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ACT 4000, FINAL EXAMINATIONADVANCED ACTUARIAL TOPICS

APRIL 24, 20079:00AM - 11:OOAM

University Centre RM 210- 224 (Seats 266- 304)Instructor: Hal W. Pedersen

You have 120 minutes to complete this examination. When the invigilator instructsyou to stop writing you must do so immediately. If you do not abide by this instructionyou will be penalised.

Each question is worth 10 points. If the question has multiple parts, the parts areequally weighted unless indicated to the contrary: Provide sufficient reasoning to backup your answer but do not write more than necessary.

This examination consists of 12 questions. Answer each question on a separate pageof the exam book. Write your name and student number on each exam book thatyou use to answer the questions. Good luck!

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Suppose call and put prices are given by

Strike 80 100

Call premium 22 9Put premium 4 21

(IJ Find the convexity violations.

(1.-) What spread would you use to effect arbitrage?

105

524.80

(Q t:y-I A New York finn is offering a new financial instrument called a "happy

calL" It has a payoff function at time T equal to max(.5S, S - K), where S is the priceof a stock and K is a fixed strike price. You always get something with a happy call. Let

P be the price of the stock at time t = 0 and let C, and C2 be the prices of ordinary calIswith strike prices K and 2K, respectively. The fair price of the happy call is of the fonn

CH = exP + fJC. + yC2.

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Find the constants ex, fJ, and y. C '..L. iLv",( ••• IL •....,..~....'c..••(fIJ

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You are interested in stock that will either gain 30% thisyear or lose 20% this year. The one-year annual effective rate of interestis 10%. The stock is currently selling for $10.

(1) (4 points) Compute the price of a European call option on thisstock with a strike price of $11.50 which expires at the end of the year.

(2) (4 points) Compute the hedge portfolio (i.e. the amount of stockand one-year bonds to hold that replicate the option's payoffs) for thisEuropean call option.

(1) (2 points) Consider a European put option on this stock with astrike price of $11.50 which expires at the end of the year. ft is possibleto structure just the right amount of European call options on this stockwith a strike price of $11.50 expiring at the end of the year, togetherwith one-year bonds, and shares of the stock so as to replicate thepayoffs from the put option. Determine how many shares of stock,how many call options, and how many one-year bonds are needed toreplicate the payoffs from the put option and use this to price the putoption.

A non-dividend-paying stock has a current price of 800p. In any unit of time (t, t + 1)the price of the stock either increases by 25% or decreases by 20%. £1 held in cashbetween times t and t + 1 receives interest to become £1.04 at time t + 1. The stock

price after t time units is denoted by St.

(i) Calculate the risk-neutral probability measure for the model.

(ii) Calculate the price (at t = 0) of a derivative contract written on the stock withexpiry date t = 2 which pays 1,000p if and only if S2 is not 800p (and

otherwise pays 0).

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For a stock index, S = $100, a = 30%, r = 5%, (5 = 3%, and T = 3. Let

n = 3. [n refers to the number of binomial periods]

( I ) What is the price of a European put option with a strike of $95?

( 1-) What is the price of all American call option with a strike of $95?

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S = $40, a = 30%, r = 8%, and () = O.

Suppose you sell a 40-strike put with 91 days to expiration.

( , ) What is delta?

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(' 1..-) If the option is on 100 shares, what investment is required for a delta-hedged portfolio?

The Black-Scholes price of a three-month European call with strike price 100 on a

stock that trades at 95 is 1.33, and its delta is 0.3. The price of a three-month pure discount

risk-free bond (nominal 100) is 99. You sell the option for 1.50 and hedge your position.

, One month later (the hedge has not been adjusted), the price of the stock is 97, the market

. price of the call is 1.41, and its delta is 0.36. You liquidate the portfolio (buy the call and

undo the hedge). Assume a constant, continuous risk-free interest rate and compute the net

profit or loss resulting from the trade.

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Let S = $40, K = $45, (J = 0.30, r = 0.08, T = I, and () = O.

a. What is the price of a standard call?

b. What is the price of a knock-in call with a barrier of $44. Why?

c. What is the price of a knock-out call with a barrier of $44? Why?

