Jessica Wachter

Class notes for Finance 911

The Martingale Representation Approach

1. Notation and definitions

2. Martingale property of prices and no-arbitrage
3. Individual optimization
4. Asset pricing: Binomial model

Notation and definitions

Assume T periods. Let

S1t

St = ... ex-dividend prices of risky securities at time t

SM t

X1t

Xt = ... dividends at time t

XM t
Sj,t+1 + Xj,t+1
rj,t+1 =
1
return on security j from time t to t + 1
Sjt
rf,t+1 = return on the riskfree security from time t to t + 1
Let Bt be the price of the riskfree security (assume B0 > 0). Note: Bt+1 = (1 + rf,t+1 )Bt .
Let

D1t

Dt = ... ,
DM t

where
Djt = ts=0 Xjs ,
be the accumulated dividend process.
Define the share process:

t+1

1,t+1

..
=
,
.
M,t+1
1

Jessica Wachter

Class notes for Finance 911

where
j,t+1 = shares in security j chosen at time t
and
t+1 = shares in riskfree security.
Recall:
Definition. A consumption plan c is financed by (, ) if
>
St = t Bt + t> (St + Xt ) ct
t+1 Bt + t+1

for t = 0, 1, ..., T 1 and

CT = T BT + T> (ST + XT ).
Note: We assume that ct is adapted to F, and t predictable by F.
Now we will extend our definition of arbitrage to multiple periods:
Definition. An arbitrage opportunity is a consumption plan c financed by (, ) s.t.
1. ct 0

t = 0, ..., T .

2. ct > 0 for at least one event at Ft for some t.

3. 0 B0 + 0> (S0 + X0 ) 0.
LHS of last item is the cost of setting up the consumption plan. Note that this has the
same interpretation as in the 2-period case: the payoff is non-negative and must be
strictly positive at some state of the world.
Definition. A process Yt , adapted to F, is a martingale under a probability if
Et [Ys ] = Yt

s t

where Et [] denotes expectation taken w.r.t. conditional on information available up

to time t.

Jessica Wachter

Class notes for Finance 911

Define prices normalized by the bond price Bt :

Sjt
Bt
Bt
=1
=
Bt
ct
=
Bt
Xjt
=
Bt

Sjt
=

Bt
ct

Xjt

Djt
= ts=0 Xjs
= ts=0

Xjs
Bs

Martingale properties of prices and no-arbitrage

The following theorem is the multiperiod version of the Fundamental Theorem of Asset
Pricing (FTAP).
Theorem. No arbitrage a probability measure such that

Sjt
+ Djt
= Et [Sj,t+1
+ Dj,t+1
]

where Et [] denotes expectations taken under .

Notes:
1. is called the equivalent martingale measure.
2. The theorem says that normalized prices plus normalized accumulated dividends
are a martingale under .

3. Subtracting Djt
from both sides implies

Sjt
= Et [Sj,t+1
+ Xj,t+1
].

Solving forward (and noting that no arbitrage implies that ST = ST = 0) implies

that

Sjt
= Et [Ts=t+1 Xjs
].

Jessica Wachter

Class notes for Finance 911

4. is also called the risk-neutral measure because these relations would hold under
E[] for the risk-neutral agent.
Case of a risk-neutral agent:
Consider the Euler equation for an investor with period utility u:

 0
u (ct+1 )
(1 + rj,t+1 ) = 1 j.
Et 0
u (ct )
Consider a risk-neutral agent i.e. u00 = 0 u0 constant. The Euler equation for
the riskfree asset therefore implies
Et [(1 + rf,t+1 )] = 1.
Because the rf,t+1 is known at time t,
1
= .
1 + rf,t+1
Substituting in for in the Euler equation implies


1 + rj,t+1
Et
= 1.
1 + rf,t+1
If we have a risk-neutral agent, expected returns on all assets are equal!
Applying the definition of a return and dividing by Sjt


Sj,t+1 + Xj,t+1
= Sjt .
Et
1 + rf,t+1
Note:
Bt+1 = (1 + rf,t+1 )Bt .
Dividing by Bt gives us


Et


Sj,t+1 + Xj,t+1
Sjt
=
Bt (1 + rf,t+1 )
Bt

and therefore that




Et Sj,t+1
+ Xj,t+1
= Sjt
.

Adding Djt
implies




.
Et Sj,t+1
+ Dj,t+1
= Sjt
+ Djt

Therefore Sjt
+ Djt
is a martingale, as required.

