1 Parameterization of Binomial Models and Derivation of the Black-Scholes PDE.

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1 1 Parameterization of Binomial Models and Derivation of the Black-Scholes PDE. Previously we treated binomial models as a pure theoretical toy model for our complete economy. We turn to the issue of how to estimate parameters in the model and demonstrate the power of the binomial model as a pricing tool we will derive the famous Black-Scholes equation in this binomial setting. 1.1 The Market: Consider our stock: Time at t, its value is S 0 Time at t + δ, its pdf is p (S) Here, we consider a very short time interval δ, i.e., δ 1. therefore, at t = δ, the expected value of the stock price is Z E (S) = Sp(S) ds and its variance is Var(S) = E (S E (S)) 2 Z = (S E (S)) 2 p (S) ds Next, we prescribe a model for our stock price movement: 1. Assumption 1: i.e., the rate of the expected return is µ. µ S E = e µδ S 0 1

2 2. Assumption 2: where σ is the volatility. µ S Var = σ 2 δ S 0 N.B. Assumption 2 is a consequence of the following theorem, roughly stated, that if the total volatility over time is bounded below away from zero and bounded above, and if there is always some (finite) volatility over any time interval, then (see Neftci s book, pp ). Consider the binomial model Var(S) δ where p is a subjective probability. 1.2 Parameterization: We want to match the mean and variance of this binomial model with those of our market, i.e., mean: ps 0 u +(1 p) S 0 d = S 0 e µδ var: N.B. Var(X) =E (X 2 ) [E (X)] 2. Therefore, ps 2 0u 2 +(1 p) S 2 0d 2 [ps 0 u +(1 p) S 0 d] 2 = S 2 0σ 2 δ pu +(1 p) d = e µδ (1) pu 2 +(1 p) d 2 [pu +(1 p) d] 2 = σ 2 δ (2) thus, we have 2 equations, 3 unknowns: p, u, d. There is a freedom of imposing constraints in our parameterization. Eq. (1) gives p = eµd d u d Substitute this p into Eq. (2) yields e µδ (u + d) ud e 2µδ = σ 2 δ (3) Next we are going to impose some specific constraints to obtain expressions for u, d, p. 2

3 1.2.1 The First Choice: Impose To the accuracy of O (δ), we have ud =1. u = e +σ δ d = e σ δ and the corresponding risk-neutral probability is which is independent of µ. q = erδ d u d e rδ e σ δ = e +σ δ e σ δ The no-arbitrage principle dictates the following risk-neutral valuation of a contingent claim with payoff f (S T ): The present value f 0 = e rδ [qf (S 0 u, t + δ)+(1 q) f (S 0 d, t + δ)] in which the rate of the expected market return µ does not appear. ADeepLookatThisPrice: Since f (S 0 u, t + δ) = f (S 0 + S 0 u S, t + δ) = f (S 0 + S 0 (u 1),t+ δ) ³ = f ³S 0 + S 0 e +σ δ 1,t+ δ, = ³ f 0 = e nqf ³S rδ 0 + S 0 e +σ δ 1 +(1 q) f ³ o ³S 0 + S 0 e σ δ 1 in which, for simplicity of notation, we have dropped t + δ. Taylor-expansion to O (δ) : f 0 = e ½q rδ f (S 0 )+ f ³ S 0 e +σ δ f ³ 2 S 2 S 0 2 S0 2 0 e +σ δ 1 +(1 q) f (S 0 )+ f ³ S 0 e σ δ f ³ 2 ¾ S0 2 e σ δ 1 S 0 2 S 2 0 in which all the derivatives are evaluated at time t + δ. Note that 3

4 1. Since e rδ [qs 0 u +(1 q) S 0 d]=s 0, we have qe +σ δ +(1 q) e σ δ = e rδ 2. ³ 2 e +σ δ 1 ³ 2 e σ δ 1 σ 2 δ σ 2 δ Therefore, accurate to O (δ), we obtain f 0 = e rδ f (S 0,t+ δ)+ f S 0 e rδ f S S 0 S 0 2 σ2 δs0 2 2 f S 2 0 Using e rδ 1 rδ + O δ 2 e +rδ 1+rδ + O δ 2 leads to f f 0 = f (S 0,t+ δ)+δ rf (S 0,t+ δ)+rs S 0 2 σ2 S0 2 2 f (4) S0 2 accurate upto O (δ). This formula gives the relation between the values of the contingent claim at time t and time t + δ Black-Scholes Equation: Finally, we note that Eq. (4) can be rewritten as f (S 0,t+ δ) f 0 δ f rf (S 0,t+ δ)+rs S 0 2 σ2 S0 2 2 f S 2 0 =0 As δ 0, we arrive at the famous Black-Scholes PDE: f t where the subscript 0 is dropped. rf + rs f S σ2 S 2 2 f S 2 =0 4

