微信客服：xiaoxionga100
微信客服：ITCS521
证明题代写-MATH 11158
时间：2022-02-23
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
MATH 11158: Optimization Methods in Finance
Arbitrage and Asset Pricing1
Akshay Gupte
School of Mathematics, University of Edinburgh
Week 1 and 2 : 20 – 27 January, 2022
1Sections 4.1–4.4 in the textbook
Email: akshay.gupte@ed.ac.uk
1 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Arbitrage LP
2 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Arbitrage
A trading strategy is a sequence of investment (buying, selling, borrowing
of securities) decisions over time.
For example: borrow now x amount of money to buy certain assets. At
some (determined) future time sell the assets and repay the debt.
Arbitrage is a trading strategy on an asset that results in net profit due
to different prices of the asset in different markets or in different forms. It
exists because of inefficiencies in the financial markets
3 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Example 1
Example 1 : A bank accepts deposits yielding an interest rate of 5%.
Also, it lends money at the same interest rate. You have an opportunity
to invest for 1 year into a government bond that costs £500, yielding a
10% interest.
— Is there arbitrage (i.e. can you make money without risk)?
Solution: The “situation” is easy: one should “borrow from bank and
buy bonds!”.
• We can borrow £ 1, 047.62 from the bank and use this to buy 2
bonds: Cash flow is
£1, 047.62− 2× £500 = £47.62
We now have £47.62, two bonds and debts to the bank
• At time 1, we sell the bonds and repay our debts to the bank
2× 1.10× 500− 1.05× 1047.62 = 0
We sell the bonds (at 10% profit), repay our debts to the bank (at
5% interest) and keep the £47.62!
4 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Example 2
Example 2 : Suppose we start with 2,000 USD currency. There are three
currency exchange places with the following conversion rates
• 1 USD ≡ 0.894 Euro
• 1 GBP ≡ 1.276 Euro
• 1 GBP ≡ 1.432 USD
Is there arbitrage ?
The following cycle of conversions : USD → Euro → GBP → USD, for
the initial amount of 2,000 USD leads to a net amount of 2,006.59 USD
2006.59 =
(
2000 ∗ 0.894
1.276
)
∗ 1.432,
resulting in an arbitrage profit of 6.59 (= 2006.59 - 2000) USD
5 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Arbitrage Types
Definition (Arbitrage Type A)
A trading strategy that has
• positive initial cash flow and
• no risk of a loss in the future.
Definition (Arbitrage Type B)
A trading strategy that has
• nonnegative initial cash flow,
• no risk of a loss in the future and
• a positive probability of a profit in the future.
6 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Arbitrage Detection: Mathematical Model
1. Securities: S0,S1,S2, . . . ,Sn (S0 is risk free)
2. Modeling future uncertainty:
2.1 Consider 2 time periods: “now” (time 0) and “future” (time 1)
2.2 States of the world at maturity (time 1) is described by m scenarios
that are indexed by the set Ω = {ω1, ω2, . . . , ωm}, where ωi denotes
scenario i
2.3 Each scenario occurs with nonzero probability
We do not know the probabilities, we do not need to know them!
3. Prices: For i = 0, . . . , n,
3.1 S i0 = price of S
i at present (time 0)
3.2 S i1(ωj) = price of S
i at maturity under scenario ωj
4. Risk-free security: Suppose we earn interest rate r on security S0
4.1 S0 is the risk-free security (e.g., cash in savings account)
4.2 S00 = 1 (price of risk-free security at present is 1)
4.3 S01 (ωj) = 1 + r = R for all j
Note that it is indeed risk free as its price at time 1 does not depend
on ωj . We can also borrow money at the risk free rate r .
7 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Type-A Arbitrage and Linear Inequalities
portfolio x = (x0, x1, . . . , xn), where xi is the amount (number of units),
either positive or negative, of security S i .
At least x0 can be negative (borrow money at risk free rate)
Note: since we are not starting with an initial amount of money, we are
building a portfolio as opposed to an investment portfolio as we will see
later in portfolio theory
• Cost of portfolio x at time 0 (= negative of initial cash flow)
P0 = P0(x) :=
n∑
i=0
S i0xi
• No risk of a loss in the future
n∑
i=0
S i1(ωj)xi ≥ 0, j = 1, 2, . . . ,m. (1)
Type-A arbitrage exists ⇐⇒ there is a portfolio x such that P0(x) < 0
and inequalities (1) hold (these are linear inequalities in x)
8 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Arbitrage: ... back to our example
Example (Arbitrage detection II). A bank accepts deposits yielding an
interest rate of 5%. Also, it lends money at the same interest rate. You
have an opportunity to invest for 1 year into a government bond that
costs £500, yielding a 10% interest.
1. Is there a type-A arbitrage opportunity?
2. Is there a type-B arbitrage opportunity?
Solution: The answer is easy, borrow from bank and buy bonds!
However, this simple problem lets us understand the concepts
• There are 2 assets: S0 (cash/bank) and S1 (bond); so n = 1
• There are 2 time periods: now (time 0) and a year after (time 1)
• Only one scenario: m = 1 and Ω = {ω1}, since bank and
government bonds can be seen as risk-free
• Cash: S00 = 1, R = 1.05, S01 (ω1) = S00 × R = 1.05
• Bond: S10 = 500, S11 (ω1) = S10 × 1.1 = 550
9 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Example (Cont.)
We will build a portfolio x = (x0, x1)
T . Some things we can do:
Strategy 1.
We can borrow £1, 047.62 from the bank (x0 = −1, 047.62) and use this
to buy 2 bonds (x1 = 2).
• At time 0, cost of building our portfolio (net cash outflow) is
P0(x) =
n∑
i=0
S i0xi = S
0
0 x0+S
1
0 x1 = 1×(−1, 047.62)+500×2 = −47.62
• At time 1, we have only one possible outcome (ω1), and the cost
(=value) of our portfolio (we own it now, so positive value means
we are good!) is:
n∑
i=0
S i1(ωj)xi = 1.05× (−1, 047.62) + 550× 2 = 0
ie. we sell the bonds (at 10% profit), repay our debts to the bank
(at 5% interest) and keep the £47.62!
This is type-A arbitrage!
10 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Example (cont): Strategy 2
Do the same as in Strategy 1, except scale your position x by constant
c > 0. That is, choose: x0 = c × (−1, 047.619) (borrow c times more
cash) and x1 = c × 2 (buy c times as many bonds as before).
• At time 0, the incurred cost of building our portfolio (i.e., net cash
outflow) is
P0(x) =
n∑
i=0
S i0xi = S
0
0 x0 + S
1
0 x1 = c × (−47.619)
• At time 1, we have only one possible outcome (ω1), and the cost
(=value) of our portfolio (we own it now, so positive value means
we are good!) is:
n∑
i=0
S i1(ωj)xi = c × 0
So we earn c times as much as with Strategy 1! For c →∞, we earn as
much as we want.
11 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Example (cont): Strategy 3
We can borrow £1, 000 from the bank (x0 = −1, 000) and use this to buy
2 bonds (x1 = 2).
• At time 0, the incurred cost of building our portfolio (i.e., net cash
outflow) is
P0(x) =
n∑
i=0
S i0xi = S
0
0 x0 + S
1
0 x1 = 1×−1, 000 + 500× 2 = 0
• At time 1, we have only one possible outcome (ω1), and the cost
(=value) of our portfolio (we own it now, so positive value means
we are good!) is:
n∑
i=0
S i1(ωj)xi = 1.05× (−1, 000) + 550× 2 = 50
This is type-B arbitrage!
12 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
ADLP
13 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Arbitrage-Detecting Linear Program
Consider now the linear program
OPT := min
n∑
i=0
S i0xi
subject to
n∑
i=0
S i1(ωj)xi ≥ 0, j = 1, 2, . . . ,m.
(ADLP)
Define OPT = −∞ if the problem is unbounded
• There is type-A arbitrage ⇐⇒ OPT < 0
• (ADLP) is homogeneous (feasible region is a polyhedral cone): if x
is feasible then λ · x is also feasible for any λ > 0
• If there is a feasible x with OPT < 0 then (ADLP) is unbounded
(OPT = −∞).
14 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Detecting Type-A Arbitrage
Fact (1)
The following statements are equivalent:
(a) there is no type-A arbitrage,
(b) OPT = 0,
(c) (ADLP) is bounded.
• To detect arbitrage: solve (ADLP) and check whether OPT = 0
• If OPT 6= 0, we must have OPT = −∞ (since by Fact (1), (ADLP)
must be unbounded!). So, there must exist a feasible portfolio x
(i.e., one that does not pose any risk of a future loss) with negative
cost at time 0, i.e.,
∑n
i=0 S
i
0xi < 0
• LP solvers detect unboundedness by finding such x (a direction of
unboundedness).
This x corresponds to an arbitrage opportunity (of type-A)!
15 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Proof of Fact (1)
• Observe that OPT ≤ 0. Indeed, the empty portfolio (y = 0, i.e., we
do not own any assets) is feasible and costs nothing at time 0.
• We now claim that either OPT = 0 or OPT = −∞. Let us show
this. Assume now that y is a feasible portfolio for which the cost of
which at time 0 is negative and finite. Then the portfolio 2y is also
be feasible, with a smaller cost at time 0. So, OPT cannot be
negative and finite.
• In particular, the above observations tell us that
OPT < 0 ⇔ OPT = −∞ ⇔ (ADLP) is unbounded.
• Further, note that OPT < 0 if and only if there exists type-A
arbitrage. Indeed, OPT < 0 means that there is a portfolio whose
payoff at time 1 is nonnegative but whose cost at present is negative
(i.e., it generates positive cashflow at time 0).
We have thus shown that
OPT = −∞ ⇔ type-A arbitrage exists ⇔ (ADLP) is unbounded
The logical negations of each of these three statements must also be
equivalent, proving the statement.
16 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Example (cont): Detecting type-A Arbitrage by LP
The LP (ADLP) detecting type-A arbitrage is:
minimize P0(x) :=
n∑
i=0
S i0xi
subject to
n∑
i=0
S i1(ωj)xi ≥ 0, j = 1, 2, . . . ,m
In our earlier example (n = 1, S00 = 1, S
1
0 (ω1) = 500, m = 1,
S01 (ω1) = 1.05 and S
1
1 (ω1) = 550), this takes the form
minimize x0 + 500x1
subject to 1.05x0 + 550x1 ≥ 0
• By solving this LP in MATLAB using CVX, we find that it is
unbounded. Hence, by FACT (1), there is type-A arbitrage.
• Following the thought process of Strategy 2 we find a feasible
portfolio cy for all c > 0 such that P0(cy)→ −∞ as c →∞. This
is a “proof by hand” that the LP is unbounded (and type-A
arbitrage exists). 17 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Type-B Arbitrage detection
Type-B arbitrage in our model exists if there exists portfolio
x = (x1, . . . , ym) satisfying the following three conditions:
1. (we do not lose anything at time 0)
P0(x) ≡
n∑
i=0
S i0xi ≤ 0,
2. system of inequalities (1): (there is no risk of loss at time 1)
n∑
i=0
S i1(ωj)xi ≥ 0, j = 1, 2, . . . ,m.
3. (positive probability of positive payoff at time 1)
n∑
i=0
S i1(ωj)xi > 0 for at least one index j
18 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Detecting Type-B Arbitrage with (ADLP) - Theory
Fact (2)
Assume that no type-A arbitrage exists (ie. OPT=0). Then the following
statements are equivalent:
(a) no type-B arbitrage exists,
(b) for all optimal solutions y∗ of (ADLP), all constraints of (ADLP) are
tight. That is,
∑n
i=0 S
i
1(ωj)x
∗
i = 0 for all j = 1, 2, . . . ,m.
• If there is type-A arbitrage, Fact (2) does not tell us anything
• If there isn’t type-A arbitrage, Fact (2) tells us how to detect type-B
arbitrage in in theory. Check if the constraints are tight for all
possible solutions x∗ of (ADLP). Unfortunately it is not easy to find
all solutions to an LP, so this is not much use in practice.
Exercise
What if there is type-A arbitrage?
→ If there is type-A arbitrage then there is also type B-arbitrage.
