1. No category

Advanced Financial Models
Example sheet 2 - Michaelmas 2014
Michael Tehranchi
Problem 1. Consider a binomial two-asset model with prices
(B, S)
(8, 12)
:
3/4 vvv
v
v
vv
vv
(6, 8)
HH
HH
HH
H
1/4 HH$
(4, 5)
Introduce a call option with payout ξ1 = (S1 − 10)+ . Find its unique no-arbitrage price ξ0 .
What is the replicating portfolio for this option?
Solution 1. A replicating portfolio for the option can be found by solving
8φ + 12π = 2 = (12 − 10)+
4φ + 5π = 0 = (5 − 10)+
with solution φ = −5/4 and π = 1. The time-0 value of this portfolio is (−5/4)(6) + (1)(8) =
1/2. Hence, there is no arbitrage if and only if the time-0 price of the option is the cost of
replication ξ0 = 1/2.
The no-arbitrage price can also be found by identifying the martingale deflator Y = Y1 /Y0 ,
which solves E(Y B1 ) = B0 and E(Y S1 ) = S0 ; that is,
(3/4)8a + (1/4)4b = 6
(3/4)12a + (1/4)5b = 8
so that Y = (a, b) = (1/3, 4). There is no arbitrage iff
ξ0 = E(Y ξ1 ) = (3/4)(1/3)(2) + (1/4)(4)(0) = 1/2
as before.
Problem 2. Consider a trinomial two-asset model with B0 = B1 = 1 and S given by
S
?3





/2
2?
?1/4
??
??
1/4 ?
1/2
1.
Find all equivalent martingale measures for this model. Now introduce a call option with
payout ξ1 = (S1 − 2)+ . Show directly that the payout ξ1 cannot be replicated by trading in
the stock and bond. Prove that there is no arbitrage in the augmented market if and only
if 0 < ξ0 < 12 .
1
Solution 2. An equivalent martingale measure solves EQ (S1 ) = S0 , (no discounting is needed
since the num´eraire is cash)
3p + 2q + r = 2
p+q+r = 1
and hence (p, q, r) = (p, 1 − 2p, p) for 0 < p < 1/2, where p = Q{S1 = 3}, etc.
Denoting by φ and π the holding in the bond and stock that would replicate the payout
ξ1 , we need the three conditions to hold simultaneously:
φ + 3π = 1
φ + 2π = 0
φ+π = 0
Impossible!
There will be no arbitrage iff ξ0 = EQ (ξ1 ) for some equivalent martingale measure. Evaluating the expectation in terms of Q introduced earlier gives that there is no arbitrage iff
ξ0 = p for some p ∈ (0, 12 ).
Alternatively, let (φ, π, ψ) be a candidate arbitrage in the sense that
(A)
(B)
(C)
(D)
φ
φ
φ
φ
+ 2π + ψξ0
+ 3π + ψ
+ 2π
+ π
=
≥
≥
≥
0
0
(B+D-2A)
⇒
0
(A-C)
0
(1 − 2ξ0 )ψ ≥ 0
ξ0 ψ
≤ 0
If 0 < ξ0 < 1/2 the only solution to the system of inequalities is (φ, π, ψ) = 0. Otherwise,
note that if ξ0 ≥ 1/2 then the portfolio (−ξ0 , +ξ0 , −1) is an arbitrage and if ξ0 ≤ 0 then the
portfolio (+ξ0 , −ξ0 , +1) is.
Problem 3. (Call surface) Let (Bt , St )t≥0 be a model of an arbitrage-free complete financial
market with two assets and suppose that Bt+1 ≥ Bt > 0 for all t ≥ 0.
(a) Let C(T, K) be the initial replication cost of a European call option with strike K and
maturity T written on the asset with price S. Show that T 7→ C(T, K) is increasing, and
that K 7→ C(T, K) is decreasing and convex.
(b) Show that it never optimal to exercise an American call option early.