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i Let S = $40, a = 0.30, r = 0.08, T = I, and 8 = O. Also let Q = $60,, a Q = 0.50, 8Q = 0.04, and p = 0.5.

What is the price of an exchange option with S as the underlying asset

and 0.667 x Q as the strike price?

A portfolio of derivatives on a stock has a delta of 2400 and a gamma of -100.(9)' What position in the stock would create a delta-neutral portfolio? ...

(b) An option on the stock with a delta of 0.6 and a gamma of 0.04 can be traded.What position in the option and the stock creates a portfolio that is both gammaand delta neutral? ...

( t) A portfolio of derivatives on an asset is worth $10,000 and the risk-free interest rate

r> f'~J is 5%. The delta and gamma of the portfolio is zero. What is the theta? ...

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The delta of a European call option on a non-dividend-paying stock is 0.6, its gammais 0.04 and its vega is 0.1

(i) What is the delta of a European put option with the same strike price and timeto maturity as the call option? ...

, (ii) What is the gamma of a European put option with the same strike price and timeto maturity as the call option? ...

(ill) What is the vega of a European put option with the same strike price and timeto maturity as the call option? ...

A stock price S is governed by

dS = as dt + bS dz

where z is a £> ("0;...,,-,11"; ~.t..." _process. Find the process that governs

G(t) = SI/2(t) .

Of f'1&ed vL t~«- 50 b )

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([)NORMAL DISTRIBUTION TABLE

Entries represent the area under the standardized normal distribution from -00 to z, Pr(Z<z)

The value of z to the fIrSt decimal is given in the left column. The second decimal place is given in the top row.z

