276 Mathematics for Finance
relation 1 +K(0,2) = (1 +K)^2 (assuming that 1 +K>0):
Scenario K(0,2) K(1) =K(2)
ω 1 17 .14% 8 .23%
ω 2 − 8 .57% − 4 .38%
ω 3 − 20 .00% − 10 .56%
3.7The formula
1+K(0,2) = (1 +K(1)) (1 +K(2))
can be used to findK(2). For example, the following scenarios and values of
K(2) are consistent with the conditions of Exercise 3.7:
Scenario K(0,2) K(1) K(2)
ω 1 21 .00% 10 .00% 10.00%
ω 2 10 .00% 10 .00% 0 .00%
ω 3 − 1 .00% − 10 .00% 10.00%
This is not the only possible solution. Another one can be obtained from the
above by changing scenarioω 2 to
Scenario K(0,2) K(1) K(2)
ω 2 10 .00% − 10 .00% 22.22%
with the other two scenarios unaltered.
3.8For the three scenarios in Example 3.2 we find
Scenario k(1) k(2) k(0,2)
ω 1 5 .31% 3 .39% 8 .70%
ω 2 5 .31% − 10 .92% − 5 .61%
ω 3 − 5 .61% 1 .90% − 3 .70%In all three casesk(0,2) =k(1) +k(2).
3.9LetKdenote the return in the third scenario. If the expected return is equal
to 6%, then
1
2 ×(−5%) +1
4 ×6% +1
4 ×K=6%.
Solving forK, we find that the return in the third scenario must be 28%.
3.10First, compute the returnsK(1),K(2) andK(0,2) in each scenario:
Scenario K(1) K(2) K(0,2)
ω 1 10 .00% 9 .09% 20.00%
ω 2 5 .00% − 4 .76% 0 .00%
ω 3 − 10 .00% 11 .11% 0 .00%
It follows that
E(K(1))∼= 0. 25 × 10 .00% + 0. 25 × 5 .00%− 0. 5 × 10 .00%∼=− 1 .25%,
E(K(2))∼= 0. 25 × 9 .09%− 0. 25 × 4 .76% + 0. 5 × 11 .11%∼= 6 .64%,
E(K(0,2))∼= 0. 25 × 20 .00% + 0. 25 × 0 .00% + 0. 5 × 0 .00%∼= 5 .00%.
Clearly,
(1 +E(K(1)))(1 +E(K(2)))∼= 1. 0530 =1. 0500 ∼=1+E(K(0,2)).