ARBITRAGE AND GEOMETRY Arbitrage is a fundamental notion in mathematical finance, and ...

16 Pages·2006·0.17 MB·English

ARBITRAGE AND GEOMETRY
DANIELQ.NAIMANANDEDWARDR.SCHEINERMAN
1. INTRODUCTION
Arbitrage is a fundamental notion in mathematical finance, and making the “no free lunch” as-
sumption,thatarbitrageopportunitiesinthemarketplaceareunavailable,hasplayedafundamental
roleinfinancialeconomics. In1958,ModiglianiandMiller[11]usedtheprincipletoarguethatthe
wayacompanyfinancesitself,eitherbyuseofbondsorbyissuingadditionalstockisirrelevantin
determining its value. The principle appears in the options pricing formula due to Merton, Black,
and Scholes [10, 4]. The principle was placed on a solid theoretical footing in the work of Ross
[16, 17]. A gentleintroductionto thearbitrageprincipleis[20].
Thisarticleintroducesthenotionofarbitrageforasituationinvolvingacollectionofinvestments
and a payoff matrix describing the return to an investor of each investment under each of a set of
possible scenarios. We explain the Arbitrage Theorem, discuss its geometric meaning, and show
its equivalence to Farkas’ Lemma. We then ask a seemingly innocent question: given a random
payoff matrix, what is the probability of an arbitrage opportunity? This question leads to some
interestinggeometryinvolvinghyperplanearrangementsand related topics.
1.1. Payoff matrices. A dizzying array of investment opportunities is available to someone who
has money to invest. There are stocks, bonds, commodities, foreign currency, stock options, fu-
tures, real estate, and of course, savingsaccounts. We makethe simplifyingassumptionthat there
are n possible choices of investments available, and that at this moment of time (today) one may
investanyamountofone’sfinancesineach. Attheendofafixedtimeperiod(tomorrow),asingle
unit of cash (a dollar, a euro, a yen, etc.) invested in investment i will be worth some amount of
money,so that changes in valueof theinvestmentscan berepresented as an n-dimensionalvector.
A further simplifyingassumption is that there are finitely many mutually exclusivescenarios that
can occur, and thateach scenario leadsto aspecificgain(orloss)foreach investment.
As a consequence of these assumptions, we represent the changes in value, be they gains or
losses,inanm×npayoffmatrix. Thisisamatrixwitharowcorrespondingtoeachscenarioanda
column corresponding to each investment. The j,i entry, a , gives the change in value at the end
ji
of the timeperiod based on a unit invested in investmenti under scenario j. See Figure 1. Entries
inthismatrixare positiveiftheinvestmentgainsvalueand negativeifit losesvalue.
1.2. Risk-freeRate. Itistypicaltoassumethatthereisarisk-freeinvestment(e.g. U.S.Treasury
Bills) available to the investor that yields a payoff tomorrow of 1+r units for each unit invested.
Thus, a single unit today is guaranteed to be worth 1+r units tomorrow. We express gains or
lossestomorrowin presentdayunits,bymultiplyingby adiscountfactor of1/(1+r).
Thus,ifweinvestoneunitofcashtodayinaninvestmentanditbecomesworthxunitstomorrow,
therelevantentry in thepayoffmatrixisitspresent value−1+x/(1+r).
1
2 DANIELQ.NAIMANANDEDWARDR.SCHEINERMAN
I I ··· I ··· I
1 2 i n
Scenario a a ··· a ··· a
1 11 12 1i 1n
Scenario a a ··· a ··· a
2 21 22 2i 2n
. . . . .
. . . . .
. . . . .
Scenario a a ··· a ··· a
j j1 j2 ji jn
. . . . .
. . . . .
. . . . .
Scenario a a ··· a ··· a
m m1 m2 mi mn
FIGURE 1. The j,i entry of payoff matrix A is the gain (or loss) tomorrow under
scenario j foraunitinvestedin investmenti today.
