Solutions
1.
a. What is the defining characteristic of the binomial model, which also gives its name?
There are only two possibilities for the future price: to go up or down (hence the
name binomial model)
[10%].
b. Which of the following are not assumed about the binomial model: ”no bid-ask
spread”, ”infinite fungibility”, ”iterated conditioning”, ”utility maximisation”, ”no
storage cost”?
Iterated conditioning and utility maximisation.
[20%]
c. Describe briefly how a replicating portfolio can be used to price a financial derivative
of European type.
Set up a replicating portfolio by starting with an amount X0 and investing in the
stock and the bank account at each node on the tree so that at maturity we end
up with the same amount as the payoff of the European derivative (the replicating
condition). Then the no-arbitrage price of the option has to be equal to X0 . [30%]
d. Compute the time 0 price of a European put option with strike price 6 and maturity
three in a three-period binomial model. Assume the interest rate is zero, the initial
stock price is S0 = 8, and with each head the stock price increases by 1, and with
each tail the stock price decreases by 2 (so S1 (H) = 9, S1 (T ) = 6, S2 (HH) = 10,
etc).
[40%]
0
11
0
10
9
S0 = 8
1
9
8
V0 =
7
6
10
27
2
1
3
3
8
9
5
0
3
2
3
1
1
2
4
3
2
1
4
Use V3 = (K − S3 )+ to obtain the last column in the second tree.
Sn+1 (. . . H) = Sn + 10 = un Sn , so un = 1 +
Sn+1 (. . . T ) = Sn − 20 = dn Sn , so dn = 1 −
10
Sn
20
Sn
Risk-neutral probabilities are independent of n:
1
[10%]
(up factor).
[5%]
(down factor).
[5%]
p̃n =
1+rn −dn
un −dn
= 32 , q̃n =
1
3
(since rn = 0).
[10%]
Fill the rest of the tree using: Vn = Ẽn [Vn+1 ] (rn = 0),
get V0 =
2.
10
27
= 0.3703.
[10%]
In a binomial asset model the price of stock starts at time t = 0 with S0 = 8 and each
time it doubles (i.e. moves up by a factor u = 2) with probability p = 2/3 or it halves
(i.e. moves down by a factor d = 1/2) with probability q = 1/3. Let the interest rate be
r = 1/4 and the risk-neutral probabilities be p̃ = q̃ = 1/2 (they are known, you do not
have to compute them).
Consider an American call option expiring at time 2 with strike price K = 10.
a. Compute the intrinsic value process and display it in a tree diagram.
Gn = Sn − 10
[20%]
[10%]
22
22
6
Gn : −2
8.8
−2
Vn : 3.52
0
−6
0
−8
0
Values and diagram
[10%]
b. Compute the price process and display it in a tree diagram.
[30%]
VN = (SN − K)+ gives: V2 (HH) = 22, V2 (HT ) = V2 (T H) = V2 (T T ) = 0.
[10%]
1
American alg.: Vn = max{Sn − K, 1+r
[p̃Vn+1 (. . . H) + q̃Vn+1 (. . . T )]}
[10%]
Computing V1 (H) = 8.8, V1 (T ) = 0, V0 =
88
25
= 3.52 and diagram.
c. Give the formula for and compute the optimal exercise time.
Optimal exercise time: τ ∗ = min{n : Gn = Vn }.
Computing τ ∗ (HH) = 2, τ ∗ (HT ) = ∞, τ ∗ (T H) = ∞, τ ∗ (T T ) = ∞.
[10%]
[20%]
[10%]
[10%]
d. Without doing any calculation what do you expect about the price of a European
call option with the same strike and maturity, and why?
[30%]
The price should be 3.52. It is never optimal to exercise the American call option
before maturity. Since the American call option brings no benefits over the European
call option, they should have the same price.
2
3.
Let Mn be a symmetric random walk, and let the stock price be modelled by:
Sn := S0 · 2Mn , ∀n ≥ 0.
a. Define a symmetric random walk.
[20%]
The process M0 , M1 , M2 , . . . is a symetric random walk if: M0 = 0 and Mn =
n
X
Xj
j=1
for n ≥ 1, where Xj = 1 with probability
1
2
and Xj = −1 with probability 12 .
b. Find the value of r such that the discounted stock price process
a martingale under the real-world probability P .
