#Nathan Fox
#Homework 13
#Please Do NOT Post!

#Read packages
with(`LinearAlgebra`):

#Read procedures from class
read(`C13.txt`):

#Help procedure
Help:=proc():
    print(` Problem1() , Problem2() , FindStateVector(S0, D1) `):
    print(` ArrowDebreu(D1, i) , Problem5() `):
end:

##PROBLEM 1##

#Scenario:
#At time 0,
#
#The price of a Microsoft stock is 1000
#The price of a Google stock is 2000
#The price of a Yahoo stock is 1500
#
#There is also a Government bond avialable with interest rate
#(compounded continuously) 3 percent per annum.
#
#After exactly one year, the economy is predicted to be in one of
#four scenarios, with the following stock prices (in dollars) in
#each possible scenario
#
#Scenario A: Microsoft: 700 ; Google: 1700; Yahoo:1200;
#Scenario B: Microsoft: 900 ; Google: 2010; Yahoo:1550;
#Scenario C: Microsoft: 1100 ; Google: 2300; Yahoo:1600;
#Scenario D: Microsoft: 1500 ; Google: 2800; Yahoo:1900;
#
#The seller promises the buyer to have a choice of either buying a
#Microsoft stock with strike price 1200, a Google stock with strike
#price 2300, or a Yahoo stock with strike price 1700. The buyer of
#the option can only do one of these, so naturally he would pick
#the most lucrative one.
Problem1:=proc() local S0, D1, C:
    S0:=[1000, 2000, 1500, 1]:
    D1:=[[700, 900, 1100, 1500],
        [1700, 2010, 2300, 2800],
        [1200, 1500, 1600, 1900],
        [exp(0.03)$4]]:
    C:=[0, 50, 300, 800]:
    return OptionPrice(S0, D1, C), HedgingPortfolio(D1, C):
end:
#This returns a price of 372.7025517, and a hedging portfolio of
#[3.305555556, -1.111111111, -0.8888888889, 622.7025507]

##PROBLEM 2##

#Same scenario as problem 1, but now the buyer can exercise any
#number of strike purchases
Problem2:=proc() local S0, D1, C:
    S0:=[1000, 2000, 1500, 1]:
    D1:=[[700, 900, 1100, 1500],
        [1700, 2010, 2300, 2800],
        [1200, 1500, 1600, 1900],
        [exp(0.03)$4]]:
    C:=[0, 60, 500, 1700]:
    return OptionPrice(S0, D1, C), HedgingPortfolio(D1, C):
end:
#This returns a price of 1122.298229, and a hedging portfolio of
#[11.08333333, -5.666666667, -1.333333333, 3372.298229]
#The price went up

##PROBLEM 3##

#FindStateVector(S0, D1): inputs a list, S0, of current stock
#prices (as above), a list of lists of possible stock prices, D1,
#(with n=N) and outputs a state-price vector if it exists, and
#otherwise returns FAIL.
ScenD1:=[[700, 900, 1100, 1500],
        [1700, 2010, 2300, 2800],
        [1200, 1500, 1600, 1900],
        [exp(0.03)$4]]:
ScenS0:=[1000, 2000, 1500, 1]:

FindStateVector:=proc(S0, D1)
    return MatrixVectorMultiply(MatrixInverse(Matrix(D1)), Vector(S0)):
end:
#What the state-price-vector in the above data is
#<0.829, 1.617, -2.946, 1.470>

##PROBLEM 4##

#ArrowDebreu(D1, i): inputs a list of lists D1 (with n=N) as above,
#and an integer i between 1 and n, and outputs the ith Arrow-Debreu
#security defined on p. 11.
ArrowDebreu:=proc(D1, i):
    return MatrixInverse(Transpose(Matrix(D1)))[i]:
end:
#The A-D securities for the above scenarios are
#[0.0102777777777778, 0.00555555555555554, -0.0313888888888888, 0.0155555555555555]
#[-0.00555555555555553, -0.0111111111111111, 0.0277777777777777, -0.0111111111111111]
#[-0.00444444444444445, 0.0111111111111111, -0.00777777777777776, 0.00111111111111111]
#[8.32965749591560, 1.61740922250788, -15.4462580749503, 6.46963689003152]

##PROBLEM 5##

Problem5:=proc() local S0, D1, C, s0, su, sd, K, r, T:
    s0:=160:
    su:=200:
    sd:=140:
    K:=180:
    T:=0:
    S0:=[1, s0]:
    D1:=[[exp(r*T), exp(r*T)], [su, sd]]:
    C:=[20, 0]:
    return OptionPrice(S0, D1, C):
end:
#The option price should be 20/3 Euros