1.3. Example: The Bernoulli model. The following is a simple example to give the reader a
taste for how options are priced. Assume that when we invest one unit in a certain stock today,
currently priced at S, its value tomorrow either be one of two possibilities,either Su or Sd, where
d < 1+r < u, so that the stock under- or over-performs the risk-free rate. Consequently, there
are two possible payoffs tomorrow of a single unit investment in the stock today, u or d, and
discountinggiveseitheravaluechangeof−1+u/(1+r)or−1+d/(1+r)in today’sterms.
Anotherpossibilityistopurchaseastockoptionreferredtoasacall: wepaysomeoneanamount
P today for the right to buy from them a share of the stock at a predetermined price K (called the
strike price) tomorrow (no matter what happens to the market price), where Sd < K <Su. If the
stock goes up, then we would exercise our right to buy it at K and then immediately sell it at the
marketpriceofSu,instantlynettingSu−K.Inthiscase, thenetgainforbuyingtheoption(taking
intoaccount discounting)is
−P+(Su−K)/(1+r).
In otherwords,investingasingleunitofmoneyinsuch an optionresults inavaluechangeof
−1+(Su−K)/(P(1+r)).
Ontheotherhand,ifthestockgoesdownwewouldnotwanttoexercisetheoptionsincewewould
be better off purchasing the stock at the lower market price, and in this case we have simply lost
thecostoftheoption.
Thus,wecanwritedowna2×2payoffmatrixcorrespondingtothetwoinvestmentopportunities
(stock,option)and thetwo possiblescenarios (stockgoes up,stockgoes down)as inFigure2.
stock option
up −1+u/(1+r) −1+(Su−K)/(P(1+r))
down −1+d/(1+r) −1
FIGURE 2. The payoff matrix for a stock whose price today is S and whose price
tomorrowtakes one of two values Su or Sd, and for an option whoseprice today is
Pandgivesonetherighttopurchasethestockat astrikepriceofK tomorrow. The
risk-freeinterestrate forthisoneday period isr.
ARBITRAGEANDGEOMETRY 3
1.4. Arbitrage. Infinance,anarbitrageopportunityisacombinationofinvestmentswhosereturn
isguaranteedtooutperformtherisk-freerate. Ifsuchanopportunityweretoexistwecouldborrow
money at the lower rate, get the higher return, repay our debt, and pocket the difference. Since
our initial investment is unlimited, so is the amount of money we could make. In modeling the
behavior of markets, a common approach is to assume that an equilibrium condition exists under
whichsuchopportunitiesarenotavailable;iftheyweretobecomeavailable,theycouldonlyexist
foranextremelybriefperiodoftime. Almostassoonastheyarediscovered,theyarewipedoutas
aresultoftheresponseofinvestors.
The no arbitrage assumption is as fundamental a principle for finance as Newton’s first and
secondlawsofmotionareforphysics. Anotherimportantprinciplebearssemanticresemblanceto
Newton’sthirdlaw: foreveryactionthereisanequalandoppositereaction. Infinance,everything
thatcan beboughtcan also besold.
A consequence of this is that for every availablepayoff column there is an investmentopportu-
nity that achieves exactly the opposite effect, that is, changes the signs of the payoffs. This may
seem counterintuitive. For example, what investment leads to a payoff that is complementary to
the purchase of a share of stock? The answer is to short sell it, that is, in essence, borrow it, sell
it, and later buy it back and subsequently return it to the lender.1 Consequently, a column in the
payoff matrix may be replaced by any nonzero multiple of the column, and the modified matrix
represents thesameinvestmentopportunities.
GivenaspecificpayoffmatrixAthatgivesthebehaviorofallpossibleinvestmentsandscenarios,
anarbitrageopportunityissaidtoexistifthereisalinearcombinationofcolumnsofAallofwhose
entries are strictly positive. Thus, an arbitrage opportunity is a combination of buys and sells of
thebasicinvestmentsthatyieldsanet gainunderall scenarios.
