Solving this equation for f , f (x, T ) is what V0 should be, i.e., the cost of setting up the

equivalent portfolio. Equation (22.5) shows what the trading strategy should be.

92

23. The fundamental theorem of ¬nance.

In Section 19, we showed there was a probability measure under which Pt = e’rt St

was a martingale. This is true very generally. Let St be the price of a security in today™s

dollars. We will suppose St is a continuous semimartingale, and can be written St =

M t + At .

Arbitrage means that there is a trading strategy Hs such that there is no chance that

we lose anything and there is a positive pro¬t with positive probability. Mathematically,

arbitrage exists if there exists Hs that is adapted and satis¬es a suitable integrability

condition with

T

Hs dSs ≥ 0, a.s.

0

and

T

Hs dSs > b > µ

P

0

for some b, µ > 0. It turns out that to get a necessary and su¬cient condition for St to be

a martingale, we need a slightly weaker condition.

The NFLVR condition (“no free lunch with vanishing risk”) is that there do not

exist a ¬xed time T , µ, b > 0, and Hn (that are adapted and satisfy the appropriate

integrability conditions) such that

T

1

Hn (s) dSs > ’ , a.s.

n

0

for all t and

T

Hn (s) dSs > b > µ.

P

0

Here T, b, µ do not depend on n. The condition says that one can with positive

probability µ make a pro¬t of b and with a loss no larger than 1/n.

Two probabilities P and Q are equivalent if P(A) = 0 if and only Q(A) = 0,

i.e., the two probabilities have the same collection of sets of probability zero. Q is an

equivalent martingale measure if Q is a probability measure, Q is equivalent to P, and St

is a martingale under Q.

Theorem 23.1. If St is a continuous semimartingale and the NFLVR conditions holds,

then there exists an equivalent martingale measure Q.

The proof is rather technical and involves some heavy-duty measure theory, so we

will only point examine a part of it. Suppose that we happened to have St = Wt + f (t),

where f (t) is a deterministic increasing continuous function. To obtain the equivalent

martingale measure, we would want to let

1

t t

(f (s))2 ds

’ f (s)dWs ’ 2

Mt = e .

0 0

93

In order for Mt to make sense, we need f to be di¬erentiable. A result from measure

theory says that if f is not di¬erentiable, then we can ¬nd a subset A of [0, ∞) such

t

that 0 1A (s)ds = 0 but the amount of increase of f over the set A is positive. This last

statement is phrased mathematically by saying

t

1A (s)df (s) > 0,

0

where the integral is a Riemann-Stieltjes (or better, a Lebesgue-Stieltjes) integral. Then

if we hold Hs = 1A (s) shares at time s, our net pro¬t is

t t t

Hs dSs = 1A (s)dWs + 1A (s) df (s).

0 0 0

The second term would be positive since this is the amount of increase of f over the set

t t

A. The ¬rst term is 0, since E ( 0 1A (s)dWs )2 = 0 1A (s)2 ds = 0. So our net pro¬t is

nonrandom and positive, or in other words, we have made a net gain without risk. This

contradicts “no arbitrage.” See Note 1 for more on this.

Sometime Theorem 23.1 is called the ¬rst fundamental theorem of asset pricing.

The second fundamental theorem is the following.

Theorem 23.2. The equivalent martingale measure is unique if and only if the market is

complete.

We will not prove this.

Note 1. We will not prove Theorem 23.1, but let us give a few more indications of what is

going on. First of all, recall the Cantor set. This is where E1 = [0, 1], E2 is the set obtained

from E1 by removing the open interval ( 3 , 2 ), E3 is the set obtained from E2 by removing

1

3

the middle third from each of the two intervals making up E2 , and so on. The intersection,

E = ©∞ En , is the Cantor set, and is closed, nonempty, in fact uncountable, yet it contains

n=1

no intervals. Also, the Lebesgue measure of A is 0. We set A = E. Let f be the Cantor-

Lebesgue function. This is the function that is equal to 0 on (’∞, 0], 1 on [1, ∞), equal to

1 12 1 12 3 78

2 on the interval [ 3 , 3 ], equal to 4 on [ 9 , 9 ], equal to 4 on [ 9 , 9 ], and is de¬ned similarly on

each interval making up the complement of A. It turns out we can de¬ne f on A so that it is

1

continuous, and one can show 0 1A (s)df (s) = 1. So this A and f provide a concrete example

of what we were discussing.

94

24. American puts.

The proper valuation of American puts is one of the important unsolved problems

in mathematical ¬nance. Recall that a European put pays out (K ’ ST )+ at time T ,

while an American put allows one to exercise early. If one exercises an American put at

time t < T , one receives (K ’ St )+ . Then during the period [t, T ] one receives interest,

and the amount one has is (K ’ St )+ er(T ’t) . In today™s dollars that is the equivalent of

(K ’ St )+ e’rt . One wants to ¬nd a rule, known as the exercise policy, for when to exercise

the put, and then one wants to see what the value is for that policy. Since one cannot look

into the future, one is in fact looking for a stopping time „ that maximizes

E e’r„ (K ’ S„ )+ .

There is no good theoretical solution to ¬nding the stopping time „ , although good

approximations exist. We will, however, discuss just a bit of the theory of optimal stopping,

which reworks the problem into another form.

