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INTRODUCTION TO THE THEORYOF RANDOM PERMUTATIONS

Eugenijus Manstavicius

ERASMUS lectures, ELTE, Budapest, 2010

Contents

Introduction 1

I. ELEMENTARY THEORY 31.1 Cyclic structure of a permutation . . . . . . . . . . . . . . . . . . . . 31.2 Distribution of the cycle structure vector . . . . . . . . . . . . . . . . 61.3 The number of cycles . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.4 Feller’s coupling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.5 Estimation of the total variation distance . . . . . . . . . . . . . . . . 171.6 Estimates of conditional probabilities . . . . . . . . . . . . . . . . . . 191.7 Additive functions and the tail probabilities . . . . . . . . . . . . . . 241.8 Moments of additive functions . . . . . . . . . . . . . . . . . . . . . . 271.9 The law of large numbers . . . . . . . . . . . . . . . . . . . . . . . . . 321.10 The central limit theorem . . . . . . . . . . . . . . . . . . . . . . . . 341.11 Mean values of multiplicative functions . . . . . . . . . . . . . . . . . 381.12 The three series theorem . . . . . . . . . . . . . . . . . . . . . . . . . 47

iii

Introduction

A permutation of the set 1, 2, . . . , n = Nn can be understood as a combinatorialword, that is, an n sequence of different letters. Then appearing inversions, descents,monotone subsequences, alternating runs, patterns et cetera become very interestingobjects for deep investigations. References in this direction can be found in theseminal book by M. Bona 1. Nevertheless, we will not go along this path.

Our interests lay in permutations as bijective mappings on Nn. From such pointof view, a permutation can be imagined very visually as a labelled digraph comprisedof disjoint cycles. This exposition just mirrors the fact that each permutation is aproduct of independent cyclic permutations or cycles. The decomposition into theproduct is unique up to the order. As a labelled decomposable combinatorial struc-ture, a permutation is the simplest instance; it hides many secrets to be opened,however. Mathematical innovations created for Sn are indispensable for more com-plicated problems on general classes of combinatorial structures such as abelianpartitional complexes2, called also assemblies3.

Maybe, the strongest motivation do deal with permutations comes from algebra.All bijective mappings on Nn with respect to multiplication comprise the symmetricgroup Sn of order n!. By Cayley’s theorem, each group of the order n can be isomor-phically imbedded into Sn. This fact gives a possibility for computer calculationsof finite groups. For large n, this is not a panacea, however. The main obstacle isa great complexity and unclear asymptotic behavior of properties of the very Sn asn→∞. H. Weil4 writes:

This group has a different constitution for each individual number n. The ques-tion is whether there are nevertheless some asymptotic uniformities prevailing forlarge n . . . .

Our experience, which is also based on many investigations including those car-ried out by the Hungarian mathematicians P. Turan, P. Erdos, M. Szalay and others,suggests the answer that the asymptotic uniformities do exist for almost all permu-tations. That motivates us to start with the phrase: Take a permutation at random!

1M. Bona, Combinatorics of Permutations, Chapman and Hall/CRC Press, Boca Raton, 20042D. Foata, La serie generatrice exponentielle dans les problemes d’enumeration, Seminaire de

Mathematiques Superieures. Les Presses de l’Universite de Montreal, Quebec, 19743R. Arratia, A. D. Barbour and S. Tavare, Logarithmic Combinatorial Structures: a Probabilistic

Approach, EMS Monographs in Mathematics, EMS Publishing House, Zurich, 20034H. Weil, Philosophy of Mathematics and Natural Sciences, 1949

1

2 INTRODUCTION

I. ELEMENTARY THEORY

1.1 Cyclic structure of a permutation

Recall that the set of permutations σ : Nn → Nn with respect to multiplicationcomprises the symmetric group which we denote by Sn. Let

σ = κ1 · · ·κw , (1.1)

be the canonical decomposition of σ ∈ Sn into the product of independent cycles.Here w = w(σ) denotes their number. For instance,

σ =(1 2 3 4 5 6 7

2 3 1 4 6 5 7

)= (1, 2, 3)(4)(5, 6)(7) .

There is no need to recall more detailed definition of the product (1.1); instead ofit, we may imagine the labelled digraph being an unordered collection of w cycles.For the instance above, we have the directed edges:

1 7→ 2 7→ 3 7→ 1, 4 7→ 4, 5 7→ 6 7→ 5, 7 7→ 7 .

Let kj(σ) ≥ 0 be the number of cycles of length 1 ≤ j ≤ n in (1.1). Then

k(σ) = (k1(σ), . . . , kn(σ))

is called the cycle structure vector. For the instance, k(σ) = (2, 1, 1, 0, 0, 0, 0). Thenthe set of permutations with the same cycle structure

Sn(s) = σ ∈ Sn : k(σ) = s ,

where s = (s1, . . . , sn) ∈ Zn+ and `(s) := 1s1 + · · · + nsn = n agrees with a classof conjugate permutations. Such classes play important role in algebra. For thecombinatorial purposes, we need to know their cardinalities. It is worth to recalltwo simple propositions.

1.1.1 Theorem. If s = (s1, . . . , sn) ∈ Zn+ and `(s) = n, then

|Sn(s)| = n!n∏j=1

1

sj!jsj.

3

4 I. ELEMENTARY THEORY

Proof. According to the vector s, make sj boxes of size j, 1 ≤ j ≤ n:

s1︷︸︸︷(·) . . . (·) . . .

sj︷︸︸︷(·, . . . , ·)︸︷︷︸

j

. . . (·, . . . , ·)︸︷︷︸j

. . .

sn︷︸︸︷(·, . . . , ·)︸︷︷︸

n

. . . (·, . . . , ·)︸︷︷︸n

.

Exploit all n! possibilities to distribute the numbers of Nn into the boxes and considerthe latter as cycles. In each such a case, we get a decomposed permutation with thecycle structure vector s. Since the order of cycles is irrelevant, we have repeated thesame permutation

s1! . . . sj! . . . sn!

times. Moreover, a cycle in the box of size j is repeatedly exposed j times. Becauseof this reason, the same permutation has been repeated

1s1 . . . jsj . . . nsn

times. Dividing n! by the last two products, we obtain the number of σ ∈ Sn withthe cycle vector s.

Cauchy’s equality. For every n ∈ N,∑`(s)=n

n∏j=1

1

jsjsj!= 1 ,

where the summation is extended over s = (s1, . . . , sn) ∈ Znn such that `(s) = n.

The First Proof. Sum up the cardinalities found in the last theorem over thedisjoint classes indexed by s with `(s) = n.

The Second Proof is based on the fundamental property of exponential generatingfunctions corresponding to decomposable combinatorial structures and to the classof their components. In our case, stressing that there are n! permutations of ordern and (j − 1)! cycles of length j, we rewrite it as follows:∑

n≥0

n!

n!xn =

1

1− x= exp− log(1− x) = exp

∑j≥1

(j − 1)!

j!xj.

Now, it suffices to perform simple formal calculations with the right-hand serieswhich equals ∏

j≥1

∑s≥0

xjs

jss!=∑n≥0

( ∑`(s)=n

∏j≤n

1

jsjsj!

)xn.

Comparing the coefficients at xn we complete the proof. With some effort, the cardinalities |Sn(s)| can be examined elementarily.

