24 2 Models of Financial Markets on Finite Probability Spaces
For givenf∈L∞(Ω,F,P), we calla∈Ranarbitrage-free price,ifin
addition to the financial marketS, the introduction of the contingent claimf
at priceadoes not create an arbitrage possibility. How can we mathematically
formalise this economically intuitive idea? We enlarge the financial marketS
by introducing a new financial instrument which can be bought (or sold) at
priceaat timet= 0 and yields the random cash flowf(ω)attimet=T.
We don’t postulate anything about the price of this financial instrument at
the intermediate timest=1,...,T−1. The reader might think of an “over
the counter” option where the two parties agree on certain payments at times
t=0andt=T. So if we look at the linear space generated byKand the
vector (f−a) we obtain an enlarged spaceKf,aof attainable claims. The price
ashould be such that arbitrage opportunities are inexistent. Mathematically
speaking this means that we still should haveKf,a∩L∞+ ={ 0 }.Inthiscase
we say thatais anarbitrage free pricefor the contingent claimf.
Theorem 2.4.1 (Pricing by no-arbitrage).Assume thatSsatisfies (NA)
and letf∈L∞(Ω,F,P). Define
π(f)=inf{EQ[f]|Q∈Me(S)},
π(f)=sup{EQ[f]|Q∈Me(S)}, (2.8)Eitherπ(f)=π(f),inwhichcasefis attainable at priceπ(f):=π(f)=
π(f), i.e.f =π(f)+(H·S)Tfor someH ∈Hand thereforeπ(f)is the
unique arbitrage-free price forf.
Orπ(f)<π(f),inwhichcase
]π(f),π(f)[={EQ[f]|Q∈Me(S)}andais an arbitrage-free price forfiffalies in theopeninterval]π(f),π(f)[.
Proof.The caseπ(f)=π(f) follows from corollary 2.2.11 and so we only have
to concentrate on the caseπ(f)<π(f). First observe that the set{EQ[f]|
Q∈Me(S)}forms a bounded non-empty interval inR, which we denote byI.
We claim that a numberais inI iffais an arbitrage-free price forf.
Indeed, supposing thata∈Iwe may findQ∈Me(S) s.t.EQ[f−a]=0and
thereforeKf,a∩L∞+(Ω,F,P)={ 0 }.
Conversely suppose thatKf,a∩L∞+={ 0 }. Then exactly as in the proof
of the Fundamental Theorem 2.2.7, we find a probability measureQso that
EQ[g] = 0 for allg∈Kf,aand so thatQis equivalent toP. This, of course,
implies thatQ∈Me(S)andthata=EQ[f].
Now we deal with the boundary case: suppose thataequals the right
boundary ofI, i.e.,a=π(f)∈I, and consider the contingent claimf−π(f).
By definition we haveEQ[f−π(f)]≤0, for allQ∈Me(S), and therefore
by Proposition 2.2.9, thatf−π(f)∈C. We may findg∈K such that
g≥f−π(f). If the sup in (2.8) is attained, i.e., if there isQ∗∈Me(S)such
thatEQ∗[f]=π(f), then we have 0 =EQ∗[g]≥EQ∗[f−π(f)] = 0 which in