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Elect. Comm. in Probab. 11 (2006), 266–277

ELECTRONIC
COMMUNICATIONS
in PROBABILITY

EXPONENTIAL TAIL BOUNDS FOR MAX-RECURSIVE
SEQUENCES
LUDGER RÜSCHENDORF
Department of Mathematical Stochastics, University of Freiburg, Eckerstraße 1, 79104 Freiburg
i. Br. (Germany)
email: ruschen@stochastik.uni-freiburg.de
EVA-MARIA SCHOPP
Department of Mathematical Stochastics, University of Freiburg, Eckerstraße 1. 79104 Freiburg
i. Br. (Germany)
email: schopp@stochastik.uni-freiburg.de
Submitted 16 December 2005, accepted in final form 30 October 2006
AMS 2000 Subject classification: 60705, 68Q25, 68W40
Keywords: recursive algorithm, exponential bounds, divide and conquer algorithm, probabilistic analysis of algorithms
Abstract
Exponential tail bounds are derived for solutions of max-recursive equations and for maxrecursive random sequences, which typically arise as functionals of recursive structures, of

random trees or in recursive algorithms. In particular they arise in the worst case analysis of
divide and conquer algorithms, in parallel search algorithms or in the height of random tree
models. For the proof we determine asymptotic bounds for the moments or for the Laplace
transforms and apply a characterization of exponential tail bounds due to Kasahara (1978).

1

Introduction

Stochastic recursive equations of max-type arise in a great variety of problems with a recursive stochastic component as in the probabilistic analysis of algorithms or in combinatorial
optimization problems. For a list of examples in this area see the survey paper of Aldous
and Bandyopadhyay (2005). We consider in this paper random sequences (Xn ), satisfying
recurrences of the type
d

Xn =

K 
_

Ar (n)X

r=1

(r)
(n)
Ir


+ br (n) , n ≥ n0 ,

(1)
(n)

which fit with the general divide and conquer paradigma. The Ir ∈ {0,. . . ,n − 1} are
subgroup sizes of the K subproblems in which a problem of size n is split, br (n) are random
toll terms arising from the splitting process, Ar (n) is a random weighting term for subproblem r
266

Exponential tail bounds

267

(r)

and (Xn ) are independent copies of (Xn ) describing the parameter of the r-th subproblem. It
(1)
(K)
(n)
(n)
is assumed that (Xn ), . . . , (Xn ), (A(n) = (A1 (n), . . . , AK (n)), I (n) = (I1 , . . . , IK ), b(n) =
(b1 (n), . . . , br (n))) are independent while the coefficients and subgroup sizes A(n), I (n) , b(n)
d

may be dependent. = denotes equality in distribution. Xn coincides with the maximum (worst
case) of the weighted parameters of the subproblems 1, . . . , K in distribution.
A general distributional limit theorem for this type of max-recurrences was given in Neininger
and Rüschendorf (2005) , see also [9], [10] by means of the contraction method. The limit of
Xn after normalization is characterized as unique solution of a fixpoint equation of the form
d

X=

K
_

(Ar Xr + br ) ,

(2)

r=1

where (Ar , b) are limits in L2 of the coefficients (Ar (n), br (n)) and Xr are independent copies
of X.
In the present paper we study some conditions on the coefficients such that the normalized
recursive sequences
Xn − EXn
(3)
Yn =
sn

for some scaling sequences sn have exponential tails. We also give conditions which imply
exponential tails of solutions of the fixpoint equation (2). In section 2 we consider the case of
solutions X of the max-recursive equation in (2). For this case some existence and uniqueness
results have been obtained in [8], [10]. Some results on tail bounds in particular for the worst
case of FIND equation have been given in Grübel and Rösler (1996), Devroye (2001) and
Janson (2004). We derive bounds for the moments of X and obtain by Kasahara’s theorem
(1978) exponential tail bounds for X. In section 3 we consider the case of max recursive
sequences (Xn ). We establish various conditions which imply bounds for the asymptotics of
moments and Laplace transforms which again lead by Kasahara’s theorem to exponential tail
bounds for max-recurrences (Xn ). As example we discuss the worst case of FIND sequence.

2

Exponential tail bounds for max-recursive equations

Exponential tail bounds are not easily directly accessible for max-recursive sequences (Xn )
as in (1) or for solutions of max-recursive equations in (2). It will however turn out to be
possible to get suitable bounds on the moments or on the Laplace transforms of (Xn ) resp. X.
These imply exponential tail bounds by the following lemma, which is a consequence of a more
general theorem of Kasahara (1978).

