)

Preface


P-8

Module 5

Module 5: Interest Rate Models Lesson 1: Binomial Interest Rate Trees

5.1.1 Scattered in Chapter 7, 25.1 (p.752) 5.1.2 25.4 (up to the middle of p.769)

5.1.3 25.4 (from the middle of p.769 to the middle of p.773), Appendix 25.A

Lesson 2: The Black Model 5.2.1 9.1 (p.269 “Options on bonds”) 5.2.2 12.2 (p.355 “Options on futures”) 5.2.3 25.1 (p.754), 25.5 (from p.780 to the middle of p.781)

Lesson 3: An Equilibrium Equation for Interest Rate Derivatives 5.3.1 25.2 (p.757) 5.3.2 25.2 (p.758) 5.3.3 25.2 (p.759) 5.3.4 25.3 (p.794 to the end)

Lesson 4: The Rendleman-Bartter, Vasicek and Cox-Ingersoll-Ross Model 5.4.1 25.3 (p.760 − p.761) 5.4.2 25.3 (p.761 − p.762) 5.4.3 25.3 (p.762 − p.764) 5.4.4 25.2 (p.756 and p.760)


M1-1 Module 1 : Risk-neutral Valuation in Discrete-time

Lesson 1 : Introduction to Binomial Trees

Lesson 1 Introduction to Binomial Trees

1. To calculate an option price with a one-period binomial tree

2. To identify an arbitrage opportunity when an option is mispriced 3. To have a basic understanding of risk-neutral probabilities Risk-neutral pricing in discrete time is heavily based on binomial trees. By working through the examples provided in this lesson, you will pick up the concept of binomial trees very quickly. The examples given here may seem too simple to be realistic, but the purpose of the lesson is just to give you an understanding of the basic principles of binomial pricing. In later lessons, we will discuss more generalized versions of binomial trees, and explain how they can be applied to different types of derivatives. Consider a binomial model for a stock which pays dividends continuously at rate d:

– at time 0, the price is S0

– at time h, its price Sh can be equal to S0u or S0d (u > 1; d < 1)

– the continuously compounded risk-free interest rate is r

– at time h, the payoff of a derivative is Cu or Cd, depending on the state of the stock The situation can be presented graphically as follows:

OBJECTIVES

1. 1. 1 A One-Period Binomial Tree

Module 1 : Risk-neutral Valuation in Discrete-time



M1-2

The diagram above is called a binomial tree. Since it involves only a single time period, we refer this tree to a one-period binomial tree. Our goal is to price the derivative (i.e., to calculate C0). How may we achieve this goal? The fundamental idea is replication. More specifically, to price the derivative, we replicate the cash flow from the derivative by forming a replicating portfolio using the underlying stock and a risk-free bond. Let us consider the following strategy:

(a) Invest ∆ number of shares in stock and B dollars in risk-free bond at time 0.

(b) Reinvest all dividends of the stock by buying additional shares. This gives ∆edh shares at h.

(c) The replicating portfolio at time h is worth ∆edhSh + Berh. We want the replicating portfolio to generate the same cash flow as the derivative does, no matter which state it turns out to be. As such, we require

δ0

δ0

,

.

h rhu

h rhd

e S u Be Ce S d Be C

∆ + =

∆ + =

Solving for ∆ and B, we obtain

δ

0

.( )

.

h u d

rh d u

C CeS u duC dCB e

u d

−

−

−∆ =

−−

=−

At time 0, the replicating portfolio is worth ∆S0 + B. By the law of one price, the time-0 price of the option must also be ∆S0 + B.

S0u Cu

S0d Cd

S0 C0




We can interpret ∆ as the sensitivity of the option to a change in the stock price. For call options ∆ is positive, and for put options, ∆ is negative. Let us illustrate this idea with an example. Consider the following one period binomial stock price model for a nondividend-paying stock:

You are given:

(i) The period is 1 year.

(ii) The continuously compounded risk-free interest rate is 0.02.

Consider a 1-year 100-strike European call on the stock.

(a) Construct a replicating portfolio for the call option.

(b) Show that, in 1 year, the payoff from the replicating portfolio is the same as that from the call option.

(c) Calculate the time-0 price of the call option. Solution

Notice that we have u = 105/100 = 1.05 and d = 95/100 = 0.95. If the stock price becomes 105 at time 1, the payoff from the call is Cu = (105 – 100)+ = 5. If the stock price becomes 95 at time 1, the payoff from the call is Cd = (95 – 100)+ = 0. (a) The replicating portfolio should consist of ∆ units of the stock and an amount of B dollars in

risk-free bond. We calculate ∆ and B as follows:

F O R M U L A

Option Pricing Formula Based on a One-Period Binomial Tree

−

−+

−−

=+∆=−−

−d

hr

u

hrrh C

dueuC

dudeeBSC

)δ()δ(

00 (1.1.1)

Example 1.1.1

105 100 95




M1-4

δ

0

5 0 0.5,( ) 100(1.05 0.95)

h u dC CeS u d

− − −∆ = = =

− −

0.02 1 1.05 0 0.95 5 46.5594.1.05 0.95

B e− × × − ×= = −

−

In other words, the replicating portfolio consists of 0.5 purchased shares and 46.5594 borrowed at the risk-free rate.

(b) The payoff from the replicating portfolio at time 1 is as follows:

Stock price in 1 year 105 95

0.5 purchased shares 52.5 47.5 Repay loan of 46.5594 −47.5 −47.5

Total payoff 5 0

From the table above, we can see the replicating portfolio and the call option provide the same payoff, no matter if the stock price in 1 year is 105 or 95.

(c) By the law of one price, the call option and the replicating portfolio must have the same

time-0 price, which is given by ∆S0 + B = 0.5 × 100 – 46.5594 = 3.44.

Or we can calculate the call price directly from equation (1.1.1):

(0.02 0) 1 (0.02 0) 10.02 1

00.95 1.055 0 3.44

1.05 0.95 1.05 0.95e eC e

− × − ×− × − −

= × + × = − − .

[ END ] What if the price of an option is not the same as that implied by the pricing formula? If an option is mispriced, we can arbitrage, that is, we can make profit without taking any risk. So how to arbitrage if an option is mispriced? The key is ‘buy low, sell high.’

If an option is overpriced:

− The price of the option is greater than that of the replicating portfolio, ∆S0 + B.

− To arbitrage, we sell the option and buy the replicating portfolio.

If an option is underpriced:

− The price of the option is lower than that of the replicating portfolio, ∆S0 + B.

− To arbitrage, we buy the option and sell the replicating portfolio.

To illustrate, consider the following example.

1. 1. 2 Arbitraging a Mispriced Option




For a nondividend-paying stock, you are given:

(i) The current stock price is 40.

(ii) At the end of one month the stock price will be either 42 or 38.

Assume that the continuously compounded risk-free interest rate is 0.08.

(a) Calculate the current price of a 1-month 39-strike European call.

(b) Identify an arbitrage strategy if the market price of the option is 1.5. Solution

(a) As we need to use ∆ and B in part (b), it would be more convenient to calculate the call price by considering a replicating portfolio. Given Cu = 3, Cd = 0, u = 42/40 = 1.05 and d = 38/40 = 0.95, d = 0, and r = 0.08, we have

δ

0

3 0 0.75( ) 40(1.05 0.95)

h u dC CeS u d

− − −∆ = = =

− −,

10.0812 1.05 0 0.95 3 28.3106.

1.05 0.95B e

− × × − ×= = −

−

The current price of the option is ∆S0 + B = 0.75 × 40 – 28.3106 = 1.6894. (b) If the market price is 1.5, then the call is cheaper than the replicating portfolio. To capture

the arbitrage profit, we buy the option and sell the replicating portfolio. This means we buy one call, short sell 0.75 shares, and invest 28.3106 in a risk-free bond.

