Options 1 OPTIONS. Introduction

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1 Options 1 OPTIONS Introduction A derivative is a financial instrument whose value is derived from the value of some underlying asset. A call option gives one the right to buy an asset at the exercise or strike price. A put option gives one the right to sell the asset at the exercise price.

2 Options 2 An option has an exercise date, also called the strike date, maturity, or expiration date. American options can be exercised at any time up to their exercise date. European options can be exercised only at their exercise date.

3 Options 3 Complex types of derivatives cannot be priced by a simple formula such as the Black-Scholes formula. In this chapter: heuristic derivation of Black-Scholes formula

4 Options 4 Why do companies purchase options and other derivatives? Answer: to manage risk In 2000 Annual Report, the Coca Cola Company wrote: Our company uses derivative financial instruments primarily to reduce our exposure to adverse fluctuations in interest rates and foreign exchange rates and, to a lesser extent, adverse fluctuations in commodity prices and other market risks.

5 Options 5 We do not enter into derivative finanicial instruments for trading purposes. As a matter of policy, all our derivative positions are used to reduce risk by hedging an underlying economic exposure.

6 Options 6 Because of the high correlation between the hedging instrument and the underlying exposure, fluctuations in the value of the instruments are generally offset by reciprocal changes in the value of the underlying exposure. The derivatives we use are straightforward instruments with liquid markets. Derivatives can and have been used to speculate. But that should not be their primary purpose.

7 Options 7 Call options Suppose you purchased a European call on 100 shares of Stock A with exercise price of $70. At expiration, suppose Stock A is selling at $73. The option allows you to purchase the 100 shares for $70 and to immediately sell them for $73. gain of $300. Net profit isn t $300 since you paid for the option. If the option cost $2/share, then you paid $200

8 Options 8 Moreover, you paid the $200 up front but only got the $300 at the expiration date. Suppose expiration date was 3 months after purchase and r is 6% per annum or 1.5% for 3 months. The dollar value of your net profit is at t = 0 and is at t = T. exp(.015) = exp(.015)200 = 96.98

9 Options 9 We will use the notation (x) + = x if x > 0 and = 0 if x 0. With this notation, the value of a call at exercise date is (S T K) +, where K is the strike price and S T is stock s price on the exercise date, T.

10 Options 10 A call is not exercised if strike price is greater than stock price. If a call is not exercised, then one loses the cost of buying the option. One can lose money on an option even if it is exercised. In example, if Stock A were selling for $71 at T, then one would exercise and gain $100. This is less than the $200 paid for the option.

11 Options 11 Law of one price The law of one price if two financial instruments have exactly the same payoffs, then they will have the same price. This principle is used to price options. Find a portfolio or a self-financing trading strategy with a known price and same payoffs as the option in all situations. Price of the option is the same as the portfolio or trading strategy. Self-financing means requires only an initial investment and no money is withdrawn

12 Options 12 Example: Company A sells at $100/share. r is 6% compounded annually. Consider futures contract obliging party 1 to sell to party 2 one share a year from now at price P. No money changes hands now. What is the fair market price P? Not an option sale must take place. Should P depend on expected price of company A in one year? NO!!!

13 Options 13 Consider the following strategy. Party 1 can borrow $100 and buy one share now. A year from now Party 1 sells the share for P dollars Pays back $106. The profit is P 106. Therefore, P should be $106. Consider what would happen if P were not $106: Any other value of P besides $106 would lead to unlimited risk-free profits. Market would immediately correct.

14 Options 14 Arbitrage Arbitrage means a guaranteed risk-free profit with no invested capital. In other words, arbitrage is a free lunch. The arbitrage price of a security is the price that guarantees no arbitrage opportunities. The law of one price is equivalent the market being free of arbitrage. The price of $106 that we just derived is the arbitrage price.

15 Options 15 Time value of money and present value Time is money A dollar a year from now is worth less than a dollar now. We must be able to convert future values to their present values, or vice versa. Example: The arbitrage enforced future price of a stock is the present price converted into a future value.

