这个作业是完成风险股票和储蓄资产的收益分析代写，以下是作业具体内容：
练习1考虑一个单周期三州市场模型M =（B，S）日期0和T =1。我们假设有两种资产：储蓄帐户B的初始值为B0 = 1，而风险股票的初始值为价格S0 =4。无风险简单利率r等于10％。假设股票价格S1满足（S1（ω1），S1（ω2），S1（ω3））=（8、5、3）和真实世界概率P满足P（ω1）= 0.3，P（ω2）= 0.3，P（ω3）= 0.4。
（a）直接证明模型M =（B，S）是无套利的，即没有这种模型中存在套利机会。不要在这里使用FTAP
（定理2.2.1），但请参考课程说明中的定义2.2.3。
（b）考虑到期日T = 1和行使价的看涨期权K =4。检查此选项是否存在复制策略。
（c）明确找到所有可获得的或有债权的类别。
（d）找出所有measures度量Q =（q1，q2，q3）的M类对于模型M，空间Ω=（ω1，ω2，ω3）。
（e）查找所有期望值情商</ s> </ s> </ s>（S1-4-）1 + r</ s> </ s> </ s>其中Q超过所有风险中性概率测度的M类对于型号M。
（f）（MATH3975）查找X的对冲价格，即最小存在投资组合（x，φ）的初始end赋x不平等V1（x，φ）（ω）≥（S1（ω）− 4）+每个ω∈Ω成立。
Exercise 2 Consider a single-period market model M = (B, S) on the sample space Ω = {ω1, ω2, ω3}. Assume that the savings account equals B0 = 1, B1 = 1.1 and the stock price equals S0 = 5 and S1 = (S1(ω1), S1(ω2), S1(ω3)) = (7.7, 5.5, 4.4).
The real-world probability P is such that P(ωi) > 0 for i = 1, 2, 3.
(a) Find the class M of all martingale measures for the model M. Is this market model complete?
(b) Show that the claim X = (X(ω1), X(ω2), X(ω3)) = (5.5, 3.3, 2.2) is attainable and compute its arbitrage price π0(X) using the replicating strategy for X.
(c) Consider the contingent claim Y = (3, 1, 0). Show that the expected value
EQYB1
does not depend on the choice of a martingale measure Q ∈ M. Is this claim attainable?
(d) Consider the contingent claim Z = (4.4, 0, −3.3). Find the range of arbitrage prices
π0(Z) = EQZB1
where Q ∈ M. Is this claim attainable?
(e) Find the unique martingale measure Qe for the extended model Mf =(B, S1, S2) in which S1 = S and the risky asset S2is defined as the claim Z traded at its initial price π0(Z) = −0.5, that is, S20 = −0.5 and S2
1 = Z. Is the market model Mf complete?
Exercise 3 (MATH3975) Let Ω = {ω1, ω2}. We consider a single-period model M = (S1, S2) with two risky assets with prices S1 and S2 given by S1
0 = s0 > 0, S2
0 = z0 > 0 and S11(ωi) = si, S21(ωi) = zifor i = 1, 2 where 0 < s1 < s2 and 0 < z1 < z2. There are two traded assets,S1 and S2, so the wealth of a strategy φ equals Vt(φ) = φ1tS1t + φ2tS2tfort = 0, 1.
It should be stressed that the existence of the savings account B is not
postulated. Hence the process B should not be used at all in your solution.