Second Class Particles and Cube Root Asymptotics for Hammersley's Process
Author(s): Eric Cator and Piet Groeneboom
Source: The Annals of Probability, Vol. 34, No. 4 (Jul., 2006), pp. 1273-1295
Published by: Institute of Mathematical Statistics
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TheAnnals of Probability
2006, Vol. 34, No. 4, 1273-1295
DOI: 10.1214/009117906000000089
? InstituteofMathematical Statistics, 2006
SECOND CLASS PARTICLES AND CUBE ROOT ASYMPTOTICS
FOR HAMMERSLEY'S
PROCESS
Eric
By
University
Delft
We
show
on
of a longest
of the length
from (0,0) to (f, t) is equal to 2E(f - X(f))+,
of a second
the flux
and the path
and Groeneboom
contexts
particle,
33 (2005)
continuing
and
the
between
the approach
of
879-903].
[8] show that in
paper Kim and Pollard
are confronted with estimators which converge at rate
rate nl//2 and that in this situation the limit distribution
A prototype
this phenomenon
"cube root asymptotics."
is the maximum
likelihood
estimator of a decreasing
density,
to the (almost surely unique)
locally at rate nI//3 after rescaling
They
of such an estimator
location
class
[Ann. Probab.
t)
path L(t,
is the location of
we
of the usual
is nonnormal.
which
North-East
where X(t)
the positive
In an influential
1. Introduction.
many
n1/3 instead
weakly
with
process,
on
class
variance
statistical
Amsterdam
?
at time t. This
that both E(?
X(i))+
particle
implies
.
are based on the relation
of L(t,
Proofs
t) are of order i2/
a second
Cator
Groeneboom
for a stationary
of Hammersley's
version
jt-axis
and Poisson
sinks
the positive
that,
the variance
j-axis,
Piet
and Vrije Universiteit
of Technology
sources
Poisson
and
Cator
call
converges
of the maximum
motion minus a parabola. The characteriza
in terms of Airy functions was given in [6].
that the asymptotics
for longest
subse
conjectured
increasing
can be analyzed
North-East
Ham
of
by studying
longest
paths
of Brownian
tion of this limit distribution
It has
been
which
quences,
in estimation
is related to these cube root phenomena
process,
mersley's
theory
to derive the asymptotics
and, in particular, that it should be possible
along similar
lines. However,
up till now, the cube root limit theory for longest increasing subse
and
quences
longest North-East
paths has been based on certain analytic relations,
see, for example,
[2] and [3].
involving Toeplitz determinants;
In this paper we will work with Hammersley's
process with sources and sinks,
as defined
in [4]. We will give a short description
1.We
here, based on Figure
the space-time
on
as
of
consider
that
started
the
jc-axis
sources,
paths
particles
to a Poisson distribution with parameter X, and we consider
distributed
according
as a time axis.
In the positive quadrant we have a Poisson process of what
we call a-points
1 by x), which will have intensity
(denoted in Figure
1, unless
otherwise
On
the
i-axis
also
sometimes
will
be
called
(which
specified.
j-axis) we
the i-axis
Received
AMS
2000
Key words
process,
cube
July 2005;
revised
September
subject
classifications.
and phrases.
Longest
root convergence,
2005.
Primary
60C05,
60K35;
increasing
subsequence,
second
class particles,
Burke's
1273
secondary
Ulam's
theorem.
60F05.
problem,
Hammersley's
1274
E. CATOR AND P. GROENEBOOM
(0,*) (M)
(0,0)
FlG.
1.
a Poisson
Space-time
(3,0)
paths
of the Hammersley'
s process,
with
sources
and
sinks.
of what we
call sinks of intensity
1/?. The three Poisson
are independent.
processes
to the right of it jumps
At the time an a-point
appears, the particle immediately
the leftmost particle
At the time a sink appears,
to the location of the a-point.
we
at time s,
intersect a line at time
disappears. To know the particle configuration
have
process
at
the space-time
paths. The counting process of the particle configuration
time t is denoted by Lx(-, t), where we start counting at the first sink on the ?-axis
t) equals
(0, oo) x {t}, so Lx(x,
up to (0, t), and continue counting on the halfline
x
the total number of sinks in the segment
[0, t] plus the number of crossings
{0}
s with
of space-time
[0, x] x {t}.
paths of the segment
The total number of space-time
paths in [0, jc] x [0, t] is called theflux at (jc, t).
= 1, we will denote L\(x,
X
If
It is in fact equal to L\(x,
t) just by L(jc, t),
t).
The flux Lx(x, t) equals the length of a longest
unless this can cause confusion.
that
to (x,t), where
"weakly" means
path from (0,0)
weakly NE (North-East)
are allowed
sources
or sinks from
the ?-axis,
1, trace back a longest
up of-points. To see this from Figure
(jc, t) to (0,0), and note that one will pick up exactly one
weakly
< x <
or
one sink from each space-time
or
one
source
path. Note that, if 0
a-point
?
of
(or crossings
t) is the number of particles
paths)
t)
Lx(x,
y, Lx(y,
space-time
on the segment
[x, y] x {t}.
we
to pick up either
from
the x-axis
before we
start picking
NE path from
of the length
A heuristic argument for the cube root behavior of the fluctuation
runs as fol
of a longest weakly NE path for the stationary Hammersley
process
=
with
NE
A
?
1.
that
for simplicity,
lows. Suppose,
length
path
longest weakly
=
t) can pick up points from either x- or j-axis before starting on a
L\(t,
L(t, t)
let, for -t <z<t,
strictly NE path to (t, t). Furthermore,
(1.1)
N(z) -
?number of sources in [0, z] x {0},
in {0} x [0 |z|j
j number of sinks
if z > 0,
if z < 0,
1275
CUBE ROOT ASYMPTOTICS
and
. ;
I(12)
A ru;
( )=
length of longest strictlyNE path from (z, 0) to (t, t),
if z > 0,
length of longest stricliy NE path from (0, |z|) to (t, t),
ifz<0.
Note
that the processes
At and N
are independent
L(t, t) = sup{N(z) +Mz)
(1.3)
The process
ogy of uniform
-
and that
:-t<z<t}.