16 Options 16 Let r be the risk-free annual interest rate. present value of $D one year from now is $D/(1 + r) (simple interest) or $D exp( r) (continuous compounding). Thus, $D now is worth $(1+r)D dollars a year from now without compounding, or ${exp(r)} D dollars under continuous compounding. When $D is a future cash flow, then its present value is also called a discounted value and r is the discount rate.

17 Options 17 Distinction between simple interest and compounding is not essential an interest rate of r without compounding is equivalent to an interest rate of r with continuously compounding where so that 1 + r = exp(r ) r = exp(r ) 1 or r = log(1 + r). We will work with both simple and compound interest, whichever is most convenient.

18 Options 18 Examples If r = 5%, then r = log(1.05) =.0488 or 4.88%. If r = 4%, then r = exp(.04) 1 = or 4.08%. In general, r > r Occasionally, we make the unrealistic assumption that r = 0. This allows us to focus on other concepts.

19 Options 19 Pricing calls a simple binomial example Suppose stock is selling for $80. At the end of one time period it can either have increased to $100 or decreased to $60. What is the current value of a call with strike price = $80 and T = 1?

20 Options 20 At T = 1, call option will be worth $20 ($100 $80) if the stock has gone up $0 dollars if the stock has gone down. However, the question is what is the option worth now?

21 Options 21 Stock Option? Given: exercise price = 80 hedge ratio = 1/2 buy 1/2 share borrow $30 initial value of portfolio = (1/2)(8) 30 = $10 Example of one-step binomial option pricing.

22 Options 22 The current value of the option depends only on r. Start with r = 0. The value of the option is $10. How did I get this value?

23 Options 23 Consider the following investment strategy. Borrow $30 and buy one-half of a share of stock. The cost upfront is $40 $30 = $10. If stock up, then portfolio worth 100/2 30 = 20 (same as option). If stock down, then portfolio worth 60/2 30 = 0 (same as option). By law of one price: present value of the call = present value of the portfolio = $10.

24 Options 24 Let s summarize what we have done. We have found a portfolio of stock and risk-free that replicates the call option. The current value of the portfolio is easy. And, the option must have the same value.

25 Options 25 Suppose we have just sold a call option. By purchasing this portfolio we have hedged the option. By hedging is meant that we have eliminated all risk: the net return of selling the option and purchasing the portfolio is exactly 0, no matter what.

26 Options 26 How did I know that the portfolio should be 1/2 share of stock and $30 in cash? The volatility of one share of the stock is $100 $60 = $40. The volatility of the option is $20 $0 = $20. The ratio of volatilities is 1/2 (called the hedge ratio). Portfolio must have same volatility as option so portfolio must have one-half share.

27 Options 27 Stock Option? Given: exercise price = 80 hedge ratio = 1/2 buy 1/2 share borrow $30 initial value of portfolio = (1/2)(8) 30 = $10 Example of one-step binomial option pricing.

28 Options 28 Key point: The number of shares in the portfolio must equal volatility of option hedge ratio = volatility of stock. So now we know why the portfolio holds 1/2 share. How was it determined that $30 should be borrowed? If stock down, portfolio worth $30 minus amount borrowed and option worth $0. Thus, the amount borrowed is $30.

29 Options 29 Key point: When stock down value of portfolio = value of the option. Alternatively, when stock up value of portfolio = value of the option. So $50 minus the amount borrowed = $20.

30 Options 30 Stock Option? Given: exercise price = 80 hedge ratio = 1/2 buy 1/2 share borrow $30 initial value of portfolio = (1/2)(8) 30 = $10 Example of one-step binomial option pricing.

31 Options 31 Suppose interest rate is 10%. Then borrow $30/(1.1) = $27.27 so amount owed after one year is $30. Cost of portfolio is 40 30/1.1 = $ So value of option is $ if the risk-free rate is 10%. Higher than when the risk-free rate is 0, because the initial borrowing is more expensive.

