Yuval Peres
Lecture notes edited by B
alint Vir
ag and Elchanan Mossel
Preface. These notes record lectures I gave at the Statistics Department, University of
California,BerkeleyinSpring1998. Iamgratefulto thestudentswhoattendedthecourse
and wrotetherstdraft ofthenotes: DiegoGarcia,Yoram Gat,Diogo A.Gomes,Charles
Holton, Frederic Latremoliere, Wei Li, Ben Morris, Jason Schweinsberg, Balint Virag, Ye
XiaandXiaowenZhou. ThedraftwaseditedbyBalintViragandElchananMossel. Ithank
Pertti Mattila for the invitation to lecture on thismaterial at the joint summerschool in
Chapter 1. Brownian Motion 1
1. Motivation {Intersectionof Brownianpaths 1
2. Gaussianrandomvariables 1
3. Levy's construction of Brownian motion 4
4. Basic properties ofBrownianmotion 6
5. Hausdor dimensionand Minkowskidimension 11
6. Hausdor dimensionof theBrownian pathand theBrowniangraph 13
7. Onnowheredierentiability 17
8. Strong Markovpropertyand thereectionprinciple 18
9. Localextrema ofBrownianmotion 19
10. Area ofplanarBrownianmotion paths 20
11. Zeros of theBrownian motion 21
12. Harris'Inequalityand itsconsequences 26
13. Pointsof increaseforrandomwalksand Brownian motion 29
14. Frostman's Lemma, energy,and dimensiondoubling 33
15. Skorokhod'srepresentation 38
16. Donsker'sInvariancePrinciple 43
17. Harmonicfunctions and Brownian motioninR d
46
18. Maximumprincipleforharmonicfunctions 49
19. The Dirichletproblem 50
20. Polar pointsand recurrence 51
21. Capacityand harmonicfunctions 53
22. Kaufman's theoremon uniformdimension doubling 56
23. Packingdimension 58
24. Hausdor dimensionand randomsets 59
25. An extensionof Levy's modulusof continuity 60
Brownian Motion
1. Motivation { Intersection of Brownian paths
ConsideranumberofBrownianmotionpathsstartedat dierentpoints. Saythatthey
intersect if there is a point which lies on all of the paths. Do the paths intersect? The
answerto thisquestiondepends onthedimension:
InR 2
,anynitenumberofpathsintersectwithpositiveprobability(thisisatheorem
of Dvoretsky,Erd}os,Kakutani inthe1950's),
InR 3
,two pathsintersect withpositiveprobability,butnotthree(thisis atheorem
of Dvoretsky,Erd}os,Kakutani andTaylor),
In R d
for d4,no pairof paths intersect withpositiveprobability.
The principle we willuse to establish these results is intersection equivalencebetween
BrownianmotionandcertainrandomCantor-typesets. Herewewillintroducetheconcept
for R 3
only. Partition the cube [0;1] 3
in eight congruent sub-cubes, and keep each of the
sub-cubeswithprobability 1
2
. Foreachcubethatremainedattheendofthisstage,partition
it into eight sub-cubes, keeping each of them with probability 1
2
, and so on. Let Q 3; 1
2
denote thelimitingset| that is,theintersection of thecubesremaining at allsteps. This
set is not empty with positive probability, since, if we consider that the remaining
sub-cubes of a given cube as its \children"in a branching process, then the expected number
of ospringsis four,sothisprocesshaspositiveprobabilitynotto dieout.
Onecanprove that,thereexiststwopositiveconstantsC
1 ;C
2
suchthat,ifisaclosed
subset of [0;1] 3
, and fB
t
g is a Brownian motion started at a point uniformly chosen in
[0;1] 3
,then:
C
1 P
Q
3; 1
2
\6=;
P( 9t0 B
t
2)C
2 P
Q
3; 1
2
\6=;
The motivation is that, though the intersection of two independent Brownian paths is
a complicatedobject, theintersection of two sets of theform Q 3; 1
2
is a set of the same
kind|namely,Q 3; 1
4
. The previouslydescribedbranchingprocessdiesoutassoon aswe
intersect more than two of these Cantor-type sets|hence the result about intersection of
paths inR 3
.
2. Gaussian random variables
Brownianmotionisat themeetingpointofthemostimportant categoriesofstochastic
processes: it is a martingale, a strong Markov process, a process with independent and
stationary increments, and a Gaussian process. We will construct Brownian motion as a
Definition 2.1. Areal-valuedrandomvariableXonaprobabilityspace(;F;P)has
a standard Gaussian (orstandard normal) distributionif
P(X>x)= 1
p
2 Z
+1
x e
u 2
=2
du
A vector-valued random variable X has an n-dimensional standard Gaussian
dis-tributionifits ncoordinatesare standardGaussianand independent.
A vector-valued random variable Y : ! R p
is Gaussian if there exists a
vector-valuedrandomvariableXhavingann-dimensionalstandardGaussiandistribution,apn
matrixA and ap-dimensionalvector b such that:
Y =AX+b (2.1)
We arenowreadyto denetheGaussianprocesses.
Definition 2.2. A stochasticprocess (X
t )
t2I
issaid to bea Gaussian processiffor
all k and t
1 ;:::;t
k
2I thevector (X
t1
;:::;X
t
k )
t
is Gaussian.
Recallthat thecovariance matrixof a randomvector isdened as
Cov(Y)=E
(Y EY)(Y EY) t
Then,bythelinearityofexpectation,theGaussian vector Y in(2.1) has
Cov(Y)=AA t
:
Recallthatan nnmatrixAis saidto beorthogonalifAA t
=I
n
. Thefollowinglemma
shows thatthedistributionofaGaussianvector isdeterminedbyits meanand covariance.
Lemma 2.3.
(i) If is an orthogonal nn matrix and X is an n-dimensional standard Gaussian
vector, then X is also an n-dimensionalstandard Gaussian vector.
(ii)If Y andZ areGaussianvectorsin R n
such that EY =EZ andCov(Y)=Cov(Z),
then Y and Z have the same distribution.
Proof.
(i) As the coordinates of X are independent standard Gaussian, X has density given
by:
f(x)=(2) n
2
e kxk=2
where k k denotes Euclidean norm. Since preserves this norm, the density of X is
invariant under.
(ii)ItissuÆcienttoconsiderthecasewhenEY =EZ =0. Then,usingDenition2.1),
there exist standardGaussianvectors X
1 ,X
2
andmatrices A;C sothat
Y =AX
1
and Z =CX
2 :
By adding some columnsof zeroesto A or C if necessary, we can assume that X
1 , X
2 are
bothk-vectors forsome k and A,C arebothnk matrices.
form a basis for the space A. For any matrix M let M
i
denote the ith row vector of M,
and denethelinear mapfrom A to C by
A
i =C
i
fori=1:::`.
Wewanttocheckthatisanisomorphism. Assumethatthereisavectorv
1 A
1
+:::+v
` A
`
whose imageis 0. Then the k-vector v =(v
1 ;v
2 ;:::;v
`
;0;:::;0) satises v t
C =0,and so
kv t
Ak 2
= v t
AA t
v = v t
CC t
v = 0, giving v t
A = 0. This shows that is one-to-one, in
particular dimA dimC. By symmetry A and C must have the same dimension, so is
an isomorphism.
AsthecoeÆcient(i;j)ofthematrixAA t
isthescalarproductofA
i andA
j
,theidentity
AA t
=CC t
impliesthat is an orthogonal transformation from A to C. We can extend
itto maptheorthocomplement ofA totheorthocomplementof Corthogonally,getting an
orthogonal map:R k
!R k
. Then
Y =AX
1
; Z =CX
2
=AX
2 ;
and (ii)follows from (i).
Thus,thersttwomomentsofaGaussianvectoraresuÆcienttocharacterizeits
distri-bution,hencetheintroductionofthenotationN(;)todesignatethenormaldistribution
withexpectation and covariance matrix. Ausefulcorollary of thislemmais:
Corollary 2.4. Let Z
1 ;Z
2
beindependent N(0; 2
) random variables. Then Z
1 +Z
2
and Z
1 Z
2
aretwo independent random variables havingthe same distributionN(0;2 2
).
Proof. 1
(Z
1 ;Z
2
) isa standardGaussianvector,and so,if:
= 1
p
2
1 1
1 1
thenis an orthogonal matrixsuch that
( p
2) 1
(Z
1 +Z
2 ;Z
1 Z
2 )
t
=
1
(Z
1 ;Z
2 )
t
;
and ourclaim followsfrom part (i)of theLemma.