<
z h> t~l/3{N(zt2/3)
\z\
\z\t2/3},
on
to
two-sided
convergence
compacta
in the topol
tl/3, converges
Brownian motion,
originat
the expectation
of t~l^3At(zt2^3)
has an
ing from zero. As will be shown below,
upper bound (as t -> oo) of the form
asymptotic
2,2/3 _k|/l/3_lz2f
is seen by taking expectations
and optimizing
the choice of X in the inequal
4.1. This suggests that the distance to zero of the exit point where the
longest path leaves either jc- or y-axis cannot be of larger order than r2/3, since oth
cannot cope with the downward parabolic drift
erwise the Brownian motion
?\z2,
the
that
fluctuation
of At{z)
is of order
The latter
assuming
temporarily
Op(t1/3).
fact we know to be true from the analytic approach, not used in our probabilistic
approach. On the other hand, we will derive this in Section 7; see (7.7).
The limit behavior
of the exit point can be compared
to the behavior
of the
which
ity in Lemma
location of the maximum
of Brownian motion minus a parabola, which plays a key
role in the asymptotics
for the cube root estimation
above. The
theory, mentioned
crucial difference,
is that the exit point is the location of the maximum
of
however,
the sum of two independent processes
instead of the maximum
of just one process.
We note here that the n]//3 convergence
in estimation
theory (so slower con
than the usual n1^2-convergence)
to
the t2^3 order of the
vergence
corresponds
to zero of the exit point, which
distance
(after a time-reversal
argument, based on
Burke's
theorem for Hammersley's
can
be
called
be
process)
"super-diffusive"
of a second class particle. The slower convergence
in estimation
theory is
caused by the fact that the estimators have an interpretation
in terms of the location
of amaximum,
just as the exit point for a longest weakly NE path inHammersley's
havior
process.
The key relation which allows us to make the heuristic argument above
rigorous
is Theorem
2.1 of Section 2, which,
in combination
with (time-reversal)
results
=
of Section
where Z(t)
is the rightmost
3, tells us that Var(L(i,
t))
2EZ(f)+,
see (3.2), Section 3. It is
point where a longest weakly NE path leaves the x-axis;
=
shown in Section 4 that this implies EZ(i)+
In
Section 5 we compare
0(t2^3).
NE
with
NE
and
obtain a bound on the
longest strictly
paths
longest weakly
paths
difference
between
the lengths of these paths. This allows us to also obtain a lower
bound
for EZ(i)+
in Section
6. Finally,
in Section
7 we
discuss
tightness
results
1276
E. CATOR AND P. GROENEBOOM
for the original Hammersley
results of Sepp?l?inen
[9].
sources
without
process
and sinks,
in connection
with
in par
Our methods
in [4], which concern,
heavily rely on the ideas developed
in behavior below and above the path of a second class parti
ticular, the difference
cle and Burke's
theorem for Hammersley's
process, which enables us to use time
reversal
and reflection.
2. Variance
the concept's
an important
of the flux and
location
second
class particle
role in later sections.
of a second
and dual
class particle.
We will need
class particle, which also play
second class particle
is created by
second
A "normal"
the first sink), and a dual
putting an extra source at (0,0) (thus effectively
removing
an
extra
at
is
sink
thus effectively
second class particle
created by putting
(0,0),
as the location at time t of a second class
the first source. Define X(t)
removing
a
in
with source and sink intensity equal
process
particle
stationary Hammersley
to 1 (the symmetric case),
particle for this case.
as the location
and Xf(t)
at time t of a dual second
class
in [4], a "normal" second class particle X(t)
jumps to the previ
explained
ous position of the ordinary ("first class") particle that exits through the first sink
at the time of exit, and successively
of particles
jumps to the previous positions
a
to
to
at
to the
the
of
times
where
these
it,
directly
right
particles jump
position
a
of
The
dual
left
the second class particle.
second class particle was
concept of
a
in [4], but there it is seen as
second class particle for the process
also considered
As
"moving from left to right." Figure 2 shows the trajectories of a second class parti
cle and a dual second class particle. Note that we always have X(t) < Xf(t), which
is evident from Figure 2.
.o.
-O?O?O?<8>?O???
(X(t\t)
.
(x,t)
>??
o
o
5S?e??
o
o
o
o
<t>?o?o?o?0?<$>
*
*
*
*
*
<&?*?*?*?*?*
FIG.
2.
Trajectories
of(X(t),
t) and
(X'(f),
t).
(X'(t),t)
1277
CUBE ROOT ASYMPTOTICS
1, source
process with ?-intensity
Hammersley
>
0 and consider
the flux L\(x,t).
1/?. Fix x,t
intensity ? and sink-intensity
and X'x(t) as the locations at time t of a second class particle and
Denote X\(t)
a dual second class particle, respectively. We use the subscript ? to indicate that
of the location of the (dual) second class particle depends on ?. If
the distribution
=
?
result:
1, the subscript is suppressed. We have the following
Now
a stationary
consider
Theorem
2.1.
Var(L?(jc,
Remark
A
2.1.
location
of a second
exclusion
processes
Remark
2.2.
t))
= -kx
similar
+
+
A
2?E(jc
between
relation
class particle has been
(TASEP) in [5].
the variance
proved
that taking ? = JTfx
Note
Var(L?(x, t))
=
Xx(t))+.
for totally
of
the flux
asymmetric
and
the
simple
yields
2?E(jc
Xx(t))+.
clarity we use the four wind direc
tions N, E, S and W to denote the number of crossings of the four respective
sides
= N +
= N + W.
of the rectangle
[0, x] x [0, t] [so L?(jc, t)
W]. Clearly, S + E
We also know from Burke's
theorem for Hammersley's
process
(see [4]) that N
Proof
of Theorem
2.1.
and E are independent,
For notational
just like 5 and W. This means
that
= Var(W + N)
= Var(W) + Var(A0 + 2Cov(W, N)
(2.1) = Var(W) + Var(A0 + 2Cov(5 + E-N,N)
Var(L?(jc, 0)
=
=
We
want
to investigate
Var(W)
--Xx
A
-
+
Var(A0
2Cov(S,
N)
+ 2Cov(S,N).
A^). It turns out that we can do this by varying the
For s > 0, we define a source-intensity
of ? + s.
source-intensity
appropriately.
The sinks remain a Poisson
with
We
denote
process
1/?.
intensity
expectations
with respect to this new source intensity by Ee. Define
Cov(5,
an=Ee(N\S
Note
=
n).
that an does, in fact, not depend on e, since we condition
on the number
in [0, x], and the sources outside this interval do not influence N. Then
sources
3s ?=0
ds
8=0^
n\
of
1278
E. CATOR AND P. GROENEBOOM
=
1 ^?v-? (xX)
->
e
_ ^
^-^
-?e
XKan-n-x}
This
shows
-E(NS)-xE(N).