19 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Detecting Type-B Arbitrage with (ADLP) - Practice
For type-B arbitrage the following conditions need to be satisfied:
P0(y) ≡
n∑
i=0
S i0xi ≤ 0, (i)
n∑
i=0
S i1(ωj )xi ≥ 0, j = 1, 2, . . . ,m, (ii)
n∑
i=0
S i1(ωj )xi > 0 for at least one index j (iii)
We can find such an investment x (if one exists) by solving
OPT := min −
m∑
j=1
1
m
n∑
i=0
S i1(ωj )xi
subject to
n∑
i=0
S i1(ωj )xi ≥ 0, j = 1, 2, . . . ,m.
n∑
i=0
S i0xi ≤ 0
(ADLP)
Define OPT = −∞ if the problem is unbounded
20 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Proof of Fact (2)
• Since we assume that no type-A arbitrage exists, from Fact (1) we
know that OPT = 0.
• Assume (a) holds, i.e no type-B arbitrage. This means that there
cannot exist y satisfying (i), (ii) and (iii).
• Pick any y∗ optimal for (ADLP). Then y∗ clearly satisfies (ii) (since
it is feasible for (ADLP)) and )(i) (since OPT = 0). Hence, it must
be the case that (iii) does not hold for y∗.
• Since (ii) holds, this means that (b) must hold.
We have shown that (a) implies (b). It remains to show the converse.
Exercise
Use similar arguments to finish the proof of Fact (2). That is, show that
if there is no type-A arbitrage, then (b) implies (a).
21 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Dualising (ADLP)
Multiply both the objective and the constraints of (ADLP) by −1 and
use that for any function f , −min f = max(−f )), to get
−OPT =max
n∑
i=0
(−S i0)xi
subject to
n∑
i=0
(−S i1(ωj))xi ≤ 0, j = 1, 2, . . . ,m.
(D)
Note that (D) is in standard form of the dual problem. Indeed, if we now
let b = (−S00 ,−S10 , . . . ,−Sn0 )T ∈ Rn+1, c = 0 ∈ Rm and A ∈ R(n+1)×m
with Aij = −S i1(ωj), then (D) can be written as
−OPT =max bT x
subject to AT x ≤ c . (D)
22 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Problem dual to (D)
The dual of (D) is the standard primal problem:
−OPT =min cT y
subject to Ay = b
y ≥ 0,
(P)
which, after substitution of c , b and A, can be written as
−OPT =min
m∑
i=1
0yj
subject to
m∑
i=1
−S i1(ωj)yj = −S i0, i = 0, 1, . . . , n
yj ≥ 0, j = 1, 2, . . . ,m.
(P)
• All feasible points are also optimal! (Why?)
• (P) does not necessarily have to be feasible
23 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
We would like to understand the meaning of the dual solution y (if it
exists!)
−OPT =min
m∑
i=1
0 · yj
subject to
m∑
i=1
S i1(ωj)yj = S
i
0, i = 0, 1, . . . , n
yj ≥ 0, j = 1, 2, . . . ,m.
(P)
24 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Sports Betting Example
For an upcoming football match the odds offered by a bookmakers are:
home win: 4/11, draw: 7/2, away win 19/2. Is there an arbitrage
opportunity?
Note that odds of a/b mean that if you were to bet amount b on an
event, if the event happens you get a payoff of a (and additionally your
stake b), if the event does not happen, you loose your stake. I.e. cost of
investment is b. In case of win the payoff is a+b, in case of loss the
payoff is 0.
→ Show how this fits into the arbitrage detection setup from the lecture:
Assume there are 4 investments: bet on home win, bet on away win, bet
on draw and cash. Assume that the return (or interest rate) on cash is 1,
i.e. you can borrow unlimited amount of money (to use for bets) without
needing to pay interest. There are three possible outcomes: Home win,
draw and away win. You do not need to know the actual probabilities of
these outcomes.
We have 4 investments, S0 is cash and S1 to S3 are the three bets –
home win, draw and away win. We have 3 scenarios for outcome of the
match – home win, draw or away win. Thus, n = 3 and m = 3. 25 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
→ Work out the values for S i0 and S i1(ωj) and set up the Arbitrage
Detection LP in CVX. Note that you can only place positive amounts as
bets. Is there type-A or type-B arbitrage?
The initial cost of the investments is the vector S0 = (1, 11, 2, 2) (given by
denominators in betting odds). For writing the ADLP constraints which are
one for each scenario, we will need the following matrix
A =
1 15 0 01 0 9 0
1 0 0 21
Each row corresponds to a scenario and each column corresponds to an
investment. So, A12 is the future value of investment S
1 in scenario 1, which is
S11 (ω1) = 11 + 4 = 15. Solve the ADLP as follows. Remember to add
nonnegativity on the three investments (bets) because we can only buy bets
(whereas yi < 0 for i ∈ {1, 2, 3} would mean that we are selling bet i) but for
cash we are allowed to borrow and so it can be negative.
26 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
n = 3;
S0 = [1 11 2 2];
A = [1 15 0 0;1 0 9 0;1 0 0 21];
cvx_begin
variable x(n+1);
minimize (S0’*x)
subject to
A*x >= 0;
x(2:n+1) >= 0;
cvx_end
The optimal LP value is 0 and so there is no type-A arbitrage.
Detecting type-B arbitrage : Refer to previous slide for detecting in
practice. Type-B arbitrage exists if and only if the value OPT = −∞
27 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Arbitrage in Derivatives
28 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Finite vs. Infinite Scenarios
ADLP can be used in all situations with a finite number of scenarios, i.e.,
when Ω = {ω1, ω2, . . . , ωm}
Challenges :
• Finite scenarios may not describe reality well in all situations
• Larger the m, larger (and possibly harder) to solve the ADLP
• Scenarios may need to be generated if they are not available
naturally, or if an infinite set Ω is discretised to obtain finite
scenarios. Scenario generation is not a trivial task in general
Infinitely many scenarios2 (indexed as ω ∈ Ω) can be dealt with in special
cases. One such case is a portfolio of derivatives that are
• written on the same underlying security
• with the same maturity date (time 1)
Value of the security at maturity is uncertain/random
2Ω can be countable or uncountable set
29 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Finite vs. Infinite Scenarios
ADLP can be used in all situations with a finite number of scenarios, i.e.,
when Ω = {ω1, ω2, . . . , ωm}
Challenges :
• Finite scenarios may not describe reality well in all situations
• Larger the m, larger (and possibly harder) to solve the ADLP
• Scenarios may need to be generated if they are not available
naturally, or if an infinite set Ω is discretised to obtain finite
scenarios. Scenario generation is not a trivial task in general
Infinitely many scenarios2 (indexed as ω ∈ Ω) can be dealt with in special
cases. One such case is a portfolio of derivatives that are
• written on the same underlying security
• with the same maturity date (time 1)
Value of the security at maturity is uncertain/random
2Ω can be countable or uncountable set
29 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Finite vs. Infinite Scenarios
ADLP can be used in all situations with a finite number of scenarios, i.e.,
when Ω = {ω1, ω2, . . . , ωm}
Challenges :
• Finite scenarios may not describe reality well in all situations
• Larger the m, larger (and possibly harder) to solve the ADLP
• Scenarios may need to be generated if they are not available
naturally, or if an infinite set Ω is discretised to obtain finite
scenarios. Scenario generation is not a trivial task in general
Infinitely many scenarios2 (indexed as ω ∈ Ω) can be dealt with in special
cases. One such case is a portfolio of derivatives that are
• written on the same underlying security
• with the same maturity date (time 1)
Value of the security at maturity is uncertain/random
2Ω can be countable or uncountable set
29 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Portfolio of Derivatives
Single underlying security under infinitely many scenarios
S0 = known price at time 0, S1(ω) = uncertain price at maturity
Derivatives S i for i = 1, . . . , n written on the security, with known prices
S i0 at time 0 and random prices at maturity
S i1 := S
i
1(ω) = fi (S1(ω))
where function fi is known for all i . Think of S1 as a random variable
S1 : ω ∈ Ω 7→ S1(ω) ∈ R
Portfolio x = (x1, x2, . . . , xn) of derivatives (xi is investment in S
i )
Cost at time 0 : P0(x) :=
n∑
i=1
S i0xi
Value at maturity : P1(x ,S1) := P1(x , ω) =
n∑
i=1
S i1xi =
n∑
i=1
fi (S1(ω))xi
This can be used to detect arbitrage
30 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Portfolio of Derivatives
Single underlying security under infinitely many scenarios
S0 = known price at time 0, S1(ω) = uncertain price at maturity
Derivatives S i for i = 1, . . . , n written on the security, with known prices
S i0 at time 0 and random prices at maturity
S i1 := S
i
1(ω) = fi (S1(ω))
where function fi is known for all i .
Think of S1 as a random variable
S1 : ω ∈ Ω 7→ S1(ω) ∈ R
Portfolio x = (x1, x2, . . . , xn) of derivatives (xi is investment in S
i )
Cost at time 0 : P0(x) :=
n∑
i=1
S i0xi
Value at maturity : P1(x ,S1) := P1(x , ω) =
n∑
i=1
S i1xi =
n∑
i=1
fi (S1(ω))xi
This can be used to detect arbitrage
30 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Portfolio of Derivatives
Single underlying security under infinitely many scenarios
S0 = known price at time 0, S1(ω) = uncertain price at maturity
Derivatives S i for i = 1, . . . , n written on the security, with known prices
S i0 at time 0 and random prices at maturity
S i1 := S
i
1(ω) = fi (S1(ω))
where function fi is known for all i . Think of S1 as a random variable
S1 : ω ∈ Ω 7→ S1(ω) ∈ R
Portfolio x = (x1, x2, . . . , xn) of derivatives (xi is investment in S
i )
Cost at time 0 : P0(x) :=
n∑
i=1
S i0xi
Value at maturity : P1(x ,S1) := P1(x , ω) =
n∑
i=1
S i1xi =
n∑
i=1
fi (S1(ω))xi
This can be used to detect arbitrage
30 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Portfolio of Derivatives
Single underlying security under infinitely many scenarios
S0 = known price at time 0, S1(ω) = uncertain price at maturity
Derivatives S i for i = 1, . . . , n written on the security, with known prices
S i0 at time 0 and random prices at maturity
S i1 := S
i
1(ω) = fi (S1(ω))
where function fi is known for all i . Think of S1 as a random variable
S1 : ω ∈ Ω 7→ S1(ω) ∈ R
Portfolio x = (x1, x2, . . . , xn) of derivatives (xi is investment in S
i )
Cost at time 0 : P0(x) :=
n∑
i=1
S i0xi
Value at maturity : P1(x ,S1) := P1(x , ω) =
n∑
i=1
S i1xi =
n∑
i=1
fi (S1(ω))xi
This can be used to detect arbitrage
30 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Portfolio of Derivatives
Single underlying security under infinitely many scenarios
S0 = known price at time 0, S1(ω) = uncertain price at maturity
Derivatives S i for i = 1, . . . , n written on the security, with known prices
S i0 at time 0 and random prices at maturity
S i1 := S
i
1(ω) = fi (S1(ω))
where function fi is known for all i . Think of S1 as a random variable
S1 : ω ∈ Ω 7→ S1(ω) ∈ R
Portfolio x = (x1, x2, . . . , xn) of derivatives (xi is investment in S
i )
Cost at time 0 : P0(x) :=
n∑
i=1
S i0xi
Value at maturity : P1(x ,S1) := P1(x , ω) =
n∑
i=1
S i1xi =
n∑
i=1
fi (S1(ω))xi
This can be used to detect arbitrage
30 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Type-A Arbitrage : Semi-Infinite LP
Find portfolio x ∈ Rn that is cheapest portfolio at time 0 and has no risk
of loss in future (value at maturity is ≥ 0 for all scenarios)
OPT = min
x
P0(x) s.t. P1(x , ω) ≥ 0, ∀ω ∈ Ω
• Type-A arbitrage does not exist ⇐⇒ OPT = 0
• Semi-infinite LP P0(x) = S>0 x and P1(x , ω) =
∑n
i=1 fi (S1(ω))xi are
linear functions of x , but there are infinitely many constraints since
Ω is an infinite set
• Standard LP solvers (e.g., CVX) cannot solve this as it is
Constraints can be reformulated when fi is piecewise linear so that this
can be solved as an LP
31 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Type-A Arbitrage : Semi-Infinite LP
Find portfolio x ∈ Rn that is cheapest portfolio at time 0 and has no risk
of loss in future (value at maturity is ≥ 0 for all scenarios)
OPT = min
x
P0(x) s.t. P1(x , ω) ≥ 0, ∀ω ∈ Ω
• Type-A arbitrage does not exist ⇐⇒ OPT = 0
• Semi-infinite LP P0(x) = S>0 x and P1(x , ω) =
∑n
i=1 fi (S1(ω))xi are
linear functions of x , but there are infinitely many constraints since
Ω is an infinite set
• Standard LP solvers (e.g., CVX) cannot solve this as it is
Constraints can be reformulated when fi is piecewise linear so that this
can be solved as an LP
31 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Reformulating Constraints
Suppose that for each j , fi (S1) := fi (S1(ω)) is a piecewise linear function
of S1 with a breakpoint
3 at Kj
Lemma∑n
i=1 fi (S1(ω)) is a piecewise linear function of S1 with breakpoints
K1,K2, . . . ,Kn.
