Solution to Final Exam 423-F17 1. In the following question, the market model is NOT necessarily the Black-Scholes model. (a) A European call option with strike price K = $20 and exercise date in six months is traded at C E (0) = $2. The current price of the underlying stock, which pays no dividends, is S(0) = $25 and the continuously compounded annual interest rate is r = 3%. Find an arbitrage opportunity. Solution: Notice that S(0)−Ke−r/2 = 25−20e−0.015 = 5.30 > 2 = C E (0). An arbitrage opportunity is as follows: At time t = 0: 1) short-sell 1 share of the stock (collect S(0)); 2) deposit Ke−r/2 in the bank; 3) buy the European call. By doing so, we make a profit of S(0) − Ke−r/2 − C E (0) > 0. In six months we can exit the short position without investing any money, so we have an arbitrage. At time t = 1/2: 1) withdraw K dollars from the bank; 2) exercise the European call and buy the stock at price K; 3) return the stock to the original owner, exiting the short position. (b) Prove that an American call option on a stock which pays no dividends should never be exercised before maturity. Solution: Since the stock pays no dividends, we have that C A (t) = C E (t) for every t ∈ [0, T ]. If we exercise the option at time t the payoff is (S(t) − K)+ . It is clear then that the option should not be exercised when S(t) ≤ K. When S(t) > K, we can do better by trading the option in the market, since C E (t) ≥ S(t) − Ke−r(T −t) ≥ S(t) − K. 2. In this problem, we consider a binomial tree model and r denotes the discrete time interest rate per period. (a) Prove that in the N -step binomial model, for every t ∈ {0, . . . , N − 1} the stock price satisfies " # ∗ S(t + 1) S(t) = E S(t) . 1+r Solution: Since p∗ = " E∗ r−d u−d , we have # S(t + 1) (1 + u)S(t) (1 + d)S(t) S(t) ∗ S(t) = p∗ +(1−p∗ ) = [p (1 + u) + (1 − p∗ )(1 + d)] = S(t). 1+r 1+r 1+r 1+r Consider a two-step Binomial model. You are given the following incomplete tree, corresponding to a European put option with strike price K = 65. 0 2.6545 ? 14.6 17.09 35.6 Figure 1: European put with K = 65. (b) Compute the discrete time interest rate per period r and the risk-neutral probability p∗ . Hint: Write an appropriate system of equations and solve it. Solution: The pricing formulas in the binomial tree give us 2.6545 = 14.6(1 − p∗ ) 14.6p∗ + 35.6(1 − p∗ ) , 17.09 = , 1+r 1+r whose solution is p∗ = 0.8 and r = 0.1. (c) Find the price of the put option at t = 0 and draw the complete binomial tree for the stock price. Solution: The price of the European put at time 0 is given by P E (0) = 2.6545p∗ + 17.09(1 − p∗ ) = 5.04. 1+r The terminal values of the stock when there is at least one jump down can be deduced from the payoff of the European put. We get that S ud = 50.4 and S dd = 29.4. Then, we can compute Sd = 50.4p∗ + 29.4(1 − p∗ ) = 42. 1+r Now, since S ud = (1 + u)S d and S dd = (1 + d)S d , we have u = 50.4 42 − 1 = 0.2 and d = We can then complete the binomial tree for the stock price, which is the following 86.4 72 60 50.4 42 29.4 Figure 2: Stock price. 29.4 42 − 1 = −0.3. 3. A chooser option is a contract that allows its holder to decide at a later (fixed) time t1 whether the option will be a call or a put. The value V (t1 ) of the chooser option at time t1 is the maximum between the values of the call and put options, that is V (t1 ) = max{C E (t1 ), P E (t1 )}. In this problem we assume that the choices for the put and call have the same maturity T and the same strike price K. (a) Prove that V (t1 ) = C E (t1 ) + max{0, Ke−r(T −t1 ) − S(t1 )}. Solution: From the put-call parity we have V (t1 ) = max{C E (t1 ), C E (t1 ) + Ke−r(T −t1 ) − S(t1 )} = C E (t1 ) + max{0, Ke−r(T −t1 ) − S(t1 )}. (b) Prove that the price of chooser option at time 0 must be given by V (0) = C E (0) + PbE (0), where C E denotes a European call option with maturity T