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MATH3075 Assignment 2: Solutions
1. CRR model: American call option. [5 marks] Assume the CRR model with T = 2, the stock
price S0 = 45, S1u = 49.5, S1d = 40.5 and the interest rate r = 0.05. Consider the American call
option with the reward process g(St, t) = (St Kt)+ for t = 0, 1, 2 where the random strike price
satisfies K0 = 40, K1(ω) = 35.5 for ω ∈ {ω1, ω2}, K1(ω) = 38.5 for ω ∈ {ω3, ω4} and K2 = 36.45.
(a) Find the parameters u and d, compute the stock price at time t = 2, and find the unique
martingale measure P .
Answer: We have
u =
49.5
45
= 1.1, d =
40.5
45
= 0.9.
Hence the stock price equals
S0 = 45, S1 = (49.5, 40.5), S2 = (54.45, 44.55, 36.45).
Since u = 1.1, d = 0.9 and r = 0.05, we obtain
p =
1 + r d
u d =
0.95 0.9
1.1 0.9 =
0.05
0.2
= 0.25.
(b) Compute the price process Ca for this option using the recursive relationship
Cat = max
{
(St Kt)+, (1 + r) 1 EP
(
Cat+1 | Ft
)}
with the terminal condition Ca2 = (S2 K2)+.
Answer: We have
Ca2 = (S2 K2)+ =
(
(54.45 36.45)+, (44.55 36.45)+, (36.45 36.45)+) = (18, 8.1, 0).
Consequently,
Cau1 = max
(
(49.5 35.5)+, (0.95) 1(0.25 · 18 + 0.75 · 8.1))
)
= max (14, 11.132) = 14,
Cad1 = max
(
(40.5 38.5)+, (0.95) 10.25 · 8.1
)
= max (2, 2.132) = 2.132,
Ca0 = max
(
(45 40)+, (0.95) 1(0.25 · 14 + 0.75 · 2.132)
)
= max (5, 5.367) = 5.367.
(c) Find the rational exercise time τ 0 for the holder of this option.
Answer: Exercise the option at time t = 1 if the stock price rises during the first
period. Otherwise, do not exercise before time 2. Hence
τ 0 (ω) = 1 for ω ∈ {ω1, ω2},
τ 0 (ω) = 2 for ω ∈ {ω3, ω4}.
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(d) Find the issuer’s replicating strategy for the option up to the rational exercise time τ 0
and show that the wealth of replicating strategy matches the price computed in part (b).
Answer: At time t = 0, we need to solve
0.95 00 + 49.5
1
0 = 14,
0.95 00 + 40.5
1
0 = 2.132.
Hence ( 00, 10) = ( 53.975, 1.319).
If the stock price rises during the first period, the option is exercised at time t = 1. There-
fore, there is no need to compute the strategy at time 1 for ω ∈ {ω1, ω2}. If the stock price
declines during the first period, we need to solve
0.95 01 + 44.55
1
1 = 8.1,
0.95 01 + 36.45
1
1 = 0.
Hence ( 01, 11) = ( 38.368, 1) if the stock price has declined during the first period, that is,
for ω ∈ {ω3, ω4}.
We will check that the wealth of the replicating strategy matches the price computed in
part (b).
V0( ) = 53.989 + 1.319 · 45 = 5.366.
At time 1, if the price declined during the first period, that is, for ω ∈ {ω3, ω4}
V1( )(ω) = 38.368 + 1.319 · 40.5 = 2.132.
(e) Compute the profit of the issuer at time T if the holder decides to exercise the option at
time T .
Answer: If the option is not exercised by its holder at time t = on the event A1 = {ω1, ω2}
then the profit of the issuer at time T = 2 is given by
(1 0.05)(14 11.132) = 0.95 · 2.868 = 2.7246.
The issuer’s profit on the event A 2 = {ω3, ω4} equals zero.
2. Black-Scholes model: European claim [5 marks] We place ourselves within the setup
of the Black-Scholes market modelM = (B,S) with a unique martingale measure P . Consider
a European contingent claim X with maturity T and the following payoff
X = max (K,ST ) LST
where K = erTS0 and L > 0 is an arbitrary constant. We take for granted the Black-Scholes
pricing formulae for the call and put options.