We define for functions f, g on R+ → R+
f (x) ≤as g(x)

if

f (x) ∼as g(x)

if

lim

f (x)
≤1
g(x)

(4)

lim

f (x)

= 1.
g(x)

(5)

x→∞

and
x→∞

Lemma 2.1 (Kasahara (1978)) Let X be a random variable p, a > 0

268

Electronic Communications in Probability
 1/p
1
and b := pea
1) For X ≥ 0 are equivalent:
a)

and
b)
2) For general X and p ≥ 1
c)
−1

−(q−1)

− ln P (X > x) ≤as axp
kXkq ≤as bq 1/p

(6)

for q → ∞, q ∈ 2N

(7)

a) is further equivalent to:
ln EetX ≤as ctq
1

p

(8)

1
q

where c = q (pa)
and + = 1,
3) The statements in 1),2) remain valid also if ≤as is replaced by asymptotic equivalence ∼as .
Remark: In the paper of Kasahara (1978) the statement of Lemma 2.1 was given for the
asymptotic equivalence case (as in part 3)). The method of proof in that paper however also
allows to cover the ≤as -bounds as in parts 1), 2) of Lemma 2.1.


Theorem 2.2 Consider the max recursive equation (2) and assume that E
q ≥ q0 . If for some p > 0
W
k br kq
f (q) :=

≤as bq 1/p ,
PK
1 − (E r=1 Aqr )1/q

PK

r=1

Aqr < 1 for

(9)

then (2) has a unique solution X in Mq0 , the class of all distributions with finite moments of
order q0 . Further X has moments of any order and
p

P (X > x) ≤as e−ax
where a =

1

pebp .

Proof: The existence and uniqueness of a solution X (in distribution) of the max recursive
equation (2) follows from Neininger and Rüschendorf (2005), Theorem 5. For the proof of (9)
we establish bounds for the moments of X:
kXkq

=k

K
_

(Ar Xr + br )kq

r=1

≤k

K
_

Ar Xr kq + k

E

(Ar Xr )q

r=1

=

K
_

r=1

≤

K
X

EAqr EXrq

r=1

≤

K
X
r=1

EAqr

!1/q

K
_

br kq

r=1
!1/q

!1/q

+k

+k

K
_

br kq

r=1
K
_

br kq

r=1

kXkq + k

K
_

r=1

br kq .

Exponential tail bounds

269

This implies that
kXkq

≤

W
k br kq
P
1/q = f (q) ≤as bq 1/p .
K
q
1−
r=1 EAr

Thus the tail estimate follows from an application of Lemma 2.1.

(10)



Remark: Condition (9) can arise in various different ways. Two typical of these are
a) (9) holds if
eb := sup k
q

and

k
X

EAqr

r=1

_

br kq = k

!1/q

_



≤as

br k∞ < ∞

1−


eb
q −1/p .
b

(11)

(12)

Condition (12) implies in particular that sup kAr k∞ ≤ 1.
r≤K

b) If
δ := lim sup
q→∞

and
k

K
_

k
X

EAqr

r=1

!1/q

0 and let kYi kr ≤as
cr1/p as r → ∞ for i = 0, . . . , n0 −1. Further we assume


K

1 _
≤as br1/p

)
(b
(n)
−
EX
+
A
(n)EX
(n)
r
n
r

sn
Ir

a)
and

K

(n)

1 X s(Ir )
|Ar (n)| ≤ 1,
+
η r=1 sn
or b)

n ≥ n0 , for some η > 1

(19)


_

K
≤as br1/p

)
(b
(n)
−
EX
+
A
(n)EX
(n)
r
n
r


Ir

(20)

r

r=1

and

(18)

r

r=1

K

(n)

X s(Ir )
1
+
|Ar (n)| ≤ 1,
sn r=1 sn

n ≥ n0 ,

(21)

Then there exists some function h(x), such that
p

h(x) ≤as e−ax

(22)

and
P (Yn ≤ x) ≤ h(x),
where a =

∀n ∈ N0 .

(23)

1
1
in case a) and a =
in case b).
2e(max{ηb, e})p
2e(max{b, e})p

Thus we obtain exponential tail bounds of order p uniformly for all Yn .
Proof: We establish by induction uniformly in n ∈ N0 moment estimates for (Yn ) which
by Lemma 2.1 correspond to the exponential tail bounds in (23). For n = 0, . . . , n0 −1 these
bounds are given by assumption. The normalized sequence (Yn ) satisfies the modified recursive
equation
!
K
(n)
_
s(Ir )
d
(n)
Ar (n)
(24)
Yn =
YI (n) + br
r
sn
r=1
where
b(n)
:=
r

1
(br (n) − EXn + Ar (n)EXI (n) ).
r
sn

(25)

We denote by Υn the distribution of (A(n), I (n) , b(n) ) where
(n)

(n)

(n)

(n)

A(n) = (A1 (n), . . . , AK (n)), I (n) = (I1 , . . . , IK ) and b(n) = (b1 , . . . , bK ).

