Lecture 2
One Step Securities Markets
Trading is done almost continuously in the financial markets, so the stock and bond
prices vary continuously. But in order to better understand their behavior, we will analyze
in greater detail the discrete model. And to make the model even clearer, we consider a
single period model, in which there is only one step in time.
Here is the setting in which we will be working:
 The initial state of the system is at time t  0 , and the terminal time is t  1 . Trading
and consumption can be done only at these times.
   {1 ,  2 ,...,  d } will denote the possible states of the world.
 A probability measure P on  with P( )  0 for all    . Of course since P is a
probability measure,
 P( )  1 .

A bank account process B  {B0 , B1}, B0  1, B1  B0  r . It can be viewed as a bond
earning the risk-free interest rate r.
 On the market there are N risky securities, or stocks, S1 , S 2 ,..., S N . For each security,
the price at time t will be denoted S (t ) .
An investor’s portfolio will be made of the bond B and several risky securities.
Definition 1. A trading strategy (or dynamic portfolio) is a vector H  ( H 0 , H 1 ,..., H N )
describing the investor’s portfolio between t  0 and t  1 . H 0 is the amount of dollars
invested in the ban account, and for n  1,..., N , H n is the number of shares of stock S n .
Observation 1.You do not have to think of H n neither as nonnegative nor as an integer.
Define the value process V  {Vt , t  0,1} as the total worth of the portfolio at times
t  0 and t  1 . Then

N
Vt  H 0 Bt   H n S n (t ), t  0,1 .
n 1
Define the gains process
N
N
n 1
n 1
G  H 0 r   H n ( S n (1)  S n (0))  H 0 r   H n S n .
As you can see, V1  V0  G so any change in the value process is due to changes in the
values of the stocks and not due to additional investments.
Since between times t  0 and t  1 the value of the money may not stay the same, we
compare V0 and V1 by looking at their values at the same moment in time, particularly
t=0. We do this by computing the discounted price process
S (t )
S * (t ) 
.
Bt
S(1)
It follows then that S * (0)  S (0), and S* (1) 
.
1 r
Similarly, we define the discount value process V * (t ) 
process G * (t ) 
V (t )
and the discount gains
Bt
G (t )
.
Bt
Example 1. Let d  2, N  1, r  1/10, S (0)  6, S (1, 1 )  11/ 2, S (1, 2 )  77 /10 .
11
For any strategy, we have V0  H 0  6 H1  V0* , and V1  H 0  H1S (1) , but since there
10
are two states of the system, we have
11
11
11
77
V1 (1 )  H 0  H1 , and V1 (2 )  H 0 
H1 .
10
2
10
10
Accordingly, we get V1* (1 )  H 0  5H1 , and V1* (2 )  H 0  7 H1 .
33
Example 2. With the data as above, but the state space   {1 , 2 , 3}, S (1, 3 ) 
, we
10
get V1* (1 )  H 0  5H1 , V1* (2 )  H 0  7 H1 and V1* (3 )  H 0  3H1.
Example 3. Let d  3, r  1/10 and the stock prices given by the table
n S n (0)
S n (1)
1
1
2
6
5
2
3
33/10 22/10 88/10
66/10 44/10 22/10
We obtain the discounted stock price process below and the value process can be easily
calculated.
n S n* (0)
S n* (1)
1
2
3
8
2
1
6
3
2
2
5
6
4
The next example is left to the reader.
Example 4. Consider a state space with four states and a market with 2 securities, and the
stock values as in Example 3, and in addition S1 (1,  4 )  22 /10, S2 (1,  4 )  77 /10.
Compute the value and gains processes, and their discounted values as well.
These four examples will be used to present some anomalies which must not happen in a
reasonable economic model.
Definition 2. A trading strategy Ĥ is said to be dominant is there exists another trading
strategy, say H such that Vˆ0  V0 , and Vˆ1 ()  V1(ω) for all  .
Existence of dominant strategies does not make economic sense. In particular, if a
dominant strategy exists, then a person with initial zero wealth can end with a positive
amount of money, as the following theorem shows:
Theorem 1. There exists a dominant strategy if and only if there exists a trading strategy
satisfying V0  0, and V1 ( )  0 for all   .
Proof
We have to prove two statements in this theorem since it expresses a necessary and
sufficient condition for existence of a dominant strategy.
Let us assume first that a dominant strategy, Ĥ exists. Then, according to Definition 2,
there must be another strategy, H such that Vˆ0  V0 , and Vˆ1 ()  V1(ω) for all  . Consider
then the strategy H  Hˆ  H . For this strategy we have
V  Vˆ  V , and Vˆ  Vˆ ()  V (ω)  0 for all  ,
0
0
0
1
1
1
which is what we had to prove.
Conversely, assume that a trading H strategy satisfies V0  0, and V1 ( )  0 for all
  . Then H is a dominant strategy since it dominates the strategy for which
V0  0, and V1 ( )  0 for all  .
QED
Existence of dominant trading strategies leads to illogical pricing. The interpretation is
the following: if a strategy Ĥ dominates a strategy H, then at time t  0, Vˆ0  V0 , which
can be seen as the initial price of the contract, but at time t  1, strategy Ĥ is sure to be
more profitable, having a higher payoff in all the states. This is not in accordance with
what happens in reality.
One way to avoid illogical pricing is introducing the notion of linear pricing measure.
Definition 3. A linear pricing measure is a non-negative vector
  ( (1 ),  ( 2 ),...,  ( d )) , such that for every trading strategy H, we have
V1 ( )
.
(1)
B1 ( )


