Appendix B
Fundamental Theorems of
Asset Pricing - Proofs
B.1
B.1.1
Proof of the FFTAP
The finite market model is arbitrage-free “if ”
there exists an EMM
Assume there exists at least an EMM, and choose one, say Q. Suppose
that an arbitrage opportunity ↵ exists, then its discounted value process
V ⇤ must satisfy V0⇤ = 0, VT⇤
0 and EP [VT⇤ ] > 0. Since Q ⇠ P, the
latter is equivalent to EQ [VT⇤ ] > 0. But we also know, by Lemma 3.27, that
EQ [VT⇤ ] = V0⇤ = 0, hence a contradiction. So, we proved that if an equivalent
martingale measure for S ⇤ exists, then the market is arbitrage-free.
B.1.2
The finite market model is arbitrage-free “only
if ” there exists an EMM
Conversely, we assume that the market is arbitrage-free and we prove the
existence of an EMM for S ⇤ , that is some probability measure Q ⇠ P, under
which S ⇤ is a martingale. For this, by Proposition A.41, it is enough to prove
124
B.1 Proof of the FFTAP
125
that, for all bounded R-valued predictable processes ,
EQ [GT ( , S ⇤ i )] = 0 8i = 1, . . . , d
where we recall that
⇤i
GT ( , S ) =
T
X
⇤i
t (S t
S ⇤ it 1 ).
t=1
But, by Lemma 3.23, that is equivalent to show that the discounted gain
process of any self-financing trading strategy ↵ = (↵0 , ↵1 , . . . , ↵d ) at the
terminal trading date has null expectation under Q, i.e.
" T
#
X
EQ [G⇤T (ˆ
↵)] = EQ
↵
ˆ t · (Ŝt⇤ Ŝt⇤ 1 ) = 0,
t=1
where we denote ↵
ˆ := (↵1 , . . . , ↵d ) and Ŝ ⇤ := (S ⇤,1 , . . . , S ⇤,d ).
Note that in a finite space of dimension N , we can identify both random variables and probability measures by vectors of RN . Denoted ⌦ =
{!1 , . . . , !N }, we identify a R-valued random variable X on (⌦, F), X(!i ) =
xi for i = 1, . . . , N , with the vector (x1 , . . . , xN ) of RN , and a probability measure Q on (⌦, F), Q({!i }) = qi for i = 1, . . . , N , with the vector
(q1 , . . . , qN ) of RN .
By Definition 3.28, the assumpion equates to
V \ RN
+ = {0},
where
N
RN
+ := {x = (x1 , . . . , xN ) 2 R | x1
0, x2
0, . . . , xN
0},
V := {G⇤T (ˆ
↵) | ↵
ˆ Rd -valued F-predictable process}.
We would like to (strictly) separate the disjoint convex sets V and RN
+ \ {0}
by a hyperplane induced by a linear functional on RN , from which we can
then construct an EMM. Unfortunately, the standard separation theorems
do not directly apply. However, in finite dimension, we can overcome this
difficulty by introducing a convex compact subset of RN
+ \ {0} and using a
separation result obtained from the following lemma.
126
B. Fundamental Theorems of Asset Pricing - Proofs
Lemma B.1. Let C be a convex closed subset of RN that does not contain
the origin. Then, there exists an element ⇠ 2 C such that
|⇠|2  x · ⇠
8x 2 C.
Proof. Since C is closed, it contains an element ⇠ such that |⇠| = dist(C, 0).
Indeed, dist(C, 0) = inf{|x|, x 2 C}, which means there exists a bounded
sequence {xn }n2N of elements of C such that |xn |
! dist(C, 0), with a sub-
n!1
N
sequence {xnk }k2N converging to a vector ⇠ of R such that |⇠| = dist(C, 0),
i.e. |⇠|  |x| for all x 2 C, and ⇠ 2 C because C is closed. Moreover, since C
is convex, for all x 2 C and all t 2 [0, 1], ⇠ + t(x
|⇠|2  |⇠ + t(x
⇠) 2 C, and so
⇠)|2 = |⇠|2 + 2t⇠ · (x
Thus, for all t 2]0, 1], 0  2⇠ · (x
⇠) + t2 |x
⇠|2 .
⇠|2 , and going to the limit
⇠) + t|x
for t converging to 0 from the right, we get ⇠ · (x
⇠)
0, which ends the
proof.
Corollary B.2. Let K be a convex compact subset of RN and V a vector
subspace of RN such that K \ V = ;. Then, there exists an element ⇠ 2 RN
such that
x·⇠ =0
8x 2 V,
Proof. We consider the set C = K
y·⇠ >0
V := {y
8y 2 K.
x|x 2 V, y 2 K}, which does
not contain the origin, because K and V are disjoint. We would like to apply
Lemma B.1, so we have to check if C is convex and closed. We can easily
show that C is convex: for all x1 , x2 2 V, y1 , y2 2 K, for all t 2 [0, 1],
t(y1
x1 ) + (1
t)(y2
x2 ) = (ty1 + (1
t)y2 )
(tx1 + (1
t)x2 ) ,
which is in C because both K and V (as any vector space) are convex. Moreover, we can prove that C is also closed: let {zn }n2N be a sequence of ele-
ments of C converging to z 2 RN , for all n 2 N there must be two elements
xn 2 V, yn 2 K such that zn = yn
xn ; since K is compact, there is a
subsequence {ynk }k2N converging to y 2 K; since a vector space is always
B.1 Proof of the FFTAP
closed, we also have xnk
as well as zn
!y
127
! x 2 V, hence znk = ynk
xnk
n!1
x = z 2 C.
n!1
!y
n!1
x 2 C,
Finally, by Lemma B.1, there exists ⇠ 2 C such that
|⇠|2  (y
x) · ⇠ = y · ⇠
Since this holds for all
y·⇠
2
|⇠| > 0.