1.1.2 Theorem. Assume n ≥ 3. Then

max |Sn(s)| : s ∈ Zn+, `(s) = n = n!/(n− 1) .

Moreover, the maximum is achieved for the unique vector s = (1, 0, . . . , 0, 1, 0).

1.1. CYCLIC STRUCTURE OF A PERMUTATION 5

Proof. DenoteΠ(s): = 1s12s2 · · ·nsns1!s2! · · · sn!

and find its minimum under the extra condition `(s) = n. Set

r := r(s) = maxj : sj ≥ 1 .

Firstly, we claim that the vector s with r ≥ 3 minimizing the product Π(s) satisfiesthe condition

s2 = · · · = sr−1 = 0. (1.2)

Indeed, assuming the converse statement, we had an index i, 2 ≤ i ≤ r − 1, suchthat si ≥ 1 and

n =n∑j=1

jsj ≥ isi + rsr ≥ i+ r .

Consequently, the vector has the (i+ r)-th coordinate and we may redistribute twoof the summands in n so that

isi + rsr = i(si − 1) + r(sr − 1) + (i+ r)1 .

Now, instead of the vector s, we can take

s(1): = (s1, . . . , si − 1, . . . , sr − 1, . . . , 1, 0, . . . 0)

obtained from s by changing the i-th, r-th, and (i+ r)-th coordinates. Let they beequal to si − 1, sr − 1, and 1, respectively. The new vector also satisfies `(s(1)) = nand

Π(s(1))

Π(s)=

i+ r

isi rsr≤ i+ r

ir=

1

i+

1

r≤ 1

2+

1

3=

5

6< 1 .

We have succeeded in finding a smaller value of the product. The contradictionproves our claim (1.2). Thus, if r ≥ 3, we may proceed with the vectors minimizingΠ(s) and having the form

s = (s1, 0, . . . , 0, sr, 0, . . . , 0) .

Assume now that r ≥ 2. We claim that s1 = 1. If s1 ≥ 2, then, as in the proofof the first claim, we can exploit the relations

n ≥ 1s1 + rsr ≥ 2 + r, 1s1 + rsr = 1(s1 − 1) + r(sr − 1) + (1 + r)1 .

We can take the vector s(2) obtained from s by subtracting 1 from the first and ther-th coordinates but, substituting 1 for 0 on the (1 + r)-th place. Then

Π(s(2))

Π(s)=

1 + r

s1 rsr≤ 1 + r

2r=

1

2

(1 +

1

r

)≤ 3

4< 1 .

We have again a contradiction showing that s1 = 1 or s1 = 0. In the second case,we had the vector

s = (0, . . . , 0, s, 0, . . . , 0) ,


where s ≥ 1 is on the r-th position. Then rs = n and

Π(s) = rss! = (n/s)ss! = n(n/s)s−1(s− 1)! ≥ n > n− 1 .

This value is greater than expected. Check the remaining case if r ≥ 2. Then

s = (1, 0, . . . , 0, sr, 0, . . . , 0) .

If r = 2, then, of course, there are no zeros between 1 and sr. The value sr satisfiesthe equation 1s1+rsr = 1+rsr = n. If sr ≥ 2, then 2r < n and we have a possibilityto use the (2r)-th coordinate.

Let r ≥ 2 and sr ≥ 2. Then

rsr + (2r) · 0 = r(sr − 2) + (2r) · 1.

Introduce the vector

s(3) = (1, 0, . . . , 0, sr − 2, 0, . . . , 1, 0, . . . , 0) ,

here sr − 2 stands on the r-th place and 1 is on the (2r)-th one. We see that

Π(s(3))

Π(s)=

(2r)rsr−2(sr − 2)!

rsrsr!=

2

rsr(sr − 1)≤ 1

2< 1 .

That is impossible. Consequently, if r ≥ 2, then the vector minimizing the productsatisfies sr = 1. From the equality 1 + rsr = 1 + r = n, we find r = n−1. The valueof the product equals n− 1.

If r = 1, we obtain the vector s = (s1, 0, . . . , 0) and s1 = n. For it, the value ofproduct is n, that is, larger than in the previous case.

The theorem is proved. Exercise. Find the second and the third largest classes of permutations and the

appropriate cycle structure vectors. Which class is the smallest in cardinality?

1.2 Distribution of the cycle structure vector

To start a probabilistic theory on permutations, we introduce the measure

νn(A) :=|A|n!, A ⊂ Sn .

Since now σ ∈ Sn is an elementary event and every mapping f : S → R is a ran-dom variable (r.v.). Emphasizing the combinatorial content and transgressing theprobability tradition we will write f(σ) instead of f . Later on, we will also usesome abstract probability space Ω,F , P and r.vs defined on it η, ξ, . . . withoutthe argument ω ∈ Ω.

Firstly, we examine the distribution of coordinates kj(σ), 1 ≤ j ≤ n, of thecycle structure vector k(σ). Observe that the relation `(k(σ)) = n, satisfied for eachσ ∈ Sn, implies the stochastic dependence of them.

1.2. DISTRIBUTION OF THE CYCLE STRUCTURE VECTOR 7

Recall the folklore Chinese restaurant problem asking to find the probability thatexactly m from n, 1 ≤ m ≤ n, hats of n gentlemen were returned to their ownersafter a dinner. It is assumed that each of the possibilities is equally like. In ournotation, solution to the problem is

νn(k1(σ) = m) =1

m!

n−m∑s=0

(−1)s

s!→ 1

m!

∞∑s=0

(−1)s

s!= e−1 1m

m!,

if n → ∞. Here m ∈ Z+ is a fixed number. To extend this formula, we need thefollowing elementary inclusion-exclusion lemma.

1.2.1 Lemma. Let A1, . . . , AJ ⊂ X be finite sets. The number of elements a ∈ Xbelonging to exactly m sets Ai, 1 ≤ i ≤ J , equals

Σm −(m+ 1

m

)Σm+1 +

(m+ 2

m

)Σm+2 ∓ · · ·+ (−1)J−m

(J

m

)ΣJ . (1.3)

Here

Σk =∑

1≤i1<...<ik≤J

∑x∈X

1x ∈ Ai1 ∩ . . . ∩ Aik

and 1· denotes the indicator of the set in the braces.

The next result belongs to V.L. Goncharov [?].

1.2.1 Theorem. Let 1 ≤ j ≤ n and m ∈ Z+. Then

νn(kj(σ) = m) =1

jmm!

∑0≤s≤n/j−m

(−1)s

jss!.

Moreover, if j and m are fixed and n→∞, then

νn(kj(σ) = m) =e−1/j

jmm!+ o(1) .

Proof. Let κi bet an arbitrary cycle of length j with vertexes from Nn. Assumenow that Ai ⊂ Sn is the subset of permutations with the cycle κi. We have

J : = n(n− 1) · · · (n− j + 1)/j

different cycles of length j with vertexes in Nn and the same number of the subsetsAi. It remains to find the number of permutations belonging to exactly m of them.The lemma above is at our disposal.

Calculating Σk, we observe that, if

σ ∈ Ai1 ∩ . . . ∩ Aik ,


these k cycles κi1 , . . . ,κik must be independent; therefore, this determines jk ≤ nvertexes of σ, while the remaining n− jk vertexes can be distributed in other cyclesarbitrarily. Hence the number of such permutations equals

(n− jk)!1jk ≤ n .