The absence-of-arbitrage assumption leads to constraints on payoff matrices. To illustrate this,
consider the stock/option example above. If the two columns of the matrix in Figure 2 are not
scalar multiples of each other, then they form a basis of R2 and so we can find a combination of
buys and sells of the two investments to yield any payoff vector we wish. Thus, the absence of
arbitrageimpliesthat thetwo columnsare multiplesofeach other,and thisgives
(Su−K)(1+r−d)
P= ,
(1+r)(u−d)
so that once the various parameters involved, including the interest rate, the possible factors by
which the stock price can change, the price of the stock today, and the strike price K, are set, the
priceoftheoptioncan becalculated.
2. THE ARBITRAGE THEOREM
The Arbitrage Theorem provides an interesting and convenient characterization of the no arbi-
tragecondition. Givenanm×npayoffmatrixA,andann-vectorx,theproductAxgivesthepayoff
vector that results from investing x in investment i, for i =1,...,n. Consequently, we define the
i
payoff space P(A) for a given payoff matrix A to be the column space of A, which we can view
as theset ofpayoffvectorscorrespondingto all possiblecombinationsofinvestments. This forms
asubspaceofRm ofdimensionat mostn.
Theorem 1 (Arbitrage). Given an m×n payoff matrix A, exactly one of the following statements
holds:
1
Infact, short-sellingtypicallyrequiresthatone putaside assets ascollateral, andpracticallyspeaking,investors
areconstrainedbytheiractions.
4 DANIELQ.NAIMANANDEDWARDR.SCHEINERMAN
(A1) Somepayoffvector inP(A)hasall positivecomponents,i.e. Av>0forsomev∈Rn.
(A2) There existsa probabilityvector p =[p ,...,p ]t thatisorthogonaltoeverycolumnof A,
1 m
i.e. p tA=0,where p t ≥0 andp t1=1.
In otherwords,theabsenceof arbitrageisequivalenttotheexistenceofan assignmentofprob-
abilitiesto scenariosunderwhich everyinvestmenthas an expected returnofzero.
Foraconcrete example,considerthesituationdescribed abovein which ourpayoffmatrixcon-
tainsacolumnoftheform
−1+u/(1+r)
.
−1+d/(1+r)
(cid:20) (cid:21)
Theabsenceofarbitragemeansthat thereisaprobabilityvector
p
u
p
d
(cid:20) (cid:21)
orthogonal to it and every other column of the payoff matrix. It follows immediately that p =
u
1+r−d and p =1−p = u−1−r.
u−d d u u−d
These probabilities do not have the familiar interpretation as relative frequencies. However,
the probability assignment is a practical one in that under the assumption of no arbitrage, any
investmentshouldhaveazeroexpectedpayoff. Thus,theseprobabilitiesprovideuswithamethod
for establishing the price of any security whose payoff depends on the behavior of the stock. For
example,supposewehavetheopportunitytoinvestinasecuritywhosepriceisPtodayandwhose
payoff tomorrow is r if the stock goes up and r if the stock goes down. Then we can add a
u d
columntoourpayoffmatrixdescribingthepresent valuecorrespondingtoaunitinvestment
−1+r /P(1+r)
u
.
−1+r /P(1+r).
d
(cid:20) (cid:21)
Absenceofarbitrageleadsto theconclusionthat
p (−1+r /P(1+r))=p (−1+r /P(1+r))
u u d u
and solvingforP weobtain
P=p r /(1+r)+p r /(1+r),
u u d d
thatis, theno-arbitragepriceoftheinvestmentis theexpected valueofitsdiscountedpayoff.
The Arbitrage Theorem is but one exampleof a theorem of the alternative appearing in convex
analysis in which one asserts the existence of a vector satisfying exactly one of two properties.
One of the more fundamental results of this type is Farkas’ Lemma [7], treatments of which may
be found in many books on finite-dimensional optimization, including [3, 13, 19] (see also [2]).
The Arbitrage Theorem is frequently presented as a simple consequence of Farkas’ Lemma, for
example,see[3, 15].
Lemma 2 (Farkas). Given an m×n matrix A˜ and an n-vector b˜, exactly one of the following
statementsholds:
(i) There existsx≥0 suchthatA˜x=b˜.
(ii) There existsysuchthatytA˜ ≥0 andytb˜ <0.