Jessica Wachter

Class notes for Finance 911

A look back at FTAP in the 2 period model

In fact weve seen the equivalent martingale measure before.
Consider the 2 period model with state of nature . We showed that there is no
arbitrage if and only if there exists a random variable such that
X
Sj =
Xj j.

Define a new probability measure on :

= P

Note that

= 1 and > 0 . Further note that

B1
1
.
=P
B0

Therefore
Sj =

 
Sj
B1 X Xj X Xj X

Xj = E Xj .
=
=
=
B0
B0
B1
B1

This shows that in the 2-period case, there is a straightforward mapping between
and the equivalent martingale measure.
Proof of the multiperiod FTAP
We will assume something slightly stronger than no-arbitrage, namely that one
individual i with u0i > 0 such that
max
Ei
i i
c ,

T
X

it ui (cit )

t=0

has a solution given by the first order conditions (where Ei denotes expectations under
this individuals probability beliefs), and such that the individual assigns strictly
positive probabilities to each state. We will construct under this assumption. For a
proof of both the if and only if that uses only no-arbitrage, see Duffie chapter 2.

Jessica Wachter

Class notes for Finance 911

Recall that the probability of event at is equal to

X
at =

at

while the probability of as conditional on at is given by:

T
 as
if as at
P (as at )
at
as (at ) =
=
P (at )
0 otherwise
Given , define at , as (at ) similarly.
We will work with the FOC for a single agent. Accordingly, drop the i subscript. Recall
the Euler equation for this agent:
u0 (ct ) = Et [u0 (ct+1 )(1 + rj,t+1 )]
Rewriting:


u0 (ct+1 )
(Sj,t+1 + Xj,t+1 ) .
Sjt = Et 0
u (ct )


(1)

We can also apply this equation to the riskfree rate:

u0 (ct ) = Et [u0 (ct+1 )(1 + rf,t+1 )]
which implies that


u0 (ct+1 )
Bt+1 .
Bt = Et 0
u (ct )


(2)

Note that successive substitution and the law of iterated expectations implies that


0
T t u (cT )
Bt = Et
BT .
u0 (ct )

(3)

Finally, note that no-arbitrage implies that Bt > 0 for all t.

Proof:
The proof has two steps: (1) Construct a probability measure and (2) show that this
measure satisfies the martingale property.
1. We need to define such that > 0 for all and
Define
= T

= 1.

u0 (cT ()) BT ()
.
u0 (c0 )
B0
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Jessica Wachter

Class notes for Finance 911

BT ()
B0

Because > 0, u0 > 0 and

> 0, > 0.

Applying (3) at t = 0 implies

B0 =

u0 (cT ())
BT ().
u0 (c0 )

Dividing by B0 implies
X

u0 (cT ()) BT ()
= 1.
u0 (c0 )
B0

and therefore that

X

= 1.

2. Now we need to show that this implies the martingale property. It suffices to
show that



Sjt
= Et Sj,t+1
+ Xj,t+1
(4)
We first need to calculate the conditional probability at+1 (at ). By definition of at ,
at =

at

u0 (cT ()) BT ()
=

u0 (c0 )
B0
at
X
1
T u0 (cT ())BT ()
= 0
u (c0 )B0 a
t
t 0
at u (ct (at ))Bt (at ) X T t u0 (cT ())BT ()
=

u0 (c0 )B0
at
u0 (ct (at ))Bt (at )
a
X

= at t

u (ct (at ))Bt (at )

u0 (c0 )B0

The last equality follows from (3).

Applying the definition of conditional probability:
(
a
at+1
u0 (ct+1 (at+1 )) Bt+1 (at+1 )
= t+1

if at+1 at

u0 (ct (at ))
Bt (at )
at
at
at+1 (at ) =
0 otherwise

Jessica Wachter

Class notes for Finance 911

We now show (4). In event at , the LHS of (4) equals

Sjt (at )
Bt (at )
X
1
=
Bt (at ) a F

Sjt
(at ) =

t+1

at+1 (at )

t+1

u0 (ct+1 (at+1 ))
(Sj,t+1 (at+1 ) + Xj,t+1 (at+1 ))
u0 (ct (at ))

X at+1 u0 (ct+1 (at+1 ))

1

(Sj,t+1 (at+1 ) + Xj,t+1 (at+1 ))

=
Bt (at ) a a at
u0 (ct (at ))
t+1

X at+1 Bt (at )
1
(Sj,t+1 (at+1 ) + Xj,t+1 (at+1 ))
=
Bt (at ) a a at Bt+1 (at+1 )
t+1

