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MASM22 - Lecture Notes in Financial Mathematics

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					Lecture Notes in Financial Mathematics
0     Introduction to the module
From the point of view of mathematics, we divide our world into a deterministic one, where
something happens with certainty (e.g. £100 put into a bank for a year with interest rate
3% will become 103 after a year), and in a random one, where something happens with
uncertainty (like a future price of a share of stock of some company). To describe random
phenomena, one uses probability theory, and in particular, theory of stochastic processes
and stochastic analysis. It is doubtless that financial markets belong to the random world.
     A lot of rigorous mathematics will be needed and learned. Below, we provide a list
of mathematical tools and some financial concepts.
     Mathematical terms: probability space, filtration, random variables, expectation and
conditional expectation, risk neutral probability measures, martingale and martingale
measure, stochastic process of discrete time.
     Financial terms: real world with uncertainty, information, stocks, bank accounts
(interest rate), discounted price, gain process, arbitrage and no arbitrage opportunities,
contingent claims (financial derivatives), European options, Americam options, complete
and incomplete markets. Financial market can be divided into two types.
     Underlying stocks–stock market, bond market,commodities markets, foreign curren-
cies market.

    • Stock markets, such as those in New York, London and Tokyo;

    • Bond markets, which deal in goverment and other bonds;

    • Currency markets, where currencies are bought and sold;

    • Commodity markets, where physical assets such as oil, gold, copper, wheat or
      electricity are traded.

     Derivatives–such as options and futures.
     Definition A European call option gives its holder the right, but not the obligation,
to purchase from the writer a prescribed asset for a prescribed price at a prescribed time
in the future.                                                                           2
     The prescribed time in the future is known as the expiry date or the exercise date or
the maturity. The prescribed purchase price is known as the strike price or the exercise
price.
     Example Today (1st October 2004) Professor Truman (the writer) writes a European
call option that gives you (the holder) the right to buy 200 shares in BT for £1.20 each
on 1st December 2004. On 1st December 2004 you would then take one of two actions:
     a if the actual value of a BT’s share turns out to be more than £1.20 you would
exercise your right to buy the shares from Prof. Truman–for you could immediately sell
them for a profit.
     b if the actual value of a BT’s share turns out to be less than £1.20 you would not
exercise your right to buy the shares from Prof. Truman–the deal is not worthwhile.
Note that because you are not obliged to purchase the shares, you do not lose money (in
case (a) you gain money and in case (b) you neither gain nor lose). Professor Truman
on the other hand will not gain any money on December 1st, and may lose an unlimited
amount. To compensate for this imbalance, when the option is agreed on October 1st you

                                            1
would be expected to pay Prof. Truman an amount of money known as the value of the
option.
     The key question that we address in this class is
     How much should the holder pay for the privilege of holding the option?
(In other words, how do we compute a fair price for the value of the option?)
     The direct opposite of a European call option is a European put option.
     Definition A European put option gives its holder the right, but not the obligation,
to sell to the writer a prescribed asset for a prescribed price at a prescribed time in the
future.                                                                                  2
     It is useful to visualize in terms of payoff diagrams. We let E denote the exercise
price and S denote the asset price at expiry date. (Of course, S is not known at the time
when the option is taken out.) At the expiry date, sell it for S, gaining an amount S − E.
On the other hand, if E ≥ S then the holder gains nothing. Hence, we say that the value
of the European call option at the expiry date, denote C, is
                                   C = max(S − E, 0).
Plotting S on the x-axis and C on the y-axis gives the payoff diagram in Figure 1.

                                           Call Option




               C




                0                               S
                         E


                       Figure 1: Payoff diagram for European call.


    Consider now a European put option. If, at the expiry date, E > S then the holder
may buy the asset at S and exercise the option by selling it at E, gaining an amount
E − S. On the other hand, if S ≥ E the holder should do nothing. Hence, the value of
the European put option at the expiry date, denote P , is
                                   P = max(E − S, 0).

                                            2
The corresponding payoff diagram is plotted in Figure 2.

               E




                       P




                   0                               S
                             E


                           Figure 2: Payoff diagram for European put.


     It is possible to plot payoff diagrams for combinations of options. For example,
suppose you hold a call option and a put option on the same asset with the same expiry
date and the same exercise price E. Then the overall value at expiry date is the sum of
max(S − E, 0) and max(E − S, 0), which is equivalent to |S − E|.
     Another possibility is to hold a call option with exercise price E1 and, for the same
asset and expiry date, to write a call option with excise price E2 , where E2 > E1 . At the
expiry date, the value of the first option is max(S − E1 , 0) and the value of the second is
− max(S − E2 , 0). Hence, the overall value at expiry is max(S − E1 , 0) − max(S − E2 , 0).
This combination gives an example of a bull spread. The corresponding payoff diagram
is plotted in Figure 3.

                              B = max(S − E1 , 0) − max(S − E2 , 0)

 Options have become extremely popular in recent times. They can be used for speculation
and for hedging. For example if you believe that Marks and Spencer shares are due to
increase then you may speculated buy becoming the holder of a suitable call option.
(Typically, you can make a greater profit relative to your original payout than you would
do by simply purchasing Marks and Spencer shares.) On the other hand, if you are the
owner of a company that is committed to purchasing a factory in Germany for an agreed
price in German Marks in three months’ time, then you may wish to hedge some risk by
taking out an option that makes some profit in the event that the UK pound drops in the
value against the German Mark.


                                               3
                                                Bull Spread    E
                                                                2
                E2−E1




                        0
                                                    S
                                      E
                                      1



                            Figure 3: Payoff diagram for a bull spread.


     To describe these markets, probability theory mainly uses two approaches. In the
first one, one uses continuous time (t ∈ [0, T ]) stochastic process for modeling financial
markets. Here, stochastic analysis, as a research field in mathematics, is needed to study
                                                                   o
complex real world problems. One uses Brownian motion, Itˆ calculus, measure theory,
stochastic differential equations and stochastic partial differential equations. An advanced
level mathematics is needed here
     In this module, we will consider the second approach, where one uses discrete time
(t = 0, 1, 2, . . . , T ) stochastic processes. This is a simplified version of continuous time
models. So, a pre-knowledge of measure theory and stochastic analysis can be avoided
in this module. But this module still offers rigorous mathematical approach to simplified
financial markets. All the important concepts will be covered in this module. Last but
not least, discrete time financial markets are heavily used by practitioners.
     Finally, we mention that in 1997, Myron Scholes and Robert Merton were awarded
the Nobel Prize in Economics for their work related to the celebrated Black–Scholes option
pricing formula (Fischner Black died in 1995).
     Exercises
     1) Insert the word “rise” or “fall” to complete the following sentences:
     The holder of a European call option hopes that the asset price will...
     The writer of a European call option hopes the the asset price will...
     The holder of European put option hopes the the asset price will...
     The writer of a European put option hopes the the asset price will...
     2) Convince yourself that max(S − E, 0) + max(E − S, 0) is equivalent to |S − E|.

                                                4
1     Single period security markets
1.1     Mathematical model and some basic concepts
We consider the simplest case for introductory purposes. So, we only consider a two stage
model. We have time t = 0 (now). We denote by S(0) the price of a particular stock at
t = 0. Next, we have time t = 1 (some point of time in the future), a terminal date. We
denote by S(1) the price of the stock at t = 1. There is no trading at times in between. At
time t = 0, S(0) is known, but the price of the stock at the future time t = 1 is unknown,
actually it is a random variable.
    We assume that we have a finite sample space

                               Ω = {ω1 , ω2 , . . . , ωK },         K < ∞.

So, there are K possible values for S(1):

                                 S(1, ω1 ), S(1, ω2 ), . . . , S(1, ωK ).

    There is a probability measure P on Ω which describes with which probability an
event ωi happens. We suppose that P (ωi ) > 0 for all i (i.e., each ωi may happen with a
positive probability) and the word probability means that
                                             K
                                                  P (ωi ) = 1.
                                            i=1


    We have a bank account process B = {B(0), B(1)}, where B(0) = 1 is a deposit of
1(bond) into a bank at time t = 0, and B(1) is then the value of the bank account at time
t = 1. We will suppose that B(1) is a given deterministic (non-random) value. Usually,
we require that B(1) ≥ 1. We denote

                                          γ:=B(1) − B(0),

which is called interest rate.
      A price process is S = {S(0), S(1)}, where S(t) is the price of a security (e.g. a share
of stock) at time t, t = 0, 1. As we said, S(0) is known and S(1) is a random variable,
i.e., a function of ω ∈ Ω. In general, we will have N different types of securities, so that
we will use a vector-valued price process S = {S(t), t = 0, 1}, where

                         S(t) = (S1 (t), S2 (t), . . . , SN (t)),      N < ∞,

and Sn (t) is the time t price of security n.
    A trading strategy
                                    H = (H0 , H1 , . . . , HN )
is an investor’s portfolio as carried forward from time t = 0 to time t = 1. In particular,
H0 is the number of pounds invested into the bank account, and Hn , n = 1, 2, . . . , N , is
the number of units of security n (e.g. shares of stock n) hold between time t = 0 and
time t = 1. Generally speaking Hn , n = 0, 1, . . . , N , can be positive or negative. Here,
Hn < 0 means borrowing, or selling short in financial terms.
     For example, H0 = −5 means that you borrowed from the bank 5 at time t = 0 and
must return 5B(1) to the bank at time t = 1. Or H3 = −2 means that you borrowed 2

                                                     5
shares of stock 3 at time t = 0, and at time t = 1, you must return the amount of money
equal to 2 times price of stock 3 at time t = 1.
    The value process V (t) is the total value of the portfolio at time t. i.e.,
                                                    N
                        V (t):=H0 B(t) +                  Hn Sn (t),           t = 0, 1.
                                                    n=1

Note that V (1) is a random variable.
   The gain process is the random variable describing the total profit or loss:

                   G:=V (1) − V (0)
                                           N                                     N
                     =H0 B(1) +                 Hn Sn (1) − H0 B(0) −                 Hn Sn (0)
                                       n=1                                      n=1
                                 N
                     =H0 γ +           Hn ∆Sn ,
                                 n=1

where
                                           ∆Sn :=Sn (1) − Sn (0)
is the gain or loss of the price of security n.
     But 1 at time t = 0 is not worth 1 at time t = 1 (due to inflation). Therefore, we
need to normalize the prices. We introduce the discounted price process by

                                                  ∗           Sn (t)
                                                 Sn (t):=            .
                                                              B(t)

In particular,
                   ∗                            ∗         Sn (1)
                  Sn (0) = Sn (0),             Sn (1) =          ,          n = 1, 2, . . . , N.
                                                          B(1)
    The discounted value process is therefore defined by
                                                          N
                        ∗     V (t)                ∗
                      V (t):=       = H0 +     Hn Sn (t),                        t = 0, 1,
                              B(t)         n=1

and the discounted gain process is given by
                                                                   N
                             ∗         ∗             ∗                        ∗
                            G :=V (1) − V (0) =                          Hn ∆Sn (t).
                                                                  n=1

Here
                                             ∗    ∗        ∗
                                           ∆Sn :=Sn (1) − Sn (0)
is the discounted gain or loss of the price of security n.

Example 1.1 Assume K = 2, N = 1, γ = 1/9, S(0) = 5, S(1, ω1 ) = 20/3, S(1, ω2 ) =
40/9. Then

                                       B(1) = 1 + γ = 10/9,

                                                          6
                                            S(1, ω1 )   20   10
                             S ∗ (1, ω1 ) =           =         = 6,
                                             B(1)       3    9
                                            S(1, ω2 )   40   10
                             S ∗ (1, ω2 ) =           =         = 4.
                                             B(1)       9    9

      For an arbitrary trading strategy H = (H0 , H1 ), we have

                                      V (0) = H0 + 5H1 ,
                                10       20
                   V (1, ω1 ) = H0 + H1 ,          V ∗ (1, ω1 ) = H0 + 6H1 ,
                                9        3
                                10       40
                   V (1, ω2 ) = H0 + H1 ,          V ∗ (1, ω2 ) = H0 + 4H1 ,
                                9        9
                                    1        5
                           G(ω1 ) = H0 + H1 ,        G∗ (ω1 ) = H1 ,
                                    9        3
                                   1        5
                          G(ω2 ) = H0 − H1 ,        G∗ (ω2 ) = −H1 ,
                                   9        9

Example 1.2 With everything else as in the example above, let K = 3 and set S(1, ω3 ) =
30/9. Then,
                                   S ∗ (1, ω3 ) = 3
and
                               10     30
                   V (1, ω3 ) =   H0 + H1 ,           V ∗ (1, ω3 ) = H0 + 3H1 ,
                               9      9
                                  1    15
                        G(ω3 ) = H0 − H1 ,             G∗ (ω2 ) = −2H1 .
                                  9     9

Example 1.3 Let K = 3, N = 2, γ = 1/9. Let

             S1 (0) = 5, S1 (1, ω1 ) = 60/9, S1 (1, ω2 ) = 60/9, S1 (1, ω3 ) = 40/9,
             S2 (0) = 10, S2 (1, ω1 ) = 40/3, S2 (1, ω2 ) = 80/9, S2 (1, ω3 ) = 80/9.

    Exercise. Compute the discounted price process. Specify V , V ∗ , G, G∗ for any given
trading strategy H = (H0 , H1 , H2 ). Compute V , V ∗ , G, G∗ for H = (100, 200, 300).