Equivalently,thefollowingtwo statementsareequivalent:
(F1) There existsx≥0 suchthatA˜x=b˜.
(F2) ytA˜ ≥0 impliesytb˜ ≥0.
ARBITRAGEANDGEOMETRY 5
Definition 3. Thepolyhedralconvexcone generated by a(i), i=1,...,n∈Rm is thesubsetofRm
defined by
n
(cid:229) xa(i) : x ≥0 .
i i
(i=1 )
Wedenotethissetby C(a(i),i=1,...,n)and werefer tothea(i), i=1,...,nas generatorsofthe
cone.
Farkas’ Lemma has an intuitive geometric interpretation. Let the columns of A be denoted by
a(i), i=1,...,n,and take
C :=C(a(i),i=1,...,n).
A
Condition (F1) says that b ∈C . On the other hand, assuming b 6=0, (F2) says that if y makes a
A
non-obtuse angle with every non-zero cone generator a(i) then the angle y makes with b is non-
obtuse.
We show how the Arbitrage Theorem follows from Farkas’ Lemma and, conversely, how to
prove Farkas’ Lemma from the Arbitrage Theorem. Farkas’ Lemma is central to the theory of
linearprogrammingand,inthesamespirit,[14]showshowvariouscombinatorialdualitytheorems
can bederivedfromoneanother.
2.1. Proofofthe ArbitrageTheorem using Farkas’Lemma.
Proof. First, if(A1)and(A2)are bothsatisfied,then wehave
p tAv=0v=0,
buton theotherhand Av>0 so p tAv>0 sincetheentries inp are nonnegativeand sumtoone. It
followsthat (A1)and(A2)are mutuallyexclusive.
Nextweshowthat(A1)and (A2)cannotbothfail. If(A2)fails,then thereisno solutionto
At 0
p =
1 1
(cid:20) (cid:21) (cid:20) (cid:21)
At 0
with p ≥0. It follows that condition (F1) in Farkas’ Lemma fails for A˜ = and b˜ = ,
1 1
(cid:20) (cid:21) (cid:20) (cid:21)
soFarkas’Lemmaguaranteesthat(F2)alsofails. ThismeansthatytA˜ ≥06⇒ytb˜ ≥0,thatis,there
v
exists an n+1-vector y such that ytA˜ ≥0 and ytb˜ <0. Writing y= where v =[v ,...,v ]t
s 1 n
ands∈RtheseconditionssaythatAv+s1≥0ands<0,andwecon(cid:20)clud(cid:21)ethatAv>0,so(A1)is
valid. (cid:3)
2.2. ProvingFarkas’sLemmafromtheArbitrageTheorem. Asitturnsout,theArbitrageThe-
orem together with some work leads to a proof of Farkas’ Lemma. Note first that the implication
(F1) ⇒ (F2) is immediate: if we can find x≥0 such that Ax=b, then assuming that ytA≥0 we
haveytb=ytAx≥0.
Our focus is on the more difficult implication (F2) ⇒ (F1) in Farkas’ Lemma. It is instructive
to first prove this implication under a certain technical assumption. For a polyhedral convex cone
C ⊆Rm,wedefine
L(C):=C ∩−C.
This is easily seen to form a subspace of Rm and consists of all lines contained in C passing
throughtheorigin.
6 DANIELQ.NAIMANANDEDWARDR.SCHEINERMAN
Definition4. ApolyhedralconvexconeC ispointedifL(C)={0}.Inotherwords,ifitcontains
no linesthroughtheorigin,thatis, v,−v∈C impliesv=0.
a(1)
a(2)
a(3)
0
FIGURE 3. Thevectorsa(i),i=1,2,3generateapointedconeinR3.
a(1)
a(2)
L
a(3)
0
a(4)
FIGURE 4. Thevectorsa(i),i=1,2,3,4generateanon-pointedconeinR3.
