Solutions to Homework Set 1
Math 456
Topics in Financial Mathematics
Prof. Wickerhauser
In these solutions, "plus(X)" refers to the positive part of X:
plus(X) = max(X,0) = { X, if X>0, else 0 }
Thus plus(1)=1 but plus(-99)=0.
Also, PDF files mentioned here are viewable and downloadable from the
class website at links near the link to this solutions file.
Exercise 2.36.
In Example 2.10 show that 0 ≤ pi ≤ 1. The solution is contained in
the text so this exercise merely asks you to write it out.
Equation 2.11 gives
pi = [RS(0)-S(1,down)]/[S(1,up)-S(1,down)]
We have chosen the up and down states so that Equation 2.2 holds:
S(1,up) > S(1,down)
Hence the denominator S(1,up)-S(1,down) is positive. Also from Equation 2.2,
S(1,up) > RS(0) > S(1,down)
so by subtracting S(1,down) from all three terms we get
S(1,up)-S(1,down) > RS(0)-S(1,down) > 0
and then dividing by the positive quantity S(1,up)-S(1,down) gives
1 > pi=[RS(0)-S(1,down)]/[S(1,up)-S(1,down)] > 0
Exercise 2.39. Leave K as a parameter in your graph and your answer.
See the graph at hw1ex239.pdf
Conclude that the profit will be positive if S(T)>K+C(0).
Exercise 2.40. Again, leave K as a parameter in your graph and your answer.
See the graph at hw1ex240.pdf
Conclude that the profit will be positive if S(T) < K-P(0). For the
AOL/AUG03/16.00/PUT example with K=16.00 and premium P(0)=0.45, the
profit will be positive if, at T (the third Friday of August, 2003)
the price of a share of AOL is less than 16.00-0.45 = 15.55.
Buying this Put option insures the holder of AOL shares against loss
if the share price drops below 16.00, and even pays back part or all
of the premium if the share price drops below 15.55.
Exercise 2.41.
See the graph at hw1ex241.pdf
Let S(T) be the stock price at expiry (T = 3 months). Let X be the
share purchase asset and Y be the options asset. Then the values
of the two strategies are:
[X] Purchase 100 shares for S(0)=$94:
X(0) = 100*S(0) = $9400,
Xprofit = X(T)-X(0) = 100*S(T)-$9400.
[Y] Purchase 2000 contracts with K=$95 for C(0)=$4.70
Y(0) = 2000*C(0) = 2000*$4.70 = $9400
Yprofit = Y(T)-Y(0) = 2000*plus(S(T)-K)-$9400
Profit and loss from either strategy is determined by the price S(T).
If Xprofit and Yprofit are plotted together as functions of S(T), as
in hw1ex241.pdf, their graphs will intersect at S(T)=0, where
Yprofit=Xprofit=-$9400, and again at S(T)=$100, Yprofit=Xprofit=+$500.
Solve for the right-hand intersection point, which will occur at some
S(T)>K, using the equation
0 = Yprofit-Xprofit = 2000*plus(S(T)-K)-100*S(T) =
= 2000*(S(T)-K)-100*S(T) = 1900*S(T)-2000*K,
so S(T) = 2000*K/1900 = 2000*$95/1900 = $190000/1900 = $100.
For 0~~$100 the Y strategy will yield higher profit with the
difference increasing rapidly as S(T) increases.
If one believes that S(T)>$100 is likely, then one should choose
option strategy Y to obtain a higher expected profit.
Exercise 2.43.
Claim: Unless S(T+)=S(T-)-D, there is an arbitrage opportunity.
Proof: Otherwise, construct an arbitrage (of type one) as follows:
If S(T+) > S(T-)-D, then
buy S at time T- : -S(T-)
collect D at time T: +D
sell S at T+: +S(T+)
----------------------------
net: S(T+)+D-S(T-) > 0
Else S(T+) < S(T-)-D, so
sell S short at time T- : +S(T-)
pay out D at time T: -D
buy S to cover at T+: -S(T+)
----------------------------
net: S(T-)-D-S(T+) > 0
In either case there is profit with no risk. Here we assume that any
borrowing is cost-free since T+, T-, and T are nearly equal. Note
that the borrower of shares sold short is responsible for paying the
dividend to the lender.
Exercise 2.44.