                                                7
1.2    Arbitrage opportunities
The model is unreasonable if investors had a possibility to make a profit on a transaction
without being exposed to the risk of incurring a lost.
     An arbitrage opportunity is a trading strategy H such that
     (a) V (0) = 0,
     (b) V (1, ω) ≥ 0 for all ω ∈ Ω,
     (c) EV (1) > 0.
     (Note: EV (1) = K P (ωi )V (1, ωi ).)
                         i=1
     Equivalently to (b) and (c), we can demand that (b) holds and V (1, ω) > 0 for at
least one ω ∈ Ω.
     Note that an arbitrage opportunity is a riskless way of making money: you start with
nothing and without any chance of going into a debt, there is a chance of ending up with
a positive amount of money. If such a situation existed, then everybody would use this
trading strategy, effecting the prices of the securities. So, if the model is a sensible one,
there cannot exist any arbitrage opportunity.

Lemma 1.1 There exists an arbitrage opportunity H if and only if there exists a trading
         ˆ           ˆ                             ˆ                      ˆ
strategy H such that G∗ (ω) ≥ 0 for all ω ∈ Ω and EG∗ > 0 (i.e., ∃ω ∈ Ω : G(ω) > 0).

                        ˆ
Proof. ⇒ Trivial: define H:= H. Then

                    ˆ        ˆ            ˆ                      V (1, ω)
                    G∗ (ω) = V ∗ (1, ω) − V ∗ (0) = V ∗ (1, ω) =          ,
                                                                  B(1)
which is always ≥ 0 and > 0 for at least one ω ∈ Ω.
            ˆ                                ˆ        ˆ
    ⇐ Let H be trading strategy such that G∗ ≥ 0 and EG∗ > 0. Define
                                                 ˆ            ˆ
                                        H:=(H0 , H1 , . . . , HN )

with
                                                         N
                                        H0 := −              ˆ ∗
                                                             Hn Sn (0).
                                                       n=1

Then
                                N                            N                   N
               ∗
             V (0) = H0 +           ˆ ∗
                                    Hn Sn (0) = −                  ˆ ∗
                                                                   Hn Sn (0) +         ˆ ∗
                                                                                       Hn Sn (0) = 0
                            n=1                              n=1                 n=1

and
                                                   N
                            ∗
                          V (1) = H0 +                   ˆ ∗
                                                         Hn Sn (1)
                                                   n=1
                                             N                        N
                                    =−             ˆ ∗
                                                   Hn Sn (0) +              ˆ ∗
                                                                            Hn Sn (1)
                                             n=1                      n=1
                                         N
                                    =         ˆ    ∗   ˆ
                                              Hn ∆Sn = G∗ ,
                                        n=1

which is ≥ 0 for all ω ∈ Ω and > 0 for at least one ω ∈ Ω.

                                                         8
Example 1.4 Consider a market model with K = 3, N = 2, γ = 0 and the following
security prices:

                  S1 (0) = 4, S1 (1, ω1 ) = 8, S1 (1, ω2 ) = 6, S1 (1, ω3 ) = 3,
                  S2 (0) = 7, S2 (1, ω1 ) = 10, S2 (1, ω2 ) = 8, S2 (1, ω3 ) = 4,

      There is an arbitrage opportunity

                                         H = (3, 1, −1)

Indeed,

                                    V ∗ (0) = 3 + 4 − 7 = 0,
                                 V ∗ (1, ω1 ) = 3 + 8 − 10 = 1,
                                 V ∗ (1, ω2 ) = 3 + 6 − 8 = 1,
                                 V ∗ (1, ω3 ) = 3 + 3 − 4 = 2.

So, H is an arbitrage opportunity.

     One says that the law of one price holds for a security market model if there do not
                          ˆ     ˜
exist two strategies, say H and H, such that
                              ˆ            ˜
                              V ∗ (1, ω) = V ∗ (1, ω) for all ω ∈ Ω

but
                                        ˆ         ˜
                                        V ∗ (0) > V ∗ (0).
                                                                              ˆ     ˜
That is to say that the law of one price holds if, for two trading strategies H and H such
that
                             ˆ           ˜
                            V ∗ (1, ω) = V ∗ (1, ω) for all ω ∈ Ω
we have
                                        ˆ         ˜
                                        V ∗ (0) = V ∗ (0).

Proposition 1.1 If there are no arbitrage opportunities, then the law of one price holds.

Proof. Assume the law of one price does not hold. Then there exist two trading strategies
ˆ       ˜     ˆ        ˜           ˆ         ˜
H and H with V ∗ (1) = V ∗ (1) and V ∗ (0) > V ∗ (0). Therefore,
                                ˆ        ˜
                                G∗ (ω) < G∗ (ω) for all ω ∈ Ω.

Define a new trading strategy
                                  ˜    ˆ
                             Hn :=Hn − Hn ,         n = 0, 1, . . . , N.

Then
                                  ˜        ˆ
                         G∗ (ω) = G∗ (ω) − G∗ (ω) > 0 for all ω ∈ Ω.
Therefore, by Lemma 1.1, there exists an arbitrage opportunity.
   Note: The converse to Proposition 1.1 is not true.


                                                9
Example 1.4 (continued) We saw that there exists an arbitrage opportunity in this
                                                         ˆ     ˜
model. Still, the law of one price now holds. Indeed, if H and H are trading strategies
                       ˜
          ˆ ∗ (1, ω) = V ∗ (1, ω) for all ω ∈ Ω, then
such that V
                         ˆ      ˆ     ˆ    ˜     ˜       ˜
                         H0 + 8H1 + 10H2 = H0 + 8H1 + 10H2 ,
                          ˆ      ˆ    ˆ    ˜     ˜      ˜
                          H0 + 6H1 + 8H2 = H0 + 6H1 + 8H2 ,
                          ˆ      ˆ    ˆ    ˜     ˜      ˜
                          H0 + 3H1 + 4H2 = H0 + 3H1 + 4H2 ,

i.e.,                               ˆ
                           
                            1 8 10        ˜  0
                                     H0 − H0
                                      ˆ   ˜
                           1 6 8  H1 − H1  = 0 .                                  (1.1)
                            1 3 4     ˆ   ˜
                                     H2 − H2      0
But                                        
                                     1 8 10
                                det 1 6 8  = 2 = 0.
                                     1 3 4
So, the linear equation (1.1) has the unique solution
                                        ˆ    ˜
                                        H0 − H0 = 0,
                                        ˆ    ˜
                                        H1 − H1 = 0,
                                        ˆ    ˜
                                        H2 − H2 = 0,

           ˆ   ˜                     ˆ       ˜
Therefore, H = H, and in particular, V (0) = V (0), i.e., the law of one price holds.




                                            10
1.3    Risk neutral probability measures
A probability measure Q on Ω is said to be a risk neutral probability measure if
                                    Q(ω) > 0 for all ω ∈ Ω,
                                      ∗
                                 EQ ∆Sn = 0,     n = 1, . . . , N.
                 ∗      K            ∗                 ∗      ∗         ∗      ∗
(Recall: EQ ∆Sn =       i=1 Q(ωi )∆Sn (ωi ).) Since ∆Sn = Sn (1) − Sn (0) and Sn (0) is a
constant (i.e., independent of ω), the latter condition is equivalent to
                                 ∗        ∗
                             EQ Sn (1) = Sn (0),      n = 1, . . . , N,             (1.2)
i.e., under the risk neutral probability measure, the expected discounted price of each
                                          ∗
security is equal to its original price (Sn (0) = Sn (0)).

Theorem 1.1 There are no arbitrage opportunities if and only if there exists a risk neu-
tral probability measure.

Example 1.5 Consider Example 1.1. Recall that N = 1, K = 2, S ∗ (0) = 5, S ∗ (1, ω1 ) =
6, S ∗ (1, ω2 ) = 4.
     Let us find a risk neutral probability measure. By (1.2) we should have
                         EQ S ∗ (1) = 6Q(ω1 ) + 4Q(ω2 ) = 5 = S ∗ (0),
                                     Q(ω1 ) + Q(ω2 ) = 1,
                                    Q(ω1 ) > 0, Q(ω2 ) > 0.
It is easy to see that
                                   Q(ω1 ) = 1/2, Q(ω2 ) = 1/2
is the unique solution. By the no-arbitrage theorem, the market has no arbitrage oppor-
tunities.

Example 1.6 Consider Example 1.2. Recall that N = 1, K = 3, S ∗ (1, ω1 ) = 6,
S ∗ (1, ω2 ) = 4, S ∗ (1, ω3 ) = 3. We search for a risk neutral probability measure. We
get the following system of equations:
                              6Q(ω1 ) + 4Q(ω2 ) + 3Q(ω3 ) = 5,                      (1.3)
                               Q(ω1 ) + Q(ω2 ) + Q(ω3 ) = 1,                        (1.4)
                             Q(ω1 ) > 0, Q(ω2 ) > 0, Q(ω3 ) > 0.                    (1.5)
By (1.3), (1.4), we easily get
                                      Q(ω2 ) = 2 − 3Q(ω1 ),
                                      Q(ω3 ) = 2Q(ω1 ) − 1.
Therefore
                         (Q(ω1 ), Q(ω2 ), Q(ω3 )) = (λ, 2 − 3λ, 2λ − 1),
where λ is a real parameter. To get (1.5), we thus have to demand that
                                           1    2
                                             <λ< .                                  (1.6)
                                           2    3
Thus, for each λ as in (1.6), we get a risk neutral probability measure. According to the
no-arbitrage theorem, there are no arbitrage opportunities in the market.

                                                11
Example 1.7 Consider Example 1.3. We have
                   ∗          ∗                ∗                ∗
                 S1 (0) = 5, S1 (1, ω1 ) = 6, S1 (1, ω2 ) = 6, S1 (1, ω3 ) = 4,
                 ∗            ∗                  ∗                ∗
                S2 (0) = 10, S2 (1, ω1 ) = 12, S2 (1, ω2 ) = 8, S2 (1, ω3 ) = 8.

If a risk neutral probability measure exists, it should satisfy

                             6Q(ω1 ) + 6Q(ω2 ) + 4Q(ω3 ) = 5,
                            12Q(ω1 ) + 8Q(ω2 ) + 8Q(ω3 ) = 10,
                              Q(ω1 ) + Q(ω2 ) + Q(ω3 ) = 1,

which has a unique solution

                         Q(ω1 ) = 1/2, Q(ω2 ) = 0, Q(ω3 ) = 1/2.

Since Q(ω2 ) is not positive, a risk neutral probability measure does not exist. Therefore,
there is an arbitrage opportunity in the market.
    Exercise. (i) Find an arbitrage opportunity. (ii) Show that the law of one price holds.

Proof of the theorem. “If” part. Assume there is a risk neutral probability measure Q.
Then, for any trading strategy H, we have
                                                     N
                                       ∗                        ∗
                                 EQ G = EQ                 Hn ∆Sn
                                                     n=1
                                               N
                                                             ∗
                                           =         Hn EQ ∆Sn = 0.
                                               n=1

If an arbitrage opportunity H exists then, for this H, G∗ (ω) ≥ 0 for all ω ∈ Ω and for at
least one ω0 , G∗ (ω0 ) > 0. Since Q(ω) > 0 for all ω ∈ Ω, this gives

                       EQ G∗ =         Q(ω)G∗ (ω) ≥ Q(ω0 )G∗ (ω0 ) > 0,
                                 ω∈Ω

which is a contradiction. Therefore, an arbitrage opportunity does not exist.
    “Only if” part. Consider

W :={X ∈ RK : X = G∗ = (G∗ (ω1 ), G∗ (ω2 ), . . . , G∗ (ωK )) for some trading strategy H}.

Then W is a linear subspace of RK , i.e., if X, Y ∈ W and a, b ∈ R, then aX + bY ∈ W
(check this).
    Consider

                  A:={X ∈ RK : X1 ≥ 0, X2 ≥ 0, . . . , XK ≥ 0, X = 0},

i.e., A is the nonnegative orthant of RK without origin. By Lemma 1.1, there exists an
arbitrage opportunity if and only if W ∩ A = ∅, and for an arbitrage opportunity, the
discounted gain G∗ belongs to W ∩ A. Therefore, there are no arbitrage opportunities if
and only if W ∩ A = ∅.
      Consider the orthogonal subspace of W in RK :

                      W ⊥ = {Y ∈ RK : X · Y = 0 for all X ∈ W },

                                                     12
where X · Y = X1 Y1 + X2 Y2 + · · · + XK YK . Since W ∩ A = ∅, there exists a ray in W ⊥
along which the components of every point not at the origin are strictly positive (check
this e.g. for K = 2 and K = 3). Take the point Q = (Q1 , . . . , QK ) on this ray satisfying

                                  Q1 + Q2 + · · · + QK = 1

(note that we also have Qi > 0, i = 1, . . . , K). Set

                                           Q(ωi ):=Qi .

Since Q ∈ W ⊥ , we have, for any trading strategy H,
                                     K
                                          Q(ωi )G∗ (ωi ) = 0,
                                    i=1

i.e.,
                                           EQ G∗ = 0.
Thus, Q is a risk neutral probability measure.

Example 1.7 (continued) We already know that there is an arbitrage opportunity. We
              ∗
can compute ∆Sn as follows:
                           ∗              ∗            ∗
                       ∆S1 (ω1 ) = 1, ∆S1 (ω2 ) = 1, ∆S1 (ω3 ) = −1,
                         ∗              ∗                ∗
                       ∆S2 (ω1 ) = 2, ∆S2 (ω2 ) = −2, ∆S2 (ω3 ) = −2.

Therefore, the subspace W from the proof of Theorem 1.1 looks as follows:

        W = {X ∈ R3 : X = (H1 + 2H2 , H1 − 2H2 , −H1 − 2H2 ) for some H1 , H2 ∈ R}.