As a conclusion of this section, we state the following tail estimate for the standard
Gaussiandistribution:
Lemma 2.5. LetZ be distributed as N(0;1). Then for all x0:
x
x 2
+1 1
p
2 e
x 2
=2
P( Z >x) 1
x 1
p
2 e
x 2
=2
Proof. Theright inequalityisobtained bythe estimate:
P(Z >x) Z
+1
x u
x 1
p
2 e
u 2
=2
du
since, intheintegral, ux. The left inequalityisproved asfollows: Let usdene
f(x):=xe x
2
=2
(x 2
+1) Z
+1
e u
2
=2
We remarkthat f(0)<0 and lim
x!+1
f(x)=0:Moreover,
f 0
(x) = (1 x 2
+x 2
+1)e x
2
=2
2x Z
+1
x e
u 2
=2
du
= 2x
Z
+1
x e
u 2
=2
du 1
x e
x 2
=2
;
so theright inequality impliesf 0
(x)0 for all x 0. This impliesf(x)0, provingthe
lemma.
3. Levy's construction of Brownian motion
3.1. Denition. StandardBrownian motion on an intervalI =[0;a] orI =[0;1) is
denedbythefollowingproperties:
Definition 3.1. A stochastic process fB
t g
t2I
isa standard Brownian motionifit
isa Gaussianprocess suchthat:
(i) B
0 =0,
(ii) 8k naturaland 8t
1
<:::<t
k
in I: B
t
k B
t
k 1
;:::;B
t
2 B
t
1
areindependent,
(iii) 8t;s2I with t<s B
s B
t
hasN(0;s t) distribution.
(iv) Almost surely,t7!B
t
is continuouson I.
Asa corollaryof thisdenition,one can already remarkthat forallt;s2I:
Cov(B
t ;B
s
)=s^t:
Indeed, assume that t s. Then Cov(B
t ;B
s
) = Cov (B
t B
s ;B
s
) +Cov(B
s ;B
s ) by
bilinearityof the covariance. The rst term vanishes by the independence of increments,
andthesecondtermequalssbyproperties(iii)and(i). ThusbyLemma2.3wemayreplace
properties (ii)and (iii)inthedenitionby:
Forall t;s2I, Cov (B
t ;B
s
)=t^s.
Forall t2I,B
t
hasN(0;t) distribution.
orby:
Forall t;s2I with t<s, B
t B
s and B
s
areindependent.
Forall t2I,B
t
hasN(0;t) distribution.
Kolmogorov's extension theorem implies the existence of any countable time set
sto-chasticprocessfX
t
gifweknowitsnite-dimensionaldistributionsandtheyareconsistent.
Thus, standard Brownian motion could be easily constructed on any countable time set.
HoweverknowingnitedimensionaldistributionsisnotsuÆcienttogetcontinuouspaths,
asthefollowingexampleshows.
Example 3.2. Suppose that standard Brownian motion fB
t
g on [0;1] has been
con-structed, and consider an independent random variable U uniformlydistributed on [0;1].
Dene:
~
B
t =
B
t
ift6=U
0 otherwise
Thenite-dimensionaldistributionsoff ~
B
t
garethesameastheonesoffB
t
g. However,the
process f ~
3. LEVY'SCONSTRUCTION OFBROWNIANMOTION 5
Inmeasuretheory,oneoftenidentiesfunctionswiththeirequivalenceclassfor
almost-everywhere equality. As the above example shows, it is important not to make this
iden-tication in the study of continuous-time stochastic processes. Here we want to dene a
probabilitymeasureon theset ofcontinuous functions.
3.2. Construction. Thefollowingconstruction,duetoPaulLevy,consistofchoosing
the \right" values for the Brownian motion at each dyadic point of [0;1] and then
inter-polating linearly between these values. This construction is inductive, and, at each step,
a process is constructed, that hascontinuous paths. Brownian motion isthen theuniform
limitof these processes|hence its continuity. We willusethe followingbasiclemma. The
proof can be found,forinstance, inDurrett (1995).
Lemma 3.3 (Borel-Cantelli). Let fA
i g
i=0:::1
be a sequence of events, andlet
fA
i
i.o. g=limsup
i!1 A
i =
1
\
i=0 1
[
j=i A
j ;
where\i.o." abbreviates \innitely often".
(i) If P
1
i=0 P(A
i
)<1, then P(A
i
i.o. )=0.
(ii) If fA
i
g are pairwise independent, and P
1
i=0 P(A
i
)=1, then P(A
i
i.o.)=1.
Theorem 3.4 (Wiener1923). Standard Brownian motion on [0;1) exists.
Proof. (Levy 1948)
We rstconstructstandardBrownianmotionon[0;1]. Forn0,letD
n
=fk=2 n
:0
k 2 n
g, and let D = S
D
n
. Let fZ
d g
d2D
be a collection of independent N(0;1) random
variables. We willrst construct the valuesof B on D. Set B
0
=0, and B
1 = Z
1
. In an
inductive construction,foreach nwe willconstruct B
d
foralld2D
n
sothat
(i) Forallr <s<t inD
n
,theincrement B
t B
s
hasN(0;t s)distributionand is
independentofB
s B
r .
(ii) B
d
ford2D
n
areglobally independent oftheZ
d
ford2DnD
n .
These assertions hold for n = 0. Suppose that they hold for n 1. Dene, for all
d2D
n nD
n 1
arandomvariable B
d by
B
d =
B
d +B
d +
2
+ Z
d
2 (n+1)=2
(3.1)
where d +
= d+2 n
, and d = d 2 n
, and both are in D
n 1
. Since 1
2 [B
d
+ B
d ] is
N(0;1=2 n+1
)byinduction,and Z
d =2
(n+1)=2
isanindependentN(0;1=2 n+1
),theirsumand
theirdierence,B
d B
d
andB
d+ B
d
arebothN(0;1=2 n
)andindependentbyCorollary
2.4. Assertion(i) follows fromthisand theinductivehypothesis,and (ii)is clear.
HavingthuschosenthevaluesoftheprocessonD,wenow\interpolate"betweenthem.
Formally,let F
0
(x)=xZ
1
,and forn1,let let usintroducethefunction:
F
n (x)=
8
<
: 2
(n+1)=2
Z
x
forx2D
n nD
n 1 ;
0 forx2D
n 1 ;
linear betweenconsecutive pointsinD :
Thesefunctions arecontinuouson [0;1], andforall n andd2D n B d = n X i=0 F i (d)= 1 X i=0 F i (d): (3.3)
Thiscan be seenbyinduction. Supposethat itholdsforn 1. Letd2D
n D
n 1 . Since
for0in 1 F
i
islinear on [d ;d +
],weget
n 1 X i=0 F i (d)= n 1 X i=1 F i
(d )+F
i (d + ) 2 = B d +B d + 2 : (3.4) Since F n
(d)=2
(n+1)=2
Z
d
,comparing(3.1) and (3.4) gives(3.3).
Ontheother hand, we have,bydenitionof Z
d
and byLemma 2.5:
P jZ
d jc
p n exp c 2 n 2
so theseries P 1 n=0 P d2Dn P(jZ d jc
p
n) converges assoon asc> p
2log2. Fix such a c.
By the Borel-Cantelli Lemma 3.3 we concludethat there exists a random but niteN so
thatforall n>N and d2D
n
we have jZ
d j<c
p
n,and so:
kF n k 1 <c p n2 n=2 : (3.5)
This upper bound implies that the series P
1
n=0 F
n
(t) is uniformly convergent on [0;1],
and so it has a continuous limit, which we call fB
t
g. All we have to check is that the
increments of this process have the right nite-dimensional joint distributions. This is a
directconsequenceof thedensityof thesetDin[0;1] andthecontinuityof paths. Indeed,
let t 1 > t 2 > t 3
be in [0;1], then they are limits of sequences t
1;n , t 2;n and t 3;n in D,
respectively. Now
B t 3 B t 2 = lim k!1 (B t 3;k B t 2;k )
is a limit of Gaussian random variables, so itself is Gaussian with mean 0 and v
ari-ance lim n!1 ( t 3;n t 2;n ) = t
3 t
2
. The same holds for B
t2 B
t1
, moreover, these two
random variables are limit of independent random variables, since for n large enough,
t 1;n > t 2;n > t 3;n
. Applying this argument for any number of increments, we get that
fB
t
ghasindependentincrementssuchthatandforalls<tin[0;1]B
t B
s
hasN(0;t s)
distribution.
We have thusconstructed Brownian motion on [0;1]. To conclude, if fB n
t g
t
forn 0
areindependentBrownianmotionson [0;1], then
B t =B btc t btc + X 0i<btc B i 1
meetsour denitionof Brownianmotionon [0;1).
