The Mathematics of Arbitrage

184 9 Fundamental Theorem of Asset Pricing

atom{ 3 n+1, 3 n+2,...}.S 0 =0andSn−Sn− 1 is defined as the variable
gn(3(n−1) + 1) =n;gn(3(n−1) + 2) = 1;gn(3n)=− 1 .The processSis
not bounded but a normalisation of the functionsgnallows us to replaceS
by a bounded process. To keep the notation simple we prefer to continue with
the locally bounded processSgiven above. For the measurePwe choose any
measure that gives a strictly positive mass to all natural numbers and such
that for allnwe haveEP[gn]=0.ThespaceW^0 is precisely the set:
{
∑

n≥ 1

angn|(nan)n≥ 1 is bounded

}

.

Take now forfthe function defined as

for alln≥1:f(3(n−1) + 1) = 0 ;f(3(n−1) + 2) = 1 andf(3n)=0.

From the description ofW^0 it follows that forhinW^0 andx∈Rthe
random variablex+hcan only dominatefifx≥1. The constant function 1
clearly dominatesf. This shows that

inf{x| there ish∈W^0 withx+h≥f}=1.

On the other hand ifQis a local martingale measure forSthennQ[3(n−
1)+ 1] +Q[3(n−1)+ 2] =Q[3n] implies thatQ[3(n−1)+ 2]≤^12 Q[{3(n−1)+
1 ,3(n−1) + 2, 3 n}], with strict inequality ifQis equivalent toP. Therefore
EQ[f]≤^12 with strict inequality forQinMe(P). If we take any measureQ
such thatQ[3(n−1) + 1] = 0 andQ[3(n−1) + 2] =Q[3n]thenQis inM(P)
andEQ[f]=^12. It is now clear that maxQ∈M(P)EQ[f]=^12.
This example also shows that in Theorem 9.5.2 (ii), the conditionQ∈
Me(P) may not be replaced by the conditionQ∈M(P). Referring to the
proof of [D 92, Lemma 5.7], we remark that in this example the functionf
is not in^12 +W^0 −L∞+ but it is in the weak-star-closure of it. To see this
letfnbe the function defined asfn(3(k−1) + 2) = 1 for allk≤nand 0
elsewhere. The functionsfnare smaller than^12 +

∑n

k=1(

1
2 gk) and therefore
are in^12 +W^0 −L∞+, they converge weak-star tof.ThesetW^0 −L∞+is not
even norm closed as the following reasoning shows. An elementhinW^0 −L∞+
is of the form

∑

n≥ 1 angn−kwherekis inL

∞

• and|nan|is bounded, say by
  m.Ifanis positive thenh(3(n−1)+ 2)≤angn≤mnand ifanis negative then
  h(3(n−1) + 2)≤0. In any caseh(3(n−1) + 2)≤mn. Take now the function
  pdefined asp(3(n−1) + 1) = 0,p(3(n−1) + 2) =√^1 nandp(3n)=√−^1 n.It

is easy to see thatpis in the norm closure ofW^0 −L∞+ but it cannot be in
W^0 −L∞+since the converge ofp(3(n−1) + 2) to 0 is too slow. This reasoning
also shows that the elementf, described above, cannot be in the norm closure
of the setx+W^0 −L∞+ for anyx<1.

To remedy this “gap” phenomenon, well-known ininfinitedimensional
linear programming, we will use another set to calculate the infimum. The