Notice that X1 + X3 = 0 for each X ∈ W . Conversely, given any vector X ∈ R3 with
X1 + X3 = 0, one can find a unique strategy (H1 , H2 ) with G∗ = X (check this). Hence,

             W = {X ∈ R3 : X1 + X3 = 0} = {X = (X1 , X2 , −X1 ), X1 , X2 ∈ R},

which yields
                               W ∩ A = {(0, X2 , 0), X2 > 0}.
Thus, for any X2 > 0, there exists a trading strategy H which gives rise to the discounted
gain process (0, X2 , 0).
     Exercise. Find an arbitrage strategy H which gives rise the discounted gain process
(0, X2 , 0). (check Exercise 2 in Example Sheet 1).




                                                13
1.4       Contingent claims: valuation and hedging
A contingent claim is a random variable (function of ω ∈ Ω) X representing a pay-off at
time t = 1. A contingent claim is part of a contract that a buyer and a seller agree at
time t = 0. The seller promises to pay the buyer the amount X(ω) at time t = 1.

Example 1.8 European call option is the right, but not the obligation to buy one stock
of a specified type at a specified price K at terminal time t = 1.
     Assume the stock price at t = 1 is S(1, ω), ω ∈ Ω. There are two possibilities at time
t = 1 of maturity.
     (i) If S(1, ω) > K, then the owner of the option will buy the stock for the price S(1, ω)
and immediately sell it on the open market for the price S(1, ω), therefore obtaining the
pay-off S(1, ω) − K.
     (ii) If S(1, ω) ≤ K, then the owner of the option will not exercise the option and the
pay-off is 0. Hence, the pay-off of the option at time t = 1 is

                                                               S(1, ω) − K, if S(1, ω) > K,
    X(ω) = (S(1, ω) − K)+ = max{0, S(1, ω) − K} =
                                                               0,           otherwise.

K is called the exercise price or the striking price.

Example 1.9 European put option is the right, but not the obligation to sell one stock of
a specified type at a specified price K at terminal time t = 1. Again assume that at t = 1
the stock price is S(1, ω). If K > S(1, ω), then the owner of the option will exercise the
option, buying the stock for the price K and immediately sell it on the open market for
the price K, therefore obtaining the pay-off K − S(1, ω). If K ≤ S(1, ω), then the owner
of the option will not exercise the option and the pay-off is 0. Hence, the pay-off of the
European put option is

                                                              K − S(1, ω), if S(1, ω) < K,
   X(ω) = (K − S(1, ω))+ = max{0, K − S(1, ω)} =
                                                              0,           otherwise.

    Question: What is the “fair” price that the buyer should pay the seller at time t = 0
in order that the two parties to be happy with their contract?
    A contingent claim is said to be attainable or marketable if there exists a trading
strategy H, called the replicating or hedging portfolio, such that

                               V (1, ω) = X(ω) for all ω ∈ Ω,

i.e.,
                                         N
                      X(ω) = H0 B(1) +         Hn Sn (1, ω) for all ω ∈ Ω.
                                         n=1

        Assume a contingent claim X is attainable. Let
                                                    N
                                  V (0) = H0 +          Hn Sn (0)
                                                 n=1

be the time t = 0 value for a fixed replicating strategy H.

                                               14
    Consider the three following possibilities.
    a) Assume that at t = 0 the price of the claim p > V (0). We can sell the claim
for p at t = 0 and follow the trading strategy H at a cost of V (0), having put the
difference p − V (0) > 0 in our pocket. At t = 1, V (1, ω) = X(ω) can be used to settle
the contingent claim’s pay-off at time t = 1. This is a riskless way to make profit (an
arbitrage opportunity).
    b) Assume p < V (0). We can buy the contingent claim and follow the trading strategy
−H, having put the difference

                              (−p) − (−V (0)) = V (0) − p > 0

in our pocket. At t = 1,
                                                   N
                           X(ω) + (−H0 B(1) −           Hn Sn (1)) = 0.
                                                  n=1

So, our contingent claim can be used in order to settle the trading strategy −H at time
t = 1. This is again an arbitrage opportunity.
                                                               ˆ                  ˆ
    c) Assume p = V (0). If there is another trading strategy H such that X = V (1), but
ˆ                                                     ˆ
V (0) = V (0), then we can use the trading strategy H to lock a riskless profit (compare
                                                      ˆ
with the cases a) and b) and use the trading strategy H instead of H). Therefore, we have
                 ˆ
to require that V (0) = V (0). Recall that, by Proposition 1.1, if there are no arbitrage
                                                                   ˆ
opportunities, then the law of one price holds, and in particular, V (0) = V (0).
    Thus, if there are no arbitrage opportunities on the market and if a contingent claim
X is attainable, then
                                                  N
                                 V (0) = H0 +          Hn Sn (0)
                                                n=1

is the fair price of the contingent claim, where H is an arbitrary replicating portfolio of
X.
     Furthermore, we have

Theorem 1.2 If the market has no arbitrage opportunities and a contingent claim X is
attainable, then the fair price of X is
                                                              V (1)       X
                  p = V (0) = V ∗ (0) = EQ V ∗ (1) = EQ             = EQ      ,
                                                              B(1)       B(1)
where Q is a risk neutral probability measure (which exists according to Theorem 1.1).

Remark 1.1 1) p can be computed without computing H
  2) p is independent of the real world probability measure P .

Proof. It remains to prove that EQ V ∗ (1) = V ∗ (0). We have
                                                        N
                                ∗                                ∗
                            EQ V (1) = EQ H0 +               Hn Sn (1)
                                                       n=1
                                              N
                                                          ∗
                                    = H0 +         Hn EQ Sn (1)
                                             n=1


                                             15
                                                  N
                                                          ∗
                                     = H0 +           Hn Sn (0)
                                               n=1
                                     = V ∗ (0).

Example 1.10 Consider Example 1.1 (as well as Example 1.5). We have proved that
Q(ω1 ) = 1/2, Q(ω2 ) = 1/2 is a risk neutral probability measure. Let X(ω1 ) = 7 and
X(ω2 ) = 2. If X is attainable, then the fair price p of X is
                                X     1 7     1 2     81
                      p = EQ        =       +       =    = 4.05.
                               B(1)   2 10/9 2 10/9   20
To check that the claim X is attainable, we can solve H = (H0 , H1 ) from
                       X(ω) = H0 B(1) + H1 S(1, ω),               ω = ω1 , ω2 .
We thus have
                                     10        20
                                        H0 +      H1 = 7,
                                     9         3
                                     10        40
                                        H0 +      H1 = 2,
                                     9         9
yielding H0 = −7.2, H1 = 9/4. (Notice that indeed V (0) = −7.2 + 9 · 5 = 4.05 = p.)
                                                                        4
     The solution tells us that, if you are a seller of the contingent claim, you borrow £
7.2 from the bank and use £ (p − H0 ) = 4.05 + 7.2 = 11.25 to purchase 11.25 : S(0) =
11.25 : 5 = 2.25 = H1 shares of the risky asset at time t = 0. At time t = 1, you have
to pay back £7.2 · 10 = 8 for the £7.2 loan. Then, at state ω1 , the shares are worth
                        9
S(1, ω1 ) · H1 = 20 · 2.25 = 15. So, V (1, ω1 ) = 15 − 8 = 7 = X(ω1 ). At state ω2 , the shares
                  3
are worth S(1, ω2 ) · H1 = 40 · 2.25 = 10. So, V (1, ω2 ) = 10 − 8 = 2 = X(ω2 ).
                             9

Example 1.11 In the example above, consider the European call option with striking
price K = 5. Then
                                                           5
                                                             if ω = ω1 ,
                                                             ,
                        X(ω) = (S(1, ω) − K)+ =            3
                                                          0, if ω = ω2 .
If X is attainable, then the price of the call at time t = 0 is
                                   X    5 9 1     9 1  3
                         p = EQ        = ·  · +0·   · = .
                                  B(1)  3 10 2    10 2 4
To check that X is attainable, assume H = (H0 , H1 ) is a hedging portfolio, so that
                                    10      20     5
                                       H0 +    H1 = ,
                                    9       3      3
                                     10     40
                                        H0 + H1 = 0,
                                     9       9
which yields H0 = −3, H1 = 3/4.

    Exercise. Describe the replicating portfolio (and how to hedge it) in your own words.
    Exercise. Consider the European put option for K = 5 in Example 1.10. Calculate
the price of the option and the replicating portfolio. Describe the portfolio in your own
words.

                                               16
Example 1.12 Consider Example 1.2 as well as Example 1.6 and an arbitrary contingent
claim X. It is attainable if there are H0 , H1 such that

                    H0 B(1) + H1 S(1, ω) = X(ω) for ω = ω1 , ω2 , ω3 ,

i.e.,
                                 10         20
                                    H0 +       H1 = X(ω1 ),                       (1.7)
                                 9          3
                                 10         40
                                    H0 +       H1 = X(ω2 ),                       (1.8)
                                 9          9
                                 10         30
                                    H0 +       H1 = X(ω3 ).                       (1.9)
                                 9          9
We have three equalities and two unknowns. These equations have a solution under some
additional condition. From (1.9), we have

                                       9               30
                               H0 =         X(ω3 ) −      H1 .                   (1.10)
                                       10              9

Substituting (1.10) into (1.7) gives
                                       30      20
                            X(ω3 ) −      H1 +    H1 = X(ω1 ),
                                       9       3
which yields
                                       9
                                H1 =      (X(ω1 ) − X(ω3 )).                     (1.11)
                                       30
Substituting (1.10) into (1.8) gives
                                       30      40
                            X(ω3 ) −      H1 +    H1 = X(ω2 ),
                                       9       9
which yields
                                      9
                                H1 =    (X(ω2 ) − X(ω3 )).                       (1.12)
                                     10
By (1.11) and (1.12), if a hedging portfolio exists, we must have
                       9                     9
                          (X(ω1 ) − X(ω3 )) = (X(ω2 ) − X(ω3 )),
                       30                    10
so that
                             X(ω1 ) − 3X(ω2 ) + 2X(ω3 ) = 0.
Otherwise, the contingent claim would not be attainable.
   Notice that the risk neutral probability measure is not unique in this case (see Ex-
ample 1.6).




                                              17
1.5       Complete and incomplete markets
Example 1.12 shows that, even if there is no arbitrage in the market (which is equivalent to
the existence of a risk neutral probability measure), there is no guarantee that a contingent
claim is attainable.
    The model is said to be complete if every contingent claim X can be generated by
some trading strategy (i.e., is attainable). Otherwise the model is said to be incomplete.
    If a trading strategy H = (H0 , H1 , . . . , HN ) generates a contingent claim X, we must
have
                                   N
                      B(1)H0 +          Sn (1, ω)Hn = X(ω) for all ω ∈ Ω.
                                  n=1
Here, Ω = {ω1 , . . . , ωK }. That is
          B(1)H0 + S1 (1, ω1 )H1 + S2 (1, ω1 )H2 + · · · + SN (1, ω1 )HN = X(ω1 ),
          B(1)H0 + S1 (1, ω2 )H1 + S2 (1, ω2 )H2 + · · · + SN (1, ω2 )HN = X(ω2 ),
                                              .
                                              .
                                              .
       B(1)H0 + S1 (1, ωK )H1 + S2 (1, ωK )H2 + · · · + SN (1, ωK )HN = X(ωK ).       (1.13)
Consider H as                                    
                                               H0
                                              H1 
                                           H= . 
                                              
                                              . 
                                                .
                                                    HN
and define the K × (N + 1) matrix
                                                                             
                      B(1) S1 (1, ω1 ) S2 (1, ω1 ) . . .          SN (1, ω1 )
                    B(1) S1 (1, ω2 ) S2 (1, ω2 ) . . .           SN (1, ω2 ) 
                A= .
                                                                             
                                 .          .        .                .
                     .          .          .        .                .       
                        .        .          .        .                .       
                      B(1) S1 (1, ωK ) S2 (1, ωK ) . . .          SN (1, ωK )
and let
                              X = (X(ω1 ), X(ω2 ), . . . , X(ωK ))T .
Then we can write down (1.13) in the form
                                             AH = X.                                  (1.14)

    By the definition of completeness, the model is complete if and only if the equation
(1.14) has a solution H for each X. Recall the result from the linear algebra saying that
an equation AH = X has a solution for each right-hand side X if and only if the matrix
A has rank K, i.e., A has K independent rows. Thus, we get

Proposition 1.2 The model is complete if and only if the number of states in Ω equals
the rank of the matrix A.

Example 1.13 Consider Example 1.1. The matrix
                                                 10   20
                                           A=     9    3
                                                 10   40
                                                  9    9

has rank 2, so that the model is complete.

                                                18
Example 1.14 Consider Example 1.12. The matrix
                                     10 20 
                                              9     3
                                       A =  10
                                              9
                                                   40 
                                                    9
                                              10   10
                                               9    3

has rank 2, whereas K = 3. Therefore, the model is incomplete.

Proposition 1.3 Assume there are no arbitrage opportunities. Then a contingent claim
                                   X
X is attainable if and only if EQ B(1) takes the same value for all risk neutral probability
measures Q.

Remark 1.2 We know that no arbitrage is equivalent to the existence of at least one
risk neutral probability measure.

Proof of the proposition. ⇒ Assume the contingent claim is attainable. Then, by Theorem
1.2,
                                                X
                                   V (0) = EQ
                                              B(1)
for each risk neutral probability measure. Here, V (0) is the initial value of the replicating
portfolio, which is independent of the choice of Q.
     ⇐ See Appendix.