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Electronic Communications in Probability

Then conditioning by the vector (A(n), I (n) , b(n) ) we obtain in the induction step with β :=
WK (n)
r=1 br

!q
K
(n)

_
s(Ir ) (r)

q
(n) 
Y (n) + br
E|Yn |
=E
Ar (n)

Ir


sn
r=1
!q
K
(n)
X
s(Ir ) (r)
(n)
|Ar (n)|
≤E
|Y (n) | + |br |
Ir
sn
r=1
!l
Z X
q  
K
X
s(ir ) (r)
q
|Yir | |β|q−l dΥn
|ar |
=
E
sn
l
r=1
l=0
!l
Z X
q  
K
l X
1
q
s(ir )
p
cl (1 + o(1))
(∗) ≤
|ar | |β|q−l dΥn
l
s
n
r=1
i=0
Further in case a) we obtain

q
q 
1
≤ max{ηb, c}q p (1 + o(1)) E η1 + 1 − η1
q

1
= max{ηb, c}q p (1 + o(1)) .

E|Yn |q

In case b) we obtain


q
q 
1
≤ max{b, c}q p (1 + o(1)) E s1n +1− s1n
q

1
= max{b, c}q p (1 + o(1)) .

E|Yn |q

(26)

(27)

For the proof of (∗) we consider
E

K
X
s(ir )

sn

r=1

!l

(r)
|ar ||Yir |

X

=

j1 +...+jK =l
ir x) = ax2 (1 + o(1)) as x → ∞, for i = 0, . . . , n0 −1,
and
b)

sup
r≤K,n≥n0

Ar (n) < 1.

(29)
(30)

Then
− ln P (Xi > x) ≤as ax2 uniformly in i ∈ N0 .

(31)

Proof: The proof is by induction. For i ≤ n0 − 1 (31) holds by assumption (29). For the
induction step we obtain by conditioning as in section 2 for x ≥ 1




!2q
!2q
Z
K
K
_
_
(n)
(r)
(r)
Ar (n)X (n)
ar Xir
= P
P
≤ x
≤ x dP (A(n),I )
Ir

r=1

r=1

=

Z Y
K

1

(r)
X ir

P

r=1

x 2q
≤
ar


Z Y
K
−a

=
1 − e

1
x 2q
ar

!

!2

(1+o(1))

r=1

With Υn = P (A(n),I
EXn2q

(n)

)

∞

1

=

Z Z

=

1



1−
1−

∞

K 
Y

−a

1−e

1
x 2q
ar

!2

(1+o(1))

r=1

∞

1

Z Z

)



 (A(n),I (n) )
 dP

Ir

r=1

≤

(n)

we thus obtain from majorized convergence

= EXn2q 1{Xn x) = ax2 (1 + o(1))
=: ax (1 + gi (x), i ≤ n − 1, gi (x) → 0 as x → ∞
2

and let g(x) := supi≤n−1 gi (x). Choosing K0 ≥ 1 such that g(x) ≤ 1 − c for x ≥ K0 , 0 < c < 1,
we obtain by some calculation
Z ∞
Z ∞
2
2
e−ajcx x2i−1 dx
e(1+g(x)) e−ajx x2i−1 dx
≤ Cj
0

1

and thus we obtain
EXn2q

=

K
X

K
j

j=1

≤

K
X

K
j

j=1

≤







(−1)j+1 E(Aj (n)2q−1 )2q

r≤K,n≥n0

E(Ar (n)2q−1 )2q q!

sup
r≤K,n≥n0

≤



≤



1

2

e−ajx

(1+o(1)) 2q−1

x

dx + O(1)

+ O(1)
C j |E(Aj (n)2q−1 )|2q (2q−2)(2q−4)···2
(2acj)q

sup

≤

R∞

2q
 2+ε

q




K
j

C j + O(1)

1
(1 + C)K + O(1)
((2 + ε)ae)q
 2+ε
q


2
1
(2 + ε)e
2q
.
(2 + ε)ae
c2q

sup E(Ar (n)2q−1 )q!(1 + C)K + O(1)

|E(Ar (n)2q−1 )|

1
2q
((2 + ε)ae)



K
1 X
(2ac)q j=1

e(2 + ε)
c

q!

r≤K,n≥n0

1
((2 + ε)ae)

2q
 2+ε
(1 + o(1)).
(2q)

Lemma 2.1 implies
2+ε

(1+o(1))

2+ε

+o(x2+ε )

P (Xn > x) ≤ e−ax
= e−ax

2

= e−ax

(xε (1+|e
c|))

2+ε

+|e
c|x2+ε

<

e−ax

<

e−ax for x ≥ 1.