4. Basic properties of Brownian motion
Let fB(t)g
t0
be a standard Brownian motion, and let a6= 0. The following scaling
relation isa simpleconsequenceof thedenitions.
Also,dene thetime inversion of fB
t gas
W(t)=
0 t=0;
tB( 1
t
) t>0.
We claim thatW isa standardBrownian motion. Indeed,
Cov(W(t);W(s))=tsCov(B( 1
t ;
1
s
))=ts( 1
t ^
1
s
)=t^s;
so W and B have the same nite dimensional distributions, and they have the same
dis-tributions as processes on the rational numbers. Since the paths of W(t) are continuous
exceptmaybe at 0,we have
lim
t#0
W(t)= lim
t#0;t2Q
W(t)=0 a.s.
sothepaths of W(t)are continuous on[0;1) a.s. As acorollary,weget
Corollary 4.1. [Law of Large Numbers for Brownian motion]
lim
t!1 B(t)
t
=0 a:s:
Proof. lim
t!1 B(t)
t
=lim
t!1 W(
1
t
)=0a.s.
Exercise 4.2. Provethisresultdirectly. UsetheusualLawofLargeNumbersto show
that lim
n!1 B(n)
n
= 0. Then show that B(t) does not oscillate too much between n and
n+1.
Remark. Thesymmetryinherentinthetimeinversionpropertybecomesmore
appar-ent ifone considerstheOrnstein-Uhlenbeckdiusion,whichis given by
X(t)=e t
B(e 2t
)
This is a stationary Markov chain with X(t) N(0;1) for all t. It is a diusion with
a drift toward the origin proportional to the distance from the origin. Unlike Brownian
motion, the Ornstein-Uhlenbeck diusion is time reversible. The time inversion formula
givesfX(t)g
t0 d
=fX( t)g
t0
. Fort near 1, X(t) relates to theBrownianmotion near
0,and fortnear 1,X(t)relates to theBrownianmotion near1.
OneoftheadvantagesofLevy's constructionofBrownianmotionisthatiteasilyyields
a modulusofcontinuityresult. FollowingLevy,we dened Brownianmotion asan innite
sum P
1
n=0 F
n
, where each F
n
is a piecewise linear function given in (3.2). Its derivative
existsalmost everywhere, andbydenitionand (3.5)
kF 0
n k
1
kF
n k
1
2 n
C
1 (!)+
p
n2 n=2
(4.1)
TherandomconstantC
1
(!)isintroducedtodealwiththenitelymanyexceptionsto(3.5).
Nowfort;t+h2[0;1], we have
jB(t+h) B(t)j X
n jF
n
(t+h) F
n (t)j
X
hkF 0
n k
1 +
X
2kF
n k
1
By (3.5) and (4.1) if`>N forarandomN,then theabove is boundedby
h(C
1 (!)+
X
n` c
p
n2 n=2
)+2 X
n>` c
p
n2 n=2
C
2 (!)h
p
`2 `=2
+C
3 (!)
p
`2 `=2
Theinequalityholdsbecauseeachseriesisboundedbyaconstanttimesitsdominantterm.
Choosing`=blog
2
(1=h)c, andchoosingC(!)to take careof thecases when`N,weget
jB(t+h) B(t)jC(!) r
hlog
2 1
h
: (4.3)
The result is a (weak) form of Levy's modulusof continuity. This is not enough to make
fB
t
g adierentiablefunctionsince p
hh forsmallh. But we stillhave
Corollary 4.3. For every < 1
2
, Brownian paths are - Holder a.s.
Exercise 4.4. Show thata Brownian motionisa.s. not 1
2
- Holder.
Remark. Theredoesexistat=t(!)suchthatjB(t+h) B(t)jC(!)h 1
2
foreveryh,
a.s. However, such thavemeasure 0. Thisistheslowestmovement that islocallypossible.
Solution. LetA
k ;n
betheeventthatB((k+1)2 n
) B(k2 n
)>c p
n2 n=2
. Then
P(A
k ;n
)=P(B(1)>c p
n ) c
p
n
c 2
n+1 e
c 2
n=2
:
If0<c< p
2log(2)thenc 2
=2<log(2)and2 n
P(A
k ;n
)!1. Therefore,
P( 2
n
\
k =1 A
c
k ;n
)=[1 P(A
k ;n )]
2 n
e 2
n
P(A
k ;n )
!0:
Thelastinequalitycomesfrom thefact that1 xe x
for all x. Byconsideringh=2 n
;one
canseethat
P
8h B(t+h) B(t)c p
hlog
2 h
1
=0 ifc< p
2log2 :
We remark that the paths are a.s. not 1
2
-Holder. Indeed, the log
2 (
1
h
) factor in (4.3) cannot be
ignored. WewillseelaterthattheHausdordimensionofthegraphofBrownianmotionis 3
2 a.s.
Having proven that Brownian paths are somewhat \regular", let us see why they are
\bizarre". One reason is that the paths of Brownian motion have no intervals of
mono-tonicity. Indeed, if [a;b] is an interval of monotonicity, then dividing it up into n equal
sub-intervals[a
i ;a
i+1
]eachincrement B(a
i
) B(a
i+1
) hasto have thesamesign. Thishas
probability22 n
, and taking n! 1 shows that theprobabilitythat [a;b]is an interval
of monotonicity must be 0. Taking a countable union gives that there is no interval of
monotonicitywith rationalendpoints, buteach monotoneinterval wouldhave a monotone
rationalsub-interval.
We willnowshowthatforanytimet
0
,Brownianmotionisnotdierentiableatt
0 . For
this, weneed asimpleproposition.
Proposition 4.5. A.s.
limsup B(n)
p
n
=+1; liminf
n!1 B(n)
p
n
Remark. Comparing thiswithCorollary 4.1, it isnatural to ask what sequence B(n)
shouldbe dividedby to geta limsupwhich isgreater than 0but lessthan1. An answer
isgiven bytheLawof theiterated logarithm inalater section.
The proof of (4.4) relieson the followingstandard fact, whose proof can be found, for
example,inDurrett (1995). Consideraprobabilitymeasureon thespace ofreal sequences,
andlet X
1 ;X
2
;::: bethesequenceof randomvariablesitdenes. An event,thatisa
mea-surablesetofsequences,AisexchangeableifX
1 ;X
2
;::: satisfyAimpliesthatX
1 ;X
2 ;:::
satisfy A for all nite permutations . Here nite permutation means that
n
=n for all
suÆcientlylargen.
Proposition 4.6 (Hewitt-Savage 0-1 Law). If A is an exchangeableevent for an i.i.d.
sequencethen P(A) is0 or 1.
Proof of Proposition 4.5.
P(B(n)>c p
n i.o. )limsup
n!1
P(B(n)>c p
n)
Bythescalingproperty,theexpressioninthelimsupequalsP(B(1)>c),whichispositive.
LettingX
n
=B(n) B(n 1)theHewitt-Savage 0-1lawgivesthatB(n)>c p
ninnitely
often. Takingtheintersectionoverallnaturalcgivestherstpart of(4.4),and thesecond
isproved similarly.
ThetwoclaimsofProposition4.5togethermeanthatB(t)crosses0forarbitrarilylarge
values of t. If we use time inversion W(t)=tB( 1
t
), we get that Brownian motion crosses
0 for arbitrarily small values of t. Letting Z
B
= ft : B(t) = 0g, this means that 0 is an
accumulation pointfrom theright forZ
B
. Butweget even more. Fora functionf,dene
theupperand lower right derivatives
D
f(t) = limsup
h#0
f(t+h) f(t)
h
;
D
f(t) = liminf
h#0
f(t+h) f(t)
h
:
Then
D
W(0)limsup W(
1
n
) W(0)
( 1
n )
limsup p
nW( 1
n
)=limsup B(n)
p
n
which is innite by Proposition 4.5. Similarly, D
W(0) = 1, showing that W is not
dierentiableat 0.
Corollary 4.7. Fix t
0
0. Brownian motion W a.s. satises D
W(t
0
) = +1,
D
W(t
0
) = 1, and t
0
is an accumulation point from the right for the level set fs :
W(s)=W(t
0 )g.
Proof. t!W(t
0
+t) W(t
0
)is a standardBrownian motion.
