Model Solutions to Homework Set 3
Math 456
Topics in Financial Mathematics
Prof. Wickerhauser
Read Chapters 5 and 6 of the textbook, "Binomial Models In Finance" by
John van der Hoek and Robert J. Elliott
NOTE: When asked to produce a spreadsheet, you may instead implement
the model in Octave or another system. For full credit you must to
translate the algorithm into a computer program and produce output
from several examples. Include your code
Do the following exercise from the textbook Chapter 5.2, p.88:
Exercise 5.3.
See the file "ch5eg51.ods" on the class website for a LibreOffice
spreadsheet for Example 5.1 on page 83 of the textbook.
Do the following exercises from the textbook Chapter 6.3, p.96:
Exercise 6.7.
See the file "ch6eg63.ods" on the class website for a LibreOffice
spreadsheet for Example 6.3 on page 92 of the textbook.
Exercise 6.8. Generalize to N>2 and prove the result in a
model-independent manner from the no arbitrage axiom.
First derive some general results valid for all N>1:
Since P(N,j)=1 for all j, use pi=pi(N-1,j) to compute
P(N-1,j) = [pi*P(N,j+1) + (1-pi)*P(N,j)]/R(N-1,j)
= [ pi*1 + (1-pi)*1 ]/R(N-1,j)
= 1/R(N-1,j)
Since G(N,j)=S(N,j) for all j, compute
G(N-1,j) = [pi*G(N,j+1) + (1-pi)*G(N,j)]
= [pi*S(N,j+1) + (1-pi)*S(N,j)]
= R(N-1,j)*[pi*S(N,j+1) + (1-pi)*S(N,j)]/R(N-1,j)
= R(N-1,j)*S(N-1,j)
Likewise F(N,j)=S(N,j) for all j, so compute
F(N-1,j) = S(N-1,j)/P(N-1,j)
= R(N-1,j)*S(N-1,j)
= G(N-1,j)
Now fix N=2 and suppose that (R(1,1)-R(1,0))*(S(1,1)-S(1,0))>0.
Without loss we may assume that R(1,1)>R(1,0)>0 and S(1,1)>S(1,0)>0,
since these quantities must be positive and since we may relabel the
up and down states. But then with the above equations we have
G(1,1) = R(1,1)*S(1,1) > R(1,0)*S(1,0) = G(1,0)
Reset pi=pi(0,0) to compute
G(0,0) = pi*G(1,1) + (1-pi)*G(1,0)
Finally, by backward induction with that same pi compute
F(0,0) = S(0,0)/P(0,0)
= [pi*S(1,1)+(1-pi)*S(1,0)]/[pi*P(1,1)+(1-pi)*P(1,0)]
= [pi*S(1,1)+(1-pi)*S(1,0)]/[pi/R(1,1)+(1-pi)/R(1,0)]
= [pi*G(1,1)/R(1,1)+(1-pi)*G(1,0)/R(1,0)]/[pi/R(1,1)+(1-pi)/R(1,0)]
= X*G(1,1) + (1-X)*G(1,0),
where
X = [pi/R(1,1)] / [pi/R(1,1)+(1-pi)/R(1,0)]
= pi*R(1,0) / [pi*R(1,0)+(1-pi)*R(1,1)]
< pi, since the denominator is between R(1,0) and R(1,1)>R(1,0),
and so
1-X > 1-pi.
Note that X,(1-X) is a probability function with a lesser weight X1-pi for G(1,0). But since
G(1,1)>G(1,0)>0, it follows that
F(0,0) = X*G(1,1) + (1-X)*G(1,0) < pi*G(1,1) + (1-pi)*G(1,0) = G(0,0),
so G(0,0) > F(0,0), as claimed.
Finally, consider the following generalization to N>2:
Theorem. If N>2 and (R(N-1,j+1)-R(N-1,j))*(S(N-1,j+1)-S(N-1,j))>0 for
all j={0,1,...,N-2}, then G(0,0)>F(0,0).
Proof.
Apply the N=2 result to conclude that G(N-2,j)>F(N-2,j) for all j. Now
suppose that F(0,0)>=G(0,0). Then there is an arbitrage:
At time 0, sell one forward contract at strike price F(0,0) and buy
one future contract at strike price G(0,0). Total expense is $0.
At time N-2, in whatever state j, the long margin account will
contain (by the calculation on p.94 of the textbook)
M(N-2,j) = M(0,0)+Interest + [G(N-2,j)-G(0,0)]
> M(0,0)+Interest + [F(N-2,j)-F(0,0)]
since G(N-2,j) > F(N-2,j) and G(0,0) =< F(0,0) by hypothesis.
