242 - Year ±rom 3-year zero ±rom 7-year zero 1 2 3 39 ....

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Example of Duration Matching Suppose that you have liability of $100 million that will come due in 5 years. You wish to invest today to protect against interest rate variability. You must invest in a combination of 3 and 7 year zero coupon bonds. The term structure is currently Fat at 5%. How much should you invest in each bond today? The present value of the liability is: 100 1 . 05 5 = 78 . 35 Let x be the fraction invested in the three-year zero coupon bond. x must satisfy: 3 x + 7(1 - x ) = 5 which implies that x is 50%. This means that 0 . 5 × 78 . 35 = 39 . 18 is invested in each bond. This yields the following cash Fows:
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Unformatted text preview: Year ±rom 3-year zero ±rom 7-year zero 1 2 3 39 . 18 × 1 . 05 3 = 45 . 35 4 5 6 7 39 . 18 × 1 . 05 7 = 55 . 13 1 Let’s look at what happens to values at year 5 when yields change: Yield Liability Investment in 3-Year Investment in 7-year Total r 100 45 . 35 × (1 + r ) 2 55 . 13 / (1 + r ) 2 – 4% 100 49.05 50.97 100.02 5% 100 50.00 50.00 100.00 6% 100 50.96 49.06 100.02 You do slightly better than matching perfectly! Why? Convexity! 2...
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This note was uploaded on 11/15/2008 for the course MS&E 242 taught by Professor Primbs during the Fall '06 term at Stanford.

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242 - Year ±rom 3-year zero ±rom 7-year zero 1 2 3 39 ....

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