5 1.2.3 The Second Choice: p = 1 2 To the order of O (δ), thesolutionsare Now Eqs. (1) and (2) become u + d = 2e µδ u 2 + d 2 = 2σ 2 δ +2e 2µδ u = e ( µ 1 2 σ2 )δ+σ δ d = e ( µ 1 2 σ2 )δ σ δ For this binomial model, the associated risk-neutral probability is q = erδ d u d = e( 1 r µ+ 2 σ2 )δ e σ δ e +σ δ e σ δ It appears that all parameters, u, d, q depend on µ. Does the risk-neutral pricing also depend on µ? Let us examine this question. Consider a contingent claim with payoff f (S T ), according to the no-arbitrage valuation, its present value is f 0 = e rδ [qf (S 0 u, t + δ)+(1 q) f (S 0 d, t + δ)] ³ = e nqf ³S rδ 0 + S 0 e (µ 1 2 σ2 )δ+σ δ 1 +(1 q) f ³ ³S 0 + S 0 e (µ 1 2 σ2 )δ σ δ o 1. Again, for simplicity of notation, t + δ is dropped from the second line above. Taylor expansion yields, up to O (δ), f 0 = e ½q rδ f (S 0,t+ δ)+ f ³ S 0 e (µ 1 2 σ2 )δ+σ δ f ³ 2 S 0 2 S2 0 e (µ 1 S0 2 2 σ2 )δ+σ δ 1 2 ¾ +(1 q) f (S 0,t+ δ)+ f S 0 S 0 ³ e ( µ 1 2 σ2 )δ σ δ f 2 S2 0 in which all the derivatives are evaluated at time t + δ. Note that 1. Again because e rδ [qs 0 u +(1 q) S 0 d]=s 0, we have S 2 0 qe (µ 1 2 σ2 )δ+σ δ +(1 q) e (µ 1 2 σ2 )δ σ δ = e rδ ³ e ( µ 1 2 σ2 )δ σ δ 1 (5) 5

6 2. = ³ e (µ 1 2 σ2 )δ±σ δ µµ 12 σ2 δ ± σ δ ³ µδ ± σ δ = σ 2 δ + O 2 ³δ 3/ σ2 δ 1 Using the results in points 1-2 above, accurate to O (δ), we obtain f 0 = (1 rδ) f (S 0,t+ δ)+ f S 0 rδ + 1 S 0 2 σ2 δs0 2 2 f S0 2 f = f (S 0,t+ δ)+δ rf (S 0,t+ δ)+rs S 0 2 σ2 S0 2 2 f S0 2 Note that (6) 1. Due to the order shown in point 2 above, we only need to evaluate q upto O (1) to ensure theentireexpression(5) is valid up to the order O (δ). Therefore, q = e( 1 r µ+ 2 σ2 )δ e σ δ e +σ δ e σ δ δ 1 (µ r) σ = 1 2 = 1 ³ 1/2 2 + O δ + O (δ) 2. Eq. (6) isexactlythesamepriceasweobtainedinthecaseofthefirst choice, i.e., ud =1. In particular, we note that, although the parameters, u,d,q, now depend on µ, the rate of the expected market returned. The risk-neutral pricing is still independent of µ it disappears into higher order terms, which will have no bearings on our no-arbitrage price. 3. We can again obtain the same Black-Scholes PDE, which describe how the value of a contingent claim will evolve in time. 4. A real miracle: different parameterizations lead to the same Black-Scholes PDE under the no-arbitrage principle for a given market. 5. For different markets (described by different values of µ), as long as the market volatility σ is the same, we will get the same price for the option. The rate of the expected market return is irrelevant. This is a far stronger result than the case where, for a given set of u, d, no-arbitrage pricing is independent of p as in our theoretical binomial toy model of economy. 6

7 1.3 The Simplest Binomial Model Note that 1 q = e(r µ+ 2 σ2 )δ e σ δ e +σ δ e σ δ δ = (µ r) σ + O (δ). If we choose then µ = r q = O (δ) The O (δ)-term in q = 1 + O (δ) will not affect our risk-neutral pricing or the derivation of 2 the Black-Scholes PDE because its combined contribution with f (Su,t + δ) and f (Sd,t + δ) to the price is higher order than O (δ). Since p =1/2, we can simply choose p = q =1/2. For risk-neutral valuation, the simplest binomial model for a short duration δ with a price accuracy within O (δ) is 7