32 Options 32 Here s how to valuate one-step binomial options for other values of the parameters. Suppose current price is s 1 after one time period the stock either goes up to s 3, or down to s 2 strike price is K risk-free rate is r

33 Options 33 s 2 < K < s 3, so the option is exercised if and only if the stock goes up. This is a reasonable assumption for the following reasons: If s 2 < s 3 K, then the option will never be exercised its price must be 0 If K s 2 < s 3, then the option will always be exercised it is a futures contract.

34 Options 34 Therefore, the hedge ratio is δ = s 3 K s 3 s 2. (1) This is the number of shares purchased cost is δ s 1

35 Options 35 amount borrowed is and δs r, (2) amount paid back will be δs 2. price of the option (cost of portfolio) is { δ s 1 s } 2 = s { 3 K s 1 s } 2. (3) 1 + r s 3 s r

36 Options 36 If the stock goes up, then the option is worth (s 3 K) the portfolio is also worth (s 3 K). If the stock goes down, both the option and the portfolio are worth 0. Thus, the portfolio does replicate the option.

37 Options 37 Example In the example analyzed before, s 1 = 80 s 3 = 100 s 2 = 60 K = 80 Therefore, δ = The price of the option is = 1 2. { r }, which is $10 if r = 0 and $ if r = 0.1.

38 Options 38 The amount borrowed is which is δs r = (1/2) r = r, $30 if r = 0 $27.27 if r =.1

39 Options 39 Two-step binomial option pricing

40 Options 40 A one-step binomial model may be reasonable for very short maturities. For longer maturities, multiple-step models needed. A multiple-step model can be analyzed by doing the individual steps going backwards in time.

41 Options 41 Where we are headed eventually: As the number of steps increases and the holding period decreases accordingly stock price follows a geometric Brownian motion price of option converges to Black-Scholes formula

42 Options 42 Stock Option ? 80 80? 0 70? 60 0 Exercise price = $80 Two-step binomial model for option pricing.

43 Options 43 Assume r = 0 for now. Using the pricing principles just developed and working backwards, we can fill in the question marks:

44 Options 44 Stock Option /2 share borrow $35 1 share borrow $80 5 A 10 B 20 0 E D Exercise price = $80 0 shares borrow $0 Pricing the option by backwards induction. 0 C 0 F

45 Options 45 Let s show that our trading strategy is self-financing. To do this we need to show that we invest no money other than the initial $5. Suppose that the stock is up on the first step, so we are at node B. Then our portfolio is worth $90/2 $35 or $10. At this point we borrow $45 and buy another half-share for $45; this is self-financing. If the stock is down on the first step, we sell the half share of stock for $35 and buy off our debt; again the step is self-financing.

46 Options 46 Arbitrage pricing by expectation Options are priced by arbitrage, The expected value of the option is not relevant. In fact, we have not even considered the up and down probabilities.

47 Options 47 However, there is a remarkable result showing that arbitrage pricing can be done using expectations. There exists probabilities of the stock moving up and down such that the arbitrage price is equal to the expected value. Whether these are the true probabilities is irrelevant. These probabilities do give the correct arbitrage price. In fact, they are the easiest means for computing arbitrage derived prices.

48 Options 48 Let now be time 0 and one-step ahead be time 1. Present value of $D dollars at time 1 is $D/(1 + r). Let f(2) = 0 and f(3) = s 3 K be the values of option if stock moves up or down.

49 Options 49 There is a value of q between 0 and 1, such that the present value of the option is 1 {qf(3) + (1 q)f(2)}. (4) 1 + r (4) is the present value of the expectation of the option at time 1 when q is up probability.

50 Options 50 How do we find this magical value of q? q must satisfy r {qf(3) + (1 q)f(2)} = s 3 K s 3 s 2 { s 1 s } r Substituting f(2) = 0 and f(3) = s 3 K into equation and solving for q: q = (1 + r)s 1 s 2 s 3 s 2.