Let Gt denote the amount you will receive at time t. For American puts, we set

Gt = e’rt (K ’ St )+ .

Our problem is to maximize E G„ over all stopping times „ .

We ¬rst need

Proposition 24.1. If S and T are bounded stopping times with S ¤ T and M is a

martingale, then

E [MT | FS ] = MS .

Proof. Let A ∈ FS . De¬ne U by

S(ω) if ω ∈ A,

U (ω) =

T (ω) if ω ∈ A.

/

It is easy to see that U is a stopping time, so by Doob™s optional stopping theorem,

E M0 = E MU = E [MS ; A] + E [MT ; Ac ].

Also,

E M0 = E MT = E [MT ; A] + E [MT ; Ac ].

Taking the di¬erence, E [MT ; A] = E [Ms ; A], which is what we needed to show.

Given two supermartingales Xt and Yt , it is routine to check that Xt § Yt is also a

n n

supermartingale. Also, if Xt are supermartingales with Xt “ Xt , one can check that Xt

95

is again a supermartingale. With these facts, one can show that given a process such as

Gt , there is a least supermartingale larger than Gt .

So we de¬ne Wt to be a supermartingale (with respect to P, of course) such that

Wt ≥ Gt a.s for each t and if Yt is another supermartingale with Yt ≥ Gt for all t, then

Wt ¤ Yt for all t. We set „ = inf{t : Wt = Gt }. We will show that „ is the solution to the

problem of ¬nding the optimal stopping time. Of course, computing Wt and „ is another

problem entirely.

Let

Tt = {„ : „ is a stopping time, t ¤ „ ¤ T }.

Let

Vt = sup E [G„ | Ft ].

„ ∈Tt

Proposition 24.2. Vt is a supermartingale and Vt ≥ Gt for all t.

Proof. The ¬xed time t is a stopping time in Tt , so Vt ≥ E [Gt | Ft ] = Gt , or Vt ≥ Gt . so

we only need to show that Vt is a supermartingale.

Suppose s < t. Let π be the stopping time in Tt for which Vt = E [Gπ | Ft ].

π ∈ Tt ‚ Ts . Then

E [Vt | Fs ] = E [Gπ | Fs ] ¤ sup E [G„ | Fs ] = Vs .

„ ∈Ts

Proposition 24.3. If Yt is a supermartingale with Yt ≥ Gt for all t, then Yt ≥ Vt .

Proof. If „ ∈ Tt , then since Yt is a supermartingale, we have

E [Y„ | Ft ] ¤ Yt .

So

Vt = sup E [G„ | Ft ] ¤ sup E [Y„ | Ft ] ¤ Yt .

„ ∈Tt „ ∈Tt

What we have shown is that Wt is equal to Vt . It remains to show that „ is optimal.

There may in fact be more than one optimal time, but in any case „ is one of them. Recall

we have F0 is the σ-¬eld generated by S0 , and hence consists of only … and „¦.

96

Proposition 24.4. „ is an optimal stopping time.

Proof. Since F0 is trivial, V0 = sup„ ∈T0 E [G„ | F0 ] = sup„ E [G„ ]. Let σ be a stopping

time where the supremum is attained. Then

V0 ≥ E [Vσ | F0 ] = E [Vσ ] ≥ E [Gσ ] = V0 .

Therefore all the inequalities must be equalities. Since Vσ ≥ Gσ , we must have Vσ = Gσ .

Since „ was the ¬rst time that Wt equals Gt and Wt = Vt , we see that „ ¤ σ. Then

E [G„ ] = E [V„ ] ≥ EVσ = E Gσ .

Therefore the expected value of G„ is as least as large as the expected value of Gσ , and

hence „ is also an optimal stopping time.

The above representation of the optimal stopping problem may seem rather bizarre.

However, this procedure gives good usable results for some optimal stopping problems. An

example is where Gt is a function of just Wt .

97

25. Term structure.

We now want to consider the case where the interest rate is nondeterministic, that

is, it has a random component. To do so, we take another look at option pricing.

Accumulation factor. Let r(t) be the (random) interest rate at time t. Let

t

r(u)du

β(t) = e 0

be the accumulation factor. One dollar at time T will be worth 1/β(T ) in today™s dollars.

Let V = (ST ’ K)+ be the payo¬ on the standard European call option at time T

with strike price K, where St is the stock price. In today™s dollars it is worth, as we have

seen, V /β(T ). Therefore the price of the option should be

V

.

E

β(T )

We can also get an expression for the value of the option at time t. The payo¬, in terms

of dollars at time t, should be the payo¬ at time T discounted by the interest or in¬‚ation

rate, and so should be

T

’ r(u)du

(ST ’ K)+ .

e t

Therefore the value at time t is

β(t) V

T

’ r(u)du

(ST ’ K)+ | Ft = E V | Ft = β(t)E | Ft .

Ee t

β(T ) β(T )

From now on we assume we have already changed to the risk-neutral measure and

we write P instead of P.

Zero coupon. A zero coupon bond with maturity date T pays $1 at time T and nothing

before. This is equivalent to an option with payo¬ value V = 1. So its price at time t, as

above, should be

1 T

’ r(u)du