This is the value of the inner sum in Σk provided that κi1 , . . . ,κik are independent.For the other possibilities of indexes i1, . . . , ik, the inner sum is zero.

How many nonzero summands are in Σk? As we have noted, these summandscorrespond to independent cycles κi1 , . . . ,κik of length j. To find the number of suchfamilies of cycles, we can collect numbers from Nn to kj positions taking accountinto their order. For this, there are n(n − 1) · · · (n − jk + 1) possibilities. Thesenumbers must comprise k cycles of length j. Since the cyclic change of the ordergives the same cycle, we must divide the latter number by jk. Moreover, the order ofthe very cycles is irrelevant; therefore, we have again to divide by k!. Consequently,in the outer sum of Σk, there are

n(n− 1) · · · (n− jk + 1)1

jkk!

nonzero summands. Hence

Σk = n(n− 1) · · · (n− jk + 1)1

jkk!· (n− jk)!1jk ≤ n =

n!

jkk!1jk ≤ n .

Inserting these values into formula (1.3), we obtain

|σ ∈ Sn : kj(σ) = m| =J∑

k=m

(−1)k−m(k

m

)Σk

=J∑

k=m

(−1)k−m(k

m

)n!

jkk!1jk ≤ n

=n!

m!

∑m≤k≤n/j

(−1)k−m

jk(k −m)!.

The substitution k = m+s and division by n! yield the first claim of Theorem 1.2.1.The second statement is now evident.

Let ξj be a Poisson r.v. with Eξj = 1/j. The latter theorem can be rewrittenas L(kj(σ))→ L(ξj) as n→∞ provided that j is fixed. Here and afterwards L(X)denotes the distribution of a r.v. X. Can we extend this to a many-dimensionalrelation? The answer will be given in the sequel. The next relations explain thedependence of coordinates in k(σ).

1.2.2 Theorem. Let ξ1, . . . , ξn be mutually independent Poisson r.vs given on somespace, Eξj = 1/j for 1 ≤ j ≤ n, ξ = (ξ1, . . . , ξn), and ζ := `(ξ). Then

νn(k(σ) = s) = 1`(s) = nn∏j=1

1

jsjsj!= P (ξ = s| ζ = n) .

1.2. DISTRIBUTION OF THE CYCLE STRUCTURE VECTOR 9

Proof. The first equality follows from Theorem 1.1.1. The proof of the secondone is an easy exercise.

More useful is the following infinite version of the last theorem.

1.2.3 Theorem. Let 0 < x < 1 and ξ(x)1 , . . . , ξ

(x)n , . . . be mutually independent

Poisson r.vs given on some space with Eξ(x)j = xj/j for j ≥ 1. Denote

ξ(x) = (ξ(x)1 , . . . , ξ(x)

n , ξ(x)n+1, . . . ) ,

ζ(x) := 1ξ(x)1 + · · ·+ nξ(x)

n + (n+ 1)ξ(x)n+1 + · · · ,

andk(σ) = (k1(σ), . . . , kn(σ), 0, . . . ).

Thenνn(k(σ) = s) = P (ξ(x) = s| ζ(x) = n) (1.4)

for each s = (s1, . . . , sn, sn+1, . . .) ∈ Z∞+ .

Proof. Firstly, check that

∞∑j=1

P (ξ(x)j 6= 0) =

∞∑j=1

(1− exp

− xj

j

)<∞∑j=1

xj

j<∞

if 0 < x < 1. By the Borel-Cantelli lemma, the series of r.vs ξ(x) converges withprobability one and determines a r.v. Secondly, we see that only the vectors s withsj = 0 for j ≥ n + 1 and such that `(s) = n can give give nonzero probabilities inthe both sides of (1.4). Afterwards, only such the vectors s are taken into account.

Further, we find

P (ζ(x) = n) = P (ξ(x)n+1 = ξ

(x)n+2 = . . . = 0)P (1ξ

(x)1 + · · ·+ nξ(x)

n = n)

= exp−

∞∑j=n+1

xj

j

∑`(s)=n

n∏j=1

P (ξ(x)j = sj)

= exp−∞∑j=1

xj

j

∑`(s)=n

n∏j=1

xjsj

jsjsj!

= (1− x)xn. (1.5)

In the last step, we applied Cauchy’s equality. Summation above has run over s ∈ Zn+satisfying `(s) = n. We have also

P (ξ(x) = s, ζ(x) = n) =∞∏

j=n+1

P (ξ(x)j = 0)

n∏j=1

P (ξ(x)j = sj)

= exp−∞∑j=1

xj

j

n∏j=1

xjsj

jsjsj!= (1− x)xn

n∏j=1

1

jsjsj!.


The ratio of the last two probabilities gives the already known value of the frequencyon the left-hand side in (1.4).

The theorem is proved. In the sequel, we denote En the mean value with respect νn and use E to denote

the mean value in the space Ω,F , P in which the r.vs ξ(x)j , j ≥ 1, are defined.

1.2.4 Theorem. Let 0 < x < 1 and Ψ:Z∞+ → C be a function such that EΨ(ξ(x))exists. Then

EnΨ(k(σ)) = E(Ψ(ξ(x))| ζ(x) = n) .

Moreover,

(1− x)−1EΨ(ξ(x)) =∞∑n=0

EnΨ(k(σ))xn . (1.6)

Proof. The first claim follows from (1.4) in the previous theorem.Further, we use the well-known probabilistic equality

EX = E(E(X|Y )) =∞∑n=0

E(X|Y = n)P (Y = n)

provided that the mean value EX exists. Here Y is an arbitrary r.v. taking valuesin Z+. Applying the latter, we just take Y = ζ(x), X = Ψ(ξ(x)), and combine (1.5)with the already proved formula.

The mean value EΨ(ξ(x)) is a function defined in the interval (0, 1). By theprinciple of analytic continuation, one can use (1.6) for complex values x belongingto the open unit disk. This relation is fundamental calculating the mean valuesEnΨ(k(σ)) as the Taylor coefficients. We now demonstrate such a possibility.

Set(u)r = u(u− 1) · · · (u− r + 1), r = 0, 1, . . .

and recall that the factorial moment of order r of a Poisson r.v. ξ with Eξ = λ is

E(ξ)r = λr . (1.7)

1.2.5 Theorem. Let r1, . . . , rs ≥ 0, m: = 1r1 + · · · + srs, and 1 ≤ s ≤ n be fixed.Then the mixed factorial moment

Enm : = En

((k1(σ))r1 · · · (ks(σ))rs

)= 1m ≤ n

s∏j=1

1

jrj= 1m ≤ n

s∏j=1

E(ξj)rj .

Proof. If m > n ≥ s, then one of the differences (kj(σ)− i), i ≤ rj is zero. HenceEnm = 0 as well as 1m ≤ n = 0.

Assume m ≤ n and apply the formula in Theorem 1.2.4 for

Ψ(k) = (k1)r1 · · · (ks)rs .

1.3. THE NUMBER OF CYCLES 11

We have

(1− x)−1E( s∏j=1

(ξj)rj

)=∑n≥0

Enmxn , 0 < x < 1 .