51 Options 51 From previous slide: q = (1 + r)s 1 s 2 s 3 s 2. (5) We want q to be between 0 and 1. From (5) one can see that 0 q 1 if s 2 (1 + r)s 1 s 3.

52 Options 52 s 2 (1 + r)s 1 s 3 is required for market to be arbitrage-free. If we invest s 1 in a risk-free asset at time 0, then value of our holdings at time 1 will be (1 + r)s 1. If we invest s 1 in stock, then value of our holdings at time 1 will be either s 2 or s 3. If s 2 (1 + r)s 1 s 3 were not true, then there would be an arbitrage opportunity.

53 Options 53 For example, if (1 + r)s 1 < s 2 s 3, then could borrow at rate r and invest in stock at t = 1 we pay back (1 + r)s 1 and receive at least s 2 > (1 + r)s 1. Thus, the requirement that the market be arbitrage-free ensures that 0 q 1.

54 Options 54 A general binomial tree model Consider a possibility non-recombinant tree as seen in the following figure.

55 Options 55 7 q(1) 3 q(3) 1 q(3) q(1) 5 2 q(2) 1 q(2) 4 Two-step non-recombinant tree.

56 Options 56 Assume that: At the jth node the stock is worth s j and the option is worth f(j). The jth node leads to either the 2j + 1th node or the 2jth node after one time tick. The actual time between ticks is δt. Interest is compounded continuously at a fixed rate r so that B 0 dollars now is worth exp(rn δt)b 0 dollars after n time ticks.

57 Options 57 Then at node j: The value of the option is { } f(j) = exp( r δt) q j f(2j + 1) + (1 q j )f(2j). where The arbitrage determined q j is q j = er δt s j s 2j s 2j+1 s 2j. (6) The number of shares of stock to be holding is φ j = f(2j + 1) f(2j) s 2j+1 s 2j = hedge ratio.

58 Options 58 Denote the amount of capital to hold in the risk-free asset by ψ j. typically ψ j is negative because money has been borrowed. Since portfolio replicates option, f(j) = s j φ j + ψ j. Therefore, ψ j = {f(j) φ j s j }. (7) (ψ j changes in value to e r δt {f(j) φ j s j } after one time tick). The probability of any path is product of q j s along the path.

59 Options 59 An example The tree for the previous example is shown in the next figure. Because r = 0 and stock moves up or down same amount ($10), the q j all equal 1/2. It follows from q j = er δt s j s 2j s 2j+1 s 2j. that whenever r = 0 and the up moves = s 2j+1 s j and down moves = s j s 2j, are length, then q j = 1/2 for all j.

60 Options

61 Options 61 The probability of each full path from node 1 to one of nodes 4, 5, 6, or 7 is 1/4. Given the values of the option at nodes 4, 5, 6, and 7, it is easy to compute the expectations of the option s value at other nodes. The path probabilities are independent of the exercise price, since they depend only on the prices of stock at the nodes and on r. Therefore, it is easy to price options with other exercise prices.

62 Options 62 Exercise Assuming the same stock price process as in the figure, price the call option with an exercise price of $70. Answer: Given this exercise price, it is clear that the option is worth $0, $10, $10, and $30 dollars at nodes 4, 5, 6, and 7, respectively. Then we can use expectation to find that the option is worth $5 and $20 at nodes 2 and 3, respectively. Therefore, the option s value at node 1 is $12.50; this is the price of the option.

63 Options 63 Martingales Martingale: probability model for a fair game, that is, a game where the expected changes in one s fortune are always zero. More formally,..., Y 0, Y 1, Y 2,... is a martingale if for all t. E(Y t+1 Y t,..., Y 1 ) = Y t In most cases that we will study E(Y t+1 Y t,..., Y 1 ) = E(Y t+1 Y t ) (Markov)

64 Options 64 Example On each toss of a fair coin we wager half of our fortune on heads. Our fortune at time t is Y t. We win or loss Y t /2 with probability 1/2. Our fortune at time t + 1 is either Y t /2 or (3/2)Y t, each with probability 1/2. Therefore, E{Y t+1 Y t } = (1/2)(Y t /2) + (1/2)(3/2)Y t = Y t. our sequence of fortunes is a martingale.

