Electron. Commun. Probab. 17 (2012), no. 27, 1–8.
DOI: 10.1214/ECP.v17-2063
ISSN: 1083-589X
ELECTRONIC
COMMUNICATIONS
in PROBABILITY
On a concentration inequality for
sums of independent isotropic vectors∗
Michael C. Cranston†
Stanislav A. Molchanov‡
Abstract
We consider a version of a classical concentration inequality for sums of independent,
isotropic random vectors with a mild restriction on the distribution of the radial part
of these vectors. The proof uses a little Fourier analysis, the Laplace asymptotic
method and a conditioning idea that traces its roots to some of the original works on
concentration inequalities.
Keywords: concentration inequality; isotropy.
AMS MSC 2010: Primary 60F05, Secondary 60E15.
Submitted to ECP on August 18, 2009, final version accepted on April 23, 2012.
1
Introduction
The purpose of this paper is to give a multi-dimensional version of a concentration
inequality that traces its origins to P. Lévy [14]. These Lèvy type concentration results
assert a rate of decay on the maximal amount of mass that the distribution of a sum
of independent, nondegenerate, random variables can place in an arbitrary interval.
We remark that this is in contrast to another version of the concentration phenomena,
see [13] and the vast literature listed there, where measures concentrate most of their
mass on particular subsets of a measure space. However, the proof of our result relies
on an early version of the latter type of concentration phenomena due to Bernstein and
Hoeffding. Motivation to study the concentration inequality in the context of isotropic
random vectors arises from a variety of sources. Sums of independent isotropic vectors
are the so-called isotropic random flights which arise in problems in astronomy, [2], [4].
Such sums appear in random searches in the context of biological encounters, [3]. They
are also used as polymer models in chemistry, [6], [8]. Yet another example of isotropic
random vectors arises in a discrete model for the theory of magnetic fields generated
by a turbulent media, [15], where random vectors of the form M ξ appear where M is a
random element of the orthogonal group, O(3), and the random vector ξ is independent
of M.
Early works on the concentration inequalities in the real valued case are due to
Döblin, [5], Kolmogorov [11], Lèvy [14] and Rogozin [17]. The higher dimensional case
has been treated in Kanter, [10]. In [10], the author considered independent, RN valued, symmetric random vectors X1 , · · · , Xn . Then if C ⊂ RN is a convex set and
∗ Research supported by grants from NSF, DMS-1007176 and DMS-0706928.
† M. Cranston, University of California Irvine, USA. E-mail: mcransto@math.uci.edu
‡ S. Molchanov, University of North Carolina at Charlotte, USA. E-mail: smolchan@uncc.edu
On a concentration inequality for sums of independent isotropic vectors
λ=
Pn
i=1
P (Xi 6∈ C) it was proved that
P (X1 + · · · + Xn + x ∈ C) ≤ Φ(λ)
where Φ(s) = e−s (I0 (s) + I1 (s)) and I0 and I1 are the modified Bessel functions given by
Ik (s) =
∞
X
s 2m+k
1
.
m!(m + k)! 2
m=0
Our work considers the convex set C = Q(l), the cube centered at Q(l) = [0l]d . As
remarked in [10], Φ( n2 ) ≥ cn−1/2 . Thus, the Kanter result doesn’t contain the dimensional dependence which does appear in our bound on the concentration function in
Theorem 2.1.
The authors wish to thank the referee for pointing out the Lèvy decoupage technique
[1], used in our conditioning argument at (3.6) below, which greatly improved an earlier
version of the paper.
2
Statement of Result
Let us give a precise statement of the result proven by Kolmogorov [11]. Suppose
{ξk }k≥1 are independent random variables defined on some probability space (Ω, F, P ).
Pn
Set Sn = k=1 ξk and define for l > 0, the concentration function
Qk (l) = sup P (x ≤ ξk ≤ x + l)
x
and
QSn (l) = sup P (x ≤ Sn ≤ x + l).
x
Kolmogorov’s version of the concentration inequality states that there exists a constant
C > 0 such that if
L ≥ l, L2 ≥ l2 log s
where
s=
n
X
(1 − Qk (l))
k=1
then
CL
QSn (L) ≤ √ .
l s
We prove a higher dimensional version of this result. Due to possible degeneracies, the
higher dimensional result will not hold in general, but a natural condition to impose
on random vectors under which the result turns out to be true is isotropy. We say an
Rd valued random vector, X, on a probability space (Ω, F, P ) is isotropic if P X −1 =
P (U X)−1 for every U ∈ O(d), the group of orthogonal matrices. With this definition we
have,
Theorem 2.1. Let X1 , X2 , · · · be independent, isotropic random vectors with values in
Rd , d ≥ 2 and put Sn = X1 + X2 + · · · + Xn . Let l > 0 and L > 0 be given. Define, for
a > 0, the cube
Q(l) = [0, a]d ⊂ Rd .