3This means that for S1 ≤ Kj it is a linear function with one slope and for S1 ≥ Kj it is a linear
function with a different slope
32 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Reformulating Constraints
Theorem
Suppose the breakpoints are ordereda as : 0 ≤ K1 ≤ K2 ≤ · · · ≤ Kn.
Then, the function P1(x ,S1) := P1(x ,S1(ω)) is ≥ 0 for all ω ∈ Ω if and
only if all the following conditions are satisfied,
1. P1(x , 0) ≥ 0
2. P1(x ,Kj) ≥ 0 for all j = 1, . . . , n
3. Right-derivative of P1(x ,S1) at S1 = Kn is ≥ 0
aThis assumption is without loss of generality since we can sort the values
Both the piecewise linear functions illustrated on the previous slide satisfy
above conditions
33 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Type-A Arbitrage : LP
Semi-infinite ADLP minx
∑n
i=1 S
i
0xi s.t.
∑n
i=1 fi (S1(ω))xi ≥ 0,∀ω ∈ Ω
becomes a linear program
Corollary
Under the piecewise linear assumptions made so far on a portfolio of
derivatives, type-A arbitrage can be detected by solving
OPT = min
x
n∑
i=1
S i0xi
s.t.
n∑
i=1
fi (0)xi ≥ 0
n∑
i=1
fi (Kj )xi ≥ 0, j = 1, . . . , n
n∑
i=1
(
fi (Kn + 1)− fi (Kn)
)
xi ≥ 0.
34 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Options
35 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
What are Options ?
An option is
• the right to buy (call option) or sell (put option) an underlying
security (e.g., bond, stock, commodity)
• at a certain price (strike)
• in a certain time frame
• at any point before expiration date: [0,T ] (American option)
• at expiration date T (maturity) (European option)
You have the right to buy/sell but no obligation to do the transaction
36 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
What are Options ?
An option is
• the right to buy (call option) or sell (put option) an underlying
security (e.g., bond, stock, commodity)
• at a certain price (strike)
• in a certain time frame
• at any point before expiration date: [0,T ] (American option)
• at expiration date T (maturity) (European option)
You have the right to buy/sell but no obligation to do the transaction
Example : A call option to buy 1 oz. of gold on 1 March for £900.
• If on 1 March the market price of gold is £800/ounce you would just buy
the gold for the market price (and bin the option). The value of the
option on 1 March under this scenario is £0
• If on 1 March the market price of gold is £1000/ounce you would exercise
the option and buy gold for £900 (and sell it on the market). The value
of your option on 1 March under this scenario was £100.
36 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
What are Options ?
An option is
• the right to buy (call option) or sell (put option) an underlying
security (e.g., bond, stock, commodity)
• at a certain price (strike)
• in a certain time frame
• at any point before expiration date: [0,T ] (American option)
• at expiration date T (maturity) (European option)
You have the right to buy/sell but no obligation to do the transaction
Example : A put option to sell an ounce of gold on 1 March for £900.
• If on 1 March the market price of gold is £800/ounce you would just buy
the gold from the market for £800 and exercise your option to sell the gold
for £900. The value of the option on 1 March under this scenario is £100
• If on 1 March the market price of gold is £1000/ounce you would just sell
the gold for the market price (and bin the option). The value of your
option on 1 March under this scenario was £0.
36 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
What are Options ?
An option is
• the right to buy (call option) or sell (put option) an underlying
security (e.g., bond, stock, commodity)
• at a certain price (strike)
• in a certain time frame
• at any point before expiration date: [0,T ] (American option)
• at expiration date T (maturity) (European option)
You have the right to buy/sell but no obligation to do the transaction
There are other exotic options: Payoff at maturity determined by
• any price in [0,T ] (lookback option)
• average price in [0,T ] (Asian option)
where T = maturity/expiration date
36 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
What are Options ?
An option is
• the right to buy (call option) or sell (put option) an underlying
security (e.g., bond, stock, commodity)
• at a certain price (strike)
• in a certain time frame
• at any point before expiration date: [0,T ] (American option)
• at expiration date T (maturity) (European option)
You have the right to buy/sell but no obligation to do the transaction
Two questions
• What is the value (=fair price) of the option now?
→ This is option pricing. Can use RNPM for that!
• There are several options (different strike prices, different prices
now) on the same security on the market. How do we detect if there
is arbitrage?
→ We’ll answer that now!
36 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
European Call and Put Options
EU Call Option: The right to buy a certain underlying security at
expiration date at an agreed (strike) price. Its value at maturity is
f (S1) = (S1 − strike)+ := max{S1 − strike, 0}
S1
C1
strike
EU Put Option: The right to sell a certain underlying security at
expiration date at an agreed (strike) price. Its value at maturity is
f (S1) = (strike− S1)+ := max{strike− S1, 0}
S1
C1
strike
In both options, the option value at maturity is a piecewise linear
function with breakpoint at the strike price
37 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Arbitrage Detection in EU Call Options
Suppose we have n derivatives S1, . . . ,Sn of EU Call Options on the
same underlying security, with the same maturity, with strike prices
0 < K1 < K2 < · · · < Kn, respectively. Then
fi (S1) = (S1 − Ki )+ := max{S1 − Ki , 0}
fi (Kj) = (Kj − Ki )+ =
{
Kj − Ki , if j > i
0, otherwise.
The ADLP in the Corollary is minx{c>x : Ax ≥ 0}, where
c = (S10 ,S
2
0 , . . . ,S
n
0 )
>
A =
K2 − K1 0 · · · 0 0
K3 − K1 K3 − K2 · · · 0 0
...
...
...
...
Kn − K1 Kn − K2 · · · Kn − Kn−1 0
1 1 · · · 1 1
38 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
The dual to the ADLP is the feasibility LP
max
y
0>y s.t. A>y = c , y ≥ 0
No type-A arbitrage is equivalent to each of the following
1. ADLP is bounded
2. Dual to ADLP is feasible
No type-A or type-B arbitrage is equivalent to each of the following
1. ADLP is bounded with all optimal solutions x having Ax = 0
2. Dual to ADLP is strictly feasible, i.e., there exists y > 0 with
A>y = c
39 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
The dual to the ADLP is the feasibility LP
max
y
0>y s.t. A>y = c , y ≥ 0
No type-A arbitrage is equivalent to each of the following
1. ADLP is bounded
2. Dual to ADLP is feasible
No type-A or type-B arbitrage is equivalent to each of the following
1. ADLP is bounded with all optimal solutions x having Ax = 0
2. Dual to ADLP is strictly feasible, i.e., there exists y > 0 with
A>y = c
39 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
The dual to the ADLP is the feasibility LP
max
y
0>y s.t. A>y = c , y ≥ 0
No type-A arbitrage is equivalent to each of the following
1. ADLP is bounded
2. Dual to ADLP is feasible
No type-A or type-B arbitrage is equivalent to each of the following
1. ADLP is bounded with all optimal solutions x having Ax = 0
2. Dual to ADLP is strictly feasible, i.e., there exists y > 0 with
A>y = c
39 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
The system of equations A>y = c is
K2 − K1 k3 − K1 · · · Kn − K1 1
0 K3 − K2 · · · Kn − K2 1
...
...
...
...
0 0 · · · Kn − Kn−1 1
0 0 · · · 0 1
y1
y2
...
yn−1
yn
=
S10
S20
...
Sn−10
Sn0
which can be solved by backward substitution to get
yn = S
n
0 , yn−1 =
Sn−10 − Sn0
Kn − Kn−1
yi =
S i0 − S i+10
Ki+1 − Ki −
S i+10 − S i+20
Ki+2 − Ki+1 , i = 1, . . . , n − 2
This implies that there is no arbitrage ⇐⇒ all the yi > 0.
40 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Main Theorem on Arbitrage in EU Call Options
Theorem
For a derivative of EU Call Options S1, . . . ,Sn written on the same
underlying security, with the same maturity, with strike prices
0 < K1 < K2 < · · · < Kn, there is no arbitrage if and only if the following
three conditions hold:
1. S i0 > 0 for all i = 1, 2, . . . , n (only required for i = n)
2. S i0 > S
i+1
0 for all i = 1, 2, . . . , n − 1 (only required for i = n − 1)
3. The piecewise linear function mapping Ki to S
i
0, for i = 1, . . . , n, is
convex and the slopes of its linear pieces are strictly increasing.
41 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Example
Consider 4 call options with current prices (spot values) S0 = (10, 8, 3, 1),
and strikes K = (10, 20, 30, 60). Do we have arbitrage ?
Plot spot values S i0 against strikes Ki
10 20 30 40 50 60
10
8
3
1
strikes
spot values
Si0−S
i+1
0
Ki+1−Ki = negative of slope of piece i
Properties 1) and 2) are satisfied
3) is not satisfied since the
slopes are: −2, −5 and −2/3, and
−2 < −5 < −2/3 is not true
Theorem =⇒ arbitrage exists
How to detect/find x detecting the arbitrage opportunity?
42 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Example
Consider 4 call options with current prices (spot values) S0 = (10, 8, 3, 1),
and strikes K = (10, 20, 30, 60). Do we have arbitrage ?
Plot spot values S i0 against strikes Ki
10 20 30 40 50 60
10
8
3
1
strikes
spot values
Si0−S
i+1
0
Ki+1−Ki = negative of slope of piece i
Properties 1) and 2) are satisfied
3) is not satisfied since the
slopes are: −2, −5 and −2/3, and
−2 < −5 < −2/3 is not true
Theorem =⇒ arbitrage exists
How to detect/find x detecting the arbitrage opportunity?
42 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
How to find x?
Consider a situation where successive slopes are not strictly increasing:
K1 K2 K3
S0
1
S0
2
S0
3
S
S2 is overpriced. Sell S2 and use money
to buy a combination of S1 and S3
Let α ∈ (0, 1) s.t. K2 = αK1 + (1−α)K3
Take portfolio x = (α, 0, 1− α)
• Cost at time 0 < cost of 1 unit of
S2:
x1S
1
0 + x3S
3
0 = S︸ ︷︷ ︸
cost of portfolio at time 0
< S20︸︷︷︸
cost of S2 at time 0
=⇒ Positive initial cash-flow
• Payoff at maturity is
P1(x , S1) = (S1 − K1)+x1 + (S1 − K3)+x3
≥ (x1(S1 − K1) + x3(S1 − K3))+
= (S1 − K2)+, ∵K2 = x1K1+x3K3
So, payoff of portfolio cannot be
worse than payoff of S2
=⇒ No risk of loss in the future
43 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
How to find x?
Consider a situation where successive slopes are not strictly increasing:
K1 K2 K3
S0
1
S0
2
S0
3
S
S2 is overpriced. Sell S2 and use money
to buy a combination of S1 and S3
Let α ∈ (0, 1) s.t. K2 = αK1 + (1−α)K3
Take portfolio x = (α, 0, 1− α)
• Cost at time 0 < cost of 1 unit of
S2:
x1S
1
0 + x3S
3
0 = S︸ ︷︷ ︸
cost of portfolio at time 0
< S20︸︷︷︸
cost of S2 at time 0
=⇒ Positive initial cash-flow
• Payoff at maturity is
P1(x , S1) = (S1 − K1)+x1 + (S1 − K3)+x3
≥ (x1(S1 − K1) + x3(S1 − K3))+
= (S1 − K2)+, ∵K2 = x1K1+x3K3
So, payoff of portfolio cannot be
worse than payoff of S2
=⇒ No risk of loss in the future
43 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Risk Neutral Probability
44 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Risk Neutral Probability Measure (RNPM)
Definition (Def 4.2)
A risk neutral probability measure on Ω = {ω1, . . . , ωm} is a vector
p = (p1, . . . , pm)
T ∈ Rm satisfying
1.
