Question

6. Consider the following risk-free securities available to buy or sell to all investors in the market: Security Price (t=0) Cash flow (t=1) Cash flow (t=2) Cash flow (t=3) 76 120 68 216 102 Investments and Securities Markets FIN 320 a) Compute the YTM offered by securities A and C. b) Write down the equation that you would need to solve to find the YTM of security D. c) Compute the no-arbitrage price of security B. What is its YTM? d) Compute the missing cash flow paid by security E at t = 3 so that its no-arbitrage price at t = 0 is $102. e) If security D was suddenly trading for $205, but all the other prices stayed the same, is there an arbitrage opportunity that could be exploited? If so, clearly explain how you would implement the strategy and the profit that you expect to make.

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Answer 1

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Answer (a) For Security A: Price at t = 0 will be = 76 = (Cash flow at t = 1) / (1 + YTM) = 80 / (1 + YTM) Hence, YTM = 80 / 76 - 1= 5.26% For security C: 68 = 80 / (1 + YTM)); Hence, YTM = (80 /68)13 - 1 = 5.57% Answer (b) Let y be the YTM of the security D then, we need to solve the equation: 216 = 80 / (1 + y) + 80 / (1 + y)2 + 80 / (1 + y) = 80 /y x [1 - ( 1+y)-?]; Or, 8° [1 – (1+y)-3) = 216 Answer (c) Price of security D = Present value of all its future cash flows Hence, 216 = 80 / (1 + sl) + 80 / (1 + S2)2 + 80 / (1 + 53)3 = 80 / (1 + 5.26%) + 80 / (1 + s2)2 + 80 / (1 + 5.57%)3 = 144.00 + 80 / (1 + s2)2 Hence, 80 / (1 + s2)2 = 216 - 144 = 72.00 Hence, s2 = (80 / 72)1/2 - 1 = 5.41% Hence, no arbitrage price for security B = 120 / (1 + s2)2 = 120 / (1 + 5.41%)2 = $ 108.00 Answer (d) E is same as C in terms of the cash flows. E = 102 / 68 x C = 1.5C Hence, the missing cash flow of Eat t = 3 will be = 1.5 x corresponding cash flow of C = 1.5 x 80 = 120

Answer (e) Security D can be expressed as a liner combination of A, B and C. Create a security A + 2/3B + C It will have a cash flows of 80 at t = 1; 2/3 x 120 = 80 at t = 2 and 80 at t = 3 Hence its cash flows are identical to that of D. Hence, price of D = Price of A + 2/3B + C = 76 + 2/3 x 108 + 68 = 216