MATH3075/3975 Financial Derivatives
Tutorial 3: Solutions
Exercise 1 Let (x, ϕ) be the replicating strategy for the contingent claim X = g(S1) = S1, that
is,
V1(x, ϕ) := (x− ϕS0)(1 + r) + ϕS1 = S1.
More explicitly, for every ωi ∈ Ω = {ω1, ω2},
V1(x, ϕ)(ωi) = (x− ϕS0)(1 + r) + ϕS1(ωi) = S1(ωi).
The unique solution reads: ϕ = 1 and
x =
1− ϕ
1 + r
S1(ω) + ϕS0 = S0.
Alternatively, we may argue that if a contingent claim X is offered at a price different from S0 then
an arbitrage opportunity arises. Hence the unique fair value of the claim X = S1 at time 0 equals
S0.
Exercise 2 We fix K > 0 and we write C = C(K) and P = P (K). We start by observing that
CT − PT = (ST −K)+ − (K − ST )+ = ST −K. (1)
The arbitrage price π0 is an additive map on the space of contingent claims, so that
C0 − P0 = π0(CT )− π0(PT )
additive
= π0(CT − PT )
(1)
= π0(ST −K)
additive
= π0(ST )− π0(K) = S0 − (1 + r)−1K,
since we know that π0(ST ) = S0 and π0(K) = (1+r)
−1K. Alternatively, we may use the risk-neutral
valuation formula
C0 − P0 = π0(CT − PT ) = EP̃
(
CT − PT
1 + r
)
= EP̃
(
ST −K
1 + r
)
= EP̃
(
ST
1 + r
)
− EP̃
(
K
1 + r
)
= S0 − (1 + r)−1K.
Exercise 3 We consider the options defined in Examples 2.1.1 and 2.1.2. Recall that T = 1 (say,
one month) and S0 = K = 1 (say, one AUD).
• For the call option, we need to solve
V1(x, ϕ) := (x− ϕS0)(1 + r) + ϕS1 = C1 = (S1 −K)+,
or, more explicitly, {
4
3
(x− ϕ) + 2ϕ = 1,
4
3
(x− ϕ) + 1
2
ϕ = 0.
We obtain (x, ϕ) = (5/12, 2/3) so that
x− ϕS0 = 5/12− 8/12 = −3/12,
meaning that after selling (aka “writing” or “issuing”) the call option at the price 5/12, in
order to buy 2/3 shares of the stock S at time 0, we need to borrow 3/12 units of cash. Notice
that here we buy shares and we borrow cash at time 0.
1
• For the put option, we solve
V1(x, ϕ) := (x− ϕS0)(1 + r) + ϕS1 = P1 = (K − S1)+,
that is, {
4
3
(x− ϕ) + 2ϕ = 0,
4
3
(x− ϕ) + 1
2
ϕ = 1
2
.
We now obtain (x, ϕ) = (1/6,−1/3). Note that x−ϕS0 = 1/6+1/3 = 1/2 meaning that after
selling the put option at the price 1/6 and short-selling of 1/3 share of the stock at time 0,
we invest 1/2 units of cash in the savings account. Notice that here we short sell shares and
we lend cash at time 0.
A. Comments on replication. We deal with two independent questions:
a) how to replicate the payoff C1 of the call option?
b) how to replicate the payoff P1 of the put option?
This means that we need to find the initial wealth of each replicating strategy as well as the number
of shares we need to buy or short sell to replicate a given option. The initial wealth should not be
confused with our “initial endowment”, which is assumed to be null. The respective answers for
the call and put are:
a) x = 5/12 (initial wealth) and ϕ = 2/3 (hence we buy shares),
b) x = 1/6 (initial wealth) and ϕ = −1/3 (hence we short sell shares).
Notice that the initial wealth of replicating portfolio gives the fair price (also called the arbitrage
price) for the option we sell, that is, C0 = 5/12 and P0 = 1/6.
B. Comments on the put-call parity. Recall that the put-call parity in a single-period case
states that C0 − P0 = S0 −K(1 + r)−1. Since we have found that C0 = 5/12 (from part (a)) and
P0 = 1/6 (from part (b)) so that
LHS = C0 − P0 = 5/12− 1/6 = 3/12 = 1/4
and the right-hand side equals
RHS = S0 −K(1 + r)−1 = 1− 1(1 + 1/3)−1 = 1− 3/4 = 1/4.
This confirms that our approach to the valuation and hedging problems (a) and (b) is consistent
with the put-call parity. However, we cannot deduce solutions to (a) or (b) above from the put-call
parity.
C. Comments on the concept of hedging. A hedging strategy (or a hedging portfolio) is a pair
(x, ϕ) and our goal was to “hedge” the risk of our short position in either a call and put option in
case (a) and (b), respectively. If we sell either a call or a put option, then our transactions can be
summarised as follows:
1. at time 0, we sell an option at some initial price,
2. then, also at time 0, we hedge the risk exposure by establishing a replicating portfolio com-
posed of our (either positive or negative) positions in shares and cash in the bank account
(or, equivalently, bonds),
3. at time T , we first liquidate (unwind) our hedging portfolio and,
4. then, again at time T , we settle the contract by delivering the payoff of the option to the holder.
