Table Of ContentA COMBINATORIAL METHOD FOR CALCULATING THE
MOMENTS OF LE´VY AREA
7
0
0
DANIELLEVIN AND MARKWILDON
2
n
Abstract. WepresentanewwaytocomputethemomentsoftheL´evy
a
J areaofatwo-dimensionalBrownianmotion. Ourapproachusesiterated
integrals and combinatorial arguments involving theshuffle product.
1
3
]
R
1. Introduction
P
. In this paper we present a new approach to the problem of finding the
h
t moments of the signed area swept out by a two-dimensional Brownian mo-
a
tion. This is a classical problem of great importance, originally solved by
m
L´evy (see [9]).
[
We begin by explaining how these moments may be defined. Given a
1 piecewise smooth path γ : [0,T] R2 we may complete it to a loop γ¯ by
t
v →
closing it with the chord from γ to γ . We may then define its signed area
2 T 0
0 to be
0
n(γ¯,x)dx
2
R2
0 ZZ
7 where n(γ¯,x) is the winding number of γ¯ about the point x R2.
∈
0 rγ
/ T
h
t
a
m
+
:
v
i −
X
r
a 0
+
γ r
0
Figure 1. Contributions to the signed area of γ : [0,T] R2.
→
Date: February 2, 2008
2000 Mathematics Subject Classification 60J65 (primary), 05A15 (secondary).
Key words and phrases. L´evy area, shuffle product,signature of a path.
The first authorissupported bytheEPSRCFellowship “Partial differential equations
— A rough path approach” GR/S18526/01.
The second author is supported by EPSRCGrant EP/D054664/1.
1
2 DANIELLEVINANDMARKWILDON
Now suppose that B : [0,T] R2 is a two-dimensional Brownian mo-
t
tion. If we complete B to a loop→B¯ and attempt to define the signed area
of B as before, then we immediately encounter the problem that almost
t
surely, n(B¯ , ) is not integrable on R2 (see [8, Theorem 55.I]). One solution
t
·
is to replace B with a sequence of piecewise linear dyadic approximations.
t
In [7, Theorem 10] and [8, Chapter 55], L´evy proved that almost surely, the
winding number integral is defined for each approximation, and that the
sequence of areas converges. This gives one possible definition of the L´evy
area of the process B .
t
We may also define the signed area corresponding to a smooth path γ :
[0,T] R2 with γ = (x ,y ) by
t t t
→
1 T
(x x )y′ (y x )x′ ds.
2 s− 0 s− s− 0 s
Z0
This observation motivate(cid:0)s the following alternative(cid:1)definition of L´evy area.
Definition 1.1. Let B = (X ,Y ) for t [0,T] be a two-dimensional Brow-
t t t
∈
nian motion starting at 0. The L´evy area of (B ) is given by the
t 0≤t≤T
stochastic integral
1 T
= (X dY Y dX ).
T s s s s
A 2 −
Z0
Remark 1.2. In the sequel, we use as a shorter notation for .
1
A A
L´evy showed in [9] that almost surely the definitions of signed area by
dyadic approximation and by stochastic integration agree. In his paper he
also gave two different ways to find the characteristic function, and hence
the moments, of when T = 2π.
T
A
Theorem 1.3 (L´evy). If T = 2π then
E(exp(iz ))= (coshπz)−1.
2π
A
L´evy’s first proof uses the definition of by dyadic approximation.
T
A
His second starts from Definition 1.1, but depends on earlier work by Kac,
Siegert, Cameron and Martin (see references cited in [9, 1.6]).
§
Inthispaperwepresentadirectandlargely self-contained proofofLevy’s
Theorem, using Definition 1.1 to define L´evy area. Our approach, which we
outline in 2 below, is based on the fact that moments of L´evy area can be
§
expressed as iterated integrals and hence calculated explicitly by exploiting
the combinatorics of shuffle products.
Itseemslikelythatourmethodscanbeappliedmorebroadly. Wemention
here that by using the multiplication
1
(x,y,a)(x′,y′,a′) = x+x′,y+y′,a+a′+ (xy′ yx′)
2 −
(cid:16) (cid:17)
wemay identifypoints (x,y,a) R3 with elements of theHeisenberggroup.