Example 1.14 (continued) We know that, in this example, a risk neutral probability
measure is not unique. In fact, there are infinitely many of them:
                                                            1    2
                      Q = (λ, 2 − 3λ, −1 + 2λ) for each       <λ< .
                                                            2    3
For any such Q,
             X       9                        9                         9
       EQ        =λ·    · X(ω1 ) + (2 − 3λ) ·    · X(ω2 ) + (−1 + 2λ) ·    · X(ω3 )
            B(1)     10                       10                        10
                  9          9            9
                 = X(ω2 ) − X(ω3 ) + λ(X(ω1 ) − 3X(ω2 ) + 2X(ω3 )).
                  5         10           10
By Proposition 1.3, X is marketable if and only if
                  9         9        9
                    X(ω2 ) − X(ω3 ) + λ(X(ω1 ) − 3X(ω2 ) + 2X(ω3 ))
                  5         10       10
takes the same value for all values of λ ∈ (1/2, 2/3). This holds if and only if

                              X(ω1 ) − 3X(ω2 ) + 2X(ω3 ) = 0.

(compare with Example 1.12 where we prove that the above condition is necessary for X
to be attainable).

Theorem 1.3 Assume there are no arbitrage opportunities. Then the model is complete
if and only if there exists exactly one risk neutral probability measure.


                                             19
Proof. ⇐ Assume there exists only one risk neutral probability measure. Take any
                                X
contingent claim X. Then EQ B(1) trivially takes the same value for any risk neutral
probability measure. By Proposition 1.3, X is attainable.
    ⇒ Suppose every contingent claim is attainable, but there are two different risk
                                          ˆ
neutral probability measures, say Q and Q. Then there exists k0 such that Q(ωk0 ) =
 ˆ
Q(ωk0 ). Define
                                       B(1), if ω = ωk0 ,
                             X(ω):=
                                       0,      otherwise.
Then
                               X                              X     ˆ
                         EQ        = Q(ωk0 ),           EQ
                                                         ˆ        = Q(ωk0 ),
                              B(1)                           B(1)
so that
                                           X         X
                                     EQ        = EQ
                                                  ˆ      .
                                          B(1)      B(1)
Hence, by Proposition 1.3, X is not attainable, which is a contradiction. Thus, a risk
neutral probability measure is unique.
    By Theorems 1.1, 1.2, and 1.3, we have:
Moral: Suppose there exists a unique risk neutral probability measure Q. Then there
are no arbitrage opportunities on the market, the market is complete and any contingent
claim X has its fair price
                                               X
                                      p = EQ      .
                                             B(1)



1.6       Appendix: Proof of the “if ” part of Proposition 1.3
Assume the contingent claim is not attainable. Consider the K × (N + 1) matrix A. Since
X is not attainable, there is no solution to AH = X. By using Farkas’ lemma from linear
algebra, there exists a row vector

                                         π = (π1 , . . . , πK )

satisfying
                                     πA = 0,            πX > 0.
                                                                           ˆ
    Take an arbitrary risk neutral probability measure, which we denote by Q. Take
λ > 0 small enough such that
                          ˆ
                  Q(ωk ):=Q(ωk ) + λπk B(1) > 0 for all k = 1, 2, . . . , K
       ˆ
(since Q(ωk ) > 0 for all k = 1, 2, . . . , K, we can always find such λ). Then
                         K               K                     K
                              Q(ωk ) =         ˆ
                                               Q(ωk ) + λ           πk B(1) = 1,
                        k=1              k=1                  k=1

since
                                             K
                                                 ˆ
                                                 Q(ωk ) = 1
                                          k=1


                                                   20
         ˆ
(because Q is a probability measure) and
                                               K
                                                     πk B(1) = 0
                                               k=1

(because πA = 0). Thus, Q is also a probability measure with Q(ω) > 0 for all ω ∈ Ω.
                                       ∗
Furthermore, for any discounted price Sn (1), n = 1, 2, . . . , N ,
                                                     K
                                     ∗                                Sn (1, ωk )
                                 EQ Sn (1) =                 Q(ωk )
                                                     k=1
                                                                        B(1)
                          K                                    K
                             ˆ      Sn (1, ωk )               Sn (1, ωk )
                       =     Q(ωk )             +     πk B(1)             .               (1.15)
                         k=1
                                      B(1)        k=1
                                                                B(1)

Since πA = 0, we get
                           K
                                πk Sn (1, ωk ) = 0,                n = 1, 2, . . . , N.   (1.16)
                          k=1

                            ˆ
By (1.15), (1.16) and since Q is a risk neutral probability measure,
                                     K
                       ∗                         ∗                ∗        ∗
                   EQ Sn (1)    =         Q(ωk )Sn (1, ωk ) = EQ Sn (1) = Sn (0).
                                                               ˆ
                                    k=1

Thus, Q is a risk neutral probability measure.
   However,
                                                         K
                                        X                              X(ωk )
                                    EQ      =                 Q(ωk )
                                       B(1)           k=1
                                                                       B(1)
                                 K                                    K
                                    ˆ      X(ωk )
                              =     Q(ωk )        +λ     πk X(ωk )
                                k=1
                                           B(1)      k=1
                                                      X
                                          = EQ
                                             ˆ            + λπX.                          (1.17)
                                                     B(1)

But λ > 0 and πX > 0, which, by virtue of (1.17), implies

                                                X         X
                                          EQ        = EQ
                                                       ˆ      ,
                                               B(1)      B(1)

which is a contradiction. Thus, each X is attainable.




                                                         21
1.7      Return
Assume Sn (0) > 0, the return Rn for the risky security n is defined by
                                   Sn (1) − Sn (0)
                            Rn =                   ,    n = 1, 2, . . . N
                                        Sn (0)
and the return of a Bank account is
                                           B(1) − B(0)
                                    R0 =               = r.
                                              B(0)
Notice
                       ∗        ∗       Sn (1) − B(1)Sn (0)
                      Sn (1) − Sn (0) =
                                                B(1)
                                        (1 + Rn )Sn (0) − (1 + R0 )Sn (0)
                                      =
                                                     1 + R0
                                               Rn − R0
                                      = Sn (0)         .
                                               1 + R0

Proposition 1.4 If Q is a probability measure with Q(ω) > 0 for all ω ∈ Ω, then Q is a
risk neutral probability measure if and only if
                                  Rn − R0
                             EQ           = 0,       n = 1, 2, . . . , N.                 (1.18)
                                  1 + R0
If the interest R0 = r is deterministic, then (1.18) is equivalent to

                                           EQ Rn = r.

      The mean return for security n defined by
                    ¯
                    Rn = ERn , ( with real world probability mesure P ).
              ¯
The difference Rn − r is called the risk premium for security n.
   Let Q be a risk neutral probability measure and define
                                                   Q(ω)
                                          L(ω) =         ,
                                                   P (ω)
then we have
                                                       ¯
                                     cov(Rn , L) = r − Rn .
Consider the return of a portfolio corresponding to a trading strategy H = (H0 , H1 , . . . , HN ).
Assume V (0) > 0 and define
                                           V (1) − V (0)
                                      R=
                                               V (0)
and
                                            ¯
                                            R = ER.
It is the same as before we have
                                     ¯
                                     R − r = −cov(R, L).


                                                22
2      Multiperiod security market
2.1       Mathematical models: probability space, filtration and stochas-
          tic processes
Suppose T ∈ N. In this section, we will consider a T -period model. So, we now have
trading times t = 0, 1, 2, . . . , T .
    As in the single period model, we have a sample space Ω = {ω1 , ω2 , . . . , ωK }, K ∈ N,
and a probability measure P on Ω such that P (ω) > 0 for each ω ∈ Ω.
    We have a bank account process B(t), t = 0, 1, 2, . . . , T , given by

                           B(t) = (1 + γ)t ,           t = 0, 1, 2, . . . , T,

so that
                                             B(t) − B(t − 1)
                                     γ=                      .
                                                B(t − 1)
(In other words, the interest rate pertaining to the time interval (t−1, t) remains constant.)
     We have N risky security processes

                                   Sn (t),       t = 0, 1, . . . , T,

where n = 1, 2, . . . , N . Each Sn (t) represents the time t price of a a risky security n. For
each n ∈ N, the price Sn (0) is known, while for t ≥ 1, Sn (t) is a random variable, i.e., a
function of ω.
     Let us consider the sample space Ω = {ω1 , . . . , ωK }. A collection F of subsets of Ω is
called a σ-algebra (or σ-field) on Ω if
     (a) Ω ∈ F,
     (b) Ac ∈ F whenever A ∈ F (recall Ac = Ω \ A),
     (c) A ∪ B ∈ F whenever A, B ∈ F.
     (Note that by (a), (b) ∅ = Ωc ∈ F.)
     The couple (Ω, F) is called a measurable space, and the triple (Ω, F, P ), where P is
a probability measure on (Ω, F), is called a probability space.

Example 2.1 The collections of sets

                                             F = {∅, Ω}

and
                                    F = {all subsets of Ω}
are trivial examples of σ-algebras.

      A partition P of Ω is a collection {Ai } of non-empty subsets of Ω such that

                                  Ai = Ω,      Ai ∩ Aj = ∅ if i = j.
                              i


     From a partition P of Ω one can construct a σ-algebra by taking all possible unions
of sets from P (including the empty set).

                                                  23
     On the other hand, given a σ-algebra F, there exists a partition P of Ω which gives
rise to the σ-algebra F as described above.
     A random variable X (a function of ω) is called measurable with respect to a σ-algebra
F if, for any real a ∈ R,
                                {ω ∈ Ω : X(ω) = a} ∈ F.
Equivalently, X is measurable with respect to F if X takes a constant value on each set
from the partition P corresponding to F (check this!).

Example 2.2 Let
                                    Ω = {ω1 , ω2 , ω3 , ω4 , ω5 }
and consider the σ-algebra

                   F = {∅, {ω1 , ω2 }, {ω3 , ω4 , ω5 }, {ω1 , ω2 , ω3 , ω4 , ω5 }}.

The corresponding partition is

                                 P = {{ω1 , ω2 }, {ω3 , ω4 , ω5 }}.

Then the random variable
                                           6, if ω = ω1 , ω2 ,
                              X(ω) =
                                           10, if ω = ω3 , ω4 , ω5 ,

is measurable with respect to F, while the random variable

                                            1, if ω = ω1 , ω3 , ω5 ,
                               Y (ω) =
                                            0, if ω = ω2 , ω4 ,

is not measurable with respect to F,

    The σ-algebra generated by a random variable X is defined to be the smallest σ-algebra
with respect to which X is measurable. It is denoted by F(X).
    Since X takes on only a finite number of values, say {x1 , x2 , . . . , xM }, we set

                          Ai :={ω : X(ω) = xi },             i = 1, . . . , M.

Then
                                                   M
                                            Ω=          Ai
                                                  i=1

and
                                    Ai ∩ Aj = ∅,             i = j.
Thus, {Ai }M is a partition of Ω. One can easily show that the σ-algebra F(X) corre-
           i=1
sponds to this partition {Ai }M .
                              i=1


Example 2.2 (continued) We have F = F(X), while F(Y ) corresponds to the parti-
tion
                          {{ω1 , ω3 , ω5 }, {ω2 , ω4 }},
so that
                 F(Y ) = {∅, {ω1 , ω3 , ω5 }, {ω2 , ω4 }, {ω1 , ω2 , ω3 , ω4 , ω5 }}.

                                                 24
    The σ-algebra F(X, Y ) generated by the random variables X and Y is defined as
the smallest σ-algebra with respect to which both X and Y are measurable. (Note:
F(X) ⊂ F(X, Y ), F(Y ) ⊂ F(X, Y )).
    Suppose X takes on values {x1 , . . . , xM } and Y takes on values {y1 , . . . , yL }. Define

                             Aij :={ω : X(ω) = xi , Y (ω) = yj }.

Then, analogously to the above, {Aij , i = 1, . . . , M, j = 1, . . . , L} is a partition of Ω
which corresponds to the σ-algebra F(X, Y ). (If some of the sets Aij are empty, they
must be omitted from the partition)
      Analogously, we also define a σ-algebra F(X1 , . . . , XM ) generated by random variables
X1 , . . . , XM .

Example 2.2 (continued) To find F(X, Y ), we construct the following sets

                         A11 :={ω : X(ω) = 6, Y (ω) = 1} = {ω1 },
                         A12 :={ω : X(ω) = 6, Y (ω) = 0} = {ω2 },
                         A21 :={X(ω) = 10, Y (ω) = 1} = {ω3 , ω5 },
                          A22 :={X(ω) = 10, Y (ω) = 0} = {ω4 }.

Taking all possible unions of these sets, we get the σ-algebra F(X, Y ).

     A stochastic process X is a real-valued function X(t, ω) of both t ∈ {0, 1, . . . , T } and
ω ∈ Ω = {ω1 , . . . , ωK }. For each fixed ω ∈ Ω, t → X(t, ω), being a function of t only, is
called a sample path of the process. For each t fixed, ω → X(t, ω) is a random variable.
     Now, for each n ∈ {1, . . . , N }, we can consider the price process {Sn (t)}T as a
                                                                                       t=0
stochastic process.
     Let {X(t)}T be a stochastic process on a probability space (Ω, F, P ). Let {Ft }T
                 t=0                                                                          t=0
be a sequence of σ-algebras on Ω such that

                                     F0 ⊂ F1 ⊂ · · · ⊂ FT .