Exploiting the independence of r.vs, from (1.7), we obtain

( l∏j=1

1

jrj

)xm (1 + x+ · · ·) =

∑n≥0

Enmxn ,

if 0 < x < 1. Comparing the coefficients at xn, we complete the proof.

1.2.1 Corollary. The finite-dimensional distributions of the vector

(k1(σ), . . . , kn(σ), 0, . . . )

converge to that of ξ(1)) which has been defined in Theorem 1.2.3. Moreover, allmixed moments also converge to the appropriate moments of the limit r.vs.

Proof. It suffices to recall that the many-dimensional Poisson distribution isuniquely determined by its moments. The result follows from the last theorem.

Exercise. Using Theorem 1.2.4, find the sum∑σ∈Sn

zw(σ), z ∈ C .

1.3 The number of cycles

As we have remarked, the number of cycles

w(σ) = k1(σ) + · · ·+ kn(σ)

is the sum of dependent r.vs, nevertheless, we can examine value distribution. Thetwo-dimensional sequence

c(n,m) := |σ ∈ Sn : w(σ) = m|, 0 ≤ m ≤ n ,

where c(0, 0) := 1, had been dealt with long before the probabilistic combinatoricsstarted.

1.3.1 Theorem. We have

c(n+ 1,m) = c(n,m− 1) + nc(n,m)

for 1 ≤ m ≤ n.


Proof. We split all permutations defined on Nn+1 = 1, . . . , n, n + 1 into twoclasses. To the first class we ascribe a permutation if the number n+ 1 comprises acycle of length one in it. The remaining permutations comprise the second class.

The cardinality of the first class coincides with the number of permutations inSn having m− 1 cycle, therefore it equals c(n,m− 1).

Each permutation of the second class can be obtained from σ ∈ Sn having mcycles. For this, we can consecutively insert n + 1 into each of the cycles of σ. Wehave j positions in a cycle of length j and altogether n positions. Going along thispath, from each of σ ∈ Sn having m cycles, we construct n permutations of thesecond class. Consequently, the cardinality of the latter is nc(n,m). Adding theobtained cardinalities, we complete the proof of Theorem 1.3.1.

Recall that the Stirling numbers of the first kind s(n,m), 0 ≤ m ≤ n, are definedby the equalities

(x)n =n∑

m=0

s(n,m)xm . (1.8)

It is an easy exercise to verify that they satisfy the recurrence relation

s(n+ 1,m) = s(n,m− 1)− ns(n,m), 1 ≤ m ≤ n ,

where s(n, 0) = 0 if n ≥ 1 and s(0, 0) = 1. The substitution of −x in the place of xin (1.8) and Theorem 1.3.1 imply the following relation.

1.3.1 Corollary. For 0 ≤ m ≤ n,

c(n,m) = (−1)n−ms(n,m) = |s(n,m)| .

The asymptotical behavior as n→∞ of the local probabilities

νn(w(σ) = m) =c(n,m)

n!

is fairly involved. The integral limit theorem, established in 1942 by V.L. Goncharovand presented below, is historically notable.

Let

Φ(x): =1√2π

∫ x

−∞e−u

2/2du

be the distribution function of the standard normal law. For brevity, in the proofof the next theorem, we will use the classical asymptotic estimate(

z + n− 1

n

)=nz−1

Γ(z)(1 + O(n−1)) , (1.9)

if n→∞ uniformly in |z| ≤ 1. Here Γ(z) is the Euler gamma-function.

Goncharov’s Theorem. If n→∞, then

νn(w(σ)− log n < x√

log n) = Φ(x) + o(1)

uniformly in x ∈ R.

1.3. THE NUMBER OF CYCLES 13

Proof. Firstly, we find the moment generating function

ψn(z): =1

n!

∑σ∈Sn

zw(σ) =1

n!

∑`(s)=n

∑σ∈Sn(s)

zw(σ), z ∈ C .

Here, as previously, s = (s1, . . . , sn) ∈ Zn+. The summands in the inner sum areequal to zs1+···+sn and their number is |Sn(s)|. Theorem 1.1.1 implies

ψn(z) =1

n!

∑`(s)=n

zs1+···+sn |Sn(s)| =∑`(s)=n

n∏j=1

(zj

)sj 1

sj!.

This recalls the second proof of Cauchy’s equality. The same argument now leadsto the equality

(1− y)−z =∞∏j=1

expzyj

j

=∞∏j=1

∞∑s=0

(zyj)s

jss!=∞∑n=0

ψn(z)yn .

Here y is a formal unknown. The n-th coefficient of the series on the left-hand sidecan be found by Newton’s formula. Hence by (1.9)

ψn(z) =

(z + n− 1

n

)=nz−1

Γ(z)(1 + O(n−1)), |z| ≤ 1 , (1.10)

as n→∞.Now, the characteristic function of the law under investigation

ϕn(t): = e−it√

lognψn(eit/√

logn), t ∈ R.

It remains to find the asymptotic formulae of each of the factors on the right-handside. We have

Γ(eit/√

logn) = 1 + o(1)

uniformly in |t| ≤ T for every T > 0. The asymptotic expansions of the exponentialfunction imply

ϕn(t) = exp − it√

log n+ (eit/√

logn − 1) log n(1 + o(1))

= e−t2/2(1 + o(1))

as n→∞ with the same uniformity in t. This completes the proof of the Goncharovtheorem.

Observe an interesting relation. The first equality in (1.10) can be rewritten as

ψn(z) =(

1 +z − 1

n

)(1 +

z − 1

n− 1

)· · ·(

1 +z − 1

2

)z .

This shows that the distribution of w(σ) coincides with that of the sum of theindependent Bernoulli r.vs ηj, 1 ≤ j ≤ n, defined by

P (ηj = 1) = 1− P (ηj = 0) =1

j.

This raises a suspicion that there exists a probability space carrying kj(σ) and ηj,j ≥ 1, where the first r.vs can be expressed in terms of the second ones. This leadsto the theory of coupling of probability spaces.


1.4 Feller’s coupling

Let Ω,F , P be a probability space and ηj, j ≥ 1, be the independent Bernoullir.vs defined on it. As previously, P (ηj = 1) = 1/j for j ≥ 1. We also assumethat the space is rich enough and carries the choices of elements from finite sets asrandom events independently of the family ηj, j ≥ 1. Afterwards a number taken atrandom will mean that it is taken from an appropriate set with equal probability.W. Feller has proposed the probabilistic algorithm to generate a random σ.

Feler’s algorithm:

Initialization. Open the permutation σ starting with the cycle containing 1;that is, write

σ: = (1

and define the set X1: = Nn \ 1.Step 1. If ηn = 1, then close the just opened cycle and start the next one with

the least number of the set X1; that is, write

σ = (1)(2

and define X2: = X1 \ 2.If ηn = 0, then continue the opened cycle with a number s taken at random from

X1; that is, write

σ = (1, s

and set X2: = X1 \ s.Assume that we did 1 ≤ i < n steps and already have the combination

σ = κ1 · · ·κj(a, . . . , r ;

where κk, 1 ≤ k ≤ j, are independent cycles and the involved numbers from Nn

comprise the set Ai. Set Xi: = Nn \ Ai.Step i + 1. If ηn−i = 1, then we close the cycle, open the next one with the

least number of Xi (say, t); that is, write

σ = κ1 · · ·κj(a, . . . , r)(t

and set Xi+1: = Xi \ t.If ηn = 0, then we continue the started cycle with a number q taken at random

from Xi; that is, write

σ = κ1 · · ·κj(a, . . . , r, q

and set Xi+1: = Xi \ q.The end. Since P (η1 = 1) = 1, we close the last opened cycle and return the

generated σ.