X at+1

=
S
(a
)
+
X
(a
)
t+1
t+1
j,t+1
j,t+1
at
at+1 at
X

=
at+1 (at ) Sj,t+1
(at+1 ) + Xj,t+1
(at+1 )
at+1 Ft+1




The last summation is equal Et Sj,t+1

+ Xj,t+1
in event at .
2

Link to prices of state contingent claims

Define
=

B0
,
BT ()

at = at
and

B0
,
Bt (at )

as Bt (at )
as
as (at ) =
=
at
at Bs (as )

if as at and 0 otherwise. Note that this latter definition implies

as (at ) = as (at )

Bt (at )
Bs (as )

because of the definition of conditional probability applied to the risk-neutral measure.

8

Jessica Wachter

Class notes for Finance 911

We can interpret these as state prices. Why?

FTAP implies

Sjt

Et

T
X

Xjs

s=t+1

using the definition of S and X :

Sjt = E

T
X

Xjs

s=t+1

X X

X X

Bt
Bs

as (at )

as Fs

Bt (at )
Xjs
Bs (as )

as (at )Xjs .

as Fs

Therefore the price is given by the payoff in each state, multiplied by the value of a
dollar in that state. There is a one-to-one mapping between the equivalent martingale
measure and a system of state prices. Note that we do not need to take a stand on the
true probabilities to define either state prices or the equivalent martingale measure.
Now we ask, when is the equivalent martingale measure unique? Clearly this will be the
case if and only if this process is unique, and it shouldnt be a surprise, given our
2-period work, that this occurs when markets are complete.
Theorem. Assume no arbitrage. Then the equivalent martingale measure is unique if
and only if markets are dynamically complete.
The proof is very similar to the 2-period case.

Individual optimization

Consider problem P1:

max E
c,,

T
X

t u(ct )

(P1)

t=0

such that
>
t+1 Bt + t+1
St = t Bt + t> (St + Xt ) ct

(1a)

T BT + T> (ST + XT ) = cT

(1b)

0 B0 + 0> (S0 + X0 ) = W0 .

(1c)
9

Jessica Wachter

Class notes for Finance 911

Recall that if c, , and satisfy the first 2 of these conditions, then we say that (, )
finances c. Note that P1 is the standard dynamic portfolio choice and investment
problem.
Consider problem P2:
max E
c

T
X

t u(ct )

(P2)

t=0

T
X

ct = W0 ,

(2)

t=0

where E denotes expectation written w.r.t. the equivalent martingale measure.

0
ct = Bctt W0 = W
B0
Theorem. Assume no arbitrage and dynamically complete markets. Then P1 is
equivalent to P2, i.e. (c, , ) solves P1 iff (c) solves P2.
Lemma If c is financed by (, ) then
E

T
X

ct = 0 + 0> (S0 + X0 )

t=0

Think of wealth as an asset that pays consumption as its dividend. Then this statement
has to hold because the LHS is the total dividend, and the RHS is the value of wealth
at time 0.
Proof:
Recall homework problem:
>
t+1 Bt + t+1
St = t Bt + t> (St + Xt ) ct
T BT + T> (ST + XT ) = cT

T
X
t=0

ct = 0 B0 +

0> (S0

+ X0 ) +

T
X

t> (St

+ Xt St1 ) +

t=1

T
X

t (Bt Bt1 )

t=1

Rewrite the b.c. in term of quantities deflated by the bond price:

>
t+1 + t+1
St = t + t> (St + Xt ) ct

T + T> (ST + XT ) = cT
10

Jessica Wachter

Class notes for Finance 911

Then the same reasoning

T
X

ct

= 0 +

0> (S0

X0 )

T
X

)
t> (St + Xt St1

t=1

t=0

Take E of both sides, and note that

E

T
X

t> (St

Xt

)
St1

=E

T
X

)=0
[St + Xt ] St1
t> (Et1

t=1

t=1

which implies
E

T
X

ct = 0 + 0> (S0 + X0 ).

t=0

2
We now prove the theorem.
Proof:
Because the objective functions are the same, it suffices to show that constraints are the
same.
Suppose (, , c) satisfy (1). By (1a,b) and the Lemma, c satisfies
E

T
X

ct = 0 + 0> (S0 + X0 ),

t=0

and by (1c), c satisfies (2).

Now assume c satisfies (2).
By dynamic completeness, we can find a strategy , that finances c. Equations
(1a,b) must be satisfied for this strategy.
By Lemma, we must have
0 +

0> [S0

X0 ]

=E

T
X

ct = W0

t=0

and therefore, by (2), we have (1c).

