1. Drake DRAKE UNIVERSITY Fin 288 Valuing Options Using Binomial Trees

2. Drake Drake University Fin 288 Binomial Trees A binomial tree is designed to represent the possible paths the stock price might follow. To begin we want to assume that the stock will be held for one period of time with two possible outcomes for the price at the end of the period, either an increase or decrease (following a binomial probability distribution) The model will then be extended to account for multiple price movements through multiple periods of time.

3. Drake Drake University Fin 288 A simple Example Assume that the stock is currently selling for $20 and at the end of three months it will be worth either $22 or $18. Assume we want to value a European call option with an exercise price of $21 at the end of the three months. There are two possible outcomes: Stock Price = $22 and intrinsic value of option = $1 Stock Price = $18 and intrinsic value of option = $0

4. Drake Drake University Fin 288 Possible Outcomes Shown on Binomial Tree Model Stock Price =$20 Stock Price = $22 Stock Price = $18 Time 0Time 1 l Option Price = $1 l l Option Price = $0

5. Drake Drake University Fin 288 A no arbitrage solution To price the option we want to assume that we can use it to eliminate risk. Combining a short position in the call option with a spot position in the underlying stock produces a result where gains on the stock are offset by option. We want to establish a riskless portfolio so that the value is the same regardless of whether the price of the stock increases to $22 or decreases to $18 in 3 months. Given that the portfolio is riskless, the portfolio should earn a rate equal to the risk free rate.

6. Drake Drake University Fin 288 Value of Portfolio in 3 months Consider a portfolio of  shares of the stock and one option. If the stock price increases to $22 and the option has a $1 intrinsic value the portfolio is worth: $22  -$1 If the stock price decreases to $18 and the option has a $0 intrinsic value the portfolio is worth: $18

7. Drake Drake University Fin 288 Determining the number of shares in the portfolio The portfolio has two possible outcomes in three months: $22  - $1 and $18  Since the two possible outcomes have equal values it is possible to solve for  $22  - $1 = $18  $22  - $18  $1 ($22  - $18)  $1  $1/$4 =.25

8. Drake Drake University Fin 288 Time Value If the stock price increases to $22 the portfolio is worth $22  $1 = $22(.25) -1 = $4.5 If the stock price decreases to $18 the portfolio is worth $18  $18(.25) = $4.5 Since the portfolio is risk free it should have a present value at time 0 equal to $4.5 discounted at the risk free rate for 3 months. Assuming a risk free rate of 12% the PV is: $4.5e -0.12(.25) =$4.367

9. Drake Drake University Fin 288 Pricing the option Let the price of the option be f Given the current share price of $20, the value of the portfolio at time zero should be equal to $20(.25) – f Which must equal the value of either portfolio 4.367 Setting the two equal to each other an solving for f finds the value of the call. $20(.25) – f = 4.367 f =.633

10. Drake Drake University Fin 288 An overpriced call If the option price was greater than 0.633 you could make a risk free return above the risk free rate. What if the value of the call was.80 today? The current value of the portfolio = $20(.25)-.80 = $4.20 In either outcome the value of the portfolio after three months is 4.5 this implies a risk free return of 4.20e r(.25) =4.5 r =.2759 = 27.59%

11. Drake Drake University Fin 288 An overpriced call Since you can earn a return greater than the risk free rate the demand for the stock will increase causing its price to increase. Additionally the supply of the option will increase (since everyone will want to write an option to take advantage of the higher price) decreasing its price.

12. Drake Drake University Fin 288 An underpriced call What if the price of the call is less than.633? Assume it is.50, since it is cheap, you buy it and sell the stock short receiving $25(.25) = $5 your total cash flow is $5 -.50 = $4.50 In three months you need to buy the stock to close the short sale. If the price increased to $22 you need to pay -$22(.25) + 1 = -$4.50 If the price decreased to $18 you need to pay -$18(.25) = -$4.50

13. Drake Drake University Fin 288 An underpriced call In either case you essentially borrowed $4.50 at time zero and repaid $4.50 in three months. You have borrowed at a rate much less than the risk free rate. The market will react to this increasing the call price (as demand for the option increases) and decreasing the share price (as there are more people shorting the stock).