The payoff from the arbitrage strategy is summarized below:

Stock price in 1 month Cost at time 0 42 38

Buy the call 1.5 3 0

Sell 0.75 shares 0.75 × −40 = −30

0.75 × −42 = −31.5

0.75 × −38 = −28.5

Invest 28.3106 in a risk-free bond 28.3106 28.3106 × e0.08/12

= 28.5 28.5

Net Position −0.1894 0 0

With this strategy, you get 0.1894 now but need not pay anything for certain in 1 month when the option expires.

[ END ]

Example 1.1.2




M1-6

You will need to know how to arbitrage a mispriced option in Exam MFE. See the following example. You are given the following regarding stock of Widget World Wide (WWW):

(i) The stock is currently selling for $50.

(ii) One year from now the stock will sell for either $40 or $55.

(iii) The stock pays dividends continuously at a rate proportional to its price. The dividend yield is 10%.

The continuously compounded risk-free interest rate is 5%. While reading the Financial Post, Michael notices that a one-year at-the-money European call written on stock WWW is selling for $1.90. Michael wonders whether this call is fairly priced. He uses the binomial option pricing model to determine if an arbitrage opportunity exists. What transactions should Michael enter into to exploit the arbitrage opportunity (if one exists)? (A) No arbitrage opportunity exists.

(B) Short shares of WWW, lend at the risk-free rate, and buy the call priced at $1.90.

(C) Buy shares of WWW, borrow at the risk-free rate, and buy the call priced at $1.90.

(D) Buy shares of WWW, borrow at the risk-free rate, and short the call priced at $1.90.

(E) Short shares of WWW, borrow at the risk-free rate, and short the call priced at $1.90. Solution

To see if the option is mispriced, we need to calculate the price of the option. First, let us calculate ∆ and B. Given Cu = 5, Cd = 0, u = 55/50 = 1.1 and d = 40/50 = 0.8, h = 1, d = 0.1, and r = 0.05, we have

δ 0.1

0

5 0 0.3016( ) 50(1.1 0.8)

h u dC Ce eS u d

− −− −∆ = = =

− −,

0.05 1 1.1 0 0.8 5 12.6831.1.1 0.8

B e− × × − ×= = −

−

The correct price of the option should be ∆S0 + B = 0.3016 × 50 – 12.6831 = 2.40.

This means the call is underpriced. To arbitrage, Michael should buy the call low and sell the replicating portfolio, which consists of 0.3016 shares, high. The only choice that involves buying the call and selling the shares is (B). So the correct answer is (B).

[ END ]

Example 1.1.3 [MFE 09 May #3]




F O R M U L A

Notice that we can rewrite equation (1.1.1) as follows:

The pricing formula in this form can be interpreted easily. The expression p*Cu + (1 − p*)Cd may be seen as the expected payoff from the option, and the option price C0 is simply the expected payoff discounted at the risk-free interest rate. We call p* the risk-neutral probability of an increase in the stock price. Keep in mind that p* is not the true probability that the stock will go up. It is just for our convenience to express the option price as an expectation. The imaginary world in which the probability of an increase in stock price is p* is referred to as the risk-neutral world. Let E* denotes expectation in the risk-neutral world. We have

E*[Sh] = p*S0u + (1 – p*) S0d = S0e(r − d)h,

which means that, in the risk-neutral world, the return on the stock is the risk-free interest rate. This property holds in this imaginary world only. In the real world, it does not hold in general. We say we work under the risk-neutral measure when we use risk-neutral probabilities in our calculations. To apply the risk-neutral pricing formula, we follow the procedure below:

Step 1: Calculate the risk-neutral probability p*.

Step 2: Calculate the expected payoff using the risk-neutral probability.

Step 3: Discount the expected payoff with the risk-free interest rate. We illustrate the risk-neutral pricing formula with the following example.

1. 1. 3 Risk-Neutral Probabilities

Risk-Neutral Pricing Formula Based on a One-Period Binomial Tree

C0 = e−rh[p*Cu + (1 − p*)Cd] , (1.1.2)

where

dudep

hr

−−

=− )(

*d

. (1.1.3)




M1-8

Price the call option in Example 1.1.1 by considering the risk-neutral pricing formula. Solution

Recall that we u = 1.05, d = 0.95, r = 0.02, d = 0, and h = 1. Substituting, we have

.702013.095.005.1

95.0*1)002.0(

=−

−=

×−ep

The expected payoff from the option is

p*Cu + (1 − p*)Cd = 0.702013 × 5 + 0.297977 × 0 = 3.510065.

The price of the option is the expected payoff discounted at the risk-free interest rate:

C0 = e−0.02×1 × 3.510065 = 3.44.

[ END ] Now, let us see how these concepts may be examined in Exam MFE. For a one-year straddle on a nondividend-paying stock, you are given:

(i) The straddle can only be exercised at the end of one year.

(ii) The payoff of the straddle is the absolute value of the difference between the strike price and the stock price at expiration date.

(iii) The stock currently sells for $60.00.

(iv) The continuously compounded risk-free interest rate is 8%.

(v) In one year, the stock will either sell for $70.00 or $45.00.

(vi) The option has a strike price of $50.00.

Calculate the current price of the straddle. (A) $0.90 (B) $4.80 (C) $9.30 (D) $14.80 (E) $15.70

Example 1.1.5 [MFE 07 May #14]

Example 1.1.4




Solution

We have u = 70/60 and d = 45/60, r = 0.08, d = 0, and h = 1. The first step is to calculate the the risk-neutral probability of an increase in the stock price:

0.08* 60 45 0.799889 0.8.

70 45ep −

= = ≈−

The expected payoff from the straddle in the risk-neutral world is given by

p* | 70 − K | + (1 − p*)| 45 − K | = 0.8 × | 70 − 50 | + 0.2 × | 45 − 50 | = 17.

The price of the straddle is the expected payoff discounted at the risk-free interest rate:

C0 = e−0.08×1 × 17 = 15.70.

[ END ]

F U L I S T

One-Period Binomial Tree

δ

0

.( )

.

h u d

rh d u

C CeS u duC dCB e

u d

−

−

−∆ =

−−

=−

C0 = ∆S0 + B. Risk-Neutral Pricing Formula

C0 = e−rh[p*Cu + (1 − p*)Cd],

where du

dephr

−−

=− )(

*d

.

F O R M U L A




M1-10

1. For a nondividend-paying stock, you are given:

(i) The current stock price is 50.(ii) At the end of one month the stock price will be either 52 or 48.(iii) The continuously compounded risk-free interest rate is 0.02.

Consider a 1-month 51-strike European call on the stock.

(a) Construct a replicating portfolio for the call option.(b) Calculate the time-0 price of the call option.

2. For a stock, you are given:

(i) The current stock price is 50.(ii) At the end of three months the stock price will be either 45 or 55.(iii) The stock pays dividends at a rate proportional to its price. The dividend yield is 4%.(iv) The continuously compounded risk-free interest rate is 3%.

Consider a 3-month 53-strike European put on the stock.

(a) Find ∆ for the put option.(b) Calculate the time-0 price of the put option.


(i) The current stock price is 100.(ii) It is known that at the end of one year the stock price will be either 105 or 95.Assume that the continuously compounded risk-free interest rate is 0.03.(a) Calculate the risk-neutral probability of an increase in stock price in one year.(b) Calculate the time-0 price of a 1-year at-the-money European call on the stock.


(i) The current stock price is 100.(ii) It is known that at the end of one year the stock price will be either 90 or 110.(iii) The stock pays dividends at a rate proportional to its price. The dividend yield is 5%.(iv) The continuously compounded risk-free interest rate is 0.06.

(a) Calculate the price of a 1-year 105-strike European call on the stock.(b) Calculate the price of a 1-year 105-strike European put on the stock.

Exercise 1. 1

(c) Show that the prices you found in parts (a) and (b) satisfy the put-call parity.