11

Jessica Wachter

Class notes for Finance 911

2
Economic Importance
P2 decomposes the consumption and portfolio problem into 2 parts.
1. Find an optimal consumption plan ct by standard Lagrange method.
2. Implement the optimal consumption plan by dynamically trading in the
underlying securities.
How to do (1)? We need to write b.c. in terms of the physical measure.
First recall the relationship between state prices and the EMM:
B0
at = at
Bt (at )
Define the state-price density as state prices normalized by probabilities:
a
at = t
at
Replace state prices by the state-price density to obtain:
B0
at at = at
Bt (at )
It follows that
E

T
X

ct = E

t=0

X ct
Bt
t

Therefore, the budget constraint (2) can be rewritten as

X B0
E
ct = W0 .
Bt
t
We can then use the definition of at to rewrite (2) in terms of the physical measure:
"
#
X
E
t ct = W0
t

Now the expectation in the objective function and in the budget constraint are taken
under the same measure. We can therefore use standard the standard Lagrange
multiplier method to write the first-order condition for the agent as
t u0 (ct ) = t ,
where is a Lagrange multiplier determined by b.c. (2).
12

Jessica Wachter

Class notes for Finance 911

Asset Pricing: Binomial Economy

Assume 2 long-lived securities: Stock (assume dividends are zero) following a binomial
process

1 + rt+1 = SSt+1
can take one of two values u and d. Note that u(up)> d(down) and
t
p is the probability of an up move.
Bond Normalize B0 = 1. Let R = 1 + rf Bt = Rt
Note: markets are dynamically complete. No-arbitrage will give us the price of an
option (Cox, Ross & Rubinstein, 1979).
Let denote probability of u under the equivalent martingale measure.
What is in terms of the model parameters?

We know Et [St+1
] = St . Therefore,

St = St+1
(u) + (1 )St+1
(d).

Substituting in for S :
St
St+1 (u)
St+1 (d)
=
+ (1 )
.
Bt
Bt+1
Bt+1
Therefore
1
1
+ (1 )St+1 (d)
R
R
u
d

= St + (1 )St ,
R
R

St = St+1 (u)

and
1 =

u
d
+ (1 ) .
R
R
13

Jessica Wachter

Class notes for Finance 911

It follows that
=

Rd
.
ud

Note that

uR
.
ud
Given u > d, for > 0, we must have d < R. Also, for 1 > 0, we must have
u > R. This makes sense! If u, d are on the same side of R, we would have an arbitrage.


T
Prob(exactly n up moves at T ) =
( )n (1 )T n
n
T!
=
( )n (1 )T n
n!(T n)!
1 =

ST (exactly n up moves) = S0 un dT n
Consider the price of a European call option (or of an American or European call
option, assuming that rf 0) with exercise price K and time to expiration T . As
shown earlier in the course, the price (and payoff) of this option at time T is given by
CT = max[ST K, 0]. By no-arbitrage:
Ct = Et [CT ]


= Et R(T t) max[ST K, 0] .
Therefore,


Ct = Et R(T t) max[ST K, 0]

T t 
X
T t
(T t)
= R
( )n (1 )T tn max[St un dT tn K, 0]
n
n=0

T t 
X
T t
(T t)
= R
( )n (1 )T tn St un dT tn
n
n=j

KR

(T t)


T t 
X
T t
( )n (1 )T tn
n
n=j

where j = minimum # of up moves such that St un dT tn > K.

Let


T t 
X
T t
(j, T t, ) =
( )n (1 )T tn
n

n=j

14

Jessica Wachter

Class notes for Finance 911

It is convenient to define
0

u
Rd u
=
R
ud R

Then,
0

1 =
0

(u d)R (R d)u
dR + du
uR d
d
=
=
= (1 )
(u d)R
(u d)R
ud R
R
0

Note that > 0, 1 > 0. We can write the call price as follows:
0

Ct = St (j, T t, ) KR(T t) (j, T t, ).

In the limit, we can find the continous-time asset pricing result. Let r = log R. Then,
as the # of trials goes to , i.e. as T is divided into more subintervals,
0

(j, T t, ) N (d1 ),
where
d1 =

log(S/K) + (r + 2 /2)(T t)

T t

and
N (d1 ) = P (x d1 ) x N (0, 1).
Moreover,
(j, T t, ) N (d2 ),
where

d2 = d1 T t.

Therefore,
Ct = St N (d1 ) Ker(T t) N (d2 ).

15