8 2 Multiperiod Binomial Trees 2.1 Setup: 1. 2 securities: (a) a risky asset (e.g., stock without dividend); (b) a risk-free asset (e.g., bond). 2. A series of times: at which trades take place. 0,δt,2δt,,Nδt= T 3. A binomial tree of possible states for stock prices and constant interest rate r (can be easily generalized to r k for time interval kδt. 1. Since the market permits no-arbitrage, we have s 2j <e rδt s j <s 2j+1 j This is an example of dynamically complete market. 2.2 Generalization: 8

9 Recombinant Tree: Note that: At time step n, 1. There are 2 n states for the non-recombinant tree; 2. There are (n +1)states for the recombinant tree. This fact gives rise to the numerical advantage of recombinant trees. Note that the parameterization with u, d gives a naturally recombined tree: Risk-neutral valuation for each node: with the replicating portfolio: Note that φ. f now = e rδt (qf up +(1 q) f down ) q = erδt S now S down S up S down Stock : shares, = f up f down S up S down Bond : f now S now 9

10 2.2.1 Evaluation is just "working backward through the tree": Example: For simplicity, r =0, and the price movement is assumed to be such that q = 1 2 for this tree. at every node What is the value of a European call option with K =100at T =3δt? Note that the value at maturity is AtthenodewithS =140: (S T 100) + =60, 20, 0, 0 f = qf up +(1 q) f down = = 40 2 Working backward, we conclude the value of the option is 15. Why this is the correct price? Because it can be replicated at every trading time. The replication (in one possible path) Consider that a bank sells the option: 1. At time t =0, = 25 5 =0.5 shares of stock stock : S = = 50 Bond : f now S =15 50 = 35 10

11 i.e.,the back sells the option for $15 borrows $35 uses these $15 + $35 = $50 to buy 1 2 shares of the stock. 2. Next step: say, the stock goes up to 120. the the new is = =0.75 Need additional 0.25 shares of stock to be purchased at the present value of stock, $ Ifthestock,say,goesupto$140, then = Need $ = $30 borrowed Debt = $35 + $30 = $65. = =1 Need another 0.25 shares at $140/share : = $35 Debt : $35 + $65 = $ If the stock,say,goes down to $120 (at maturity), then, The debt = $100, which matches the strike K Portfolio: $120 {z} 1 Share of Stock $100 = $20, which replicates the option claim. Thus, replicating the claim. That is, the bank sells the option at t =0;Attimet = T, it could deliver one share of stock, collect K =$100, pay off the loan. No gain, no loss! Conclusion: The portfolio is self-financing at each trading, i.e., the total value of the portfolio before and after each trade are the same. This can be again seen as follows: For simplicity, assume interest rate r =0. Notation: At tick-time i, 1. Start: A portfolio Π i : i+1 stock ψ i+1 = f i i+1 S i Π 0 : 1 S 0 + ψ 1 = f 0 11

12 2. One tick: Π 1 : worth ψ S 1 = ψ S 0 1 S S 1 = f (S 1 S 0 ) = f 0 + f 1 f 0 S 1 S 0 (S 1 S 0 ) = f 1 rebalance such that f 1 = 2 S 1 + ψ 2 3. Tick-time 2: Π 2 : worth 2 S 2 + ψ 2 = f (S 2 S 1 ) = f 1 + f 2 f 1 S 2 S 1 (S 2 S 1 ) = f 2 4. At t = T : Π T : T 1 S T + ψ T 1 = f T 1 + T 1 (S T S T 1 ) = f T which produces the claim. Thus, the replicating process is self-financing. 12

13 2.3 Valuation Formula: Case of General Trees: q 1 = erδt S 1 S 2 S 3 S 2 q 2 = erδt S 2 S 4 S 5 S 4 q 3 = erδt S 3 S 6 S 7 S 6 Therefore, f (3) = e rδt [q 3 f (7) + (1 q 3 ) f (6)] f (2) = e rδt [q 2 f (5) + (1 q 2 ) f (4)] = f (1) = e rδt [q 1 f (3) + (1 q 1 ) f (2)] = e 2rδt [q 1 q 3 f (7) + q 1 (1 q 3 ) f (6) (1 q 1 ) q 2 f (5) + (1 q 1 )(1 q 2 ) f (4)] i.e., Initial value of a claim X = e rnδt (the probability of the path associated with a final state i) (payoff of state i) final states 13

14 2.3.2 Case of Recombinant Trees q = erδt d u d f (1) = e "q 2rδt 2 f (7) + 2q (1 q) f (5) +(1 q) 2 f (4) recombined paths For the N-step: The present value of an option with payoff f (S T ): e rnδt N X k=0 µ N k q k (1 q) N k f S 0 u k d N k µ N where isthenumberofwaysofhavingk steps up and N k steps down in a total N k time-steps. e.g., A European call has the present value: e rnδt N X k=0 µ N k q k (1 q) N k S 0 u k d N k K + # 14