BG_FNCE4040_HW3Sol

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Derivative Securities, Fall 2023 FNCE 4040, Homework 3 Solutions Homework 3 Solutions 1. ( 20 points ) Consider a binomial world in which a stock can go up in value by 20% or down by 10% over the next year. The stock is currently trading at $ 100. The annual risk-free rate is 5%. Consider a call option that expires in one year and has a strike price of $ 110. (a) (*) What should be the current value of the call option? Solution. Since the stock price moves proportionally to its current value S 0 , u = 1 . 2 and d = 0 . 9. Therefore, the risk-neutral probability that the stock will go up in one year is p = (1 + r f ) − d u − d = 1 . 05 − 0 . 90 1 . 20 − 0 . 90 = 0 . 5 , and the current value of the call option is given by C 0 = p × max(120 − 110 , 0) + (1 − p ) × max(90 − 110 , 0) 1 + r f = 0 . 5 × 10 1 . 05 = 4 . 76 . (b) (**) If the call option was trading for $ 3.20, can you generate an arbitrage profit? Solution. Since the no-arbitrage price of the call option is $ 4.76, the call option traded at $ 3.20 is undervalued . So, we can construct an arbitrage strategy by buying the call option for $ 3.20 and selling its replicating portfolio for $ 4.76. We can find the composition of the replicating portfolio by combining ∆ shares of the stock and B risk-free bonds (each priced at $ 1 today) to match the future payoff of the call option on both the up and the down states: ∆ × 120 + B × 1 . 05 = 10 ∆ × 90 + B × 1 . 05 = 0 yielding a replicating portfolio of (∆ = 1 / 3 , B = − 28 . 57). Therefore, the cash flows of the arbitrage strategy are as follows: In one year Today d state u state Buy call option − 3 . 20 0 10 . 00 Sell 1 / 3 shares of stock 33 . 33 − 30 . 00 − 40 . 00 Buy 28.57 risk-free bonds (Lend $ 28.57) − 28 . 57 30 . 00 30 . 00 Net Cash Flows 1 . 56 0 . 00 0 . 00 (c) (**) ADDITIONAL QUESTION (not in Canvas): If the call option was trading for $ 6.10, can you generate an arbitrage profit? Solution. Since the no-arbitrage price of the call option is $ 4.76, the call option traded at $ 6.10 is overvalued . So, we can construct an arbitrage strategy by selling the call option for $ 6.10 and buying its replicating portfolio for $ 4.76. Given the composition of the replicating portfolio (∆ = 1 / 3 , B = − 28 . 57), the cash flows of the arbitrage strategy are as follows: © Buffa-Garc´ ıa, Leeds School of Business Page 1 of 7

Derivative Securities, Fall 2023 FNCE 4040, Homework 3 Solutions In one year Today d state u state Sell call option 6 . 10 0 − 10 . 00 Buy 1 / 3 shares of stock − 33 . 33 30 . 00 40 . 00 Sell 28.57 risk-free bonds (Borrow $ 28.57) 28 . 57 − 30 . 00 − 30 . 00 Net Cash Flows 1 . 34 0 . 00 0 . 00 2. ( 20 points ) Consider a binomial world in which a stock can go up in value by 35% or down by 15% over the next year. The stock is currently trading at $ 45. The annual risk-free rate is 5%. (a) (*) What is the risk-neutral probability that the stock will go up over the next year? Solution. Since the stock price moves proportionally to its current value S 0 , u = 1 . 35 and d = 0 . 85. Therefore, the risk-neutral probability that the stock will go up in one year is p = (1 + r f ) − d u − d = 1 . 05 − 0 . 85 1 . 35 − 0 . 85 = 0 . 4 . (b) (*) What is the value of a call option that expires in one year and has a strike price of $ 55? Solution. Given the risk-neutral probability p = 0 . 4, the current value of the call option is given by C 0 = p × max(60 . 75 − 55 , 0) + (1 − p ) × max(38 . 25 − 55 , 0) 1 + r f = 0 . 4 × 5 . 75 1 . 05 = 2 . 1905 . (c) (*) What is the value of a put option that expires in one year and has a strike price of $ 55? Solution. Given the risk-neutral probability p = 0 . 4, the current value of the put option is given by P 0 = p × max(55 − 60 . 75 , 0) + (1 − p ) × max(55 − 38 . 25 , 0) 1 + r f = 0 . 6 × 16 . 75 1 . 05 = 9 . 5714 . (d) (**) Suppose we know that the true (not risk-neutral) probability q that the stock will go up is 42%, then the (annual) expected return of the stock, r S , can be computed as 45 = 0 . 42 × (45 × 1 . 35) + 0 . 58 × (45 × 0 . 85) 1 + r S ⇒ r S = 6% What is the (annual) expected return of the call option r C ? What is the (annual) ex- pected return of the put option r P ? Solution. The expected return of the call option is the rate r C such that 2 . 1905 = 0 . 42 × 5 . 75 1 + r C ⇒ r C = 10 . 25% © Buffa-Garc´ ıa, Leeds School of Business Page 2 of 7