Such a sequence of σ-algebras is called a filtration. We say that the process {X(t)}T is
                                                                                   t=0
adapted to {Ft }T if, for each t, Xt is Ft -measurable.
                t=0
    If
                    Ft = F(X(0), X(1), . . . , X(t)),   t = 0, 1, . . . , T,
then the filtration {Ft }T is said to be generated by the stochastic process {X(t)}T .
                        t=0                                                       t=0


Example 2.3 Suppose that the price S(t, ω) of a certain security is given by

                                           S(0) = 5,
                  S(1, ω1 ) = 8, S(1, ω2 ) = 8, S(1, ω3 ) = 4, S(1, ω4 ) = 4,
                  S(2, ω1 ) = 9, S(2, ω2 ) = 6, S(2, ω3 ) = 6, S(2, ω4 ) = 3.

For t = 0,
                                   F0 = F(S(0)) = {∅, Ω}.
For t = 1,
                               F1 = F(S(0), S(1)) = F(S(1)),

                                               25
the latter equality being due to the fact that S(0) is constant on Ω. The partition
corresponding to F1 is given by
                                 A1 = {S(1) = 8} = {ω1 , ω2 },
                                 A2 = {S(1) = 4} = {ω3 , ω4 }.
For t = 2,
                            F2 = F(S(0), S(1), S(2)) = F(S1 , S2 ).
We have:
                             A11 = {S(1) = 8, S(2) = 9} = {ω1 },
                             A12 = {S(1) = 8, S(2) = 6} = {ω2 },
                              A13 = {S(1) = 8, S(2) = 3} = ∅,
                              A21 = {S(1) = 4, S(2) = 9} = ∅,
                             A22 = {S(1) = 4, S(2) = 6} = {ω3 },
                             A23 = {S(1) = 4, S(2) = 3} = {ω4 }.
Thus, the partition corresponding to F2 is given by
                                    {{ω1 }, {ω2 }, {ω3 }, {ω4 }}.

     A stochastic process {X(t)}T is said to be predictable with respect to a filtration
                                       t=0
{Ft }T if, for each t = 1, 2, . . . , T , X(t) is measurable with respect to Ft−1 . (Note that if
     t=0
a process is predictable, then it is adapted.)
     Let now Ft be the σ-algebra generated by the stock prices Sn (u), u = 0, 1, . . . , t,
n = 1, 2, . . . , N . Thus, Ft describes the information the the investor possesses at time t.
We will also suppose that
                                      FT = {all subsets of Ω}.

Example 2.3 (continued) Suppose t = 1. Then the investor knows the values of S0
and S(1). Suppose that S(0) = 5, S(1) = 4. Therefore
                                       ω ∈ A2 = {ω3 , ω4 },
and A2 is an element of F1 .

      A trading strategy H(t) = (H0 (t), H1 (t), . . . , HN (t)) is a vector of stochastic processes
Hn (t), t = 1, 2, . . . , T , where Hn (t) is the number of shares of stock n that the investor
owns (i.e., carries forward) from time t − 1 to time t, n = 1, 2, . . . , N , and H0 (t)(1 + γ)t−1
is the amount of money invested into the bank at time t − 1, i.e., H0 (t) is the number of
bonds that the investor owns from time t − 1 to time t.
      Thus, H(t) is the trading strategy established by the investor at time t − 1. What
the investor sees at time t − 1 is the partition corresponding to Ft−1 and Hn−1 (t) should
be a constant on each set from this partition. Thus, Hn−1 (t) should be Fn−1 -measurable,
i.e., we should assume that, for each n = 1, 2, . . . , N , {Hn (t)}T is a predictable process
                                                                        t=0
with respect to the filtration {Ft }T ..  t=0
      The value process V (t), t = 0, 1, . . . , T , is the stochastic process defined by
                                                      N
                               V (0) = H0 (1) +           Hn (1)Sn (0),
                                                   n=1


                                                 26
                                            N
                  V (t) = H0 (t)B(t) +           Hn (t)Sn (t),     t = 1, 2, . . . , T.
                                           n=1

Thus, V (0) is the initial value of the portfolio, and V (t), t = 1, 2, . . . , T , is the time t
value of the portfolio before any transactions are made at the same time.
    We denote
                                 ∆Sn (t):=Sn (t) − Sn−1 (t),
which is the change of the value of the stochastic process between times t − 1 and t.
Then Hn (t)∆Sn (t) is the one-period gain or loss due to the ownership of Hn (t) shares of
security n between times t − 1 and t, and t Hn (u)∆Sn (u) is the cumulative gain or
                                              u=1
loss through time t due to the investment in security n.
     The gain process is defined by
                     t                      N     t
           G(t):=         H0 (u)∆B(u) +                Hn (u)∆Sn (u),       t = 1, 2 . . . , T,
                    u=1                    n=1 u=1

where
            ∆B(u) = B(u) − B(u − 1) = (1 + γ)u − (1 + γ)u−1 = γ(1 + γ)u−1 .
The G(t) is the cumulative gain or loss through time t of the portfolio.

Example 2.3 (continued)Given a trading strategy H(t) = (H0 (t), H1 (t)), t = 1, 2, we
get
                                   V (0) = H0 (1) + 5H1 (1),
                                   (1 + γ)H0 (1) + 8H1 (1), ω = ω1 , ω2 ,
                     V (1, ω) =
                                   (1 + γ)H0 (1) + 4H1 (1), ω = ω3 , ω4 ,
                            
                            (1 + γ)2 H0 (2, ω) + 9H1 (2, ω), ω = ω1 ,
                            
                  V (2, ω) = (1 + γ)2 H0 (2, ω) + 6H1 (2, ω), ω = ω2 , ω3 ,
                            
                             (1 + γ)2 H0 (2, ω) + 3H1 (2, ω), ω = ω4 ,
                            

                                         γH0 (1) + 3H1 (1), ω = ω1 , ω2 ,
                          G(1, ω) =
                                         γH0 (1) − H1 (1), ω = ω3 , ω4 ,
                   
                   γH0 (1) + 3H1 (1) + γ(1 + γ)H0 (2, ω) + H1 (2, ω),
                                                                                         ω   = ω1 ,
                   
                   γH (1) + 3H (1) + γ(1 + γ)H (2, ω) − 2H (2, ω),
                      0          1                0            1                          ω   = ω2 ,
         G(2, ω) =
                   γH0 (1) − H1 (1) + γ(1 + γ)H0 (2, ω) + 2H1 (2, ω),
                                                                                         ω   = ω3 ,
                   
                    γH0 (1) − H1 (1) + γ(1 + γ)H0 (2, ω) − H1 (2, ω),
                   
                                                                                          ω   = ω4 .

    For t = 1, 2, . . . , T − 1, we defined V (t) as the time t value of the portfolio before
any transactions take place. What is, however, the time t value of the portfolio after
transactions? It is evidently
                                   N
                H0 (t + 1)B(t) +         Hn (t + 1)Sn (t),       t = 1, 2, . . . , T − 1.
                                   n=1

If the latter value coincides with V (t), then there is no infusion from outside or withdrawal
of money. In practical terms, this means that the accession of a new asset is financed
through the sale of some other assets.

                                                      27
      Thus, a trading strategy H is said to be self-financing if
                                                       N
              V (t) = H0 (t + 1)B(t) +                         Hn (t + 1)Sn (t),                 t = 1, 2, . . . , T − 1.            (2.1)
                                                      n=1

Proposition 2.1 A trading strategy H is self-financing if and only if
                                     V (t) = V (0) + G(t),                     t = 1, 2, . . . , T.                                  (2.2)

Proof. ⇒ Suppose H is self-financing, then by (2.1)
                                     t                                                       t   N
V (0) + G(t) = V (0) +                    H0 (u)(B(u) − B(u − 1)) +                                   Hn (u)(Sn (u) − Sn (u − 1))
                                    u=1                                                     u=1 n=1
              t                            t                                       t       N                        t       N
= V (0) +          H0 (u)B(u) −                 H0 (u)B(u − 1) +                              Hn (u)Sn (u) −                    Hn (u)Sn (u − 1)
             u=1                          u=1                                 u=1 n=1                              u=1 n=1
                                     t                 t
                   = V (0) +              V (u) −              V (u − 1) = V (t),                    t = 1, 2, . . . , T.
                                    u=1               u=1

      ⇐ Suppose (2.2) holds. We then have
                                                    V (t) = V (0) + G(t)
                          t                                                    t        N
        = V (0) +              H0 (u)(B(u) − B(u − 1)) +                                     Hn (u)(Sn (u) − Sn (u − 1))
                       u=1                                                    u=1 n=1
              t                            t                                    t  N                                t       N
= V (0) +          H0 (u)B(u) −                 H0 (u)B(u − 1) +                              Hn (u)Sn (u) −                    Hn (u)Sn (u − 1)
             u=1                          u=1                                 u=1 n=1                              u=1 n=1
                                t                 t                                  t           N
              = V (0) +              V (u) −           H0 (u)B(u − 1) −                               Hn (u)Sn (u − 1)
                               u=1               u=1                                        u=1 n=1
      t−1                            t                                    t    N
  =         V (u) + V (t) −               H0 (u)B(u − 1) −                              Hn (u)Sn (u − 1),               t = 1, . . . , T.
      u=0                           u=1                                u=1 n=1

Therefore,
                      t−1                   t                                          N
                              V (u) =             H0 (u)B(u − 1) +                           Hn (u)Sn (u − 1)
                      u=0                 u=1                                          n=1
                    t−1                                     N
               =              H0 (u + 1)B(u) +                   Hn (u + 1)Sn (u) ,                       t = 1, . . . , T,
                    u=0                                    n=1

and hence
                                                                 N
                  V (t − 1) = H0 (t)B(t − 1) +                        Hn (t)Sn (t − 1),                  t = 2, . . . , T,
                                                                n=1

or equivalently
                                                           N
               V (t) = H0 (t + 1)B(t) +                         Hn (t + 1)Sn (t),                 t = 1, . . . , T − 1.
                                                       n=1


                                                                     28
Thus, H is self-financing.
   We define the discounted price process

                    ∗         Sn (t)
                   Sn (t):=          ,      t = 0, 1, . . . , T, n = 1, 2, . . . , N,
                              B(t)

the discounted value process
                                    
                                                 N
                                    H0 (1) +                ∗
                                                      Hn (1)Sn (0), t = 0,
                                    
                        V (t)       
              V ∗ (t):=       =                 n=1
                                                N
                        B(t)   H0 (t) +                    ∗
                                                     Hn (t)Sn (t),     t = 1, 2, . . . , T,
                                                n=1

and the discounted gain process
                                T     N
                   G∗ (t):=                       ∗
                                          Hn (u)∆Sn (u),         t = 1, 2, . . . , T,
                               u=1 n=1

where
                                              ∗        ∗
                                    ∆Sn (u):=Sn (u) − Sn (u − 1).

    Absolutely analogously to Proposition 2.1, one can prove

Proposition 2.2 A trading strategy H is self-financing if and only if

                       V ∗ (t) = V ∗ (0) + G∗ (t),           t = 1, 2, . . . , T.




                                                  29
2.2    Conditional expectations and martingales
Let Ω = {ω1 , . . . , ωK } be a sample space and let P be a probability measure on Ω such
that P (ωi ) > 0 for each i = 1, 2, . . . , K (P being defined on the σ-algebra of all subsets
of Ω). Let F be a σ-algebra on Ω with the corresponding partition

                                       P = {B1 , B2 , . . . , Bn }.

    Let Y be a random variable on Ω taking values {y1 , y2 , . . . , ym }. For each Bj ∈ P,
we define an elementary conditional probability
                                                       P ({Y = yi } ∩ Bj )
                           P (Y = yi | Bj ):=                              ,
                                                             P (Bj )

which is the probability that Y = yi under the condition Bj (i.e., the probability that
Y (ω) = yi if we know that ω ∈ Bj ). We now define the elementary conditional expectation
                                                  m
                            E(Y | Bj ):=               yi P (Y = yi | Bj ),
                                                 i=1

which is the expectation of Y under the condition Bj .
    But different Bj ’s can happen, so that it is reasonable to define the conditional ex-
pectation as the random variable

                          E(Y | F)(ω):=E(Y | Bj )                     if ω ∈ Bj ,

which is called the conditional expectation of Y with respect to the σ-algebra F. Note that
E(Y | F) is, by definition, constant on each Bj ∈ F. Therefore, E(Y | F) is measurable
with respect to F.

Proposition 2.3 We have
                                       E(E(Y | F)) = E(Y ).

Proof. Let P = {Bj }n be the minimal partition corresponding to F. Then, for ω ∈ Bj ,
                    j=1

                                                               m
                    E(Y | F)(ω) = E(Y | Bj ) =                      yi P (Y = yi | Bj ).
                                                              i=1

Therefore, by the definition of the expectation,
                                                       n
                            E(E(Y | F)) =                   E(Y | Bj )P (Bj )
                                                    j=1
                                   n     m
                               =               yi P (Y = yi | Bj )P (Bj )
                                j=1 i=1
                               n   m
                                               P ({Y = yi } ∩ Bj )
                           =              yi                       P (Bj )
                               j=1 i=1
                                                     P (Bj )
                                    n  m
                               =                yi P ({Y = yi } ∩ Bj )
                                   j=1 i=1


                                                       30
                                  m          n
                              =         yi         P ({Y = yi } ∩ Bj )
                                  i=1        j=1
                                    m
                              =          yi P ({Y = yi }) = EY.
                                   i=1


Proposition 2.4 If F1 ⊂ F2 , then

                              E(E(Y | F2 ) | F1 ) = E(Y | F1 ).