2 R, y 2 K.
(x · ⇠) 8x 2 V,
2 R, we necessarily have x · ⇠ = 0, hence also
Now, we denote by K the convex hull of the set of random variables
1{!n }
n2N
, that is
K :=
=
(
(
N
X
n=1
µn 1{!n } µn
0 for n = 1, . . . , N,
x = (x1 , . . . , xN ) 2 RN
N
X
xn = 1
n=1
)
N
X
n=1
µn = 1
)
,
which is a convex, compact subset of RN
+ . Since K \ V = ; and V is a vector
space, we can apply Corollary B.2 and find an element ⇠ = (⇠1 , . . . , ⇠N ) 2 RN
such that
(i) ⇠ · x = 0 for all x 2 V;
(ii) ⇠ · y > 0 for all y 2 K.
Since the canonical base of RN is contained in K, we get that ⇠n = ⇠ · ✏n > 0
for all n = 1, . . . , N , i.e. ⇠ has strictly positive components. Thus, we can
normalize ⇠ to obtain a probability measure equivalent to P. Precisely, we
define a probability measure Q on (⌦, F) as associated to a vector (q1 , . . . , qN )
of RN , in the following way:
Q({!n }) = qn ,
where
PN
n=1 qn
qn :=
⇠n
⇠n
= PN
> 0,
|⇠|
⇠
n
n=1
n = 1, . . . , N,
= 1 for all n = 1, . . . , N . Then, Q is a probability measure
equivalent to P; moreover, (i) translates into EQ [G⇤ (ˆ
↵)] = 0 for all Rd -valued
predictable processes ↵
ˆ = (↵1 , . . . , ↵d ), which ends the proof of the FFTAP.
128
B. Fundamental Theorems of Asset Pricing - Proofs
B.2
B.2.1
Proof of the SFTAP
An arbitrage-free finite market model is complete “only if ” there is a unique EMM
Assume that the market is arbitrage-free, i.e. satisfies NA, and complete.
By contradiction, suppose that there are two di↵erent equivalent martingale
measures for S ⇤ , Q1 , Q2 . Now, consider any event A 2 F and take the
European contingent claim which takes value 1 on A and 0 otherwise, X =
1A . Since the market is complete, X is attainable, thus by Theorem 3.34 its
arbitrage-free price at time 0 is unique and equal to
EQ1 [1A ] = EQ2 [1A ].
Since A was arbitrarily chosen in F, the equation above translates into
Q1 (A) = Q2 (A)
8A 2 F,
therefore Q1 = Q2 , that is there cannot be two distinct EMMs for S ⇤ .
B.2.2
An arbitrage-free finite market model is complete “if ” there is a unique EMM
Assume that the equivalent martingale measure for S ⇤ , Q, is unique. By
contradiction, suppose that the (arbitrage-free) market is not complete.
As in the proof of the FFTAP, we identify random variables with vectors
in RN . We denote by
V := { V ⇤,↵ | ↵ self-financing strategy}
the set of all discounted values of self-financing strategies, which is a vector
space, and remark that the market is not complete if and only if V ( RN .
Let us define a scalar product h·,·iQ in RN as follows:
Q
hX, Y iQ = E [XY ] =
N
X
i=1
xi yi Q({!i }),
B.2 Proof of the SFTAP
129
for all X = (x1 , . . . , xN ), Y = (y1 , . . . , yN ) 2 RN . Then, there exists a vector
⇠ 2 RN \ {0} orthogonal to V, i.e. such that
h⇠, XiQ = EQ [⇠X] = 0 8X 2 V.
In particular, as every constant belongs to V (it can be reached by taking
initial investment equal to the final constant discounted value and ↵1 = . . . =
↵d = 0), by taking X = 1 we get EQ [⇠] = 0.
Now, chosen arbitrarily
as
where k⇠k1
> 1, we define another probability measure Q
✓
◆
⇠i
Q ({!i }) = 1 +
Q({!i }), i = 1, . . . , N,
k⇠k1
:= max |⇠i |. If we can prove that Q is another EMM for S ⇤ ,
1iN
then we have a contradiction, as well as the end of the proof.
First, Q is a probability measure, since
Q (⌦) =
=
N
X
i=1
N
X
i=1
Q ({!i })
N
1 X
Q({!i }) +
⇠i Q({!i })
k⇠k1 i=1
1
EQ [⇠] = 1.
k⇠k1
Moreover, Q ⇠ P because, by definition, it assigns strictly positive proba=1+
bility to each elementary event. It remains to be proved that S ⇤ is a Q martingale. Again, by Proposition A.41, this is equivalent to prove that, for
all bounded R-valued predictable processes ,
EQ [GT ( , S ⇤ i )] = 0 8i = 1, . . . , d.
For all i = 1, . . . , d, we have
EQ [GT ( , S ⇤ i )] =
N
X
i=1
GT ( , S ⇤ i )Q({!i }) +
= EQ [GT ( , S ⇤ i )] +
N
1 X
⇠i GT ( , S ⇤ i )Q({!i })
k⇠k1 i=1
1
EQ [⇠GT ( , S ⇤ i )]
k⇠k1
= EQ [GT ( , S ⇤ i )] = 0,
since GT ( , S ⇤ i ) 2 V and Q is an EMM for S ⇤ .