?
that
-
Cov(N, S) = E(NS)
(2.2)
= X
E(N)E(S)
d
de
where
^
XKan
n=0
n=0
=
(xX)
E?(N),
e=0
we use that E(S) = ?jc.
will calculate
this derivative
in the following manner. Fix, independently,
a Poisson
1 of or-points
in (0, oo)2, a Poisson
of
of intensity
process
process
on
sources of intensity X on the jc-axis and a process of sinks of intensity
the
l/X
/-axis. Now we add an independent Poisson process of intensity s to the process of
We
sources. Define N? as the number of crossings
of the North-side
[i.e., (0, jc) x {t}]
for the process with the added sources.
Note
that if we add an extra source at (z, 0), then N increases by 1 if and only
if Xx(t; z) < jc, where Xx(t; z) is the location of a second class particle at time t,
that
which
started at (z, 0). We denote Xx(t) = Xx(t; 0). This means
E(N?) = E(N0) + s f
(2.3)
Therefore,
by using
the stationarity
E(t{Xx(t;z)<x})dz + 0(e2).
of the Hammersley
Cov(N, 5) = ??I
de
= \(
process,
E(N?)
\e=o
E(tlXx(t;z)<x])dz
(2.4) =X fX F(Xk(t) <x-z)dz
Jo
= X
=
fXF(x-Xx(t)>z)dz
Jo
XE(x-Xx(t))+.
Combining thiswith (2.1) gives
Var(L?(jc,
Now
sity 1.We
t))
=
-Xx+t-
a stationary Hammersley
denoted the flux of this process
consider
+
2?E(jc
process
Xk(t))+.
with
at (jc, t) by L(x,
source
D
(and sink) inten
t), and the location of a
1279
CUBE ROOT ASYMPTOTICS
started at (0,0),
at time t, which
(dual) second class particle
Note that under the map
by X(t)
i-> (x/X,Xt),
(x,t)
source intensity
1 gets transformed
process with
source intensity X (and corresponding
sink intensity
shows
ing argument
a stationary
process with
XXk(t) = X(t/X)
(2.5)
where
=
and
XX[(t)
=
into a stationary
l/X). This rescal
X\t/X),
equality in distribution.
like to bound Var(L^(i,
t)) in the case where
t)) in terms of Var(L(i,
2.1 and (2.5), we get, using the inequality
Theorem
(A + B)+ <
denotes
We would
X > 1. Using
A+ + B+,
Var(L?(i, t))
If we
[resp. Xf(t)].
= -Xt +
t/X + 2E{Xt t/X + t/X X(t/X))+
<(X\/X)t + 2E(t/X X(t/X))+
= (X\/X)t + Var(L(//?, t/X)).
the intuitively
show
clear result that, for X >
1,
(2.6)Var(L(i?X, t/X)) < Var(L(i, t)),
we have proved
that
(2.7)
Var(L?(i, t)) <(XWe
can show
(2.8)
that Theorem
(2.6) by noting
V<ir(L(x,t))
l/X)t + Var(L(i, t)).
2.1 for X =
1 is equivalent
to
= -x + t+ 2 ?
Jo F(X(t)<z)dz.
Define
(j)(x,t)
=
Var(L(jc,i))
source
the
and sink intensities
since
(p is symmetric,
Clearly,
reflection
symmetry of the process.
Furthermore,
(2.8) shows that 0
is a continuously
are equal, which
differentiable
function,
=
gives
with
di(?(x,t)
-\+2F{X(t)<x).
< t) >
can show that F(X(t)
have proved
(2.6), since
1/2, we would
> 0 for a
that
is
function
in t.
d\(t)(t, t)
0 implies
(j)(t, t)
symmetric
increasing
Since reflecting Hammersley's
in the diagonal preserves
the distribu
process
If we
tion, while
interchanging
the trajectories
of X' and X, we know
that
=
<
?(X(t) > jc) P(X'(jc) < t) P(X(jc) < t).
Choosing
/=i,we
see that
F(X(t) >t)<
which
shows
that ?(X(t)
<t)>
F(X(t) <t)
=
F(X(t)
1/2. As noted before,
<
t),
this proves
(2.6).
1280
E. CATOR AND P. GROENEBOOM
3. Connection
between
second
class
we
and
exit points.
As has al
the flux L\(x,
t) in two ways:
particles
can view
ready been noted in the Introduction,
it is the number of space-time
paths in the square [0, jc] x [0, t], but is also the
to (jc, t), where
of
NE
the
length
longest weakly
path from (0,0)
"weakly NE"
means
as long
that we are allowed to pick up sources or sinks, as well as a-points,
we
as we are going North-East.
To work with this latter representation,
will
which
use in the symmetric
case when both the source- and the sink-intensity
mainly
are 1, we define, for ? t < z < t, N(z)
and At(z) by (1.1) and (1.2). Remember
are
TV
that the processes
and
and that
At
independent
L(t, t) = sup{W(z) +Mz)
(3.1)
:-t<z<t}.
important aspect of this representation
leaves either the x-axis or the y-axis. Define
Another
path
at which
is the location
a longest
:
+ At(z) = L(t, t)}
Z(t) = sup{z e [-t, t] N(z)
(3.2)
and
Z\t) = inf{z
(3.3)
We
call Z(t)
and Z'(t)
:
+Mz)
[-/, t] N(z)
exit points
for a longest path,
that leave the axis on (Z(t), 0) [or (0, -Z(f))]
From
this definition
and using
the symmetry
= L(t, t)}.
since
there exist
longest
of the situation,
we
can see that
and Z(t) = -Z\t).
(3.4)Z\t)<Z(t)
We have defined
will need another link between
the two representations.
as the position
at time ? of a second class particle,
and X\t)
respectively
second class particle, that starts at (0, 0). Now define
We
Y(t) =
X(t)
dual
if X(t)<t,
t-X(t),
M{s
paths
or on (Z'(i), 0) [or (0, -Z'(i))L
>0:X(s)
>t}-t,
ifX(t) > t,
>t}-t,
ifX\t)
and
y\t) =
Since X'{t)
X and Y.
It also
shows
>
X(t),
mf{s >0:X'(s)
<
we have Y'(t)
two relations
a<t
if X'(t)<t,
t-Xf(t),
=?
which
we will
(3.5)
=?