1.4.1 Lemma. Feller’s algorithm generates a random σ with the probability 1/n!.

1.4. FELLER’S COUPLING 15

Proof. Observe that, during the initialization and in n − 1 steps, new num-bers from Nn are included into σ. Hence we obtain a permutation in its canonicalrepresentation.

The probability of the combination obtained in the first step is 1/n.Assume that the probability of the combination obtained in the i-th step is

1

n· 1

n− 1· · · 1

n− i+ 1.

In different steps, only independent r.vs and events are used, therefore we maymultiply the probabilities. In the first case of Step (i + 1), the combination isobtained with the probability

1

n· 1

n− 1· · · 1

n− i+ 1· 1

n− i,

while in the second case when ηn−i = 0, this probability equals

1

n· 1

n− 1· · · 1

n− i+ 1·(

1− 1

n− i

) 1

n− i− 1.

Consequently, in either of the cases the probability is the same. By the inductionprinciple, we obtain the expected probability 1/n! of σ.

In what follows, we can imagine that the generated σ = σ(ω), where ω ∈ Ω isan elementary event, is defined in the space Ω,F , P. This allows to express kj(σ)in terms of ηi, i ≥ 1. We will denote by

X =d Y

the coincidence of distributions of the r.vs X and Y .

1.4.1 Theorem. We have

n∑j=1

ηj =d

n∑j=1

kj(σ) = w(σ) .

Proof. It suffices to observe that, during the generation of a σ by Feller’s algo-rithm, a cycle is finished if and only if ηj = 1, 1 ≤ j ≤ n. Hence the sum of theser.vs is equal to the number of cycles w(σ).

1.4.2 Theorem. Define r.vs

Xnj : = ηn−j+1(1− ηn−j+2) · · · (1− ηn)

+

n−j∑i=1

ηi(1− ηi+1) · · · (1− ηi+j−1)ηi+j (1.11)

in the space Ω,F , P. Then

kj(σ) =d Xnj, 1 ≤ j ≤ n .


Proof. The first cycle of length j appears in the generated σ if and only if(ηn, . . . , ηn−j+1) = (0, . . . , 0, 1). This is equivalent to the equality

ηn−j+1(1− ηn−j+2) · · · (1− ηn) = 1 .

Later such a cycle is formed up if and only if the value of vector (ηn, . . . , η1) is(1, 0, . . . , 0, 1). Here, between ones, there are j − 1 zeros. The summand in the sumover i = 1, . . . , n− j of (1.11) is one if and only if such a run occurs. In other words,the sum counts the cycles of length j starting with the second one.

The theorem is proved.

1.4.3 Theorem. Denote

Xj =∞∑i=1

ηi(1− ηi+1) · · · (1− ηi+j−1)ηi+j .

If n→∞, then P (Xnj → Xj) = 1 for each fixed j ∈ N.

Proof. First of all, we check that the r.v. Xj is correctly defined. Indeed,

∞∑i=1

P(ηi(1− ηi+1) · · · (1− ηi+j−1)ηi+j 6= 0

)≤

∞∑i=1

P (ηi = 1, ηi+j = 1) =∞∑i=1

1

i(i+ j)<∞ .

Consequently, by the Borel-Cantelli lemma, the infinite series of r.vs converges al-most surely (a.s.). Hence

Rnj: =∑i>n−j

ηi(1− ηi+1) · · · (1− ηi+j−1)ηi+j → 0 (a.s.)

Moreover, ifQnj: = ηn−j+1(1− ηn−j+2) · · · (1− ηn) ,

thenP (Qnj = 1) ≤ P (ηn−j+1 = 1) = 1/(n− j + 1)→ 0 ,

for every j, as n→∞. This implies that Qnj → 0 a.s. Altogether,

|Xj −Xnj| ≤ Rnj +Qnj → 0 (a.s.)

for every fixed j ≥ 1.The theorem is proved. The next representation of the Poisson r.vs in terms of the Bernoulli ones is

interesting from the probabilistic point of view.

1.4.1 Corollary. In the notation above

(ξ1, ξ2, . . .) =d (X1, X2 . . .) .

Proof. It suffices to apply Theorems 1.2.1, 1.4.2, and 1.4.3. Exercise. Using Theorem 1.4.1 and the Berry–Esseen bound for the conver-

gence rate in the central limit theorem, find the convergence rate in the Goncharovtheorem.

1.5. ESTIMATION OF THE TOTAL VARIATION DISTANCE 17

1.5 Estimation of the total variation distance

Recall the definition of the total variation distance between the distributions of r.vsX and Y which are (maybe) defined on different probability spaces Ω1,F1, P1 andΩ2,F2, P2. If their values range in a metric space M and M is the Borel class ofits subsets, then this distance is

d(X, Y ): = d(L(X),L(Y )): = supB∈M

|P1(X ∈ B)− P2(Y ∈ B)| .

For discrete r.vs taking the values in Zr+, the distance equals

d(X, Y ) =1

2

∑s∈Zr+

|P1(X = s)− P2(Y = s)| .

It appears that the distributions of the vectors

kr(σ): = (k1(σ), . . . , kr(σ))

and

ξr: = (ξ1, . . . , ξr)

are close if r = o(n) as n → ∞. As previously, ξj, 1 ≤ j ≤ r ≤ n, are mutuallyindependent Poisson r.vs with Eξj = 1/j. Set

dr(n): = d(L(kr(σ)),L(ξr)) .

Seeking an estimate for dr(n), we involve another metrics.

1.5.1 Lemma. Let Xnj and Xj are the r.vs defined in the previous section. Then

dr(n) ≤∑j≤r

E|Xnj −Xj|: = dr(n).

Proof. Denote

Xnr = (Xn1, . . . , Xnr).

By Theorem 1.4.2, the distribution of this vector coincides with that of the vectorkr(σ). Moroever, Corollary 1.4.1 allows us to substitute Xr := (X1, . . . , Xr) for ξr.Hence

dr(n) = supA⊂Zr+

∣∣∣P (Xnr ∈ A)− P (Xr ∈ A)∣∣∣ ≤ P (Xnr 6= Xr)

≤ P(∑j≤r

|Xnj −Xj| ≥ 1)≤ dr(n) .

The lemma is proved. The next result is fundamental for the theory of random permutations.


Fundamental Lemma. If 1 ≤ r ≤ n, then

dr(n) ≤ 2r

n− r + 1.

Proof. By the just proved lemma, it suffices to deal with the metrics dr(n). Inthe notation of previous section, we have

Xj −Xnj = Rnj −Qnj =∑i>n−j

ηi(1− ηi+1) · · · (1− ηi+j−1)ηi+j

−ηn−j+1(1− ηn−j+2) · · · (1− ηn) .

Exploiting the independence of the factors and omitting some of them, we obtain

E|Xj −Xnj| ≤∑i>n−j

EηiEηi+j + Eηn−j+1

=∑i>n−j

1

l(i+ j)+

1

n− j + 1

=1

j

∑i>n−j

(1

i− 1

i+ j

)+

1

n− j + 1

=1

j

∑n−j<i≤n

1

i+

1

n− j + 1

≤ 1

n− j + 1+

1

n− j + 1≤ 2

n− r + 1,

for j ≤ r. Adding these inequalities, we complete the proof. Of course, the just applied argument does not give an optimal estimate of dr(n).

