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We continue, using (N = k) ∈ Fk ‚ Fk+1 ‚ Fk+2 , and we obtain

E [MN ; N = k] = E [Mk ; N = k] = E [Mk+1 ; N = k] = · · · = E [Mn ; N = k].



If we change the equalities in the above to inequalities, the same result holds for sub-
martingales.
As a corollary we have two of Doob™s inequalities:
Theorem 5.4. If Mn is a nonnegative submartingale,
1
P(maxk¤n Mk ≥ ») ¤ » E Mn .
(a)
2 2
E (maxk¤n Mk ) ¤ 4E Mn .
(b)
For the proof, see Note 2 below.
Note 1. We prove Proposition 5.1. If g is convex, then the graph of g lies above all the
tangent lines. Even if g does not have a derivative at x0 , there is a line passing through x0
which lies beneath the graph of g. So for each x0 there exists c(x0 ) such that

g(x) ≥ g(x0 ) + c(x0 )(x ’ x0 ).

Apply this with x = X(ω) and x0 = E [X | G](ω). We then have

g(X) ≥ g(E [X | G]) + c(E [X | G])(X ’ E [X | G]).

If g is di¬erentiable, we let c(x0 ) = g (x0 ). In the case where g is not di¬erentiable, then we
choose c to be the left hand upper derivate, for example. (For those who are not familiar with
derivates, this is essentially the left hand derivative.) One can check that if c is so chosen,
then c(E [X | G]) is G measurable.
Now take the conditional expectation with respect to G. The ¬rst term on the right is
G measurable, so remains the same. The second term on the right is equal to

c(E [X | G])E [X ’ E [X | G] | G] = 0.




Note 2. We prove Theorem 5.4. Set Mn+1 = Mn . It is easy to see that the sequence
M1 , M2 , . . . , Mn+1 is also a submartingale. Let N = min{k : Mk ≥ »} § (n + 1), the ¬rst
time that Mk is greater than or equal to », where a § b = min(a, b). Then

P(max Mk ≥ ») = P(N ¤ n)
k¤n


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and if N ¤ n, then MN ≥ ». Now

MN
P(max Mk ≥ ») = E [1(N ¤n) ] ¤ E ;N ¤ n (5.1)
»
k¤n
1 1
= E [MN §n ; N ¤ n] ¤ E MN §n .
» »
Finally, since Mn is a submartingale, E MN §n ¤ E Mn .
We now look at (b). Let us write M — for maxk¤n Mk . If E Mn = ∞, there is nothing
2

to prove. If it is ¬nite, then by Jensen™s inequality, we have

E Mk = E [E [Mn | Fk ]2 ] ¤ E [E [Mn | Fk ] ] = E Mn < ∞
2 2 2



for k ¤ n. Then
n
—2 2 2
¤E Mk < ∞.
E (M ) = E [ max Mk ]
1¤k¤n
k=1

We have

E [MN §n ; N ¤ n] = E [Mk§n ; N = k].
k=0

Arguing as in the proof of Theorem 5.3,

E [Mk§n ; N = k] ¤ E [Mn ; N = k],

and so

E [MN §n ; N ¤ n] ¤ E [Mn ; N = k] = E [Mn ; N ¤ n].
k=0

The last expression is at most E [Mn ; M — ≥ »]. If we multiply (5.1) by 2» and integrate over
» from 0 to ∞, we obtain
∞ ∞

E [Mn : M — ≥ »]
2»P(M ≥ »)d» ¤ 2
0 0

= 2E Mn 1(M — ≥») d»
0
M—
= 2E Mn d»
0

= 2E [Mn M ].