14. Drake Drake University Fin 288 Generalizing the Model Let the stock price at time 0 be S 0 with two possible outcomes: The stock price increases by a factor greater than 1, u The resulting stock price is then S 0 u The stock price decrease by a factor less than 1, d. The resulting stock price is then S 0 d If the stock price increases the value of the option is f u If the stock price decreases the value of the option is f d

15. Drake Drake University Fin 288 Binomial Tree Model S0fS0f S0ufuS0ufu S0dfdS0dfd Time 0Time 1 l l l

16. Drake Drake University Fin 288 The generalization Setting the value of the two portfolios at time 1 equal produces S 0 u  -f u = S 0 d  -f d Solving for  produces: Delta is thus the change in the option price divided by the change in the stock price

17. Drake Drake University Fin 288 The PV of the portfolio Since the value of the portfolio at time 1 is the same regardless of whether the price increased or decreased either price can be used to find the value of the portfolio at time 0 Using a stock price increase the value of the portfolio at time 0 is (S 0 u  -f u )e -rT

18. Drake Drake University Fin 288 The value of the option The PV of the portfolio has to equal the cost of setting up the option at time 0: S 0  f = (S 0 u  -f u )e -rT The goal is to solving for f, the current value of the option. f = S 0  (S 0 u  -f u e -rT f = S 0  S 0 u  e -rT -f u e -rT f = S 0  ue -rT )-f u e -rT

19. Drake Drake University Fin 288 Solving for the price of the option, f substitute f = S 0  ue -rT )-f u e -rT Simplifying produces f = e -rT [pf u +(1-p)f d ] Where :

20. Drake Drake University Fin 288 Returning to the Example: In the previous example we had a current price of $20 with the possibility of an increase to $22 or $20u=$22 solving for u, u =1.1 Similarly $20d=$18, solving for d, d=.9 Given the inputs from before u = 1.1 and d=.9 r=.12, T =.25, f u =1, and f d =0 Plugging into the equations on the previous slide p =.6523 and f =.6333 Which is the same as before

21. Drake Drake University Fin 288 A Quick Question Does the probability of an increase in the stock price impact the value of the option? NO! regardless of the probability the value of the option is the same – the probability of an increase or decrease is already incorporated in the price of the stock. Since the option price is based on the stock price, the probability does not impact the option price.

22. Drake Drake University Fin 288 Risk Preferences Review Risk Loving vs. Risk Neutral vs. Risk Adverse Assume you are faced with two equally likely outcomes: A gain of $10 and a loss of $5. How much would you be willing to pay to accept the risk of the possible loss?

23. Drake Drake University Fin 288 Risk Preferences Review Risk Neutral: If you are willing to pay $2.50, you are willing to pay a “fair price” to accept the risk. (If you repeated the event over and over on average you would receive your $2.50) Risk Averse: If you are willing to pay less than 2.50 lets say 2.00, you are risk averse. The $.50 represents a risk premium, the additional return you expect to earn for accepting the risk. The lower the amount you are willing to pay the more risk averse you are.

24. Drake Drake University Fin 288 Risk - Neutral Valuation Now assume that p is the probability of an increase in the stock price and 1-p is the probability of a decrease in the price of the stock. This would imply that the option price f = e -rT [pf u +(1-p)f d ] is the expected payoff from owning the option discounted at the risk free rate.

25. Drake Drake University Fin 288 Risk Neutral Valuation Using p in the same way produces the Expected value of the stock at time t E(S T )=pS 0 u+(1-p)S 0 d E(S T )=pS 0 (u-d)+S 0 d E(S T )=S 0 e rT The stock price grows on average at the risk free rate! Substituting

26. Drake Drake University Fin 288 Risk Neutral Valuation In a risk neutral world investors do not get compensated for risk. Implying the expected return on all securities is the risk free rate of interest. In the case of assuming p = the probability of an increase in the stock price, the option is equal to its expected payoff discounted at the risk free rate. This is based upon our portfolio assumptions that both return the risk free return. The risk neutral valuation is the same as the no arbitrage valuation.

27. Drake Drake University Fin 288 No Arbitrage Stock Price =$20 Stock Price = $22 Stock Price = $18 Time 0Time 1 l Option Price = $1 l l Option Price = $0

28. Drake Drake University Fin 288 No arbitrage Previously we found that the no arbitrage value of the option was.633 assuming that the we used the risk free rate of 12% in the valuation.