∈
It is known (see [12, Section 3.2.2]) that the process (X ,Y , ) is a Brow-
t t t
A
nian motion on this group. Computing aspects of the joint distribution of
THE MOMENTS OF LE´VY AREA 3
(X ,Y , ) is a hard problem, involving Harish-Chandra formulae (see [2]);
t t t
A
it is in effect aimed at understanding the heat kernel on the Heisenberg
group (see [4]). It seems likely that our approach may shed some light on
thesequestions. Moreover, itshouldbepossibletouseourmethodstostudy
the signed areas obtained when we replace Brownian motion by measures
related to higher order PDEs.
2. Outline
The outline of our proof is as follows. In 3 we use a simple scaling
§
argument to show that Theorem 1.3 is implied by the following assertion
about the moments of .
A
Theorem 2.1. E n = 2−nE .
n
A
Here E is the nth Euler number, as defined by the generating function
n
∞ zn
E = (cosz)−1.
n
n!
n=0
X
The first few non-zero Euler numbers are E = 1, E = 1, E = 5, E = 61.
0 2 4 6
Of course all odd-numbered Euler numbers are zero. (Correspondingly, one
can easily see that E n = 0 if n is odd.) We then show that the moments
A
of can be expressed using iterated integrals.
A
In 4 we introduce the shuffle product on the tensor algebra of a vector
space,§and use it to give an expression for E n as a certain coefficient in
A
the expansion of a shuffle product. In 5 we use a combinatorial argument
§
to determine this coefficient, thereby proving Theorem 2.1 and hence L´evy’s
Theorem.
3. The moments of L´evy’s area for Brownian motion
We first show that L´evy’s theorem (Theorem 1.3) follows from Theo-
rem 2.1. If we scale the Brownian path B = (X ,Y ) defined for 0 t 1
t t t
≤ ≤
by setting B˜ = √TB then we obtain a new Brownian path defined for
s s/T
0 t T. As before, is the L´evy area of B at time 1 and is the L´evy
t T
ar≤ea o≤f B˜ at time T.AIt follows easily from Definition 1.1 thAat = T .
s T 1
A A
Hence, assuming that Theorem 2.1 holds, the moments of L´evy area at
time T are given by
E n = Tnn!E .
AT n
In particular, by setting T = 2π we find that the characteristic function of
L´evy area at time 2π is
∞ (iz)n
E(exp(iz )) = πnE = (cosπiz)−1 = (coshπz)−1,
2π n
A n!
n=0
X
where we have absolute convergence of the series for z < 1/2.
| |
Therefore, to prove Theorem 1.3, we may concentrate on finding the mo-
ments of . For this we shall need the following algebraic definition.
A
4 DANIELLEVINANDMARKWILDON
Definition 3.1. Let V be a real vector space. Let
∞
T((V)) = V⊗k.
k=0
Y
Thus a typical element of T((V)) is a formal infinite sequence (a ,a ,a ,...)
0 1 2
where a V⊗k. (By convention V⊗0 = R.) Clearly T((V)) is a real vector
k
∈
space. It is easy to check that T((V)) becomes an algebra with unit if we
define the product of a = (a ,a ,a ,...),b = (b ,b ,b ,...) T((V)) by
0 1 2 0 1 2
∈
k
a b = ..., a b ,... .
j k−j
⊗ ⊗
j=0
X
We also define the exponential of a tensor a to be the formal sum
∞ a⊗n
exp(a)= .
n!
n=0
X
We can now define the signature of a Brownian motion. As motivation,
we first recall this definition for a path. If γ :[0,T] R2 is a path of finite
→
length (see [5]) then its signature is the formal infinite sum
∞
X (γ) = dγ dγ T((R2))
s,t t1 ⊗···⊗ tk ∈
k=0Zs<t1<···<tk<t
X
defined for 0 s< t T.