Using Cauchy-Schwarz, this is bounded by

2(E Mn )1/2 (E (M — )2 )1/2 .
2



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On the other hand,
∞ ∞

2»P(M ≥ »)d» = E 2»1(M — ≥») d»
0 0
M—
2» d» = E (M — )2 .
=E
0
We therefore have
E (M — )2 ¤ 2(E Mn )1/2 (E (M — )2 )1/2 .
2


Recall we showed E (M — )2 < ∞. We divide both sides by (E (M — )2 )1/2 , square both sides,
and obtain (b).

Note 3. We will show that bounded martingales converge. (The hypothesis of boundedness
can be weakened; for example, E |Mn | ¤ c < ∞ for some c not depending on n su¬ces.)
Theorem 5.5. Suppose Mn is a martingale bounded in absolute value by K. That is,
|Mn | ¤ K for all n. Then limn’∞ Mn exists a.s.

Proof. Since Mn is bounded, it can™t tend to +∞ or ’∞. The only possibility is that it
might oscillate. Let a < b be two rationals. What might go wrong is that Mn might be larger
than b in¬nitely often and less than a in¬nitely often. If we show the probability of this is 0,
then taking the union over all pairs of rationals (a, b) shows that almost surely Mn cannot
oscillate, and hence must converge.
Fix a < b, let Nn = (Mn ’ a)+ , and let S1 = min{k : Nk ¤ 0}, T1 = min{k > S1 :
Nk ≥ b ’ a}, S2 = min{k > T1 : Nk ¤ 0}, and so on. Let Un = max{k : Tk ¤ n}. Un
is called the number of upcrossings up to time n. We want to show that maxn Un < ∞ a.s.
Note by Jensen™s inequality Nn is a submartingale. Since S1 < T1 < S2 < · · ·, then Sn+1 > n.
We can write
n+1 n+1
2K ≥ Nn ’ NSn+1 §n = (NSk+1 §n ’ NTk §n ) + (NTk §n ’ NSk §n ).
k=1 k=1
Now take expectations. The expectation of the ¬rst sum on the right and the last term are
greater than or equal to zero by optional stopping. The middle term is larger than (b ’ a)Un ,
so we conclude
(b ’ a)E Un ¤ 2K.
Let n ’ ∞ to see that E maxn Un < ∞, which implies maxn Un < ∞ a.s., which is what we
needed.

Note 4. We will state Fatou™s lemma in the following form.
If Xn is a sequence of nonnegative random variables converging to X a.s., then E X ¤
supn E Xn .
This formulation is equivalent to the classical one and is better suited for our use.

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6. The one step binomial asset pricing model.
Let us begin by giving the simplest possible model of a stock and see how a European
call option should be valued in this context.
Suppose we have a single stock whose price is S0 . Let d and u be two numbers with
0 < d < 1 < u. Here “d” is a mnemonic for “down” and “u” for “up.” After one time unit
the stock price will be either uS0 with probability P or else dS0 with probability Q, where
P + Q = 1. We will assume 0 < P, Q < 1. Instead of purchasing shares in the stock, you
can also put your money in the bank where one will earn interest at rate r. Alternatives
to the bank are money market funds or bonds; the key point is that these are considered
to be risk-free.
A European call option in this context is the option to buy one share of the stock
at time 1 at price K. K is called the strike price. Let S1 be the price of the stock at time
1. If S1 is less than K, then the option is worthless at time 1. If S1 is greater than K, you
can use the option at time 1 to buy the stock at price K, immediately turn around and
sell the stock for price S1 and make a pro¬t of S1 ’ K. So the value of the option at time
1 is
V1 = (S1 ’ K)+ ,