Solution 3. Let Y be the unique martingale deflator such that Y0 = 1. We will show that Y
is a supermartingale. Since the market is complete, there is no problem with integrability.
Hence
Yt Bt
|Fs
E(Yt |Fs ) ≤ E
Bs
1
=
E (Yt Bt |Fs )
Bs
1
=
(Ys Bs ) = Ys .
Bs
2
Next we show that the process defined by Yt (St − K)+ = (Yt St − Yt K)+ is a submartingale.
Jensen’s inequality and the martingale property of Y S together imply
E[(Yt St − Yt K)+ |Fs ] ≥ (E[Yt St − Yt K|Fs ])+
= (Ys Ss − E[Yt |Fs ]K)+
≥ (Ys Ss − Ys K)+
where the supermartingale property of Y is used in the last line.
(a) We have the formula
C(T, K) = E[YT (ST − K)+ ].
That K 7→ C(T, K) is decreasing and convex is immediate from the same properties of
K 7→ (ST − K)+ . That T 7→ C(T, K) is increasing is a consequence of the submartingale
property of Y (S − K)+ .
Alternatively, if Ct (T, K) is the time t value of the replication strategy for the call option
with maturity T and strike K, we know from lectures that the augmented market with prices
(B, S, C(T1 , K1 ), . . . , C(Tn , Kn )
has no arbitrage.
Convexity in K: consider the wealth
Xt = pCt (T, K1 ) + (1 − p)Ct (T, K0 ) − Ct (T, Kp )
associated with buying p calls with strike K1 , buying 1 − p calls with strike K0 and selling
one call of strike Kp = pK1 + (1 − p)K0 , where 0 < p < 1. Since
XT = p(ST − K1 )+ + (1 − p)(ST − K0 )+ − (ST − Kp )+ ≥ 0 a.s.
it follows from no-arbitrage that Xt ≥ 0 a.s. for all 0 ≤ t ≤ T . A similar argument shows
K 7→ Ct (T, K) is non-increasing.
Increasing in T : consider the wealth at time t associated with buying one call of maturity
T and selling one of maturity T − 1, so that
Xt = Ct (T, K) − Ct (T − 1, K)
for 0 ≤ t ≤ T − 1. At time T − 1, if ST −1 > K, trade the expired call for a portfolio long
one stock and short K/BT −1 units of the bank account. The time T wealth is then
XT = (ST − K)+ − (ST − KBT /BT −1 )1{ST −1 >K} ≥ 0
so by no-arbitrage Xt ≥ 0 a.s. for all t.
(b) Problem 7 below shows that the Snell envelope of the submartingale (Yt (St − K)+ )0≤t≤T
is just Ut = E[YT (ST − K)+ |Ft ]. Therefore U is a martingale so U = U0 + M , and hence the
optimal stopping time τ ∗ defined in lectures is identically T .
Problem 4. Let M be a martingale and K a predictable process, and let
Xt =
t
X
Ks (Ms − Ms−1 ).
s=1
Let T > 0 be a non-random time horizon. Show that if XT is integrable then (Xt )0≤t≤T is a
true martingale. Hint: Show that XT −1 is integrable.
3
Solution 4. Let τN = inf{t ≥ 0 : |Kt+1 | > N }. Note Xs 1{t≤τN } is integrable for all 0 ≤ s ≤ t,
since M is integrable by definition of martingale, and Ks is bounded on {t ≤ τN }. Hence we
have
E[XT 1{τN ≤T } |FT −1 ] = E[XT −1 1{T ≤τN } + KT 1{T ≤τN } (MT − MT −1 )|FT −1 ]
= XT −1 1{T ≤τN } + KT 1{T ≤τN } E[MT − MT −1 |FT −1 ]
= XT −1 1{T ≤τN } .
Now, we have assumed that XT is integrable, and since |XT 1{τN ≤T } | ≤ |XT | we can apply
the dominated convergence theorem
E[XT |FT −1 ] = E[lim XT 1{τN ≤T } |FT −1 ]
N
= lim XT −1 1{T ≤τN }
N
= XT −1 .
