Question

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The following options prices were observed for calls and puts on Lannister Ltd for the trading...

The following options prices were observed for calls and puts on Lannister Ltd for the trading day of July 6 2019. Use this information in Questions 3-8. The stock was priced at $163.37. The expirations were July 17, August 21 and October 16. The continuously compounded risk-free rates associated with the three expirations were 0.0517, 0.0542 and 0.0565, respectively. The options have European expiries.

Lannister Ltd CALLS
STRIKE JUL AUG OCT
150 9.50 11.25 13.61
155 5.70 7.96 10.88
160 2.23 5.01 8.04
165 0.77 2.79 6.90
Lannister Ltd PUTS
STRIKE JUL AUG OCT
150 0.17 1.18 2.69
155 0.71 2.66 4.44
160 2.22 4.63 6.60
165 5.61 7.42 8.81

Question: Showing all formula and workings where applicable; Let the standard deviation of the continuously compounded return on the stock be 20 percent. Ignore dividends. Respond to the following:

  1. What is the theoretical fair value of the October 165 call? Calculate this answer by hand and then re-calculate it using BlackScholesMertonBinomial10e.xlsm.
  2. Based on your answer in part a, recommend a riskless strategy.
  3. If the stock price decreases by $1, how will the option position offset the loss on the stock?
  4. Use the Black-Scholes-Merton European put option pricing formula for the October 160 put option. Repeat parts a, b and c of Question 3 with respect to the put.
  5. Buy 100 shares of Lannister Ltd at $163.37 and short one October 165 call. Hold the position until expiration. Determine the profits and graph the results. Identify the strategy, breakeven stock price at expiration, the maximum profit, and the maximum loss. Discuss any special considerations associated with this strategy. Note: use the OptionStrategyAnalyzer10e.xlsm to obtain the required payoff diagram.
  6. Buy 100 shares of Lannister Ltd at $163.37 and go long one October 160 put. Hold the position until expiration. Determine the profits and graph the results. Identify the strategy, breakeven stock price at expiration, the maximum profit, and the maximum loss. Discuss any special considerations associated with this strategy. Note: use the OptionStrategyAnalyzer10e.xlsm to obtain the required payoff diagram.
  7. Construct an options strategy by going short one October 160 call and long one October 165 call using Lannister Ltd options. Hold the position until expiration. Determine the profits and graph the results. Identify the strategy, breakeven stock price at expiration, the maximum profit, and the maximum loss. Discuss any special considerations associated with this strategy. Note: use the OptionStrategyAnalyzer10e.xlsm to obtain the required payoff diagram.
  8. Construct an options strategy by going long one October 165 put and long one October 165 call using Lannister Ltd options. Hold the position until expiration. Determine the profits and graph the results. Identify the strategy, breakeven stock price at expiration, the maximum profit, and the maximum loss. Discuss any special considerations associated with this strategy. Note: use the OptionStrategyAnalyzer10e.xlsm to obtain the required payoff diagram

Solutions

Expert Solution

a. We know that,

Black Scholes Merton formula for call options is given by:

where,

Here - Underlying price, K - Strike Price, r - risk- free rate, - standard deviation, T - Time to expiration  

From the data given we know that for Oct 165 Call,

S0 = 163.37, K = 165, r = 0.0565,  %, T = 103/365 = 0.282192

From this we get, = 0.109746 and = 0.003502

Substituting and we get, N( ) = 0.543694 , N(- ) = 0.456306, N() = 0.501397, N(-) = 0.498603

From this we get, c = 7.4014

d.  We know that,

Black Scholes Merton formula for put options is given by:

where,

Here - Underlying price, K - Strike Price, r - risk- free rate, - standard deviation, T - Time to expiration  

From the data given we know that for Oct 160 Put,

S0 = 163.37, K = 160, r = 0.0565,  %, T = 103/365 = 0.282192

From this we get, = 0.109746 and = 0.003502

Substituting and we get, N( ) = 0.543694 , N(- ) = 0.456306, N() = 0.501397, N(-) = 0.498603

From this we get, c = 4.248751


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