65 Options 65 Let P t, t = 0, 1,... be the stock price at the end of the tth step. Then Pt process. := exp( rt δt)p t is the discounted price

66 Options 66 Key fact: Under the {q j } probabilities, the discounted price process Pt is a martingale. Recall: q j = er δt s j s 2j s 2j+1 s 2j. To see that Pt is a martingale, we calculate using the definition (6) of q j : E(P t+1 P t = s j ) = q j s 2j+1 + (1 q j )s 2j = s 2j + q j (s 2j+1 s 2j ) = s 2j + {exp(r δt)s j s 2j } = exp(r δt)s j. This holds for all values of s j.

67 Options 67 Therefore, so that or E(P t+1 P t ) = exp(r δt)p t, E{exp( r(t + 1) δt)p t+1 P t )} = exp( rt δt)p t, This shows that P t E(P t+1 P t ) = P t. is a martingale.

68 Options 68 Martingale or risk-neutral measure Any set of path probabilities, {p j }, is called a measure of the process. The measure {q j } is called the martingale measure or the risk-neutral measure. We will also call {q j } the risk-neutral path probabilities.

69 Options 69 The risk-neutral world If all investors were risk-neutral, that is, indifferent to risk, then there would be no risk premiums and all expected asset prices would rise at the risk-free rate. Therefore, all asset prices discounted at the risk-free rate would be martingales. We know that we do not live in such a risk-free world But expectations with respect to a risk-neutral model give arbitrage-free prices.

70 Options 70 Example It was argued that if a stock is selling at $100/share and r is 6%, then the correct (= arbitrage-free) future price of a share one year from now is $106. We can now calculate this value using the risk-neutral measure in the risk-neutral world, the expected stock price will increase to exactly $106 one year from now.

71 Options 71 Trees to random walks to Brownian motion Getting more realistic Binomial trees are useful because they illustrate several important concepts, in particular: arbitrage pricing self-financing trading strategies hedging computation of arbitrage prices using the risk-neutral measure

72 Options 72 However, binomial trees are not realistic: Stock prices are continuous, or at least approximately continuous. This lack of realism can be alleviated by increasing the number of steps. One can increase the number of steps without limit to derive the Black-Scholes formula. The present section will get us closer to that goal.

73 Options 73 A three-step binomial tree The next figure is a three-step tree where at each step the stock price either goes up $10 or down $10. Assume that the risk-free rate is r = 0. Strike price = $80

74 Options 74 All q s are 1/ t = 0 t = 1 t = 2 t = 3

75 Options 75 P t is stock price at time t Risk-neutral path probabilities are each 1/2 Using risk neutral probabilities P t is a stochastic process, P t+1 equals P t ± $10 process is a random walk.

76 Options 76 We have P t = P 0 + ($10){2(W W t ) t} (8) where W 1,, W 3 are independent and W t equal 0 or 1, each with probability 1/2. If W t is 1, then 2W t 1 = 1 so price jumps up $10. If W t is 0, then 2W t 1 = 1 so the price jumps down $10.

77 Options 77 The random sum W W t is Binomial(t, 1/2) mean = t/2 (using formula: np) variance = t/4. (using formula: np(1 p)) The value of the call option is where x + equals x if x 0 and equals 0 otherwise. E{(P 3 K) + } (9) The expectation in (9) is with respect to the risk-neutral probabilities.

78 Options 78 Since W 1 + W 2 + W 3 is Binomial(3, 1/2) it equals 0, 1, 2, or 3 with probabilities 1/8, 3/8, 3/8, and 1/8, respectively. Using formula: ( ) n p x (1 p) n x x Therefore, E{(P 3 K) + } = 1 8 [ {P 0 30 K + (20)(0)} + + 3{P 0 30 K + (20)(1)} + + 3{P 0 30 K + (20)(2)} + + {P 0 30 K + (20)(3)} + ].