Figure
3 shows
be important
{X(t) <a} = {Y(t)>t-a}
{X'(t)<a}
a>t
Y(t).
{X(t)>a}
{X\t)>a}
=
=
=
> t.
the relation
later:
and
{Y'(t)>t-a},
{Y(a) <t-a)
{Y'(a)<t-a}.
and
between
1281
CUBE ROOT ASYMPTOTICS
-Y(a)
-Y(t)
A
t
FIG.
3.
Relating
X
and
a
Y.
Figure 4. The left picture shows a realization of the Hammersley
to Z(t) and Z'(t). The
two
and
process
longest weakly NE paths, corresponding
now
same
but
reflected in the point
realization,
right picture shows the
(r\t, ^t).
of a second class and a dual
Note
that the longest paths become
trajectories
Now
consider
to Y(t),
and that Z(t) corresponds
process,
to
in
states
the
reflected
Burke's
theorem
that
[4]
Y'(t).
Zf(t) corresponds
so that we can
is also a realization of the stationary Hammersley
process,
process
indeed conclude
that
second
in the reflected
class particle
while
Z(t) = Y(t)
In particular,
that
and Z'(t) = Yf(t).
using Theorem
this means,
(3.6)
Var(L(t,t))
2.1 and noting
=
that (t
?
X(t))+
o
<|>
<>
o
I Y'(t)
o
o
4>
<>
o
o
?
o
<>
<!>-?
o
z\t)
o
<t>O O O O O O b o o o1 o o
Z(t)t
FlG.
4.
Longest
path
is distributed
Y(t)+,
2EZ(t)+.
<J>-O0-0-0-?
<>
<>
=
as
trajectory
of a second
class
particle.
1282
E. CATOR AND P. GROENEBOOM
is of order 0(t2/3).
4. EZ(t)+
We wish to control the exit point Z(t). We
an auxiliary Hammersley
to the
do this by considering
process L\, coupled
> 1
sources
a
to
X
the
Poisson
of
process
original one, by thickening
intensity
a
and thinning the sinks to Poisson process of intensity
l/X. The process At then
satisfies the following
inequality.
will
LEMMA
Let X>\
4.1.
and define
t) as the flux
L\(x,
of L\
at (jc, t). Then,
forO<z<t,
Mz)<Lx(t,t)-Lx(z,0).
that a strictly NE path from (z, 0) to (t, t) is shorter than a
to either
this path is allowed
longest weakly
path from (z, 0) to (t,t), where
x
on
or
sources
of
with
of
L\
L\
[z, t]
{z} x [0, t].
{0},
pick up crossings
pick up
this longest weakly NE path is equal to the number of space-time
However,
paths
in turn, is equal to L\{t,
the number of
in [z, t] x [0, t] of Lx, which,
t) minus
PROOF.
It is clear
NE
sources
We
on [0, z] x
can now
there exists
Theorem
{0}.
D
theorem. We use the notation a(x)
the following
<
such that, for all parameters x,a(x)
Mb(x).
show
<b(x)
a constant M
4.2.
Let0<c<
Then
t/EZ(t)+.
,iz"i>tEz('ws(i4?(?+?)'
PROOF.
that, for any ? >
Note
1,
= L(t, t)}
?{Z(t) > u) = ?{3z > u :N(z) + At(z)
< P{3z > u :N(z) + Lx{t, t) - Lx(z, 0) > L(t, t)}
=
F{3z
> u :N(z)
-
Lx(z,
0) > L(t,
t)
Lx(-, 0) is a thickening of L(-, 0), we get that ?X-\(z)
? 1.
This means
is in itself a Poisson process with intensity ?
Since
P{Z(0 >u}<
To have a useful
bound
E?x-i(u)
is maximal.
-
This means
F{?x-\
for all 0 < u <
E{Lx(t, t)
-
(w)< Lk(t, t)
\t,
we
L(t, /)} = (A
that we choose
xu
=
choose
(i-u/trl/2.
-
-
Lx(t,
:= Lx(z,
that
t)}.
0)
L(t, /)}.
X such that
\)u
-
tik +
- - 2\
-
N(z)
if
CUBE ROOT AS YMPTOTICS
useful
Some
that hold for all 0 < u <
inequalities,
elementary
1283
\t,
are
K<2,
(4.1)
-
E?^-ii?)
E{Uu(t,
t)
-
L(t, 0} > \u2/t,
k?-\/k?<2u/t.
Note
that, due to (2.7) and (3.6),
<
2(Var{L??(i, t)} + Var{L(i, t)})
Var{L??(f, r) L{t, t)\
Now
we
can use Chebyshev's
?{Zit) >u)<
<?EZit)+
+ 2tiku-l/ku)
<8EZ(i)+
+ 4w.
inequality:
<
Pj?^-iOi)
Lk?it,t)
-
Lit, t)\
<P|Vi(?)<EJVi,_i(H)-?2/(8/)}
+ ?{LXuit, t) Lit, t) > E?^-iiu)
<
(4.2)
64i2(?M
-
\)u
t2
t)
-
we
?r
see that
3 )
P{Z(0>h}<p{z(?)>-?}<^3
we
now follows
use
(4.2). This means
u=
from choosing
4.3.
limsup
that (4.2)
Let Z(t)+
f_>oor
f2/3
=
hmsup
,-W
t. The
0(t2/3).
t) be defined
Var(L(M))
EZ(?)+ = r
?^?
+ ?2EZ(0+
M4
D
that EZ(t)+
and L(t,
i2
is true for all 0 < u <
cEZ(t)+.
can show
this theorem we
COROLLARY
Lit, t)) + u2/iSt)}
t2EZ(t)+
u?
|f,
-
u*
^ t2
If t > u >
Lit, t) > E{LXuit, t)
64i2(8EZ(r)+ + 4w)
U^
With
h2/(80}
?4
+ ?{LKit,
where
-
??^-<
2?2/3
as in (3.6). Then
,
+oo.
theorem
1284
E. CATOR AND P. GROENEBOOM
Proof.
there exists
(3.6), we only have to prove
Using
a sequence
such that
tn f +00
EZfe)+
r
hm -r??
n-+oo
Using
4.2, we
Theorem
=
the statement
)+. Suppose
+00.
tn1'5
see that
+
P(2??,+
;l)Al.
>?Z(,?)+l<5^-?(-l
Using dominated
this shows that
for EZ(f
[note that t2/(EZ(tn)+)3
convergence
is a bounded
sequence],
POO
/ F{Z(tn)+>cEZ(tn)+}dc^O,
Jo
which
would
the absurd
imply
assertion
that
lEZ(i?)+J
As
a corollary
we get the following:
4.4.
