dstefan
06-06-2008, 12:25 AM
Chapter 2, Problem 9
Assume that the continuously compounded instantaneous interest curve has the form
r(t) ~=~ 0.05 ~+~ 0.005 \ln(1+t), ~~\forall~ t \geq 0.
(i) Find the corresponding zero rate curve;
(ii) Compute the 6-month, 12-month, 18-month, and 24-month discount factors;
(iii) Find the price of a two year semiannual coupon bond with coupon rate 5%.
Solution:
(i) The zero rate curve r(0,t) is
r(0,t) = \frac{1}{t} ~\int _0^t r(\tau) ~d \tau = \frac{1}{t} ~\int _0^t 0.05 ~+~ 0.005 \ln(1+\tau) ~d \tau
= \frac{1}{t} ~\left( 0.05 t ~+~ 0.005~(~(1+t) \ln(1+t) - t~) \right) = 0.045 ~+~ 0.005 (1+t) \frac{\ln(1+t)}{t}
(ii) The 6-month, 12-month, 18-month, and 24-month discount factors are, respectively,
disc(1) = e^{-r(0,0.5) 0.5} = 0.97478242; disc(2) = e^{-r(0,1)} = 0.94939392;
disc(3) = e^{-r(0,1.5) 1.5} = 0.92408277; disc(4) = e^{-r(0,2) 2} = 0.89899376
(iii) The price of the two year semiannual coupon bond with coupon rate 5% is
B = \frac{0.05}{2} ~100 ~e^{-r(0,0.5) 0.5} ~+~ \frac{0.05}{2} ~100 ~e^{-r(0,1)} ~+~ \frac{0.05}{2} ~100 ~e^{-r(0,1.5) 1.5} +~ \left( 100 ~+~ \frac{0.05}{2} ~100 \right) ~e^{-r(0,2) 2}
= 2.5 ~\mbox{disc}(1) ~+~ 2.5 ~\mbox{disc}(2) ~+~ 2.5 ~\mbox{disc}(3) ~+~ 102.5 ~\mbox{disc}(4) = 99.267508
Assume that the continuously compounded instantaneous interest curve has the form
r(t) ~=~ 0.05 ~+~ 0.005 \ln(1+t), ~~\forall~ t \geq 0.
(i) Find the corresponding zero rate curve;
(ii) Compute the 6-month, 12-month, 18-month, and 24-month discount factors;
(iii) Find the price of a two year semiannual coupon bond with coupon rate 5%.
Solution:
(i) The zero rate curve r(0,t) is
r(0,t) = \frac{1}{t} ~\int _0^t r(\tau) ~d \tau = \frac{1}{t} ~\int _0^t 0.05 ~+~ 0.005 \ln(1+\tau) ~d \tau
= \frac{1}{t} ~\left( 0.05 t ~+~ 0.005~(~(1+t) \ln(1+t) - t~) \right) = 0.045 ~+~ 0.005 (1+t) \frac{\ln(1+t)}{t}
(ii) The 6-month, 12-month, 18-month, and 24-month discount factors are, respectively,
disc(1) = e^{-r(0,0.5) 0.5} = 0.97478242; disc(2) = e^{-r(0,1)} = 0.94939392;
disc(3) = e^{-r(0,1.5) 1.5} = 0.92408277; disc(4) = e^{-r(0,2) 2} = 0.89899376
(iii) The price of the two year semiannual coupon bond with coupon rate 5% is
B = \frac{0.05}{2} ~100 ~e^{-r(0,0.5) 0.5} ~+~ \frac{0.05}{2} ~100 ~e^{-r(0,1)} ~+~ \frac{0.05}{2} ~100 ~e^{-r(0,1.5) 1.5} +~ \left( 100 ~+~ \frac{0.05}{2} ~100 \right) ~e^{-r(0,2) 2}
= 2.5 ~\mbox{disc}(1) ~+~ 2.5 ~\mbox{disc}(2) ~+~ 2.5 ~\mbox{disc}(3) ~+~ 102.5 ~\mbox{disc}(4) = 99.267508