Solutions
1.
Toss a fair coin repeatedly (fair means that the probability of each side is 0.5). Let Xj = 1
if the j-th toss is head and Xj = −1 if the j-th toss is tail. Consider the stochastic process
M0 , M1 , M2 , . . . defined by M0 = 0 and Mn =
n
X
Xj for n ≥ 1.
j=1
a. Show that M0 , M1 , M2 , . . . is a martingale.
[20%]
Mn = X0 + . . . + Xn depends on the first n tosses −→ adapted.
[8%]
Martingale condition (1-step ahead):
En (Mn+1 ) = En (Mn + Xn+1 ) = En (Mn ) + En (Xn+1 ) (linearity)
= Mn + E(Xn+1 ) (taking out what’s known + independence)
= Mn + 1 · 0.5 + (−1) · 0.5 = Mn .
[12%]
Therefore the process is a martingale.
b. Let σ and k be positive constants and define Sn = k n eσMn . Find the constant k
such that the process S0 , S1 , S2 , . . . is a martingale.
[20%]
Sn = k n eσMn depends on the first n tosses via Mn −→ adapted.
[8%]
Check the martingale condition (1-step ahead):
En (Sn+1 ) = En (k n+1 eσMn+1 ) = k n+1 En (eσ(Mn +Xn+1 ) )
= k n+1 eσMn En (eσXn+1 ) (taking out what’s known)
= k n+1 eσMn E(eσXn+1 ) (independence)
= k n+1 eσMn (eσ · 0.5 + e−σ · 0.5)
This is equal to Sn = k n eσMn iff:
k=
c. Define I0 = 0 and In =
n−1
X
eσ
2
.
+ e−σ
[12%]
Mk (Mk+1 − Mk ) for n ≥ 1. Show that:
k=0
1
n
In = Mn2 − .
2
2
1
[20%]
Induction over n (direct method also accepted):
n=1:
[5%]
I1 = M0 (M1 − M0 ) = 0(M1 − M0 ) = 0
1 2 1
1
1
M1 − = X12 − = 0 since X12 ≡ 1,
2
2
2
2
so the two are equal.
n→n+1:
[5%]
1
n
In+1 = In + Mn (Mn+1 − Mn ) = ( Mn2 − ) + Mn Xn+1
2
2
1
n
2
= ( (Mn+1 − Xn+1 ) − ) + (Mn+1 − Xn+1 )Xn+1
2
2
n 1 2
1 2
= Mn+1 − − Xn+1 ,
2
2 2
2
= 1).
and the result follows by induction (use the fact that Xn+1
[10%]
d. Show that the process I0 , I1 , I2 , . . . 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 (In+1 )] = g(In ), 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).
[40%]
From:
1 2
n+1
1
n+1
f (In+1 ) = f ( Mn+1
−
) = f ( (Mn + Xn+1 )2 −
)
2
2
2
2
we see that f (In+1 ) is a function f1 of Mn and Xn+1 , i.e. f (In+1 ) = f1 (Mn , Xn+1 ),
where f1 (m, x) = f ( 21 (m + x)2 − n+1
) (here f1 depends on f and n).
[5%]
2
Since Mn depends on tosses 1, 2, . . . , n and Xn+1 depends on toss n+1 we can apply
independence lemma:
[5%]
∃g1 (m) = E(f1 (m, Xn+1 )) s.t.: En (f1 (Mn , Xn+1 )) = g1 (Mn ).
n+1
1
Take equation above: g1 (m) = E(f1 (m, Xn+1 )) = E(f ( (m + Xn+1 )2 −
))
2
2
1
n+1 1
1
n+1 1
= f ( (m + 1)2 −
) + f ( (m − 1)2 −
) .
2
2 2
2
2 2
From this and part c. we obtain g1 (Mn ) = f (In + Mn ) 21 + f (In − Mn ) 12 , and since
√
g1 (Mn ) = g1 (−Mn ) replacing Mn = ± 2In + n gives g1 (Mn ) = g(In ), with g(·)
√
√
given explicitly by: g(x) = f (x + 2x + n) 12 + f (x − 2x + n) 21 .
