27.11.2012 Views

Lecture Notes for Finance 1 (and More).

Lecture Notes for Finance 1 (and More).

Lecture Notes for Finance 1 (and More).

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

3.1. FINANCIAL MARKETS AND ARBITRAGE 13<br />

The interpretation is that it should not be possible to <strong>for</strong>m a portfolio at<br />

zero cost which delivers non-negative payments at all future dates <strong>and</strong> even<br />

gives a strictly positive payment at some date. And it should not be possible<br />

to <strong>for</strong>m a portfolio at negative cost (i.e. a portfolio which gives the owner<br />

money now) which never has a negative cash flow in the future.<br />

Usually type 1 arbitrages can be trans<strong>for</strong>med into type 2. arbitrages, <strong>and</strong><br />

vice versa. For instance, if the exists a ci > 0, then we easily get from 2 to 1<br />

But there is not mathematical equivalence (take π = 0 or C = 0 to see this).<br />

Definition 4 The security market is arbitrage-free if it contains no arbitrage<br />

opportunities.<br />

To give a simple characterization of arbitrage-free markets we need a<br />

lemma which is very similar to Farkas’ theorem of alternatives proved in<br />

Matematik 2OK using separating hyperplanes:<br />

Lemma 1 (Stiemke’s lemma) Let A be an n × m−matrix: Then precisely<br />

one of the following two statements is true:<br />

1. There exists x ∈ Rm ++ such that Ax = 0.<br />

2. There exists y ∈ R n such that y ⊤ A > 0.<br />

We will not prove the lemma here (it is a very common exercise in convexity/linear<br />

programming courses, where the name Farkas is encountered).<br />

But it is the key to our next theorem:<br />

Theorem 2 The security market (π, C) is arbitrage-free if <strong>and</strong> only if there<br />

exists a strictly positive vector d ∈ RT ++ such that π = Cd.<br />

In the context of our security market the vector d will be referred to as<br />

a vector of discount factors. This use of language will be clear shortly.<br />

Proof. Define the matrix<br />

⎛<br />

⎞<br />

−π1 c11 c12 · · · c1T<br />

⎜ −π2 c21 c22 · · · c2T<br />

⎟<br />

A = ⎜<br />

⎝<br />

.<br />

. . . ..<br />

⎟<br />

. ⎠<br />

−πN cN1 cN2 · · · cNT<br />

First, note that the existence of x ∈ R T+1<br />

++ such that Ax = 0 is equivalent<br />

to the existence of a vector of discount factors since we may define<br />

di = xi<br />

x0<br />

i = 1, . . .,T.<br />

separating<br />

hyperplane<br />

Stiemke’s lemma<br />

discount factors

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!