Derivative Securities, Fall 2023 FNCE 4040, Homework 3 Solutions The expected return of the put option is the rate r P such that 9 . 5714 = 0 . 58 × 16 . 75 1 + r P ⇒ r P = 1 . 50% Note that the expected return of a call option is higher than the expected return of the underlying asset, whereas the expected return of a put option is lower than the expected return of the underlying asset. This is because call options are levered long positions in the underlying asset, so they should be more risky than pure investments in the underlying (they have higher betas). Put options, on the other hand, are short positions in the underlying asset coupled with lending. A trading strategy that is short an asset ought to have low expected returns since it will be a negative beta security (as long as the underlying asset has expected returns above the risk-free rate). 3. ( 35 points ) Consider pricing a call option on the S&P 500 index with the expiration date in 4 months and a strike price of $ 100. The S&P 500 can move up by 35% or down by 20% every 2 months, and is currently trading for $ 100. The annual risk-free rate is 5%. (a) (**) What would you estimate the value of the call option to be today? What would you estimate the value of the call option to be in two months if the S&P 500 goes up in value? What if it goes down? Solution. The up and down movements of the S&P 500 in the binomial tree are captured by u = 1 . 35 and d = 0 . 8. Moreover, since 2 months correspond to one period in the binomial model, (1 + r f ) = (1 + 5%) 2 / 12 = 1 . 00816. From these one can back out the risk-neutral probabilities as p = (1 + r f ) − d u − d = 1 . 00816 − 0 . 8 1 . 35 − 0 . 8 = 0 . 3785 (1 − p ) = 0 . 6215 as well as the binomial evolution of the S&P500 over the next four months. Evolution of S&P500 today in 2 months in 4 months 182.25 135.00 100.00 108.00 80.00 64.00 Using these, one can work backwards through the tree and solve for the value of the call option at each point in time and in each state of the tree. Evolution of Call option price today in 2 months in 4 months 82 . 25 = max(182 . 25 − 100 , 0) 35 . 81 = 0 . 3785 × 82 . 25+0 . 6215 × 8 1 . 00816 15 . 30 = 0 . 3785 × 35 . 81+0 . 6215 × 3 1 . 00816 8 . 00 = max(108 − 100 , 0) 3 . 00 = 0 . 3785 × 8+0 . 6215 × 0 1 . 00816 0 . 00 = max(64 − 100 , 0) © Buffa-Garc´ ıa, Leeds School of Business Page 3 of 7

