MATH3075 Assignment 1: Solutions
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MATH3075 Assignment 1: Solutions
1. Single-period market model [12 marks] Consider a single-period market model M = (B, S) on a finite sample space Q = {^1,^2,仙}. We assume that the money market account B equals Bo = 1 and Bi = 4 and the stock price S satisfies So = 2.5 and Si = (18,10,2). The real-world probability P is such that P(她)=p乞 > 0 for i = 1, 2, 3.
(a) Find the class M of all martingale measures for the model M. Is the market model M arbitrage-free? Is this market model complete?
Answer: [2 marks] We need to solve: qi + % + 如=1, 0 < q乞 < 1 and (since 1 + r = 4)
Eq(Si) = 18qi + 10q2 + 2q3 = (1 + r)So = 10
or, equivalently,
Eq(Si - (1 + r)So) = 8qi 8q3 = 0.
Let q2 = a. Then qi = q3 = where 0 < a < 1. Hence
M = {(qi,q2,qs) | qi = q3 = , q2 = a, 0 < a < 1}.
The market model M is arbitrage-free since M = 0. Moreover, it is incomplete since the uniqueness of a martingale measure for M fails to hold.
(b) Find the replicating strategy (90,9O) for the claim X = (5,1, -3) and compute the arbitrage price no(X) at time 0 through replication.
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Answer: [2 marks] First solution. The wealth process of a portfolio (^甘，#o)satisfies for t = 0,1
4旎 + 18^o = 5,
4旎 + 10^o = 1,
4旎 + 2°o = -3.
We obtain (况,9o) = (—1, 0.5) and thus no(X) = ^Bo + ^So = —1 + 0.5(2.5) = 0.25. Hence at time 0 we buy 0.25 shares of stock. For this purpose, after receiving 0.25 units of cash from the buyer of the claim X, we borrow one unit of cash in the money market.
Second solution. Alternatively, one may use a portfolio (x, 9) e R2 and represent the wealth as follows: K(x,9)= x and
Vi (x, 9) = (x — 9So)(1 + r) + 9Si = x(1 + r) + 9(Si — So(1 + r)) = xBi + 9(Si — S°Bi).
Then we solve the following equations
4x + 89 = 5,
4x + 09 = 1, 4x — 89 = —3.
From the second equation, we obtain no(X) = x = 0.25 and thus 9 = 0.5.
(c) Compute the arbitrage price n°(X) using the risk-neutral valuation formula with an arbi trary martingale measure Q from M.
Answer: [2 marks] For any 0 < q2 = a < 1 and qi = q3 = , the risk-neutral valuation
formula yields
1 / 1 — a 1 — a \
no(X) = Eq(X/Bi)=项5 ―a + a 3 ― ) = 0.25.
(d) Show directly that the contingent claim Y = (Y(wi),Y(s),Y(仙))=(10, 8, —2) is not at tainable, that is, no replicating strategy for Y exists in M.
Answer: [2 marks] To find a replicating strategy, we need to solve the following equations
4妒 + 1891 = 10,
4妒 + 1091 = 8,
4妒 + 2寸1 = —2.
The strategy (", 91) = (*,扌)is a unique solution to the first two equations, but it fails to satisfy the last one. Hence no replicating strategy for Y exists in M.
(e) Find the range of arbitrage prices for Y using the class M of martingale measures for the model M.
Answer: [2 marks] We compute the range of prices for Y consistent with the no-arbitrage principle. We have
1 / 1 — a 1 — a\
no (Y) = Eq(Y/Bi) = —(10 +8a 2 ) =1 + a.
Since from part (c) we know that a e (0,1), it is clear the range of prices no(Y) consistent with the no-arbitrage principle is the open interval (1, 2).
(f) Suppose that you have sold the claim Y for the price of 3 units of cash. Show that you may find a portfolio (x, 9) with the initial wealth x = 3 such that Vi (x, 9) > Y.