This shows that XT −1 is integrable, and by induction (Xt )0≤t≤T is integrable. An integrable
local martingale in discrete time is a true martingale.
Problem 5. Let (Xk )k∈K be a collection of real-valued random variables, where K is an
arbitrary (possibly uncountable) index set. Our aim is to show there exists a random variable
ˆ taking values in R ∪ {+∞} such that
X
ˆ ≥ Xk almost surely for all k ∈ K, and
• X
ˆ almost surely.
• if Y ≥ Xk almost surely for all k ∈ K then Y ≥ X
ˆ = ess supk Xk exists.
This will show that the X
(a) Show that there is no loss assuming that |Xk (ω)| ≤ 1 for all (k, ω). Hint: Consider
x
Xk0 = f (Xk ) where f (x) = 1+|x|
.
Let C be the collection of countable subsets of K. Let
x = sup E[sup Xk ]
C∈C
k∈C
ˆ = sup ˆ Xk .
Let Cn ∈ C be such that E[supk∈Cn Xk ] > x − 1/n and let Cˆ = ∪n Cn . Let X
k∈C
ˆ a random variable, i.e. measurable? Show that E(X)
ˆ = x.
(b) Why is X
ˆ < Xk ) > 0 for some k ∈ K. Show that E(X
ˆ ∨ Xk ) > x. Why is this
(c) Suppose that P(X
a contradiction?
ˆ a.s.
(d) Let Y be a random variable such that Y ≥ Xk a.s. for all k ∈ K. Prove that Y ≥ X
Solution 5. (a) Given the family (Xk )k∈K let Xk0 = f (Xk ). Suppose we know that there
exists a random variable Xˆ 0 with the properties characterising the essential supremum of the
ˆ = f −1 (Xˆ 0 ). Since f is strictly increasing
family (Xk0 )k of bounded random variables. Let X
ˆ has the properties characterising the essential supremum of (Xk )k .
it is easy to see that X
ˆ n = supk∈C Xk is measurable for each n. Also
(b) Since Cn is countable, the function X
n
ˆ is also a random variable. (For instance, note {X
ˆ >
∪n Cn = Cˆ is countable and hence X
x} = ∪k∈Cˆ {Xk > x}. ) Now, we there is no loss supposing Cn−1 ⊆ Cn for all n ≥ 1, and
ˆ = supn X
ˆ n = limn X
ˆ n . The result follows from the bounded convergence
hence we have X
theorem.
4
ˆ < Xk ) > 0. Then X
ˆ ∨ Xk ≥ X
ˆ a.s. with strict inequality with positive
(c) Suppose that P(X
ˆ
ˆ
probability. Hence E(X ∨ Xk ) > E(X) = x. This is a contradiction since this would show
that
E( sup Xh ) > sup E[sup Xh ].
C∈C
ˆ
h∈C∪{k}
h∈C
(d) If P(Y ≥ Xk ) = 1 for all k ∈ K, then
ˆ = P(∩ ˆ {Y ≥ X
ˆ k }) = 1
P(Y ≥ X)
k∈C
since Cˆ is countable.
Problem 6. (Optional sampling theorem) Let (Xt )0≤t≤T be a discrete time submartingale,
and let σ and τ be stopping times such that 0 ≤ σ ≤ τ ≤ T almost surely. Show the
inequality
E(Xσ ) ≤ E(Xτ ).
Furthermore, prove that X is a martingale if and only if E(XT ) = X0 .
Solution 6. Write the telescoping sum
E(Xτ − Xσ ) =
T
X
E[1{σ<s≤τ } (Xs − Xs−1 )]
s=1
=
T
X
E[E(1{σ<s≤τ } (Xs − Xs−1 )Fs−1 )].
s=1
Notice that {σ < s} = {σ ≤ s − 1} ∈ Fs−1 and {τ ≥ s} = {τ ≤ s − 1}c ∈ Fs−1 . Therefore,
1{σ<s≤τ } is Fs−1 -measurable, and we can take it out of the expectation with respect to Fs−1 :
E(Xτ − Xσ ) =
T
X
E[1{σ<s≤τ } E(Xs − Xs−1 Fs−1 )].
s=1
Each term in the above sum is non-negative since X is a submartingale.