Proof. See Appendix.

Proposition 2.5 If X is measurable with respect to F, then

                                  E(XY | F) = XE(X | F).

Proof. Let P = {Bj }n be the minimal partition corresponding to F. Then, since X is
                    j=1
measurable with respect to F, X(ω) = xj for all ω ∈ Bj . For any ω ∈ Bj , by definition

                                  E(XY | F) = E(XY | Bj )
                                            = E(xj Y | Bj )
                                            = xj E(Y | Bj ).

Therefore, E(XY | F) = XE(Y | F).

Corollary 2.1 If X is measurable with respect to F, then E(X | F) = X.

Proof. Set Y = 1. Since E(1 | F) = 1, we conclude the statement form Proposition 2.5.


Proposition 2.6 We have

                       E(aX + bY | F) = aE(X | F) + bE(Y | F)

for any constants a, b ∈ R.

Proof. Trivial.

Example 2.4 Let Ω = {ω1 , ω2 , ω3 , ω4 } and let X be a random variable with X(ωi ) = xi
(xi = xj if i = j). Consider the σ-algebra F0 = {∅, Ω}. The corresponding minimal
partition is P0 = {Ω}. So, according to the definition of the conditional expectation, we
have for all ω ∈ Ω
                                 E(X | F0 )(ω) = E(X).
    Consider now the σ-algebra F1 on Ω = {ω1 , ω2 , ω3 , ω4 } whose corresponding minimal
partition is
                              P = {{ω1 , ω2 }, {ω3 , ω4 }}.
Then
                                                              P (ω1 )
                        P (X = x1 | {ω1 , ω2 }) =                           ,
                                                          P (ω1 ) + P (ω2 )

                                                    31
                                                           P (ω2 )
                             P (X = x2 | {ω1 , ω2 }) =                   ,
                                                       P (ω1 ) + P (ω2 )
                                                           P (ω3 )
                             P (X = x3 | {ω3 , ω4 }) =                   ,
                                                       P (ω3 ) + P (ω4 )
                                                           P (ω4 )
                             P (X = x4 | {ω3 , ω4 }) =                   .
                                                       P (ω3 ) + P (ω4 )

Hence, for ω ∈ {ω1 , ω2 },

                  E(X | F1 )(ω) = E(X | {ω1 , ω2 })
                                          P (ω1 )               P (ω2 )
                                = x1                   + x2
                                     P (ω1 ) + P (ω2 )      P (ω1 ) + P (ω2 )
                                  x1 P (ω1 ) + x2 P (ω2 )
                                =                         ,
                                     P (ω1 ) + P (ω2 )

and or ω ∈ {ω3 , ω4 },

                  E(X | F1 )(ω) = E(X | {ω3 , ω4 })
                                          P (ω3 )               P (ω3 )
                                = x3                   + x4
                                     P (ω3 ) + P (ω4 )      P (ω4 ) + P (ω3 )
                                  x3 P (ω3 ) + x4 P (ω4 )
                                =                         .
                                     P (ω3 ) + P (ω4 )

Thus, E(X | F1 ) is the random variable given by
                                           x1 P (ω1 )+x2 P (ω2 )
                                              P (ω1 )+P (ω2 )
                                                                 ,   if ω ∈ {ω1 , ω2 },
                     E(X | F1 )(ω) =       x3 P (ω3 )+x4 P (ω4 )
                                              P (ω3 )+P (ω4 )
                                                                 ,   if ω ∈ {ω3 , ω4 }.

    By a trivial generalization of the above example, one proves the following proposition.

Proposition 2.7 Let Ω = {ω1 , . . . , ωK }, let F be a σ-algebra on Ω and let P be the
corresponding partition. Let X : Ω → R be a random variable. Then for each ω ∈ Ω, we
have
                                  X(ωi1 )P (ωi1 ) + · · · + X(ωil )P (ωil )
                   E(X | F)(ω) =                                            ,
                                          P (ωi1 ) + · · · + P (ωil )
where {ωi1 , . . . , ωil } is the element of the partition P that contains ω.

     Now, suppose that P is a probability measure on a sample space Ω and suppose
{Ft }T is a filtration on Ω (in particular, F0 ⊂ F1 ⊂ · · · ⊂ FT ).
     t=0
     A stochastic process {Zt }T adapted to the filtration {Ft }T is called a martingale
                               t=0                                        t=0
with respect to {Ft }T if for any t1 , t2 ∈ {0, 1, . . . , T }, t1 < t2 , we have
                     t=0

                                         E(Zt2 | Ft1 ) = Zt1 .

Proposition 2.8 A stochastic process {Zt }T adapted to the filtration {Ft }T is a mar-
                                          t=0                             t=0
tingale if and only if

                             Zt = E(ZT | Ft ),           t = 0, 1, . . . , T − 1.

                                                    32
Proof. ⇒ Follows from the definition with t1 = t and t2 = T .
    ⇐ For any t1 < t2 < T , by Proposition 2.4, we have

               E(Zt2 | Ft1 ) = E(E(ZT | Ft2 ) | Ft1 ) = E(ZT | Ft1 ) = Zt1 .

Example 2.5 Let Ω = {ω1 , ω2 , ω3 , ω4 }, P (ω1 ) = P (ω2 ) = P (ω3 ) = P (ω4 ) = 1/4 and let
us consider the following filtration:

                         F0 = {∅, Ω},
                         F1 is generated by P1 = {{ω1 , ω2 }, {ω3 , ω4 }},
                         F2 = {all subsets of Ω}.

Consider the stochastic process

                                  S(0, ω) = 5/2 for all ω,
             S(1, ω1 ) = 3/2, S(1, ω2 ) = 3/2, S(1, ω3 ) = 7/2, S(1, ω4 ) = 7/2,
                  S(2, ω1 ) = 1, S(2, ω2 ) = 2, S(2, ω3 ) = 3, S(2, ω4 ) = 4.

Then {S(t)}t=0,1,2 is a martingale with respect {Ft }t=0,1,2 . Indeed,

                                                     1                  5
                         E(S(2) | F0 ) = E(S1 ) =      (1 + 2 + 3 + 4) = ,
                                                     4                  2
and for ω ∈ {ω1 , ω2 }
                                                        1       1
                                                     1· 4 +2·   4    3
                              E(S(2) | F1 )(ω) =        1           = ,
                                                        4
                                                          +1
                                                           4
                                                                     2
and for ω ∈ {ω3 , ω4 }
                                                    1           1
                                                 3· 4 +4·       4    7
                              E(S(2) | F1 )(ω) =    1               = .
                                                    4
                                                      +1
                                                       4
                                                                     2
Hence, E(S(2) | F1 ) = S(1).




                                                33
2.3    Risk neutral probability measures (martingale probability
       measures) and no-arbitrage opportunities
An arbitrage opportunity is a trading strategy H such that
(a) V (0) = 0,
(b) V (T ) ≥ 0 for all ω ∈ Ω,
(c) EV (T ) > 0,
(d) H is self-financing,
or equivalently,
(a) V ∗ (0) = 0,
(b) V ∗ (T ) ≥ 0 for all ω ∈ Ω,
(c) EV ∗ (T ) > 0,
(d) H is self-financing,
or equivalently,
(a) V ∗ (0) = 0,
(b) V ∗ (T ) ≥ 0 for all ω ∈ Ω,
(c) ∃ ω0 ∈ Ω: V ∗ (T, ω0 ) > 0,
(d) H is self-financing.

Lemma 2.1 There exists an arbitrage opportunity H if and only if there exists a trading
         ˆ           ˆ             ˆ                ˆ
strategy H such that G∗ (T ) ≥ 0, EG∗ (T ) > 0, and H is self-financing.

    The proof of Lemma 2.1 can be carried out analogously to the proof of this result for
the single period model (Lemma 1.1).

Example 2.6 Consider Example 2.3. Recall:

                                           S(0) = 5,
                  S(1, ω1 ) = 8, S(1, ω2 ) = 8, S(1, ω3 ) = 4, S(1, ω4 ) = 4,
                  S(2, ω1 ) = 9, S(2, ω2 ) = 6, S(2, ω3 ) = 6, S(2, ω4 ) = 3.

The filtration F0 , F1 , F2 is given by

          P0 = {Ω}, P1 = {{ω1 , ω2 }, {ω3 , ω4 }}, P2 = {{ω1 }, {ω2 }, {ω3 }, {ω4 }}.

H0 must be measurable with respect to F0 . Therefore, H0 = (H0 (1), H1 (1)) is a deter-
ministic vector.
    Let now γ ≥ 1/8 and consider H(1) = (0, 0) and

                                           8
                                          1+γ
                                              , −1   , ω = ω1 , ω2 ,
                           H(2, ω) =
                                         (0, 0),        ω = ω3 , ω4 .

Then H(2) is measurable with respect to F1 , V (0) = 0, V (1, ω) = 0,
                                                8
      H0 (2, ω)(1 + γ) + H1 (2, ω)S(1, ω) =        (1 + γ) − 8 = 0         for ω = ω1 , ω2
                                               1+γ
and
                 H2 (2, ω)(1 + γ) + H1 (2, ω)S(1, ω) = 0       for ω = ω3 , ω4 .

                                               34
So, H is self-financing strategy. Moreover,
                               
                                  8
                                1+γ · (1 + γ)2 − 9 ≥ 0, ω = ω1 ,
                               
                                  8
                     V (2, ω) = 1+γ · (1 + γ)2 − 6 > 0, ω = ω2 ,
                               
                                 0,                      ω = ω3 , ω4 .
                               

Thus, this trading strategy H is an arbitrage opportunity.

    We define a risk neutral probability measure (martingale measure) as a probability
measure Q on Ω such that
(a) Q(ω) > 0 for all ω ∈ Ω,
                             ∗
(b) The discounted price {Sn (t)}T is a martingale under Q for every n = 1, 2, . . . , N ,
                                    t=0
with respect to the filtration {Ft }T , where each Ft is the σ-algebra generated by the
                                      t=0
                    ∗
random variables, Sn (u), u = 1, . . . , t, n = 1, . . . , N .
    Thus, we have
                        ∗                  ∗
                   EQ (Sn (t2 ) | Ft1 ) = Sn (t1 ),        t1 < t2 , n = 1, 2, . . . , N.

The latter equation is equivalent to

            EQ (Sn (t2 ) | Ft1 ) = (1 + γ)t2 −t1 Sn (t1 ),        t1 < t2 , n = 1, 2, . . . , N,

or by Proposition 2.4, to

              EQ (Sn (T ) | Ft ) = (1 + γ)T −t Sn (t),           t < T, n = 1, 2, . . . , N.       (2.3)

Example 2.6 (continued)Let us try to calculate a risk neutral probability measure for
this model (if it exists). For t = 0, we have F0 = {∅, Ω}, so that E(· | F0 ) = EQ (·).
Therefore, setting in (2.3) t = 0, T = 2,

                      9Q(ω1 ) + 6Q(ω2 ) + 6Q(ω3 ) + 3Q(ω4 ) = 5(1 + γ)2 .                          (2.4)

Next, for t = 1, P1 = {{ω1 , ω2 }, {ω3 , ω4 }} and

                                                         Q(ω1 )
                           Q(S(2) = 9 | {ω1 , ω2 }) =                 ,
                                                      Q(ω1 ) + Q(ω2 )
                                                         Q(ω2 )
                           Q(S(2) = 6 | {ω1 , ω2 }) =                 ,
                                                      Q(ω1 ) + Q(ω2 )
                           Q(S(2) = 3 | {ω1 , ω2 }) = 0,
                           Q(S(2) = 9 | {ω3 , ω4 }) = 0,
                                                         Q(ω3 )
                           Q(S(2) = 6 | {ω3 , ω4 }) =                 ,
                                                      Q(ω3 ) + Q(ω4 )
                                                         Q(ω4 )
                           Q(S(2) = 3 | {ω3 , ω4 }) =                 .
                                                      Q(ω3 ) + Q(ω4 )

    So, for ω ∈ {ω1 , ω2 },

                                                                       9Q(ω1 ) + 6Q(ω2 )
              EQ (S(2) | F1 ) (ω) = EQ (S(2) | {ω1 , ω2 }) =                             .
                                                                        Q(ω1 ) + Q(ω2 )

                                                      35
Thus, setting in (2.3) t = 1, T = 2, we get

                               9Q(ω1 ) + 6Q(ω2 )
                                                 = (1 + γ)8,
                                Q(ω1 ) + Q(ω2 )

so that
                      9Q(ω1 ) + 6Q(ω2 ) = 8(1 + γ)(Q(ω1 ) + Q(ω2 )).                  (2.5)
    Analogously, for ω ∈ {ω3 , ω4 },

                                                               6Q(ω3 ) + 3Q(ω4 )
             EQ (S(2) | F1 ) (ω) = EQ (S(2) | {ω3 , ω4 }) =                      .
                                                                Q(ω3 ) + Q(ω4 )

Hence
                               6Q(ω3 ) + 3Q(ω4 )
                                                 = (1 + γ)4,
                                Q(ω3 ) + Q(ω4 )
so that
                      6Q(ω3 ) + 3Q(ω4 ) = 4(1 + γ)(Q(ω3 ) + Q(ω4 )).                  (2.6)
From (2.5) and (2.6), we easily obtain
                                              2 + 8γ
                                     Q(ω1 ) =        Q(ω2 ),                          (2.7)
                                              1 − 8γ
                                              1 + 4γ
                                     Q(ω3 ) =        Q(ω4 ).                          (2.8)
                                              2 − 4γ

    Substituting (2.7), (2.8) into (2.4), we get

           9(2 + 8γ)                    6(1 + 4γ)
                     Q(ω2 ) + 6Q(ω2 ) +           Q(ω4 ) + 3Q(ω4 ) = 5(1 + γ)2 ,
            1 − 8γ                       2 − 4γ
so that
                      24(1 + γ)          12(1 + γ)
                                Q(ω2 ) +           Q(ω4 ) = 5(1 + γ)2 .               (2.9)
                       1 − 8γ             2 − 4γ
    Substituting (2.7), (2.8) into

                           Q(ω1 ) + Q(ω2 ) + Q(ω3 ) + Q(ω4 ) = 1,

we have
                                3               3
                                    Q(ω2 ) +        Q(ω4 ) = 1.                      (2.10)
                             1 − 8γ          2 − 4γ
(2.9) minus 4 times (2.10) gives
                                       12
                                            Q(ω2 ) = 1 + 5γ,
                                     1 − 8γ
so that
                                            (1 + 5γ)(1 − 8γ)
                                Q(ω2 ) =                     .                       (2.11)
                                                   12
By (2.7) and (2.11)
                                            (1 + 4γ)(1 + 5γ)
                                Q(ω1 ) =                     .                       (2.12)
                                                   6

                                                36
By (2.10) and (2.11),
                                            (1 − 2γ)(3 − 5γ)
                                 Q(ω4 ) =                    ,                           (2.13)
                                                   6
and by (2.8) and (2.13),
                                            (1 + 4γ)(3 − 5γ)
                                 Q(ω3 ) =                    .
                                                   12
    If 0 ≤ γ < 1/8, all Q(ωi )’s are positive, so that we have a martingale measure. On
the other hand, if γ ≥ 1/8, we get Q(ω2 ) ≤ 0, so that there is no martingale measure.
    Notice that we have found before that, for γ ≥ 1/8, there is an arbitrage opportunity.