[15%]
We have shown ∃g(x) s.t. En (f (In+1 )) = En (f1 (Mn , Xn+1 )) = g1 (Mn ) = g(In ), so
the process has the Markov property.
[15%]
2
2.
a. Define the following terms:
i. payoff of an option
ii. European put option
iii. American put option
iv. perpetual American put option
[40%]
i. = amount of money earned when exercising the option
ii. = contract paid for at time 0 giving the owner the right, but not the obligation, to sell 1 share of the stock for a fixed price K at maturity time
iii. = contract paid for at time 0 giving the owner the right, but not the
obligation, to sell 1 share of the stock for a fixed price K at any time up to
and including the maturity time
iv. = American put option with infinite maturity (N = ∞)
b. Suppose the initial stock price is S0 = 80, with each head the stock price increases by
10, and with each tail the stock price decreases by 20. In other words, S1 (H) = 90,
S1 (T ) = 60, S2 (HH) = 100, etc. Assume the interest rate is always zero.
Consider a European put option with strike price 60, expiring at time three. What
is the price of this put at time zero? (Hint: compute first the up and down factors,
and then the risk-neutral probabilities.)
[60%]
0
110
0
100
90
S0 = 80
10
9
80
V0 =
70
60
100
27
2
10
3
80
9
50
0
3
1
3
2
3
10
20
40
1
20
3
40
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
(up factor).
[5%]
(down factor).
[5%]
Risk-neutral probabilities are independent of n: p̃n =
rn = 0).
1+rn −dn
un −dn
Fill the rest of the tree using: Vn = Ẽn [Vn+1 ] (rn = 0), get V0 =
3
[10%]
= 23 , q̃n =
100
27
1
3
(since
[10%]
= 3.703. [30%]
3.
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 agent who borrows V0 at time 0 and buys an American put option expiring
at time 2 with strike price K = 10. To pay back the loan the agent trades in the stock
and the money markets and optimally exercises the put. (By doing this the agent is
actually setting up a replicating portfolio.)
a. Compute the price V0 = v0 (S0 ) of this option at time 0.
[30%]
VN = (K − SN )+ gives: V2 (HH) = 0, V2 (HT ) = V2 (T H) = 2, V2 (T T ) = 8. [10%]
1
[p̃Vn+1 (. . . H) + q̃Vn+1 (. . . T )]}
American alg.: Vn = max{K − Sn , 1+r
Computing V1 (H) = 0.8, V1 (T ) = 6 and V0 =
68
25
= 2.72
b. Compute the optimal exercise time for the holder of this option.
Optimal exercise time: τ ∗ = min{n : Gn = Vn }.
−22
[30%]
[10%]
0.8
2
2.72
6
2
6
8
Gn
[10%]
0
−6
2
[10%]
8
Vn
Tree diagrams of intrinsic value Gn = K − Sn and option price Vn
Computing τ ∗ (HH) = ∞, τ ∗ (HT ) = 2, τ ∗ (T H) = 1, τ ∗ (T T ) = 1
c. What assets are in the replicating portfolio and what is replicated?
Replicating portfolio: American put option, stock, money markets (bank)
Growth in the bank (loan) is replicated.
[10%]
[10%]
[20%]
[10%]
[10%]
d. How much is worth the agent’s position in the stock and money market at time 1
if the first stock move was down?
[20%]
If first stock move is down agent optimally exercises put and gets 6.
[10%]
This combined with the stock and money markets allows the agent to offset the loan
V0 (1 + r) = 3.4, so the stock and money market worth is 3.4 − 6 = −2.6. [10%]
4
4.
Consider a two-period binomial model with S0 = 4, 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. 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
P (HH) = 49 , P (HT ) = 29 , P (T H) =
p̃ =
1+r−d
u−d
[5%]
2
9
, P (T T ) =
1
9
[5%]
= 0.5 = q̃
[5%]
P̃ (HH) = P̃ (HT ) = P̃ (T H) = P̃ (T T ) = 0.25
[5%]
b. Compute the Radon-Nikodym derivative random variable Z.