Pn
Assume pi = P (Xi ∈
/ Q(l)), satisfies i=1 pi → ∞. Then given any ∈ (0, 1), for every
x ∈ Rd ,
Pn
22 ( i=1 pi )2
P (Sn ∈ x + Q(L)) ≤2 exp −
n
!d
!− d2
(2.1)
r
n
X
d L
+(1 + o(1))
(1 − )
pi
.
2π l
i=1
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On a concentration inequality for sums of independent isotropic vectors
There various immediate corollaries that may be derived from this. We sate two
such.
Corollary 2.2. Assume in addition to the conditions above that
n
X
r
pi > 2−2/d
i=1
d L
2π l
!2
and
d
Pn
2 21 i=1 pi 2
1
n
ln q
> Pn
d .
4
2( i=1 pi )2
d L
2π l
Then there is a positive constant c independent of n, l and L such that,
P (Sn ∈ x + Q(L)) ≤ c
!− d2 d
L
.
pi
l
i=1
n
X
Proof. The conditions of the corollary ensure that the solution of
2 =
2(
n
Pn
i=1
pi )2
ln
2 ((1 − )
q
Pn
i=1 pi )
d
d L
2π l
d
2
satisfies 0 < < 21 . This follows since the left hand side minus the right hand side is
increasing in and this forces the solution to be less than 14 . Since is then bounded
from 1 we can take this in (2.1) and find the c as claimed by using Theorem 2.1.
Another possibility is the following.
Corollary 2.3. Let X1 , X2 , · · · be independent, isotropic random vectors with values in
Rd , d ≥ 2, for which
1
P (Xi ∈
/ Q(1)) ≥ , for all i ≥ 1.
2
There is a constant c(d) such that for every x we have
P (Sn ∈ x + Q(1)) ≤
c(d)
.
nd/2
Pn
n
Proof. In this case,
i=1 pi ≥ 2 and L = l = 1. Taking =
corollary holds with a suitable choice of constant c.
3
1
2
in Theorem 2.1 the
Proof of the Concentration Inequality
We commence with two lemmas, the first is a local central limit theorem. This is
likely a known result and as the proof is short we include it for completeness.
e1 , · · · , Yem are iid uniformly distributed on the unit sphere
Lemma 3.1. Suppose that Y
d
d
e1 + · · · + Yem then
in R . If pem denotes the density of Y
r
pedm (0)
∼
d
2πm
!d
, m → ∞.
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On a concentration inequality for sums of independent isotropic vectors
Proof. If Y is uniformly distributed on the unit sphere in Rd , then for k ∈ Rd , we first
compute
ψd (k) = E[eihk,Y i ].
Given k, one selects coordinates on the sphere φ1 , φ2 , · · · , φd−1 , φi ∈ [0, π), for i =
1, · · · , d − 2, and φd−1 ∈ [0, 2π), so that k = |k| cos φ1 . The volume form on the sphere,
normalized to have total mass one, is given in these coordinates by
dV =
Γ( d2 + 1)
d
dπ 2
sind−2 φ1 sind−3 φ2 · · · sin φd−2 dφ1 dφ2 · · · dφd−1 .
Then since the characteristic function of a random vector Y which is uniformly distributed on the sphere of radius 1 in Rd is real,
E[eihk,Y i ] = cd
π
Z
cos(|k| cos φ1 ) sind−2 φ1 dφ1 ,
0
−1
Rπ
where cd = 0 sind−2 φ1 dφ1
.
For d = 2, by [12] page 115,
Z
1 π
cos(|k| cos φ)dφ
π 0
Z
1 π
=
cos(|k| sin φ)dφ
π 0
ψ2 (k) =
=J0 (|k|),
where J0 (z) is the 0th order Bessel function of the first kind given by
(z/2)4
(z/2)6
(z/2)2
+
−
+ ··· .
(1!)2
(2!)2
(3!)2
J0 (z) = 1 −
For d = 3,
ψ3 (k) =
Z
1
2
π
cos(|k| cos φ1 ) sin φ1 dφ1
0
Z
1 1
cos(|k|w)dw
2 −1
sin(|k|)
=
.
|k|
=
For d > 3, we have by [12] page 114,
Z
π
cos(|k| cos φ1 ) sind−2 φ1 dφ1
d
d
d
|k|− 2 +1 J d −1 (|k|),
=2 2 −1 Γ
2
2
ψd (k) =cd
0
where J d −1 is the order
2
Since for any d ≥ 2,
d
2
− 1 Bessel function of the first kind.
pedm (0) =
1
(2π)d
Z
m
(ψd (k)) dk,
(3.1)
Rd
we need to check the asymptotics of the right hand side.