Better estimates can be found in [?]. For many purposes, it suffices to have onlydr(n) = o(1) for r = o(n) as n→∞. Can one extend this bound of r?

1.5.1 Theorem. Let n→∞. The estimate of the total variation distance dr(n) =o(1) holds if and only if r = o(n).

Proof. The sufficiency is established in the Fundamental lemma. To prove thenecessity, we will use the expression

dr(n) = supA⊂Zr+

∣∣∣νn(kr(σ) ∈ A)− P (ξr ∈ A)∣∣∣

= supA⊂Zr+

∣∣∣P (ξr ∈ A| ζ = n)− P (ξr ∈ A)∣∣∣ .

Here as previously, ζ = 1ξ1 + · · · + nξn. It remains to choose a bad set A ⊂ Zr+.Take

kr ∈ Zr+: 1k1 + · · ·+ rkr > n .The conditional probability now equals zero. Hence

dr(n) ≥ P(∑j≤r

jξj > n)≥ P

(r2

∑r/2<j≤r

ξj > n)

= P (Y > 2n/r) ,

1.6. ESTIMATES OF CONDITIONAL PROBABILITIES 19

where Y is the Puasson r.v. with the parameter∑r/2<j≤r

1/j ∼ log 2, r →∞ .

If r = 2εn and ε > 0 is fixed, then

P (Y > ε−1) ≥ c(ε) > 0 .

The theorem is proved. Resuming this section, we stress once more that the first coordinates in the cycle

structure vector kr(σ), if r ≤ εn for every ε > 0, allow an approximation by thevector ξr with independent coordinates. What could we do with the remainingcomponents corresponding to the long cycles? In the next section we include anapproach proposed by the author in [?]. The very idea goes back to I.Z. Ruzsa’spaper in probabilistic number theory.

1.6 Estimates of conditional probabilities

The purpose of this section is to find upper estimates of the conditional probabilitiesappearing in the relation

νn(k(σ) ∈ A) = P (ξ ∈ A| `(ξ) = n), A ⊂ Zn+ ,

which has been proved in Theorem 1.2.2 in terms of unconditional probabilities.Here, as previously, `(k) = 1k1 + · · ·+ nkn, k = (k1, . . . , kn) ∈ Zn+. It is essential toconsider all components of the vector k(σ). The idea is to extend the set A but notup to the whole Zn+ which would be trivial.

We will exploit geometric interpretation of the semi-lattice Zn+; therefore, it isconvenient to examine a probability space Zn+,Zn, µ, where µ is a probability mea-sure defined on all subsets of Zn+ comprising Zn. Later on, µ will be the distributionof the poissonian vector ξ, and we will return to the space Ω,F , P just applyingthe relation

µ(A) = P (ω: ξ(ω) ∈ A), A ⊂ Zn+. (1.12)

Let us introduce some notation. Set

ki = (ki1, ki2, . . . , kin) ∈ Zn+, i = 1, 2, . . . ,

Ωm: = k ∈ Zn+: `(k) = m, 0 ≤ m ≤ n ,

and

ej: = (0, . . . , 1, . . . , 0) ,

where 1 stands at the j-th place, j = 1, . . . , n. The relation

k11k21 + k12k22 · · ·+ k1nk2n = 0 ,


will be denoted by k1 ⊥ k2; the very vectors will be called orthogonal. We will writek1 ≤ k2 if k1j ≤ k2j for each 1 ≤ j ≤ n. We will use the notation k1 ‖ k2 for tworelations:

k1 ≤ k2, k1 ⊥ k2 − k1 .

In the last case, we will say that k1 exactly enters into the vector k2.Let us a bit specify the measure µ. Take arbitrary distributions 0 ≤ pj(k) ≤ 1,

where k ≥ 0 and pj(0) + pj(1) + · · · = 1, on Z+ and define µ on Zn+ by

µ(k) := µ(k) =n∏j=1

pj(kj) .

In this notation, our purpose is to estimate the conditional probabilities µ(U |Ωn)by unconditional ones. Observe that only the case Pn := µ(Ωn) = o(1) as n→∞ isnot trivial. A delicate extension of the event U ⊂ Zn+ is as follows:

V = V (U) := k = k1 + k2 − k3 : k1, k2, k3 ∈ U, k1 ⊥ k2 − k3, k3 ‖ k2.

For an arbitrary event U ⊂ Zn+, set U = Zn+ \ U . For brevity, the next result isformulated for the complementary events.

1.6.1 Theorem. Let n ≥ 1 and let there exist positive constants c, c1, C1, and C2

such that

(i) pj(0) ≥ c for all 1 ≤ j ≤ n;

(ii) µ(Ωm) ≤ C1 Pn for all 0 ≤ m ≤ n− 1;

(iii) Pn ≥ c1n−1;

(iv) ∑k≥1,≤nkj=m

pj(k)

pj(0)≤ C2

m

for all 1 ≤ m ≤ n.

Thenµ(V |Ωn) ≤ Cµ(U).

Here C > 0 is a constant depending on the constants in the conditions.

We start with the following technical lemma.

1.6.1 Lemma. Let condition (i) be satisfied. Denote π = kej with an arbitraryk ≥ 1 and j ≤ n, q(π) = pj(k)/pj(0),

Qn = π : `(π) = jk ≤ n ,

Q′ = π ∈ Qn : ∃k ∈ U, k ⊥ π, π + k ∈ U ,and Q′′ = Qn \Q′. If µ(U) < c2/32, then∑

π∈Q′′q(π) ≤ 4c−1µ(U) .


Proof. The essence of the claim is rather clair. If the probability µ(U) is large,then the set Q′ is also large. Hence the compliment Q′′ should be small. The claimquantitatively expresses this observation.

Let π = kej ∈ Q′′, where k ≥ 1 and jk ≤ n, and define

Wπ = l = π + k: k ∈ U, k ⊥ π ⊂ U .

If π ⊥ k, then kj = 0. Hence

µ(π + k) = pj(k)∏i≤ni6=j

pi(ki) =(pj(k)

pj(0)

)(pj(0)

∏i≤ni 6=j

pi(ki))

= q(π)µ(k) . (1.13)

Consequently,

µ(U) ≥ µ(Wπ) = q(π)∑

k∈U,k⊥π

µ(k)

≥ q(π)(∑k∈U

µ(k)−∑k∈Ωkj≥1

µ(k))

= q(π)(

1− µ(U)− (1− pj(0)))

≥ cq(π)(1− c/32) > q(π)c/2 (1.14)

andq(π) ≤ 2c−1µ(U) ≤ c/8 . (1.15)

If π1 = rei 6= π, π1 ∈ Q′′, then Wπ ∩Wπ1 = ∅ for i = j and

Wπ ∩Wπ1 ⊂ l = π + π1 + k : k ∈ Ω, k ⊥ π, k ⊥ π1 ,

for i 6= j. Hence in either of the cases, from (1.13), we obtain

µ(Wπ ∩Wπ1) ≤ q(π)q(π1)∑k∈Ω

µ(k) = q(π)q(π1) . (1.16)

Take an arbitrary subset Q ⊂ Q′′ and denote

W =⋃π∈Q

Wπ, a := a(Q) =∑π∈Q

q(π) .