≤ ≤
Example 3.2. For example, if e ,e R2 and γ = e +te, then
0 t 0
∈
X (γ) = exp(t s)e.
s,t
−
One of the most important properties of the signature is that it is multi-
plicative (for the proof of the following theorem see [1]).
Theorem 3.3. For any 0 r < s < t T
≤ ≤
X (γ) = X (γ) X (γ).
r,t r,s s,t
⊗
Definition 3.4. Let B : [0,T] R2 be a two-dimensional Brownian mo-
→
tion. For 0 s < t T we define the signature of B to be the formal
≤ ≤
infinite sum
∞
X (B) = dB dB .
s,t t1 ⊗···⊗ tk
k=0Zs<t1<...<tk<t
X
Remark 3.5. Up till now, the signature of a Brownian motion and other
auxiliary objects could be interpreted either in the Stratonovich or Itoˆ sense.
However, the next theorem holds only for the Stratonovich integral. For its
proof see either Fawcett [3] or Lyons–Victoir [10].
THE MOMENTS OF LE´VY AREA 5
Theorem 3.6. If B is a Brownian motion in R2 then
1
(1) E X (B) = exp (e e +e e )
0,1 1 1 2 2
2 ⊗ ⊗
(cid:18) (cid:19)
(cid:0) (cid:1)
where e and e are any two orthogonal vectors in R2.
1 2
In particular, (1) implies that
1
(2) E X2n(B) = (e e +e e )⊗n
0,1 2nn! 1⊗ 1 2⊗ 2
where X2n is the c(cid:0)omponent(cid:1)of X lying in (R2)⊗2n.
0,1 0,1
4. Shuffle products and other combinatorial objects
We now introduce an important combinatorial object which will be used
in the sequel.
Definition 4.1. We define the set S of (m,n) shuffles to be the subset
m,n
of permutations in the symmetric group S defined by
m+n
S = σ S : σ(1) < < σ(m), σ(m+1) < < σ(m+n) .
m,n m+n
{ ∈ ··· ··· }
Remark 4.2. The term “shuffle” is used because such permutations arise
when one riffle shuffles a deck of m+n cards cut into one pile of m cards
and a second pile of n cards.
Let V = R2 with the orthogonal basis e and e . Let V∗ be the space
1 2
dual to V and let e1 and e2 be its dual basis. Let n N. The elements
∈
e e , where each i 1,2 , for k = 1,...,n, form a basis of V⊗n.
Tih1e⊗c·o·r·r⊗espionndingdualbasiks∈of{(V∗)}⊗n isgivenbytheelementsei1 ein.
⊗···⊗
There is a natural duality , :V⊗n (V⋆)⊗n R defined by:
h i × →
ei1 ein,e e = δ ...δ .
⊗···⊗ j1 ⊗···⊗ jn i1j1 injn
Definition 4.(cid:10)3. Set (k ,...k ) = (i ,.(cid:11)..,i ,j ,...,j ). The shuffle
1 m+n 1 m 1 n
product of two tensors eI = ei1 eim and eJ = ej1 ejn, is the
⊗···⊗ ⊗···⊗
tensor eI eJ defined by
eI eJ = ekσ−1(1) ekσ−1(m+n).
⊗···⊗
σ∈XSm,n
Let T(V⋆) = ∞ (V⋆)⊗k be the ordinary tensor algebra on V⋆. The shuffle
k=0
product extends to a bilinear map
L
T(V⋆) T(V⋆) T(V⋆).
× →
For example, the reader may check that
(e f) g = e f g+e g f +g e f
⊗ ⊗ ⊗ ⊗ ⊗ ⊗ ⊗
for any e,f,g (R2)⋆. For an alternative description of the shuffle product,
∈
see Definition 5.14 in [5].
6 DANIELLEVINANDMARKWILDON
Remark 4.4. It follows easily from the definition that the shuffle product
is commutative and associative. We shall use the following notation for the
shuffle product applied N times:
a N = a ... a.
N
Later in 5 we shall also use the analogous version of the shuffle product
§ | {z }
defined on the tensor powers of V.