Sn
depends on the first n tosses −→ adapted.
(1 + r)n
Sn
, n ≥ 0 is
(1 + r)n
[30%]
[10%]
Note that Sn+1 = Sn · 2Xn+1 . Martingale condition (1-step ahead):
En (
Sn+1
Sn · 2Xn+1
)
=
E
(
)
n
(1 + r)n+1
(1 + r)n+1
Sn
En (2Xn+1 ) (taking out what’s known)
=
(1 + r)n+1
Sn
E(2Xn+1 ) (independence)
=
(1 + r)n+1
1
1
Sn
Sn
(2 + 2−1 ) =
=
n+1
(1 + r)
2
2
(1 + r)n
The last equality holds iff r = 1/4, so the process is a martingale iff r = 1/4. [20%]
c. Under the risk-neutral probability a discounted American option price process is a
supermartingale, so it cannot be used for pricing. How can this process be transformed into a martingale that can then be used for pricing?
[20%]
By stopping it at the optimal stopping time.
d. Use the independence lemma to show that the process S0 , S1 , S2 , . . . is a Markov
process, (i.e. show that for any nonnegative integer n and any arbitrary function
f (·) there exists another function g(·) such that En [f (Sn+1 )] = g(Sn ), where En [X]
is the conditional expectation of X based on the information at time n).
Construct the function g(·) explicitly. (It may depend on n and f).
[30%]
From b. we know that Sn+1 = Sn ·2Xn+1 , so: f (Sn+1 ) = f (Sn ·2Xn+1 ) = f1 (Sn , Xn+1 ),
where f1 (s, x) = f (s · 2x ) (here f1 depends on f ).
[10%]
Sn depends on tosses 1, 2, . . . , n and Xn+1 depends on toss n + 1 we can apply independence lemma: ∃g1 (s) = E(f1 (s, Xn+1 )) s.t. En (f1 (Sn , Xn+1 )) = g(Sn ). [10%]
Take the equation above and derive g(s) explicitly: g(s) = E(f1 (s, Xn+1 )) = E(f (s·
2Xn+1 )) = f (2s) 21 + f (s/2) 12 .
[5%]
3
We have shown ∃g(x) s.t. En (f (Sn+1 )) = En (f1 (Sn , Xn+1 )) = g(Sn ), so the process
has the Markov property.
[5%]
4
4.
Consider a two-period binomial model with S0 = 8, up factor u = 2, down factor d = 0.5
and interest rate r = 0. Assume the probability that the stock price moves up is p = 32 .
a. Does this binomial model satisfy the no-arbitrage condition? Justify your answer.
Check d < 1 + r < u, or 0.5 < 1 < 2, yes.
[10%]
b. Compute the actual probabilities: P (HH), P (HT ), P (T H), P (T T ), and the riskneutral probabilities: P̃ (HH), P̃ (HT ), P̃ (T H), P̃ (T T ).
[20%]
Multiply probabilities on each branch
4
2
2
1
P (HH) = , P (HT ) = , P (T H) = , P (T T ) =
9
9
9
9
1+r−d
1 + 0 − 0.5
1
2
p̃ =
=
= , q̃ =
u−d
2 − 0.5
3
3
2
4
1
P̃ (HH) = , P̃ (HT ) = P̃ (T H) = , P̃ (T T ) =
9
9
9
[5%]
[5%]
[5%]
[5%]
c. Compute the Radon-Nikodym derivative process Zn , n = 0, 1, 2, and display it in a
tree diagram.
[20%]
Zn = En (Z)
[5%]
1
Z(HH) = , Z(HT ) = Z(T H) = 1, Z(T T ) = 4
4
1
Z1 (H) = E1 (Z)(H) = pZ(HH) + qZ(HT ) =
2
Z1 (T ) = E1 (Z)(T ) = pZ(T H) + qZ(T T ) = 2
2
1
4 1 2
Z0 = E(Z) = · + · 1 + · 1 + · 4 = 1
9 4 9
9
9
Tree diagram
Z2 = Z =
P̃ (ω)
,
P (ω)
[5%]
[2%]
[2%]
[4%]
[2%]
d. Compute the state price densities: ζ2 (ω), ω ∈ {HH, HT, T H, T T }.