Now let Yt,T = E(XT |Ft ) − Xt for 0 ≤ t ≤ T . Since X is a submartingale, we have Yt,T ≥ 0
a.s. However,
E(Yt,T ) = E(XT ) − E(Xt ) = X0 − E(Xt ).
by the tower property of conditional expectation and the assumption that E(XT ) = X0 .
However, by the submartingale property X0 − E(Xt ) ≤ 0. Hence Yt,T = 0 a.s. for all t,
implying that X is a martingale. (Note this argument works for continuous time also.)
Problem 7. Let (Yt )0≤t≤T be a given adapted, integrable process, and let (Ut )0≤t≤T be its
Snell envelope.
(a) Show that if Y is a supermartingale then Ut = Yt for all t, and if Y is submartingale,
then Ut = E(YT |Ft ).
(b) Let τ be any stopping time taking values in {0, . . . , T }. Show that the process (Ut∧τ )0≤t≤T
is a supermartingale.
(c) Define the random time τ∗ by
τ∗ = min {t ∈ {0, . . . , T } : Ut = Yt } .
5
Show that τ∗ is a stopping time. Furthermore, show that the process (Ut∧τ∗ )t∈{0,...,T } is a
martingale and, in particular, U0 = E(Yτ∗ ). (That is, τ∗ is an optimal stopping time, possibly
different than τ ∗ defined in lectures.)
Solution 7. (a) In both cases we proceed by induction. First suppose that Y is a supermartingale, and that Ut+1 = Yt+1 for some t < T . Then
Ut = max{Yt , E(Ut+1 |Ft )} = max{Yt , E(Yt+1 |Ft )} = Yt
since Yt ≥ E(Yt+1 |Ft ) by assumption, completing the induction. Similarly, suppose that Y
is a submartingale, and that Ut+1 = E(YT |Ft+1 ). Then
Ut = max{Yt , E(Ut+1 |Ft )} = max{Yt , E[E(YT |Ft+1 )|Ft ]} = E(YT |Ft )
by the tower property and the assumption Yt ≤ E(YT |Ft ), and we’re done.
(b) Since U is a supermaringale and the event {τ ≥ t + 1} = {τ ≤ t}c is in Ft , we have
E[U(t+1)∧τ − Ut∧τ |Ft ] = E[1{t+1≤τ } (Ut+1 − Ut )|Ft ]
= 1{t+1≤τ } E[Ut+1 − Ut |Ft ]
≤ 0
so the stopped process (Ut∧τ )0≤t≤T is also supermartingale.
(c) Now, the event
{τ∗ > t} = {Y0 < U0 , . . . , Yt < Ut }
is Ft -measurable since both Y and U are adapted, hence τ∗ is a stopping time.
Since Ut = E(Ut+1 |Ft ) on the event {t + 1 ≤ τ∗ } we have
U(t+1)∧τ∗ − Ut∧τ∗ = 1{t+1≤τ∗ } (Ut+1 − Ut )
= 1{t+1≤τ∗ } [Ut+1 − E(Ut+1 |Ft )].
In particular E[Yτ∗ ] = E[Uτ∗ ] = E[UT ∧τ∗ ] = U0 .
Problem 8. Consider the adapted process Y given by
Y
? 10
~~
~
~
~~
~~
8
A @@@
@@
@@
1/3
@
3/5 7
7;
;;
;;
;;
;;
;
6
2/5 ;;
~?
;;
1/4 ~~
~
;;
; ~~~~
4@
@@
@@
@
3/4 @@
2/3
2
6
where the above diagram should be read as P(Y1 = 8) = 3/5, P(Y2 = 10|Y1 = 8) = 2/3, etc.