Lemma 5. TheconeC ispointedifandonlyifwhenever(cid:229) n xa(i)=0forx ≥0, i=1,...,nit
A i=1 i i
followsthatxa(i) =0 foralli.
i
Proof. If C is not pointed, then v,−v∈C for some v6=0, and we can write v=(cid:229) n l a(i) and
A i=1 i
−v=(cid:229) n w a(i) where l ,w ≥0. In addition, not all l a(i) vanish and not all w a(i) vanish. We
i=1 i i i i i
thenhave(cid:229) n x a(i)=0 wherex =l +w ≥0 and somexa(i) are nonvanishing.
i=1 i i i i i
Ontheotherhand,suppose(cid:229) n xa(i)=0wherex ≥0, i=1,...,nandsomex a(k) 6=0.Then
i=1 i i k
wecan write−x a(k) =(cid:229) n xa(i) ∈C, and x a(k) ∈C,C isnot pointed. (cid:3)
k i6=k i k
Lemma6. Theimplication(F2)⇒(F1)holdswhenAisanm×nmatrixsuchthatC isapointed
A
cone.
ARBITRAGEANDGEOMETRY 7
Proof. Observethat(F2)holdsforamatrixA,thenitalsoholdsforthematrixobtainedbyremov-
ing any zero columnsof A. Also, if (F1) holds for the matrix with the zero columns of A removed
thenitalsoholdsfortheoriginalmatrix. Itfollowsthatwithoutlossofgeneralityofwecanassume
thatall ofthecolumnsofA arenonzero.
Itfollowseasilyfrom (F2)that
At
y6>0,
−bt
(cid:20) (cid:21)
forally.Thus,condition(A1)fails,andfromtheArbitrageTheoremweconcludethat(A2)holds,
u
thatis, thereexistsaprobabilityvector withu an n-vector,and s ascalar, such that
s
(cid:20) (cid:21)
At
[ut,s] =0.
−bt
(cid:20) (cid:21)
Expanding and transposing, we obtain Au = sb. We can assume s > 0 since otherwise the as-
sumption of pointedness is violated. It follows that we can divide both sides by s and we obtain
(F1). (cid:3)
To establish the general case we need some results concerning the structure of non-pointed
polyhedralconvexcones.
Lemma 7. Given a polyhedral convex cone C =C(a(i),i=1,...,n)⊆Rm, let L =L(C) and
let L⊥ denote its orthogonal complement. Then any given x ∈ C can be expressed uniquely as
x=u+vwhere u∈C ∩L⊥,v∈L. Weexpress thisstatementsymbolicallybywriting
C =(C ∩L⊥)⊕L.
Proof. Since L ⊆C it is clear that (C ∩L⊥)+L ⊆C. On theother hand, if x∈C, there exist
uniqueu∈L⊥,andv∈L andx=u+v.Sincev∈L ⊆C wehave−v∈C,sou=x−v∈C. (cid:3)
Lemma8. ForC,L =L(C)andL⊥ asinLemma7,leta˜(i)=T (a(i))denotetheprojection
L⊥
ofa(i) ontoL⊥, fori=1,...,n.Then
C ∩L⊥ =C(a˜(i),i=1,...,n).
Proof. ByLemma7,wecanwritea˜(i)−a(i)=v∈L ⊆C,soa˜(i)=a(i)+v∈C,andwehavea˜(i)∈
C ∩L⊥,fori=1,...,n.ItfollowsthatC(a˜(i),i=1,...,n)⊆C ∩L⊥.Fortheotherinclusion,if
x∈C ∩L⊥ then we can write x=(cid:229) n l a(i) where l ≥0, i=1,...,n. Using linearitywe have
i=1 i i
x=T (x)=(cid:229) n l a˜(i), and weconcludethat x∈C(a˜(i),i=1,...,n). (cid:3)
L⊥ i=1 i
Lemma 9. IfC is apolyhedralconvexconeandL =L(C)then theconeC ∩L⊥ ispointed.
Proof. If v,−v∈C ∩L⊥ thenv∈L(C),butthen v⊥vsoweconcludethatv=0. (cid:3)
Armed with these results we are now prepared to use the Arbitrage Theorem to show (F2) ⇒
(F1).