(a) Sketch the profile of the payoff X as a function of the stock price ST at time T and show
that X admits the following representation
X = K + CT (K) LST
where CT (K) denotes the payoff at time T of the European call option with strike K.
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Answer: We assume that the stock price is nonnegative and thus we need to examine
the function g : lR+ —+ JR given by
g(Sr) = max(K,Sr) – LSr
where L > 0 is a strictly positive constant.
If Sr SK then max (K, Sr)= Kand thus g(Sr) = K – LSr. If Sr 2′. K then max (K, Sr)=
Sr and thus g(Sr) = (1- L )Sr. We thus see that g is a piecewise linear continuous function
with g(O) =Kand its derivative satisfies g'(Sr) =-Lon (O,K) and g'(Sr) = 1- L on
( K, oo). It is easy to sketch the graph of the function g : lR+ —+ JR. by considering the
following cases: (a) 0 < L < 1, (b) L = 1 and (c) L > 1. (a) (b)
K (c) 0 LL 1 K L ==- I ,_I-::: 0 ….J_— ============–7 ST
It is clear that
max (K,ST ) = K + (ST K)1{ST>K} = K + (ST K)+ = K + CT (K)
and thus
X = max (K,ST ) LST = K + CT (K) LST = X1 +X2 +X3
(b) Find an explicit expression for the arbitrage price pit(X) at time 0 ≤ t < T in terms of Ft := ertS0, St and S0. Then compute the price pi0(X) in terms of S0 and use the equality N(x) N( x) = 2N(x) 1 to simplify your result. Answer: For every 0 ≤ t < T , we have pit(X) = pit(X1) + pit(X2) + pit(X3) where pit(X1) = pit(K) = BtB 1 T K = e r(T t)K = e r(T t)erTS0 = ertS0 = Ft and pit(X3) = pit( LST ) = LSt For X2 = CT (K) we use the Black-Scholes formula pit(X2) = pit(CT (K)) = Ct(K) = StN(d+(St, T t)) Ke r(T t)N(d (St, T t)) where K = erTS0. Notice that Ke r(T t) = erTS0e r(T t) = S0ert = Ft and d+(St, T t) = ln(St/K) + (r + 0.5σ 2)(T t) σ √ T t = ln(St/K) + ln(e r(T t)) + 0.5σ2(T t) σ √ T t = ln(Ste r(T t)/K) + 0.5σ2(T t) σ √ T t = ln(Ste r(T t)/S0erT ) + 0.5σ2(T t) σ √ T t = ln(St/Ft) + 0.5σ 2(T t) σ √ T t so that d (St, T t) = d+(St, T t) σ √ T t = ln(St/Ft) 0.5σ 2(T t) σ √ T t . We conclude that pit(X) = Ft + Ct(K) LSt = Ft + StN ( ln(St/Ft) + 0.5σ 2(T t) σ √ T t ) FtN ( ln(St/Ft) 0.5σ2(T t) σ √ T t ) LSt. Hence for t = 0 we obtain pi0(X) = S0 + S0N ( 0.5σ √ T ) S0N( 0.5σ√T ) LS0 = S0(2N(0.5σ√T ) L). 4 (c) Find the limit limT→0 pi0(X). Answer: Since lim x→0 N(x) = N(0) = 0.5, we have that lim T→0 N ( 0.5σ √ T ) = 0.5 and thus lim T→0 pi0(X) = lim T→0 [ S0 ( 2N ( 0.5σ √ T ) L)] = S0(2 lim T→0 N ( 0.5σ √ T ) L) = (1 L)S0. (d) Find the limit limσ→∞ pi0(X). Answer: Since lim x→∞ N(x) = 1, we have that lim σ→∞N ( 0.5σ √ T ) = 1 and thus lim σ→∞pi0(X) = limσ→∞ [ S0 ( 2N ( 0.5σ √ T ) L)] = S0(2 lim σ→∞N ( 0.5σ √ T ) L) = (2 L)S0. (e) Explain why the price of pi0(X) is positive for L = 1 by analysing the payoff X when L = 1. Answer: [2 marks] For L = 1, the payoff X admits the following representation X = PT (K) = (K ST )+ (see the graph (b)) and thus the claim X represents one long position in the put option with strike K. Hence pi0(X) = P0(K) > 0 since the price of a put option
in the Black-Scholes model is strictly positive for every strike K > 0.
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