and strike price K and PbE denotes a European put option with maturity t1 and strike price Ke−r(T −t1 ) . Solution: The term max{0, Ke−r(T −t1 ) − S(t1 )} = (Ke−r(T −t1 ) − S(t1 ))+ corresponds to the payoff of a European put option with maturity t1 and strike price Ke−r(T −t1 ) . Thus, we can replicate the chooser option buying one share of this put and the European call option with maturity T and strike price K. By the non-arbitrage principle, V (0) must be equal to C E (0) + PbE (0). (c) Consider a Black-Scholes model with r = 0.1, σ = 0.3, S(0) = 80, and no dividends. Compute the price of the chooser option, for which the holder can choose at time t1 = 0.25 years whether to hold the call or put option with maturity T = 0.5 years and strike price K = 85. Solution: First, we find the price of the European call with maturity T and strike price K. We have √ S(0) 1 1 d1 = √ ln + (r + σ 2 )T = 0.06, d2 = d1 − σ T = −0.16, K 2 σ T and therefore C E (0) = S(0)Φ(d1 ) − Ke−r T Φ(d2 ) = 6.62. For the European put with maturity t1 and strike price Ke−r(T −t1 ) we have √ 1 2 1 S(0) b d1 = √ ln + (r + σ )t1 = 0, db2 = db1 − σ t1 = −0.15, −r(T −t ) 1 2 σ t1 Ke and PbE (0) = Ke−r(T −t1 ) Φ(−db2 )e−rt1 − S(0)Φ(−db1 ) = 5.25. We have finally that V (0) = 6.62 + 5.25 = 11.87. 4. Consider a stock in the Black-Scholes model with volatility σ = 20%, initial price S(0) = $100 and suppose that the continuously compounded interest rate is r = 10%. You sell 300 six-months European call options on the stock with strike price K = $102. (a) Compute the delta and the price of the call option. Solution: We can compute d1 (0) = 0.28 and d2 (0) = 0.14. Then we have from the normal table ∆C E = Φ(d1 (0)) = 0.6103, C E (0) = S(0)Φ(d1 (0)) − Ke−rT Φ(d2 (0)) = 7.1120. (b) Find the zero-valued delta-neutral portfolio. Solution: In this case the delta-neutral portfolio contains z = −300 calls. Since 0 = ∆V = x + z∆C E , we have x = −z∆C E = 300 × (0.6103) = 183.09 shares of the stock. Finally, since 0 = V (0) = xS(0)+y+zC E (0), the risk-free position is given by y = −xS(0)−zC E (0) = −16174.23. (c) The price of the stock after 5 days is $98. Compare the values of the zero-valued delta-neutral and the zero-valued unhedged portfolios. Assume you don’t make changes to both portfolios in these 5 days. Solution: The unhedged portfolio U contains x = 0 shares of the stock, z = −300 calls and a risk-free position of y = 300 × 7.1120 = 2133.60. Let t = 5/356. We can compute d1 5 365 = 0.13, d2 5 365 = −0.01, and from the Black-Scholes formula we get CE (t) = 5.8561. We compute the values of the delta-neutral and the unhedged portfolios 0.1×5 5 5 5 E 365 = −16174.23e + 183.09S − 300C = −11.56, V 365 365 365 0.1×5 5 5 E U = 2133.60e 365 − 300C = 379.69. 365 365 (d) Find the zero-valued delta-vega neutral portfolio by using an additional European put with expiration in six months and strike price $105. Show that the portfolio is also delta-gamma neutral. Solution: For this European call option, we have calculated in part (a) that ∆C E = 0.6103, C E (0) = 7.1120. The gamma and vega of the call option are ΓC E = 0.0271 and VC E = 27.1251. b = 105 is different and T = 0.5. We can compute then For the European put option PbE , note that K b b d1 (0) = 0.08 and d2 (0) = −0.06 which leads to PbE (0) = 5.5170, ∆PbE = −0.4681, ΓPbE = 0.0281 and VPbE = 28.1194. The delta-vega neutral portfolio contains z1 = −300 calls. V Since 0 = VV = z1 VC E + z2 VPbE , it contains z2 = −z1 VCbE = 289.39 put options. PE Since 0 = ∆V = x + z1 ∆C E + z2 ∆PbE , it contains x = −z1 ∆C E − z2 ∆PbE = 318.55 shares of the stocks. Since 0 = V (0) = xS(0) + y + z1 C E (0) + z2 PbE (0), the risk-free position is given by y = −xS(0) − z1 C E (0) − z2 PbE (0) = −31317.79. Finally, we compute gamma of the portfolio ΓV = z1 ΓC E + z2 ΓPbE = −300 × 0.0271 + 289.39 × 0.0281 = 0.