Key Idea of Replication
Introduction to Mathematical Finance:
Part I: Discrete-Time Models
Conclusion: Replication
We are able to price an arbitrary security with random value
C1 by the relation
AIMS and Stellenbosch University
April-May 2012
C0 =
1
EQ [C1 ]
1+r
because there is a strategy for trading the stock and bond at
1
time t = 0 that requires an initial investment of 1+r
EQ [C1 ]
and that has a random terminal value of C1 .
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One-Period Binomial Model (OPBM): A Complete Market
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One-Period Binomial Model (OPBM): A Complete Market
We can directly verify completeness by showing that a given
contingent claim C is attainable if No Arbitrage:
Sd < S0 (1 + r ) < Su holds.
The solution is
Observe that the condition
C1 (ω) = ∆S1 (ω) + b(1 + r )
for all
ω ∈ {up, down}
∆=
is a system of two linear equations for the two real variables ∆ and
b
C1 (up) − C1 (down)
Su − Sd
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C1 (down)Su − C1 (up)Sd
(Su − Sd )(1 + r )
Therefore, the unique arbitrage-free price of C is the cost of the
portfolio at time t = 0 given by
C0 = ∆S0 +b =
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and b =
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C1 (up) S0 (1 + r ) − Sd C1 (down) Su − S0 (1 + r )
·
+
·
.
1+r
Su − Sd
1+r
Su − Sd
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Delta: interpretation
Martingale Representation Theorem
Remark (Martingale Representation Theorem: Special case)
Thanks to the arbitrage principle, from the 3 equations below


 C1 (up) = ∆Su + b(1 + r )
C1 (down) = ∆Sd + b(1 + r )


C0 = ∆S0 + b
Definition (Delta)
The Delta
∆=
C1 (up) − C1 (down)
Su − Sd
measures sensitivity of option price to stock movements:
∆=
δC1
Change in option price
=
δS
Change in stock price
we remark that
e1 − C
e0 = ∆(S
e1 − S
e0 )
C
This equality is a special case of the Martingale Representation
Theorem in the Binomial model, which states:
e is a Q -MG, any other Q -MG M satisfies
Since by construction, S
Remark
This parameter play an important role in options pricing
models.
e1 − S
e0 ) ,
M1 − M0 = c 0 (S
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One Period Trinomial Model
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for some c0 ∈ R
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Adding One More State to the Binomial
The Market:
B
M=
S
t=0
1
S0
and
D=
B
S
up
1+r
Su
mid
1+r
Sm
down
p > 0 pm > 0
u
1+r
1+r
1+r
=
Sd
uS0
mS0
pd > 0
1+r
dS0
We force throughout d < m < u and pd = 1 − pu − pm .
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Lemma (Arbitrage-Free Condition for the 1-Period 3-nomial
Model)
Theorem
A necessary and sufficient condition for no-arbitrage in the
one-period trinomial model is
In the one-period trinomial model, the following statements
are equivalent
(1) the one-period trinomial model is arbitrage-free.
d < 1+r < u.
(2) the risky asset satisfy Sd < S0 (1 + r ) < Su , equivalently
Sd
Su
1+r < S0 < 1+r .
Proof. Indeed, we need that Sd < (1 + r )S0 , i.e. d < 1 + r , the
risky asset must be able to underperform the risk-free return at
at least one state of the economy. Similarly, we need that
Su > (1 + r )S0 , i.e. u > 1 + r , the risky asset must be able to
outperform the risk-free return at at least one state of the
economy. It is easy to see that if this condition is violated then
an arbitrage is possible.
(3) there exists an equivalent martingale measure.
Proof. We prove (2) ⇐⇒ (3) . The case (3) =⇒ (2) is
immediate since
Sd
Sm
Sd
Su
Su
< qu
+ qm
+ qd
= S0 <
1+r
1+r
1+r
1+r
1+r
for an EMM Q = (qu , qm , qd ). By construction Sd < Sm < Su .
Remark
The inclusion of a middle state does not change our
arbitrage-free criterion from the Binomial model.
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Proof. Now we deal with the case (2) =⇒ (3). More delicate
part since it is a “constructive proof”. Our assumptions are
Sd
Su
< S0 <
1+r
1+r
We have no information where is
2 cases
Case A:
Case B:
In conclusion, we have
and Sd < Sm < Su
Sm
1+r
compared to S0 . We have
S0 = α
Thus, we can take
Sm
Su
Sd
< S0 ≤
<
,
1+r
1+r
1+r
Sd
Sm
Su
<
≤ S0 <
.
1+r
1+r
1+r
qu =
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(1 − α)
(1 − α)
, qm =
, qd = α.
2
2
This is a equivalent martingale measure 0 < qu , qm , qd < 1 and
qu + qm + qd = 1.
I deal with case A (HWK you do case B). Let x be the centre
Sd
Sm
Su
Sm
Su
of [ 1+r
, 1+r
], i.e. x = 21 1+r
+ 12 1+r
. It satisfies 1+r
< S0 < x.
Sd
Therefore S0 is a convex combination of 1+r and x. Meaning
Sd
that we have 0 < α < 1 s.t. S0 = α 1+r
+ (1 − α)x.
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Sd
Sd
(1 − α) Sm
(1 − α) Su
+ (1 − α)x = α
+
+
1+r
1+r
2
1+r
2
1+r
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1-Period 3nomial Model: When a Contingent Claim is
Replicable?
Example
M=
B
S
t=0
1
100
D=
B
S
up
1.01
105
A cc C is replicable, if there exists a trading strategy a, b, s.t.
mid down
1.01 1.01
102
99
auS0 + b(1 + r ) = C1 (up) = Cu
amS0 + b(1 + r ) = C1 (mid) = Cm
We consider a European Call option on the risky asset with strike
price K = 101. Can we replicate this claim? i.e. find (a, b) ∈ R 2
s.t.