Observation 2. Notice that if in (1) we consider the strategy that invests only in the bank
account and nothing in the risky assets, we get
H B ( )
V0*   ( )V1* ( )   ( ) 0 1
 H 0  ( ), but V0*  V0  H 0 , which implies
B1 ( )



V0*   ( ) V1* ( )   ( )
 ( )  1 , therefore 


must be a probability measure. Moreover, if in (1) we consider
the strategy that invests in just one particular asset S n (thus H j  0 for all j  n ), then
S n* (0) 
 ( ) S


*
n
(1,  ), for all n  1,..., N .
(2)
Notice that the discounted price of any asset at time t  0 is its initial price. Equation (2)
says that the initial price of the asset equals the expected value of its discounted future
price, under the linear pricing measure  .
Theorem 2. The vector  is a linear pricing measure is and only if it is a probability
measure satisfying (2).
Proof Again this statement gives a necessary and sufficient condition for a d-dimensional
vector to be a linear measure. Compared to Theorem 1, the equivalent statement is
actually a system of equations to be checked.
Now, from Observation 2, it has been already proved that if  is a linear pricing measure,
then it is a probability measure satisfying (2).
We have to prove the converse statement. If  is a probability measure satisfying (2),
then  is a non-negative vector, and
N
N
V0*  V0  H 0   H n S n (0)  H 0   ( )   H n   ( ) S n* (1,  )   ( ) V1* ( )
n 1

n 1
therefore (1) holds, and  is a linear pricing measure.


