108156.pdf

272 Mathematics for Finance

2.30a) If the continuous compounding rate is 8%, then the price of the bond will

be

5e−^0.^08 +5e−^2 ×^0.^08 +5e−^3 ×^0.^08 + 105e−^4 ×^0.^08 ∼= 89. 06

dollars.

b) If the rate is 5%, then the price of the bond will be

5e−^0.^05 +5e−^2 ×^0.^05 +5e−^3 ×^0.^05 + 105e−^4 ×^0.^05 ∼= 99. 55

dollars.

2.31The price of the bond as a function of the continuous compounding ratercan

be expressed as 5e−r+5e−^2 r+5e−^3 r+ 105e−^4 r. The graph of this function is

shown in Figure S.1. Whenr↘0, the price approaches 5 + 5 + 5 + 105 = 120

dollars. In the limit asr↗∞the price tends to zero.

Figure S.1 Bond price versus interest rate in Exercise 2.31

2.32The timetprice of the coupon bond in Examples 2.9 and 2.10 is

10er(t−1)+ 10er(t−2)+ 10er(t−3)+ 10er(t−4)+ 110er(t−5) if 0≤t< 1 ,

10er(t−2)+ 10er(t−3)+ 10er(t−4)+ 110er(t−5) if 1≤t< 2 ,

10er(t−3)+ 10er(t−4)+ 110er(t−5) if 2≤t< 3 ,

10er(t−4)+ 110er(t−5) if 3≤t< 4 ,

110er(t−5) if 4≤t< 5.

The graph is shown in Figure S.2.

2.33In Figure S.2 we can see that the bond price will reach $95 for the first time dur-

ing year one, when the bond price is given by 10er(t−1)+10er(t−2)+10er(t−3)+

10er(t−4)+ 110er(t−5). Puttingr=0.12, we can find the desired timetwhen

the price will reach $95 by solving the equation

10er(t−1)+ 10er(t−2)+ 10er(t−3)+ 10er(t−4)+ 110er(t−5)=95.

This givest∼= 0 .4257 years or 155.4days.

2.34Since the bond is trading at par, its initial price is the same as the face value

F= 100. The implied continuous compounding ratercan be found by solving

the equation

8e−r+8e−^2 r+ 108e−^3 r= 100.

This givesr∼= 0 .0770 or 7.70%.