Does this implythat each t
0
is an accumulation point from the right for the level set
fs:W(s)=W(t
0
)ga.s.? Certainlynot;consider,forexamplethelast0offB
t
gbeforetime
t
0
-smusthave measure0. This istrue ingeneral. SupposeAis ameasurable event (set of
paths)such that
8t
0
P(t!W(t
0
+t) W(t
0
)satisesA)=1:
Let
t
be the operator that shifts paths by t. Then P( T
t
0 2Q
t0
(A)) = 1: In fact, the
Lebesguemeasureofpointst
0
sothatW doesnotsatisfy
t0
(A)is0a.s. Toseethis,apply
Fubinito thedoubleintegral
Z Z
1
0
1(W 2=
t
0
(A))dt
0
dP(W)
Exercise 4.8. Showthat8t
0 P(t
0
isalocalmaximumforB)=0;buta.s.localmaxima
area countable dense setin(0;1).
NowheredierentiabilityofBrownianmotionthereforerequiresamorecarefulargument
thannon-dierentiabilityat axed point.
Theorem 4.9 (Paley,Wiener,Zygmund1933). A.s. Brownian motion is nowhere
dif-ferentiable. Furthermore, almost surely for all t either D
B(t)=+1 or D
B(t)= 1.
Remark. For local maxima we have D
B(t) 0, and for local minima, D
B(t)0,
soit isimportantto have theeither-or inthestatement.
Proof. (Dvoretsky, Erd}os, Kakutani 1961) Suppose that there is a t
0
2 [0;1] such
that 1<D
B(t
0 )D
B(t
0
)<1. Then forsome M we wouldhave
sup
h2[0;1] jB(t
0
+h) B(t
0 )j
h
M: (4.5)
If t
0
is containedinthe binary interval[(k 1)=2 n
;k=2 n
]forn>2, thenforall 1j n
thetriangleinequalitygives
jB((k+j)=2 n
) B((k+j 1)=2 n
) jM(2j+1)=2 n
: (4.6)
Let
n;k
be theevent that (4.6) holdsfor j=1,2,and 3. Then bythescaling property
P(
n;k )P
jB(1)j7M= p
2 n
3
;
which isat most(7M2 n=2
) 3
,sincethe normaldensity islessthan1/2. Hence
P 2
n
[
k=1
n;k !
2 n
(7M2 n=2
) 3
=(7M) 3
2 n=2
;
whosethe sumoverall nis nite. By theBorel-Cantellilemma:
P( (4.5)is satisfed)P 2
n
[
k=1
n;k
forinnitelymanyn !
=0:
That is,forsuÆcientlylargen, there arenothree goodincrementsinarow so(4.5) isnot
satised.
Exercise 4.10. Let>1=2. Showthata.s.forallt>0;9h>0suchthatjB(t+h)
Solution. Supposethatthereisat
0
2[0;1]suchthat
sup
h2[0;1]
B(t0+h) B(t0)
h
1; and
inf
h2[0;1] B(t
0
+h) B(t
0 )
h
1
Ift
0 2
k 1
2 n
; k
2 n
for n>2,thenthetriangleinequalitygives
jB( k+j
2 n
) B(
k+j 1
2 n
)j2( j+1
2 n
)
:
Fixl1=( 1
2
)andletn;k betheevent
jB( k +j
2 n
) B( k +j 1
2 n
)j2 h
(j+1)
2 n
i
for j=1,2 ::: l
Then
P(
n;k
)[P(jB(1)j2 n=2
2( l+1
2 n
)
] l
[2 n=2
2( l+1
2 n
)
] l
sincethenormaldensityislessthan1/2. Hence
P( 2
n
[
k =1
n;k )2
n
[2 n=2
2( l+1
2 n
)
] l
=C[2
(1 l( 1=2))
] n
;
whichsums. Thus
P(limsup
n!1 2
n
[
k =1
n;k)=0:
Exercise 4.11. (Hard.) A.s.ifB(t
0
)=max
0t1
B(t) thenD
B(t
0
)= 1.
5. Hausdor dimension and Minkowski dimension
Definition 5.1 (Hausdor1918). LetAbeasubsetofametricspace. For >0,and
a possiblyinnite >0 denea measure ofsizeforA by
H
()
(A)=inff X
j jA
j j
:A [
j A
j ;jA
j j<g;
wherejjappliedtoasetmeansitsdiameter. ThequantityH
(1)
(A)iscalledtheHausdor
content of A.Dene the-dimensionalHausdor measureof Aas
H
(A)=lim
#0 H
()
(A)=sup
>0 H
() (A):
H
() is a Borel measure. Sub-additivity is obvious since H
()
(A[D) H
() (A)+
H
()
(D). Countable additivitycan beshownwithmore eort. The graphof H
(A)versus
shows that there is a critical value of where H
(A) jumps from 1 to 0. This critical
valueis calledtheHausdor dimensionof A.
Definition 5.2. The Hausdor dimension ofA is
dim (A)=inff: H
Notethat given >>0and H
()
(A)<1,we have
H
()
(A) ( )
H
() (A):
where the inequality follows from the fact that jA
j j
( )
jA
j j
for a covering set A
j
of diameter . Since is arbitrary here, we see that if H
(A) < 1, then H
(A) = 0:
Therefore, we can also denetheHausdor dimensionof Aas
supf:H
(A)=1g:
Nowlet uslookat anotherkindof dimensiondened asfollows.
Definition 5.3. The upper and lower Minkowski dimension (also called\Box"
dimension)ofA is denedas
dim
M
(A) = lim
#0 logN
(A)
log (1=) ;
dim
M
(A) = lim
#0 logN
(A)
log (1=) ;
whereN
(A)istheminimumnumberofballsofradiusneededto coverA. Ifdim
M (A)=
dim
M
(A), we callthecommonvaluetheMinkowski dimensionofA,denoted bydim
M .
Exercise 5.4. Show thatdim
M ([0;1]
d
)=d.
Proposition 5.5. For any set A, we have dim
H
(A)dim
M (A).
Proof. Bydenition,H
(2) N
(A)(2)
. If >dim
M
(A),thenwehavelimN
(A)(2)
=
0. It thenfollows thatdim
H
(A)dim
M (A).
We nowlookat some examplestoillustratethedenitionsabove. LetA beall therational
numbersintheunit interval. Since forevery set Dwe have dim
M
(D)=dim
M
(D),we get
dim
M
(A) = dim
M
(A ) =dim
M
([0;1]) = 1. On the other hand, a countable set A always
hasdim
H
(A)=0. Thisisbecause forany;>0,wecan coverthe countable pointsinA
byballsof radius=2, =4, =8, and soon. Then
H
() (A)
X
j jA
j j
X
j (2
j
)
<1:
Anotherexampleistheharmonicsequence A=f0g S
f1=ng
n1
. Itis acountable set, soit
hasHausdordimension0. Itcan be shownthatdim
M
(A)=1=2. (ThosepointsinAthat
arelessthan p
can becovered by1= p
ballsofradius. Therest canbecoveredby2= p
ballsof radius, one oneach point.)
We haveshown inCorollary 4.3that Brownianmotionis -Holderforany<1=2a.s.
This will allow us to infer an upper bound on the Hausdor dimension of its image and
graph. Denethe graphG
f
ofa functionastheset ofpoints(t;f(t))astrangesoverthe
domainof f.
Proposition 5.6. Let f :[0;1]!R bea -Holder continuous function. Then
Proof. Since f is -Holder, there exists a constant C
1
such that, if jt sj , then
jf(t) f(s)jC
1
=C
1 (1=
1
). Hence, theminimumnumberof ballsof radiusto
coverG
f
satisesN
(G
f )C
2 (1=
2
)forsome other constant C
2 .
Corollary 5.7.
dim
H (G
B
)dim
M (G
B
)dim
M (G
B
)3=2: a.s.
A functionf :[0;1]! R is called\reverse" -Holderifthere exists aconstant C>0 such
that foranyinterval [t;s], there is asubinterval[t
1 ;s
1
][t;s],suchthat jf(t
1
) f(s
1 )j
Cjt sj
.
Proposition 5.8. Let f : [0;1] ! R be -Holder and \reverse" -Holder. Then
dim
M (G
f
)=2 .
Remark. It followsfrom thehypotheses oftheabovepropositionthatsuchafunction
f hasthe propertydim
H (G
f
)>1:(Przytycki-Urbansky,1989.)
Exercise 5.9. Prove Proposition5.8.
Example 5.10. TheWeierstrassnowheredierentiablefunction
W(t)= 1
X
n=0 a
n
cos(b n
t);
ab>1, 0<a<1 is -Holder and \reverse" -Holderforsome 0< <1. For example,if
a=1=2, b=4,then =1=2.
Lemma 5.11. Suppose X isa complete metric space. Let f :X !Y be -Holder. For
any A X, we have dim
H
f(A) dim
H
(A)=: Similar statements for dim
M
and dim
M
are also true.