∑
j pj = 1; pj > 0 ∀j , (i.e., p is a vector of positive probabilities)
2.
S i0 =
1
R
m∑
i=1
pjS
i
1(ωj) :=
1
R
E[S i1], i = 0, 1, . . . , n. (2)
Here R = 1 + r where r = return rate of risk-free asset S0
• Given m, n, R, S i0 and S i1(ωj) for all i = 0, 1, . . . , n and
j = 1, . . . ,m, a RNPM does not have to exist
• We will see that existence of RNPM ⇐⇒ non-existence of arbitrage
• If RNPM exists, it allows us to see current prices S i0 as discounted
expected values of future prices, where E is taken wrt a RNPM
45 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Risk Neutral Probability Measure (RNPM)
Definition (Def 4.2)
A risk neutral probability measure on Ω = {ω1, . . . , ωm} is a vector
p = (p1, . . . , pm)
T ∈ Rm satisfying
1.
∑
j pj = 1; pj > 0 ∀j , (i.e., p is a vector of positive probabilities)
2.
S i0 =
1
R
m∑
i=1
pjS
i
1(ωj) :=
1
R
E[S i1], i = 0, 1, . . . , n. (2)
Here R = 1 + r where r = return rate of risk-free asset S0
• Given m, n, R, S i0 and S i1(ωj) for all i = 0, 1, . . . , n and
j = 1, . . . ,m, a RNPM does not have to exist
• We will see that existence of RNPM ⇐⇒ non-existence of arbitrage
• If RNPM exists, it allows us to see current prices S i0 as discounted
expected values of future prices, where E is taken wrt a RNPM
45 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Further Insights
• Note that while pj is NOT EQUAL TO the probability that ωj
happens, yet it acts as if it was (in the above sense)
• It is natural to require that if two securities, say S i and Sk , have the
same future payoffs (S i1(ωj) = S
k
1 (ωj) for all j), then they should
have the same price at present: S i0 = S
k
0 . If RNPM exists, then (2)
gives a formula for computing this price.
• RNPM can be used to price new securities using a replication
technique.
• If a RNPM exists, it need not be unique. However, (2) will hold for
all such p, so it does not matter which one is used in the calculation
of the present price.
Exercise 0. Construct an example (i.e., find some concrete numbers for
m, n, R, S i0 and S
i
1(ωj)) such that a RNPM does not exist. Hint: use
m = 2, n = 1.
46 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
First Fundamental Theorem of Asset Pricing
Theorem (Thm 4.2)
A RNPM exists if and only if there is no arbitrage (of either type).
Proof ⇐: assume there is no arbitrage.
• In particular, since there is no type-A arbitrage, applying Fact (1) we
know that (ADLP) (and hence (D)) has an optimal solution.
• Using strong duality, we conclude that (P) – the dual of ADLP – is
feasible (since (D) has an optimal solution).
• In particular, both (P) and (D) are feasible. We can thus apply
Goldman-Tucker complementary slackness theorem for LPs, which
says that there exists strictly complementary optimal solutions x∗
and (y∗, s∗):
x∗ + s∗ > 0. (3)
• Now we utilize the assumption that there is no type-B arbitrage.
Applying Fact 2: all constraints are tight, we get
s∗ = c − AT y∗ = 0. Plugging this into (3), we conclude that x∗ > 0
(that is x∗i > 0,∀i).
47 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Proof: (Continued)
We now claim that the vector p = (p1, p2, . . . , pm)
T , where pj = Rx
∗
j , is
the vector of risk-neutral probabilities. For this we need to check that p
satisfied the three properties in the definition.
• First, clearly pj > 0 for all j (since x∗j > 0 for all j).
• Second, note that the first constraint (i.e., the one corresponding to
i = 0: cash) in (P) looks as follows:
m∑
i=1
S01 (ωj)︸ ︷︷ ︸
=R
x∗j = S
0
0 = 1,
which implies that
∑
j pj = 1.
• Finally, let us look at the i-th constraint in (P) (for any i):
m∑
i=1
S i1(ωj)x
∗
j = S
i
0.
It can be written as S i0 =
1
R
∑m
i=1 S
i
1(ωj)Rx
∗
j =
1
R
∑m
i=1 S
i
1(ωj)pj ,
recovering (2). Hence, p is a RNPM on Ω.
48 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Example
Example : Show directly from definition that in the previous exercise
there is no risk neutral probability measure (RNPM). Then argue this via
the Fundamental Thm of Asset Pricing.
Solution. Since m = 1 (we have one scenario only), the only possible
probability measure on Ω is p1 = 1. Let us check whether it satisfies the
conditions in the definition of a RNPM:
1.
∑
j pj = 1 and pj > 0 for all j trivially holds since m = 1 and p1 = 1,
2. We further need the following identities to hold:
S i0 =
1
R
m∑
i=1
pjS
i
1(ωj), i = 0, 1.
This holds for i = 0 since 1 = 11.05p11.05 = 1. It does not hold for
i = 1 since 500 6= 11.05p1550. Hence, RNPM does not exist.
We reach the same conclusion by applying the Fundamental Theorem of
Asset Pricing and the previous Exercise which says that arbitrage exists.
49 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Pricing Securities using RNPM
50 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Binomial Model
n = 2 (3 securities) and m = 2 scenarios ω1 and ω2
• S0: risk-free security/cash (yielding return R = 1 + r)
• S1: underlying security
• S2: derivative of underlying security
Assume U > D > 0 are constants such that
S11 =
{
U × S10 , under ω1 (“UP ′′)
D × S10 , under ω2 (“DOWN ′′).
price of the derivative at maturity (time 1) is a function of the price of
the underlying security at maturity:
S21 = f (S
1
1 )
Price the derivative : Assuming there is no arbitrage, find a formula for
the spot price (i.e., price at time 0) S20
51 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
We can use the RNPM to find the price at time 1 of the derivative S2:
• Start with the two securities S0 (cash) and S1 (underlying)
Assume no arbitrage when only these two securities are present
• According to the Fundamental Thm of Asset Pricing, there exists a
RNPM, i.e. “probabilities” p1, p2 for ω1, ω2, such that
S00 =
1
R
[
p1S
0
1 (ω1) + p2S
0
1 (ω2)
]
S10 =
1
R
[
p1S
1
1 (ω1) + p2S
1
1 (ω2)
]
• Using S00 = 1,S01 (ω1) = S01 (ω2) = R, the first equation gives
p1 + p2 = 1
• while using S11 (ω1) = US10 ,S11 (ω2) = DS10 , the second equation gives
p1U + p2D = R.
52 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
We can use the RNPM to find the price at time 1 of the derivative S2:
• Start with the two securities S0 (cash) and S1 (underlying)
Assume no arbitrage when only these two securities are present
• According to the Fundamental Thm of Asset Pricing, there exists a
RNPM, i.e. “probabilities” p1, p2 for ω1, ω2, such that
S00 =
1
R
[
p1S
0
1 (ω1) + p2S
0
1 (ω2)
]
S10 =
1
R
[
p1S
1
1 (ω1) + p2S
1
1 (ω2)
]
• Using S00 = 1,S01 (ω1) = S01 (ω2) = R, the first equation gives
p1 + p2 = 1
• while using S11 (ω1) = US10 ,S11 (ω2) = DS10 , the second equation gives
p1U + p2D = R.
52 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
We can use the RNPM to find the price at time 1 of the derivative S2:
• Start with the two securities S0 (cash) and S1 (underlying)
Assume no arbitrage when only these two securities are present
• According to the Fundamental Thm of Asset Pricing, there exists a
RNPM, i.e. “probabilities” p1, p2 for ω1, ω2, such that
S00 =
1
R
[
p1S
0
1 (ω1) + p2S
0
1 (ω2)
]
S10 =
1
R
[
p1S
1
1 (ω1) + p2S
1
1 (ω2)
]
• Using S00 = 1,S01 (ω1) = S01 (ω2) = R, the first equation gives
p1 + p2 = 1
• while using S11 (ω1) = US10 ,S11 (ω2) = DS10 , the second equation gives
p1U + p2D = R.
52 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
• Using S00 = 1,S01 (ω1) = S01 (ω2) = R, the first equation gives
p1 + p2 = 1
• while using S11 (ω1) = US10 ,S11 (ω2) = DS10 , the second equation gives
p1U + p2D = R.
• Solving for p1, p2 we get
p1 =
R − D
U − D , p2 =
U − R
U − D
This is the RNPM.
• We can now use the RNPM to price the derivative S2:
Under the no arbitrage condition, the RNPM measure must hold for
S2 as well, hence
S20 =
1
R
[p1S
2
1 (ω1) + p2S
2
1 (ω2)]
53 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
• Using S00 = 1,S01 (ω1) = S01 (ω2) = R, the first equation gives
p1 + p2 = 1
• while using S11 (ω1) = US10 ,S11 (ω2) = DS10 , the second equation gives
p1U + p2D = R.
• Solving for p1, p2 we get
p1 =
R − D
U − D , p2 =
U − R
U − D
This is the RNPM.
• We can now use the RNPM to price the derivative S2:
Under the no arbitrage condition, the RNPM measure must hold for
S2 as well, hence
S20 =
1
R
[p1S
2
1 (ω1) + p2S
2
1 (ω2)]
53 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
• Using S00 = 1,S01 (ω1) = S01 (ω2) = R, the first equation gives
p1 + p2 = 1
• while using S11 (ω1) = US10 ,S11 (ω2) = DS10 , the second equation gives
p1U + p2D = R.
• Solving for p1, p2 we get
p1 =
R − D
U − D , p2 =
U − R
U − D
This is the RNPM.
• We can now use the RNPM to price the derivative S2:
Under the no arbitrage condition, the RNPM measure must hold for
S2 as well, hence
S20 =
1
R
[p1S
2
1 (ω1) + p2S
2
1 (ω2)]
53 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
• We have worked with 2 scenarios and 3 securities because we can
deal with this by hand.
Effectively we have solved a 2× 2 system of linear equations to
obtain the RNPM and use the solution to “price” the 3rd
(derivative) security.
Alternatively we could have obtained the RNPM from the dual
solution of the ADLP.
• However, the technique can easily be adapted to our general model
with arbitrary finite number of scenarios (ω1, . . . , ωn) and n + 1
securities S0, . . . ,Sn.
In this case we would have to computationally solve an n× n system
(or again find the duals of ADLP)
54 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Pricing of Securities via Replication
Approach: We will try to form a portfolio y = (y0, y1) of cash (S0) and
the underlying (S1) such that its price P1(y) at time 1 matches the price
of the derivative, under both scenarios:
P1(y ;ωj) := S
0
1 (ωj)y0 + S
1
1 (ωj)y1 = S
2
1 (ωj), j = 1, 2.
• The above is a system of 2 linear equations in 2 unknowns (y0, y1).
• The solution is:
y0 =
US21 (ω2)− DS21 (ω1)
R(U − D) , y1 =
S21 (ω1)− S21 (ω2)
S10 (U − D)
, (4)
• We then say that since the portfolio has the same future
behaviour/payoff/value as the derivative S2, the spot price of the
derivative must be equal to the spot price of the portfolio:
S20 = P0(y) = S
0
0 y0 + S
1
0 y1. (5)
55 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
• By plugging (4) into (5), we obtain:
S20 =
1
R
R − DU − D︸ ︷︷ ︸
:= p1
S21 (ω1) +
U − R
U − D︸ ︷︷ ︸
:= p2
S21 (ω2)
. (6)
• If D < R < U, then p1, p2 as defined in (6) satisfy
0 < p1, p2 < 1, p1 + p2 = 1 that is they act like a probability.
Indeed this is the Risk Neutral Probability Measure (RNPM)!
• (6) says that the spot value (S20 ) of the derivative security is equal to
its present (i.e., discounted) expected value, where expectation is
taken with respect to the RNPM.
• However, for (p1, p2) to be a RNPM we still need to check that (6)
holds for securities S0 and S1 as well.
Exercise
Check that (6) holds for securities S0 and S1 as well, thus completing the
verification of the fact that p is a RNPM.