Corollary
Let c >
1. Then
nz(t)>ct2'3}<^.
This
PROOF.
corollary.
Remark.
is an immediate
consequence
c?
of Theorem
4.2 and the previous
D
to a result on transversal fluctuations
result can be compared
path in [7]. He shows that all longest stricly NE paths from (0,0)
in a strip along the diagonal of width
ty, with probability
tending
This
of a longest NE
to (t,t) remain
to 1, as t ?> 00, for any y > 2/3. To this end, he uses the analytic results in [2].
with Corollary 4.4, shows that the transversal fluctuation
Section 3, in combination
?
<
s
a
t
at time
of
s)2?3.
longest weakly NE path from (0,0) to (t, t) is of order (t
the reflected
This is due to the fact that any longest weakly NE path lies within
see Figure 4.
trajectories of a second class particle and a dual second class particle;
same
NE
for
on
result
the
NE
Our result
strictly
paths, as the
weakly
paths implies
the longest strictly NE path that is
short argument will show: consider
following
to the right of all other longest strictly NE paths and suppose, at time s < t, this
of
Since the sources and sinks are independent
path is to the right of the diagonal.
a
at
of
is
least
we
there
still
this
have that, given
this event,
event,
1/2
probability
that Z(t) > 0. The corresponding
longest weakly NE path (so the weakly NE path
that is most
to the right)
cannot
be to the left of the considered
longest
strictly NE
1285
CUBE ROOT ASYMPTOTICS
to the right of a
they cannot cross), so the order of the fluctuations
same
cannot
than
the
order
for
NE
be
longest weakly
higher
path
longest strictly
to the left, a similar argument holds. The remark at the
NE paths. For fluctuations
lower bound result.
the corresponding
end of Section 6 discusses
(because
path
To get a lower bound
5. Strictly NE paths and restricted
weakly NE paths.
a strictly NE path and
on EZ(t)+,
we need to control the difference
between
a weakly NE path in a stationary Hammersley
process L\ (with source inten
sity 1), where the weakly NE path is only allowed to pick up sources in an interval
[0, st2/3] x {0}. To do this, we consider another independent Hammersley
process
1; for
Lx on [0, t]2 with source intensity X, sink intensity
l/X and a-intensity
are
of
the
this process,
the corresponding
sources, sinks and a-points
independent
to this process Li, we consider Lq as the correspond
for L\. Coupled
processes
but has no
that uses the same a-points,
process
ing (nonstationary)
Hammersley
or sinks.
sources
We
denote
space-time
segment
L$(x,
t) as the number
of the Hammersley
process
that, for 0 < z < t,
paths)
[0, x] x {t}. Note
-
(5.1)
AM
This
of particles
At(z)
=
L0(r,
t)
the number
(i.e.,
without
-
-
L0(t
of crossings
or sinks with
sources
of
the
z, t).
?
from the fact that Lo(t
z,t) equals the length of the longest (strictly)
NE path from (0,0) to (t
z, t). Also note that {N(z) :z e (0, t)} [the number of
sources of the process L\ in the interval (0, t)] and {Lx(t, t) ? Lx(t ? z,t):z
e
are
two
Poisson
(0, t)}
processes.
independent
follows
Define
process
(5.2)
as the position
at time t of a dual second
X[(t)
Lx, that started in (0, 0). Then we know that
x < y <
X[(t)
=?
Lx(y,
t)
-
Lx(x,
t)
<
L0(y,
class
t)
-
particle
L0(x,
of the
t).
to the fact that if we leave out all the sources of Lx, the space-time
not
do
paths
change above the trajectory of X'x (this is one of the key ideas in [4];
the reader might want to check this fact by looking at Figure 2). This means
that if
we define the process Lasa
uses
a
same
that
the
process
Hammersley
-points and
we
sinks as L^, but starts without
that
have
sources,
any
This
is due
x <
X'k{t)
=?
Lx(x,
t)
=
L(x,
t).
from the fact that the set of particles of L is at all
(5.2) now follows
Inequality
times a subset of the set of particles
of the Hammersley
process Lo, since this
no
has
L
whereas
does
have
sinks.
sinks,
process
Theorem
limsupPJ
5.1.
sup
Fix L > 0. Then
{N(z) + At(z)}
-
At(0) > Ltl/3\ = 0(e3),
e?0.
E. CATOR AND P. GROENEBOOM
1286
PROOF.
We will
use the auxiliary
constructed
process
Hammersley
above. De
fine
X=\-rrx'3.
If
Xfx{t)
> t, (5.2) tells us that, for 0 < z < t,
Lk(t,
t)
-
Lk(t -z,t)<
L0(t,
t)
-
L0(t
-
z, t).
Using (5.1) and defining
Nk(z)
we
see that (remember
that N
=
Lx(t,t)-Lk(t-z,t),
and At are independent)
sup {N(z)+Mz)}~M0)>Lt1^3}
(
J
lze[0,e?2/3]
^[0,??2/3]
(5.3)
<
The
second
sup {N(z) Nk(z)} > Ltl/3\ + F{X[(t) < t).
p{he[0,8t2/3]
i
term on the right-hand
4.4:
side of
can be bounded
(5.3)
using
(2.5),
(3.5) and Corollary
=
F{X'(t/k) < Xt}
P{X'k(t) <t}
=
?{Zf(t/X)>t(l/X-X)}
(5.4)
<F{Z(t/X)>t(l/X-X)}
=
P{Z(i/(l
> rt2/3(2
rt'l/3))
-
r/"1/3)/(l
rt~l/3)}
<P{Z(f)>rf2/3}<r~3,
t = t/(\ ? ri~1/3).
4.4 with argument
for all r e [1, i1/3), applying Corollary
side of (5.3) concerns a hitting time for the dif
The first term on the right-hand
this can be written
After rescaling,
ference of two independent Poisson processes.
as
P{30<z<,
: ?~1/3{7V(z?2/3)
-
The process z ?- t~xl3{N{zt2/3)
Brownian motion
process
(5.5) Wr(z)
-
converges,
?k{zt2/3)}
=
W(2z)
+
?x{zt2'3)}
>
L).
as t -> oo, to the drifting
z > 0,
rz,
on compacta, where W is standard Brown
in the topology of uniform convergence
of Donsker's
on R+. Hence, we get, by a standard application
theorem,
ian motion
lim P{3o<z<, r:
l/3{N(zt2/3)
(5.6)
= P
sup Wr(z)>L\.