where x+ is max(x, 0). The principal question to be answered is: what is the value V0 of
the option at time 0? In other words, how much should one pay for a European call option
with strike price K?
It is possible to buy a negative number of shares of a stock. This is equivalent to
selling shares of a stock you don™t have and is called selling short. If you sell one share
of stock short, then at time 1 you must buy one share at whatever the market price is at
that time and turn it over to the person that you sold the stock short to. Similarly you
can buy a negative number of options, that is, sell an option.
You can also deposit a negative amount of money in the bank, which is the same
as borrowing. We assume that you can borrow at the same interest rate r, not exactly a
totally realistic assumption. One way to make it seem more realistic is to assume you have
a large amount of money on deposit, and when you borrow, you simply withdraw money
from that account.
We are looking at the simplest possible model, so we are going to allow only one
time step: one makes an investment, and looks at it again one day later.
Let™s suppose the price of a European call option is V0 and see what conditions
one can put on V0 . Suppose you start out with V0 dollars. One thing you could do is
buy one option. The other thing you could do is use the money to buy ∆0 shares of
stock. If V0 > ∆0 S0 , there will be some money left over and you put that in the bank. If
V0 < ∆0 S0 , you do not have enough money to buy the stock, and you make up the shortfall
by borrowing money from the bank. In either case, at this point you have V0 ’ ∆0 S0 in

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the bank and ∆0 shares of stock.
If the stock goes up, at time 1 you will have

∆0 uS0 + (1 + r)(V0 ’ ∆0 S0 ),

and if it goes down,
∆0 dS0 + (1 + r)(V0 ’ ∆0 S0 ).

We have not said what ∆0 should be. Let us do that now. Let V1u = (uS0 ’ K)+
and V1d = (dS0 ’ K)+ . Note these are deterministic quantities, i.e., not random. Let

V1u ’ V1d
∆0 = ,
uS0 ’ dS0
and we will also need
1 + r ’ d u u ’ (1 + r) d
1
W0 = V1 + V1 .
u’d u’d
1+r
In a moment we will do some algebra and see that if the stock goes up and you had
bought stock instead of the option you would now have

V1u + (1 + r)(V0 ’ W0 ),

while if the stock went down, you would now have

V1d + (1 + r)(V0 ’ W0 ).

Let™s check the ¬rst of these, the second being similar. We need to show

∆0 uS0 + (1 + r)(V0 ’ ∆0 S0 ) = V1u + (1 + r)(V0 ’ W0 ). (6.1)

The left hand side of (6.1) is equal to

V1u ’ V1d
∆0 S0 (u ’ (1 + r)) + (1 + r)V0 = (u ’ (1 + r)) + (1 + r)V0 . (6.2)
u’d
The right hand side of (6.1) is equal to

1 + r ’ d u u ’ (1 + r) d
V1u ’ V1 + V1 + (1 + r)V0 . (6.3)
u’d u’d

Now check that the coe¬cients of V0 , of V1u , and of V1d agree in (6.2) and (6.3).
Suppose that V0 > W0 . What you want to do is come along with no money, sell
one option for V0 dollars, use the money to buy ∆0 shares, and put the rest in the bank

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(or borrow if necessary). If the buyer of your option wants to exercise the option, you give
him one share of stock and sell the rest. If he doesn™t want to exercise the option, you sell
your shares of stock and pocket the money. Remember it is possible to have a negative
number of shares. You will have cleared (1 + r)(V0 ’ W0 ), whether the stock went up or
down, with no risk.
If V0 < W0 , you just do the opposite: sell ∆0 shares of stock short, buy one option,
and deposit or make up the shortfall from the bank. This time, you clear (1 + r)(W0 ’ V0 ),
whether the stock goes up or down.
Now most people believe that you can™t make a pro¬t on the stock market without
taking a risk. The name for this is “no free lunch,” or “arbitrage opportunities do not
exist.” The only way to avoid this is if V0 = W0 . In other words, we have shown that the
only reasonable price for the European call option is W0 .
The “no arbitrage” condition is not just a re¬‚ection of the belief that one cannot get
something for nothing. It also represents the belief that the market is freely competitive.
The way it works is this: suppose W0 = $3. Suppose you could sell options at a price
V0 = $5; this is larger than W0 and you would earn V0 ’ W0 = $2 per option without risk.
Then someone else would observe this and decide to sell the same option at a price less

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