Proofof Farkas’Lemma bywayof theArbitrageTheorem. Assume (F2) holds for a given m×n
matrix A with columns a(i),i = 1,...,n. Take C = C(a(i),i = 1,...,n), L = L(C), L⊥, and
a˜(i) = T (a(i)),i = 1,...,n as in Lemmas 7 and 8. We can write a(i) = a˜(i)+aˆ(i), and where
L⊥
aˆ(i) ∈L. Also,wecan define b˜ =T (b)and writeb=b˜+bˆ,wherebˆ ∈L.
L⊥
8 DANIELQ.NAIMANANDEDWARDR.SCHEINERMAN
a(1)
a(2)
L
C ∩L⊥
a(3)
0
a(4)
FIGURE 5. SlicingtheconeinFigure4 yieldsapointedcone.
Nowsupposey∈L⊥⊆Rm andyta˜(i)≥0fori=1,...,n.Thenyta(i)=yta˜(i)+ytaˆ(i)=yta˜(i)≥
0, for i = 1,...,n and we can conclude from (F2) that ytb ≥ 0. It follows that ytb˜ = yt(b−bˆ) =
ytb≥0.
Summarizing,wehaveshownthat
(F2′) y∈L⊥ ⊆Rm and yta˜(i) ≥0 fori=1,...,n ⇒ ytb˜ ≥0.
Since the a˜(i),i=1,...,n generate a pointed cone in L⊥ and b˜ ∈L⊥, Lemma 6 allows us to
concludethat b˜ ∈C(a˜(i),i=1,...,n),i.e. we can write b˜ =(cid:229) n xa˜(i) for somex ≥0. It follows
i=1 i i
that
n n n n
b=b˜+bˆ = (cid:229) xa˜(i)+bˆ = (cid:229) x(a(i)+aˆ(i))+bˆ = (cid:229) x a(i)+ (cid:229) xaˆ(i)+bˆ ∈C,
i i i i
i=1 i=1 i=1 (i=1 )
wherewehaveusedthefact that L ⊆C to concludethat theterm inbraces liesinC. (cid:3)
3. GEOMETRY OF GENERIC PAYOFF MATRICES
Fora givenm×n payoff matrix, theexistence ofarbitrage amounts to the statement that P(A)
intersects the positive orthant O+ := x∈Rm :x >0, j =1,...,m . Thus the probability that
j
a random payoff matrix exhibits arbitrage equals the probability that its column space meets the
positive orthant which, in turn, is the(cid:8)probability that a random n-d(cid:9)imensional subspace of Rm
intersectsO+.
This leads us to ask: How many orthants does an n-dimensional subspace of Rm intersect? For
the most part, the answer is independent of the choice of the subspace and is closely related to
other geometric counting problems. See [1], [5] pages 285–290, [6] page 73, [9], [12], [21], [22],
[23], and [24].
Lete ,e ,...,e denotethestandard orthonormalbasisofRm.
1 2 m
ARBITRAGEANDGEOMETRY 9
Definition 10. Let V be an n-dimensional subspace of Rm. We say that V is generic if for
some (and therefore for any) basis {v ,v ,...,v } of V and for any m−n standard basis vec-
1 2 n
tors e ,...,e , the vectors {v ,v ,...,v ,e ,...,e } are linearly independent. Likewise, an
i1 im−n 1 2 n i1 im−n
m×n matrixA iscalled genericifthecolumnspaceofAis agenericsubspaceofRm.
Notethatforn=mgenericitysimplysaysthatdetA6=0.Clearly,agenericmatrixhasfull-rank.
Next we present some notation for the orthants of Rm. Let P = e⊥. The subspaces P are
j j j
the coordinate hyperplanes and these separate Rm into orthants, i.e., the connected components
of Rm− P . Two vectors v and w (none of whose coordinates is zero) are in the same orthant
j
providedthesignofv equals thesignofw forall i.
i i
S
Denote by S the set of m-vectors d = (d ,...,d ) where each d is ±1. Multiplication of
m 1 m i
membersofS is defined coordinatewise. TheorthantO isdefined by
m d
Od ={x∈Rm :d ixi >0, i=1,...,m}.