5. For a one-year strangle on a nondividend-paying stock, you are given:

(i) The strangle can only be exercised at the end of one year.(ii) Let S(1) be the stock price at the end of one year. The payoff from the straddle is as

follows: Range of S(1) Payoff

S(1) ≤ 60 60 – S(1) 60 < S(1) < 70 0

S(1) ≥ 70 S(1) – 70

(iii) The stock currently sells for 60.00. (iv) The continuously compounded risk-free interest rate is 8%. (v) In one year, the stock will either sell for 75.00 or 45.00.

Calculate the current price of the strangle.


(i) The current stock price is 25.(ii) It is known that at the end of two months the stock price will be either 23 or 27.Assume that the continuously compounded risk-free interest rate is 0.10. Let S(1/6) be thestock price at the end of two months. Calculate the price of a derivative that pays offS2(1/6) at this time.

7. For a 6-month butterfly spread on a stock, you are given:(i) The butterfly spread can only be exercised at the end of 6 months.(ii) Let S(0.5) be the stock price at the end of 6 months. The payoff from the butterfly

spread is as follows: Range of S(0.5) Payoff

S(0.5) ≤ 80 0 80 ≤ S(0.5) < 100 S(0.5) – 80 100 ≤ S(0.5) < 120 120 − S(0.5)

S(0.5) ≥ 120 0

(iii) The stock currently sells for 100. (iv) In 6 months, the stock will either sell for 90 or 110.

(v) The stock pays dividends at a rate proportional to its price. The dividend yield is d. (vi) The continuously compounded risk-free interest rate is 5%. Calculate the current price of the butterfly spread.




M1-12

8. Show that for the single-period binomial tree to make sense, we must have d < e(r – d)h < u. 9. Suppose that the market price of the call option in Question 1 is 0.50. Construct an

arbitrage strategy. 10. Suppose that the market price of the put option in Question 2 is 4.60. Construct an

arbitrage strategy. 11. You are given the following regarding stock of Czar:

(i) The stock is currently selling for $100.

(ii) One year from now the stock will sell for either $90 or $110.


The continuously compounded risk-free interest rate is 5%.

Your friend, Gilbert, notices that a one-year at-the-money European put written on the stock is selling for $4.00. Gilbert wonders whether this put is fairly priced. He uses the binomial option pricing model to determine if an arbitrage opportunity exists.

What transactions should Gilbert enter into to exploit the arbitrage opportunity (if one exists)? (A) No arbitrage opportunity exists.

(B) Short shares of Czar, lend at the risk-free rate, and buy the put priced at $4.00.

(C) Buy shares of Czar, borrow at the risk-free rate, and buy the put priced at $4.00.

(D) Buy shares of Czar, borrow at the risk-free rate, and short the put priced at $4.00.

(E) Short shares of Czar, borrow at the risk-free rate, and short the put priced at $4.00. 12. For a nondividend-paying stock, you are given:


(ii) It is known that at the end of one year the stock price will be either 100u or 92.

(iii) The continuously compounded risk-free interest rate is 0.05.

(iv) The price of a 1-year at-the-money call option on the stock is 1.78.

Calculate u. 13. Which of the following statement(s) is/are true regarding a single-period binomial model

with a time period h?

(I) If X is the expected payoff of a European option in the real world, then the price of the option can be obtained by discounting X at the risk-free interest rate.

(II) To calculate the risk-neutral probability of an increase in stock price, the risk premium of the stock is needed.




(III) In the risk-neutral world, the expected stock price at time h is the forward price for the stock delivering at time h.

(A) (I) only. (B) (III) only. (C) (I) and (II) only. (D) (I) and (III) only. (E) None of the statement is true.




M1-14

Solutions to Exercise 1.1 1. We have u = 52/50 = 1.04 and d = 48/50 = 0.96. The payoff is Cu = (52 – 51)+ = 1 and Cd =

(48 – 51)+ = 0 if the stock price rises and falls, respectively.

(a) The replicating portfolio should consist of ∆ units of the stock and an amount of B dollars risk-free bond. We calculate ∆ and B as follows:

δ

0

1 0 0.25( ) 50(1.04 0.96)

h u dC CeS u d

− − −∆ = = =

− −,

=−

×−×= −

96.004.1196.0004.112/02.0eB −11.9800.

The replicating portfolio consists of 0.25 purchased shares and $11.9800 borrowed at the risk-free rate.

(b) By the law of one price, the call option and the replicating portfolio must have the same time-0 price, which is given by ∆S0 + B = 0.25 × 50 – 11.9800 = 0.5200.

2. We have u = 55/50 = 1.1 and d = 45/50 = 0.9. The payoff is Cu = (53 – 55)+ = 0 and Cd =

(53 – 45)+ = 8 if the stock price rises and falls, respectively.

(a) We calculate ∆ as follows:

0.04 0.25 0 8 0.792050(1.1 0.9)

e− × −∆ = = −

−,

(b) To calculate the put price, we need B, which is given by

0.03 0.25 1.1 8 0.9 0 43.67121.1 0.9

B e− × × − ×= =

−.

The put price is ∆S0 + B = −0.7920 × 50 + 43.6712 = 4.07.

3. (a) We are given u = 105/100 = 1.05, d = 95/100 = 0.95, r = 0.03, d = 0, and h = 1. Substituting, we have

p* =95.005.1

95.01)003.0(

−−×−e

= 0.804545.

(b) The expected payoff from the option is

p*Cu + (1 − p*)Cd = 0.804545 × 5 + 0 = 4.022725.

The price of the option is the expected payoff discounted at the risk-free interest rate:

C0 = e−0.03×1 × 4.022725 = 3.9038.




4. (a) We are given u = 110/100 = 1.1, d = 90/100 = 0.9, r = 0.06, d = 0.05, and h = 1.Substituting, we have

p* = 9.01.1

9.01)05.006.0(

−−×−e

= 0.550251.

The expected payoff from the call option is

p*Cu + (1 − p*)Cd = 0.550251 × 5 + 0.449749 × 0 = 2.751255.

The price of the call option is the expected payoff discounted at the risk-free interest rate:

C0 = e−0.06×1 × 2.751254 = 2.5910.

(b) The expected payoff from the put option is

p*Cu + (1 − p*)Cd = 0.550251 × 0 + 0.449749 × 15 = 6.746235.

The price of the put option is the expected payoff discounted at the risk-free interest rate:

C0 = e−0.06×1 × 6.746235 = 6.3534.

(c) The difference between the call price and put price is 2.5910 − 6.3534 = −3.7624, which is the same as S0e−d×1 − Ke−r×1 = 100e−0.05 – 105e−0.06 = −3.7623. Therefore the put-call parity holds (the very slight difference is due to rounding).

5. We are given u = 75/60 = 1.25, d = 45/60 = 0.75, r = 0.08, d = 0, and h = 1.Substituting, we have

p* = 75.025.1

75.01)008.0(

−−×−e

= 0.666574.

If the stock price rises in 1 year, the payoff from the strangle will be Cu = 75 – 70 = 5; otherwise, the payoff will be Cd = 60 – 45 = 15. So the expected payoff from the strangle is

p*Cu + (1 − p*)Cd = 0.666574 × 5 + 0.333426 × 15 = 8.33426. The price of the strangle is the expected payoff discounted at the risk-free interest rate:

C0 = e−0.08×1 × 8.33426 = 7.69.

6. We are given u = 27/25 = 1.08, d = 23/25 = 0.92, r = 0.10, d = 0, and h = 1/6.Substituting, we have

p* = 92.008.1

92.06/1)01.0(

−−×−e

= 0.605040.

Also, we have Cu = 272 = 729 and Cd = 232 = 529. So the price of the derivative is given by

e−r × 1/6[p*Cu + (1 − p*)Cd] = e−0.1/6 × 650.0079 = 639.26.




M1-16

7. We have Cu = 120 – 110 = 10 and Cd = 90 – 80 = 10 (same as Cu). While it is not possibleto compute p* because the value of d is not given, the risk-neutral expectation is

p* × 10 + (1 − p*) × 10 = 10.