Proof. Ifdim
H
(A)< <1,thenthereexistsacoverfA
j
gsuchthatA S
j A
j and
P
j jA
j j
< . Then ff(A
j
)g is a cover for f(A), and jf(A
j
)j CjA
j j
, where C is the
constant from the-Holdercondition. Thus,
X
j jf(A
j )j
=
C =
X
j jA
j j
<C =
!0
as!0,and hence dim
H
f(A)=:
Corollary 5.12. For A[0;1), we have dim
H
B(A)2dim
H
(A)^1 a:s:
6. Hausdor dimension of the Brownian path and the Brownian graph
The nowhere dierentiability of Brownian motion established in the previous section
suggests that its graph has dimension higher than one. Recall that the graph G
f of a
function f is the set of points (x;f(x)) where x ranges over the set where f is dened.
Taylor(1953) showed thatthegraph ofBrownianmotion hasHausdor dimension3=2.
Denethed-dimensionalstandardBrownianmotionwhosecoordinatesareindependent
one dimensional standardBrownian motions. This measure is invariant under orthogonal
transformations of R d
transforma-We willseethat planarBrownian motionisneighborhoodrecurrent,that is,itvisitsevery
neighborhood in the plane innitely often. In this sense, the image of planar Brownian
motion is comparable to the plane itself; another sense in which this happens is that of
Hausdor dimension: the image of planar and higher dimensional Brownian motion has
Hausdor dimensiontwo. Summingup,we willprove
Theorem 6.1 (Taylor 1953). Let B be d dimensional Brownian motion dened on
the time set [0;1]. Then
dim
H G
B
=3=2 a.s.
Moreover, if d2, then
dim
H
B[0;1]=2 a.s.
HigherdimensionalBrownianmotion therefore doublesthedimension ofthe timeline.
Naturally, the question arises whether this holds for subsets of the time line as well. In
certain sense,thiseven holdsford=1: notethe\^d"inthe followingtheorem.
Theorem 6.2 (McKean1955). For every subset A of [0;1), the image of A under d
dimensional Brownian motionhas Hausdor dimension 2dim
H
A^d a.s.
Theorem 6.3 (UniformDimensionDoubling(Kaufman 1969)).
Let B beBrownian motionin dimensionat least 2 A.s, for any A[0;1), we have
dim
H
B(A)=2dim
H (A).
Notice the dierence between the last two results. Kaufman's theorem has a much
stronger claim: itstatesdimensiondoublinguniformlyforallsets. Forthistheorem,d 2
is a necessary condition: we will see later that the zero set of one dimensional Brownian
motion hasdimension half,whileitsimage is thesingle point 0. We willprove Kaufman's
theorem inalater section. ForTheorem(6.1) weneedthe followingfact.
Theorem 6.4 (Mass DistributionPrinciple). If A X supports a positive Borel
mea-suresuch that (D)CjDj
for anyBorelsetD,then H
1 (A)
(A)
C
. Thisimpliesthat
H
(A) (A)
C
, and hencedim
H
(A):
Proof. IfA S
j A
j ,then
P
j jA
j j
C 1
P
j (A
j )C
1
(A):
Example 6.5. Consider themiddlethirdCantor setK
1=3
. We willseethat
dim
M (K
1=3
)=dim
H (K
1=3
)=log2=log3:
At thenth levelof theconstruction of K
1=3
,wehave 2 n
intervalsof length3 n
. Thus,
if3 n
<3 n+1
we haveN
(K
1=3 )2
n
. Therefore: dim
M (K
1=3
)log2=log3. Onthe
otherhand,let =log2=log3and betheCantor-Lebesguemeasure,whichassignsmass
2 n
foranylevel ninterval. LetD bea subsetof K
1=3
,and let nbethe integer such that
3 n
jDj3 1 n
. TheneverysuchDiscoveredbyatmostfourshorterterracingintervals
of theform[k3 n
;(k+1)3 n
]forsome k. We thenhave
(D)42 n
4jDj
;
and so
H
(K
1=3 )H
1 (K
1=3
)1=4;
which impliesthat dim
H
Next,we introducethe energymethoddueto Frostman.
Theorem 6.6 (Frostman 1935). Givena metricspace(X;), ifisaniteBorel
mea-sure supported on AX and
E
()
def
= ZZ
d(x)d(y)
(x;y)
<1;
then H
1
(A)>0, andhencedim
H
(A).
It can be shown that under the above conditions H
(A) = 1. The converse of this
theorem is also true. That is, for any <dim
H
(A), there exists a measure on A that
satisesE
()<1.
Proof. Given a measure ,dene thefunction
(;x)
def
= Z
d(y)
(x;y)
;
sothatE
()=
R
(;x)d(x). LetA[M]denotethesubsetofAwhere
(;)isatmost
M. There existsa numberM such thatA[M]haspositive-measure, sinceE
()isnite.
Let denote the measure restricted to theset A[M]. Then forany x 2A[M], we have
( ;x)
(;x) M. Now let D be a bounded subset of X. If D\A[M] = ; then
(D) = 0. Otherwise, take x 2 D\A[M]. Let m be the largest integer such that D is
containedintheopen ballofradius 2 m
aboutx. Then
M
Z
d(y)
(x;y)
Z
D
d(y)
(x;y)
2 m
(D):
The last inequality comes from the fact that (x;y) 2 m
for each y in D. Thus, we
have (D) M2 m
M(2jDj)
. This also holds when D\A[M] = ;. By the Mass
DistributionPrinciple,we concludethatH
1
(A)>H
1
(A[M])>0.
Nowweare readyto prove thesecond partof Taylor'stheorem.
Proof of Theorem6.1, Part2. From Corollary 4.3 we have that B
d
is Holder
forevery <1=2 a.s. Therefore, Lemma5.11 impliesthat
dim
H B
d
[0;1]2:
a.s. Fortheotherinequality,wewilluseFrostman'senergymethod. Let<2. Weusethe
occupationmeasure
B def
= LB 1
,whichmeansthat
B
(A)=LB 1
(A), forall measurable
subsets Aof R d
,or, equivalently,
Z
f(x)d
B (x)=
Z
1
0 f(B
t )dt
forall measurablefunctions f. We want to showthat
E Z
R d
Z
R d
d
B (x)d
B (y)
jx yj
=E Z
1
0 Z
1
0
dsdt
jB(t) B(s)j
<1: (6.1)
Letusevaluatethe expectation:
EjB(t) B(s)j
=E(jt sj 1=2
jZj)
=jt sj =2
Z
R c
d
jzj
e jzj
2
=2
HereZ denotesthed-dimensionalstandardGaussianrandomvariable. Theintegralcanbe
evaluatedusingpolarcoordinates, butallweneedisthatitisaniteconstant cdepending
on d and only. Substitutingthisexpressioninto (6.1) and usingFubini'stheoremwe get
EE
(
B )=c
Z
1
0 Z
1
0
dsdt
jt sj =2
2c Z
1
0 du
u =2
<1: (6.2)
ThereforeE
(
B
)<1 a.s.
Remark. Levy showed earlierin1940 that, when d=2, we have H 2
(B[0;1])=0 a.s.
The statement isactuallyalso true foralld 2.
NowletusturntothegraphG
B
ofBrownianmotion. Wewillshowaproof oftherst
halfof Taylor'stheorem forone dimensionalBrownianmotion.
Proof of Theorem6.1, Part1. We have shown inCorollary5.7 that
dim
H G
B
3=2:
Fortheotherinequality,let<3=2 andletAbethesubsetofthegraph. Deneameasure
on thegraphusingprojectionto thetimes axis:
(A) def
= L(f0t1:B(t)2Ag):
Changing variables, the energyof can be writtenas:
ZZ
d(x)d(y)
jx yj
= Z
1
0 Z
1
0
dsdt
(jt sj 2
+jB(t) B(s)j 2
) =2
:
Boundingtheintegrand,taking expectations, and applyingFubiniwe getthat
EE
()2 Z
1
0 E
(t 2
+B(t) 2
) =2
dt: (6.3)
Letn(z) denotethe standardnormal density. By scaling,theexpectedvalue above can be
writtenas
2 Z
+1
0 (t
2
+tz 2
) =2
n(z)dz: (6.4)
Comparing the size of the summands in the integration suggests separating z p
t from
z> p
t. Thenwecan bound(6.4) above bytwice
Z p
t
0 (t
2
) =2
dz+ Z
1
p
t (tz
2
) =2
n(z)dz=t 1
2
+t =2
Z
1
p
t z
n(z)dz:
Furthermore, weseparate thelastintegralat 1. We get
Z
1
p
t z
n(z)dz c
+
Z
1
p
t z
dz:
The laterintegralis oforder t (1 )=2
. Substitutingthese resultsinto (6.3),wesee thatthe
expected energy is nite when < 3=2. The claim now follows form Frostman's Energy
7. On nowhere dierentiability
Levy (1954) askswhether itis truethat
P[ 8tD
B(t)2f1g]=1?