79 Options 79 Examples If P 0 = 100 and K = 80, then P 0 30 K = 10 and E{(P 3 K) + } = 1 8 { ( ) + + 3( ) ( ) + + ( ) + } ( ) = = as seen in previous figure, which is shown again on the next page.

80 Options 80 All q s are 1/ t = 0 t = 1 t = 2 t = 3

81 Options 81 Similarly, if P 0 = 100 and K = 100, then P 0 30 K = 30 and E{(P 3 K) + } = 1 8 { ( ) + + 3( ) ( ) + + ( ) + } = 1 60 ( ) = 8 8 = 7.5

82 Options 82 More time steps Let s consider a call option with T = 1. Take time interval [0, 1] divide into n steps, each of length 1/n. Suppose that stock price goes up or down σ/ n at each step.

83 Options 83 Then after m steps (0 m n) t = m/n price is P m/n = P 0 + σ {2(W W m ) m}. n

84 Options 84 From last slide: P m/n = P 0 + σ n {2(W W m ) m}. Since, W W m is Binomial(m, 1/2) E(P t P 0 ) = P 0. and Var(P t P 0 ) = 4σ2 m n 4 = m n σ2 = tσ 2. By the CLT, as n, P t converges to N(P 0, tσ 2 ).

85 Options 85 K = exercise price. Value of option is expectation with respect to risk-neutral measure of present value at expiration. Therefore, as the number of steps goes to, price of option converges to where Z is N(0, 1) E{(P 0 + σz K) + } (10) so that P 1 = P 0 + σz is N(P 0, σ 2 ).

86 Options 86 P t is a discrete time stochastic process since t = 0, 1/n, 2/n,..., (n 1)/n, 1. In fact, P t is a random walk. As n, P t becomes a continuous time stochastic process. Limit process is called Brownian motion.

87 Options 87 Brownian motion Brownian motion is the limit of random walks as the frequency of steps increases. B t starting at B 0 = 0 has the following properties: 1. E(B t ) = 0 for all t. 2. Var(B t ) = tσ 2 for all t. 3. If t 1 < t 2 < t 3 < t 4, then B t2 B t1 and B t4 B t3 are independent. 4. B t is normally distributed for any t.

88 Options 88 Geometric Brownian motion & Black-Scholes s exp(µ/n ± σ/ n) = (s up, s down ). q = s exp(r/n) s down s up s down exp(r/n) exp(µ/n σ/ n) = exp(µ/n + σ/ n) exp(µ/n σ/ n) 1 ( 1 µ r + ) σ2 /2 2 σ. n { P t = P m/n = P 0 exp µt + σ m } (2W i 1). n i=1

89 Options 89 W i is either 0 or 1 (so 2W i 1 = ±1) { m } σ σm(2q 1) E 2W i 1 = n i=1 n σ ( ) r µ σ 2 /2 m n σ n ) = (r µ σ2 m 2 n = t(r µ σ2 /2). Var { σ m n i=1 2W i 1 since q 1/2 as n. } = 4σ2 mq(1 q) n tσ 2,

90 Options 90 In the risk-neutral world P t P 0 exp{(r σ 2 /2)t + σb t }. (11) (11) does NOT depend on µ, only on σ. Payoff at time T is [ P0 exp{(r σ 2 /2)T + σb T } K ] +. (12) Since B T N(0, T ), we can write B T = T Z where Z N(0, 1). The discounted value of (12) is [ { } P 0 exp σ2 T 2 + σ T Z ] exp( rt )K +. (13)

91 Options 91 From previous slide: Discounted value is [ { } P 0 exp σ2 T 2 + σ T Z ] exp( rt )K +. Therefore, call price is [ { } C = P 0 exp σ2 T 2 + σ T z ] exp( rt )K + φ(z)dz,

92 Options 92 Computing integral leads to: Black-Scholes formula C = Φ(d 1 )S 0 Φ(d 2 )K exp( rt ) where Φ is the standard normal CDF, d 1 = log(s 0/K) + (r + σ 2 /2)T σ T, and d 2 = d 1 σ T.