For each pathγ , s [0,T] of finitelength wenow introduceareal-valued
s
∈
function
ϕ(γ) : T(V⋆) R
→
defined on the tensor e = ei1 ein by
⊗···⊗
ϕe(γ) = ei1 ⊗···⊗ein, dγt1 ⊗···⊗dγtn
(cid:28) Z0<t1<···<tn<T (cid:29)
(3) = dγi1...dγin.
t1 tn
Z0<t1<···<tn<T
The fundamental property of ϕe(γ) (see [12, Theorem 2.15]) is that
(4) ϕe(γ)ϕf(γ) = ϕe f(γ).
From now on we change slightly notations for e ,e and e1,e2. Let x, y
1 2
be a basis for R2 and let x∗, y∗ be the dual basis of (R2)⋆. We use these
techniques together to prove the following theorem.
Theorem4.5. ThenthmomentofL´evyareaattime1foratwo-dimensional
Brownian motion B starting at zero is the signature X (B) contracted with
0,1
the shuffle powers of the dual tensor 1 (x∗ y∗ y∗ x∗). That is,
2 ⊗ − ⊗
(x x+y y)⊗n
E( n)= 2−2n (x∗ y∗ y∗ x∗) n, ⊗ ⊗ .
A ⊗ − ⊗ n!
(cid:28) (cid:29)
Proof. Let B = (X ,Y ), 0 s 1 be a Brownian path in R2 starting at
s s
≤ ≤
zero. Then
1 1 1
= (X dY Y dX ) = (dX dY dY dX ).
s s s s t s t s
A 2 − 2 −
Z0 Z0<t<s<1
In a more canonical notation as in (3) we may write this down as
1
= (dX dY dY dX ).
A 2 t1 t2 − t1 t2
Z0<t1<t2<1
So A = ϕ12(x∗⊗y∗−y∗⊗x∗)(B) = 21(ϕx∗⊗y∗(B)−ϕy∗⊗x∗(B)). Further, us-
ing (4) we have
1 n
n = (x∗ y∗ y∗ x∗), dB dB
A 2n ⊗ − ⊗ t1 ⊗ t2
D Z0<t1<t2<1 E
1
= (x∗ y∗ y∗ x∗) n, X2n (B) .
2n ⊗ − ⊗ 0,1
D E
THE MOMENTS OF LE´VY AREA 7
Taking the expectation of n we have by (2),
A
E( n) = 2−n (x∗ y∗ y∗ x∗) n, E X2n (B)
A ⊗ − ⊗ 0,1
D ((cid:0)x x+y(cid:1)E y)⊗n
= 2−2n (x∗ y∗ y∗ x∗) n, ⊗ ⊗ . (cid:3)
⊗ − ⊗ n!
(cid:28) (cid:29)
HencetoproveTheorem2.1itissufficienttoprovethefollowing theorem.
Theorem 4.6.
(x∗ y∗ y∗ x∗) n, (x x+y y)⊗n = 2nn!E .
n
⊗ − ⊗ ⊗ ⊗
D E
5. Proof of Theorem 4.6
5.1. From now on it will often be convenient to use a shorter notation for
elements in thestandardbasis of the tensor algebra T(V), in which wewrite
xy rather than x y, x2 rather than x x, and so on. Using this notation
⊗ ⊗
the standard basis elements of T(V) are simply the words in the letters x
and y.
We shall say that a word in the letters x and y is even if (i) it is of the
form z2...z2 where each z x,y , and (ii) there are equal numbers of x’s
1 n i ∈ { }
and y’s. When we expand
(x x+y y)⊗n
⊗ ⊗
weobtainthesumofallwordsinxandyoflength2nsatisfyingcondition(i).
If such a word is not killed by (x⋆ y⋆ y⋆ x⋆) n then clearly it must
⊗ − ⊗
also satisfy (ii). Hence Theorem 4.6 is equivalent to the following assertion.
