Binomial TreesA useful and very popular technique for pricing a stock option involves constructing a binomial tree .1. A one-step binomial modelConsider a European call with $21,1/4K T == and 0$20S =. To simplify analysis, we assume that at the end of three months, we have $22$18T S or =. Hence, the option value is either $1 or $0. The option specified in such a way does not involveuncertainty about the value of the portfolio, hence it’s return must equal the risk-free interest rate.Consider a portfolio consisting of a long position in ∆ shares of the stock and a short position in one call option. Ifa.Stock price goes up from $20 to $22, the share are worth 22∆ and value of the option is $1, so the portfolio value is 22∆ - 1;b.Stock price goes down from $20 to $18, the portfolio is worth 18∆, therefore∆−=∆22118hence 0.25∆=. A riskless portfolio therefore consists of long 0.25 shares and short 1 option. In both cases, the value of the portfolio is $4.5. If the risk-free interest rate is 12%/pa, the present value of the portfolio is0.121/4e−×=4.5 4.367Let f denotes the option price, so the cost of the portfolio today is×−=−200.255f ftherefore f = 0.633. This shows that in the absence of arbitrage, the current value of the option must be $0.633.A GeneralisationNow we assume that at the end of T, S T can move up to S0u, u>1, which give payoff f u, or move down to S0d, d<1, which gives payoff f dS0 fS0u f u S0d f dAt the end of the life of the option, the value of the portfolio is either S 0u ∆ - f u or S 0d ∆ - f d , but they must the sameS 0u ∆ - f u = S 0d ∆ - f d or00u df f S u S d −∆=− (1)In this case the portfolio is riskless and must be equal the risk-free interest rate. The present value of the portfolio is(S 0u ∆ - f u )e -rTand the cost of setting up the portfolio is S 0∆ - f , it then follows:S 0∆ − f = (S 0u ∆ − f u )e -rTorf = S 0∆ − (S 0u ∆ − f u )e -rTSubstituting for ∆ from equation (1), we havef = e -rT [p f u + (1 − p ) f d ) (2) whererT e dp u d −=−For instance, in the previous example, u = 1.1, d = 0.9, r = 0.12, T = 0.25, f u = 1, and f d = 0, we can calculate:0.121/40.90.65231.10.9e p ×−==−therefore:()0.120.250.652310.347700.633f e −×=×+×=which is the same as before.It is important to note that in this discussion we do not involve the probabilities of the stock price moving up or down.2Risk-neutral ValuationThough equation (2) is derived without assumption about probabilities of up and down movements, it is natural to interpret p and (1 −p) as the probabilities of an up and down movements in stock price. Thereforep f u+ (1 −p) f dis the expected payoff from the option. With this interpretation of p, equation (2) then states that the value of the option today is its expected future value discounted at the risk-free rate (see the derivation of Black-Scholes solution).On the other hand, with p being the probability of moving up, the expected stock price at time T, E(S T), is given byE(S T) = p S0u + (1 −p) S0dSubstituting from equation (2) for p, this reduces toE(S T) = S0e rT (3) showing that the stock price grows, on average, at the risk-free rate. Setting the probability of the up movement equal to p is equivalent to assuming that the return on the stock equals the risk-free rate. This result is based on risk-neutral principle, which states that expected return on all securities is the risk-free interest rate.Returning to the previous example, let p be the probability of moving up, then p must satisfy22p + 18(1 −p) = 20e0.12×0.25or 4p = 20e0.12×0.25− 18, and hence, p = 0.6523.At the end of the three months, the call option has a 0.6523 probability of being worth 1 and 0.3477 probability of being worth 0, so its expected value is0.6523×1 + 0.3477×0 = 0.6523Discounting at the risk-free rate, the present value of the option is$0.6523e−0.12×0.25 = $0.633.Hence, no-arbitrage and risk-neutral give the same answer.3 Two-step Binomial TreesThe upper number is S; the lower number is f . The start priceis $20 and in each of two time steps may go up or down byDEF24.2 3.219.8 0.0 16.2 0.010%. Each time step is three months long, and the risk-free interest rate is 12%/pa, and K = $21. Clearly, at node D the option price is $24.2 − $21 = $3.2, but at nodes E and F the option is out of the money and its value is zero.At node C, the option price is 0, using the notation introduced before, u = 1.1, d = 0.9, r = 0.12, and T = 0.25, so that p = 0.6523, and from equation (2) at B:e−0.12×0.25(0.6523×3.2 + 0.3477×0) = 2.0257We now calculate the option price at the initial node A. Similarly, we havee−0.12×0.25(0.6523×2.0257 + 0.3477×0) = 1.2823A GeneralisationWe now assume that S0 is the initial price and the time step is δt years. The following notations are self-explained:f u = e-rδt[p f uu+ (1 −p) f ud) (4)f d = e-rδt[p f ud+ (1 −p) f dd] (5)f = e-rδt[p f u+ (1 −p) f d] (6) Substituting (4) and (5) into (6), we havef = e-2rδt[p2f uu+ 2p(1 −p) f ud+ (1 −p) f ud] (7) We explain this in the following diagram:Stock and option prices in a general two-step treeS 0u 2f uuS 0udf udS 