Therefore the price of the butterfly spread is simply e−0.05×0.5 × 10 = 9.753.

8. Consider the following two portfolios:

Payoff at time h Position Cost down state up state

(1) Tailing position for stock S0e−dh S0d S0u (2) S0e−dh in bank account S0e−dh S0e(r – d)h S0e(r – d)h

If S0e(r – d)h ≥ S0u, or equivalently e(r – d)h ≥ u, then in the up state portfolio (1) cannot be better than (2), and in the down state portfolio (1) must be more inferior. Thus no one would like to own the stock.

If S0e(r – d)h ≤ S0d, or equivalently e(r – d)h ≤ d, then in the down state portfolio (1) cannot be worse than (2), and in the up state portfolio (1) must be superior. Thus no one would like to invest in the bank account.

9. The binomial model implies that the call price should be 0.5200. Therefore, the call isunderpriced. To arbitrage, we buy the call low and sell the replicating portfolio high.Using the values of ∆ and B from Question 1, the arbitrage strategy is as follows:

− Buy 1 call.− Sell 0.25 shares of the stock.− Invest $11.9800 at the risk-free interest rate.

10. The binomial model implies that the put price should be 4.07. Therefore, the put isoverpriced. To arbitrage, we sell the put high and buy the replicating portfolio low. Usingthe values of ∆ and B from Question 2, the arbitrage strategy is as follows:

− Sell 1 put.− Sell 0.7920 shares of the stock.− Invest $43.6712 in a risk-free bond.

11. To see if the option is mispriced, we need to calculate the price of the option. First, let uscalculate ∆ and B. Given Cu = 0, Cd = 10, u = 110/100 = 1.1 and d = 90/100 = 0.9, h = 1, d = 0.05, and r = 0.05, we have

δ 0.05

0

0 10 0.475615( ) 100(1.1 0.9)

h u dC Ce eS u d

− −− −∆ = = = −

− −,




0.05 1 1.1 10 0.9 0 52.317618.1.1 0.9

B e− × × − ×= =

−

The correct price of the option is ∆S0 + B = −0.475615 × 100 + 52.317618 = 4.756.

This means the put is underpriced. To arbitrage, Gilbert should buy the put low and sell the replicating portfolio. Note that the replicating portfolio consists of a short position of the stock and therefore selling the replicating portfolio would involve a long position of the stock.

The only choice that involves buying the put and buying the shares is (C). So the correct answer is (C).

12. We have d = 92/100 = 0.92, r = 0.05, d = 0, Cu = 100(u −1), Cd = 0, K = 100, and S0 = 100.

Using equation (1.1.1), we obtain

0.05 1 0.05 10.05 1 0.92 100( 1) 0 1.78

0.92 0.92e u ee u

u u

× ×− × − −

× − + × = − − .

Solving the above, we obtain u = 1.0133. However, this violates the condition that u > exp(rh) = 1.0513. As such, there is no valid solution to the problem.

13. Statement (I): It is true if the expectation is taken under the risk-neutral world, but not the

real world. So Statement (I) is false.

Statement (II): From equation (1.1.3), it is easy to see that the risk premium of the stock is not needed. So Statement (II) is false.

Statement (III): Let E* denotes expectation in the risk-neutral world. We have

E*(Sh) = p* S0u + (1 – p*) S0d = S0e(r−d)h,

which is the just the forward price. So Statement (III) is true.

Therefore, the answer is (B).


T10-1 Mock Test 10

Mock Test 10

1. For a nondividend-paying stock whose time-t price is S(t), the true stochastic process is

dS(t) = 0.25S(t)dt − 0.3S(t)dZ(t),

where Z(t) is a standard Brownian motion.

To estimate the price of a 1-year 30-strike call on the stock, Jason simulates the following 7 time-1 stock prices that follow the stochastic process above:

39.2293, 20.7411, 46.6881, 22.2922, 35.2482, 24.8668, 33.5350.

Issac then tells Jason that he should simulate using the risk-neutral stochastic process, using a continuously compounded risk-free interest rate of 0.12. Jason adjusts the 7 stock prices accordingly to obtain the Monte-Carlo estimate.

Calculate Jason’s Monte-Carlo estimate for the call price. (A) 2.077

(B) 2.342

(C) 4.847

(D) 5.465

(E) There is not enough information to determine the answer.

Mock Test 10


T10-2

2. You are given:

(i) The following binomial tree for a nondivdend-paying stock:

Each period of the tree is 2 months. The values of u and d are constant throughout the tree.

(ii) The risk-free interest rate is 2% for every 2 months. That is, 1 dollar would accumulate to 1.02 after 2 months.

Consider a 6-month American cash-or-nothing call that pays 1 as long as the asset overshoots 13.

Calculate the price of the call. (A) 0.1538

(B) 0.1569

(C) 0.1665

(D) 0.1752.

(E) 0.1803

17.28 14.4 12 12.96 10 10.8 9 9.72 8.1 7.29


T10-3 Mock Test 10

3. You are given:

(i) The following incomplete binomial interest rate trees: The rates are continuously compounded.

(ii) The risk-neutral probability of an up-move is always 0.6.

(iii) The continuously compounded 1-year deferred 2-year forward rate is 8.92%. Calculate ruu. (A) 10.8%

(B) 11.1%

(C) 11.4%

(D) 11.7%

(E) 12.0% 4. For a stock that is going to pay a $2 dividend of 3 months from now, you are given:


(ii) The volatility of the prepaid forward price for the delivery of 1 share 5 months from now is 0.38.

(iii) The continuously compounded risk-free rate is 8%.

Calculate the price of a European 5-month at-the-money put on the stock. (A) 8.20

(B) 8.43

(C) 8.86

(D) 10.18

(E) 10.70

ruu 10% r0 9% 6.5% 5.5%

Mock Test 10


T10-4

5. Let Z(t) be a standard Brownian motion.

For t > 0, evaluate ∫

−

tsZssZ

0 )(d

2)(exp .

(A)

−

2)(exp ttZ

(B)

−

2)(exp ttZ − t

(C)

−

2)(exp ttZ − 2t

(D) 42

)(exp2tttZ −

−

(E) None of the above. 6. You are given that

(i) the spot price of 1 Canadian dollar is 0.8 US dollars;

(ii) the Canadian-dollar US-dollar exchange rate has a volatility of 15%;

(iii) the continuously compounded risk-free interest rates of Canada and the United States are 11% and 5%, respectively.

Consider a Canadian-dollar denominated call option to buy 75 USD with 90 Canadian dollars three months from now. By constructing a one-period binomial tree, calculate the delta of the option.

(A) 0

(B) 50.24

(C) 51.00

(D) 64.24

(E) 65.21


T10-5 Mock Test 10

7. Assume the Black-Scholes framework. For a stock that pays dividends continuously at a rate proportional to its price, you are given:


(ii) The stock’s volatility is 0.2.

(iii) The continuously compounded expected rate of stock-price appreciation is 9%.

Let )4()2( SSG = .

Calculate the mean of G. (A) 5e0.21

(B) 5e0.24

(C) 5e0.25

(D) 5e0.26

(E) 5e0.30 8. You have going to sell a Rolls Royce car.

(i) The price of the car is 250,000 British pounds, which you will receive in two months.

(ii) The current exchange rate is 1.35 US dollars per British pound.

(iii) You now buy 2-month at-the-money currency options of the European type to cover foreign currency risk.

(iv) The continuously compounded risk-free interest rates in the US and the UK are 5% and 9%, respectively.

(v) The natural logarithm of the pound-per-dollar exchange rate is an arithmetic Brownian motion with variance per year being 5%.

Determine total cost of the currency options in US dollars, to the nearest dollar. (A) $1,748

(B) $3,972

(C) $7,478

(D) $11,065

(E) $13,289

Mock Test 10


T10-6

9. For 0 ≤ t ≤ T, let P(r, t, T ) be the price at time t of a zero-coupon bond that pays 1 at time T, if the short-rate at time t is r. You are given:

(i) The time-0 short rate is 5%.