The followingpropositiongivesanegative answerto thisquestion.
Proposition 7.1. A.s there is an uncountable set of times t at which the upper right
derivative D
B(t) iszero.
We sketch a proof below. Stronger and more general results can be found in Barlow
and Perkins(1984).
(SKETCH). Put
I=
B(1); sup
0s1 B(s)
;
and denea functiong:I ![0;1]bysetting
g(x)=supfs2[0;1]:B(s)=xg:
It is easy to check that a.s. the interval I is non-degenerate, g is strictly decreasing, left
continuous and satises B(g(x)) = x. Furthermore, a.s. the set of discontinuitiesof g is
dense in I since a.s. B has no interval of monotonicity. We restrict our attention to the
event of probability1 on which these assertionshold. Let
V
n
=fx2I :g(x h) g(x)>nh forsome h2(0;n 1
)g:
Sincegisleftcontinuousandstrictlydeceasing,onereadilyveriesthatV
n
isopen;itisalso
denseinI aseverypointofdiscontinuityofg isalimitfromtheleftofpointsofV
n
:Bythe
Bairecategorytheorem,V := T
n V
n
isuncountableanddenseinI:Nowifx2V thenthere
is a sequence x
n
"x such that g(x
n
) g(x) >n(x x
n
):Setting t=g(x) and t
n =g(x
n )
we have t
n
#tand t
n
t>n(B(t) B(t
n
));fromwhichitfollowsthatD
B(t)0:Onthe
other handD
B(t)0 sinceB(s)B(t)forall s2(t;1);bydenitionof t=g(x):
Exercise 7.2. Letf 2C([0;1]):ProvethatB(t)+f(t)isnowheredierentiablealmost
surely.
Isthe\typical"functioninC([0;1]) nowhere dierentiable? Itis aneasy applicationof
theBaire category theoremto show thatnowheredierentiabilityisa genericpropertyfor
C([0;1]): This result leaves something to be desired, perhaps, as topological and measure
theoreticnotions ofa \large"set neednotcoincide. Forexample,the setof pointsin[0;1]
whosebinaryexpansionhaszeroswithasymptoticfrequency1=2 isameagerset, yetithas
Lebesgue measure 1. We consider a related idea proposed by Christensen (1972) and by
Hunt, Sauer and Yorke (1992). Let X be a separable Banach space. Say that A X is
prevalent if there exists a Borel probabilitymeasure on X such that (x+A)=1 for
every x2X. A set iscallednegligible ifits complement isprevalent.
Proposition 7.3. If A
1 ;A
2
;::: are negligible subsets of X then S
i1 A
i
negli-Proof. Foreachi1let
A
i
beaBorelprobabilitymeasuresatisfying
A
i (x+A
i )=0
for all x 2 X: Using separability we can nd for each i a ball D
i
of radius 2 i
centered
at x
i
2 X with
A
i (D
i
) > 0: Dene probabilitymeasures
i
; i 1; by setting
i (E) =
A
i
(E +x
i jD
i
) for each Borel set E; so that
i
(x+A
i
) = 0 for all x and for all i: Let
(Y
i
;i 0) be a sequence of independent random variables with dist(Y
i ) =
i
: For all i
we have
i [jY
i j 2
i
] = 1. Therefore, S = P
i Y
i
converges almost surely. Writing
for the distribution of S and putting
j
= dist(S Y
j
); we have =
j
j
; and hence
(x+A
j )=
j
j (x+A
j
)=0forallx andforallj:Thus(x+[
i1 A
i
)=0forallx:
Proposition 7.4. A subsetA of R d
is negligiblei L
d
(A)=0:
Proof. ()) Assume A is negligible. Let
A
be a (Borel) probability measure such
that
A
(x+A)=0 for all x 2 X:Since L
d
A =L
d
(indeed L
d
=L
d
for any Borel
probabilitymeasure on R d
)wehave 0=L
d
A
(x+A)=L
d
(x+A)forall x2X:
(() If L
d
(A)= 0 then therestriction of L
d
to the unit cube is a probability measure
which vanisheson every translateof A:
Remark. It follows from Exercise 7.2 that the set of nowhere dierentiable functions
isprevalentin C([0;1]):
8. Strong Markov property and the reection principle
For each t 0 let F
0
(t)=fB(s) :stg be the smallest-eld making every B(s);
st;measurable,and setF
+
(t)=\
u>t F
0
(u)(theright-continuousltration). Itisknown
(see, for example, Durrett (1996), theorem 7.2.4) that F
0
(t) and F
+
(t) have the same
completion. A ltration fF(t)g
t0
is a Brownian ltration if for all t 0 the process
fB(t+s)g
s0
isindependentofF(t)andF(t)F
0
(t):Arandomvariable isastopping
time fora Brownianltration fF(t)g
t0
iff tg2F(t) forall t: For any randomtime
we denethepre- -eld
F():=fA:8tA\f tg2F(t)g:
Proposition 8.1. (Markov property) For every t0 the process
fB(t+s) B(t)g
s0
is standard Brownian motionindependent of F
0
(t) andF
+ (t):
It is evident from independence of increments that fB(t+s) B(t)g
s0
is standard
Brownian motionindependent of F
0
(t): That thisprocess is independent of F
+
(t) follows
from continuity;see, e.g.,Durrett (1996, 7.2.1)for details.
Themainresultof thissectionisthestrongMarkov propertyforBrownianmotion,
establishedindependentlybyHunt (1956) and Dynkin(1957):
Theorem 8.2. Supposethat isastoppingtime forthe Brownian ltration fF(t)g
t0 .
Then fB(+s) B()g
s0
is Brownian motion independent of F():
Sketch of Proof. Supposerstthat isanintegervaluedstoppingtimewithrespect
to aBrownianltrationfF(t)g
t0
:Foreach integer j theevent f =jgisinF(j) andthe
process fB(t+j) B(j)g
t0
" > 0; and approximating by such discrete stopping times gives the conclusion in the
generalcase. See, e.g., Durrett (1996,7.3.7) formore details.
Oneimportant consequenceof thestrongMarkov propertyis thefollowing:
Theorem 8.3 (Reection Principle). If is a stopping time then
B
(t):=B(t)1
(t)
+(2B() B(t))1
(t>)
(Brownian motion reected attime ) isalso standard Brownian motion.
Proof. We shalluse anelementaryfact:
Lemma 8.4. LetX;Y;Z berandom variables withX;Y independent andX;Z
indepen-dent. If Y d
Z then (X;Y) d
(X;Z):
The strong Markov property states that fB( +t) B()g
t0
is Brownian motion
independentofF();andbysymmetrythisisalsotrueoff (B(+t) B())g
t0
:Wesee
from thelemma that
(fB(t)g
0t
;fB(t+) B()g
t0 )
d
(fB(t)g
0t
;f(B() B(t+))g
t0 );
and thereectionprinciplefollows immediately.
Remark. Consider = infft : B(t) = max
0s1
B(s)g: Almost surely fB( +t)
B()g
t0
isnon-positive on some right neighborhoodof t=0, and hence is not Brownian
motion. ThestrongMarkovpropertydoesnotapplyherebecause is notastoppingtime
for any Brownian ltration. We willlater see that Brownian motionalmost surelyhas no
point of increase. Since is a point of increase of the reected process fB
(t)g; it follows
thatthe distributionsof Brownianmotionand of fB
(t)gare singular.
Exercise 8.5. Prove thatifA isa closedset then
A
isa stoppingtime.
Solution. fAtg= T
n1 S
s2[0;t]\Q
fdist (B(s);A) 1
n
g2F0(t):
More generally, if A is a Borel set then the hitting time
A
is a stopping time (see Bass
(1995)).
SetM(t)= max
0st
B(s):Our next resultsays M(t) d
jB(t)j:
Theorem 8.6. If a>0 then P[ M(t)>a]=2P[ B(t)>a]:
Proof. Set
a
=minft0:B(t)=agand let fB
(t)g be Brownianmotion reected
at
a
: Then fM(t) > ag is the disjoint union of the events fB(t) > ag and fM(t) >
a;B(t)ag; and sincefM(t)>a;B(t)ag=fB
(t)agthe desiredconclusion follows
immediatelyfrom thestrongMarkov property.