+

a105 + b1.01 = 4 = (Su − 101)
a102 + b1.01 = 1 = (Sm − 101)+


a99 + b1.01 = 0 = (Sd − 101)+
adS0 + b(1 + r ) = C1 (down) = Cd
NB: we have a system of 3 linear equations and 2 unknowns
We can now express a in two ways, namely
a =
Impossible!
a =
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1-Period 3nomial Model: When a Contingent Claim is
Replicable?
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1
(Cu − Cm )
(u − m)S0
1
(Cm − Cd )
(m − d)S0
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We always work under the No Arbitrage (NA) condition
d < 1 + r < u. Let us find an EMM Q = {qu , qm , qd } with
Q (up) = qu , Q (mid) = qm , Q (down) = qd s.t. by definition:
S0 = E Q [
Cu − Cm
Cm − Cd
=
u−m
m−d
S1
]
1+r
We need
Therefore, a necessary and sufficient condition is that


 dqd + mqm + uqu = 1 + r
qd + qm + qu = 1


qu > 0, qm > 0, qd > 0
(m − d)Cu − (u − d)Cm + (u − m)Cd = 0
This specifies a two-dimensional space of
C = (Cu , Cm , Cd ) ∈ R + × R + × R + . Therefore “most”
contingent claims are not replicable, as a consequence the
3-nomial market is not complete.
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One Period Trinomial Model: EMMs
It follows that we have
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A solution exists but is not unique. Moreover it is given by

(λm+(1−λ)u)−(1+r )
)

= u−(1+r
− λ u−m
 qu (λ) =
u−d
u−d
u−d
)−d u−(1+r )
qm = λ with 0 < λ < min{ (1+r
,
m−d
u−m }


(1+r )−(λm+(1−λ)d)
1+r −d
m−d
qd (λ) =
= u−d − λ u−d
u−d
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One Period Trinomial Model: EMM
One Period Trinomial Model: Replicable Claim
For a replicable CC X , it follows:
λ
EQ [
Example (Continued)
=
Continuing the example above, where
1 + r = 1.01, d = 0.99, m = 1.02, u = 1.05 we find that
3
X
Q λ (ωi )
i=1
=
2 1
1 1
(qu (λ), qm , qd (λ)) = ( − λ, λ, − λ)
3 2
3 2
with 0 < λ < 2/3.
=
X
]
1+r
X (ωi )
1+r
ω1 = up, ω2 = mid, ω3 = down
1
((1 + r ) − d)X (ω1 ) + (u − (1 + r ))X (ω3 )
(1 + r )(u − d)
−λ (m − d)X (ω1 ) − (u − d)X (ω2 ) + (u − m)X (ω3 )
(1 −
u
+ ( 1+r
− 1)X (up)
(u − d)
d
1+r )X (down)
The above expression is independent of λ! and does not depend
explicitly from m and X (mid) (it is implicit in
(m − d)X (ω1 ) − (u − d)X (ω2 ) + (u − m)X (ω3 ) = 0.
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One Period Trinomial Model: Replicable Claim
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One Period Trinomial Model: EMM
Suppose that a claim C with terminal payoff C1 is perfectly
replicable. This means that
Conclusion:
(x − aS0 )(1 + r ) + aS1 = C1
holds for some initial capital x and an investment in a numbers
of stock at time zero:
a time t = 0
aS0 + b = x
=⇒
b = x − aS0
and a time t = 1
λ
x = EQ [

