dstefan
03-22-2008, 01:36 PM
Chapter 8, Problem 8:
Use bootstrapping to obtain a zero rate curve given the prices of the following semiannual coupon bonds:
\begin{array}{lcl} \mbox{Maturity} & \mbox{Coupon Rate} & \mbox{Price} \\ 6 ~\mbox{months} & 0 & 97.5 \\ 1 ~\mbox{year} & 5 & 100 \\ 20 ~\mbox{months} & 6 & 103 \\ 40 ~\mbox{months} & 5 & 102 \\ 5 ~\mbox{years} & 4 & 103 \\ \end{array}
Assume that the overnight rate is 5% and that the zero rate curve is linear on the following time intervals:
\left[0, 0.5 \right]; \left[0.5, 1 \right]; \left[1, \frac{5}{3} \right]; \left[\frac{5}{3}, \frac{10}{3} \right]; \left[\frac{10}{3}, 5 \right].
Solution:
Hint: Obtain the zero rate curve r(0,t) for 0 \leq t \leq 1 from the prices of the first two bonds, and using the fact that r(0,0) = 0.05. The third bond pays coupons in 2, 8, 14, and 20 months, when it also pays the face value of the bond. From the pricing formula (\ref{bond_price_semiannual_bootstrap}), it follows that
103~=~ 3 ~e^{-\frac{1}{6} r(0,\frac{1}{6})}~+~ 3 ~e^{-\frac{2}{3}r(0,\frac{2}{3})}
~+~ 3 ~e^{-\frac{7}{6} r(0,\frac{7}{6})}~+~103 ~e^{-\frac{5}{3} r(0,\frac{5}{3})}.
The zero rates r(0,\frac{1}{6}) and r(0,\frac{2}{3}) are already known. Since the zero rate curve is assumed to be linear for t \in \left[1, \frac{5}{3} \right], we find that
r(0,t)~=~ \frac{3}{2} \left( \frac{5}{3} - t \right)~r(0,1)~+~ \frac{3}{2} \left( t - 1 \right)~r\left(0,\frac{5}{3}\right).
Therefore,
r\left(0,\frac{7}{6}\right)~=~ \frac{r(0,1)}{3} ~+~ \frac{r\left(0,\frac{5}{3}\right)}{4}.
Set x = r(0,\frac{5}{3}) and use Newton's method to solve for x in (\ref{r_5_3_first}).
Use bootstrapping to obtain a zero rate curve given the prices of the following semiannual coupon bonds:
\begin{array}{lcl} \mbox{Maturity} & \mbox{Coupon Rate} & \mbox{Price} \\ 6 ~\mbox{months} & 0 & 97.5 \\ 1 ~\mbox{year} & 5 & 100 \\ 20 ~\mbox{months} & 6 & 103 \\ 40 ~\mbox{months} & 5 & 102 \\ 5 ~\mbox{years} & 4 & 103 \\ \end{array}
Assume that the overnight rate is 5% and that the zero rate curve is linear on the following time intervals:
\left[0, 0.5 \right]; \left[0.5, 1 \right]; \left[1, \frac{5}{3} \right]; \left[\frac{5}{3}, \frac{10}{3} \right]; \left[\frac{10}{3}, 5 \right].
Solution:
Hint: Obtain the zero rate curve r(0,t) for 0 \leq t \leq 1 from the prices of the first two bonds, and using the fact that r(0,0) = 0.05. The third bond pays coupons in 2, 8, 14, and 20 months, when it also pays the face value of the bond. From the pricing formula (\ref{bond_price_semiannual_bootstrap}), it follows that
103~=~ 3 ~e^{-\frac{1}{6} r(0,\frac{1}{6})}~+~ 3 ~e^{-\frac{2}{3}r(0,\frac{2}{3})}
~+~ 3 ~e^{-\frac{7}{6} r(0,\frac{7}{6})}~+~103 ~e^{-\frac{5}{3} r(0,\frac{5}{3})}.
The zero rates r(0,\frac{1}{6}) and r(0,\frac{2}{3}) are already known. Since the zero rate curve is assumed to be linear for t \in \left[1, \frac{5}{3} \right], we find that
r(0,t)~=~ \frac{3}{2} \left( \frac{5}{3} - t \right)~r(0,1)~+~ \frac{3}{2} \left( t - 1 \right)~r\left(0,\frac{5}{3}\right).
Therefore,
r\left(0,\frac{7}{6}\right)~=~ \frac{r(0,1)}{3} ~+~ \frac{r\left(0,\frac{5}{3}\right)}{4}.
Set x = r(0,\frac{5}{3}) and use Newton's method to solve for x in (\ref{r_5_3_first}).