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274 Mathematics for Finance


2.39Because the bond is trading at par and the interest rates remain constant,
the price of the bond at the beginning of each year will be $100. The sum of
$1,000 will buy 10 bonds at the beginning of year one. At the end of the year
the coupons will pay 10×8 = 80 dollars, enough to buy 0.8 bonds at $100
each. As a result, the investor will be holding 10 + 0.8=10.8 bonds in year
two. At the end of that year the coupons will pay 10. 8 ×8=86.4 dollars and
0 .864 additional bonds will be purchased at $100 each. The number of bonds
held in year three will be 10.8+0.864 = 11.664, and so on.
In general, the number of bonds held in yearnwill be

10

(
1+ 1008

)n− 1
,

which gives 10.0000, 10.8000, 11.6640, 12.5971, 13.6049 bonds held in years
one to five.

Chapter 3


3.1The tree representing the scenarios and price movements in Example 3.1 is
shown in Figure S.3.

Figure S.3 Tree of price movements in Example 3.1

3.2The tree representing the scenarios and price movements is shown in Fig-
ure S.4. There are altogether eight scenarios represented by the paths through
the tree leading from the ‘root’ on the left towards the rightmost branch tips.
3.3We can use (3.1) to find
Scenario S(0) S(1) S(2) S(3)
ω 1 45 .00 49.50 51.98 46. 78
ω 2 45 .00 47.25 51.98 57. 17
ω 3 45 .00 47.25 42.53 46. 78
The tree is shown in Figure S.5.
3.4When dividends are payable, formula (3.1) becomes
S(n)=S(n−1)(1 +K(n))−div(n),
which gives
Scenario S(0) S(1) S(2) S(3)
ω 1 45 .00 48.50 49.93 43. 93
ω 2 45 .00 46.25 49.88 53. 86
ω 3 45 .00 46.25 40.63 43. 69
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