Theorem 2.1 There are no arbitrage opportunities if and only if there exists a martingale
measure.

Proof of the “if ” part of the theorem. We first need the following lemma.

Lemma 2.2 If {Z(t)}T is a martingale with respect to a filtration {Ft }T and {H(t)}T
                        t=0                                           t=0         t=0
is a predictable process with respect to the same filtration, then
                   t
         G(t) =         H(u)∆Z(u),      t = 1, 2, . . . , T, ∆Z(u) = Z(u) − Z(u − 1),
                  u=1

is also a martingale.

Proof. Since Z(t) and H(t) are adapted to Ft , so is G(t). For t2 > t1 ≥ 1, we have, by
Propositions 2.4, 2.5, 2.6 and Corollary 2.1

                     E(G(t2 ) | Ft1 ) = E(G(t2 ) − G(t1 ) + G(t1 ) | Ft1 )
                            = E(G(t2 ) − G(t1 ) | Ft1 ) + G(t1 )
   = E(H(t1 + 1)(Z(t1 + 1) − Z(t1 )) + · · · + H(t2 )(Z(t2 ) − Z(t2 − 1)) | Ft1 ) + G(t1 )
= E(H(t1 + 1)(Z(t1 + 1) − Z(t1 )) | Ft1 ) + · · · + E(H(t2 )(Z(t2 ) − Z(t2 − 1)) | Ft1 ) + G(t1 )
                     = E(E(H(t1 + 1)(Z(t1 + 1) − Z(t1 )) | Ft1 ) | Ft1 )
                + E(E(H(t1 + 2)(Z(t1 + 2) − Z(t1 + 1)) | Ft1 +1 ) | Ft1 )
             + · · · + E(E(H(t2 )(Z(t2 ) − Z(t2 − 1)) | Ft2 −1 ) | Ft1 ) + G(t1 )
                     = E(H(t1 + 1)E((Z(t1 + 1) − Z(t1 )) | Ft1 ) | Ft1 )
                + E(H(t1 + 2)E((Z(t1 + 2) − Z(t1 + 1)) | Ft1 +1 ) | Ft1 )
             + · · · + E(H(t2 )E((Z(t2 ) − Z(t2 − 1)) | Ft2 −1 ) | Ft1 ) + G(t1 )
                                          = G(t1 ),

where the latter formula holds since Z(t) is a maringale, so that all the corresponding
conditional expectations become 0.

Lemma 2.3 If Q is a martingale measure and H is a self-financing strategy, then V ∗ —
the discounted value process corresponding to H—is a martingale under Q.




                                               37
Proof. By Proposition 2.2, H is self-financing if and only if

                                    V ∗ (t) = V ∗ (0) + G∗ (t).

Since V ∗ (0) is F0 -measurable (i.e., a constant), it is enough to show that G∗ (t) is a
martingale. By definition, we have:
                                            N    t
                                ∗                             ∗
                              G (t) =                 Hn (u)∆Sn (u).
                                            n=1 u=1

Now, by Lemma 2.2, for each n ∈ {1, . . . , N },
                                        t
                                                     ∗
                                             Hn (u)∆Sn (u)
                                       u=1

is a martingle. Thus, G∗ (t) is a sum of N martingales, and hence G∗ (t) is a martingale.


     Suppose Q is a martingale measure. Suppose H is an arbitrage opportunity. Then
V0∗ = 0, VT∗ ≥ 0 and, at least for one ω ∈ Ω, V ∗ (T, ω ) > 0. Since Q(ω) > 0 for all ω ∈ Ω,
this implies that EQ VT∗ > 0. But by Lemma 2.3, Vt∗ is a martingale under Q, so that

                                       V0∗ = EQ VT∗ > 0,

which is a contradiction. Hence, there are no arbitrage opportunities.
Proof of the “only if ” part of the theorem. See Appendix.




                                                 38
2.4    Binomial model
Let us consider a multi-period model with a bank account with interest rate γ and one
security. We suppose that, for each moment of time t = 0, 1, . . . , T − 1, if S(t) is the
current price of the security, then at time t + 1, there are only two possibilities for the
price S(t + 1): either, with probability p, the price S(t) goes up by a factor u > 1, and
thus S(t + 1) becomes S(t)u, or with probability 1 − p the price S(t) goes down by a
factor d ∈ (0, 1), and thus S(t + 1) becomes S(t)d. We assume that the moves over time
are independent of each other.
     So, we define a sample space Ω as the set of all sequences of T digits each of which is
either 0 or 1, i.e.,
                                 ω = (0, 1, 0, 0, 1, . . . , 0, 1),
                                             T digits

where the t-th digit 1 means that the price goes up between the times t − 1 and t, and
the t-th digit 0 means that the price goes down between the times t − 1 and t. Thus,
                              S(t, ω) = S(0)un(ω,t) dt−n(ω,t) ,
where n(ω, t) denotes the number of 1’s in the first t digits of ω. We then obviously define
the probability measure P by setting
                                P (ω) = pn(ω) (1 − p)T −n(ω) ,                      (2.14)
where n(ω):=n(ω, T ) is the number of 1’s in ω.
   We note that
      P (S(t) = S(0)un dt−n ) = P (exactly n moves “up” and t − n moves “down”)
                  t n
              =     p (1 − p)t−n .
                 n
    Let us consider any underlying single period model. So, at time t − 1 the security
costs S(t−1) and then, at time t S(t) = S(t−1)u with probability p, and S(t) = S(t−1)d
with probability 1 − p. Let us require that 1 + γ < u. Since γ ≥ 0, we have
                                      d < 1 + γ < u.                                (2.15)

    Absolutely analogously to the case S(0) = 1, which was discussed at Example Classes,
we conclude that this model has a unique martingale measure, which is given as follows:
the probability to go up is equal to
                                           1+γ−d
                                      q=         ,                                  (2.16)
                                            u−d
and the probability to go down is equal 1 − q. Since the moves over the time are inde-
pendent, one may guess that the measure Q on Ω defined by
                                Q(ω) = q n(ω) (1 − q)T −n(ω)                        (2.17)
is a martingale measure for this model (compare with (2.14)). In fact, one can directly
prove this. Thus, by Theorem 2.1, we have

Proposition 2.9 There are no arbitrage opportunities for the binomial model, and the
measure Q defined by (2.17) is a martingale measure for this model.


                                             39
2.5      Options
Contingent claim

Similarly to the single period model, a contingent claim X is a random variable, i.e., a
function X : Ω → R which represents the time T pay-off from a “seller” to the “buyer”.
If, as usual, {S(t)}T is a price process, then
                    t=0

                                   X(ω):=(S(T, ω) − K)+

is a European call option, which is a right, but not an obligation to buy a share of the
stock at the terminal time T for the price K.
     The pay-off
                                 X(ω):=(K − S(T, ω))+
is a European put option, which is a right, but not an obligation to sell a share of stock
at the terminal time T for the price K.
     In both cases, K is called a striking price.
     The pay-off
                                                                         +
                                S(0, ω) + S(1, ω) + · · · + S(T, ω)
                       X(ω):=                                       −K
                                             T +1

is called an Asian, or averaging option.
     We will always assume that there is a risk neutral probability measure Q for the
market, and hence there are no arbitrage opportunities.
     Similarly to the single period case, a contingent claim is said to be marketable, or
attainable if there exists a self-financing trading strategy H such that

                                VT (ω) = X(ω) for all ω ∈ Ω.

The corresponding portfolio is said to generate, or replicate X.
   By Lemma 2.3, under the measure Q, V ∗ (t) is a martingale. Therefore,

                                V ∗ (t) = EQ (V ∗ (T ) | Ft )
                                                    X
                                        = EQ                  Ft .
                                               (1 + γ)T

Thus,


                            V (t) = V ∗ (t)B(t)
                                              X
                                  = EQ            T
                                                    Ft (1 + γ)t .
                                           (1 + γ)

      In particular,
                                 p = V (0) = EQ (X)(1 + γ)T
is the fair price of the continent claim X.



                                               40
Example 2.7 (Example 2.6 continued) Let us suppose that γ = 0. Then, as proved
above,
                                 1 1 1 1
                           Q=     ,   , ,
                                 6 12 4 2
is a risk neutral probability measure. Consider the             European call option with striking
price K = 5. Then
                                               
                                               4,
                                                                ω = ω1 ,
                                           +
                       X(ω) = (S(2, ω) − 5) = 1,                 ω = ω2 , ω3 ,
                                               
                                                 0,              ω = ω4 .
                                               


    Let us assume, for a moment, that this call is attainable. Then
                                       1     1       1     1
                        p = EQ (X) =     ·4+    · 1 + · 1 + · 0 = 1.
                                       6     12      4     2
So, p = 1 is the price of the contingent claim.
    Next, for ω = ω1 , ω2 ,

                                  V ∗ (1, ω) = EQ (X | F1 )(ω)
                                                1           1
                                                6
                                                    ·4+    12
                                                                ·1
                                            =        1      1
                                                     6
                                                       +   12
                                            = 3,

and for ω = ω3 , ω4 ,

                                  V ∗ (1, ω) = EQ (X | F1 )(ω)
                                                1       1
                                                4
                                                    ·1+ 2 ·0
                                            =        1
                                                     4
                                                       +1
                                                        2
                                             1
                                            = .
                                             3

   Let us now prove that X is indeed attainable and let us find the hedging portfolio
H(t) = (H0 (t), H1 (t)), t = 1, 2. First, we note that the measurability condition implies
H(1) is constant, and H(2) is constant on {ω1 , ω2 } and on {ω3 , ω4 }.
   We next have:

                           V (2, ω1 ) = 4 = H0 (2, ω1 ) + 9H1 (2, ω1 ),                     (2.18)
                           V (2, ω2 ) = 1 = H0 (2, ω1 ) + 6H1 (2, ω1 ).                     (2.19)

(2.18) and (2.19) imply

                  H0 (2, ω1 ) = H0 (2, ω2 ) = −5, H1 (2, ω1 ) = H1 (2, ω2 ) = 1.            (2.20)

    Similarly,

                           V (2, ω3 ) = 1 = H0 (2, ω3 ) + 6H1 (2, ω3 ),                     (2.21)
                           V (2, ω4 ) = 0 = H0 (2, ω3 ) + 3H1 (2, ω3 ).                     (2.22)


                                                41
(2.21) and (2.22) imply
                                             1
                  H0 (2, ω3 ) = H0 (2, ω4 ) = , H1 (2, ω3 ) = H1 (2, ω4 ) = −1.   (2.23)
                                             3

    Next, for ω = ω1 , ω2 ,

                               V (1, ω) = 3 = H0 (1) + 8H1 (1),                   (2.24)

and for ω = ω1 , ω2 ,
                                            1
                               V (1, ω) =     = H0 (1) + 4H1 (1).                 (2.25)
                                            3
(2.24) and (2.25) imply
                                            7          2
                                  H0 (1) = − , H1 (1) = .                         (2.26)
                                            3          3
    (2.20), (2.23), (2.26) define a strategy.
    Exercise. Check that this strategy is self-financing. Explain the hedging portfolio
in your own words if you a seller of the option.




                                               42
2.6    Complete markets
Analogously, to the single period market, we will say that a (multi-period) model is
complete if every contingent claim X is attainable. Otherwise the model is said to be
incomplete.
    Without proof, we state the following theorem, whose formulation is identical with
the corresponding result for the single period market.

Theorem 2.2 Assume there are no arbitrage opportunities, or equivalently, there exists
a martingale measure. Then the model is complete if and only if the martingale measure
is unique.

    For the proof of this theorem, see e.g. the book “Mathematics of Financial Markets”
by R.J. Elliot and P.E. Kopp.