[10%]
Z
1
ζ2 =
[6%], ζ2 (HH) = , ζ2 (HT ) = ζ2 (T H) = 1, ζ2 (T T ) = 4 [4%]
2
(1 + r)
4
e. Using the numbers computed in part d. find the time-zero price of an Asian option
whose payoff at time 2 is
( 31 Y2
+
− 8) , where Yn =
n
X
Sk is the sum of stock prices
k=0
between time 0 and time n.
V0 = E[ζVN ] =
X
[40%]
ζ(ω)VN (ω)P (ω) (here ζ = ζ2 )
[10%]
ω∈Ω
1
32
8
V2 = ( Y2 − 8)+ : V2 (HH) = , V2 (HT ) = , V2 (T H) = V2 (T T ) = 0
3
3
3
1 32 4
8 2
2
1
48
16
V0 = ·
· +1· · +1·0· +4·0· =
= .
4 3 9
3 9
9
9
27
9
5
[10%]
[20%]
5.
a. Define the following terms:
i. interest-rate swap
ii. forward contract
iii. m-period swap rate
iv. interest rate caplet
[40%]
i. = contract paying Sn at time 1 ≤ n ≤ m, with Sn = K − Rn−1
ii.= contract agreed today to buy at time m asset Sm for fixed price K
iii.= value of K that makes the price of the interest rate swap at time 0 equal to 0
iv.= contract that makes a single payment (Rn−1 − K)+ at time n
b. In the Black-Derman-Toy model the interest rate at time 0 ≤ n ≤ N is given by:
1 ω2 ...ωn )
Rn (ω1 ω2 . . . ωn ) = an · b#H(ω
,
n
where the power of bn is the number of heads in the path ω1 ω2 . . . ωn (i.e. the number
0.05
of up steps). Consider a two-period model (N = 2) with an =
, bn = 1.44. The
1.2n
risk-neutral probabilities are p̃ = q̃ = 1/2.
i. Consider at time n = 0 the zero-coupon bonds with maturities n = 1 and
n = 2. Compute the prices of these bonds at times n = 0 and n = 1. [30%]
ii. Compute the price at time n = 0 of the two-period interest rate cap with
K = 0.05.
[30%]
i. R0 = 0.05, R1 (H) = 0.05 · 1.441 = 0.072, R1 (T ) = 0.05 · 1.440 = 0.05.
1
1
Recall D0 = 1, D1 = 1+R
and Bn,m = D1n Ẽn [Dm ]. Compute:
, D2 = (1+R0 )(1+R
0
1)
1
1
ω1
D1
D2
P̃
1+R0
1+R1
1
1
1
1
H 1.05 = 0.95238 1.072 = 0.932835 1.05 1.05·1.072 = 0.888415 21
1
1
1
1
T 1.05
= 0.95238
= 0.95238
= 0.907029 21
1.05
1.05
1.05·1.05
Using D0 = 1 we have: B0,1 = Ẽ[D1 ] = D1 = 0.95238,
B0,2 = Ẽ[D2 ] = D2 (H) 12 + D2 (T ) 21 = 0.897722, B1,1 = 1
D2 (H)
B1,2 (H) = D11(H) Ẽ1 [D2 ](H) = D
= 1+R11 (H) = 0.932835
1 (H)
B1,2 (T ) =
1
Ẽ [D2 ](T )
D1 (T ) 1
=
D2 (T )
D1 (T )
=
1
1+R1 (T )
+
= 0.95238
[15%]
[15%]
+
ω1 R0 (R0 − K)
R1
(R1 − K)
ii. Use the table: H 0.05
0
0.072
0.022
T 0.05
0
0.05
0
Cap = Caplet0 + Caplet1
[5%]
+
caplet0 = Ẽ[D1 (R0 − K) ] = 0
[10%]
+
+
Caplet1 = Ẽ[D2 (R1 − K) ] = D2 (H)(R1 (H) − K) 0.5 + D2 (T )(R1 (T ) − K)+ 0.5
= 0.907029 · 0.022 · 0.5 + 0 = 0.009977319
[10%]
Thus: Cap = 0.009977319.
[5%]
6