QED
The next theorem relates the concept of dominant strategy with the linear pricing
measure. We will state this theorem without proof. Its proof involves linear optimization
principles that are beyond the scope of this course.
Theorem 3. There exists a linear pricing measure if and only if there are no dominant
strategies.
Now, Theorem 3 states that if we want to avoid models that have dominant strategies, we
need to concentrate on those models that have linear pricing measures. Two questions can
be asked at this moment:
1. What is easier to check: the existence of dominant strategies or the existence of a
linear pricing measure?
2. Are the models that admit a linear pricing measure satisfying from the economic
point of view?
Assume that a model does not have dominant strategies, this it is not possible to start with
zero wealth and end with positive wealth in all the states is the space. But what if it is
possible to start with zero wealth and end with positive wealth in at least one state? This
brings us to the notion of arbitrage.
Definition 4. An arbitrage opportunity is some trading strategy H such that
(a) V0  0
(b) V1 ( )  0, and V1 ( j )  0 for at least one j  1,..., d
An economic model is said to be in equilibrium if there are no arbitrage opportunities.
For short period of times there may exist arbitrage opportunities but they do not last,
since if they did, everybody would try to make profits without any risk.
Theorem 4. If there exists a dominant strategy, then there exists an arbitrage
opportunity.
Proof. The proof is immediate from Theorem 1.
QED
The converse of the above statement is not true, as the next example will show.
Example 5. Consider a market model with two states and a single security, with
50
d  2, N  1, r  1 / 9, S (0)  5, S (1, 1 ) 
, S (1,  2 )  7 .
9
Let us consider a trading strategy: H  ( H 0  5, H 1  1) . We have
V0  H 0  H 1 S (0)  0, and
1 50
1
13
V(1, 1 )  5  (1  ) 
 0, V(1,  2 )  5  (1  )  7  .
9
9
9
9
Therefore H is an arbitrage opportunity. However, this model does not admit dominant
trading strategies since there exists a linear pricing measure  (1 )  1,  ( 2 )  0 .
The definition of arbitrage involves the values of the portfolio at times t  0 and t  1 .
There is an equivalent definition involving the discounted values, and yet another one
involving the discounted gains process: H is an arbitrage opportunity if and only if
(a) V0*  0
(b) V1* ( )  0, and V1* ( j )  0 for at least one j  1,..., d
or, equivalently
(a) V0*  0
(b) G * ( )  0, and G * ( j )  0 for at least one j  1,..., d
Unfortunately, it is not easy to check directly whether a model has any arbitrage
opportunities, at least when there are two or more risky securities.
This brings us to the notion of risk neutral probability measure. Notice than in
Example 5 the pricing measure was defined in such a way that  ( )  0 in the sate in
which the value of the portfolio at time t  1 was positive.
Definition 5. A probability measure Q on  is a risk neutral probability measure if
(a) Q ( )  0 for all   , and
(b) EQ (S n* )  0, for all n  1,2..., N .
Notice that condition (b) in the definition is the same as the equation (2) satisfied by the
linear pricing measure. So a risk neutral probability measure is nothing but a linear
pricing measure that assigns positive probabilities to every state of the space.
Theorem 5. A probability measure Q on  is a risk neutral probability measure if and
only if
(3) Q ( )  0 for all   , and
(4) EQ (G* )  0, for any discounted gains process .
Proof.
Notice that (a) in Definition 5 and (3) are equivalent, so we only need to show that (b)
and (4) are equivalent.
N
If Q is a risk neutral probability measure, and if G * ( )   H n S n* ( ) is the discounted
n 1
gains process associated with a trading strategy, then
N
EQ (G* )   G* ( )Q( )   H n Sn* ( )Q( )