(ii) The short rate process is

dr(t) = 0.05[1 − 4r(t)]dt + 0.2dZ(t),


(iii) The Sharpe ratio of interest rate risk is 0.1.

(iv) P(r, t, T) = A(t, T)exp[–B(t, T)r] for some functions A(t, T) and B(t, T).

(v) P(0.05, 0, 2) = 0.847435.

Calculate the time-0 theta of a 2-year zero-coupon bond with a face value of 100. That is, compute 1000Pt(0.05, 0, 2).

(A) 78

(B) 80

(C) 83

(D) 87

(E) 92 10. Which of the following statements concerning barrier options is / are correct?

(i) An up-and-in put becomes more valuable as we increase the frequency with which we observe the asset price in determining if the barrier has been crossed.

(ii) A down-and-out call becomes more valuable as we increase the frequency with which we observe the asset price in determining if the barrier has been crossed.

(iii) A down-and-out call becomes more valuable as we raise the barrier. (A) (i) only

(B) (ii) only

(C) (iii) only

(D) (i) and (ii) only

(E) (i) and (iii) only


T10-7 Mock Test 10

11. Consider a two-period binomial tree for a futures index. You are given:

(i) The current futures price is 50.

(ii) The time step over each period is 3 months.

(iii) u / d = 1.6.

(iv) The risk-neutral probability of an up move is 5 / 9.

(v) The continuously compounded risk-free interest rate is 10%.

Consider an option on futures index that pays the value of the futures index at the end of 6 months if the futures price is less than 40.

Calculate the price of the option. (A) 1.1

(B) 2.2

(C) 3.2

(D) 4.3

(E) 5.3

Mock Test 10


T10-8

12. Suppose that the short rate process r(t) follows the Cox-Ingersoll-Ross model.

Let y(r, t) be continuously compounded yield to maturity of a t-year zero-coupon bond, when the initial short rate is r. Which of the following figures show(s) the possible yield curve?

Figure (i) Figure (ii)

Figure (iii) Figure (iv) (A) (i) only

(B) (i) and (ii) only

(C) (iii) only

(D) (iii) and (iv) only

(E) (i), (ii), (iii) and (iv)

y(r, t)

t

y(r, t)

t

r r

y(r, t)

t

y(r, t)

t

r r


T10-9 Mock Test 10

13. You are given:

(i) The current USD-Euro exchange rate is 0.75 $/Euro.

(ii) The USD-denominated continuously compounded interest rate is 4.5%.

(iii) The Euro-denominated continuously compounded interest rate is 4%.

(iv) The price of a USD-denominated 6-month 0.75-strike call to buy 1 Euro is 0.026.

Find the price of a Euro-denominated 6-month 1/0.75-strike put to sell 1 USD. (A) 0.024

(B) 0.037

(C) 0.043.

(D) 0.046

(E) 0.053 14. For a nondividend-paying stock, you are given:

(i) The current price of the stock is 100.

(ii) The volatility of the stock is 100%.

(iii) The continuously compounded risk-free interest rate is 0%.

Consider a 1-year European gap call option on the stock. The strike price and the payment trigger of the gap call option is 130 and 100, respectively. Calculate the Black-Scholes price of the gap call option.

(A) 21

(B) 23

(C) 25

(D) 27

(E) 29

Mock Test 10


T10-10

15. Enrica purchases a forward start option which will give her 1 year from today a 1-year at-the-money European call option on a stock. You are given:

(i) The stock pays no dividend.

(ii) The volatility of the stock is 0.3.


(iv) The current Black-Scholes price of the forward start option is 7.5.

Calculate the forward price for delivery of 1 share of the stock 1 year from today. (A) 54.5

(B) 54.7

(C) 56.5

(D) 56.7

(E) 58.5 16. For a nondividend-paying stock, you are given:

(i) The current price of the stock is 18.5.

(ii) In 6 months, the stock will either go up to 22.5 or down to 15.


(iv) The current price of a 6-month 20-strike European call option on the stock is 1.55.

Which of the following is true? (A) There is no arbitrage opportunity.

(B) You can make an arbitrage profit of 0.24 if you buy a call option, sell 0.33 shares of the stock and lend 4.85.

(C) You can make an arbitrage profit of 0.24 if you sell a call option, buy 0.33 shares of the stock and borrow 4.85.

(D) You can make an arbitrage profit of 0.12 if you buy a call option, sell 0.66 shares of the stock and lend 10.78.

(E) You can make an arbitrage profit of 0.12 if you buy a call option, sell 0.66 shares of the stock and lend 10.78.


T10-11 Mock Test 10



(ii) The volatility of the stock is 0.2.


(iv) The expected rate of return on the stock is 12%


Consider a 3-month 34-strike call on the stock. The price of this call is 2.024. Find the Sharpe ratio for this call.

(A) 0.3

(B) 0.4

(C) 0.5

(D) 0.6

(E) It cannot be determined from the information given in the question. 18. Consider a two-period binomial model for a nondividend-paying stock. You are given:


(ii) The time step over each period is 6 months.

(iii) u = 1.181, where u is one plus the rate of capital gain on the stock per period if the price goes up.

(iv) d = 0.890, where d is one plus the rate of capital loss on the stock per period if the price goes down.

(v) The continuously compounded risk-free interest rate is 0.05.

Find the price of a 1-year 75-strike American put on the stock. (A) 4.96

(B) 5.46

(C) 5.96

(D) 6.46

(E) 6.96

Mock Test 10


T10-12

19. For a nondividend-paying stock whose time-t price is S(t), you are given:


(ii) The true stochastic process for S is

)(d25.0d2.0)()(d tZt

tStS

+=


Compute E[S 2(1) | S(1) < 1].

(A) 0.191

(B) 0.250

(C) 0.752

(D) 0.763

(E) 0.868 20. Assume the Black-Scholes framework for a stock. You are given:

(i) The stock pays dividends continuously at a rate that is proportional to its price.

(ii) Under the risk-neutral measure, there is a probability of 0.5517 that the stock price 1 year will be lower than the current price.

(iii) The option elasticity for a 1-year at-the-money European call on the stock is 2.9.

The continuously compounded risk-free interest rate is 10%.

Calculate the delta of the call. (A) 0.495

(B) 0.512

(C) 0.536

(D) 0.572

(E) 0.619


T10-13 Mock Test 10

21. You use a one-period binomial model to a nondivdiend-paying stock. Consider a European 6-month at-the-money put option on the stock. You are given:

(i) The period is six months.

(ii) u = 1.2 and d = 0.8.

(iii) The current price of the stock is 100.

(iv) The appropriate discount rate for the put option in the real world is −20%.


Find α, the continuously compounded rate of return of the stock. (A) 0.02

(B) 0.04

(C) 0.07

(D) 0.09

(E) 0.11 22. You are given the following prices of 9-month knock-out put options written on a stock S,

whose current price is 45.

Strike Down-and-out barrier Price 35 0 0.1802 35 30 0.0576 43 0 1.8477 43 30 1.6068 43 35 0.7248 43 40 0.0363

Consider a special 9-month option with the following payoff:

down-and-in barrier 40 not reached

down-and-in barrier 40 reached partial down-and-out barrier 30 not reached

partial down-and-out barrier 30 reached

0 max[43 − S(0.75), 0] max[35 − S(0.75), 0]

Calculate the price of the special option. (A) 0.12

(B) 0.81

(C) 1.42

(D) 1.69

(E) 1.91

Mock Test 10


T10-14

23. You are given

(i) P(t, T ) is the time-t price of a zero-coupon bond that pays 1 at time T.

(ii) r(t, T ) is the time-t spot interest rate that is applicable in (t, T ). The rate is continuously compounded and annualized.

(iii) The forward bond price process Ft,2[P(2, 4)]: 0 ≤ t ≤ 2 follows a geometric Brownian motion.