Find the Snell envelope U and find the decomposition Ut = U0 + Mt − At , where M is a
martingale and A is a predictable increasing process with M0 = A0 = 0. Identify the two
stopping times τ∗ = min{t ≥ 0 : Ut = Yt } and τ ∗ = min{t ≥ 0 : At+1 > 0}. Show directly
the equality U0 = E(Yτ∗ ) = E(Yτ ∗ ).
Solution 8. Let (Ft )t∈{0,1,2} be the filtration generated by the process Y. The Snell envelope
U is defined by U2 = Y2 and Ut = max{Yt , E(Ut+1 |Ft ). The Doob decomposition can be
found by M0 = A0 = 0 and Mt+1 = Mt + Ut+1 − E(Ut+1 |Ft ) and At+1 = At + Ut − E(Ut+1 |Ft ).
The values found this way can be read from the diagram
(Y, U, M, A)
(10, 10, 3, 0)
n7
2/3 nnnn
n
nnn
nnn
(8, 9, 2, 0)
w;
ww
w
ww
ww
3/5 ww
ww
ww
w
ww
ww
w
w
ww
PPP
PPP
PPP
1/3 PPP'
(7, 7, 0, 0)
(7, 7, 0, 0)
GG
GG
GG
GG
GG
GG
GG
2/5 GGG
GG
GG
GG
#
(6, 6, 0, 1)
n7
1/4 nnnn
n
nnn
nnn
(4, 4, −3, 0)
PPP
PPP
PPP
3/4 PPP'
(2, 2, −4, 1)
Then τ∗ = 0 almost surely, while
∗
τ (ω) =
2
1
if ω ∈ {Y1 = 8}
if ω ∈ {Y1 = 4}
Note
E(Yτ ∗ ) = E(Y1 1{Y1 =4} + Y2 1{Y1 =8} )
= ( 52 )(4) + ( 35 )[( 32 )(10) + ( 13 )(7)]
= 7.
Problem 9. (Binomial model) Consider the following two-period market model with two
assets. One asset is a riskless bank account with risk-free rate r = 1/4 and the other is a
7
stock with prices given by
S
45
|>
|
||
||
||
3/4
30
@ BBB
BB
BB
1/4 B!
1/2 36
15=
==
==
==
==
==
16
1/2 ===
|>
1/3 ||
==
|
==
||
||
12 B
BB
BB
BB
2/3 B!
10
Find the equivalent martingale measure Q.
(a) Consider a European put option which strike K = 15 expiring at time 2. What is the
replication cost of the option at time 0? What is the replicating strategy?
(b) What is the super-replication cost and strategy for an American put option with the
same strike and expiration date.
Solution 9. The equivalent martingale measure must satisfy EQ (1 + r)−1 St+1 |Ft = St . For
instance solving 45q + 36(1 − q) = (1 + 1/4)(30) yields q = 1/6 = Q(S2 = 45|S1 = 30).
E
Since the market is complete, the replication cost
P can be calculated inductively as
E
+
E
Q
−1 E
PT = (K − ST ) and Pt = E (1 + r) Pt+1 |Ft . The replicating strategy is found by
E
E
solving PtE (1 + r) + πt+1
[St+1 − St (1 + r)] = Pt+1
.
For the American option, the super-replication cost PA is given inductively by PTA =
A
(K − ST )+ and PtA = max{(K − St )+ , EQ (1 + r)−1 Pt+1
|Ft }. The super-replicating strategy
A
A
A
is found by solving Pt (1 + r) + πt+1 [St+1 − St (1 + r)] = Pt+1
on {t < τ ∗ } where τ ∗ is an
optimal stopping time.
8
The results can be read from the diagram, where ξt = (K − St )+ .