This corresponds to the case of cash settlement; the case of physical delivery is somewhat
different in practice, but mathematically equivalent.
2
Liquidate means to convert assets into cash or cash equivalents by selling them on the open
market. Since the wealth at time T is equal to our liability (the amount of cash we have to pay
to the option holder), we see that we will be always left with nothing. In particular, there will be
no loss from the option contract, meaning that we are perfectly “hedged” (you may wish to say
“protected,” “sheltered,” “covered” or “insured”) against an eventual loss at time T , but also we
will never make any profit. This means that arbitrage pricing hinges on a complete elimination of
profits and losses, as opposed to speculation where agents hope to make profits, but also run the
risk of a loss.
It is important to observe that analogous arguments can be applied to the buyer (holder) of
the option and thus fair valuation is bilateral: the unique amount of cash the seller is prepared
to accept for the option is equal to the amount of cash the buyer is ready to pay. This property
justifies why we can use the name: fair pricing.
Exercise 4 We assume that r = 1
4
, S0 = 1, u = 3, d =
1
3
, p = 4
5
and we consider the digital call
option with the following payoff
g(S1) =
{
1, if S1 ≥ K,
0, otherwise.
The unique risk-neutral probability measure P̃ = (P̃(ω1), P̃(ω2)) = (p̃, 1− p̃) where
p̃ =
1 + r − d
u− d
=
1 + 1
4
− 1
3
3− 1
3
=
11
32
.
Hence the price at time 0 of the digital call option equals
π0(g(S1)) = EP̃
(
g(S1)
1 + r
)
= 4
5
EP̃
(
g(S1)
)
=
4
5
, if K ≤ dS0 = 13 ,
4
5
· 11
32
= 11
40
, if 1
3
< K ≤ 3,
0, if K > uS0 = 3,
since
K ≤ dS0 ⇒ g(S1) = 1,
dS0 < K ≤ uS0 ⇒ g(S1) = 1{S1=3},
K > uS0 ⇒ g(S1) = 0.
Note that the value of p = 4
5
was not used in the computation of the price.
Exercise 5 Assume that d < 1 + r < u and consider any trading strategy of the form (0, ϕ). Then V1(0, ϕ) := −ϕS0(1 + r) + ϕS1, so that, for every ωi ∈ Ω = {ω1, ω2}, V1(0, ϕ)(ωi) = ϕ ( S1(ωi)− S0(1 + r) ) . Since S1(ω1) = uS0 and S1(ω2) = dS0, we obtain{ V1(0, ϕ)(ω1) = ϕS0 ( u− (1 + r) ) , V1(0, ϕ)(ω2) = ϕS0 ( d− (1 + r) ) , where u− (1 + r) > 0 and d− (1 + r) < 0. 3 • If we take ϕ = 0 then, obviously, V1(0, ϕ)(ωi) = 0 for i = 1, 2. • If ϕ 6= 0 then V1(0, ϕ)(ω1) and V1(0, ϕ)(ω2) have the opposite signs, namely, V1(0, ϕ)(ω1)V1(0, ϕ)(ω2) = ϕ 2S20 ( u− (1 + r) )( d− (1 + r) ) < 0. We conclude that there is no strategy (0, ϕ) satisfying Definition 2.1.3 of an arbitrage opportunity and thus the elementary market model is arbitrage free if d < 1 + r < u. Furthermore, if either 1 + r ≤ d or 1 + r ≥ u then is is easy to see that arbitrage opportunities exist – it suffices to consider either buying (when 1 + r ≤ d) or short-selling (when 1 + r ≥ u) of the stock S. Exercise 6 Note that ϕ0 = x−ϕ1S0 is the amount invested in the savings account B. The initial wealth of our portfolio is thus given by x = V0(x, ϕ) = ϕ 0 +ϕ1S0 and the portfolio’s wealth at time t = 1 equals V1(x, ϕ) = (x− ϕ1S0)(1 + r) + ϕ1S1 = ϕ0(1 + r) + ϕ1S1. (a) For the call option with the payoff C1 = (S1 − 28.5)+, the unique replicating strategy (ϕ0, ϕ1) can be found by solving the following equations:[ 1.1 31 1.1 28 ] [ ϕ0 ϕ1 ] = [ 2.5 0 ] =⇒ (ϕ0, ϕ1) = (−700/33, 5/6). For the put option with the payoff P1 = (28.5 − S1)+, the unique replicating strategy (ϕ0, ϕ1)is computed through the following system of linear equations:[ 1.1 31 1.1 28 ] [ ϕ0 ϕ1 ] = [ 0 0.5 ] =⇒ (ϕ0, ϕ1) = (155/33,−1/6). (b) Using part (a), we can compute the price of the put and the call. We know that the initial endowment x of the replicating portfolio is given by x = ϕ0 + ϕ1S0. Hence the prices of the put and call options are equal to C0 = −700/33 + (5/6)27 = 85/66, P0 = 155/33− (1/6)27 = 13/66. (c) On the one hand, we have C0 − P0 = 85/66− 13/66 = 72/66 = 12/11. On the other hand, we obtain S0 − (1 + r)−1K = 27− 28.5/1.1 = (297− 285)/11 = 12/11. Hence the put call-parity relationship holds. (d) Recall that P̃(ω1) = 1 + r − d u− d , P̃(ω2) = 1− P̃(ω1). We have u = 31/27 and d = 28/17. Hence the unique risk-neutral probability measure P̃ (also called the martingale measure) is given by P̃(ω1) = 1.1− 28 27 31 27 − 28 27 = 17 30 , P̃(ω2) = 1− P̃(ω1) = 13 30 . 