Theorem 5.1. Let
(xy yx) n = λ v
v
−
v
X
where the sum is over all words v of length 2n. Let u = λ where the
n w
sum is over all even words w of length 2n. Then
P
u = 2nn!E
n n
5.2. We begin the proof of Theorem 5.1 by noting that if n is odd then
there are no even words of length 2n, and so u = 0, as required. We may
n
therefore assume that n = 2m is even, so the even words that appear in
(xy yx) n are of length 4m, with the pairs xx and yy each appearing
−
exactly m times.
Our proof depends on counting the combinatorial objects introduced in
the next definition.
Definition 5.2. An xy-matching is a pair (w,σ) where w is an even word,
of length 2m say, and σ is a fixed-point-free involution in the symmetric
group S such that w = x if and only if w = y.
2m i σ(i)
Given an xy-matching δ = (w,σ) we define the negativity of δ by
neg(δ) = # i 1...2m :w = x and σ(i) < i .
i
{ ∈ { } }
8 DANIELLEVINANDMARKWILDON
We define the sign of δ by sgn(δ) = ( 1)neg(δ). Let N (w) be the number of
t
−
xy-matchings with underlying word w and negativity t.
It will be very useful to represent xy-matchings by diagrams such as the
one below.
x x y y x x y y
Figure 2. The xy-matching (xxyyxxyy,(13)(28)(46)(57)).
The arcs contributing to neg(δ) are those drawn below the word, thus here
neg(δ) = 1 and sgn(δ) = 1. As an exercise, the reader may check that
−
there are in total 16 xy-matchings with underlying word xxyyxxyy and
negativity 1, and so N (xxyyxxyy)= 16.
1
Proposition 5.3. Let w be an even word of length 2m and let s+t = 2m.
The coefficient of w in (xy) s (yx) t is s!t!N (w).
t
The proof of this proposition is postponed to the appendix at the end of
this paper. The idea is to associate to each xy-matching with underlying
word w and negativity t exactly s!t! ways to obtain w by expanding the
shuffleproduct(xy) s (yx) t. With the help of a formal definition of an
expansion of a shuffle product of this form, we are able to show that these
possibilitiesareexhaustive. Hereweshallillustratethecorrespondencewhen
m = 4, w = xxyyxxyy, s = 3 and t =1.
The figure below shows one way to obtain w by expanding the shuffle
prouduct xy xy xy yx.
x y x y x y y x
r r (cid:127)
r (cid:127)
r (cid:127)
r r (cid:127)
r (cid:127)
r (cid:127)
x(cid:15)(cid:15) x(cid:3)(cid:3) (cid:27)(cid:27)y y xxr r x(cid:15)(cid:15) x(cid:127)(cid:127)(cid:127) (cid:27)(cid:27)y '' y
Figure 3. An expansion of xy xy xy yx.
We obtain the corresponding xy-matching by connecting the letters com-
ing from the same xy or yx term in xy xy xy yx.
x x y y x x y y
Figure 4. The xy-matching corresponding to the expansion in Figure 3.
THE MOMENTS OF LE´VY AREA 9
Note that this matching, (xxyyxxyy,(13)(28)(46)(57)), has negativity 1,
corresponding to the single yx term.
There are in total 3!1! ways to obtain w by expanding the shuffle product
xy xy xy yx which correspond to this matching. The remaining five
are obtained by permuting identical words in the top line of Figure 3. For
example, we could just as well get the loop joining the first x to the first y
by expanding the shuffle product as shown below.
x y x y x y y x
r r (cid:127)
r (cid:127)
r (cid:127)
r
r (cid:127)
r (cid:127)
r (cid:127)
x(cid:127)(cid:127) x(cid:19)(cid:19) y(cid:11)(cid:11) y xxr r x(cid:15)(cid:15) x(cid:127)(cid:127)(cid:127) (cid:27)(cid:27)y ** y
Figure 5. Another of the 3!1! expansions of xy xy xy yx giving
the xy-matching shown in Figure 4.
Corollary 5.4. Let w be an even word of length 2m. The coefficient of w
in (xy yx) 2m is
−
(2m)! sgn(δ)
δ
X
where the sum is over all xy-matchings with underlying word w.