(S, ξ; P E , π E ; P A , π A )
(45, 0; 0, 0; 0, 0)
hh3
hhhh
h
h
h
hh
hhhh
hhhh
1/6
(30, 0; 0, −1/27; 0, −1/6)
VVVV
VVVV
VVVV
VVVV
VVV+
5/6
o7
ooo
o
o
oo
ooo
3/8 ooo
oo
ooo
o
o
oo
ooo
o
o
ooo
(36, 0; 0, 0; 0, 0)
(15, 0; 1/3, ·; 3/2, ·)
PPP
PPP
PPP
PPP
PPP
PPP
PPP
5/8
PPP
PPP
PPP
P'
(16, 0; 0, −5/6; 0, −5/6)
hhh3
hhhh
h
h
h
h
hhhh
hhhh
5/6
(12, 3; 2/3, −1/27; 3, −1/6)
VVVV
VVVV
VVVV
VVVV
VVV+
1/6
(10, 5; 5, −5/6; 5, −5/6)
Notice that for any optimal stopping time τ ∗ , we have τ ∗ = 1 on {S1 = 12}. If the holder
of the option fails to exercise optimally, the seller of the option need only to hedge the
remaining one-period European option (in the bottom right corner of the diagram) at a cost
of 2/3 and pocket the rest 3 − 2/3 = 7/3.
Problem 10 (Martingale representation). Let ζ1 , ζ2 , . . . be a sequence of independent Bernoulli
random variables such that
P(ζt = 1) = p = 1 − P(ζt = 0)
Suppose that the filtration is Ft = σ(ζ1 , . . . ζt ). Show that for every martingale M there
exists a predictable process (θt )t≥1 such that
Mt = M0 +
t
X
θs (ζs − p).
s=1
Solution 10. Since Mt is Ft -measurable for each t, there exists a function ft : {0, 1}t → R
such that
Mt = ft (ζ1 , . . . , ζt ).
We will make use of the identity
ft (ζ1 , . . . , ζt ) = ζt ft (ζ1 , . . . , ζt−1 , 1) + (1 − ζt ) ft (ζ1 , . . . , ζt−1 , 0).
Indeed, the martingale property implies
Mt−1 = E(Mt |Ft−1 )
= p ft (ζ1 , . . . , ζt−1 , 1) + (1 − p) ft (ζ1 , . . . , ζt−1 , 0)
9
so that
Mt − Mt−1 = (ζt − p)[ft (ζ1 , . . . , ζt−1 , 1) − ft (ζ1 , . . . , ζt−1 , 0)].
The desired representation follows from identifying the Ft−1 -measurable random variable
θt = ft (ζ1 , . . . , ζt−1 , 1) − ft (ζ1 , . . . , ζt−1 , 0).
Problem 11. Suppose X is a bounded martingale with Xt 6= Xt−1 a.s. and β is a bounded
predictable process. Let Y be the martingale defined by
t
X
Yt = Y0 +
βs (Xs − Xs−1 ).
s=1
Show that β can be recovered from X and Y by the regression formula
Cov(Xt , Yt |Ft−1 )
.
βt =
Var(Xt |Ft−1 )
Solution 11. Recall that for square integrable random variables U and V and a sigma-field
G the conditional variance and covariances are defined by
Var(U |G) = E{(U − E[U |G])2 G}
Cov(U, V |G) = E{(U − E[U |G])(V − E[V |G])G}.
In this case we can take U = Xt , V = Yt and G = Ft−1 . Note that both Xt and Yt
are bounded by assumption, so integrability is not a problem. Also, since X and Y are
martingales, E(Xt |Ft−1 ) = Xt−1 and E(Yt |Ft−1 ) = Yt−1 .
Therefore, since βt is bounded and Ft−1 measurable we have
Cov(Xt , Yt |Ft−1 ) = E[(Xt − Xt−1 )(Yt − Yt−1 )|Ft−1 ]
= E[βt (Xt − Xt−1 )2 |Ft−1 ]
= βt E[(Xt − Xt−1 )2 |Ft−1 ]
= βt Var(Xt |Ft−1 )
and the result follows.
10