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

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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

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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

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The Binomial Model A stock price is currently S 0 = $20 In three months it will be either S 0 u = $22 or S 0 d = $18 Stock Price = $22 Stock Price = $18 Stock price = $20

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

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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 S 0 e rT When we are valuing an option in terms of the underlying the risk premium on the underlying is irrelevant See handout on RNV

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Risk-Neutral Tree f = [ q f u + (1 – q ) f d ]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 S 0 u ƒ u S 0 d ƒ d S0ƒS0ƒ q (1 – q )

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Risk-Neutral Probabilities S 0 u = 22 ƒ u = 1 S 0 d = 18 ƒ d = 0 S 0 ƒ q (1 – q ) Since q is a risk-neutral probability, 22q + 18(1 – q) = 20e 0.12 (0.25) q =

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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)

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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.,

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Valuing the Option S 0 u = 22 ƒ u = 1 S 0 d = 18 ƒ d = 0 S0ƒS0ƒ The value of the option is e –0.12(0.25) [ ] = 0.633

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A Two-Step Example Each time step is 3 months

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Valuing a Call Option Value at node B = e – ( ) = Value at node A = e – ( ) = A B C D E F

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A Put Option Example; K= A B C D E F

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15 What Happens When an Option is American (see spreadsheet) A B C D E F

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

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17 The Probability of an Up Move

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EXOTICS PRICING (examples) a) Average price ASIAN CALL payoff = max {0, S av – 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

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Pricing an Asian Option (BOPM) Average price ASIAN CALL(T = 3) 1.Calculate stock price at each node of tree 2.Calculate the average stock price S av,i at expiry, for each of the 8 possible paths (i = 1, 2, …, 8). 3.Calculate the option payoff for each path, that is max[S av,i – K, 0] (for i = 1, 2, …, 8) The risk neutral probability for a particular path is q i * = q k (1 – q) n-k q = risk neutral probability of an up move k = number of up moves (n – k) = the number of down moves

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Pricing an Asian Option (BOPM), contd 4.Weight each of the 8 outcomes for the call payoff max[S av,i – K, 0] by the q i * to give the expected payoff: 5.The call premium is then the PV of ES*, discounted at the risk free rate, hence:

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Pricing Barrier Options (BOPM) Down-and-out call S 0 = 100. Choose K = 100 and H = 90 (barrier) Construct lattice for S Payoff at T is max {0, S T – K } Follow every path (ie DUU is different from UUD) If on say path DUU we have any value of S 0). Use BOPM risk neutral probabilities for each path and each payoff at T

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Example: Down-and-out call S 0 =100, K= 100, q = 0.857, (1 – q) = H = 90 UUU ={115, , } Payoff = (q * = , 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 q i * = q k (1 – q) n-k,

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