29. Drake Drake University Fin 288 No arbitrage vs. Risk Neutral In the risk neutral world the expected value of the stock price in one period would be also equal to the risk free return of 12% Therefore, if p is the probability of a stock price increase, the expected value of the stock in the future must equal the value of the stock today assuming it grows at the risk free rate or: 22p+18(1-p) = 20e.12(.25) 22p-18p = 20e.12(.25) -18 4p=2.6092 p=0.6523

30. Drake Drake University Fin 288 Expected Value of Option in 3 months Given the probability of a stock price increase is 0.6523, if the price increases to $22 in our example the option is worth $1. If the price decreased to $18 the option is worth $0 The expected value of the option in three months in our example would then be: 0.6523($1)+(1-0.6523)$0 =0.6523

31. Drake Drake University Fin 288 Value of Option at Time 0 In a risk neutral setting the value of the option should also increase at the risk free rate of 12% By discounting the value of the option at time = 3 months back to time =0 at the risk free rate we can find the value of the option at time 0 0.6523e -(.12)(.25) = 0.633 Which is the same value of the option in the No Arbitrage solution

32. Drake Drake University Fin 288 Risk Neutral vs. “Real World” In the “real world” the probability of an increase in the stock price is not equal to p. This occurs when the expected payoff on the stock is equal to the risk free rate, which is not usually the case. What if expected return on the stock is 16%? Let q = the probability of an increase and solve for q assuming a 16% expected return on the stock. 22q+18(1-q) = 20e.16(.25) q =.7041

33. Drake Drake University Fin 288 Real World return on the Option If q =.7041 the expected payoff on the option is.7041(1)+(1-.7041)0 =.7041 Given a risk neutral value of the call at time 0 of 0.633 the implied discount rate on the option is.633=.7041e -r(.25) R =.4258 = 42.58%

34. Drake Drake University Fin 288 Risk Neutral While the implied discount rate for the option could be found it is based on the assumption that the value of the option at time 0 is based on the risk neutral value of the option. Without the risk neutral value of the option it would not be possible to compare how much extra return the option should pay based on its increased risk compared to the stocks expected return of 16%.

35. Drake Drake University Fin 288 Extending the tree Now assume that we want to look another 6 months out. In the previous example it was assumed that the price increased by10% or decreased by 10%. Assume that at time t = 3 months this is again the case. You can build the tree out from the possible payoffs at time t= 3 months.

36. Drake Drake University Fin 288 Possible Outcomes Shown on Binomial Tree Model Stock Price =$20 Stock Price = $22 Stock Price = $18 Time 0 Time 1 (3 months) l l l

37. Drake Drake University Fin 288 Possible Outcomes after an initial price increase Stock Price =$22 Stock Price = $24.2 Stock Price = $19.8 Time 1 (3 months) Time 2 (6 months) l l l

38. Drake Drake University Fin 288 Possible Outcomes after an initial price decrease Stock Price =$18 Stock Price = $19.8 Stock Price = $16.2 Time 1 (3 months) Time 2 (6 months) l l l

39. Drake Drake University Fin 288 Combining the possible outcomes $20 Time 0 Time 1 (3 months) l l l l l l Time 2 (6 months) Stock Price =$22 Stock Price = $18 Stock Price = $24.2 Stock Price = $19.8 Stock Price = $16.2

40. Drake Drake University Fin 288 Calculating the Option price The option price at time zero now depends upon two different time periods. Assume we want to look at an option that has a strike price of $21 at time t = 6 months. The price of the option at time 6 months will equal its intrinsic value as before, but the price at time t = 3 months will need to be calculated based upon the possible outcomes at time 6 months.

41. Drake Drake University Fin 288 Possible Outcomes after an initial price increase Stock Price =$22 Stock Price = $24.2 Option Price = $3.2 Stock Price = $19.8 Option Price = $0 Time 1 (3 months) Time 2 (6 months) l l l

42. Drake Drake University Fin 288 Option price Using the option price equation from before the price of the option assuming an initial increase to time t=3 can be found. f = e -rT [pf u +(1-p)f d ] f = e -0.12(.25) [(.6523)(3.2)+(.3477)(0)] =2.2057

43. Drake Drake University Fin 288 Possible Outcomes after an initial price decrease Stock Price =$18 Stock Price = $19.8 Option Price = $0 Stock Price = $16.2 Option Price = $0 Time 1 (3 months) Time 2 (6 months) l l l

44. Drake Drake University Fin 288 Option price Using the option price equation from before the price of the option assuming an initial decrease to time t-3 can be found. f = e -rT [pf u +(1-p)f d ] f = e -0.12(.25) [(.6523)(0)+(.3477)(0)] =0