9. Local extrema of Brownian motion
Proposition 9.1. Almost surely,every local maximum of Brownian motionis a strict
Lemma 9.2. Given two disjoint closed time intervals, the maxima of Brownian motion
on them are dierent almost surely.
Proof. Fori=1;2let[a
i ;b
i ],m
i
,denotethelower,respectivelythehigherinterval,and
thecorrespondingmaximumof Brownian motion. Note thatB(a
2
) B(b
1
) isindependent
ofthepairm
1 B(b
1
)andm
2 B(a
2
). Conditioningonthevaluesoftherandomvariables
m
1 B(b
1
) and m
2 B(a
2
),theevent m
1 =m
2
can be writtenas
B(a
2
) B(b
1 )=m
1 B(b
1 ) (m
2 B(a
2 )):
Theleft handsidebeingacontinuousrandomvariable,and theright handsideaconstant,
we see thatthisevent hasprobability0.
We nowprove Proposition9.1.
Proof. Thestatementofthelemmaholdsjointlyforalldisjointpairsofintervalswith
rationalendpoints. Thepropositionfollows,sinceifBrownianmotionhadanon-strictlocal
maximum, thenthere were two disjoint rationalintervals where Brownian motion hasthe
same maximum.
Corollary 9.3. Theset M of times where Brownian motion assumes its local
maxi-mum is countable and dense almost surely.
Proof. Consider the function from the set of non-degenerate closed intervals with
rationalendpointsto R given by
[a;b]7!inf
ta:B(t)= max
asb B(s)
:
TheimageofthismapcontainsthesetM almostsurelybythelemma. ThisshowsthatM
iscountable almost surely. We already knowthatB hasnointervalofincreaseordecrease
almost surely. It follows that B almost surelyhasa localmaximum inevery intervalwith
rationalendpoints,implyingthecorollary.
10. Area of planar Brownian motion paths
We have seenthattheimageofBrownianmotionis always2dimensional,soone might
askwhat its 2dimensionalHausdormeasure is. Itturnsouttobe0 inalldimensions;we
willprove itforthe planarcase. We willneedthefollowinglemma.
Lemma 10.1. If A
1 ;A
2 R
2
are Borelsets withpositivearea then
L
2
(fx2R 2
:L
2 (A
1 \(A
2
+x))>0g)>0:
Proof. One proof of this fact relies on (outer) regularity of Lebesgue measure. The
proof belowis more streamlined.
We may assumeA
1
and A
2
arebounded. By Fubini'stheorem,
Z
R 2
1
A1 1
A2
(x)dx= Z
R 2
Z
R 2
1
A1 (w)1
A2
(w x)dwdx
= Z
R 2
1
A
1 (w)
Z
R 2
1
A
2
(w x)dx
dw
Thus1
A
1 1
A
2
(x)>0onasetofpositivearea. But1
A
1 1
A
2
(x)=0unlessA
1 \(A
2 +x)
haspositive area, sothisprovesthelemma.
ThroughoutthissectionB denoteplanarBrownianmotion. Wearenowreadytoprove
Levy's theorem on theareaof its image.
Theorem 10.2 (Levy). Almostsurely L
2
(B[0;1])=0:
Proof. LetX denote theareaofB[0;1], and M be itsexpectedvalue. Firstwecheck
thatM <1. If a1 then
P[X >a]2P[jW(t)j> p
a=2forsome t2[0;1]]8e a=8
whereW isstandardone-dimensionalBrownianmotion. Thus
M = Z
1
0
P[X >a]da8 Z
1
0 e
a=8
da+1<1:
Note thatB(3t)and p
3B(t)have thesame distribution,and hence
EL
2
(B[0;3])=3EL
2
(B[0;1])=3M:
Note that we have L
2
(B[0;3]) P
2
j=0 L
2
(B[j;j+1]) with equality if and only if for0
i<j 2 we have L
2
(B[i;i+1]\B[j;j+1])= 0. On the other hand, forj = 0;1;2; we
have EL
2
(B[j;j+1])=M and
3M =EL
2
(B[0;3]) 2
X
j=0 EL
2
(B[j;j+1])=3M;
whence the intersection of any two of the B[j;j+1] has measure zero almost surely. In
particular,L
2
(B[0;1]\B[2;3])=0 almost surely.
LetR (x)denotetheareaofB[0;1]\(x+B[2;3] B[2]+B(1)). Ifwe conditiononthe
valuesofB[0;1];B[2;3] B(2),theninordertoevaluatetheexpectedvalueofB[0;1] \ B[2;3]
we shouldintegrate R (x)wherex hasthedistributionof B(2) B(1). Thus
0=E[L
2
(B[0;1]\B[2;3])]=(2) 1
Z
R 2
e jxj
2
=2
E[R(x)]dx
where we are averaging with respect to the Gaussian distribution of B(2) B(1). Thus
R (x)=0 a.s. forL
2
-almost all x,or, byFubini, the areaof the setwhere R (x)is positive
isa.s. zero. Fromthe lemmaweget thata.s.
L
2
(B[0;1])=0 or L
2
(B[2;3])=0:
TheobservationthatL
2
(B[0;1])andL
2
(B[2;3])areidenticallydistributedandindependent
completes theproof thatL
2
(B[0;1])=0almost surely.
11. Zeros of the Brownian motion
Inthissection,westartthestudyofthepropertiesofthezerosetZ
B
ofonedimensional
Brownianmotion. Wewillprove thatthissetisan uncountable closedsetwithnoisolated
points. This is, perhaps, surprising since, almost surely, a Brownian motion has isolated
zeros from the left (for instance, the rst zero after 1=2) or from the right (the last zero
Theorem 11.1. Let B bea one dimensional Brownian motion and Z
B
beits zero set,
i.e.,
Z
B
=ft2[0;+1):B(t)=0g:
Then, a.s., Z
B
isan uncountable closed set withno isolated points.
Proof. Clearly, with probability one, Z
B
is closed because B is continuous a.s.. To
provethatnopointofZ
B
isisolatedweconsiderthefollowingconstruction: foreachrational
q 2[0;1)considertherst zero after q,i.e.,
q
=infft>q:B(t)=0g. Notethat
q <1
a.s. and, sinceZ
B
is closed,the inf is a.s. a minimum. Bythe strongMarkovpropertywe
havethatforeachq,a.s.
q
isnotanisolatedzerofromtheright. But,sincethere areonly
countablymanyrationals,weconcludethata.s.,forall q rational,
q
isnotanisolatedzero
from theright. Our next taskis to prove that theremaining pointsof Z
B
arenotisolated
from theleft. So we claimthat any0<t2Z
B
whichis dierent from
q
forall rationalq
is notan isolatedpoint from theleft. To see thistake a sequence q
n
"t. Dene t
n =
q
n .
Clearlyq
n t
n
<t(ast
n
isnotisolatedfromtheright)andsot
n
"t. Thustisnotisolated
from theleft.
Finally, recall (see, for instance, Hewitt-Stromberg, 1965) that a closed set with no
isolatedpointsis uncountable and thisnishestheproof.
11.1. General Markov Processes. In this section, we dene general Markov
pro-cesses. ThenweprovethatBrownianmotion,reectedBrownianmotionandaprocessthat
involvesthe maximumof Brownianmotion areMarkov processes.
Definition 11.2. A function p(t;x;A), p : R R d
B ! R, where B is the Borel
-algebra inR d
,isa Markov transitionkernel provided
1. p(;;A)is measurable asa functionof (t;x), foreach A2B,
2. p(t;x;) isa Borelprobabilitymeasure forall t2R and x2R d
,
3. 8A2B; x2R d
and t;s>0,
p(t+s;x;A)= Z
R d
p(t;y;A)p(s;x;dy):
Definition 11.3. A process fX(t)g isaMarkov process withkernelp(t;x;A)ifforall
t>sand Borelset A2B we have
P(X(t)2AjF
s
)=p(t s;X(s);A);
whereF
s
=(X(u); us).
The next two examples aretrivial consequences of the Markov Property for Brownian
motion.
Example 11.4. A d-dimensionalBrownian motion isa Markov process and its
transi-tion kernelp(t;x;) hasN(x;t) distributionineach component.
SupposeZ hasN(x;t) distribution. Dene jN(x;t)j to bethedistributionof jZj.
pro-Theorem 11.6 (Levy, 1948). Let M(t) be the maximum process of a one dimensional
Brownian motion B(t),i.e. M(t)=max
0st
B(s). Then, the process Y(t)=M(t) B(t)
is Markov and its transition kernel p(t;x;) has jN(x;t)j distribution.