93 Options 93 Example S 0 = 100, K = 90, σ =.4, r =.1, and T =.25. Then and d 1 = log(100/90) + {.1 + (.4)2 /2}(.25).4.25 d 2 = d = Then Φ(d 1 ) =.7739 and Φ(d 2 ) = Also, exp( rt ) = exp{(.1)(.25)} = Therefore, = C = (100)(.7739) (90)(.9753)(.7095) = 15.1.

94 Options 94 How does the option price depend on the inputs? The next figure shows the variation in the price of a call option as the parameters change. The baseline values of the parameters are S 0 = 100, K = 100 exp(rt ), T =.25, r =.06, and σ =.1.

95 Options 95 The exercise price K and initial price S 0 have been chosen so that if invested at the risk-free rate, S 0 would increase to K at expiration time. There is nothing special about this choice of S 0 and K. In each of the subplots in the figure, one of the parameters is varied while the others are held at baseline.

96 Options Price of call Price of call 10 5 Price of call σ K S 0 Price of call Price of call r T

97 Options 97 Example GE This information was taken from The Wall Street Journal, February 14, 2001 GE closed at $47.16 on February 13, so we use S 0 = month T-bill rate was 4.91%. r = /253 = , assuming a return on the 253 trading days per year. T = 3 for options expiring in February. An option expiring in March had T = 23 For June, T = 23 + (21)(3) For September, T = 23 + (6)(21).

98 Options 98 K Month of T (in Actual B&S calculated price Implied Expiration days) Price σ =.0176 σ =.025 Volatility 35 Sep Sep Mar Feb Mar Feb Mar Sep Mar Jun Actual prices and prices determined by the Black-Scholes formula for options on February 13, K is the exercise price. T is the maturity.

99 Options 99 Early exercise of calls is never optimal Example: first option in table strike price is 35 closing price of GE was if the option had been exercised: $( )= $12.16 selling on the market for $14.90 that day. gain $( ) = $2.74 more by selling

100 Options 100 Implied volatility Implied volatility is the amount of volatility the market believes to exist currently. In the following figure The price was $5.30 on February 13, The implied volatility in figure is Determined by MATLAB s interpolation function interp1.m.

101 Options 101 price sigma Calculating the volatility implied by the option with an exercise price of $42.50 expiring in March 2001.

102 Options 102 Volatility Smiles and Polynomial Regression Superscript c means center, e.g., K c = K K. Initial model: V i = β 0 + β 1 K c i + β 2 (K c i ) 2 + β 3 T c i + β 5 (T c i ) 2 + β 6 K c i T c i + ɛ i, Final model: V i = β 0 + β 1 K c i + β 2 (K c i ) 2 + β 3 T c i + β 5 (T c i ) 2 + β 6 (T c i ) 3 + ɛ i,

103 Options 103 data ImpVol ; infile C:\book\sas\ImpVolRegData.txt ; input K T ImpVol ; T = T ; K = K ; T2 = T*T ; K2 = K*K ; KT = K*T ; K3 = K*K*K ; T3 = T*T*T ; run ;

104 Options 104 proc reg; model ImpVol = K T T2 K2 KT K3 T3/ selection=rsquare adjrsq cp best=2 ; run ;

105 Options 105 The REG Procedure Dependent Variable: ImpVol R-Square Selection Method Number in Adjusted Model R-Square R-Square C(p) Variables in Model K K K T T2 K T2 K3 T K T2 T K T2 K2 T3 min C_P but no T term K T T2 K K T T2 K2 T3 model used K T2 K2 K3 T K T2 K2 KT K3 T K T T2 K2 K3 T K T T2 K2 KT K3 T3

106 Options implied volatility exercise price K varies with T fixed at mean

107 Options 107 implied volatility maturity T varies with K fixed at mean

108 Options 108 implied volatility maturity 0 35 exercise price T and K both vary 55

109 Options 109 SE of implied volatility x maturity 0 35 exercise price Standard error surface 55

110 Options 110 Puts In this example, European and American puts do NOT have the same price at all nodes. The continuously compounded rate is r = K = 110 In this example q = exp(.05) =.6282.