Z(ω) =
P̃ (ω)
P (ω)
Z(HH) =
[10%]
[6%]
9
, Z(HT )
16
= Z(T H) = 98 , Z(T T ) =
9
4
[4%]
c. Compute the Radon-Nikodym derivative process Zn , n = 0, 1, 2, and display it in a
tree diagram.
[20%]
Zn = En (Z)
[5%]
Z2 = Z computed in part b.
[5%]
Z1 (H) = E1 (Z)(H) = pZ(HH) + qZ(HT ) =
4
9
·
9
16
+ 29 · 98 + 92 · 98 + 19 ·
9
4
[2%]
3
2
[2%]
=1
[4%]
Z1 (T ) = E1 (Z)(T ) = pZ(T H) + qZ(T T ) =
Z0 = E(Z) =
3
4
Tree diagram
[2%]
d. Compute the state price densities: ζ2 (ω), ω ∈ {HH, HT, T H, T T }.
Z
ζ2 =
(1 + r)2
ζ2 (HH) =
9
,
25
ζ2 (HT ) = ζ2 (T H) =
18
,
25
ζ2 (T T ) =
36
25
[10%]
[6%]
[4%]
e. Using the numbers computed in part d. find the time-zero price of an Asian option
whose payoff at time 2 is
( 13 Y2
+
− 4) , 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
, V2 (HT ) = 34 , V2 (T H) = V2 (T T ) = 0
V2 = ( Y2 − 4)+ : V2 (HH) = 16
3
3
9 16 4 18 4 2 18
2 36
1
16
V0 =
·
· +
· · +
·0· +
·0· =
25 3 9 25 3 9 25
9 25
9
15
5
[10%]
[20%]
5.
a. Define the following terms:
i. zero-coupon bond
ii. forward contract
iii. m-forward price at time n
iv. interest rate floorlet
[40%]
i. = bond/financial contract paying 1 at maturity (no coupons)
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 forward at time n equal to 0
iv.= contract that makes a single payment (K − Rn−1 )+ at time n
b. In the Ho-Lee model the interest rate at time n = 0 is R0 = a0 , and at time
1 ≤ n ≤ N is given by:
Rn (ω1 ω2 . . . ωn ) = an + bn · #H(ω1 ω2 . . . ωn ),
where the term multiplying bn is the number of heads in the path ω1 ω2 . . . ωn (i.e.
the number of up steps). Consider a two-period model (N = 2) with a0 = 0.05,
a1 = 0.045, b1 = 0.01. The risk-neutral probabilities are taken to be 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 floor with
K = 0.05.
[30%]
i. R0 = 0.05, R1 (H) = 0.04 + 0.1 · 1 = 0.05, R1 (T ) = 0.04 + 0.1 · 0 = 0.04.
1
1
and Bn,m = D1n Ẽn [Dm ].
, D2 = (1+R0 )(1+R
Recall D0 = 1, D1 = 1+R
0
1)
ω1
Compute: H
T
1
1+R0
1
1.05
1
1.05
1
1+R0
1
1.05
1
1.04
D1
D2
P̃
1
1.05
1
1.05
1
1.052
1
1.05·1.04
1
2
1
2
Using D0 = 1 we have:
B0,1 = Ẽ[D1 ] = D1 = 0.95238, B0,2 = Ẽ[D2 ] = D2 (H) 21 + D2 (T ) 21 = 0.91139,
B1,1 = 1
[15%]
D2 (H)
1
B1,2 (H) = D1 (H) Ẽ1 [D2 ](H) = D1 (H) = 0.95238
B1,2 (T ) =
1
Ẽ [D2 ](T )
D1 (T ) 1
=
D2 (T )
D1 (T )
= 0.961538
[15%]
ω1 R0 (K − R0 )+ R1 (K − R1 )+
ii. Use the table: H 0.05
0
0.05
0
T 0.05
0
0.04
0.01
F loor = F loorlet0 + F loorlet1
[5%]
F loorlet0 = Ẽ[D1 (K − R0 )+ ] = 0
[10%]
+
+
F loorlet1 = Ẽ[D2 (K−R1 ) ] = D2 (H)(K−R1 (H)) 0.5+D2 (T )(K−R1 (T ))+ 0.5
= 0 + 0.004578
[10%]
Thus: F loor = 0.004578.
[5%]
6