Ue[0,e]
-
Nk(zt2/3)}
>
L)
1287
CUBE ROOT ASYMPTOTICS
now get, for r < L/e,
We
pj supWr(z)>L\ <PJ
=
sup W(z)>L-sr\
supW(2^)/v/2e>(L-^r)/V2e|
PJlze[0,l]
J
(5.7)
supW(z) > (L ?r)/V2??}
j e[o,n
2
e-^u2du.
Vtt/o(L-er)/>/2?
Taking
r=
L/(2s),
V*V
we get
-(1/2)?2 du
er)/j2e
2
?
du
H
L/Vs?
nr-
^-
_j2
s
==-,
|
0,
L\f2?
y/2n
in the last
ratio approximation
for the tail of a normal distribution
using Mills'
that, with this choice of r, our estimate for the second term on
step. This means
so (5.4) now proves the theorem.
the right-hand side of (5.3) is dominant,
D
that we
Note
at a certain
uniformly
could use the proof of the theorem to show that L can even go to 0
-> 0, and still the considered
speed when e
probability would go to 0,
in t.
6. Lower
bound
We wish to bound the probability
for EZ(f)+.
that Z(t) e
we
to
an
In
order
introduce
do
station
this,
[0, et2/3].
again
independent
auxiliary
ary Hammersley
process L^, but now with intensity X > 1. In fact, we will choose
-1/3
X=\+rt
to this process, we again consider
the Hammersley
process Lo without
Coupled
sources or sinks, but with the same a-points.
This time, however, we will leave out
the sinks of the stationary process;
to be more precise, we have that
(6.1)
y > x > Xk(t)
=>
L0(y, t)
-
L0(x, t) < Lx(y, t)
-
Lk(x, t).
reason is that if we consider
the stationary process Lx and leave out the sinks
of this process, below the trajectory of X\ the space-time
paths do not change. The
set
then
follows
from
of
fact
that
the
the
inequality
particles of a process which,
The
like Lo, has no sinks, but starts with
sources
and uses
the same a-points,
will
at all
E. CATOR AND P. GROENEBOOM
1288
times be a superset of the set of particles
this to the explanation
of (5.2).
We again define
=
?x(z)
and will
show
Theorem
the following
Lk(t,t)
of the Hammersley
-
-
Lk(t
process
Lq. Compare
z,t)
result.
6.1.
limlimsupP{0
t^
?40
Proof.
Let
=
< Z(t) <
et2'3} 0.
to find s > 0 such that
r\ > 0. It is enough
< Z(t) < st2'3} < 3r?.
limsupP{0
f?>-oo
For any L, r > 0 and X=
I+
rt~1'3,
we have
P{0 < Z(t) < st2'3}< PJ sup (N(z)+Mz)) < sup (N(z)+ At(z))\
h>et2/3ze[0,st2/i] J
<
(6.2)
sup {N(z) (MO) Mz))}
PJyz>?t2?
j
+
which
allows
(6.1), which
Z < rt2/3,
Ltl'3\
sup {N(z)-(M0)-At(z))}>Ltl'3\,
PJ{ze[0,et2/3]
i
since for any L e R, X < Y implies
F(X <Y)<
<
that either X < L or Y > L, so
F({X < L] U [Y > L}) < F(X < L) + F(L < Y),
over L. For the first term of (6.2), we want to use
us to "optimize"
?
on
event {Xk(t) <t
the
rt2'3}, we know that, for all
implies that,
L0(t,
t)
-
L0(t
-
z, t) < Lk(t,
t)
-
Lx(t
-
z, t).
Therefore,
P{0<Z(O<^2/3}
<
(6.3)
sup
PJhe[st2/\rt2/3]
{N(z) Nk(z)} < Ltl/3\ + F{Xk(t)> t rt2'3}
J
sup {N(z)-(At(0)-At(z))}>Ltl'3\
>
+PJhe[0,st2/3]
=F
+
sup {N(z) Nk(z)} < Ltl/31 + F{X(t/X) > Xt Xrt2'3}
jlZe[et2/\rt2/3]
J
PJ
sup {N(z) (A,(0) Mz))}
> Ltl/3\.
CUBE ROOT ASYMPTOTICS
Using
?Z'it)
1289
- krt2? >
(3.5) (note that kt
t/k when r < ?r1/3), Z'(i) < Z(f) and
=
Z(t),
we
get, for the second
term in (6.3),
> kt - krt2?] = ?{Zikt - krt2?) < (1/? - k)t + krt2?)
< ?{Z\kt - krt2?) < (1A - k)t +
krt2?}
P{X(i/?)
=
=
-
?{Zikt
>ik-
krt2?)
?{-Z'ikt
>ik-
krt2?)
second
that we
start by choosing
than r?, since
krt2?)
krt2?)
-r2/1/3
l+r?-l/3
r sufficiently
can
term is smaller
\/k)t
-
-
rt2?
= P
Z(?(l -r?r?li))>
This means
\/k)t
large
to ensure
that the
2/3
r2tl?
>{z(ta-r2r2^>-i+rt_r/3
=
+ 0(i,/3)) > rt2? +
pjz(i
where
Now
found
0(?1/3)j
= 0( r\
t
OO,
we use Corollary
4.4 in the last step.
we turn to the first term in (6.3). This
in the previous
section. We
term is very similar to the term we
as in the proof of Theorem
5.1, that the
have,
process
z^r]/3{N(zt2/3)-?x(zt2/3)}
converges,
as t -> oo, to a drifting
(6.4)
Wriz)
Brownian
=
Wi2z)-rz,
in the topology of uniform convergence
on R+
ian motion
(this time the drift
Donsker's
theorem,
get, again using
lim P
sup
motion
process
z>0,
on compacta, where W is standard Brown
is negative
instead of positive). Hence, we
1/3
{Niz)-?xiz)}<Lt
ze[st2/3,rt2/3]
sup Wr(z) <L
Ue[?,r]
(6.5)
sup Wr(z) <L
Z
<
[0,r]
sup Wr(z)<L\+]
ze[0,r]
sup Wr(z)<L,
ze[s,r]
sup Wr(z) > L
ze\0,e)
sup Wr(z)>L
ze[0,s]
1290
E. CATOR AND P. GROENEBOOM
Since
sup
Wr(z)<L\=0,
limPJ
?4?