(iv) The following table of spot bond prices:

T 1 2 3 4 P(0, T) 0.95 0.93 0.915 0.9

(iv) Eln F1,2[P(2, 4)] = –0.25 and Varln F1,2[P(2, 4)] = 0.05

Calculate the probability that r(2, 4) is negative.

(A) 0%

(B) 3.5%

(C) 7%

(D) 9.5%

(E) 12% 24. For a stock following a geometric Brownian motion, you are given that


(ii) An option on the stock has a current price of 17.056 and a gamma of −0.0163.

(iii) It is found that when the stock price decreases by 0.3, the option price becomes 17.366.

By using delta-gamma approximation, find the price of the option if the stock price is 40.5.

(A) 16.54

(B) 16.80

(C) 17.23

(D) 17.57

(E) 18.61


T10-15 Mock Test 10

25. Consider a binomial model for a futures contract. You are given:

(i) Each period is 1 year.

(ii) The price of a 1-year at-the-money put option on the futures contract is 33.2930.

(iii) The risk-neutral probability of an up move is 1/2.

(iv) The initial futures price is 100.

(v) The continuously compounded risk-free interest rate is 0.05.

Calculate u. (A) 1.3

(B) 1.4

(C) 1.5

(D) 1.6

(E) 1.7 26. Consider a one-period binomial stock price model You are given:

(i) The time period of the tree is 1.

(ii) The stock pays no dividends.

(iii) The true probability of an up move is p = 0.6.

(iv) The utility value of 1 today for the up state is 0.8.

Find the utility value of 1 today for the down state. (A) 0.4

(B) 0.6

(C) 0.8

(D) 1.2

(E) 1.4

12

4

8

Mock Test 10


T10-16

27. Find the solution of the following stochastic differential equation:

dY(t) = [0.5 − 1.5Y(t)]dt + 0.6dZ(t), Y(0) = 4.

(A) 1.5 1.5( )

0

1( ) 4 (1 ) 0.6 d ( )3

tt t sY t e e Z s− − −= + − + ∫

(B) 1.5 1.5 1.5( )

0

1( ) 4 (1 ) 0.6 d ( )3

tt t t sY t e e e Z s− − − −= + − + ∫

(C) 1.5 1.5 0.5( )

0( ) 4 (1 ) 0.6 d ( )

tt t t sY t e e e Z s− − − −= + − + ∫

(D) 1.5 0.5( )

0

1( ) 4 (1 ) 0.6 d ( )3

tt t sY t e e Z s− − −= + − + ∫

(E) 1.5 0.5 1.5( )

0

1( ) 4 (1 ) 0.6 d ( )3

tt t t sY t e e e Z s− − − −= + − + ∫

28. For a nondividend-paying stock that follows a geometric Brownian motion, you are given:

(i) The current price stock price is 100.

(ii) The volatility of the stock is σ.

(iii) The expected rate of appreciation of the stock is α.

(iii) The 95% lognormal confidence interval for price of the stock two years from now is (75, 225).

Find α + σ. (A) 0.30

(B) 0.35

(C) 0.40

(D) 0.45

(E) 0.50


T10-17 Mock Test 10

29. Let S(t) be the time-t price of a stock. You are given:

(i) The current stock price is S(0) = 10.

(ii) The true dynamics of S(t) is given by

dS(t) = 0.1S(t)dt + 0.3S(t)dZ(t).

You are also given the following 4 uniform random numbers:

0.3974, 0.5557, 0.0968, 0.6223

Compute the Monte-Carlo estimate of E[S(0.5)] by simulating four time-0.5 stock prices. (A) 9.15

(B) 9.26

(C) 9.77

(D) 10.26

(E) 10.77 30. Consider the following 3-period binomial interest rate tree for effective annual rates. You are given that the risk-neutral probability of an up-move is always 0.6.

. Find the price of a 10% interest rate cap on a 100 three-year loan with annual interest payments.

(A) 1.4

(B) 1.6

(C) 1.8

(D) 2.0

(E) 2.2

*** End of Examination ***

14% 12% 10% 10% 8% 6%

Mock Test 10


T10-18

Solutions to Mock Test 10

1. A 16. C 2. A 17. A 3. C 18. E 4. C 19. D 5. E 20. E 6. E 21. D 7. D 22. D 8. E 23. C 9. B 24. A 10. A 25. E 11. E 26. E 12. B 27. B 13. D 28. B 14. E 29. C 15. D 30. D

1. [Module 4, Lesson 3]

Answer: (A)

The risk-neutral time-1 stock price is simulated using

ZreSS σσ +−= )5.0( 2

)0()1(

while the true time-1 stock price is simulated using

ZeSS σσα +−= )5.0( 2

)0()1(ˆ .

In both cases, Z ~ N(0, 1).

To obtain the risk-neutral stock price, we note that )exp()1(ˆ)1( α−= rSS .

)1(S S(1) Call payoff 46.6881 40.9966 10.9966 39.2293 34.4471 4.4471 35.2482 30.9513 0.9513 33.5350 29.4469 0 24.8668 < 30 0 22.2922 < 30 0 20.7411 < 30 0

The Monte-Carlo estimate is .077.27

9513.04471.49966.10 12.0 =++ −e


T10-19 Mock Test 10


Answer: (A)

The values of u and d are 1.2 and 0.9, respectively. The risk-neutral probability of an up move is

=−−

=9.02.19.002.1*p 0.4.

The American call should be exercised as soon as the underlying stock overshoots 13. Failing to do so means that later on there would at least be loss of interest on the 1 dollar earned, not to mention that later the stock may go down and not be able to overshoot the level again.

We can see that Cuuu = 1, Cuud = Cudd = Cddd = 0,

Cuu = max(1, 1.02−1p*) = 1, Cud = Cdd = 0,

Cu = 1.02−1(p*Cuu + q*Cud) = 0.4/1.02, Cd = 0,

C0 = 1.02−1(p*Cu + q*Cd) = 0.42/1.022 = 0.153787. 3. [Module 5, Lesson 1]

Answer: (C)

P(0, 3) = )16.024.024.036.0( 12.0155.019.01.00 −−−−−− +++ eeeee uurr P(0, 1) = 0re− Let the forward rate be f. Then

P(0, 1)exp(−2f ) = P(0, 3)

)16.024.024.036.0( 12.0155.019.01.0)0892.0(2 −−−−−− +++= eeeee uur On solving for ruu, we get ruu = 0.1138. 4. [Module 3, Lesson 1]

Answer: (C)

The prepaid forward price of the stock is

=×−−= )12308.0exp(2100)(5.0,0 SF P 98.03960.

The prepaid forward price of the strike price is )12508.0exp(100 ×− = 96.72161.

177823.012/538.0

125

238.0

72161.9603960.98ln

2

1 =×+

=d , d2 = –0.06747

N(−d1) = 0.429431, N(−d2) = 0.526895. The put price is 96.72161 × 0.526895 − 98.03960 × 0.429431 = 8.861.

Mock Test 10


T10-20


Answer: (E)

From the form of the choices given, consider

−=

2)(exp)( ttZtY .

Let f(t, z) = exp(z − t / 2). Then ft(t, z) = −21 exp(z − t / 2),

fz(t, z) =fzz(t, z) = exp(z − t / 2) By Itô’s lemma,

)(d2

)(exp

d2

)(exp21)(d

2)(expd

2)(exp

21)(d

tZttZ

tttZtZttZtttZtY

−=

−+

−+

−−=

Replacing t by s, and then integrating both sides from 0 to t, we get

∫

−=−

tsZssZYtY

0 )(d

2)(exp)0()(

Since Y(0) = exp[Z(0)] = 1,

∫

−=−

tsZssZtY

0 )(d

2)(exp1)(

or

12

)(exp)(d2

)(exp

0 −

−=

−∫

ttZsZssZt

.


