16655 Lecture Notes - Lecture 12: Interest Rate Swap, Life Insurance, Financial Instrument
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In the above example, to calculate the bond price at a quoted yield of 5%, we have i = 5%/2 = 0. 025. V = 1/(1+i) =1/(1+. 025)=0. 9756 an = (1-vn)/i = (1-0. 9756^19)/0. 025=14. 9789 f = 181 d = 181 g = 100 x 6%/2=3 n = 19. P = 0. 9756^1(3 x (1+14. 9789) + 100 x 0. 9756^19) Swaps generate real gains and losses: an interest rate swap over identical dates as the bond with a notional principal of. The total loss and gain of the two parties sum to. What is a derivative again: a derivative is a financial instrument that is derived from some other asset, index, event, value or condition (known as the underlying). Why would an investor in bonds use rate swaps: cashflows vary with the level of interest rates. Some of bond characteristics may not always result in the desired risk profile. Investors can exchange their risk preference through rate swaps.