Proof. Let s + t = 2m. As the shuffle product is commutative, when we
expand (xy yx) 2m we obtain ( 1)t(xy) s (yx) t exactly 2m times.
− − s
Hence, by Proposition 5.3, the coefficient we seek is
(cid:0) (cid:1)
2m
2m
( 1)ts!t!N (w) = (2m)! ( 1)tN (w) = (2m)! sgn(δ).
t t
s − −
s+t=2m(cid:18) (cid:19) t=0 δ
X X X
where the final equality holds because the middle sum counts each xy-
matching exactly once, with the appropriate sign. (cid:3)
5.3. We now introduce our second and final combinatorial object.
Definition 5.5. An XY-matching is a pair (W,σ) where W is a word in
letters X and Y with equal numbers of X’s and Y’s, say m of each, and
σ S is a permutation such that W = X if and only if W = Y.
m i σ(i)
∈
Once again, it is very useful to represent XY-matchings by diagrams.
The figure below shows a typical example.
vv
## ##
A B A B B A
;; ]] 88
Figure 5. The XY-matching (XYXYYX,(1234)(56)).
As before, we need certain quantities associated with an XY-matching.
10 DANIELLEVINANDMARKWILDON
Definition 5.6. Let ∆ = (W,σ) be an XY-matching. The length of ∆ is
the length of the word W. We define the negativity of ∆ by
neg(∆)= # i : W = X and σ(i) < i or W =Y and σ(i) > i .
i i
{ }
We define the sign of ∆ by
sgn(∆)= ( 1)neg(∆).
−
We define the cycle count of ∆ by
cyc(∆) = #disjoint cycles in the permutation σ.
Finally we let e(∆) be the even word obtained from ∆ by replacing each X
with xx and each Y with yy.
For example, if ∆ is the XY-matching shown in Figure 5 above, then ∆
has length 6, neg(∆) = 3 (the 3 arcs drawn below the word contributing),
sgn(∆) = 1, cyc(∆)= 2 and e(∆) = xxyyxxyyyyxx.
−
Proposition 5.7. Let w be an even word of length 2m and let s+t = 2m.
The coefficient of w in (xy) s (yx) t is
22ms!t! 2−c(∆)
∆
X
where the sum is over all the XY-matchings ∆ such that e(∆) = w and
neg(∆) = t.
Proof. By Proposition 2.2, it is sufficient to prove that
N (w) = 22m 2−cyc(∆)
t
∆
X
where the sum is over all the XY-matchings ∆ such that e(∆) = w and
neg(∆) = t. Todothis, weshallassociate toeach XY-matching ∆ = (W,σ)
exactly 22m−cyc(∆) xy-matchings with underlying word e(∆).
Thecanonicalsuchxy-matchingisδ =(e(W),τ)whereτ istheinvolution
defined by
τ(2i 1) = 2σ(i) if 1 i m
− ≤ ≤
For example, if W = XYXYYX and σ = (1234)(56) then
τ = (14)(27)(36)(58)(9 12)(1011),
as shown below.
vv
## ##
X Y X Y Y X
(cid:31)+ (cid:31)* ;; (cid:31)+ (cid:31)* (cid:31)*^^ 88 (cid:31)+
(cid:31) + (cid:31) * (cid:31) + (cid:31) * (cid:31) * (cid:31) +
(cid:31) + (cid:31) * (cid:31) + (cid:31) * (cid:31) * (cid:31) +
(cid:31) + (cid:31) * (cid:31) + (cid:31) * (cid:31) * (cid:31) +
(cid:15)(cid:15)(cid:31) +(cid:21)(cid:21) (cid:15)(cid:15)(cid:31) *(cid:21)(cid:21) (cid:15)(cid:15)(cid:31) +(cid:21)(cid:21) (cid:15)(cid:15)(cid:31) *(cid:21)(cid:21) (cid:15)(cid:15)(cid:31) *(cid:21)(cid:21) (cid:15)(cid:15)(cid:31) +(cid:21)(cid:21)
x x y y x x y y y y x x
Figure 6. The canonical xy-matching associated to (W,σ).