Answer: [2 marks] It suffices to give any example of a portfolio (x, 9) with the initial value x = 3 such that Vi(x, 9)(她)> Y(她)for i = 1, 2, 3. We may use the representation of the wealth at time t = 1
Vi (x, 9) = (x — 9So)(1 + r) + 9S1 = x(1 + r) + 9(Si — So(1 + r)) = xBi + 9(Si — SoBi).
Since x = 3 and Bi = (1 + r) = 4 so that SoBi = 10, it is enough to find a number 9 e R such that the following inequalities are satisfied
12 + 89〉10,
12 + 09 > 8,
12 — 89 > —2.
For instance, we may take 9 = 1, that is, buy one share of stock at time 0. Then the wealth of our portfolio at time t = 1 will be Vi (3,1) = (20,12,4) so itis clear that V1(3,1)(她)> Y (她) for i = 1, 2, 3.
2. Static hedging with options [8 marks] Consider a parametrised family of European con tingent claims with the payoff X(L) at time T given by the following expression
X(L) = min(2|K — + K — ST, L)
where a real number K > 0 is fixed and L is an arbitrary real number such that L > 0.
(a) Sketch the profile of the payoff X(L) as a function of the stock price St and find a decompo sition of X(Z/) in terms of terminal payoffs of standard call and put options with expiration date T. Notice that the decomposition of X(L) may depend on the value of the variable L.
Answer: [2 marks]
It is now easy to see that for 0 < L < 3K
X(L) = 3Pt(K) - 3Pt(K — |L)+ Ct(K) - C.(K + L)
and for L > 3K X0) = 3Pr(K) + Ct(K) - S(K + L).
Notice that the first decomposition above gives X0) = 0 when L = Q and they coincide when L = 3K.
(b) Assume that call and put options are traded at time 0 at finite prices. For each value of L > 0, find a representation of the arbitrage price t?o(X(Z/)) of the claim X0) at time Z = 0 in terms of prices at time 0 of call and put options using the decompositions from part (a).
Answer: [2 marks] By the additivity property of arbitrage pricing, we obtain, for every 0 < L < 37f,
7ro(X0)) = 3F0(K) - 3Fo(K -捉)+ Co(K) - C°(K + L) (1)
and, for every L > 3K,
在o(X0)) = 3R(K) + C°(K) - C°(K + L). (2)
In particular, we deduce from (1) that 7ro(X(L)) = 0 when L = 0, which is obvious since X(L) = 0 when L = 0.
(c) Consider any arbitrage-free market model M = (B, S) defined on some finite state space Q. Show that the arbitrage price of X(L) at time t = 0 is a monotone function of the variable L 2 0 and compute the limits limLo3K no(X(L)), limLoo no(X(L)) and limLoo no(X(L)) using the representations from part (b).
Answer: [2 marks] We observe that the payoff X(L) increases when L increases. Specifi cally, if we consider the payoffs X(Li) and X(L2) corresponding to Li and L2, respectively, where Li < L2 then it is clear that X(Li) < X(L2). Consequently, the price no(X(L)) is a nondecreasing function of the variable L.
Moreover, using (1) and (2)
£虬no(X(L)) = 3Po(K) + Co(K) - Co(4K),
Furthermore, limLoo no(X(L)) = 0 since limLoo X(L)(w) = 0 for all 心 e Q and thus
0 < lim sup Eq(B-1X(L)) < lim max(B-1X(L)(s)) = 0. L—o QeM l—o
(d) For any L > 0, examine the sign of an arbitrage price of the claim X(L) in any (not nec essarily complete) arbitrage-free market model M = (B, S) defined on some finite state space Q. Justify your answer.
Answer: [2 marks] Since the payoff X (L) is strictly positive for every value of St (except for St = K) the price no(X(L)) should be strictly positive in any arbitrage-free market model M = (B, S) since otherwise an arbitrage opportunity would arise in the extended market model. You may use the argument that the range of prices for any contingent claim X(L) coincide with the range of values of the expectation Eq(B-1Y) when Q runs over the class M of all martingale measures for the model M.
2021-11-28