111 Options (13.54) (30) Pricing a put option. Amercian put in parentheses when it differs from European put.

112 Options 112 At node 2 the European option is worth (.6282)(14) + (1.6282)(46) = $ The American option can be exercised to earn $(110 80) = $30. Thus, at node 2 the European option is worth $24.63 but the American option is worth $30. The price of the European put at node 1 is e.05 {(q)(4.91) + (1 q)(24.63)} =

113 Options 113 At node 1, the American option is worth e.05 {q (1 q) 30} = 13.65, which is more than $ The American option should NOT be exercised early at node 1 would earn only $10. American option is worth more than European at node 1 because the American can be exercised early at node 2 should the stock move down at node 1.

114 Options 114 Why are puts different than calls? A put increases in value as the stock price decreases. As the stock price decreases, the size of further price changes also decreases. At the point of diminishing returns, it is better to exercise the option and invest the profits in a risk-free asset. With calls, everything is reversed.

115 Options 115 Put-call parity Put price and call price with same K and T are related: Consider P = C + e rt K S 0. first portfolio holds one call and Ke rt dollars in the risk-free asset second portfolio holds a put and one share of stock. same payoff K or S T, whichever is larger Thus, C + e rt K = P + S 0,

116 Options 116 From previous slide: P = C + e rt K S 0, (14) (14) holds only for European options American puts are worth more than European puts, so the left hand side of (14) is larger for American than for European puts.

117 Options 117 The evolution of option prices t = 0 is when the option was written t = T is the expiration date 0 < t < T Black-Scholes formula can be used to find the option price at t S 0 in the formula set equal to S t T in the formula set equal to T t

118 Options stock price call price put price time Evolution of option prices. T = 1, σ =.1, r =.06, S 0 = 100, and K = 100 for both put and call. Blue and red are two simulations.

119 Options The Greeks stock price call price put price time Evolution of option prices. T = 1, σ =.1, r =.06, S 0 = 100, and K = 100 for both put and call.

120 Options call return / stock return time Ratio of log return on a call to log return on the underlying stock. Uses blue simulation.

121 Options 121 C(S, T, t, K, σ, r) = price of an option = C(S, T, t, K, σ, r) S Delta Θ = C(S, T, t, K, σ, r) t Theta R = C(S, T, t, K, σ, r) r Rho V = C(S, T, t, K, σ, r) σ Vega

122 Options 122 Use of : change in price of option change in price of asset. Or, using math notation, P option t P option t 1 = (P asset t P asset t 1 ) Therefore, P option t P option t 1 P option t 1 = ( P asset t 1 P option t 1 ) P asset t P asset t 1 P asset t 1

123 Options 123 From last slide: Define: Then: P option t P option t 1 P option t 1 = L = ( ( P asset t 1 P option t 1 P asset t 1 P option t 1 ) ) P asset t P asset t 1 P asset t 1 return on option L return on asset and µ option L µ asset and σ option L σ asset

124 Options 124 Recall Black-Scholes formula C = Φ(d 1 )S 0 Φ(d 2 )K exp( rt ) By differentiating the Black-Scholes formula for a call: = Φ(d 1 )

125 Options 125 Example: T = 1, σ = 0.1, r = 0.06, S 0 = 100, and K = 100 At t = 0 C(100, 1, 0, 100, 0.1, 0.06) = 7.46 d 1 = 0.65 = Φ(0.65) = revenue from call option on one share = times the revenue from one share return on call is L = (0.742)(100)/7.46 = 9.95 times return on the stock

126 Options call return / stock return time Ratio of log return on a call to log return on the underlying stock 9.95 does not include the effect of t changing (by one unit). This effect is approximately Θ = C(S, T, t, K, σ, r) t