45. Drake Drake University Fin 288 Combining the possible outcomes $20 Time 0 Time 1 (3 months) l l l l l l Time 2 (6 months) Stock Price =$22 Option Price =$2.0257 Stock Price = $18 Option Price =$0 Stock Price = $24.2 Option Price = $3.2 Stock Price=$19.8 Option Price =$0 Stock Price = $16.2 Option Price =$0

46. Drake Drake University Fin 288 Option price at time 0 Using the option price equation from before the price of the option at time 0 can be found using the option prices at time t= 3 months f = e -rT [pf u +(1-p)f d ] f = e -0.12(.25) [(.6523)(2.0257)+(.3477)(0)] =1.2823

47. Drake Drake University Fin 288 Note: In our example the time period was fixed for each step of the process and the proportional up and down movement were the same for each step. Therefore the risk neutral probability p is the same at each step.

48. Drake Drake University Fin 288 Generalizing the possible outcomes f Time 0 Time 1 (3 months) l l l l l l Time 2 (6 months) Stock Price =S 0 u Option Price =f u Stock Price = S 0 d Option Price =f d Stock Price = S 0 u 2 Option Price = f uu Stock Price=S 0 ud Option Price =f ud Stock Price = S 0 d 2 Option Price =f dd

49. Drake Drake University Fin 288 Substituting for generalization of f The original pricing equation gave us f = e -r  T [pf u +(1-p)f d ] When applied to time t=3 it becomes f u = e -r  T [pf uu +(1-p)f ud ] and f d = e -r  T [pf ud +(1-p)f dd ] f = e -r  T [pf u +(1-p)f d ] f = e -2r  T [p 2 f uu +2p(1-p)f ud +(1-p) 2 f dd ] Substituting

50. Drake Drake University Fin 288 Put Options The same procedure can be used to calculate the value of a put option. Using the same problem as before but to find the price of a put option we would need to look at the tree to find the value of the option at time t = 6 months then work back.

51. Drake Drake University Fin 288 Valuing a European Put $20 1.059 Time 0 Time 1 (3 months) l l l l l l Time 2 (6 months) Stock Price =$22 Option Price = $0.4049 Stock Price = $18 Option Price = $2.3793 Stock Price = $24.2 Option Price = $0 Stock Price=$19.8 Option Price =$1.2 Stock Price = $16.2 Option Price =$4.8

52. Drake Drake University Fin 288 Generalization of f in the put option The original pricing equation gave us f = e -2r  T [p 2 f uu +2p(1-p)f ud +(1-p) 2 f dd ] f = e -2(.12)(.25) [(.6523) 2 0+2(.6523)(.3477)1.2+(.3477) 2 4.8] =1.059136

53. Drake Drake University Fin 288 American Put Options So far we have assumed that the option can only be exercised at the end of the six months. For a put option sometimes it is favorable to exercise the option early. If the option is exercised early, the payoff will equal the exercise price minus the stock price.

54. Drake Drake University Fin 288 Combining the possible outcomes $20 1.268 Time 0 Time 1 (3 months) l l l l l l Time 2 (6 months) Stock Price =$22 Option Price = $0.4049 Stock Price = $18 Option Price = $2.3793 $3 Stock Price = $24.2 Option Price = $0 Stock Price=$19.8 Option Price =$1.2 Stock Price = $16.2 Option Price =$4.8

55. Drake Drake University Fin 288 Option Pricing Equation Now the generalization for two periods does not hold. f u and f d will need to be found independently f u = e -r  T [pf uu +(1-p)f ud ] and f d = e -r  T [pf ud +(1-p)f dd ] f u = e -.12(.25) [.6523(0)+(.3477)1.2]=.4049 f d = 3 f = e -.12(.25) [(.6523)0.4049+(.3477)3]=1.268

56. Drake Drake University Fin 288 Delta In the Black Scholes introduction we defined delta  as a measure of the sensitivity of the option value to a change in the price of the underlying asset. Earlier today, we defined delta to be the change in the option price divided by the change in the stock price or

57. Drake Drake University Fin 288 Delta In either case delta is essentially a hedge ratio. Today we found delta by solving for the number of shares of stock needed to produce a risk free return in our portfolio of a short call option and a long stock. In other words, Delta is the number of shares we should hold to produce a riskless hedge (known as delta hedging). This is consistent with the Black Scholes definition.

58. Drake Drake University Fin 288 Delta The delta of a call will be positive, the delta of a put will be negative and the delta will change over time.