56 / 56
MATH 11158: Optimization Methods in Finance
Arbitrage and Asset Pricing1
Akshay Gupte
School of Mathematics, University of Edinburgh
Week 1 and 2 : 20 – 27 January, 2022
1Sections 4.1–4.4 in the textbook
Email: akshay.gupte@ed.ac.uk
1 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Arbitrage LP
2 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Arbitrage
A trading strategy is a sequence of investment (buying, selling, borrowing
of securities) decisions over time.
For example: borrow now x amount of money to buy certain assets. At
some (determined) future time sell the assets and repay the debt.
Arbitrage is a trading strategy on an asset that results in net profit due
to different prices of the asset in different markets or in different forms. It
exists because of inefficiencies in the financial markets
3 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Example 1
Example 1 : A bank accepts deposits yielding an interest rate of 5%.
Also, it lends money at the same interest rate. You have an opportunity
to invest for 1 year into a government bond that costs £500, yielding a
10% interest.
— Is there arbitrage (i.e. can you make money without risk)?
Solution: The “situation” is easy: one should “borrow from bank and
buy bonds!”.
• We can borrow £ 1, 047.62 from the bank and use this to buy 2
bonds: Cash flow is
£1, 047.62− 2× £500 = £47.62
We now have £47.62, two bonds and debts to the bank
• At time 1, we sell the bonds and repay our debts to the bank
2× 1.10× 500− 1.05× 1047.62 = 0
We sell the bonds (at 10% profit), repay our debts to the bank (at
5% interest) and keep the £47.62!
4 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Example 2
Example 2 : Suppose we start with 2,000 USD currency. There are three
currency exchange places with the following conversion rates
• 1 USD ≡ 0.894 Euro
• 1 GBP ≡ 1.276 Euro
• 1 GBP ≡ 1.432 USD
Is there arbitrage ?
The following cycle of conversions : USD → Euro → GBP → USD, for
the initial amount of 2,000 USD leads to a net amount of 2,006.59 USD
2006.59 =
(
2000 ∗ 0.894
1.276
)
∗ 1.432,
resulting in an arbitrage profit of 6.59 (= 2006.59 - 2000) USD
5 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Arbitrage Types
Definition (Arbitrage Type A)
A trading strategy that has
• positive initial cash flow and
• no risk of a loss in the future.
Definition (Arbitrage Type B)
A trading strategy that has
• nonnegative initial cash flow,
• no risk of a loss in the future and
• a positive probability of a profit in the future.
6 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Arbitrage Detection: Mathematical Model
1. Securities: S0,S1,S2, . . . ,Sn (S0 is risk free)
2. Modeling future uncertainty:
2.1 Consider 2 time periods: “now” (time 0) and “future” (time 1)
2.2 States of the world at maturity (time 1) is described by m scenarios
that are indexed by the set Ω = {ω1, ω2, . . . , ωm}, where ωi denotes
scenario i
2.3 Each scenario occurs with nonzero probability
We do not know the probabilities, we do not need to know them!
3. Prices: For i = 0, . . . , n,
3.1 S i0 = price of S
i at present (time 0)
3.2 S i1(ωj) = price of S
i at maturity under scenario ωj
4. Risk-free security: Suppose we earn interest rate r on security S0
4.1 S0 is the risk-free security (e.g., cash in savings account)
4.2 S00 = 1 (price of risk-free security at present is 1)
4.3 S01 (ωj) = 1 + r = R for all j
Note that it is indeed risk free as its price at time 1 does not depend
on ωj . We can also borrow money at the risk free rate r .
7 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Type-A Arbitrage and Linear Inequalities
portfolio x = (x0, x1, . . . , xn), where xi is the amount (number of units),
either positive or negative, of security S i .
At least x0 can be negative (borrow money at risk free rate)
Note: since we are not starting with an initial amount of money, we are
building a portfolio as opposed to an investment portfolio as we will see
later in portfolio theory
• Cost of portfolio x at time 0 (= negative of initial cash flow)
P0 = P0(x) :=
n∑
i=0
S i0xi
• No risk of a loss in the future
n∑
i=0
S i1(ωj)xi ≥ 0, j = 1, 2, . . . ,m. (1)
Type-A arbitrage exists ⇐⇒ there is a portfolio x such that P0(x) < 0
and inequalities (1) hold (these are linear inequalities in x)
8 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Arbitrage: ... back to our example
Example (Arbitrage detection II). A bank accepts deposits yielding an
interest rate of 5%. Also, it lends money at the same interest rate. You
have an opportunity to invest for 1 year into a government bond that
costs £500, yielding a 10% interest.
1. Is there a type-A arbitrage opportunity?
2. Is there a type-B arbitrage opportunity?
Solution: The answer is easy, borrow from bank and buy bonds!
However, this simple problem lets us understand the concepts
• There are 2 assets: S0 (cash/bank) and S1 (bond); so n = 1
• There are 2 time periods: now (time 0) and a year after (time 1)
• Only one scenario: m = 1 and Ω = {ω1}, since bank and
government bonds can be seen as risk-free
• Cash: S00 = 1, R = 1.05, S01 (ω1) = S00 × R = 1.05
• Bond: S10 = 500, S11 (ω1) = S10 × 1.1 = 550
9 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Example (Cont.)
We will build a portfolio x = (x0, x1)
T . Some things we can do:
Strategy 1.
We can borrow £1, 047.62 from the bank (x0 = −1, 047.62) and use this
to buy 2 bonds (x1 = 2).
• At time 0, cost of building our portfolio (net cash outflow) is
P0(x) =
n∑
i=0
S i0xi = S
0
0 x0+S
1
0 x1 = 1×(−1, 047.62)+500×2 = −47.62
• At time 1, we have only one possible outcome (ω1), and the cost
(=value) of our portfolio (we own it now, so positive value means
we are good!) is:
n∑
i=0
S i1(ωj)xi = 1.05× (−1, 047.62) + 550× 2 = 0
ie. we sell the bonds (at 10% profit), repay our debts to the bank
(at 5% interest) and keep the £47.62!
This is type-A arbitrage!
10 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Example (cont): Strategy 2
Do the same as in Strategy 1, except scale your position x by constant
c > 0. That is, choose: x0 = c × (−1, 047.619) (borrow c times more
cash) and x1 = c × 2 (buy c times as many bonds as before).
• At time 0, the incurred cost of building our portfolio (i.e., net cash
outflow) is
P0(x) =
n∑
i=0
S i0xi = S
0
0 x0 + S
1
0 x1 = c × (−47.619)
• At time 1, we have only one possible outcome (ω1), and the cost
(=value) of our portfolio (we own it now, so positive value means
we are good!) is:
n∑
i=0
S i1(ωj)xi = c × 0
So we earn c times as much as with Strategy 1! For c →∞, we earn as
much as we want.
11 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Example (cont): Strategy 3
We can borrow £1, 000 from the bank (x0 = −1, 000) and use this to buy
2 bonds (x1 = 2).
• At time 0, the incurred cost of building our portfolio (i.e., net cash
outflow) is
P0(x) =
n∑
i=0
S i0xi = S
0
0 x0 + S
1
0 x1 = 1×−1, 000 + 500× 2 = 0
• At time 1, we have only one possible outcome (ω1), and the cost
(=value) of our portfolio (we own it now, so positive value means
we are good!) is:
n∑
i=0
S i1(ωj)xi = 1.05× (−1, 000) + 550× 2 = 50
This is type-B arbitrage!
12 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
ADLP
13 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Arbitrage-Detecting Linear Program
Consider now the linear program
OPT := min
n∑
i=0
S i0xi
subject to
n∑
i=0
S i1(ωj)xi ≥ 0, j = 1, 2, . . . ,m.
(ADLP)
Define OPT = −∞ if the problem is unbounded
• There is type-A arbitrage ⇐⇒ OPT < 0
• (ADLP) is homogeneous (feasible region is a polyhedral cone): if x
is feasible then λ · x is also feasible for any λ > 0
• If there is a feasible x with OPT < 0 then (ADLP) is unbounded
(OPT = −∞).
14 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Detecting Type-A Arbitrage
Fact (1)
The following statements are equivalent:
(a) there is no type-A arbitrage,
(b) OPT = 0,
(c) (ADLP) is bounded.
• To detect arbitrage: solve (ADLP) and check whether OPT = 0
• If OPT 6= 0, we must have OPT = −∞ (since by Fact (1), (ADLP)
must be unbounded!). So, there must exist a feasible portfolio x
(i.e., one that does not pose any risk of a future loss) with negative
cost at time 0, i.e.,
∑n
i=0 S
i
0xi < 0
• LP solvers detect unboundedness by finding such x (a direction of
unboundedness).
This x corresponds to an arbitrage opportunity (of type-A)!
15 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Proof of Fact (1)
• Observe that OPT ≤ 0. Indeed, the empty portfolio (y = 0, i.e., we
do not own any assets) is feasible and costs nothing at time 0.
• We now claim that either OPT = 0 or OPT = −∞. Let us show
this. Assume now that y is a feasible portfolio for which the cost of
which at time 0 is negative and finite. Then the portfolio 2y is also
be feasible, with a smaller cost at time 0. So, OPT cannot be
negative and finite.
• In particular, the above observations tell us that
OPT < 0 ⇔ OPT = −∞ ⇔ (ADLP) is unbounded.
• Further, note that OPT < 0 if and only if there exists type-A
arbitrage. Indeed, OPT < 0 means that there is a portfolio whose
payoff at time 1 is nonnegative but whose cost at present is negative
(i.e., it generates positive cashflow at time 0).
We have thus shown that
OPT = −∞ ⇔ type-A arbitrage exists ⇔ (ADLP) is unbounded
The logical negations of each of these three statements must also be
equivalent, proving the statement.
16 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Example (cont): Detecting type-A Arbitrage by LP
The LP (ADLP) detecting type-A arbitrage is:
minimize P0(x) :=
n∑
i=0
S i0xi
subject to
n∑
i=0
S i1(ωj)xi ≥ 0, j = 1, 2, . . . ,m
In our earlier example (n = 1, S00 = 1, S
1
0 (ω1) = 500, m = 1,
S01 (ω1) = 1.05 and S
1
1 (ω1) = 550), this takes the form
minimize x0 + 500x1
subject to 1.05x0 + 550x1 ≥ 0
• By solving this LP in MATLAB using CVX, we find that it is
unbounded. Hence, by FACT (1), there is type-A arbitrage.
• Following the thought process of Strategy 2 we find a feasible
portfolio cy for all c > 0 such that P0(cy)→ −∞ as c →∞. This
is a “proof by hand” that the LP is unbounded (and type-A
arbitrage exists). 17 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Type-B Arbitrage detection
Type-B arbitrage in our model exists if there exists portfolio
x = (x1, . . . , ym) satisfying the following three conditions:
1. (we do not lose anything at time 0)
P0(x) ≡
n∑
i=0
S i0xi ≤ 0,
2. system of inequalities (1): (there is no risk of loss at time 1)
n∑
i=0
S i1(ωj)xi ≥ 0, j = 1, 2, . . . ,m.
3. (positive probability of positive payoff at time 1)
n∑
i=0
S i1(ωj)xi > 0 for at least one index j
18 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Detecting Type-B Arbitrage with (ADLP) - Theory
Fact (2)
Assume that no type-A arbitrage exists (ie. OPT=0). Then the following
statements are equivalent:
(a) no type-B arbitrage exists,
(b) for all optimal solutions y∗ of (ADLP), all constraints of (ADLP) are
tight. That is,
∑n
i=0 S
i
1(ωj)x
∗
i = 0 for all j = 1, 2, . . . ,m.
• If there is type-A arbitrage, Fact (2) does not tell us anything
• If there isn’t type-A arbitrage, Fact (2) tells us how to detect type-B
arbitrage in in theory. Check if the constraints are tight for all
possible solutions x∗ of (ADLP). Unfortunately it is not easy to find
all solutions to an LP, so this is not much use in practice.
Exercise
What if there is type-A arbitrage?
→ If there is type-A arbitrage then there is also type B-arbitrage.