ThepositiveorthantO+ issimplyO .
(1,1,...,1)
A generic subspace V of Rm intersects some subset of the orthants of Rm. Two points in V
lie in different orthants of Rm exactly when they are separated by some coordinate hyperplane(s)
P . Thus, there is a one-to-one correspondence between the orthants intersected by V and the
j
connected componentsof
m
V − P .
j
j=1
[
The intersections of V with the coordinate hyperplanes P are subspaces of V with particular
j
properties; we show that they have codimension 1 (i.e., have dimension n−1) and lie in general
position.
Definition11. SubspacesH ,...,H ofcodimension1inann-dimensionalvectorspacearesaidto
1 m
beingeneralpositionprovidedthatdim H =m−|J|forallJ⊆{1,...,m}with1≤|J|≤n.
j∈J j
For example, when n = 2, any colle(cid:0)cTtion of(cid:1)distinct lines through the origin are in general
position. When n = 3, a collection of distinct planes through the origin are in general position
providednothreeofthemintersect inaline.
For a subspace H of codimension 1 in a vector spaceV, the complement V −H consists of a
pair of half-spaces that we can label arbitrarily as H+ and H−. Given m subspaces H ,...,H , of
1 m
codimension1 whoseassociated half-spaces havebeen labeled, wecan defineafunction
m
sgn:V − H →S
i m
i=1
[
by taking
+1 ifx∈H+,
sgn(x) = i
i −1 ifx∈H−,
( i
sothat sgn(x) indicatesonwhich sideofH thepointxlies. Ford ∈S theset
i i m
m
sgn−1(d )= x∈V : x∈Hdi
i
i=1
\n o
10 DANIELQ.NAIMANANDEDWARDR.SCHEINERMAN
is an intersections of half-spaces, so it is either empty or it forms a convex polyhedron. Thus,
the subspaces H ,...,H break up the complementV − m H into connected components, each
1 m i=1 i
associatedwithsomed ∈S , whichwerefer to as cells.
m
S
Lemma 12. Let A be a generic m×n matrix and define H = P(A)∩P for i = 1,...,m. Then
i i
H ,...,H are subspaces of codimension 1 in general position in P(A). Furthermore, the con-
1 m
nected componentsofP(A)− m H correspondtotheorthantsthatP(A)intersects.
i=1 i
Proof. Let J ⊆ {1,...,m} withS1 ≤ |J| ≤ n, then |Jc| ≥ m−n so we can find distinct indices
i ,...,i ∈Jc.Thus
1 m−n
P =span{e : i∈Jc}⊇span{e ,...,e },
j i i1 im−n
j∈J
\
and consequently
P(A)+ P ⊇colspace A,e ,...,e =Rm,
j i1 im−n
j∈J
\
(cid:2) (cid:3)
sothat
dim(P(A)+ P )=m.
j
j∈J
\
It followsthat
dim(P(A)∩ H )=dim(P(A)∩ P )
j j
j∈J j∈J
\ \
=dim(P(A))+dim( P )−dim(P(A)+ P )
j j
j∈J j∈J
\ \
=dim(P(A))+(m−|J|)−m
=dim(P(A))−|J|,
sothesubspacesP(A)∩P aresubspaces inP(A)ingeneral position.
i
Thesecond claimis elementary. (cid:3)
Thus, thenumberoforthants intersected by a generic n-dimensionalsubspaceofRm equals the
numberof cells determined by m general-position, codimension-1subspaces of Rn. Our next step
isto showthat thisvalueisgivenby
m−1 m−1 m−1
Q(m,n)=2 + +···+ .
0 1 n−1
(cid:20)(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)(cid:21)
Figure6 providesasmalltableofQ(m,n)values.
Proposition13. Forpositiveintegersm and nwe have
(i) Q(m,1)=2,
(ii) Q(m,2)=2m,
(iii) form≤n, Q(m,n)=2m andso,in particular,Q(1,n)=2,and
(iv) form,n≥2, Q(m,n)=Q(m−1,n)+Q(m−1,n−1).