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On a concentration inequality for sums of independent isotropic vectors
Starting with d = 2, we observe that ψ2 (r) = J0 (r) has a unique global maximum at
r = 0 with ψ2 (0) = 1. Also, ψ200 (0) = − 12 and for any > 0, c2 () ≡ supr≥ |ψ2 (r)| < 1.
Similarly, for d = 3, ψ3 (r) = sinr r has a unique global maximum at r = 0 with ψ3 (0) = 1.
This time ψ300 (0) = − 13 and for any > 0, c3 () ≡ supr≥ |ψ3 (r)| < 1. Finally, for d > 3, we
use the the representation, from [12] page 114,
Z 1
Γ d2
d−3
d
d
d
2 2 −1 Γ
(1 − t2 ) 2 cos(rt)dt
r− 2 +1 J d −1 (r) =
2
2
Γ 12 Γ d−1
−1
2
d
to conclude that ψd (r) = 2 2 Γ
d
2
d
r− 2 +1 J d −1 (r) has a unique global maximum at r = 0
2
with ψd (0) = 1 and
ψd00 (0)
=−
Γ
=−
Γ
1
=− ,
d
Γ
1
2
Γ
1
2
d
2
d
2
(1 − t2 )
d−3
2
t2 dt
−1
2
d−1
2
1
Z
d−1
Γ
Γ
Γ
3
Γ
2
d−1
2
Γ d2
+1
and for any > 0, cd () ≡ supr≥ |ψd (r)| < 1.
In order to determine the asymptotics at (3.1), we need the r → ∞ decay of ψd (r).
For d ≥ 2 from [12] page 134 for some positive constant c,
√ −d+1
|ψd (r)| ≤ c r
, r → ∞.
d
Now, with ωd =
1
(2π)d
dπ 2
Γ( d
2 +1)
(3.2)
being the volume of S d−1 , write
Z ∞
ωd
m
(ψd (k)) dk =
rd−1 (ψd (r)) dr
d
(2π)
d
R
(3.3)
Z0 Z ∞
ωd
ωd
m
m
d−1
d−1
r
(ψ
(r))
dr
+
r
(ψ
(r))
dr.
=
d
d
(2π)d 0
(2π)d Z
m
We use the fact that in any dimension d ≥ 2, cd () ≡ supr≥ |ψd (r)| < 1 to show the
second integral dies off exponentially fast in m. In fact,
Z
∞
r
d−1
Z
(ψd (r)) dr ≤ cd ()m/2
m
∞
rd−1 |ψd (r)|
m/2
dr
(3.4)
and by (3.2), the second integral is bounded in m for
m
2
>
2d
d−1 .
Since for any d ≥ 2, ψd00 (0) = − d1 , in the first integral, we write
(ψd (r))
m
= em ln ψd (r) ∼ e−
mr 2
2d
proceeding with the Laplace asymptotic method gives
ωd
(2π)d
Z
m
rd−1 (ψd (r)) dr ∼
0
ωd
(2π)d
ωd
=
(2π)d
r
∼
Z
|r|d−1 e−
0
r
d
2πm
d
m
!d
mr 2
2d
dr
!d Z √ m
ECP 17 (2012), paper 27.
d
0
rd−1 e−
r2
2
dr
(3.5)
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On a concentration inequality for sums of independent isotropic vectors
Thus, by (3.3), (3.4) and (3.5), for d ≥ 2,
r
pedm (0) ∼
d
2πm
!d
and the lemma is proved.
Lemma 3.2. Let Y1 , Y2 , · · · , Ym be independent random vectors with Yi uniformly distributed on the sphere of radius Ri , i = 1, · · · , m. Assume there is a number l > 0 such
that each Ri ≥ l and the Ri are non-random. Let m be an even integer. Then for L > 0
and any x ∈ Rd ,
r
P (Y1 + Y2 + · · · + Ym ∈ x + Q(L)) ≤ (1 + o(1))
d L
2πm l
!d
, m → ∞.
Proof. It suffices to provide an appropriate bound on the L∞ norm of the density of
Y1 + Y2 + · · · + Ym . Notice that
E[eihk,Yi i ] = ψd (Ri k).
Then, by the independence of Y1 , Y2 , · · · , Ym ,
E[eihk,Y1 +Y2 +···+Ym i ] =
m
Y
ψd (Ri k).
i=1
By (3.2), for m >
2d
d−1 ,
this characteristic function is integrable, which means Sm =
Y1 + Y2 + · · · + Ym has a bounded density, pdm (x), x ∈ Rd . In fact, for m even, by the
arithmetic-geometric mean and Jensen’s inequalities,
||pdm ||∞
m
Y
ψd (Ri k) dk
d
R i=1
Z
m
1 1 X
m
≤
(ψd (Ri k)) dk
(2π)d m i=1 Rd
Z
m
1 1 X 1
m
=
(ψd (k)) dk
(2π)d m i=1 Rid Rd
Z
1
m
≤
(ψd (k)) dk
(2πl)d Rd
!d
r
d 1
≤(1 + o(1))
.