If the claim C is replicable, the expectation of its discounted
payoff under all EMM gives the same value, which is the cost
to set up the replicating portfolio.
aS1 + b(1 + r ) = C1
Therefore,
C1
S1
− a(
− S0 )
1+r
1+r
Taking expectation under any EMM Q λ on both side of the last
equality, we get
C1
λ
x = EQ [
]
1+r
x=
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C1
]
1+r
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Remark
The way we proved this result did not use at all the fact that we are
in the trinomial model and is thus valid for more general models.
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Methodology of arbitrage-free pricing
No Replicable Claim
Theorem
Let C be a c.c. An arbitrage-free price π0 for C is unique if and
only if there is a replicating strategy for C . In this case,
π0 = x0 ,
If a contingent claim C is not replicable then arbitrage principle
does not specify its price π0 (C ). However arbitrage
considerations still restrict its price
where x0 is the initial capital of a replicating strategy.
Remark (Main Principle:)
Without replication there can be no unique arbitrage-free
solution for the price of a derivative:
Unique Arbitrage-Free Pricing = Replication
Moreover, the replicating strategy may not be unique.
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One Period Trinomial Model: Pricing
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π0 (X ) ≤ the value of any portfolio whose payoff dominates X
)−d u−(1+r )
Λ := {λ | 0 < λ < min{ (1+r
m−d , u−m }}. More precisely:


 aSu + b(1 + r ) ≥ Xu
aSm + b(1 + r ) ≥ Xm


aSd + b(1 + r ) ≥ Xd
If the claim is not replicable, there is a range of
arbitrage-free price that is exactly given by the open interval
λ
inf E Q [
λ∈Λ
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i.e.
λ
X
The claim is replicable iff the quantity E Q [ 1+r
] is the
same for all λ ∈ Λ; in this case, this common value is the
capital required to set up the replicating portfolio.
=⇒ π0 (X ) ≤ aS0 + b
Therefore
X
X λ
], sup E Q [
] .
1 + r λ∈Λ
1+r
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If a contingent claim X is not replicable then arbitrage does not
specify its price π0 (X ). However arbitrage consideration still
restrict its price:
Consider the one period trinomial model under the no-arbitrage
condition d < 1 + r < u. Then, all arbitrage-free prices for a
claim with terminal payoff X are given by the set
λ
X
] | λ ∈ Λ}, where
{E Q [ 1+r
I
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One Period Trinomial Model: Pricing
Theorem
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π0 (X ) ≤
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min
aSi +b(1+r )≥Xi
i=u,m,d
aS0 + b
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One Period Trinomial Model: Pricing
Min-Max Duality
π0 (X ) ≥ the value of any portfolio whose payoff is dominated by X
min
aSi +b(1+r )≥Xi
i=u,m,d
i.e.


