Tutorial 7 - Department of Mathematics, CUHK

MATH4210 Financial Mathematics 2014
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Arbitrage
Arbitrage
Arbitrage
Practice of taking advantage of price difference between two or more
markets.
Divided into two types :
Type A arbitrage initial positive cash flow;
no risk of loss in future.
Type B arbitrage no initial cash investment;
no risk of future loss;
positive probability of profit in the future.
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Arbitrage
Example 1
Bank A issues loans at 5% interest rate.
Bank B offers a savings account which pays 6% interest.
⇒ Arbitrage opportunity :
Take out a loan from bank A and place the loan into the savings in
bank B.
1% of the loan amount as profit
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Arbitrage
Example 2
Assume no service charges
Convert rate is unchanged
Consider the follow currency exchanges
1
USD
EUR
GBP
JPY
USD
0.8706 1.4279 0.0075
EUR 1.1486
1.6401 0.00861
GBP 0.7003 9.6097
0.00525
JPY 133.33 116.14 190.48
Is there any arbitrage opportunities? Difficult to answer
Solution
Exchange USD 1 for EUR 1.1486. Then exchange all EUR for JPY
133.3984. Lastly exchange all JPY back to USD 1.0005.
1
Example from http://mattmcd.github.io/2013/03/30/FX-Arbitrage-CLP.html
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Arbitrage
Linear Programming
Difficult to avoild arbitrage opportunities in complex financial
situations.
More precise definition on Arbitrage
Arbitrage Theorem (First Fundamental Theorem of Finance
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Arbitrage
Linear Programming
LP standard form
minimize
subject to
cT x
(1)
Ax = b
and x ≥ 0



with x = 

x1
x2
..
.
xn



,


a11
 a21

A= .
 ..
a12
a22
..
.
...
...


a1n

a2n 


.. , b = 

. 
am1 am2 . . . amn
Tutorial 5
b1
b2
..
.






, c = 


bn
c1
c2
..
.





cn
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Arbitrage
Linear Programming
From other form to standard form
maximize
cT x
subject to
Ax ≤ b
(2)
and x ≥ 0
We first introduce slack variables:
x1k + x2k + . . . + xnk ≤ bk
and observe that
becomes x1k + x2k + . . . + xnk + xˆk = bk (3)
− maxx (−cT x) = minx (cT x)
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Arbitrage
Linear Programming
If x is unrestricted in sign, then we can divide x to be x = x+ − x− , where
(
(
x
if
x
≥
0
−x if x ≤ 0
x+ =
, x− =
(5)
0 if x < 0
0 if x > 0
So from now on, we only consider the standard form LP problem.
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Arbitrage
Dual Problem
Primal problem
minimize
subject to
cT x
(6)
Ax = b
and x ≥ 0
Dual Problem
maximize
bT y
subject to
AT y ≤ c
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Arbitrage
Primal and Dual relation
LP Weak Duality
Let x˜ be feasible for Primal and y˜ be feasible for Dual. Then we have
bT˜
y ≤ cT ˜
x
Proof: Exercise
LP Strong Duality
Suppose that Primal Problem has an optimal solution x ∗ . The the Dual
problem has an optimal solution y ∗ , and
cT x∗ = bT y∗
That is: Suppose both Primal and Dual problem are feasible. Then both
have optimal solutions, and the respective optimal values are equal
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Arbitrage
Example
Consider the LP
minimize
subject to
47x1 + 93x2 + 17x3 − 93x4


−1 −6
1
3 
−1 −2
7
1


0

3
−10 −1


−6 −11 −2 12 
1
6
−1 −3


x1


x2 
≤


x3

x4
−3
5
−8
−7
4






(8)
Prove that x = [1, 1, 1, 1]T is optimal.
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Arbitrage
Complementary Slackness
Let x and y be feasible for Primal and Dual respectively. Then, x and y
are optimal for their respective problems iff
xi (c − AT y )i = 0
for i = 1, ..., n
By this theorem, we see that one way to solve an LP is to solve the
following system of equations in (x, y , s):
xi si = 0
for i = 1, ..., n,
Ax = b,
(9)
AT y + s = c.
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Arbitrage
Farkas lemma
Theorem
Exactly one of the following systems has a solution:
Ax = b, x ≥ 0
(10)
AT y ≤ 0, b T y > 0
Exercise:
Prove the following variant:
Exactly one of the following systems has a solution:
Ax ≤ b, x ≥ 0
AT y ≥ 0, b T y < 0,
(11)
y ≥0
Farkas lemma: One of the most important fundamental theorem in LP
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Arbitrage
Arbitrage Detection
Consider a market in which n different assets are traded. Assume that
there are m possible states at the end of the period. Suppose for every
unit of asset i ∈ {1, ..., n} owned, we can receive a payoff of rsi dollars if
state s ∈ {1, ..., n} is realized at the end of the period. Mathematically, we
have the payoff matrix Rm×n :


r11 . . . r1n

.. 
..
R =  ...
(12)
.
. 
rm1 . . . rmn
Denote xi be the number of units of asset i held. Then, given an initial
portfolio x = (x1 , ..., xn ), our wealth at the end of the period is
ws =
n
X
rsi xi ,
s ∈ {1, ..., m}
(13)
i=1
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Arbitrage
Arbitrage Detection
Denote w = (w1 , ..., wm ), we have w = Rx.
Now, we want to determine the prices of the assets at the beginning of the
period. Let ci be the cost of asset i at the beginning of the period. Let
c = (c1 , ..., cn ). The cost with portfolio x is therefore c T x. In order to
prevent arbitrage opportunities, we should prevent the situation that
No non-negative payoff out of a negative investment
⇒ any portfolio that guarantees a non-negative payoff in every state
must be a valuable
⇒
if Rx ≥ 0, then c T x ≥ 0
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Arbitrage
Arbitrage Detection
Theorem
There is no arbitrage opportunity if and only if there exists q such that
ci =
m
X
qs rsi
(15)
s=1
Proof
{if Rx ≥ 0, then c T x ≥ 0} implies {Rx ≥ 0, c T x < 0} has no solution.
By Farkas’ lemma, this implies there exists q such that R T q = c.
Note : We can check arbitrage opportunity by simply solving
Rx ≥ 0,
c T x = −1
.
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Arbitrage
Arbitrage Detection
Exercise
Consider the price and payoff, Which is/are arbitrage-free?
1
1 2
c=
, R=
1
2 1


3
c =  3 ,
3
 
3
c =  3 ,
3
1 1 1
R=
0 1 1
2 2 2
R=
1 2 3
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Arbitrage
Arbitrage Theorem
Consider a collection of n stocks with prices S i for i = 1, 2, ..., n.
Denote S 0 to be the amount of cash an invester deposit in a risk-free
savings account with simple interest rate r .
After one unit of time, the stock prices will be in one of m possible states,
denoted as ω1 , ..., ωm .
Denote S i (0) be the stock prices initially and S i (ωj ) be stock price of
stock i at t = 1 in state ωj .
At t = 1, the cash deposited will become S 0 (wj ) = (1 + r )S 0 ∀j
Let p = (p1 , ..., pm ) be a vector of probability of state ωj .
Risk-Neutral probability measure
We call p a risk-neutral probability if
S i (0) =
m
1 X
pj S i (ωj )
1+r
∀i
j=1
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Arbitrage
Arbitrage Theorem
Theorem
A risk-neutral probability measure exists if and only if there is no arbitrage.
Proof : Refer to the book P.104.
Remark - Provide a necessary and sufficient condition about the
arbitrage-free situation.
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