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Z01_MCDO8122_01_ISM_App-B - Appendix B Continuous...

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Appendix B: Continuous Compounding Question B.1 Using a continuous rate of return, we have 5 67032 100000. r e × × = This implies 5 1 4918248. r e × = . Solving, we have 5 ln(1 4918248) 0 40, r × = . = . hence the continuous return is 0 40 5 8%. r = . / = Question B.2 The unknown in the continuous interest equation is the time to maturity, T . We must solve: 0 10 50 T e . × × = 100. This implies 0 10 2. T e . × = Taking the natural log of each side, we have 0.10 ln(2) 0 693147 T × = = . which implies 6 93147 T = . years. Question B.3 In 7 years, at 7.5% continuous interest, $5 grows to 0 075 7 5 $8 4523. e . × × = . Question B.4 Let 1 S be the stock price in 1 year. The continuous return is 1 0 1 1 1 ln( / ) ln( /100) ln( ) ln(100) ln( ) 4 6052 r S S S S S = = = = . Using this formula we have: 0 S r 5 2.9957323 4 3.2188758 3 3.5065579 2 3.912023 Note these are not percentages. For example, if 1 S is 5, the return is 299.57%. For a 500% return, the stock price next year must solve 1 ln( ) 4 6052 5 4 6052 0 3948 S r = + . = − + . = − . which implies 0 3948 1 $0 6738. S e − . = = .
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172 McDonald • Fundamentals of Derivatives Markets Question B.5 1. The arithmetic return for the first year is (200 100)/100 = 200% and, for the second year, it is
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