Math 225
November 17, 2004
Homework 8 (revised) (Due Mon., 11/17)
1. Consider the two-branch model shown here. In the context of
the arbitrage theorem, there are n=1 investment and m=2
states, and the profit matrix is
100
80
50
R = (rij) = ( +20, -30 ).
The theorem says that there is either a vector (x1) that defines an arbitrage opportunity, or a
probability vector (p1, p2) that “explains” the price of the stock.
(“Arbitrage opportunity”—in the case with just one row—means that x1r11>0 and x1r12>0. A probability
vector p1,p2, which must satisfy pj≥0 and p1+p2=1, “explains” the stock price if p1r11+p2r12=0. If the matrix had
more than one row, the last equation would have to hold for every row.)
Which is it? If the latter, what is the vector (p1, p2) ?
2. Let’s try the same thing with three branches: All prices are
flat dollars, and the stock price (now 50) might change at
the next time step to 60, 50, or 40.
Now n=1, m=3, and
R = ( +10, 0, -10 ).
60
50
50
40
a. Is there a vector (p1, p2, p3) that explains the stock price ? (Find an explicit vector.)
b. A certain call option pays 10 on the top branch, 0 on the other branches. Its price at
time 0 is C. If you rely on the probability vector you found in part a, what is C ?
The profit from C would be (10-C) on the top branch, (-C) on the others. So, C would have to satisfy
p1(10-C)+p2(-C)+p3(-C)=0 for your selected probability vector. Put another way, C would have to
equal 10p1+0p2+0p3.
c. Find a different probability vector from the one in part a, that also explains the stock
price.
d. If you rely on the vector in part c, what is C ?
A market model that is consistent with more than one probability vector is called incomplete. (If it is
consistent with a unique probability vector it’s complete. If it is not consistent with any probability vector,
then we reject it because it must allow arbitrages.) We have just seen that the three-branch model is
incomplete, although the two-branch model is complete.
e. Does that mean that any value of C at all can be justified using this stock price model?
If not, what are the largest and smallest values of C that you can justify with
probability vectors?
(continued next page)
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f. Suppose that in fact, the market price of the option is C=4. What probability vector is
the market using?
With the new option, there are two investments, and R is
( +10 0 -10 )
( +6 -4 -4 ).
But now there is a unique probability vector—the market is now complete—so if you found another
option contract, you would be able to determine a unique value for it.
3. Consider the following model for the price of HPQ shares during the next three months:
(1) Time is measured in months starting now, so the time interval covered is t in [0, 3].
(That is, T = 3.)
(2) S(0) = 20.0
(3) At the end of each month, price is multiplied by u = 1.10 or by d = 0.90. The first
occurs with probability p=0.60, and the latter with probability 1-p=0.40
(independently for each time step.) So, S(1) can be 20u=22.0 or 20d=18.0, S(2)
can be 20u2 or 20ud or 20d2, and S(3) can be any of four specific values.
_____
a. Draw the appropriate tree. Show the stock price at each node and show the probability
of each branch.
b. What are the four possible values of S(3)? What are the probabilities that each will
occur? (It might help to work out the probability of each of the 8 possible routes
through the tree.)
c. What is E(S(3)) ?
d. If p were 0.50 instead of 0.60, what would be E(S(3)) ?
e. An option pays $10 at time 3 if the price of HPQ is above 26.0, and zero otherwise. Let
X be a random variable that represents the payoff at time 3 (so, X is 10 or 0). What
is E(X), using p = 0.60 ?
f. What is E(X), using p = 0.50 ?
Our theory is that the value of the option is equal its expected value using the risk-neutral probabilities,
regardless of the actual probabilities. In this case, the risk-neutral probabilities are all p=0.50 (well, they’re
all based on p=0.50), so the “actual” value of p=0.60 is irrelevant to the value of the option. Your answer to
part f is the value of the option, no matter what you think about p, or there would be an arbitrage opportunity.
(end)
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