Thus, we may take some of this money to buy one share of the
underlying S for the price
S(N-2,j) = F(N-2,j)*P(N-2,j) < F(N-2,j).
At time N, deliver the share of S to the forward contract
counterparty for F(0,0). The futures contract is cash-settled
within the margin account and we need not take delivery.
If these transactions are all done through the margin account, then
after time N it contains
M(0,0)+Interest +[G(N-2,j)-G(0,0)]-[S(N-2,j)-F(0,0)]
> M(0,0)+Interest +[G(N-2,j)-G(0,0)]-[F(N-2,j)-F(0,0)]
= M(0,0)+Interest +[G(N-2,j)-F(N-2,j)] + [F(0,0)-G(0,0)]
> M(0,0)+Interest
with the surplus being the arbitrage profit.
Conclude by the no-arbitrage axiom that F(0,0) < G(0,0).
NOTE: As will be seen in Exercise 6.9, there will be an arbitrage
opportunity if F(0,0)>G(0,0) in any case. The hypothesis for this
case implies that F(0,0)=G(0,0) also provides an arbitrage
opportunity.
Exercise 6.9.
Suppose that F(0,0) > G(0,0). Then an arbitrage opportunity exists in
all cases:
Construct a contingent claim at time 0 by selling one F(0,0) and
buying one G(0,0). This costs nothing, as we assume that we already
have a margin account. Suppose that initially there is M(0,0) in our
margin account. Then at expiry time N in state j, the margin account
will contain:
M(0,0)+Interest + [G(N,j)-G(0,0)] =
= M(0,0)+Interest + [S(N,j)-G(0,0)] % since S(N,j)=G(N,j)
> M(0,0)+Interest + [S(N,j)-F(0,0)]. % since F(0,0)>G(0,0)
Thus we have the funds to buy one share of S at S(N,j), deliver it to
the forward contract counterparty, and thereby get paid F(0,0). We
need not take delivery from the futures contract as it is cash-settled
within the margin account, which now contains
M(0,0)+Interest+[G(N,j)-G(0,0)]-[S(N,j)-F(0,0)] =
= M(0,0)+Interest+[F(0,0)-G(0,0)]
> M(0,0)+Interest
The surplus F(0,0)-G(0,0) is the profit from the arbitrage.
NOTE 1: The "Interest" portion is path-dependent and may be less than
the interest that would have been earned on M(0,0) without deductions
of principal taking place along the path. This "opportunity cost"
plus the "default risk cost" of the forward contract may in practice
eliminate the expected arbitrage profit in this general case.
NOTE 2: If risk-free return rates R(n,j) are deterministic, namely
independent of state j, then by Theorem 6.5 the fair price for the
forward contract is equal to the fair price for the futures contract.
Hence F(0,0)>G(0,0) means that some party is offering mispriced assets
and there is an arbitrage opportunity:
At time 0, sell one forward contract at F(0,0) and buy one futures
contract at G(0,0). Total investment is 0, if there is already a
margin account, say containing M(0,0).
At time N, the bond matures to yield G(0,0) which, together with
[S(N,j)-G(0,0)] from the margin account (which now contains
M(0,0)+Interest+[G(N,j)-G(0,0)]=M(0,0)+Interest+[S(N,j)-G(0,0)]),
allows us to take delivery of one share of S at the strike price
G(0,0). Deliver this to the forward contract counterparty in
exchange for F(0,0).
This clears all obligations and leaves the margin account containing
M(0,0)+Interest, plus there is a surplus F(0,0)-G(0,0) > 0.
This argument is similar to the proof of Theorem 1.2 on p.2, the Law
of One Price.
NOTE 3. If the risk-free return factors R and asset prices S satisfy
(R(N-1,j+1)-R(N-1,j))*(S(N-1,j+1)-S(N-1,j)) > 0, all j,
then by the argument in Exercise 6.8 we have G(N-2,j) > F(N-2,j) in
all states j, so the contingent claim is an arbitrage opportunity
whenever F(0,0) >= G(0,0) since it costs 0 when entered at time 0 and
adds to the margin account a profit
[G(N-2,j)-G(0,0)]-[F(N-2,j)-F(0,0)] =
= [G(N-2,j)-F(N-2,j)] + [F(0,0)-G(0,0)] > 0
in all states j when it is liquidated at time N.
NOTE 4: Exercise 6.9 provides a hint for the solution of Exercise 6.8.