Answer: (E)

The domestic country is Canada, while the foreign country is US. We have r = 11%, and rf = 5%, and x0 = 1/0.8 = 1.25.

The values of u and d are

−=−×==+×=

)06.0exp()25.015.025.006.0exp()09.0exp()25.015.025.006.0exp(

du .

After 3 months, xu = 1.25e0.09 = 1.367718, xd = 1.25e−0.06 = 1.177206.

The payoff of the option is

(75US − 90CAD)+ = (75x − 90)+ CAD.

(75xu − 90)+ CAD = 12.57885, (75xd − 90)+ CAD = 0.

The delta of the call option is

206.65177206.1367718.1

057885.12)exp( 25.005.0 =−

−=

−−

− ×−exxCC

hrdu

duf .


T10-21 Mock Test 10


Answer: (D)

The expected rate of appreciation is α − δ = 0.09 (recall that α is the total return), and σ = 0.2.

S(t) = S(0)exp[(α – δ − 0.5σ 2)t + σ Z(t)] = 5exp[0.07t + 0.2Z(t)]

S(2)S(4) = 25exp[0.07 × 6 + 0.2(Z(2) + Z(4))]

G = 5exp[0.21 + 0.1(Z(2) + Z(4))]

Noticing that

Var[Z(2) + Z(4)] = Var[Z(2)] + 2Cov[Z(2), Z(4)] + Var[Z(4)] = 2 + 2(2) + 4 = 10,

we have ln G ~ N(ln 5 + 0.21, 0.12 × 10) = N(ln 5 + 0.21, 0.1)

The mean of G is E(G) = exp(ln 5 + 0.21 + 0.5 × 0.1) = 5e0.26. 8. [Module 3, Lesson 1]

Answer: (E)

Suppose the time-t value of 1 British pound in US is x(t). We have x(0) = 1.35 and the investor will receive 250,000x(1/6) two months from now.

Because the British dollar to be received may lose value relative to US dollars, the option to be bought is a put with payoff

$250,000[1.35 − x(1/6)]+.

Together with the 250,000x(1/6), the payoff after two months become

$250,000x(1/6) + 250,000[1.35 − x(1/6)]+ = 250,000max[1.35, x(1/6)],

which is at least equal to $250,000 × 1.35 = $337500.

To value 1 unit of the put,

027386.06/05.0

61)

205.009.005.0(

1 −=×+−

=d , d2 = –0.11867

N(−d1) = 0.510924, N(−d2) = 0.547233

p(1.35, 1.5, 1/6) = 1.35(e−0.05/6 × 0.547233 − e−0.09/6 × 0.510924) = 0.05315536

The price of the currency option is 0.05315536 × 250,000 = 13288.84. 9. [Module 5, Lesson 4]

Answer: (B)

Rewrite the short rate process as

dr(t) = 0.2[0.25 − r(t)]dt + 0.2dZ(t).

Mock Test 10


T10-22

We see that the short rate follows a Vasicek model with a = 0.2, b = 0.25, and σ = 0.2. As

a result, B(0, 2) = 2.0

1 22.0 ×−− e= 1.6484.

Since the bond price follows an affine structure,

− the delta of the bond is Pr = −B(0, 2)P(0.05, 0, 2),

− the gamma of the bond is Prr = B2(0, 2)P(0.05, 0, 2).

Putting the delta and gamma into the term structure equation:

Pt + [0.05(1 − 4r) + 0.2 × 0.1]Pr + 0.5(0.22)Prr = rP

Pt(0.05, 0, 2) + [0.05(1 − 0.2) + 0.2 × 0.1]Pr(0.05, 0, 2) + 0.5(0.22)Prr(0.05, 0, 2) = 0.05P(0.05, 0, 2)

Pt(0.05, 0, 2) = 0.08013. 10. [Module 4, Lesson 1]

Answer: (A)

(i) Increasing the frequency means it is easier for the put to be knocked in. This means the put would become more valuable.

(ii) Increasing the frequency means it is easier for the call to be knocked out. This means the call would become less valuable.

(iii) Raising the barrier means that it is easier for the call to be knocked out. This means the call would become less valuable.


Answer: (E)

The risk-neutral probability of an up-move is dudp

−−

=1* . Setting it to 5/9,

16.11/1

95

−−

=d

which gives d = 0.75 and hence u = 1.2. As a result,

Fdd = 50 × 0.752 = 28.125, Fud = 50 × 1.2 × 0.75 = 45 > 40.

The price of the asset-or-nothing put is e−0.05 × 28.125 × (4/9)2 = 5.2846. 12. [Module 5, Lesson 4]

Answer: (B)

The yield curve can be upward slopping, downward slopping, or slightly humped.

Figure (iii) is a sample path of the short rate but not the yield to maturity.

Figure (iv) is not even a possible path of the short rate for the CIR model because the CIR model does not permit the short rate to be negative.


T10-23 Mock Test 10


Answer: (D)

Using the relationship 1 1( (0), , ) (0) , , (0)US US US Eur

US

c x K T x Kp Tx K

=

,

we immediately obtain 1 1(0.75, 0.75, 0.5) 0.026 0.75 0.75 , , 0.5

0.75 0.751 1, , 0.5 0.046.

0.75 0.75

US Eur

Eur

c p

p

= = ×

⇒ =


Answer: (E)

The strike price is K1 = 130 and the payment trigger is K2 = 100.

The time-t price of the gap call is given by

V(S(t), t) = S(t)e–δ(T – t)N(d1) – K1e–r(T – t)N(d2),

where tT

tTrKtSd−

−+−+=

σσ ))(2/δ(]/)(ln[ 2

21 and tTdd −−= σ12 . It is important to

know that the two ds are computed using K2.

Substituting S(0) = 100, σ = 1, r = 0, δ = 0, T – t = 1, we have d1 = 0.5 and d2 = −0.5. The price of the gap call is therefore 100N(0.5) – 130N(−0.5) = 100 × 0.6915 – 130 × (1 – 0.6915) = 29.045.


Answer: (D)

We first compute the payoff of the forward start put option:

2

1

(1) 0.3ln (0.04 0 ) 1(1) 2 0.2833

0.3 1

SSd

+ − + ×= = , 2 0.2833 0.3 1 0.0167d = − = − ,

N(d1) = 0.6103, N(d2) = 0.4920,

V(S(1), 1) = c(S(1), 1; S(1), 2) = S(1) × 0.6103 – S(1)e−0.04×1 × 0.4920

= 0.137592S(1).

The time-0 price of the option is 0.137592S(0) = 7.5, which gives S(0) = 54.5091. Hence, the forward price for delivery of 1 share of the stock 1 year from today is S(0)e0.04×1 = 54.5091e0.04 = 56.7337.

Mock Test 10


T10-24


Answer: (C)

Given Cu = 2.5, Cd = 0, u = 22.5/18.5 = 1.2162 and d = 15/18.5 = 0.8108, δ = 0, and r = 0.06, we have

δ

0

2.5 0 0.3333( ) 18.5(1.2162 0.8108)

h u dC CeS u d

− − −∆ = = =

− −,

0.06 0.5 1.2162 0 0.8108 2.5 4.8522.1.2162 0.8108

B e− × × − ×= = −

−

The current price of the replicating portfolio is ∆S0 + B = 0.3333 × 18.5 – 4.8522 = 1.31, which is lower than the current price of the option by 0.24.

By the principle of “buy low, sell high”, we can obtain an arbitrage profit of 0.14 by buying the replicating portfolio (i.e., buying 0.33 shares of the stock and borrowing 4.85 at the risk-free rate) and short-selling the call.


Answer: (A)

The Sharpe ratio for the call must be the same as that for the underlying stock. Hence, the required answer is

0.12 0.06 0.30.2

rασ− −

= =

Note that the call price is irrelevant. 18. [Module 1, Lesson 2]

Answer: (E)

The risk-neutral probability for in increase in stock price is given by

p* =(0.05) 0.5 0.891.181 0.89

e × −−

= 0.4650.