19 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Detecting Type-B Arbitrage with (ADLP) - Practice
For type-B arbitrage the following conditions need to be satisfied:
P0(y) ≡
n∑
i=0
S i0xi ≤ 0, (i)
n∑
i=0
S i1(ωj )xi ≥ 0, j = 1, 2, . . . ,m, (ii)
n∑
i=0
S i1(ωj )xi > 0 for at least one index j (iii)
We can find such an investment x (if one exists) by solving
OPT := min −
m∑
j=1
1
m
n∑
i=0
S i1(ωj )xi
subject to
n∑
i=0
S i1(ωj )xi ≥ 0, j = 1, 2, . . . ,m.
n∑
i=0
S i0xi ≤ 0
(ADLP)
Define OPT = −∞ if the problem is unbounded
20 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Proof of Fact (2)
• Since we assume that no type-A arbitrage exists, from Fact (1) we
know that OPT = 0.
• Assume (a) holds, i.e no type-B arbitrage. This means that there
cannot exist y satisfying (i), (ii) and (iii).
• Pick any y∗ optimal for (ADLP). Then y∗ clearly satisfies (ii) (since
it is feasible for (ADLP)) and )(i) (since OPT = 0). Hence, it must
be the case that (iii) does not hold for y∗.
• Since (ii) holds, this means that (b) must hold.
We have shown that (a) implies (b). It remains to show the converse.
Exercise
Use similar arguments to finish the proof of Fact (2). That is, show that
if there is no type-A arbitrage, then (b) implies (a).
21 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Dualising (ADLP)
Multiply both the objective and the constraints of (ADLP) by −1 and
use that for any function f , −min f = max(−f )), to get
−OPT =max
n∑
i=0
(−S i0)xi
subject to
n∑
i=0
(−S i1(ωj))xi ≤ 0, j = 1, 2, . . . ,m.
(D)
Note that (D) is in standard form of the dual problem. Indeed, if we now
let b = (−S00 ,−S10 , . . . ,−Sn0 )T ∈ Rn+1, c = 0 ∈ Rm and A ∈ R(n+1)×m
with Aij = −S i1(ωj), then (D) can be written as
−OPT =max bT x
subject to AT x ≤ c . (D)
22 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Problem dual to (D)
The dual of (D) is the standard primal problem:
−OPT =min cT y
subject to Ay = b
y ≥ 0,
(P)
which, after substitution of c , b and A, can be written as
−OPT =min
m∑
i=1
0yj
subject to
m∑
i=1
−S i1(ωj)yj = −S i0, i = 0, 1, . . . , n
yj ≥ 0, j = 1, 2, . . . ,m.
(P)
• All feasible points are also optimal! (Why?)
• (P) does not necessarily have to be feasible
23 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
We would like to understand the meaning of the dual solution y (if it
exists!)
−OPT =min
m∑
i=1
0 · yj
subject to
m∑
i=1
S i1(ωj)yj = S
i
0, i = 0, 1, . . . , n
yj ≥ 0, j = 1, 2, . . . ,m.
(P)
24 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Sports Betting Example
For an upcoming football match the odds offered by a bookmakers are:
home win: 4/11, draw: 7/2, away win 19/2. Is there an arbitrage
opportunity?
Note that odds of a/b mean that if you were to bet amount b on an
event, if the event happens you get a payoff of a (and additionally your
stake b), if the event does not happen, you loose your stake. I.e. cost of
investment is b. In case of win the payoff is a+b, in case of loss the
payoff is 0.
→ Show how this fits into the arbitrage detection setup from the lecture:
Assume there are 4 investments: bet on home win, bet on away win, bet
on draw and cash. Assume that the return (or interest rate) on cash is 1,
i.e. you can borrow unlimited amount of money (to use for bets) without
needing to pay interest. There are three possible outcomes: Home win,
draw and away win. You do not need to know the actual probabilities of
these outcomes.
We have 4 investments, S0 is cash and S1 to S3 are the three bets –
home win, draw and away win. We have 3 scenarios for outcome of the
match – home win, draw or away win. Thus, n = 3 and m = 3. 25 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
→ Work out the values for S i0 and S i1(ωj) and set up the Arbitrage
Detection LP in CVX. Note that you can only place positive amounts as
bets. Is there type-A or type-B arbitrage?
The initial cost of the investments is the vector S0 = (1, 11, 2, 2) (given by
denominators in betting odds). For writing the ADLP constraints which are
one for each scenario, we will need the following matrix
A =
1 15 0 01 0 9 0
1 0 0 21
Each row corresponds to a scenario and each column corresponds to an
investment. So, A12 is the future value of investment S
1 in scenario 1, which is
S11 (ω1) = 11 + 4 = 15. Solve the ADLP as follows. Remember to add
nonnegativity on the three investments (bets) because we can only buy bets
(whereas yi < 0 for i ∈ {1, 2, 3} would mean that we are selling bet i) but for
cash we are allowed to borrow and so it can be negative.
26 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
n = 3;
S0 = [1 11 2 2];
A = [1 15 0 0;1 0 9 0;1 0 0 21];
cvx_begin
variable x(n+1);
minimize (S0’*x)
subject to
A*x >= 0;
x(2:n+1) >= 0;
cvx_end
The optimal LP value is 0 and so there is no type-A arbitrage.
Detecting type-B arbitrage : Refer to previous slide for detecting in
practice. Type-B arbitrage exists if and only if the value OPT = −∞
27 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Arbitrage in Derivatives
28 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Finite vs. Infinite Scenarios
ADLP can be used in all situations with a finite number of scenarios, i.e.,
when Ω = {ω1, ω2, . . . , ωm}
Challenges :
• Finite scenarios may not describe reality well in all situations
• Larger the m, larger (and possibly harder) to solve the ADLP
• Scenarios may need to be generated if they are not available
naturally, or if an infinite set Ω is discretised to obtain finite
scenarios. Scenario generation is not a trivial task in general
Infinitely many scenarios2 (indexed as ω ∈ Ω) can be dealt with in special
cases. One such case is a portfolio of derivatives that are
• written on the same underlying security
• with the same maturity date (time 1)
Value of the security at maturity is uncertain/random
2Ω can be countable or uncountable set
29 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Finite vs. Infinite Scenarios
ADLP can be used in all situations with a finite number of scenarios, i.e.,
when Ω = {ω1, ω2, . . . , ωm}
Challenges :
• Finite scenarios may not describe reality well in all situations
• Larger the m, larger (and possibly harder) to solve the ADLP
• Scenarios may need to be generated if they are not available
naturally, or if an infinite set Ω is discretised to obtain finite
scenarios. Scenario generation is not a trivial task in general
Infinitely many scenarios2 (indexed as ω ∈ Ω) can be dealt with in special
cases. One such case is a portfolio of derivatives that are
• written on the same underlying security
• with the same maturity date (time 1)
Value of the security at maturity is uncertain/random
2Ω can be countable or uncountable set
29 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Finite vs. Infinite Scenarios
ADLP can be used in all situations with a finite number of scenarios, i.e.,
when Ω = {ω1, ω2, . . . , ωm}
Challenges :
• Finite scenarios may not describe reality well in all situations
• Larger the m, larger (and possibly harder) to solve the ADLP
• Scenarios may need to be generated if they are not available
naturally, or if an infinite set Ω is discretised to obtain finite
scenarios. Scenario generation is not a trivial task in general
Infinitely many scenarios2 (indexed as ω ∈ Ω) can be dealt with in special
cases. One such case is a portfolio of derivatives that are
• written on the same underlying security
• with the same maturity date (time 1)
Value of the security at maturity is uncertain/random
2Ω can be countable or uncountable set
29 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Portfolio of Derivatives
Single underlying security under infinitely many scenarios
S0 = known price at time 0, S1(ω) = uncertain price at maturity
Derivatives S i for i = 1, . . . , n written on the security, with known prices
S i0 at time 0 and random prices at maturity
S i1 := S
i
1(ω) = fi (S1(ω))
where function fi is known for all i . Think of S1 as a random variable
S1 : ω ∈ Ω 7→ S1(ω) ∈ R
Portfolio x = (x1, x2, . . . , xn) of derivatives (xi is investment in S
i )
Cost at time 0 : P0(x) :=
n∑
i=1
S i0xi
Value at maturity : P1(x ,S1) := P1(x , ω) =
n∑
i=1
S i1xi =
n∑
i=1
fi (S1(ω))xi
This can be used to detect arbitrage
30 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Portfolio of Derivatives
Single underlying security under infinitely many scenarios
S0 = known price at time 0, S1(ω) = uncertain price at maturity
Derivatives S i for i = 1, . . . , n written on the security, with known prices
S i0 at time 0 and random prices at maturity
S i1 := S
i
1(ω) = fi (S1(ω))
where function fi is known for all i .
Think of S1 as a random variable
S1 : ω ∈ Ω 7→ S1(ω) ∈ R
Portfolio x = (x1, x2, . . . , xn) of derivatives (xi is investment in S
i )
Cost at time 0 : P0(x) :=
n∑
i=1
S i0xi
Value at maturity : P1(x ,S1) := P1(x , ω) =
n∑
i=1
S i1xi =
n∑
i=1
fi (S1(ω))xi
This can be used to detect arbitrage
30 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Portfolio of Derivatives
Single underlying security under infinitely many scenarios
S0 = known price at time 0, S1(ω) = uncertain price at maturity
Derivatives S i for i = 1, . . . , n written on the security, with known prices
S i0 at time 0 and random prices at maturity
S i1 := S
i
1(ω) = fi (S1(ω))
where function fi is known for all i . Think of S1 as a random variable
S1 : ω ∈ Ω 7→ S1(ω) ∈ R
Portfolio x = (x1, x2, . . . , xn) of derivatives (xi is investment in S
i )
Cost at time 0 : P0(x) :=
n∑
i=1
S i0xi
Value at maturity : P1(x ,S1) := P1(x , ω) =
n∑
i=1
S i1xi =
n∑
i=1
fi (S1(ω))xi
This can be used to detect arbitrage
30 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Portfolio of Derivatives
Single underlying security under infinitely many scenarios
S0 = known price at time 0, S1(ω) = uncertain price at maturity
Derivatives S i for i = 1, . . . , n written on the security, with known prices
S i0 at time 0 and random prices at maturity
S i1 := S
i
1(ω) = fi (S1(ω))
where function fi is known for all i . Think of S1 as a random variable
S1 : ω ∈ Ω 7→ S1(ω) ∈ R
Portfolio x = (x1, x2, . . . , xn) of derivatives (xi is investment in S
i )
Cost at time 0 : P0(x) :=
n∑
i=1
S i0xi
Value at maturity : P1(x ,S1) := P1(x , ω) =
n∑
i=1
S i1xi =
n∑
i=1
fi (S1(ω))xi
This can be used to detect arbitrage
30 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Portfolio of Derivatives
Single underlying security under infinitely many scenarios
S0 = known price at time 0, S1(ω) = uncertain price at maturity
Derivatives S i for i = 1, . . . , n written on the security, with known prices
S i0 at time 0 and random prices at maturity
S i1 := S
i
1(ω) = fi (S1(ω))
where function fi is known for all i . Think of S1 as a random variable
S1 : ω ∈ Ω 7→ S1(ω) ∈ R
Portfolio x = (x1, x2, . . . , xn) of derivatives (xi is investment in S
i )
Cost at time 0 : P0(x) :=
n∑
i=1
S i0xi
Value at maturity : P1(x ,S1) := P1(x , ω) =
n∑
i=1
S i1xi =
n∑
i=1
fi (S1(ω))xi
This can be used to detect arbitrage
30 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Type-A Arbitrage : Semi-Infinite LP
Find portfolio x ∈ Rn that is cheapest portfolio at time 0 and has no risk
of loss in future (value at maturity is ≥ 0 for all scenarios)
OPT = min
x
P0(x) s.t. P1(x , ω) ≥ 0, ∀ω ∈ Ω
• Type-A arbitrage does not exist ⇐⇒ OPT = 0
• Semi-infinite LP P0(x) = S>0 x and P1(x , ω) =
∑n
i=1 fi (S1(ω))xi are
linear functions of x , but there are infinitely many constraints since
Ω is an infinite set
• Standard LP solvers (e.g., CVX) cannot solve this as it is
Constraints can be reformulated when fi is piecewise linear so that this
can be solved as an LP
31 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Type-A Arbitrage : Semi-Infinite LP
Find portfolio x ∈ Rn that is cheapest portfolio at time 0 and has no risk
of loss in future (value at maturity is ≥ 0 for all scenarios)
OPT = min
x
P0(x) s.t. P1(x , ω) ≥ 0, ∀ω ∈ Ω
• Type-A arbitrage does not exist ⇐⇒ OPT = 0
• Semi-infinite LP P0(x) = S>0 x and P1(x , ω) =
∑n
i=1 fi (S1(ω))xi are
linear functions of x , but there are infinitely many constraints since
Ω is an infinite set
• Standard LP solvers (e.g., CVX) cannot solve this as it is
Constraints can be reformulated when fi is piecewise linear so that this
can be solved as an LP
31 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Reformulating Constraints
Suppose that for each j , fi (S1) := fi (S1(ω)) is a piecewise linear function
of S1 with a breakpoint
3 at Kj
Lemma∑n
i=1 fi (S1(ω)) is a piecewise linear function of S1 with breakpoints
K1,K2, . . . ,Kn.