59. Drake Drake University Fin 288 Delta in our earlier 2 period call example $20 Delta =2.0257/(22-18) =0.5964 Time 0 Time 1 (3 months) l l l l l l Time 2 (6 months) Stock Price =$22 Option Price =$2.0257 Delta =3.2/(24.2-19.8) =.7273 Stock Price = $18 Option Price =$0 Delta = 0 Stock Price = $24.2 Option Price = $3.2 Stock Price=$19.8 Option Price =$0 Stock Price = $16.2 Option Price =$0

60. Drake Drake University Fin 288 Practice Problem Assume we have a European Put option with a strike price of $52 on a stock currently selling for $50. Assume that there are 2 years prior to expiration and each year the stock price either moves up by 20% or down by 20% and the risk free rate of interest is 5% Find the value of the option at time 0 Find the delta at each node Reprice the option assuming it is an American Option

61. Drake Drake University Fin 288 Increasing the number of steps So far we have presented extremely simple examples of the binominal tree. The example can be extended to shorter time frames and an increased number of steps. Looking at the two step model there were 4 possible paths the stock price could take.

62. Drake Drake University Fin 288 Generalizing the possible outcomes $20 Time 0 Time 1 (3 months) l l l l l l Time 2 (6 months) Stock Price =S 0 u Option Price =f u Stock Price = S 0 d Option Price =f d Stock Price = S 0 u 2 Option Price = f uu Stock Price=S 0 ud Option Price =f ud Stock Price = S 0 d 2 Option Price =f dd

63. Drake Drake University Fin 288 A three step model In a thee step model there are 2 3 =8 possible paths

64. Drake Drake University Fin 288 Increasing steps With 30 time steps there would be 31 terminal stock prices resulting in 2 30 or about 1 billion possible price paths. However the equations defining the payouts do not change as the number of steps increases. The key is the ability to define, u, d, p and the size of the step

65. Drake Drake University Fin 288 Determining u and d u and d should be determined by the stock price volatility. Let  t be one time step. Then:

66. Drake Drake University Fin 288 Increasing the number of intervals The number of intervals assumed to occur prior to the expiration of the option plays a key role in price obtained from the binomial tree model. As the number of periods increases the model will converge to a single price.

67. Drake Drake University Fin 288 An Example: one step Assume as before that we have a European call option on a share of stock and the current price of the share of stock is $20, the option expires in one year and has a strike price of $21. Let the annual volatility of the stock price be.40 and the risk free rate be 10%

68. Drake Drake University Fin 288 We need to calculate u, d and p

69. Drake Drake University Fin 288 Su and Sd The values of the stock at the end of the 6 month period would be S u =S 0 (1.3269) = 26.5379 S d =S 0 (.7536) = 15.0728

70. Drake Drake University Fin 288 Possible Outcomes Shown on Binomial Tree Model Stock Price =$20 Stock Price = $ 26.5379 Option Price = 5.5379 Stock Price = $15.0728 Option Price = $ 0 Time 0 Time 1 (6 months) l l l

71. Drake Drake University Fin 288 It is then easy to solve for the call option price. f = e -rT [pf u +(1-p)f d ] f = e -.1(.5) [.5192(5.5379)+(1-.816)0] =2.735

72. Drake Drake University Fin 288 Using two steps The same problem could be done assuming a two step tree over the 6 month period. Assume as before that we have a European call option on a share of stock and the current price of the share of stock is $20, the option expires in one year and has a strike price of $21. Let the annual volatility of the stock price be.40 and he risk free rate be 10%

73. Drake Drake University Fin 288 The two step model Again we need to solve for u, d, and p

74. Drake Drake University Fin 288 The Two Step Model $20 Time 0 Time 1 (3 months) l l l l l l Time 2 (6 months) Stock Price =24.4381 Option Price =f u Stock Price =16.3746 Option Price =f d Stock Price = 29.8365 Option Price = 8.8365 Stock Price=20 Option Price =0 Stock Price = 13.4064 Option Price =0

75. Drake Drake University Fin 288 We can use the generalized approach to solve for the value of the option at time 0 since there will not be any early exercise f = e -2r  T [p 2 f uu +2p(1-p)f ud +(1-p) 2 f dd ] f = e -2(.1).25 [.513 2 (8.8365)+2(.513)(1-.513)0+(1-.513) 20 ] =2.2124

76. Drake Drake University Fin 288 The three period model You could extend the model again to the three period model. Now the change in each step is equal to 2 months or.2563