0.000.010.020.030.040.050.060.070.080.090.0

0.50000.50400.50800.51200.51600.51990.52390.52790.53190.53590.1

0.53980.54380.54780.55170.55570.55960.56360.56750.57140.57530.2

0.57930.58320.58710.59100.59480.59870.60260.60640.61030.61410.3

0.61790.62170.62550.62930.63310.63680.64060.64430.64800.65170.4

0.65540.65910.66280.66640.67000.67360.67720.68080.68440.6879

0.5

0.69150.69500.69850.70190.70540.7088. 0.71230.71570.71900.72240.6

0.72570.72910.73240.73570.73890.74220.74540.74860.75170.75490.7

0.75800.76110.76420.76730.77040.77340.77640.77940.78230.78520.8

0.78810.79100.79390.79670.79950.80230.80510.80780.81060.81330.9

0.81590.81860.82120.82380.82640.82890.83150.83400.83650.8389

1.0

0.84130.84380.84610.84850.85080.85310.85540.85770.85990.86211.1

0.86430.86650.86860.87080.87290.87490.87700.87900.88100.88301.2

0.88490.88690.88880.89070.89250.89440.89620.89800.89970.90151.3

0.90320.90490.90660.90820.90990.91150.91310.91470.91620.91771.4

0.91920.92070.92220.92360.92510.92650.92790.92920.93060.9319

1.5

0.93320.93450.93570.93700.93820.93940.94060.94180.94290.94411.6

0.94520.94630.94740.94840.94950.95050.95150.95250.95350.95451.7

0.95540.95640.95730.95820.95910.95990.96080.96160.96250.96331.8

0.96410.96490.96560.96640.96710.96780.96860.96930.96990.97061.9

0.97130.97190.97260.97320.97380.97440.97500.97560.97610.9767

2.0

0.97720.97780.97830.97880.97930.97980.98030.98080.98120.98172.1

0.98210.98260.98300.98340.98380.98420.98460.98500.98540.98572.2

0.98610.98640.98680.98710.98750.98780.98810.98840.98870.98902.3

0.98930.98960.98980.99010.99040.99060.99090.99110.99130.99162.4

0.99180.99200.99220.99250.99270.99290.99310.99320.99340.9936

2.5

0.99380.99400.99410.99430.99450.99460.99480.99490.99510.99522.6

0.99530.99550.99560.99570.99590.99600.99610.99620.99630.99642.7

0.99650.99660.99670.99680.99690.99700.99710.99720.99730.99742.8

0.99740.99750.99760.99770.99770.99780.99790.99790.99800.99812.9

0.99810.99820.99820.99830.99840.99840.99850.99850.99860.9986

3.0

0.99870.99870.99870.99880.99880.99890.99890.99890.99900.99903.1

0.99900.99910.99910.99910.99920.99920.99920.99920.99930.99933.2

0.99930.99930.99940.99940.99940.99940.99940.99950.99950.9995

3.3

0.99950.99950.99950.99960.99960.99960.99960.99960.99960.9997

3.4

0.99970.99970.99970.99970.99970.99970.99970.99970.99970.9998

3.5

0.99980.99980.99980.99980.99980.99980.99980.99980.99980.9998

3.6

0.99980.99980.99990.99990.99990.99990.99990.99990.99990.9999

3.7

0.99990.9999·0.99990.99990.99990.99990.99990.99990.99990.9999

3.8

0.99990.99990.99990.99990.99990.99990.99990.99990.99990.9999

3.9

1.00001.00001.00001.00001.00001.00001.00001.00001.00001.0000

2.5760.995

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Both equations (9.17) and (9.18) of the textbook are violated. To see this, let us calculate the values.We have:

C (K,) - C (K2)

K2- K,

22-9100 - 80 = 0.65

and C (K2) - C (K3)

K3 - K2

9-5105 - 100 = 0.8,

which violates equation (9.17) and

P (K2) - P (K[)

K2- K[

21-4100 - 80 = 0.85

and P (K3) - P (K2)

K3 - K224.80 - 21 = 0.76,105 - 100

which violates equation (9.18).

( 1-)

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We calculate lambda in order to know how many options to buy and sell when we construct thebutterfly spread that exploits this form of mispricing. Using formula (9.19), we can calculate that

lambda is equal to 0.2. To buy and sell round lots, we multiply all the option trades by 5 .

We use an asymmetric call and put butterfly spread to profit from these arbitrage opportunities.

Transaction t=OST < 8080 :S ST :S 100100 :S ST :S 105ST > 105Buy 2 80 strike calls

-440 2 x ST - 1602 x ST - 1602 x ST - 160Sell 10 100 strike calls

+900 0 1000 - 10 X ST1000 - 10 X STBuy 8 105 strike calls

-400 0 08 X ST - 840TOTAL

+60 2 x ST - 160> 0840 - 8 x ST ~ 00

Transaction

t=OST < 8080 :S ST :S 100100 :S ST :S 105ST > 105Buy 2 80 strike puts

-8160-2xST 0 00Sell 10 100 strike puts

+21010 x ST - 100010 X ST - 100000Buy 8 105 strike puts

- 198.4840 - 8 x ST840 - 8 x ST840 - 8 x ST0TOTAL

+3.60 2 x ST - 160 > 0840 - 8 x ST ~ 00

Please note that we initially receive money and have non-negative future payoffs. Therefore, wehave found an arbitrage possibility, independent of the prevailing interest rate.

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GL:~ .. The payoff is

max[.55,5 - K] = .55 + max[O, .55 - K] = .55 + .5 max[O, 5 - 2K].

Hence, by linear pricing, we add the prices of the individual pieces to obtainCH = .SP + .5Cz. Thus LX = .5, P = 0, y = .5.

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In order to hedge the short position in the call we replicate a long call. Since

the initial delta is 0.3, we start by buying 0.3 shares of the stock for 0.3·95 = 28.5 for which

we need to we borrow a total of 28.5 - 1.33 = 27.17. We deposit the margin 1.50 - 1.33 in

the bank. We next compute the constant continuous interest rate from

99 = 100e-o.25r

with solution r = 4.02%. The value of our hedging portfolio one month later is

0.3·97 - 27.17eo.o402.f2= 1.84

We buy the call for 1.41. Our total return is

(1.50 - 1.33)eo.0402.f2+ 1.84 - 1.41 = 0.60

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