Consider 3-step binomial model with one asset with initial price S(0) = 100.
(a) For D = 0.9, U = 1.1 and p = 1 2 :
(i) Determine the distribution of the price S(t) for t = 1, 2, 3.
(ii) Compute the expected price E(S(t)) for t = 1, 2, 3.
(iii) Determine the distribution of the return K(t) for t = 1, 2, 3.
(iv) Compute the expected return E(K(t)) for t = 1, 2, 3.
(b) Suppose D = 0.9 and p = 1 2 . Determine all values of U > D for which the expected return E(K(3)) is:
(i) equal to 0;
(ii) smaller than 0;
(iii) bigger than 0.
(c) Let 0 < D < U. Find p ∈ [0, 1], expressed as a formula involving D and U, for which E(S(3)) = 100.
Let S1 and S2 be two risky assets each following a 2-step binomial model with uniform probability and parameters D1 = 10/11, U1 = 1.1 and D2 = 0.95, U2 = 1.05, respectively. In which asset it is more reasonable to invest? kindly explain your choice.
The solution of this problem should contain a statement saying which of the two assets is better supported by an argument containing suitable computations. The solution may not be unique.
(a) For D = 0.9, U = 1.1 and p = 1 2 :
(i) Determine the distribution of the price S(t) for t = 1, 2, 3.
(ii) Compute the expected price E(S(t)) for t = 1, 2, 3.
(iii) Determine the distribution of the return K(t) for t = 1, 2, 3.
(iv) Compute the expected return E(K(t)) for t = 1, 2, 3.
(b) Suppose D = 0.9 and p = 1 2 . Determine all values of U > D for which the expected return E(K(3)) is:
(i) equal to 0;
(ii) smaller than 0;
(iii) bigger than 0.
(c) Let 0 < D < U. Find p ∈ [0, 1], expressed as a formula involving D and U, for which E(S(3)) = 100.
Let S1 and S2 be two risky assets each following a 2-step binomial model with uniform probability and parameters D1 = 10/11, U1 = 1.1 and D2 = 0.95, U2 = 1.05, respectively. In which asset it is more reasonable to invest? kindly explain your choice.
The solution of this problem should contain a statement saying which of the two assets is better supported by an argument containing suitable computations. The solution may not be unique.
