A given mortgage security is trading at par. The expected average price change from a projected change in a given market yield is 1 for the mortgage security and 0.4 and 2.0 for hedging instrument one and two respectively. The expected average price change from a projected twist in the yield curve is 0.4 for the mortgage security and 0.3 and 0.5 for hedging instrument one and two respectively. What positions in hedging instruments one and two should a manager take to hedge the price of the mortgage security from the projected market changes? For every dollar of face value of the mortgage security: | A) | buy $2.5 of hedging instrument one and $0.5 of hedging instrument two. |
| B) | sell $0.75 of hedging instrument one and $0.35 of hedging instrument two. |
| | C) | sell $2.5 of hedging instrument one and $0.5 of hedging instrument two. |
| | D) | sell $2.0 of hedging instrument one and $0.56 of hedging instrument two. |
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Answer and Explanation
To answer this, we set up the following two equations and two unknowns. (NH1)(0.4) + (NH2)(2.0) = -1.0
(NH1)(0.3) + (NH2)(0.5) = -0.4, where NH1 and NH2 are the positions to take in hedging instruments one and two respectively. Multiplying the second equation by 4 and subtracting it from the first gives (NH1)(-0.8)=0.6, and thus NH1=-0.75. Substituting this into either expression and solving NH2 gives NH2=-0.35. (-0.75)(0.4)+(-0.35)(2)=-1
(-0.75)(0.3)+(-0.35)(0.5)=-0.4
To answer this, we set up the following two equations and two unknowns. (N H1)(0.4) + (N H2)(2.0) = -1.0 (N H1)(0.3) + (N H2)(0.5) = -0.4, where NH1 and NH2 are the positions to take in hedging instruments one and two respectively. Multiplying the second equation by 4 and subtracting it from the first gives (NH1)(-0.8)=0.6, and thus NH1=-0.75. Substituting this into either expression and solving NH2 gives NH2=-0.35. (-0.75)(0.4)+(-0.35)(2)=-1 (-0.75)(0.3)+(-0.35)(0.5)=-0.4 To answer this, we set up the following two equations and two unknowns. (N H1)(0.4) + (N H2)(2.0) = -1.0 (N H1)(0.3) + (N H2)(0.5) = -0.4, where NH1 and NH2 are the positions to take in hedging instruments one and two respectively. Multiplying the second equation by 4 and subtracting it from the first gives (NH1)(-0.8)=0.6, and thus NH1=-0.75. Substituting this into either expression and solving NH2 gives NH2=-0.35. (-0.75)(0.4)+(-0.35)(2)=-1 (-0.75)(0.3)+(-0.35)(0.5)=-0.4 [此贴子已经被作者于2008-9-18 17:49:18编辑过] | 
发表于 2008-9-16 09:53
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