0d 2f ddS fFollowing the derivation of equation (3), we can obtain similar result based on the principle of risk-neutral valuation. The values p2, 2p(1 − p), and (1 −p)2 are the probabilities that the upper, middle, and lower final nodes will be reached. The option price is equal to its expected payoff in a risk-neutral world discounted at the risk-free interest rate given in equation (7).4 A Put Option ExampleThe procedures described above are applicable to any deriva-tive dependent on a stock whose price changes are binomial. Consider a two-year European put option on a stock with spotprice S = $50 and K = $52 and r = 5%. We suppose there are two time steps and in each case the stock price either moves up or down by a proportional amount of 20%:72 0 48 432 20So, u = 1.2, d = 0.8, the risk-neutral probability p is given by0.0510.81.20.8rTe d ep u d×−−==−−The possible final stock prices are: $72, $48 and $32, and in this case f uu = 0; f ud = 4, and f dd = 20, from equation (7):f = e−2×0.05×1(0.62822×0 + 2×0.6282×0.3718×4 + 0.37182×20) = 4.1923This result can be equally obtained using equation (6) and working back through the tree one step at a time.5 American OptionsWe now consider the above put option again with the same condi- tions. But it’s now an American. We evaluate it using a binomial72 048 432 20tree. The procedure is to work back through the tree from the end to the beginning, testing at node if early exercise being optimal. The stock prices and their probabilities are unchanged. The values for the option at the final nodes are also unchanged. At B, early exercise leads to negative payoff, so from equation (2) the value of the option is 1.4147. At C, equation (2) gives the value of the option as 9.4636. But early exercise at K = $52 leads to a payoff of $12, which is the optimal value of the option. Finally, at A, equation (2) givese−0.05×1(0.6282×1.4147 + 0.3718×12) = 5.0894But if exercise at A the payoff is $2, which is clearly not optimal. So the value of the option is $5.0894.6 DeltaIn deriving the Black-Scholes equation, we knowV S ∂∆=∂It is an important parameter in the pricing and hedging of options. In general we define the delta of a stock option is the ratio of the change in the price of the stock option to the change in the price of the underlying stock. In the following case:The delta of the above call option is:100.252218−=−Stock price = $22 Option price = $1 Stock price = $18 Option price = $0In the example of European put option, the delta at the end of the first time step is1.41479.46360.40246040−=−− and is either040.16677248−=−− or4201.00004832−=−−at the end of the second time step. Clearly, delta changes overthe two time steps. Thus, in order to maintain a riskless hedge using an option and the underlying stock, we need to adjust our holdings in the stock periodically.7Matching Volatility with u and dIn above discussion, values of u and d are given. But in practice, they are chosen to match the volatility of the stock price. We suppose that the expected return (in the real world) on a stock is µand it volatility is σ. The step length is δt, and the stock price goes up by a proportional amount u or goes down by d. The probability of an up movement (in the real world) is assumed to be q.Stock price change in (a) the real world, and (b) the risk-neutral world The expected stock price at the end of the first time step is S 0e µδt. But on the binomial tree the expected stock price at thisS 0u S 0dtime is:qS 0u + (1 − q ) S 0dEquating the two stock prices:qS 0u + (1 − q ) S 0d = S 0e µδt ort e dq u d µδ−=− (8)As stock return follows an Itô process, its variance over the period of δt is σ2δt. But on the tree in (a), the variance of the return isqu 2 + (1 − q )d 2 − [qu + (1 − q )d ]2Equating the two variancesqu 2 + (1 − q )d 2 − [qu + (1 − q )d ]2 = σ2δt (9) Substituting (8) into (9), we havee µδt (u + d ) − ud − e 2µδt = σ2δt Ignoring high order terms of δt , we can solve the above equation by,u ed e σσ−== (10)In our previous analysis we show that we can replacethe tree in (a) by that in (b) that represents the of risk-neutral. The probability p can be expressed asr t e dp u d δ−=−The expected stock price is S 0e r δt . The variance of the stockprice return ispu 2 + (1 − p )d 2 − [pu + (1 − p )d ]2 = e µδt (u + d ) − ud − e 2µδt Substituting for u and d from (10), we find this equals σ2δt when higher order terms of δt are ignored.This analysis shows that when we move from the real world to the risk-neural world the expected return on the stock changes, but its volatility remains the same (at least in the limit as δt tends to zero). This is an illustration of an important general result known as Girsanov’s theorem.The binomial models presented so far have been unrealistically simple. In real practice, the life of the option is typically divided into 30 or time steps of length δt. In each time step there is a binomial stock price movement. This means that 31 terminal stock prices and 230, or about 1 billion, possible stock prices paths are considered.。