 aSu + b(1 + r ) ≤ Xu
aSm + b(1 + r ) ≤ Xm


aSd + b(1 + r ) ≤ Xd
X
= min max aS0 + b +
πi Xi − (aSi + b(1 + r ))
a,b πi ≥0
=⇒ π0 (X ) ≥ aS0 + b
i=u,m,d
πi ≥0 a,b
i=u,m,d
π0 (X ) ≥
max
aSi +b(1+r )≤Xi
i=u,m,d
= max min a S0 −
aS0 + b
=
max
aS0 + b ≤ π0 (X ) ≤
min
aSi +b(1+r )≥Xi
i=u,m,d
aS0 + b
=
Therefore, we obtain the strongest possible consequences for
π0 (X ) by solving a pair of linearR. programming
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Min-Max Duality: Explanation
i
X
πi ≥0
P
i πi Si =S0 , i πi (1+r )=1
qi ≥0
P
Si
i qi 1+r =S0 , i qi =1
i
πi Xi
i=u,m,d
X
max
P
X
X
πi Si + b 1 −
πi (1 + r ) +
πi Xi
i
max
P
aSi +b(1+r )≤Xi
i=u,m,d
X
πi ≥0 a,b
Conclusion:
πi Xi − (aSi + b(1 + r ))
X
= max min aS0 + b +
Therefore
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aS0 + b
i=u,m,d
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qi
X
Xi
≡ max E Q [
]
Q
1+r
1+r
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Min-Max Duality: Explanation
We have the following equality
min
aSi +b(1+r )≥Xi
i=u,m,d
aS0 + b
The second equality
X
min max aS0 + b +
πi Xi − (aSi + b(1 + r ))
X
= min max aS0 + b +
πi Xi − (aSi + b(1 + r ))
a,b πi ≥0
a,b πi ≥0
i=u,m,d
i=u,m,d
= max min aS0 + b +
since
πi ≥0 a,b
(
0
max πi Xi − (aSi + b(1 + r )) =
πi ≥0
∞
if aSi + b(1 + r ) ≥ Xi
otherwise.
πi ≥0
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X
i=u,m,d
πi Xi − (aSi + b(1 + r ))
i=u,m,d
holds by the duality theorem of linear programming, which
says in this setting that
and therefore
max
X
min max = max min .
(
0 if aSi + b(1 + r ) ≥ Xi ∀i
πi Xi − (aSi + b(1 + r )) =
∞ otherwise.
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Min-Max Duality: Explanation
Min-Max Duality: Explanation
The following equality
X
X
X
max min a S0 −
πi Si + b 1 −
πi (1 + r ) +
πi Xi
πi ≥0 a,b
=
i
max
P
i
X
πi ≥0
P
i πi Si =S0 , i πi (1+r )=1
i
The last equality
πi Xi
i=u,m,d
X
max
P
is established in the same way as the 1st equality
πi ≥0
P
i πi Si =S0 , i πi (1+r )=1
=
(
X
0
πi Si =
min a S0 −
a
−∞
i
P
if
i πi Si = S0
otherwise.
X
max
qi ≥0
P
P
Si
i qi 1+r =S0 , i qi =1
πi Xi
i=u,m,d
qi
i=u,m,d
X
Xi
≡ max E Q [
]
Q
1+r
1+r
is just a change in variables.
and
min b 1 −
X
b
πi (1 + r ) =
i
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(
0
−∞
P
if
i πi (1 + r ) = 1
otherwise.
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Remark
The inequality below is easily proved directly:
min
aSi +b(1+r )≥Xi
i=u,m,d
aS0 + b ≥
X
max
πi ≥0
P
P
i πi Si =S0 , i πi (1+r )=1
πi Xi
Remark (Continued)
i=u,m,d
We established
I
I
Let a, b s.t. aSi + b(1 + r ) ≥ Xi , ∀ i = u, m, d
P
Let
P (πi )i=u,m,d s.t. πi ≥ 0, i πi Si = S0 and
i πi (1 + r ) = 1
aS0 + b ≥
Minimizing LHS (over all admissible a,b) and Max the RHS (over
all admissible (πi )i=u,m,d ) we conclude.
The opposite inequality is more subtle; the main point of Linear
Programming Duality Theorem is to prove it.
∀ i = u, m, d
=⇒
=⇒
i
=⇒
aS0 + b ≥
since πi ≥ 0
i
X
i
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πi Xi
i
Therefore
aSi + b(1 + r ) ≥ Xi
∀ i = u, m, d πi aSi + b(1 + r ) ≥ πi Xi
X
X
πi aSi + b(1 + r ) ≥
πi Xi
X
πi Xi
since
X
i
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πi Si = S0 ,
X
πi (1 + r ) = 1
i
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One Period Trinomial Model: Final Conclusion
Min-Max Duality
Remark (Lower Bound)
The lower bound is handled similarly. Conclusion:
max
aSi +b(1+r )≤Xi
i=u,m,d
aS0 + b ≤ π0 (X ) ≤
min
aSi +b(1+r )≥Xi
i=u,m,d
aS0 + b
has a dual
min E Q [
Q
X
X
] ≤ π0 (X ) ≤ max E Q [
]
Q
1+r
1+r
I
The Duality of Replicating Portfolios and Risk-Neutral
Probabilities
I
The problem of constructing the best replicating portfolio to
determine the price of a contingent claim turns out to be dual
to the problem of finding the best valuation of the contingent
claim under all risk-neutral probabilities permitted by the
market.
for Q ∈ P the set of EMM.
Remark
The One Period Trinomial Model is the simplest example of
incomplete market.
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Comments:
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