Binomial Option Pricing Model (BOPM)

1. Binomial Option Pricing Model (BOPM) References: Neftci, Chapter 11.6 Cuthbertson & Nitzsche, Chapter 8

2. Linear State Pricing • A 3-month call option on the stock has a strike price of 21. • Can we price this option? • Can we find a complete set of traded securities to price the option payoffs? • If we make the simplifying assumption that there are only 2 states of the world (up and down), then we only need the prices of two independently distributed traded assets, e.g. the underlying stock and the risk-free asset

3. Linear State Pricing • Algebraically, • If S = 2, this is a system of equations in two unknowns • To get a unique solution for it, we need at least 2 independent equations

4. The Binomial Model • A stock price is currently S0 = $20 • In three months it will be either S0u = $22 or S0d = $18 Stock Price = $22 Stock price = $20 Stock Price = $18

5. A One-Period Call Option • Option tree: Stock Price = $22 Option Price = $1 Stock price = $20 Option Price=? Stock Price = $18 Option Price = $0

6. Risk-Neutral Valuation Reminder Under the RN measure the stock price earns the risk-free rate That is, the expected stock price at time T is S0erT When we are valuing an option in terms of the underlying the risk premium on the underlying is irrelevant See handout on RNV 6

7. S0u ƒu S0 ƒ S0d ƒd Risk-Neutral Tree • f = [ qfu + (1 – q ) fd ]e-rT • The variables q and (1– q ) are the risk-neutral probabilities of up and down movements • The value of a derivative is its expected payoff in a risk-neutral world discounted at the risk-free rate q (1– q )

8. S0u = 22 ƒu = 1 q S0 ƒ S0d = 18 ƒd = 0 (1– q ) Risk-Neutral Probabilities • Since q is a risk-neutral probability, 22q + 18(1 – q) = 20e0.12 (0.25) q = 0.6523

9. RN Binomial Probabilities Formula • RN probability of up move: • RN probability of down move: 1 – q • With this probabilities, the underling grows at the risk-free rate (check it out)

10. Using the RN Binomial Probabilities Formula • In above example, u = 22/20 = 1.1 and d = 18/20 = 0.9 • So, assuming r = 12% p.a.,

11. S0u = 22 ƒu = 1 0.6523 S0 ƒ S0d = 18 ƒd = 0 0.3477 Valuing the Option The value of the option is e–0.12(0.25) [0.6523  1 + 0.3477  0] = 0.633

12. 24.2 22 19.8 20 18 16.2 A Two-Step Example • Each time step is 3 months

13. Valuing a Call Option 24.2 3.2 D • Value at node B = e–0.12´0.25(0.6523´3.2 + 0.3477´0) = 2.0257 • Value at node A = e–0.12´0.25(0.6523´2.0257 + 0.3477´0) = 1.2823 22 B 19.8 0.0 20 1.2823 2.0257 A E 18 C 0.0 16.2 0.0 F

14. 72 0 D 60 B 48 4 50 4.1923 1.4147 A E 40 C 9.4636 32 20 F A Put Option Example; K=52

15. What Happens When an Option is American (see spreadsheet) 72 0 D 60 B 48 4 50 5.0894 1.4147 A E 40 C 12.0 32 20 F 15

16. And if we did not have u and d? One way of matching the volatility of log-returns is to set where s is the volatility andΔt is the length of the time step. This is the approach used by Cox, Ross, and Rubinstein • Handout on Asset Price Dynamics

17. The Probability of an Up Move 17

18. EXOTICS PRICING (examples) a) Average price ASIAN CALL payoff = max {0, Sav– K} Remember: cheaper than an ‘ordinary’ option b) Barrier Options (e.g. up and out put) pension fund holds stocks and is worried about fall in price but does not think price will rise by a very large amount • Ordinary put? - expensive • Up and out put - cheaper

19. Pricing an Asian Option (BOPM) • Average price ASIAN CALL(T = 3) • Calculate stock price at each node of tree • Calculate the average stock price Sav,i at expiry, for each of the 8 possible paths (i = 1, 2, …, 8). • Calculate the option payoff for each path, that is max[Sav,i – K, 0] (for i = 1, 2, …, 8) The risk neutral probability for a particular path is qi* = qk(1 –q)n-k q = risk neutral probability of an ‘up’ move k = number of ‘up’ moves (n – k) = the number of ‘down’ moves

20. Pricing an Asian Option (BOPM), cont’d • Weight each of the 8 outcomes for the call payoff max[Sav,i – K, 0] by the qi* to give the expected payoff: • The call premium is then the PV of ES*, discounted at the risk free rate, hence:

21. Pricing Barrier Options (BOPM) Down-and-out call S0 = 100. Choose K = 100 and H = 90 (barrier) Construct lattice for S Payoff at T is max {0, ST– K } Follow every ‘path’ (ie DUU is different from UUD) If on say path DUU we have any value of S < 90 , then the value at T is set to ZERO (even if ST–K > 0). Use BOPM risk neutral probabilities for each path and each payoff at T

22. Example: Down-and-out call S0 =100, K= 100, q = 0.857, (1 –q) = 0.143 H = 90 UUU ={115, 132.25, 152.09} Payoff = 52.09 (q* = 0.8573, 0.629) DUU ={80, 92,105.8} Payoff = 0 NOT 5.08 (q* = 0.105) C = e-rT ‘Sum of [q* payoffs at T]’ where qi* = qk(1 – q)n-k,