2

Thus at the induction step it is possible to choose the same constants c, C.



The bounds in the following theorem are based on the Laplace transform and allow unbounded
toll terms.

Exponential tail bounds

275

Theorem 3.4 Let (Xn ) be a max-recursive sequence as in (1) and let for some nondecreasing
sequence sn ≥ 1 and q > 1
(32)
EXn = µsqn + rn , rn = o(sqn ).
Let Yn =

Xn −EXn
sn

denote the normalized sequence and assume

q

a) EeλYi ≤ ecλ for some c > 0, i = 0, . . . , n0 −1 as λ ≥ λ0
b) Eeλ

br (n)
sn

q

≤ eD1 λ

rn

λ ≥ 1, n ≥ n0 , λ → ∞, rn = max(rn , 1)

,

(n)

c) Aqr (n)s(Ir )q − sqn ≤ −δ sqn rn ,
d) Ee

λ
sn

(r)
−EXn )
(n)
Ir

(Ar (n)EX

q

≤ eD2 λ

r ≤ K, n ≥ n0 , δ > 0
rn

λ ≥ λ1 , n ≥ n0

,

Then there exists some constants L, λ2 such that for all n
E exp(λYn ) ≤ exp(Lλq (1 + o(1))), for λ ≥ λ2 ,
and

p

P (Yn > x) ≤as e−ax ,
where a =

q−1
q



1
Lq

1
 q−1

and

1
q

+

1
p

(33)

= 1.

Proof: The scaled sequence (Yn ) satisfies the recursive equation
Yn

d

=

K 
_

Ar (n)

r=1
(n)

where br

=

1
sn (br (n)

− EXn + Ar (n)EX

(r)
(n)

Ir


(n)
s(Ir ) (r)
,
Y (n) + b(n)
r
Ir
sn

). By assumption we have

E exp(λYj ) ≤ exp(cλq (1 + o(1)),

λ → ∞,

1 ≤ j ≤ no − 1

for some c > 0. We will prove by induction, that for some L ≥ c
E exp(λYj ) ≤ exp Lλq (1 + o(1)), λ ≥ λ2 ,

j ∈ N.

For the induction step let Υn denote the distribution of (I (n) , b(n) , A(n)). By conditioning and
using the induction hypothesis we obtain as in the proof of Theorem 3.2
!!
K
(n)
_
s(Ir )
(r)
(n)
E exp(λYn ) = E exp λ
Ar (n)Y (n) + br
Ir
sn
r=1


K Z
X
s(jr )
(r)
ar Yjr + λβ dΥn (j, β, a)
≤
E exp λ
sn
r=1
!q
!
K
(n)
X
s(Ir )
q
q
E exp Lλ
= exp(Lλ (1 + o(1)))
Ar (n) − 1
sn
r=1
!
.
·(1 + o(1)) + λb(n)
r

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Electronic Communications in Probability

Thus it remains to show that for λ ≥ λ2
!
!q 
#
"
(n)
s(Ir
1
q
(n)
A := sup E exp Lλ
Ar (n) − 1 (1 + o(1)) + λbr ) ≤ .
sn
K
n≥no

(34)

By assumption we obtain from the Cauchy-Schwarz inequality
(r)

E

exp(λb(n)
r )

λ br (n) + sλn (Ar (n)EX (n) − EXn )
Ir
= Ee sn

1
(r)
 12

2 sλn (Ar (n)EX (n) −EXn ) 2
b (n)
2λ rsn
I
r
Ee
≤
Ee
q

≤ eD1 λ

r n +D2 λq r n

q

= eDλ

rn

,

D := D1 + D2 .

This implies by assumption c)
q
q
A ≤ e−Lδλ rn (1 + o(1)) + Dλ rn
q
≤ e−(Lδ − D)λ rn (1 + o(1))
δ
≤ e−(L 2 + D)
1
for λ ≥ max(1, λ1 , λ2 )
≤
K

if L ≥ 2 ln K+D
using rn ≥ 1. The tail bound then follows from Lemma 2.1.
δ



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