                                          43
2.7     European call options under the binomial model
Let us recall the binomial model from subsec. 2.4, let us suppose that the condition (2.15)
is satisfied, so that there exists a risk neutral probability measure Q given by (2.17). One
may show that in this case, this model is complete. (See e.g. the book “Mathematics of
Financial Markets” by R.J. Elliot and P.E. Kopp.)
     Let us consider the European call option with striking price K. Thus,

                                                                S(0)un(ω) dT −n(ω) − K, if n ≥ n,
                                                                                               ˆ
               X(ω) = (S(T, ω) − K)+ =
                                                                0,                      otherwise,

      ˆ
where n is the minimal non-negative integer n satisfying

                                                     S(0)un dT −n > K,

or equivalently
                                                       u    n         K
                                                                >          .                                (2.27)
                                                       d            S(0)dT
Since u > d, we have log(u/d) > 0, and hence (2.27) is equivalent to,
                                                        log(K/S(0)dT )
                                                   n>                  .
                                                           log(u/d)
Therefore,
                                                      log(K/S(0)dT )
                                           ˆ
                                           n=                        + 1,                                   (2.28)
                                                         log(u/d)
where [a] denotes the integer part of a number a.
   Thus, the price of X at time t = 0 is given by

                                                   p = (1 + γ)−T EQ X(ω)
                                               T
                                      −T             T n
                     = (1 + γ)                         q (1 − q)T −n (S(0)un dT −n − K)
                                           n=ˆ
                                             n
                                                     n
                             T                                                            T
                     −T              T                                                         T n
   = S(0)(1 + γ)                       (uq)n (d(1 − q))T −n − K(1 + γ)−T                         q (1 − q)T −n
                               n
                             n=ˆ
                                     n                                                   n=ˆ
                                                                                           n
                                                                                               n
               T                           n                      T −n                   T
                     T            uq               d(1 − q)                                    T n
      = S(0)                                                             − K(1 + γ)−T            q (1 − q)T −n
                 n
               n=ˆ
                     n           1+γ                1+γ                                 n=ˆ
                                                                                          n
                                                                                               n
                         T                                                       T
                                   T n                                                 T n
             = S(0)                  q (1 − q )T −n − K(1 + γ)−T
                                     ˆ      ˆ                                            q (1 − q)T −n ,
                        n
                      n=ˆ
                                   n                                             n=ˆ
                                                                                   n
                                                                                       n

where
                                                                   uq
                                                           ˆ
                                                           q :=       ,                                     (2.29)
                                                                  1+γ
and using (2.16), it is easy to check that
                                                                    d(1 − q)
                                                     1−q =
                                                       ˆ                     .
                                                                     1+γ

      Thus, we have

                                                                  44
Proposition 2.10 (Discrete time Black–Scholes formula) Under the binomial model,
at time t = 0, the price of the European call option with striking price K is given by
                     T                                     T
                           T n                                   T n
          p = S(0)           q (1 − q )T −n − K(1 + γ)−T
                             ˆ      ˆ                              q (1 − q)T −n ,
                       n
                     n=ˆ
                           n                               n=ˆ
                                                             n
                                                                 n

      ˆ                                               ˆ
where n is given by (2.28), q is given by (2.16), and q is given by (2.29).

Remark 2.1 The classical Black–Scholes formula is derived for European call options for
a continuous time financial market. This continuous time market may be approximated
by discrete time binomial models, and the continuous time Black–Scholes formula may
then be approximated by discrete time Black–Scholes formulas.




                                            45
2.8    Appendices
Proof of Proposition 2.4

Let P1 , P2 be the minimal partitions corresponding to F1 , F2 , respectively. Denote

                                      P1 = {Aj }n1 , P2 = {Bj }n2 ,
                                                j=1            j=1

and let Y take values {y1 , . . . , ym }. As Aj ∈ F1 ⊂ F2 , so Aj is a union of some Bi ’s.
    We note that
                          E(Y | F2 )(ω) = E(Y | Bi ) if ω ∈ Bi ,
and for ω ∈ Aj

                          E(E(Y | F2 ) | F1 )(ω) = E(E(Y | F2 ) | Aj )
                                                n2
                                           =          E(Y | Bi )P (Bi | Aj )
                                                i=1
                                      n2       m
                                =                 yl P (Y = yl | Bi )P (Bi | Aj )
                                  i=1 l=1
                                 n2 m
                                                P ({Y = yl } ∩ Bi ) P (Bi ∩ Aj )
                           =
                                 i=1 l=1
                                                      P (Bi )          P (Aj )
                          m
                                                                P ({Y = yl } ∩ Bi )P (Bi )
                      =         yl
                          l=1        i=1,...,n2 : Bi ∩Aj     =∅
                                                                     P (Bi )P (Aj )
                                      (since Bi ⊂ Aj if Bi ∩ Aj = ∅)
                                m
                                                                    P ({Y = yl } ∩ Bi )
                          =           yl
                                l=1        i=1,...,n2 : Bi ∩Aj   =∅
                                                                         P (Aj )
                                                m
                                                           P ({Y = yl } ∩ Aj )
                                           =          yl
                                                l=1
                                                                 P (Aj )
                                                   m
                                            =              yl P (Y = yl | Aj )
                                                  l=1
                                                   = E(Y | Aj )
                                                = E(Y | F1 )(ω).

Proof of the “only if ” part of Theorem 2.1

Let Pt be the minimal partition corresponding to Ft . Let A ∈ Pt , t < T . Then we get a
                                                    ∗                            ∗
single period market with initial discounted price Sn (t, ω), ω ∈ A, (note that Sn (t) indeed
                                                     ∗
is a constant on A), and discounted terminal price Sn (t + 1, ω), ω ∈ A, which is a constant
on each A , where A ⊂ A, A ∈ Pt+1 .

Lemma 2.4 If the multiperiod model does not have any arbitrage opportunities, then
none of the underlying single period models has any arbitrage opportunity in the single
period sense.



                                                               46
                                                        ˆ
Proof. Suppose there exists an arbitrage opportunity H for the single period model
corresponding to some A ∈ Pt for some t < T . Therefore, by Lemma 1.1
                               ˆ    ∗                   ˆ    ∗
                               H1 ∆S1 (t + 1) + · · · + Hn ∆SN (t + 1)

is non-negative on A and is not identically equal to zero. Define
             
             0,
                                                      s ≤ t,
             ˆ
             
             Hn ,
             
                                                      s = t + 1, n = 1, 2, . . . , N, ω ∈ A,
                 ˆ1                ˆn
             −H S ∗ (t) − · · · − H S ∗ (t),
             
                    1                  N               s = t + 1, ω ∈ A, n = 0,
 Hn (s, ω) =
             0,
             
                                                      s = t + 1, ω ∈ A,
             ˆ                        ˆN
             H ∆S ∗ (t + 1) + · · · + H ∆S ∗ (t + 1), s > t + 1, ω ∈ A, n = 0,
              1 1
                                             N
             
              0,                                       otherwise.
             


    Let us show that this strategy is self-financing. Evidently, V ∗ (s) = 0 if s ≤ t. Next,
evidently
                                    N
              H0 (s + 1, ω) +                           ∗
                                          Hn (s + 1, ω)Sn (s, ω) = 0 = V ∗ (t),                    s < t.
                                 n=1

Furthermore,
                                                         N
                                                                           ∗
                             H0 (t + 1, ω) +                 Hn (t + 1, ω)Sn (t, ω)
                                                       n=1
                        N                                             N
                                           ∗                                              ∗
                  =−         Hn (t + 1, ω)Sn (t, ω) +                       Hn (t + 1, ω)Sn (t, ω)
                       n=1                                            n=1
                                                     ∗
                                        = 0 = V (t, ω),                   ω ∈ A,

and
                                N
             H0 (t + 1, ω) +                       ∗
                                     Hn (t + 1, ω)Sn (t, ω) = 0 = V ∗ (t, ω),                      ω ∈ A.
                               n=1

Therefore,
                                    N
               H0 (t + 1, ω) +                          ∗
                                          Hn (t + 1, ω)Sn (t, ω) = V ∗ (t, ω),                    ω ∈ Ω.
                                    n=1

Since
                                                                  N
                        ∗                                                        ∗
                      V (t + 1) = H0 (t + 1) +                        Hn (t + 1)Sn (t + 1),
                                                              n=1

we get, for ω ∈ A,
                                                   N                         N
                      V ∗ (t + 1, ω) = −                 ˆ ∗
                                                         Hn Sn (t) +              ˆ ∗
                                                                                  Hn Sn (t + 1)
                                                   n=1                      n=1
                                               N
                                           =         ˆ    ∗
                                                     Hn ∆Sn (t + 1),
                                               n=1


                                                             47
and for ω ∈ A,
                                             V ∗ (t + 1, ω) = 0.
For ω ∈ A,
                           N                                   N
           H0 (t + 2) +         Hn (t +       ∗
                                           2)Sn (t   + 1) =          ˆ
                                                                     Hn ∆Sn (t + 1) = V ∗ (t + 1)
                                                                          ∗

                          n=1                                 n=1

and for ω ∈ A,
                                     N
                     H0 (t + 2) +                     ∗
                                           Hn (t + 2)Sn (t + 1) = 0 = V ∗ (t + 1).
                                     n=1

For s ≥ t + 2, if ω ∈ A, we get
                          N                                               N
                 ∗
              V (s) =           ˆ    ∗
                                Hn ∆Sn (t + 1) = H0 (s + 1) +                             ∗
                                                                               Hn (s + 1)Sn (s),
                          n=1                                            n=1

and for ω ∈ A,
                                                              N
                           ∗                                                  ∗
                        V (s) = 0 = H0 (s + 1) +                   Hn (s + 1)Sn (s).
                                                             n=1

Thus, H is self-financing.
   As V ∗ (0) = 0, V ∗ (T ) ≥ 0, and
                                                      N
                                   V ∗ (T, ω) =            ˆ    ∗
                                                           Hn ∆Sn (t + 1)
                                                     n=1

is not identically equal to zero on A, we conclude that H is an arbitrage opportunity.


     Assume now that there are no arbitrage opportunities. So, for any t < T and any
                                                ∗
A ∈ Pt , we have a price process starting with Sn (t, ω), n = 1, . . . , N , (which is a constant
                                          ∗
on A for each n), with terminal price Sn (t + 1, ω), n = 1, . . . , n. According to Lemma
2.4, there are no arbitrage opportunities in this single period model. According to the
no-arbitrage theorem (Theorem 1.1), there is a risk neutral probability measure Qt,A on
the single period sample space A. It satisfies

                               Qt,A (A ) > 0 if A ∈ Pt+1 and A ⊂ A,
                                                          Qt,A (A ) = 1,
                                      A ∈Pt+1 , A ⊂A
                                        ∗
                                EQt,A ∆Sn (t + 1) = 0,            n = 1, . . . , N.

For each ω ∈ Ω, we define

                                Q(ω):=                                 Qt,A (A ).                   (2.30)
                                                  t=0,1,...,T −1
                                           A∈Pt , A ∈Pt+1 : {ω}⊂A ⊂A



    Since all Qt,A (A ) > 0, we get

                                      Q(ω) > 0 for all ω ∈ Ω.

                                                       48
Next,
   It remains to show that Q is a martingale measure.
   First, we note that the measure Qt,A was chosen in such a way that
                             ∗
                     EQt,A ∆Sn (t + 1) = 0,             n = 1, . . . , N, A ∈ Pt .        (2.31)
                                        ∗
Let A ∈ Pt and let ω ∈ A. Then, since ∆Sn (t + 1) is Ft+1 -measurable,
                            ∗                           ∗
                      EQ (∆Sn (t + 1) | Ft )(ω) = EQ (∆Sn (t + 1) | A)
                                              ∗
                             =             (∆Sn (t + 1) A )Q(A | A)
                                 A ∈Pt+1
                                               ∗
                         =                  (∆Sn (t + 1) A )Q(A | A).                     (2.32)
                             A ∈Pt+1 , A ⊂A


We claim that
                                      Q(A | A) = Qt,A (A ).                               (2.33)
Then, by (2.31) and (2.32),
                   ∗                                           ∗
             EQ (∆Sn (t + 1) | Ft )(ω) =                    (∆Sn (t + 1) A )Qt,A (A )
                                              A ∈Pt+1 , A ⊂A
                                               ∗
                                     = EQt,A ∆Sn (t + 1) = 0.                             (2.34)

Therefore,
                              ∗
                        EQ (∆Sn (t + 1) | Ft ) = 0,            n = 1, . . . , N.          (2.35)

    Now, by Proposition 2.4 and (2.35),
          ∗                        ∗         ∗                        ∗         ∗
     EQ (Sn (T ) | Ft ) = EQ (∆Sn (T ) + ∆Sn (T − 1) + · · · + ∆Sn (t + 1) + Sn (t) | Ft )
              ∗                       ∗                                 ∗              ∗
    = EQ (∆Sn (T ) | Ft ) + EQ (∆Sn (T − 1) | Ft ) + · · · + EQ (∆Sn (t + 1) | Ft ) + Sn (t)
                          ∗                                    ∗
          = EQ (EQ (∆Sn (T ) | FT −1 ) | Ft ) + EQ (EQ (∆Sn (T − 1) | FT −2 ) | Ft )
                                            ∗                           ∗
                        + · · · + EQ (EQ (∆Sn (t + 1) | Ft ) | Ft ) + Sn (t)
                                                 ∗
                                           = Sn (t).
        ∗
Thus, {Sn (t)}T is a martingale under Q.
              t=0
   It remains to show that (2.33) holds.




                                                   49

				
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posted:2/3/2011
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