4 The arbitrage prices of options can be computed using the risk-neutral valuation formula. The price of the call option satisfies C0 = EP̃ ( C1 1 + r ) = 1 1 + r EP̃ ( (S1 −K)+ ) = 1 1.1 17 30 2.5 = 85 66 . and price of the put option equals P0 = EP̃ ( P1 1 + r ) = 1 1 + r EP̃ ( (K − S1)+ ) = 1 1.1 13 30 0.5 = 13 66 . (e) Let us write r̂ = 0.05. The new replicating strategy (ϕ0, ϕ1) for the call option is given by[ 1.05 31 1.05 28 ] [ ϕ0 ϕ1 ] = [ 2.5 0 ] =⇒ (ϕ0, ϕ1) = (−200/9, 5/6). For the put option, we obtain[ 1.05 31 1.05 28 ] [ ϕ0 ϕ1 ] = [ 0 0.5 ] =⇒ (ϕ0, ϕ1) = (310/63,−1/6). The new price of the call is Ĉ0 = −200/9 + (5/6)27 = 5/18 = 35/126 < 85/66 = C0 and the new price of the put equals P̂0 = 310/63− (1/6)27 = 53/126 > 13/66 = P0.
Hence the price of the call for the interest rate r̂ = 5% is lower than for the interest rate r = 10%,
but the price of the put option is higher when the interest rate is lower. This is due to the fact
that we borrow cash when we replicate the call option, but we put cash into money market account
when we replicate the put option. As expected, the put-call parity still holds:
Ĉ0 − P̂0 = −18/126 = −1/7
and
S0 −K(1 + r̂)−1 = 27− 28.5/1.05 = (2835− 2850)/105 = −15/105 = −1/7.
Exercise 7 (MATH3975) From Proposition 2.1.4, we know that the arbitrage price x of any con-
tingent claim X = g(S1) in the elementary market model satisfies
π0(g(S1)) = EP̃
(
g(S1)
1 + r
)
where P̃ is the unique risk-neutral probability measure. We note that the probability measures P̃
and P are equivalent and thus the Radon-Nikodym density L of P̃ with respect to P is well defined,
specifically,
L(ωi) =
dP̃
dP
(ωi) =
P̃(ωi)
P(ωi)
=
{
p̃
p
, i = 1,
1−p̃
1−p , i = 2.
Consequently, if a trading strategy (x, ϕ) replicates the claim X = g(S1) then
x = π0
(
g(S1)
)
= EP̃
(
g(S1)
1 + r
)
= EP
(
L
g(S1)
1 + r
)
= EP
(
Zg(S1)
)
where we set Z := (1 + r)−1L. The pricing kernel Z can thus be used to price any contingent claim
using the real-world probability P in the elementary market model.
5
Exercise 8 (MATH3975) For S0 = 100.23, we observed the following mid-prices of European call
and put options on JPMorgan stock on 1 September (for the most current market data, enter JPM
as a stock symbol on http://www.cboe.com/delayedquote/quote-table)
Call C0(K) Strike K Put P0(K)
$3.95 $98 $2.19
$3.65 $99 $2.45
$3.12 $100 $2.91
$2.65 $101 $3.42
$2.23 $102 $4.02
If r = 0, d > 1 and 0 < d = u−1 < 1, then d < 1 + r < u and thus the martingale measure P̃ is well defined and p̃ satisfies p̃ = 1 + r − d u− d = 1− d u− d = 1− u−1 u− u−1 = u− 1 u2 − 1 = 1 u+ 1 . Hence, if the strike K satisfies dS0 < K < uS0, then C0(K) = p̃(uS0 −K) = uS0 −K u+ 1 . (2) If S0 = 100.23 and K = 100, then we obtain the following equalities C0(K) = uS0 −K u+ 1 = 100.23u− 100 u+ 1 = 3.12. It now suffices to compute u and thus also p̃ and apply (2) in order to find call prices for other strikes. We have u = 1.0619 and p̃ = 0.485. Therefore, the model price of the call with strike K = 99 equals C0(99) = p̃(uS0 −K) = 0.485(106.43− 99) = 3.60 Similarly, for K = 101 we obtain C0(101) = p̃(uS0 −K) = 0.485(106.43− 101) = 2.63. For the put option, we use the equality P0(K) = q̃(K − dS0). (3) where d = u−1 = 0.9417 and q̃ = 1− p̃ = 0.515. For instance, for K = 100 we obtain P0(100) = q̃(K − dS0) = 0.515(100− 94.3866) = 2.89. 6