Since W ⊂ U , we obtain from (1.14) and (1.16) that

µ(U) ≥ µ(W ) ≥∑π∈Q

µ(Wπ)−∑

π,π1∈Q

µ(Wπ ∩Wπ1)

≥ a(c

2− a) ≥ ca

4, (1.17)


provided that a ≤ c/4. In other words, if a(Q′′) ≤ c/4, taking Q = Q′′ we completethe proof.

Assume that a(Q′′) > c/4 and choose Q to be a maximal subset of Q′′ suchthat a(Q) ≤ c/4. Further take a π′ from the nonempty set Q \ Q′′. We haveq(π′) + a(Q) > c/4; therefore, (1.17) and (1.15) imply

µ(U) ≥ ca(Q)

4>c

4

( c4− q(π′)

)≥ c2

32.

This is a contradiction to the assumption of the lemma.Lemma 1.6.1 is proved. Proof of Theorem 1.6.1. It suffices to examine the case µ(U) ≤ c2/32. Otherwise

the claim holds with C = 32c−2.A vector l ∈ V ∩ Ωn is non zero. It allows a decomposition l = π + k with some

π = kej, 1 ≤ `(π) = kj ≤ n, where π ⊥ k. We have

`(π + k) = `(π) + `(k) = n .

Moreover, π ∈ Q′′ or k ∈ U . Indeed, in the converse case, by the definition of Q′,we had a vector k1 ∈ U such that k1 ⊥ π, π + k1 ∈ U and

l = k + (π + k1)− k1 ∈ V .

This is impossible. Now, the above notation and the equality∑π‖l

`(π) = n ,

if l ∈ Ωn, and (1.13) imply

Pnµ(V |Ωn) =1

n

∑l∈V ∩Ωn

µ(l)∑π‖l

`(π)

=1

n

∑π+k∈V ∩Ωn

π⊥k

q(π)`(π)µ(k)

≤∑π∈Q′′

q(π)∑

`(k)=n−`(π)

µ(k)

+1

n

∑k∈U

µ(k)(n− `(k))∑

`(π)=n−`(k)

q(π)

: = Σ1 + Σ2 . (1.18)

Using Lemma 1.6.1 and the conditions, we obtain

Σ1 =∑π∈Q′′

q(π)µ(Ωn−`(π)) ≤4C1

cPnµ(U)


and

Σ2 ≤C2

nµ(U) ≤ C2

c1

Pnµ(U).

Inserting the last two estimates into (1.18) and dividing byPn, we have

µ(V |Ωn) ≤(4C1

c+C2

c1

)µ(U) .

Recalling the already examined case, we see that the claim of Theorem 1.6.1 holdswith

C = max32

c2,4C1

c+C2

c1

. (1.19)


1.6.1 Corollary. Theorem 1.6.1 holds for

pj(k) = e−1/j 1

jkk!, 1 ≤ j ≤ n, k ≥ 0 . (1.20)

Proof. We have to verify the conditions of the last theorem. Condition (i) holdswith c = e−1.

The relation `(k) = m ≤ n implies kj = 0 for m+1 < j ≤ n. Hence the Cauchy’sformula yields

µ(Ωm) =∑

`(k)=m

n∏j=1

e−1/j

jkjkj!= exp

−

n∑j=1

1

j

∑`(k)=m

m∏j=1

1

jkjkj!

= exp−

n∑j=1

1

j

= Pn .

for each 0 ≤ m ≤ n. Consequently, (ii) and (iii) hold with C1 = 1 and c1 = e−1.Finally, ∑

jk=m

1

jkk!≤ 2

m+

1

m

∑k|m

2≤k≤m/2

1

(k − 1)!

( km

)k−1

≤ 2

m+

1

m

∞∑k=0

1

k!2k≤ 2

m+

e1/2

m.

We can take C2 = 4 in condition (iv).The corollary is proved. Exercises. 1. Improve the estimate of C given in (1.19). Our result for the

Poisson r.vs in the corollary gives only C = 32e2.2. Generalize conditions (1)-(iv) of Theorem 1.6.1. The attempt done in [?]

seems to be unsatisfactory.


1.7 Additive functions and the tail probabilities

In Sections 1.5 and 1.6, we have developed necessary tools to investigate the valuedistribution of mappings defined on Sn via k(σ). We now confine ourselves to oneclass of such mappings.

Given a two-dimensional real sequence hj(k), where k ∈ Z+ and j ∈ Nn, andsatisfying the condition hj(0) ≡ 0, we define by

h(σ) =n∑j=1

hj(kj(σ)) (1.21)

an additive function h : Sn → R. Evidently, the definition can be extended tothe additive functions h : Sn → G, where (G,+) is an abelian group. If hj(k) =kaj for some aj ∈ R, for each k ∈ Z+ and j ∈ Nn, such a function is calledcompletely additive. The simplest instance belonging to this class h(σ) = w(σ) hasbeen examined in the Goncharov theorem. If hj(k) = 1(k ≥ 1), the appropriatevalue of the additive function h(σ) expresses the number of cycles in σ with differentlengths.

Let σ0 denote the identical permutation in Sn. The group theoretical order ofσ ∈ Sn,

Ord(σ) := mink ∈ Z+ : σk = σ0 = L.C.M.j ≤ n : kj(σ) ≥ 1 ,

where L.C.M. denotes the least common multiplier (nor its logarithm) is not addi-tive. Nevertheless, it is known (see, for instance, [?], [?], or [?]) that log Ord(σ) beapproximated by the additive function

h(σ) =∑j≤n

1kj(σ) ≥ 1 log j

for all σ ∈ Sn but a small their subset. Even the function `(σ) ≡ n is completelyadditive in our sense.

To avoid repetitions, we now present a few corollaries of the results obtained inthe previous section applied for an additive function h : Sn → G, where (G,+) is anabelian group.

Let h(σ) be defined via hj(k) ∈ G, j, k ≥ 1. On the set Zn+, we can define anassociated additive function H:Zn+ → G by setting

H(k): =n∑j=1

hj(kj) ,

where k = (k1, . . . , kn). As in the last section, we consider them elementary eventstaken with the probabilities

µ(k) = exp−

n∑j=1

1

j

n∏j=1

1

jkk!,

Now H(k) is the sum of G-valued r.vs. Let, as previously, ξ = (ξ1, . . . , ξn) be therandom vector with independent Poisson coordinates, Eξj = 1/j for 1 ≤ j ≤ n.

1.7. ADDITIVE FUNCTIONS AND THE TAIL PROBABILITIES 25

1.7.1 Lemma. If h:Sn → G and H:Zn+ → G are additive functions, B ⊂ G andΩn = k ∈ Zn+: `(k) = n, then

νn(h(σ) ∈ B) = µ(H(k) ∈ B|Ωn) . (1.22)

Proof. Theorem 1.2.2 and relation (1.12) imply

νn(h(σ) ∈ B) =∑

H(k)∈B

νn(k(σ) = k) =∑

H(k)∈B

P (ξ = k| `(ξ) = n)

= P (H(ξ) ∈ B| `(ξ) = n) = µ(H(k) ∈ B|Ωn) .