 n 1
N
N
  H n ( Sn* ( )Q( ))  H n EQ Sn*  0
n 1

n 1
Conversely, assume Q is a probability measure satisfying (3) and (4). Take for each
security in the model the strategy that invests only in the bond and one unit of security n:
H n  1 . For this strategy the discounted gains process is G* ( )  Sn* ( ) , and for this (4)
is identical with (b).
QED
Theorem 6. There are no arbitrage opportunities if and only if there exists a risk neutral
probability measure.
We will leave this theorem without proof since some arguments involve higher linear
algebra knowledge.
Let us see the technical details involved in checking whether a model admits a risk
neutral probability measure.
Example 6. Consider the same information as in Example 1. We are given
d  2, N  1, r  1/10, S (0)  6, S (1, 1 )  11/ 2, S (1, 2 )  77 /10 . We want to check
whether there exist Q (1 )  0, Q(2 )>0 and such that
S (0)  S * (1, 1 )Q(1 )  S * (1, 2 )Q(2 ) 
S (1, 1 )Q(1 )  S (1, 2 )Q(2 )
1 r
Q(1 )  Q(2 )  1
5Q(1 )  7Q(2 )  6
We obtain the system
, which has a unique solution
Q(1 )  Q(2 )  1
1
Q(1 )  Q(2 )  .
2
So the model has a risk neutral probability measure, and therefore there cannot exist
arbitrage opportunities.
Now this one was fairly easy to solve. Let us see another one.
Example 7. Refer back to the data in Example 2:
r  1/10, S (0)  6, S (1, 1 )  11/ 2, S (1, 2 )  77 /10, S (1, 3 )  33/10 .
As before we need to solve the system:
5Q(1 )  7Q(2 )  3Q(3 )  6
Q(1 )  Q(2 )  Q(3 )  1
This system has infinitely many solutions. With Q(3 )   , we get Q(2 ) 
1  2
and
2
1  4
1
. But since Q(i )  0 , we must have   0, and   . This shows that
2
4
the system has infinitely many solutions, even under the restriction that all solutions are
positive.
So there are infinitely many risk neutral probability measures, and again, there are no
arbitrage opportunities.
Q (1 ) 
Example 8. Let us now look at the data in Example 3. Let d  3, r  1/10 and the stock
prices given by the table
n S n (0)
S n (1)
1
1
2
6
5
2
3
33/10 22/10 88/10
66/10 44/10 22/10
We obtain the discount stock price process below:
n S n* (0)
S n* (1)
1
2
3
1
6
3
2
8
2
5
6
4
2
Again, in order to check the existence of a risk neutral probability measure we need to
find a positive solution to the system
3Q(1 )  2Q(2 )  8Q(3 )  6
6Q(1 )  4Q(2 )  2Q(3 )  5
Q(1 )  Q(2 )  Q(3 )  1
As it can be seen, the system has a unique solution given by
1
1
Q(1 )  1, Q(2 )   , Q(1 )  , but it is not a risk neutral probability measure.
2
2
As it could be seen from the previous examples, the element that lead to either the
existence of several risk neutral probability measures or none at all was the complexity of
the model: more states and more securities.
As mentioned earlier, a proof of Theorem 6 will not be given. Here is however an
intuitive approach to its proof.
Denote by W  { X  R d : X  G* for some trading strategy H } . That is W is the set of
vectors in R d representing all possible discounted gains for the given model. More than
that, notice that for any two vectors in W , X1 and X 2 , and any two scalars a, b , the
vector aX 1  bX 2  W , since it represents the discounted gain aG1*  bG2* which can be
attained by using the strategy aH1  bH 2 . Therefore W is a linear subspace of R d .
Now, if arbitrage opportunities exist, we are looking for strategies that lead to
nonnegative gains. Define
A  { X  R d : X ( j )  0, X  0} ,
where X ( j ) is the j-th component of the vector X.
A model that admits arbitrage opportunities will have W  A   .
In view of Theorem 5, and risk neutral probability measure is a vector
d
Q  R d , Q( j )  0,  Q( j )  1 that satisfies (4). In the “language” of vectors this means
j 1
d
X  Q   X ( j )Q( j )  0
(5)
j 1
In the above, X  Q is nothing but the dot product of the vectors X and Q, and the
equation means that the two vectors are orthogonal in R d . But any vector satisfying
X  Q  0 is in the orthogonal complement of the subspace W , and conversely, any vector
in the orthogonal complement satisfies (5). So, if we denote by P  {Q  0,  Q( j )  1}
j
the set of all positive probability measures, the model admits a risk neutral probability
measure if and only if W   P   , where W   {Y : X  Y  0, for all X W } .
Example 9. Let us refer to Example 1 again and describe the subspace W. We have the
discounted value process V1* (1 )  H 0  5H1 , and V1* (2 )  H 0  7 H1 , so the discounted
gains process is G* (1 )   H1 , and G* (2 )  H1 , therefore W  {(1,1) H1 , H1  R} , and
it represents a line in the plane passing through the origin: x  y  0 . Then the orthogonal
complement is W   {( x, y) :  x  y  0} , another line in the plane, perpendicular onto
the above.
EXERCISES
1. Compute the value process, the discounted values process and the discounted gains
process for Example 3.
2. Compute the value process, the discounted values process and the discounted gains
process for Example 3.
3. Compute V , V * , G and G* for Example 4.
4. As seen in Example 8, for the model defined in example 3 there does not exist a risk
neutral probability measure. Therefore, according to Theorem 6, arbitrage
opportunities do exist. Find all arbitrage opportunities.
5. Does the model in Example 3 admit dominant trading strategies?
For those who loved linear algebra and still remember some of it:
6. Find W and W  for the model in Example 2.
7. Find W and W  for the model in Example 3.
8. Find W and W  for the model in Example 4.
9. Give an example of a model with two states and two securities that has arbitrage
opportunities, and find all arbitrage opportunities.