3This means that for S1 ≤ Kj it is a linear function with one slope and for S1 ≥ Kj it is a linear
function with a different slope
32 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Reformulating Constraints
Theorem
Suppose the breakpoints are ordereda as : 0 ≤ K1 ≤ K2 ≤ · · · ≤ Kn.
Then, the function P1(x ,S1) := P1(x ,S1(ω)) is ≥ 0 for all ω ∈ Ω if and
only if all the following conditions are satisfied,
1. P1(x , 0) ≥ 0
2. P1(x ,Kj) ≥ 0 for all j = 1, . . . , n
3. Right-derivative of P1(x ,S1) at S1 = Kn is ≥ 0
aThis assumption is without loss of generality since we can sort the values
Both the piecewise linear functions illustrated on the previous slide satisfy
above conditions
33 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Type-A Arbitrage : LP
Semi-infinite ADLP minx
∑n
i=1 S
i
0xi s.t.
∑n
i=1 fi (S1(ω))xi ≥ 0,∀ω ∈ Ω
becomes a linear program
Corollary
Under the piecewise linear assumptions made so far on a portfolio of
derivatives, type-A arbitrage can be detected by solving
OPT = min
x
n∑
i=1
S i0xi
s.t.
n∑
i=1
fi (0)xi ≥ 0
n∑
i=1
fi (Kj )xi ≥ 0, j = 1, . . . , n
n∑
i=1
(
fi (Kn + 1)− fi (Kn)
)
xi ≥ 0.
34 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Options
35 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
What are Options ?
An option is
• the right to buy (call option) or sell (put option) an underlying
security (e.g., bond, stock, commodity)
• at a certain price (strike)
• in a certain time frame
• at any point before expiration date: [0,T ] (American option)
• at expiration date T (maturity) (European option)
You have the right to buy/sell but no obligation to do the transaction
36 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
What are Options ?
An option is
• the right to buy (call option) or sell (put option) an underlying
security (e.g., bond, stock, commodity)
• at a certain price (strike)
• in a certain time frame
• at any point before expiration date: [0,T ] (American option)
• at expiration date T (maturity) (European option)
You have the right to buy/sell but no obligation to do the transaction
Example : A call option to buy 1 oz. of gold on 1 March for £900.
• If on 1 March the market price of gold is £800/ounce you would just buy
the gold for the market price (and bin the option). The value of the
option on 1 March under this scenario is £0
• If on 1 March the market price of gold is £1000/ounce you would exercise
the option and buy gold for £900 (and sell it on the market). The value
of your option on 1 March under this scenario was £100.
36 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
What are Options ?
An option is
• the right to buy (call option) or sell (put option) an underlying
security (e.g., bond, stock, commodity)
• at a certain price (strike)
• in a certain time frame
• at any point before expiration date: [0,T ] (American option)
• at expiration date T (maturity) (European option)
You have the right to buy/sell but no obligation to do the transaction
Example : A put option to sell an ounce of gold on 1 March for £900.
• If on 1 March the market price of gold is £800/ounce you would just buy
the gold from the market for £800 and exercise your option to sell the gold
for £900. The value of the option on 1 March under this scenario is £100
• If on 1 March the market price of gold is £1000/ounce you would just sell
the gold for the market price (and bin the option). The value of your
option on 1 March under this scenario was £0.
36 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
What are Options ?
An option is
• the right to buy (call option) or sell (put option) an underlying
security (e.g., bond, stock, commodity)
• at a certain price (strike)
• in a certain time frame
• at any point before expiration date: [0,T ] (American option)
• at expiration date T (maturity) (European option)
You have the right to buy/sell but no obligation to do the transaction
There are other exotic options: Payoff at maturity determined by
• any price in [0,T ] (lookback option)
• average price in [0,T ] (Asian option)
where T = maturity/expiration date
36 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
What are Options ?
An option is
• the right to buy (call option) or sell (put option) an underlying
security (e.g., bond, stock, commodity)
• at a certain price (strike)
• in a certain time frame
• at any point before expiration date: [0,T ] (American option)
• at expiration date T (maturity) (European option)
You have the right to buy/sell but no obligation to do the transaction
Two questions
• What is the value (=fair price) of the option now?
→ This is option pricing. Can use RNPM for that!
• There are several options (different strike prices, different prices
now) on the same security on the market. How do we detect if there
is arbitrage?
→ We’ll answer that now!
36 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
European Call and Put Options
EU Call Option: The right to buy a certain underlying security at
expiration date at an agreed (strike) price. Its value at maturity is
f (S1) = (S1 − strike)+ := max{S1 − strike, 0}
S1
C1
strike
EU Put Option: The right to sell a certain underlying security at
expiration date at an agreed (strike) price. Its value at maturity is
f (S1) = (strike− S1)+ := max{strike− S1, 0}
S1
C1
strike
In both options, the option value at maturity is a piecewise linear
function with breakpoint at the strike price
37 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Arbitrage Detection in EU Call Options
Suppose we have n derivatives S1, . . . ,Sn of EU Call Options on the
same underlying security, with the same maturity, with strike prices
0 < K1 < K2 < · · · < Kn, respectively. Then
fi (S1) = (S1 − Ki )+ := max{S1 − Ki , 0}
fi (Kj) = (Kj − Ki )+ =
{
Kj − Ki , if j > i
0, otherwise.
The ADLP in the Corollary is minx{c>x : Ax ≥ 0}, where
c = (S10 ,S
2
0 , . . . ,S
n
0 )
>
A =
K2 − K1 0 · · · 0 0
K3 − K1 K3 − K2 · · · 0 0
...
...
...
...
Kn − K1 Kn − K2 · · · Kn − Kn−1 0
1 1 · · · 1 1
38 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
The dual to the ADLP is the feasibility LP
max
y
0>y s.t. A>y = c , y ≥ 0
No type-A arbitrage is equivalent to each of the following
1. ADLP is bounded
2. Dual to ADLP is feasible
No type-A or type-B arbitrage is equivalent to each of the following
1. ADLP is bounded with all optimal solutions x having Ax = 0
2. Dual to ADLP is strictly feasible, i.e., there exists y > 0 with
A>y = c
39 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
The dual to the ADLP is the feasibility LP
max
y
0>y s.t. A>y = c , y ≥ 0
No type-A arbitrage is equivalent to each of the following
1. ADLP is bounded
2. Dual to ADLP is feasible
No type-A or type-B arbitrage is equivalent to each of the following
1. ADLP is bounded with all optimal solutions x having Ax = 0
2. Dual to ADLP is strictly feasible, i.e., there exists y > 0 with
A>y = c
39 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
The dual to the ADLP is the feasibility LP
max
y
0>y s.t. A>y = c , y ≥ 0
No type-A arbitrage is equivalent to each of the following
1. ADLP is bounded
2. Dual to ADLP is feasible
No type-A or type-B arbitrage is equivalent to each of the following
1. ADLP is bounded with all optimal solutions x having Ax = 0
2. Dual to ADLP is strictly feasible, i.e., there exists y > 0 with
A>y = c
39 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
The system of equations A>y = c is
K2 − K1 k3 − K1 · · · Kn − K1 1
0 K3 − K2 · · · Kn − K2 1
...
...
...
...
0 0 · · · Kn − Kn−1 1
0 0 · · · 0 1
y1
y2
...
yn−1
yn
=
S10
S20
...
Sn−10
Sn0
which can be solved by backward substitution to get
yn = S
n
0 , yn−1 =
Sn−10 − Sn0
Kn − Kn−1
yi =
S i0 − S i+10
Ki+1 − Ki −
S i+10 − S i+20
Ki+2 − Ki+1 , i = 1, . . . , n − 2
This implies that there is no arbitrage ⇐⇒ all the yi > 0.
40 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Main Theorem on Arbitrage in EU Call Options
Theorem
For a derivative of EU Call Options S1, . . . ,Sn written on the same
underlying security, with the same maturity, with strike prices
0 < K1 < K2 < · · · < Kn, there is no arbitrage if and only if the following
three conditions hold:
1. S i0 > 0 for all i = 1, 2, . . . , n (only required for i = n)
2. S i0 > S
i+1
0 for all i = 1, 2, . . . , n − 1 (only required for i = n − 1)
3. The piecewise linear function mapping Ki to S
i
0, for i = 1, . . . , n, is
convex and the slopes of its linear pieces are strictly increasing.
41 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Example
Consider 4 call options with current prices (spot values) S0 = (10, 8, 3, 1),
and strikes K = (10, 20, 30, 60). Do we have arbitrage ?
Plot spot values S i0 against strikes Ki
10 20 30 40 50 60
10
8
3
1
strikes
spot values
Si0−S
i+1
0
Ki+1−Ki = negative of slope of piece i
Properties 1) and 2) are satisfied
3) is not satisfied since the
slopes are: −2, −5 and −2/3, and
−2 < −5 < −2/3 is not true
Theorem =⇒ arbitrage exists
How to detect/find x detecting the arbitrage opportunity?
42 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Example
Consider 4 call options with current prices (spot values) S0 = (10, 8, 3, 1),
and strikes K = (10, 20, 30, 60). Do we have arbitrage ?
Plot spot values S i0 against strikes Ki
10 20 30 40 50 60
10
8
3
1
strikes
spot values
Si0−S
i+1
0
Ki+1−Ki = negative of slope of piece i
Properties 1) and 2) are satisfied
3) is not satisfied since the
slopes are: −2, −5 and −2/3, and
−2 < −5 < −2/3 is not true
Theorem =⇒ arbitrage exists
How to detect/find x detecting the arbitrage opportunity?
42 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
How to find x?
Consider a situation where successive slopes are not strictly increasing:
K1 K2 K3
S0
1
S0
2
S0
3
S
S2 is overpriced. Sell S2 and use money
to buy a combination of S1 and S3
Let α ∈ (0, 1) s.t. K2 = αK1 + (1−α)K3
Take portfolio x = (α, 0, 1− α)
• Cost at time 0 < cost of 1 unit of
S2:
x1S
1
0 + x3S
3
0 = S︸ ︷︷ ︸
cost of portfolio at time 0
< S20︸︷︷︸
cost of S2 at time 0
=⇒ Positive initial cash-flow
• Payoff at maturity is
P1(x , S1) = (S1 − K1)+x1 + (S1 − K3)+x3
≥ (x1(S1 − K1) + x3(S1 − K3))+
= (S1 − K2)+, ∵K2 = x1K1+x3K3
So, payoff of portfolio cannot be
worse than payoff of S2
=⇒ No risk of loss in the future
43 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
How to find x?
Consider a situation where successive slopes are not strictly increasing:
K1 K2 K3
S0
1
S0
2
S0
3
S
S2 is overpriced. Sell S2 and use money
to buy a combination of S1 and S3
Let α ∈ (0, 1) s.t. K2 = αK1 + (1−α)K3
Take portfolio x = (α, 0, 1− α)
• Cost at time 0 < cost of 1 unit of
S2:
x1S
1
0 + x3S
3
0 = S︸ ︷︷ ︸
cost of portfolio at time 0
< S20︸︷︷︸
cost of S2 at time 0
=⇒ Positive initial cash-flow
• Payoff at maturity is
P1(x , S1) = (S1 − K1)+x1 + (S1 − K3)+x3
≥ (x1(S1 − K1) + x3(S1 − K3))+
= (S1 − K2)+, ∵K2 = x1K1+x3K3
So, payoff of portfolio cannot be
worse than payoff of S2
=⇒ No risk of loss in the future
43 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Risk Neutral Probability
44 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Risk Neutral Probability Measure (RNPM)
Definition (Def 4.2)
A risk neutral probability measure on Ω = {ω1, . . . , ωm} is a vector
p = (p1, . . . , pm)
T ∈ Rm satisfying
1.
