M10_MCDO8122_01_ISM_C10

# M10_MCDO8122_01_ISM_C10 - Chapter 10 Binomial Option...

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Chapter 10 Binomial Option Pricing n Question 10.1 1. Since 25 u C = and 0 d C = we have 25 50 0.50. ∆ = = To solve the bond amount, one could use Equation (10.2); however, once we know the option’s , finding the replicating bond position is a simple algebra exercise (i.e., no memorization!). We will replicate the up-value of the call: 04 04 5 130 25 40 38 4316 e B B e . -. . × + = = - = - . The value of the option is the value of the replicating portfolio: 0 5 100 38 4316 11 5684 C = . × - . = . 2. Since 0 u P = and 25 d P = - we have 25 50 0 50. - ∆ = = - . We will replicate the up-value of the put: 04 04 5 130 0 65 62 4513 e B B e . -. -. × + = = = . The value of the option is the value of the replicating portfolio: 0 5 100 62 4513 12 4513 P = -. × + . = . n Question 10.2 1. We will use risk-neutral pricing for this problem. Our risk-neutral probability of an up movement is ( ) 04 * 8 0 4816 1 3 8 r h e d e p u d δ - . - -. = = = . - . -. The no-arbitrage value of the European call is: * * 04 0 [ (1 )] 35 0 4816 16 1958 rh u d C e C p C p e - -. = + - = × × . = . For the replicating strategy, the option’s delta is 35 50 0 70. ∆ = / = . We can use the option value result to solve for the bond amount (once again, no memorization required). The option value is the total cost of replication. The cost of 0.7 shares is \$70. Since the option value is \$16.1958, we must borrow \$70 \$16.1958 \$53.8042, - = i.e., \$53.8042. B = - 2. If we observe a price of \$17, then the option price is too high relative to its theoretical value. We sell the option for \$17 and synthetically create a call option for \$16.196. In order to do so, we buy 0.7 units of the share and borrow \$53.804. These transactions yield no risk and a profit of \$0.804.

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113 McDonald • Fundamentals of Derivatives Markets 3. If we observe a price of \$15.50, then the option price is too low relative to its theoretical value. We buy the option and synthetically create a short position in the option (we’ll receive \$16.1958). In order to do so, we short 0.7 shares and lend \$53.8042. These transactions yield no risk and a profit of \$0.696. n Question 10.3 1. We will use risk-neutral pricing for this problem. Our risk-neutral probability of an up movement is ( ) 04 * 8 0 4816 1 3 8 r h e d e p u d δ - . - -. = = = . - . -. The no-arbitrage value of the European put is: * * 04 0 [ (1 )] 15 0 5184 7 4708 rh u d P e P p P p e - -. = + - = × × . = . For the replicating strategy, the option’s delta is 15 50 0 30. ∆ = - / = - . We can use the option value result to solve for the bond amount (once again, no memorization required). The option value is the total cost of replication. By shorting 0.3 shares, we receive \$30. Since the option value is \$7.4708, we must lend \$37.4708, i.e., B = \$37.4708. 2. If we observe a price of \$8, then the option price is too high relative to its theoretical value. We sell the option for \$8 and synthetically create a long put option for \$7.471. In order to do so, we short 0.3 units of the share and lend \$37.4708. These transactions yield no risk and a profit of \$0.5292. 3. If we observe a price of \$6, then the put option price is too low relative to its theoretical value. We buy the option and synthetically create a short position in the option (we’ll receive \$7.4708). In order to do so, we buy 0.3 units of the share and borrow \$37.4708. These transactions yield no risk and a profit of \$1.4708.
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