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Derivative Securities, Fall 2023 FNCE 4040, Homework 3 Solutions (b) (**) What is the dynamic trading strategy that allows you to perfectly replicate the payoff of the option? Solution. The dynamic trading strategy that allows you to perfectly replicate the payoff of the option involves trading the underlying asset of the option and a risk-free bond at date 0, as well as at any future date depending on the value of the S&P500. The replicating portfolio at date t = 0 is such that: ∆ 0 × 135 + B 0 × 1 . 00816 = 35 . 81 , ∆ 0 × 80 + B 0 × 1 . 00816 = 3 , yielding ∆ 0 = 0 . 60 and B 0 = − 44 . 35. This portfolio would have to be rebalanced in two months time depending on whether the S&P 500 went up in value or not: • If the S&P 500 went up in value in the first two months, then in order to replicate the payoffs at maturity of the call option we would need to invest in ∆ 1 u units of the S&P500 and B 1 u dollars in cash, such that ∆ 1 u × 182 . 25 + B 1 u × 1 . 00816 = 82 . 25 , ∆ 1 u × 108 + B 1 u × 1 . 00816 = 8 , thus yielding ∆ 1 u = 1 and B 1 u = − 99 . 19. • If the S&P 500 went down in value in the first two months, then in order to replicate the payoffs at maturity of the call option we would need to invest in ∆ 1 d units of the S&P500 and B 1 d dollars in cash, such that ∆ 1 d × 108 + B 1 d × 1 . 00816 = 8 , ∆ 1 d × 64 + B 1 d × 1 . 00816 = 0 , so that ∆ 1 d = 0 . 18 and B 1 d = − 11 . 54. 4. ( 25 points ) You are an option trader and a colleague of yours working in sales asks you to price a non-standard derivative for a client. The client would like to trade a derivative that will pay the square of the stock price of AMB three weeks from now (for instance, if the stock price ends up at $ 50 in three weeks, the holder of the derivative would receive $50 2 = $2500). The current price of AMB is $ 40 and the annual volatility σ of the stock price is 68.73%. The annual risk free rate is 8%. (a) (**) Compute the evolution of the stock pice of AMB for the next three weeks, consid- ering a binomial model where a period corresponds to a week. Solution. Since σ = 0 . 6873, T = 3 / 52 and n = 3, it follows that u = e σ √ T/n = e 0 . 6873 √ 1 / 52 = 1 . 1 , and d = 1 /u = 0 . 9091 . In other words, σ p 1 / 52 represents the weekly volatility of the stock price, which we can use to infer the up and down (proportional) movements of the stock in the considered binomial model. Therefore, every week the stock price either goes up by 10% or down by 9.09%. This means that the stock price is estimated to evolve as follows: © Buffa-Garc´ ıa, Leeds School of Business Page 4 of 7

Derivative Securities, Fall 2023 FNCE 4040, Homework 3 Solutions AMB’s Stock Price today in 1 week in 2 weeks in 3 weeks 53.24 48.40 44.00 44.00 40.00 40.00 36.36 36.36 33.06 30.05 (b) (**) What is the fair price of this derivative today, and in any relevant state in the next three weeks? Solution. The simplest way to compute the price of this derivative in any node of the binomial model is to use risk-neutral pricing. To this end, we need the per-period risk-free rate r f and the risk-neutral probabilities p and (1 − p ). We obtain the weekly risk-free rate as: (1 + r f ) = (1 + 8%) 1 / 52 = 1 . 00148 ⇒ r f = 0 . 148% . Since the stock price moves up and down proportionately to the its last value, the risk- neutral probabilities remain constant over time (and over states) and are given by p = (1 + r f ) − d u − d = 1 . 00148 − 0 . 9091 1 . 1 − 0 . 9091 = 0 . 4839 ⇒ (1 − p ) = 0 . 5161 . Denoting the price of the derivative as Y , and computing the payoff of the derivative in 3 weeks, we obtain Derivative Price today in 1 week in 2 weeks in 3 weeks Y 3 uuu Y 2 uu Y 1 u Y 3 uud Y 0 Y 2 ud Y 1 d Y 3 udd Y 2 dd Y 3 ddd where the derivative payoffs in 3 weeks are given by Y 3 uuu = S 2 3 uuu = 2834 . 52 , Y 3 uud = S 2 3 uud = 1936 . 00 , Y 3 udd = S 2 3 udd = 1322 . 31 , Y 3 ddd = S 2 3 ddd = 903 . 15 , the derivative prices in 2 weeks are given by Y 2 uu = 0 . 4839 × Y 3 uuu + 0 . 5161 × Y 3 uud 1 . 00148 = 2367 . 33 , Y 2 ud = 0 . 4839 × Y 3 uud + 0 . 5161 × Y 3 udd 1 . 00148 = 1616 . 91 , Y 2 dd = 0 . 4839 × Y 3 udd + 0 . 5161 × Y 3 ddd 1 . 00148 = 1104 . 37 , © Buffa-Garc´ ıa, Leeds School of Business Page 5 of 7