J
lz [0,r]
Ue[0,r]
we
L = L(r?) > 0 sufficiently
can choose
small
to ensure
sup Wr(z)<L\<r]/2.
J
ze[0,r]
It is also clear from the argument
can next choose e > 0 sufficiently
5.1
of the proof of Theorem
small to ensure that
supWr(z) > l\ < PJ sup W(2z) >l)<
[see (5.7)]
that we
rj/2,
for this choice of L = L(rj) > 0.
It is now seen from (6.5) that this bounds the first term of (6.3) from above by rj
that we have already fixed r > 0 to bound the second term of (6.3)].
[remember
we
can
choose s > 0 so small that the third term in (6.3) is smaller than rj,
Finally,
D
5.1. This completes
the proof.
using Theorem
COROLLARY
Let Z(t)+
6.2.
.
fEZ(t)+
hminf?x-pz?
f-*oo
tn
Proof.
Using
? oo
such that
and L(t,
.
=
fVar(L(M))
hminf-^->
t-+oo
t1'5
(3.6), we only have
as in (3.6). Then
t) be defined
to prove
0.
2tll5
the statement
for EZ(i)+.
Suppose
EZfa)+ ~> U
2/3
In
Then
Since -Z'(t)
= Z(t) and Z'(t) < Z(t), we have thatF{Z(t) > 0} > 1/2, for all
t > 0. This would
mean
that, for all s > 0,
liminfP{0 < Zfo.) < ?in2/3}> i,
which
would
Remark.
contradict
We
Theorem
can again make
6.1.
D
a comparison
with
the results
in [7]. Johansson
shows that the probability that all longest strictlyNE paths from (0,0) to (t, t) stay
within a strip around the diagonal of width ty does not tend to 1, as t -> oo, for all
6.1 shows that the
results in [2]. Our Theorem
y < 2/3, again using the analytical
to
from
NE
all
that
(0, 0)
(t, t) stay within a strip
paths
longest weakly
probability
around the diagonal of width tY tends to 0, as t -> oo, for all y < 2/3 (see also the
Remark
at the end of Section
4).
1291
CUBE ROOT ASYMPTOTICS
7. Tightness
drodynamical
process, with
In the preceding
sections
it was shown, using the "hy
a
of [4] that, for
version
of Hammersley's
stationary
on
1 for the Poisson
axes and in the
the
point processes
results.
methods"
intensity
the variance of the length of a longest weakly
t2/3, in the sense that
NE
plane,
path L(t,
t) is of order
0 < liminfi"2/3 Var(L(i, *)) < limsupi~2/3 Var(L(f, *)) < oo.
(7.1)
This means,
that, for any t > 0, the sequence
in particular,
nt)
n~x/3{L(nt,
n =
2nt],
1,...,
is tight.
As noted
?
in [9], the distributional
limit result for n~l/3{Lo(nx,nt)
2n*Jxt\
sources and sinks in [2] can be translated into a
for Hammersley's
process without
limit of Yn/n1/3, where
Yn
(7.2)
=
znt([nx])-nx2/(4t),
and Znt([nx])
is the [nx]th particle at time nt, counting particles at time nt from
the left. Theorem
3.2 in [9] gives a tightness result for a more general version of
Yn, in the context of a version of Hammersley's
line, with a
process on the whole
(possibly)
initial state. The
random
result
is that, under his conditions
D and E, the
sequence
n=
Yn/(nl/3\ogn),
l,...
is tight. He conjectures
that, in fact, n1/3 logn can be replaced by n1/3. The results,
derived above, are a further indication
that indeed n{/3 logn might be replaced by
methods.
nlj/3, and that this can be derived by hydrodynamical
For
the stationary
version of Hammersley's
1 of the
process, with intensities
on
we
in
can
the
and
as the
the
define
axes,
processes
plane
znt([2nt])
location of the [2n?]th source at time nt, where we count the sources from left
to right, starting with the first source to the right of zero. Note
that at time zero
the particles are just the sources. The particles,
escaping
through a sink, are given
Poisson
location
zero at times
With
this definition,
-
(7.3)
for each
(7.4)
n~l/3{znt([2nt])
t > 0. This
t > 0, the "switch
larger than or equal to the time of escape.
our results give tightness of the sequences
can be seen
nt},
in the following
n=
1, 2,...,
way. We
have,
for M
> 0 and
relation"
n-x,3{znt([2nt])
-
nt) > M
^=?
L(nt + Mnx'3,
nt) <
[2nt].
E. CATOR AND P. GROENEBOOM
1292
Theorem
2.1 yields
= -nt - Mnl/3 +nt +
Var(L(nt + Mnx/3, nt))
2E(nt + Mn1'3
Xx(nt))+
= -Mn]/3 +
2E(nt + Mn1'3
Xx(nt))+
< Mnl/3 +
2E(nt X\(nt))+,
where
we
use
direction,
opposite
<
(Y + Z)+
1+ + Z+
in the last step. Theorem
in the
yields
-
2E(nt
=
Xi(nt))+
we get, by (7.1) and Chebyshev's
Hence,
2.1, applied
Var(L(nt,nt)).
inequality,
P{^"1/3{^([2^])-^}>M}
=
We
similarly
F{L(nt
+ Mnl/3,
nt)
-
Mnl/3 + Vcir(Li(nt,nt))
[Mnl/3 + 2nt-[2nt]}2
-
that, if nt
?
[2nt]
-
2nt
Mnl/3}
Mn1'3 + 0{(nt)2^3) _~
_{
?{M
}*
M2^3
~
nt}
<
=
-M]
0(M~l),
> 0,
Mn~1/3
-
(7.5)
<
Mnl/3
get
F{n-l/3{znt([2nt])
using
-
2nt
n-l/3{znt([2nt])
< -M
nt}
<=^
L(nt
-
Mnl/3,
nt)
>
[2nt],
which
(7.3).
proves the tightness of the sequences
the tightness of the sequence
process
Although
(Yn/n1^3) for the Hammersley
sources or sinks, as defined
in (7.2), is known from the results of [2], it
without
sections. The
is of some interest to derive this from the results of the preceding
tightness
will
(7.6)
follow
from
=
n~l/3{L0(nx,nt)-2y/xt}
n -+ oo,
Op(l),
for all x, t > 0, where Lo(nx, xt)
where the intensity of the Poisson
We
again have a "switch
[1], we know
(7.7)
show
L0(nt
+ Mnx'3,
nt) <
that
Lo(nx,
so if we
?=>
-nt}>M
n-l/3{znt([2nt])
From
is a strictly NE path from (0, 0) to (nx, xt),
in the first quadrant is equal to 1.