See the graph at hw1ex244.pdf
Without loss of generality, we may assume that the indices are chosen
such that K10 or else there would be
an arbitrage opportunity (of type two in this case, since only some
states of the world give positive profit).
Solving for C2(0), conclude that
C2(0) = (C1(0)+C3(0))/2 - C(0)/2 < (C1(0)+C3(0))/2
Exercise 2.45.
Octave commands:
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Use the given values:
Call = 3.00; Put=1.00; S0 = 10.50; K=10.00; R=1.0043;
Call-Put % left hand side is the call-put difference, ans = 2.00
S0-K/R % right-hand side is the value given by CRR, ans = 0.54282
Call-Put-S0+K/R % difference, ans = 1.4572
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Thus C(0)-P(0) > S(0)-K/R, so the parity formula is violated with
C(0)-P(0)-S(0)+K/R = 1.4572 > 0
as in Equation 2.39 on page 28 of the textbook. Exploit this by
constructing the following arbitrage, following the argument after
that equation:
At t=0,
short sell 1000 Call options to raise 1000*C(0) = $3000
buy 1000 Put options at a cost of -1000*P(0) = -$1000
borrow PV(1000K) = 1000*K/R = $9957.20
buy 1000 shares at spot for a cost of -1000*S(0) = -$10,500
This leaves a net of 3000-1000+9957.20-10500 = $1457.20, which we pocket.
At t=T (expiry),
cash settle the 1000 calls at a cost of -plus(S(T)-K)*1000
realize the value of to 1000 puts for plus(K-S(T))*1000
sell the 1000 shares of stock for S(T)*1000
repay the loan (with interest) for -K*1000
This leaves no unfunded liabilities, since for any X,
-plus(-X) + plus(X) = X,
so
-plus(S(T)-K) + plus(K-S(T)) + S(T)-K = 0,
and 1000*0 = 0.
We thus realize a type one arbitrage profit of more than $1000 using no
money of our own.
NOTE: 687 or more units would meet our $1000 profit goal.
Exercise 2.46.
Octave commands to compute the weighted linear regression:
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
N=5; % how many strike prices
xi=[17.50,18.00, 18.50, 19.00, 19.50]; % strike prices, from Table 2.1
yi=[0.36,0.19,0.12,0.06,0.06]; % C(0) premiums, from Table 2.1
ni=[171, 316, 475, 802, 594]; % open interest, from Table 2.1
n=sum(ni); % total open interest
wi=ni/n; % weights: proportional open interest
% Perform a weighted linear regression to find the best fit line y=m*x+c:
xbar=wi*xi'; ybar=wi*yi'; xybar=wi*(xi.*yi)'; xxbar=wi*(xi.*xi)';
m=(xybar-xbar*ybar)/(xxbar-xbar*xbar); m % best fit slope m = -0.12375
c=ybar-m*xbar; c % best fit y-intercept c = 2.4357
% Use the best fit line to estimate Call premium y=C(0) at strike x=K=17.00
K=17.00; C = m*K+c; C % Call premium estimate is C(0) = 0.33185
% Plot the data points in black and the best-fit line in blue:
plot(xi,yi,"k"); hold on; plot(xi,m*xi+c,"b");
% ...to see that a straight line misrepresents the relationship.
% Use the call-put parity formula C(0)-P(0)=S(0)-K/R
% to estimate P(0) = C(0)+K/R-S(0) from the estimated C(0):
S=16.96; % spot price S(0)
R=1.0042; % risk-free return factor to be used
P=C+K/R-S; P % estimated Put premium is P(0) = 0.30075
% This is very different from the market price P(0) = 0.55
% ...which confirms that the straight line approximation is misleading.
% Compute risk-neutral probability pi (call it pr) from m=-pr/R
pr=-R*m; pr, 1-pr % pr = 0.12427, so 1-pr=0.87573
% Compute up factor u from c=pr*u*S/R
u=c*R/(S*pr); u % up factor u = 1.1605
% Compute down factor d from d=(R-pr*u)/(1-pr)
d=(R-pr*u)/(1-pr); d % down factor d = 0.98202
%% EXTRA: repeat the estimates of C(0) and P(0) at K=17 using pr,u,d,
% Compute estimated P(0) from the CRR model as in Eq.2.37, p.27:
(1-pr)*(K-d*S)/R % ans = 0.30075
% Compute estimated C(0) from the CRR model as in Eq.2.33, p.26:
pr*(u*S-K)/R % ans = 0.33185
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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