Derivative Securities, Fall 2023 FNCE 4040, Homework 3 Solutions the derivative prices in 1 week are given by Y 1 u = 0 . 4839 × Y 2 uu + 0 . 5161 × Y 2 ud 1 . 00148 = 1977 . 15 , Y 1 d = 0 . 4839 × Y 2 ud + 0 . 5161 × Y 2 dd 1 . 00148 = 1350 . 41 , and the derivative price today is given by Y 0 = 0 . 4839 × Y 1 u + 0 . 5161 × Y 1 d 1 . 00148 = 1651 . 27 . 5. (***) Consider a single period trinomial model for the price of a stock. The stock price today is S 0 = 100 and it can take the following three values next period: S u = 150 , S m = 100 , S d = 60. If you can trade only the stock and a risk-free bond with a (per-period) risk-free rate of 10%, what are the upper-bound and lower-bound of the price of a put option with expiration date in one period and strike price of $ 110? Solution. This problem is about super-replication and sub-replication . With super-replication, we want to find the cheapest portfolio that dominates the payoffs of another security. With sub-replication, instead, we want to find the most expensive portfolio that that is dominated by the payoffs of another security. In our case, the security is the put option, and super- replication and sub-replication allow us to find an upper-bound and a lower-bound for the price of the put option. To find the portfolio (∆ UB , B UB ) that super-replicates the put option, we need to solve the following constrained optimization, min ∆ UB ,B UB ∆ UB 100 + B UB , subject to ∆ UB 150 + B UB 1 . 1 ≥ 0 , ∆ UB 100 + B UB 1 . 1 ≥ 10 , ∆ UB 60 + B UB 1 . 1 ≥ 50 . Solving this problem yields ∆ UB = − 0 . 5553 and B UB = 75 . 7421, which correspond to future cash flows of 0 . 026 in the u -state, 27 . 79 in the m -state, and 50 in the d -state). The cost of this portfolio, and hence the upper-bound for the put option, is $ 20.21. To find the portfolio (∆ LB , B LB ) that sub-replicates the put option, we need to solve the following constrained optimization, max ∆ LB ,B LB ∆ LB 100 + B LB , subject to ∆ LB 150 + B LB 1 . 1 ≤ 0 , ∆ LB 100 + B LB 1 . 1 ≤ 10 , ∆ LB 60 + B LB 1 . 1 ≤ 50 . © Buffa-Garc´ ıa, Leeds School of Business Page 6 of 7

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Derivative Securities, Fall 2023 FNCE 4040, Homework 3 Solutions Solving this problem yields ∆ LB = − 0 . 2 and B LB = 27 . 27, which correspond to future cash flows of 0 in the u -state, 10 in the m -state, and 18 in the d -state). The cost of this portfolio, and hence the lower-bound for the put option, is $ 7.27. Therefore, if we can only trade two assets (the underlying stock and a risk-free bond) can only conclude that the price of the put option must lie somewhere in [7 . 27 , 20 . 21]. Note that the idea of super- and sub-replication can actually be expressed in terms of risk- neutral probabilities. In particular, one can show that the above bounds are equivalent to solving min p u ,p m ,p d ≥ 0 p m 10 + p d 50 1 . 1 and max p u ,p m ,p d ≥ 0 p m 10 + p d 50 1 . 1 where the maximization and minimization problems are subject to the constraints p u 150 + p m 100 + p d 60 1 . 1 = 100 p u + p m + p d = 1 With this formulation of the problem, we look for the RNPs that maximize and minimize the value of the put option such that the RNPs price the stock correctly. The bounds using this RNPs approach are again [7 . 27 , 20 . 21]. © Buffa-Garc´ ıa, Leeds School of Business Page 7 of 7