process
relation" similar to (7.4):
that, for each
n~x/3{L0(nt,
nt)
=
Lo(nVxt,
n^/xt),
t > 0,
nt)
-
2nt}
=
Op(l),
n -> oo,
[2nt].
and
1293
CUBE ROOT ASYMPTOTICS
we get, for each e > 0 and t > 0,
?{n~l/3{znt([2nt])-nt}>M}
=
=
+ Mnl/3,
?{L0(nt
-
nt)
2nt
?{L0(ntyJl+Mn-2/3/t,
-
<
Mnl/3
[2nt]
-
2nt
Mnl/3}
+ Mn~2'3/t)
ntjl
+ Mn~2/3/t
2nt^l
-
+ 0(M2rc~1/3)
<[2rc?]-2rc?-Mrc1/3}
for sufficiently
= M
large M
>
(s) > 0 and all n
no(M,
s). Relation
(7.7) similarly
implies
<
< e,
nt}
-M)
?{n-l/3{znt([2nt])
=
> 0 and all n > no(M, s),
for sufficiently
M(s)
large M
using
< -M
^=>
nt}
L0(nt
nt) >
Mnx/3,
n-l/3{znt([2nt])
In order
to prove
(7.8)
to show
it is sufficient
(7.7),
t)
t~l/3{L0(t,
-
[2nt].
2t}
=
t -* oo.
Op(X),
Now first note that the length Lo(t, t) of a longest strictly NE path from (0,0)
?
to (t, t) is the same as At (0) in the proof of Theorem
5.1. Let, for X = 1
rt~1/3,
5.1. By (5.3), we have
Lx be defined as in the proof of Theorem
{N(z) + At(z)}
sup
z
-
MO)
> Ltl/3
[0, Kt2/3]
(7.9)
<P
We
first deal with
sup
{N(z)
-
Nx(z)} > Li1/3 + P{X^(i) < t}.
he[0,Kt2/3]
'
the second
term on the right-hand
side of (7.9). By
(5.4), we
have
=
(7.10) F{X'x(t)<t}
for r e [1,
uniformly
\tl?3].
we
note that
first
(7.9),
z^
is a zero-mean
Mr(z)
=
martingale.
To deal with
the first term on the right-hand
-
rl/3{N(zt2/3)
So we
0(r~3),
?x(zt2/3)}
-rz,
z>
0,
get
sup {N(z)-?x(z)}>Ltl/3}=?\
p( ^ze[0,Kt2/3]
J
he[0,K]
sup {Mr(z)+ rz}> l]
J
sup Mr(z)>L-rK\.
side of
E. CATOR AND P. GROENEBOOM
1294
r = L/(2K)
Taking
and using Doob's
submartingale
sup {N(z)-Nx(z)}>Ltl?\
>
we get
sup Mr{z)>L/l\
<p(
h
ze[0,KtV3]
inequality,
i
[0,K]
sup Mr(z)2>L2/4
z [0,K]
We
also have, by Corollary
-
^
4EMAK)2
4.4,
{N(z) + At(z)}?
sup
sup {N(z) + At(z)}
ze[-t,t]
ze[-Kt2?,KtW]
< 2P{Z(i) > Kt2?) <
\/K3.
So, taking K = L1/n
[note that this means
for L < t4/5], we
?i1/3,
that r = L/(2K)
=
<
jL5'n
obtain
sup {N(z) + At(z)}-At(0)>Ltl/3\
i
lze[-f,?]
= P{Li(f,
for all L<t4>5.
If L > t4/5, we first note
P{Li (/, /)
-
?) A,(0) > Li1'3} < L'5'*,
that
A,(0) > Lt1/3} < P{Li (r, i) > Li1/3}
>
>
<
<
2P{Pf
2?{P,
?Li1/3}
2L1/6?},
t. Let [x] denote the largest integer
where f, is a Poisson variable with expectation
< x and let a =
term in an expansion
the
remainder
Then,
using
Lagrange
jL1/6.
of ?, we get, for a 0
(0,1),
F{Pt
Stirling's
formula
>
?L1/6i}
<
P{/>,
for the gamma
r(a? + l)
L,
>
implies that ?{Pt
if L > t4/5, uniformly
[fli]}
function
=-?-
T(x)
t[at]e-(l-9)t
< ?-.
z
[aty.
yields
{[at]
[at]\
that, uniformly
in t >
1,
1
tat
This
>
:g-?(loga-l)f
a^oo.
V2?H7
tends to zero faster than any negative
\Ltx?)
that
in all large t and hence, we can conclude
P{Li(f, 0
-
A,(0) > Ltl/3} = 0(L_5/4),
power
of
CUBE ROOT ASYMPTOTICS
for all L >
1295
1, implying
0 < 2f - EAt(0) = E{Li(t,
t)
-
Af(0)}
(7.11)
= 0(?1/3).
<?1/3{l+^??L-5/4dL}
Thus,
-
E|At(0)
-
2t\ = E\L0(t, t)
2t\ = 0(tx'3).
This proves (7.8) and, as noted above, (7.6) now also follows. This, in turn, proves
for Yn defined by (7.2) for Hammersley's
tightness of the sequence
(Yn/n1^3),
on the axes.
process,
starting with the empty configuration
that we
Notice
at the same time
also proved
EL0(nx, nt) = EL0{nVxt, nVxi) = EAn^(0)
= 2njxt +
0(nl/3),
for all x, t > 0; see (7.11).
REFERENCES
D.
[1] Aldous,
P. (1995).
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Hammersley's
interacting
and
process
particle
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[2] Baik, J., Deift, P. and Johansson, K. (1999). On the distribution of the length of the longest
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of
subsequences
random
J. Amer.
permutations.
Math.
Soc.
12
1119-1178.
MR1682248
J. and Rains,
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external
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sources.
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E. and Groeneboom,
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Limiting
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MR1788477
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P. (2005). Hammersley's
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P. A.
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Theory
JOHANSSON,
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K.
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18
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[8] Kim,
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MR1041391
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MR2123209
of Applied
Department
Delft
University
Mekelweg
2628
The
Mathematics
of Technology
4
CD Delft
Netherlands
E-mail:
e.a.cator@ewi.tudelft.nl
p. groeneboom
@ ewi. tudelft.nl
(O?AN!)