∑
j pj = 1; pj > 0 ∀j , (i.e., p is a vector of positive probabilities)
2.
S i0 =
1
R
m∑
i=1
pjS
i
1(ωj) :=
1
R
E[S i1], i = 0, 1, . . . , n. (2)
Here R = 1 + r where r = return rate of risk-free asset S0
• Given m, n, R, S i0 and S i1(ωj) for all i = 0, 1, . . . , n and
j = 1, . . . ,m, a RNPM does not have to exist
• We will see that existence of RNPM ⇐⇒ non-existence of arbitrage
• If RNPM exists, it allows us to see current prices S i0 as discounted
expected values of future prices, where E is taken wrt a RNPM
45 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Risk Neutral Probability Measure (RNPM)
Definition (Def 4.2)
A risk neutral probability measure on Ω = {ω1, . . . , ωm} is a vector
p = (p1, . . . , pm)
T ∈ Rm satisfying
1.
∑
j pj = 1; pj > 0 ∀j , (i.e., p is a vector of positive probabilities)
2.
S i0 =
1
R
m∑
i=1
pjS
i
1(ωj) :=
1
R
E[S i1], i = 0, 1, . . . , n. (2)
Here R = 1 + r where r = return rate of risk-free asset S0
• Given m, n, R, S i0 and S i1(ωj) for all i = 0, 1, . . . , n and
j = 1, . . . ,m, a RNPM does not have to exist
• We will see that existence of RNPM ⇐⇒ non-existence of arbitrage
• If RNPM exists, it allows us to see current prices S i0 as discounted
expected values of future prices, where E is taken wrt a RNPM
45 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Further Insights
• Note that while pj is NOT EQUAL TO the probability that ωj
happens, yet it acts as if it was (in the above sense)
• It is natural to require that if two securities, say S i and Sk , have the
same future payoffs (S i1(ωj) = S
k
1 (ωj) for all j), then they should
have the same price at present: S i0 = S
k
0 . If RNPM exists, then (2)
gives a formula for computing this price.
• RNPM can be used to price new securities using a replication
technique.
• If a RNPM exists, it need not be unique. However, (2) will hold for
all such p, so it does not matter which one is used in the calculation
of the present price.
Exercise 0. Construct an example (i.e., find some concrete numbers for
m, n, R, S i0 and S
i
1(ωj)) such that a RNPM does not exist. Hint: use
m = 2, n = 1.
46 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
First Fundamental Theorem of Asset Pricing
Theorem (Thm 4.2)
A RNPM exists if and only if there is no arbitrage (of either type).
Proof ⇐: assume there is no arbitrage.
• In particular, since there is no type-A arbitrage, applying Fact (1) we
know that (ADLP) (and hence (D)) has an optimal solution.
• Using strong duality, we conclude that (P) – the dual of ADLP – is
feasible (since (D) has an optimal solution).
• In particular, both (P) and (D) are feasible. We can thus apply
Goldman-Tucker complementary slackness theorem for LPs, which
says that there exists strictly complementary optimal solutions x∗
and (y∗, s∗):
x∗ + s∗ > 0. (3)
• Now we utilize the assumption that there is no type-B arbitrage.
Applying Fact 2: all constraints are tight, we get
s∗ = c − AT y∗ = 0. Plugging this into (3), we conclude that x∗ > 0
(that is x∗i > 0,∀i).
47 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Proof: (Continued)
We now claim that the vector p = (p1, p2, . . . , pm)
T , where pj = Rx
∗
j , is
the vector of risk-neutral probabilities. For this we need to check that p
satisfied the three properties in the definition.
• First, clearly pj > 0 for all j (since x∗j > 0 for all j).
• Second, note that the first constraint (i.e., the one corresponding to
i = 0: cash) in (P) looks as follows:
m∑
i=1
S01 (ωj)︸ ︷︷ ︸
=R
x∗j = S
0
0 = 1,
which implies that
∑
j pj = 1.
• Finally, let us look at the i-th constraint in (P) (for any i):
m∑
i=1
S i1(ωj)x
∗
j = S
i
0.
It can be written as S i0 =
1
R
∑m
i=1 S
i
1(ωj)Rx
∗
j =
1
R
∑m
i=1 S
i
1(ωj)pj ,
recovering (2). Hence, p is a RNPM on Ω.
48 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Example
Example : Show directly from definition that in the previous exercise
there is no risk neutral probability measure (RNPM). Then argue this via
the Fundamental Thm of Asset Pricing.
Solution. Since m = 1 (we have one scenario only), the only possible
probability measure on Ω is p1 = 1. Let us check whether it satisfies the
conditions in the definition of a RNPM:
1.
∑
j pj = 1 and pj > 0 for all j trivially holds since m = 1 and p1 = 1,
2. We further need the following identities to hold:
S i0 =
1
R
m∑
i=1
pjS
i
1(ωj), i = 0, 1.
This holds for i = 0 since 1 = 11.05p11.05 = 1. It does not hold for
i = 1 since 500 6= 11.05p1550. Hence, RNPM does not exist.
We reach the same conclusion by applying the Fundamental Theorem of
Asset Pricing and the previous Exercise which says that arbitrage exists.
49 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Pricing Securities using RNPM
50 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Binomial Model
n = 2 (3 securities) and m = 2 scenarios ω1 and ω2
• S0: risk-free security/cash (yielding return R = 1 + r)
• S1: underlying security
• S2: derivative of underlying security
Assume U > D > 0 are constants such that
S11 =
{
U × S10 , under ω1 (“UP ′′)
D × S10 , under ω2 (“DOWN ′′).
price of the derivative at maturity (time 1) is a function of the price of
the underlying security at maturity:
S21 = f (S
1
1 )
Price the derivative : Assuming there is no arbitrage, find a formula for
the spot price (i.e., price at time 0) S20
51 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
We can use the RNPM to find the price at time 1 of the derivative S2:
• Start with the two securities S0 (cash) and S1 (underlying)
Assume no arbitrage when only these two securities are present
• According to the Fundamental Thm of Asset Pricing, there exists a
RNPM, i.e. “probabilities” p1, p2 for ω1, ω2, such that
S00 =
1
R
[
p1S
0
1 (ω1) + p2S
0
1 (ω2)
]
S10 =
1
R
[
p1S
1
1 (ω1) + p2S
1
1 (ω2)
]
• Using S00 = 1,S01 (ω1) = S01 (ω2) = R, the first equation gives
p1 + p2 = 1
• while using S11 (ω1) = US10 ,S11 (ω2) = DS10 , the second equation gives
p1U + p2D = R.
52 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
We can use the RNPM to find the price at time 1 of the derivative S2:
• Start with the two securities S0 (cash) and S1 (underlying)
Assume no arbitrage when only these two securities are present
• According to the Fundamental Thm of Asset Pricing, there exists a
RNPM, i.e. “probabilities” p1, p2 for ω1, ω2, such that
S00 =
1
R
[
p1S
0
1 (ω1) + p2S
0
1 (ω2)
]
S10 =
1
R
[
p1S
1
1 (ω1) + p2S
1
1 (ω2)
]
• Using S00 = 1,S01 (ω1) = S01 (ω2) = R, the first equation gives
p1 + p2 = 1
• while using S11 (ω1) = US10 ,S11 (ω2) = DS10 , the second equation gives
p1U + p2D = R.
52 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
We can use the RNPM to find the price at time 1 of the derivative S2:
• Start with the two securities S0 (cash) and S1 (underlying)
Assume no arbitrage when only these two securities are present
• According to the Fundamental Thm of Asset Pricing, there exists a
RNPM, i.e. “probabilities” p1, p2 for ω1, ω2, such that
S00 =
1
R
[
p1S
0
1 (ω1) + p2S
0
1 (ω2)
]
S10 =
1
R
[
p1S
1
1 (ω1) + p2S
1
1 (ω2)
]
• Using S00 = 1,S01 (ω1) = S01 (ω2) = R, the first equation gives
p1 + p2 = 1
• while using S11 (ω1) = US10 ,S11 (ω2) = DS10 , the second equation gives
p1U + p2D = R.
52 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
• Using S00 = 1,S01 (ω1) = S01 (ω2) = R, the first equation gives
p1 + p2 = 1
• while using S11 (ω1) = US10 ,S11 (ω2) = DS10 , the second equation gives
p1U + p2D = R.
• Solving for p1, p2 we get
p1 =
R − D
U − D , p2 =
U − R
U − D
This is the RNPM.
• We can now use the RNPM to price the derivative S2:
Under the no arbitrage condition, the RNPM measure must hold for
S2 as well, hence
S20 =
1
R
[p1S
2
1 (ω1) + p2S
2
1 (ω2)]
53 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
• Using S00 = 1,S01 (ω1) = S01 (ω2) = R, the first equation gives
p1 + p2 = 1
• while using S11 (ω1) = US10 ,S11 (ω2) = DS10 , the second equation gives
p1U + p2D = R.
• Solving for p1, p2 we get
p1 =
R − D
U − D , p2 =
U − R
U − D
This is the RNPM.
• We can now use the RNPM to price the derivative S2:
Under the no arbitrage condition, the RNPM measure must hold for
S2 as well, hence
S20 =
1
R
[p1S
2
1 (ω1) + p2S
2
1 (ω2)]
53 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
• Using S00 = 1,S01 (ω1) = S01 (ω2) = R, the first equation gives
p1 + p2 = 1
• while using S11 (ω1) = US10 ,S11 (ω2) = DS10 , the second equation gives
p1U + p2D = R.
• Solving for p1, p2 we get
p1 =
R − D
U − D , p2 =
U − R
U − D
This is the RNPM.
• We can now use the RNPM to price the derivative S2:
Under the no arbitrage condition, the RNPM measure must hold for
S2 as well, hence
S20 =
1
R
[p1S
2
1 (ω1) + p2S
2
1 (ω2)]
53 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
• We have worked with 2 scenarios and 3 securities because we can
deal with this by hand.
Effectively we have solved a 2× 2 system of linear equations to
obtain the RNPM and use the solution to “price” the 3rd
(derivative) security.
Alternatively we could have obtained the RNPM from the dual
solution of the ADLP.
• However, the technique can easily be adapted to our general model
with arbitrary finite number of scenarios (ω1, . . . , ωn) and n + 1
securities S0, . . . ,Sn.
In this case we would have to computationally solve an n× n system
(or again find the duals of ADLP)
54 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
Pricing of Securities via Replication
Approach: We will try to form a portfolio y = (y0, y1) of cash (S0) and
the underlying (S1) such that its price P1(y) at time 1 matches the price
of the derivative, under both scenarios:
P1(y ;ωj) := S
0
1 (ωj)y0 + S
1
1 (ωj)y1 = S
2
1 (ωj), j = 1, 2.
• The above is a system of 2 linear equations in 2 unknowns (y0, y1).
• The solution is:
y0 =
US21 (ω2)− DS21 (ω1)
R(U − D) , y1 =
S21 (ω1)− S21 (ω2)
S10 (U − D)
, (4)
• We then say that since the portfolio has the same future
behaviour/payoff/value as the derivative S2, the spot price of the
derivative must be equal to the spot price of the portfolio:
S20 = P0(y) = S
0
0 y0 + S
1
0 y1. (5)
55 / 56
Arbitrage LP Arbitrage in Derivatives Risk Neutral Probability
• By plugging (4) into (5), we obtain:
S20 =
1
R
R − DU − D︸ ︷︷ ︸
:= p1
S21 (ω1) +
U − R
U − D︸ ︷︷ ︸
:= p2
S21 (ω2)
. (6)
• If D < R < U, then p1, p2 as defined in (6) satisfy
0 < p1, p2 < 1, p1 + p2 = 1 that is they act like a probability.
Indeed this is the Risk Neutral Probability Measure (RNPM)!
• (6) says that the spot value (S20 ) of the derivative security is equal to
its present (i.e., discounted) expected value, where expectation is
taken with respect to the RNPM.
• However, for (p1, p2) to be a RNPM we still need to check that (6)
holds for securities S0 and S1 as well.
Exercise
Check that (6) holds for securities S0 and S1 as well, thus completing the
verification of the fact that p is a RNPM.
56 / 56