2πm l
1
≤
(2π)d
Z
where the last line follows from Lemma 3.1. This completes the proof.
We can now prove the main result, Theorem 2.1.
Proof. We use the Lèvy decoupage technique [1] to decompose the random variables
Xi , based on whether |Xi | ≥ l or |Xi | < l. Set
pi = P (|Xi | ≥ l).
Recall that we are assuming that pi ≥ 12 , i = 1, 2, · · · , n. For i = 1, 2, · · · , n, let the
random variables {Ui : 1 ≤ i ≤ n} and {Vi : 1 ≤ i ≤ n} be independent with
L(Ui ) = L(|Xi | | |Xi | ≥ l), L(Vi ) = L(|Xi | | |Xi | < l).
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On a concentration inequality for sums of independent isotropic vectors
Also take Bernoulli random variables {ηi : 1 ≤ i ≤ n} independent of the {Ui : 1 ≤ i ≤ n}
and the {Vi : 1 ≤ i ≤ n} with
P (ηi = 1) = 1 − P (ηi = 0) = pi .
ei : 1 ≤ i ≤ n} to be uniformly distributed on the unit sphere in
Then, as above, take {Y
d
R and independent of the previously defined random variables. In an obvious notation
we may take the probability measure P as P = P(U,V ) ×Pη ×PYe and our original random
variables may be represented as
Xi = (ηi Ui + (1 − ηi )Vi )Yei .
For 1 ≤ i ≤ n, set
Yi1 = ηi Ui Yei
and
Yi0 = (1 − ηi )Vi Yei .
Notice that {Yi1 : 1 ≤ i ≤ n, ηi 6= 0} satisfies the conditions of Lemma 3.2 with respect
to the probability measure Pη × PYe a.s.. Denote the vector of outcomes of the sequence
{ηi : 1 ≤ i ≤ n} by
ζn = (η1 , η2 , · · · , ηn ) .
Then for a given deterministic sequence (α1 , α2 , α3 , · · · , αn ) ∈ {0, 1}n , define
1−αi
i
pη (α) ≡ P (ζn = (α1 , α2 , α3 , · · · , αn )) = Πni=1 pα
.
i (1 − pi )
By Bernstein’s inequality or it’s generalization due to Hoeffding, [9], given 0 < < 1,
Pη
n
X
ηi < (1 − )
i=1
n
X
!
pi
=Pη
i=1
≤Pη
n
X
(ηi − pi ) < −
i=1
n
X
|
(ηi − pi )| > i=1
≤2 exp −
n
X
i=1
n
X
!
pi
!
(3.6)
pi
i=1
22 (
Pn
2
i=1 pi )
n
.
We then have
Pη × PYe (Sn ∈ x + Q(L)) =
X
pη (α)PYe
!
n
X
(αi Ui + (1 − αi )Vi )Yei ∈ x + Q(L)
i=1
α∈{0,1}n
X
≤
α∈{0,1}n ,
Pn
i=1
pη (α)
αi <(1−)
Pn
i=1
pi
X
+
α∈{0,1}n ,
× PYe
Pn
n
X
i=1
(3.7)
(pη (α)
αi ≥(1−)
Pn
i=1
pi
!!
(αi Ui + (1 − αi )Vi )Ye i ∈ x + Q(L)
.
i=1
It easily follows from Fubini’s theorem that if Z is independent of Sn , then for any Q
sup P (Sm + Z ∈ x + Q) ≤ sup P (Sm ∈ x + Q).
x
(3.8)
x
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On a concentration inequality for sums of independent isotropic vectors
Pn
Using Lemma 3.2 with m = [(1 − ) i=1 pi ] which we may assume is even, we conPn
Pn
clude from (3.8) that for each α ∈ {0, 1}n for which
i=1 αi ≥ (1 − )
i=1 pi we have
P(U,V ) a.s.
sup PYe
x
n
X
!
(αi Ui + (1 − αi )Vi )Yei ∈ x + Q(L)
≤ sup PYe
x
i=1
r
≤(1 + o(1))
d 1
2π l
!d
(1 − )
n
X
n
X
!
e
αi Ui Yi ∈ x + Q(L)
i=1
(3.9)
!− d2
|Q|.
pi
i=1
Pn
i=1 pi → ∞, one has P(U,V ) a.s.
P
n
22 ( i=1 pi )2
Pη × PYe (Sn ∈ x + Q(L)) ≤2 exp −
n
!d
!− d2
r
n
X
d L
+ (1 + o(1))
(1 − )
.
pi
2π l
i=1
Thus, from (3.6), (3.7) and (3.9), it follows that as
Now integrate with respect to P(U,V ) to complete the proof.
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