We can easily construct the required two-period binomial tree, which is as follows:

70 (6.9636)

82.67 (0.7429)

62.3 (12.7)

97.6333 (0)

73.5763 (1.4237)

55.447 (19.553)


T10-25 Mock Test 10

The numbers in parentheses are the option values at the corresponding nodes. They are calculated as follows:

In 12 months:

Node Option value Upper (75 – 97.6333)+ = 0 Middle (75 – 73.5763)+ = 1.4237 Lower (75 – 55.447)+ = 19.553

In 6 months:

Node Option value

Upper max[75 – 82.67, e−0.05 × 0.5(0.465 × 0 + 0.535 × 1.4237)] = max(−7.67, 0.7429) = 0.7429

Lower max[75 – 62.3, e−0.05 × 0.5(0.465 × 1.4237 + 0.535 × 19.553)] = max(12.7, 10.8483) = 12.7

At time 0, the value of the put option is given by

max[75 – 70, e−0.05 × 0.5(0.465 × 0.7429 + 0.535 × 12.7)] = max(5, 7.4765) = 6.96. 19. [Module 2, Lesson 3]

Answer: (D)

The solution of the SDE for S is S(t) = exp[0.16875t + 0.25Z(t)].

Let Y(t) = S2(t). Then Y(t) = exp[0.3375t + 0.5Z(t)], and hence

)(d5.0d4625.0)()(d tZt

tYtY

+= .

We need to compute the partial expectation E[Y(1) | Y(1) < 1].

Firstly, P(Y(1) < 1) = =−=

−+

−=− )675.0(5.0

25.04625.0

11ln

)ˆ(

2

2 NNdN 0.24984.

175.15.0675.0ˆ1 =+=d , so that 12000.0)ˆ( 1 =−dN .

Secondly, EY(1) = exp(0.4625).

Thus, the answer is 7627.0)ˆ(

)ˆ()]1([E

2

1 =−

−

dNdNY

.


Answer: (E)

From statement (ii), N(d2) = 0.4483.

Mock Test 10


T10-26

From statement (iii), )4483.0()()0(

)0(1.0

2−− −∆

∆=

−∆∆

=ΩedNKeS

SrT because K = S(0) (the

call option is at-the-money).

∆=−∆ − )]48.0([9.2 1.0e

On solving, ∆ = 619.09.1

)4483.0(9.2 1.0

=−e .


Answer: (D)

First of all, we calculate ∆ and B:

)8.02.1(100200

)(δ

−−

=−

−=∆ −

duSCC

e duh = –0.5

8.02.10202.15.005.0

−−×

=−−

= ×−− edudCuC

eB udrh = 58.518595

Thus, we have

0896.0518595.58)5.0(100

518595.58518595.58)5.0(100)5.0(100

100100100

5.005.05.05.02.0

=+−×

++−×−×

=

+∆+

+∆∆

=

××−

α

α

αγ

eee

eB

BeB

e rhhh


Answer: (D)

Let m(0.75) = minS(u): 0 ≤ u ≤ 0.75. The payoff of this option is

[43 − S(0.75)]I(S(0.75) < 43, 30 < m(0.75) < 40) + [35 − S(0.75)]I(S(0.75) < 35, m(0.75) < 30) = [43 − S(0.75)]I(S(0.75) < 43, m(0.75) > 30) − [43 − S(0.75)]I(S(0.75) < 43, m(0.75) > 40) + [35 − S(0.75)]I(S(0.75) < 35) − [35 − S(0.75)]I(m(0.75) > 30)

The price of this special option is thus 1.6068 − 0.0363 + 0.1802 − 0.0576 = 1.6931. 23. [Module 5, Lesson 2]

Answer: (C)

Let the forward bond price process be Pt(2, 4) = P0(2, 4)exp[µ t + σ Z(t)].

P0(2, 4) = P(0, 4) / P(0, 2) = 90/93. Then ln Pt(2, 4) = ln P0(2, 4) + µ t + σ Z(t) ~ N(ln P0(2, 4) + µ t, σ 2t). Putting t = 1,

ln (90/93) + µ = –0.25, so that µ = −0.25 − ln(90/93),


T10-27 Mock Test 10

and also σ 2 = 0.05.

Putting t = 2, ln P2(2, 4) = ln P(2, 4) ~ N(−0.5 − ln(90/93), 0.1)

P(r(2, 4) < 0) = P(P(2, 4) > 1) = 1 − P(Z < )1.0

)93/90ln(5.0 += 0.06978.


Answer: (A)

By delta-gamma approximation,

17.366 ≈ 17.056 + ∆(−0.3) + 2)3.0)(0163.0(21

−− ,

giving ∆ ≈ −1.03578. By using delta-gamma approximation again,

54.16)5.0)(0163.0(21)5.0)(03578.1(056.17 2 =−+−+≈V .


Answer: (E)

The price of the put option can be expressed as follows:

0.05 0.051 1(100 100 ) (100 100 ) 50 (1 )2 2

e u d e d− −+ +

− + − = − .

This is because it is required that d < 1 < u. So we have 50e−0.05(1−d) = 33.2930, which implies d = 0.3.

Also we have p* = 1 1 1 0.32 0.3

du d u

− −= =

− −, which implies u = 1.7.


Answer: (E)

We have QH = pUH = 0.6 × 0.8 = 0.48

Moreover, 8 = 12QH + 4QL. Thus QL = 0.56, and this gives

UL = 0.56 / (1 − p) = 1.4. 27. [Module 2, Lesson 2]

Answer: (B)

Rewrite the equation as dY(t) = 1.5[31 − Y(t)]dt + 0.6dZ(t).

This is an OU process with λ = 1.5, α = 31 , and σ = 0.6. Since it is also given that Y(0) = 4,

1.5 1.5 1.5( )

0

1( ) 4 (1 ) 0.6 d ( )3

tt t t sY t e e e Z s− − − −= + − + ∫ .

Mock Test 10


T10-28


Answer: (B)

From the lognormal confidence interval, we know the following:

2

2

( / 2) 2 1.96 2

( / 2) 2 1.96 2

75 100

225 100

e

e

α σ σ

α σ σ

− × −

− × +

=

=.

Dividing the second equation by the first equation, we obtain

3 = σ296.12×e ⇒ σ = 0.1981725.

Substituting σ back to the first equation, we obtain

75 = 100e2α − 0.588578 ⇒ α = 0.150448.

Hence, α + σ = 0.3486. 29. [Module 4, Lesson 3]

Answer: (C)

We can simulate Z(0.5) by using 0.5Z where Z ~ N(0, 1). So, we use the following formula to simulate the stock prices:

20.3(0.1 ) 0.5 0.3 0.5 0.0275 0.2121322(0.5) (0) 10

Z ZS S e e− × + += =

U Z S(0.25) 0.3974 –0.26 9.72724 0.5557 0.14 10.58866 0.0968 –1.3 7.80148 0.6223 0.31 10.97748

The Monte-Carlo estimate is 9.72724 10.58866 7.80148 10.977484

+ + += 9.7737.


Answer: (D) 100[r(1, 2) − 10%]+ 100[r(2, 3) − 10%]+

0 1 2 3

r(0, 1) r(1, 2) r(2, 3) time

14% 12% 10% 10% 8% 6%


T10-29 Mock Test 10

Path Probability Payoff Discounted Payoff t = 2 t = 3

↑ ↑ 0.36 2 4 2 4

1.10 1.12 1.10 1.12 1.14+

⋅ ⋅ ⋅

↑ ↓ 0.24 2 0 2

1.10 1.12⋅

↓ ↑ 0.24 0 0 0 ↓ ↓ 0.16 0 0 0

The price of the cap is

0.36 × 2 41.10 1.12 1.10 1.12 1.14

+ ⋅ ⋅ ⋅ + 0.24 × 2

1.10 1.12 ⋅

= 1.9993.

Mock Test 10


T10-30

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