Z01_MCDO8122_01_ISM_App-B

# 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 0405 8%. r =. /= ± Question B.2 The unknown in the continuous interest equation is the time to maturity, T . We must solve: 010 50 T e 100. This implies 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 0075 7 5 \$8 4523. e . ± Question B.4 Let 1 S be the stock price in 1 year. The continuous return is 10 1 1 1 ln( / ) ln( /100) ln( ) ln(100) ln( ) 4 6052 rS 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 Sr =+. = −+. = . which implies 03948 1 \$0 6738. Se −. == .

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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 (100 200)/200
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## This note was uploaded on 01/16/2011 for the course FIN 512 taught by Professor Staff during the Fall '08 term at University of Illinois, Urbana Champaign.

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Z01_MCDO8122_01_ISM_App-B - Appendix B: Continuous...

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