Proof. For t > 0, consider the two processes ^
B(t) = B(s+t) B(s) and ^
M(t) =
max
0ut ^
B(u). Dene F
B
(s) = ( B(t);0ts). To prove the theorem it suÆces to
checkthat conditionalon F
B
(s)and Y(s)=y,we have Y(s+t) d
=jy+ ^
B(t)j.
To prove theclaim notethat M(s+t)=M(s)_(B(s)+ ^
M(t)), and sowe have
Y(s+t)=M(s)_(B(s)+ ^
M(t)) (B(s)+ ^
B(t)):
Usingthefactthat a_b c=(a c)_(b c),we have
Y(s+t)=Y(s)_ ^
M(t) ^
B(t):
To nish, it suÆces to check, for every y0, that y_ ^
M(t) ^
B(t) d
=jy+ ^
B(t)j. For any
a0 write
P(y_ ^
M(t) ^
B(t)>a)=I+II;
whereI =P(y ^
B(t)>a) and
II =P(y ^
B(t)aand ^
M(t) ^
B(t)>a):
Since ^
B d
= ^
B wehave
I =P(y+ ^
B(t)>a):
To studythesecond termis usefulto denethe"time reversed" Brownianmotion
W(u)= ^
B(t u) ^
B(t);
for0ut. Note that W is also a Brownian motion for0u t sinceit iscontinuous
and its nitedimensionaldistributionsare Gaussianwiththeright covariances.
Let M
W
(t)=max
0ut
W(u). Then M
W (t)=
^
M(t) ^
B(t). Since W(t)= ^
B(t), we
have:
II =P(y+W(t)aand M
W
(t)>a):
IfweusethereectionprinciplebyreectingW(u)atthersttimeithitsawegetanother
BrownianmotionW
(u). IntermsofthisBrownianmotionwehaveII =P(W
(t)a+y).
Since W
d
= ^
B, it follows II = P(y + ^
B(t) a). The Brownian motion ^
B(t) has
continuousdistribution,and so,byadding I and II,weget
P(y_ ^
M(t) ^
B(t)>a)=P(jy+ ^
B(t)j>a):
Thisprovestheclaim and,consequently,thetheorem.
Proposition 11.7. TwoMarkov processes in R d
withcontinuouspaths, withthe same
initial distribution andtransition kernel are identical in law.
Outline of Proof. The nitedimensionaldistributionsare thesame. From thiswe
deducethattherestrictionofbothprocessesto rationaltimesagreeindistribution. Finally
SincetheprocessY(t)iscontinuousandhasthesame distributionasjB(t)j(theyhave
the same Markov transition kernel and same initial distribution) this proposition implies
fY(t)g d
=fjB(t)jg.
11.2. Hausdor dimension of Z
B
. We already know that Z
B
is an uncountable
set with no isolated points. In this section, we will prove that, with probabilityone, the
Hausdor dimension of Z
B
is 1=2. It turns out that it is relatively easy to bound from
bellow the dimension of the zero set of Y(t) (also known as set of record values of B).
Then,bytheresultsinthelastsection,thisdimensionmustbethesame ofZ
B
sincethese
two (random) sets havethe same distribution.
Definition 11.8. A time t is a recordtime for B ifY(t)=M(t) B(t)=0, i.e., ift
isa globalmaximumfromthe left.
The next lemma givesa lower bound on the Hausdor dimension of the set of record
times.
Lemma 11.9. With probability 1, dimft2[0;1]:Y(t)=0g1=2.
Proof. SinceM(t)isanincreasingfunction,wecanregarditasadistributionfunction
ofameasure,with(a;b]=M(b) M(a). Thismeasureissupportedonthesetofrecord
times. Weknowthat, withprobabilityone,theBrownianmotionisHoldercontinuouswith
anyexponent<1=2. Thus
M(b) M(a) max
0hb a
B(a+h) B(a)C
(b a)
;
where < 1=2 and C
is some random constant that doesn't depend on a or b. By the
Mass DistributionPrinciple,we get that, a.s., dimft2[0;1]:Y(t)=0g. By choosing
a sequence
n
"1=2 we nishthe proof.
RecallthattheupperMinkowskidimensionofasetisanupperboundfortheHausdor
dimension. To estimatethe Minkowskidimensionof Z
B
we willneedto know
P(9t2(a;a+):B(t)=0): (11.1)
Thisprobabilitycan becomputed explicitlyand we willleave thisasan exercise.
Exercise 11.10. Compute(11.1).
Solution. ConditionalonB(a)=x>0wehave
P(9t2(a;a+):B(t)=0jB(a)=x)=P( min
ata+
B(t)<0jB(a)=x):
Buttherighthandsideisequalto
P(max
0<t<
B(t)>x)=2P(B()>x);
usingreectionprinciple.
Byconsideringalsothecase wherexisnegativeweget
P(9t2(a;a+):B(t)=0)=4 Z
1
0 Z
1
x e
y 2
2 x
2
2a
2 p
a dydx:
Computingthislastintegral explicitly,weget
P(9t2(a;a+):B(t)=0)= 2
arctan r
However, forour purposes,thefollowingestimate willsuÆce.
Lemma 11.11. For any a;>0 we have
P(9t2(a;a+):B(t)=0)C r
a+ ;
for some appropriate positive constant C.
Proof. ConsidertheeventA givenbyjB(a+)j p
. Bythescaling propertyofthe
Brownianmotion,we can give theupperbound
P(A)=P
jB(1)j r
a+
2 r
a+
: (11.2)
However, knowing that Brownian motion has a zero in (a;a+) makes the event A very
likely. Indeed, we certainly have
P(A)P(A and 02B[a;a+]);
and thestrongMarkov propertyimpliesthat
P(A)~c P(02B[a;a+]); (11.3)
where
~
c= min
ata+
P(AjB(t)=0):
Because theminimumisachieved when t=ahave
~
c=P(jB(1)j1)>0;
byusingthescalingpropertyof theBrownianmotion.
From inequalities(11.2) and (11.3), we conclude
P(02B[a;a+]) 2
~ c
r
a+ :
Forany,possiblyrandom, closedset A[0;1]denea function
N
m (A)=
2 m
X
k=1 1
fA\[ k 1
2 m
; k
2 m
]6=;g :
Thisfunctioncountsthenumberofintervals oftheform [ k 1
2 m
; k
2 m
]intersectedbytheset A
and soisanaturalobjectifwe wantto computeMinkowskidimension. In thespecialcase
whereA=Z
B
we have
N
m (Z
B )=
2 m
X
k=1 1
f02B[ k 1
2 m
; k
2 m
]g :
The next lemma shows that estimates on the expected value of N
m
(A) will give us
boundson theMinkowskidimension (andhenceon theHausdor dimension).
Lemma 11.12. Suppose A is a closed random subset of [0;1] such that
EN
m
(A)c2 m
;
Proof. Consider
E 1
X
m=1 N
m (A)
2 m(+)
;
for>0. Then, bythe monotoneconvergence theorem,
E 1
X
m=1 N
m (A)
2 m(+)
= 1
X
m=1 EN
m (A)
2 m(+)
<1:
Thisestimate impliesthat
1
X
m=1 N
m (A)
2 m(+)
<1 a.s.;
and so, withprobabilityone,
limsup
m!1 N
m (A)
2 m(+)
=0:
From thelastequation follows
dim
M
(A)+; a.s. :
Let!0 throughsome countable sequence to get
dim
M
(A); a.s. :
Andthiscompletesthe proofof thelemma.
To getan upperboundon theHausdor dimensionof Z
B
note that
EN
m (Z
B )C
2 m
X
k=1 1
p
k
~
C2 m=2
;
sinceP 9t2
k 1
2 m
; k
2 m
:B(t)=0
C
p
k
. Thus, by thelast lemma, dim
M (Z
B
) 1=2 a.s.
Thisimpliesimmediatelydim
H (Z
B
)1=2 a.s. Combiningthisestimatewith Lemma11.9
we have
Theorem 11.13. With probability one we have
dim
H (Z
B )=
1
2 :
From this proof we can also inferthat H 1=2
(Z
B
)<1 a.s. Later in thecourse we will
prove that H 1=2
(Z
B
) =0. However, it is possibleto denea more sophisticated Hausdor
measure forwhich,withprobabilityone, 0<H (Z
B )<1.
12. Harris' Inequality and its consequences
We beginthissection byprovingHarris'inequality.
Lemma 12.1 (Harris' inequality). Suppose that
1 ;::: ;
d
are Borel probability
mea-sures on R and=
1
2
:::
d
. Let f;g:R d
!R bemeasurablefunctions that are
nondecreasing in each coordinate. Then,
Z
d
f(x)g(x)d Z
d
f(x)d
Z
d
g(x)d