The lemma is proved. Theorem 1.6.1 is applicable for the conditional probability on the right-hand side

in (1.22) For an arbitrary A ⊂ G, we denote

A+ A− A = b = a1 + a2 − a3: a1, a2, a3 ∈ A .

1.7.1 Theorem. If A ⊂ G and h(σ) is an additive G-valued function, then

νn(h(σ) 6∈ A+ A− A) ≤ Cµ(H(k) 6∈ A) = CP (H(ξ) 6∈ A) .

Here C > 0 is the constant defined in Corollary 1.6.1.

Proof. We apply Theorem 1.6.1 for

U : = k: H(k) ∈ A .

Examine the extension V = V (U), defined in Section 1.6. If

l = k1 + k2 − k3, k1, k2, k3 ∈ U, k1 ⊥ k2 − k3, k3 ‖ k2 ,

then, using the additivity, we obtain

H(l) = H(k1) + H(k2)−H(k3) .

HenceV ⊂ k: H(k) ∈ A+ A− A .

For the compliments of the sets the converse implication holds; therefore, by Theo-rem 1.6.1 and Lemma 1.7.1,

νn(h(σ) 6∈ A+ A− A) = µ(H(k) 6∈ A+ A− A|Ωn) ≤ µ(V |Ωn)

≤ Cµ(U) = Cµ(H(k) 6∈ A) .

The claim is proved. Examine the cases G = R and G = Rn.

1.7.2 Theorem. Let h:S→ R be an additive function, a ∈ R and u ≥ 0 be arbitrarynumbers, then

νn(|h(σ)− a| ≥ 3u) ≤ CP (|h(ξ)− a| ≥ u) .


Proof. The probability on the right-hand side equals

µ(|H(k)− a| ≥ u) .

IfA = x ∈ R: |x− a| < u ,

then, for every x ∈ A + A − A, we have x = x1 + x2 − x3, xi ∈ A, i = 1, 2, 3.Moreover,

|x− a| ≤ |x1 − a|+ |x2 − a|+ |x3 − a| < 3u

andA+ A− A ⊂ x: |x− a| < 3u .

We now obtain the desired inequality from Theorem 1.7.1 because of

νn(|h(σ)− a| ≥ 3u) ≤ νn(h(σ) 6∈ A+ A− A) ≤ Cµ(|H(k)− a| ≥ u) .


1.7.3 Theorem. Let ht:Sn → R, t ∈ T ⊂ R, be a family of additive functionsdefined by (1.21) via hj(k) = hjt(k) and Ht(k) be the corresponding associated familyof additive function on Zn+. For arbitrary bt ∈ R, t ∈ T , and u ≥ 0, we have

νn

(supt∈T|ht(σ)− bt| ≥ 3u

)≤ CP

(supt∈T|Ht(ξ)− bt| ≥ u

).

Proof. It suffices to repeat the argument given in the proof of previous theorem.One has just to substitute R by the additive group of real functions defined on Tand to use the supremum norm instead of the absolute value.

1.7.1 Corollary. For an arbitrary real two-dimensional array hj(k), satisfyingthe condition hj(0) = 0 and a sequence of real numbers b1, . . . , bn and u ≥ 0, wehave

νn

(max

1≤m≤n

∣∣∣∑j≤m

hj(kj(σ))− bm∣∣∣ ≥ 3u

)≤ CP

(max

1≤m≤n

∣∣∣∑j≤m

hj(ξj)− bm∣∣∣ ≥ u

).

Proof. Apply Theorem 1.7.3 to the family of truncated additive functions

h(σ,m): =∑j≤m

hj(kj(σ)), m = 1, 2, . . . , n .

The corollary is proved. Exercise. Prove an analog of Kolmogorov’s inequality:

νn

(maxm≤n

∣∣∣∑j≤m

hj(kj(σ))−∑j≤mk≤n/j

hj(k)

jkk!

∣∣∣ ≥ u

)≤ C1u

−2∑jk≤n

|hj(k)|2

jkk!,

where u > 0 is an arbitrary number and C1 > 0 is an absolute constant.

1.8. MOMENTS OF ADDITIVE FUNCTIONS 27

1.8 Moments of additive functions

Let h : Sn → R be an additive function. We now examine the moments

En|h(σ)− An|s =1

n!

∑σ∈Sn

|h(σ)− An|s ,

where s > 0 and An is an appropriate centralizing sequence. To warm up, we startwith the variance for a completely additive function

h(σ) =n∑j=1

ajkj(σ) ,

where aj ∈ R, 1 ≤ j ≤ n, are arbitrary numbers.

1.8.1 Theorem. For the just defined completely additive function h(σ), set

An: = An(h): =n∑j=1

ajj.

Then, for all n ≥ 1,

En(h(σ)− An)2 ≤ 2n∑j=1

a2j

j. (1.23)

Proof. Firstly, recall the relations obtained in Theorem 1.2.5:

Enkj(σ) =1

j

and

Enk2j (σ) = En(kj(σ))2 + Enkj(σ) = 1(2j ≤ n)

1

j2+

1

j

for each j ≤ n, and

Enki(σ)kj(σ) = 1(i+ j ≤ n)1

ij

for each pair i, j ≤ n, i 6= j. Hence

Enh(σ) =n∑j=1

ajEnkj(σ) = An .

and

Enh(σ)2 =n∑i=1

n∑j=1

aiajEn(ki(σ)kj(σ))

=n∑j=1

a2jEnkj(σ)2 +

∑i,j≤ni6=j

aiajEn(ki(σ)kj(σ))

=∑j≤n

a2j

j+∑j≤n/2

a2j

j2+∑i+j≤ni6=j

aiajij

=∑j≤n

a2j

j+∑i+j≤n

aiajij

.


Consequently,

En(h(σ)− An)2 = Enh(σ)2 − A2n

=∑j≤n

a2j

j−∑j,j≤ni+j>n

aiajij

. (1.24)

If aj, 1 ≤ j ≤ n, are nonnegative, then omitting the last term in the last inequality,we see that the claim of Theorem 1.8.1 is true even with the unit constant.

If aj take positive and negative values, we can separate them. Set u+ = u ifu ≥ 0, u+ = 0 if u < 0, and u− = u − u+. The function h(σ) can be decomposedinto the sum h+(σ) + h−(σ), where h±(σ) are completely additive functions definedby the sequences a±j respectively. After the same trick we have An = A+

n + A−n ,where A±n = An(h±). Since

En(h±(σ)− A±n )2 ≤n∑j=1

(a±j )2

j,

applying the inequality (a+ b)2 ≤ 2a2 + 2b2, a, b ∈ R, we complete the proof of thetheorem.

Inequality (1.23) may be attributed to the theory of quadratic forms. Denotingaj = xj

√j and

γ(j, σ) =√j(kj(σ)− 1), σ ∈ Sn, 1 ≤ j ≤ n ,

we can rewrite it as ∑σ∈Sn

(∑j≤n

γ(j, σ)xj

)2

≤ 2n!∑j≤n

x2j , (1.25)

where x ∈ R.To shorten, one can use the matrix exposition. Set X = (x1, . . . , xn) and define

the matrix Kn = ((γ(j, σ))), where the columns are enumerated by the permutationsσ written in an arbitrary fixed order. The dimension of Kn is (n×n!). Let ′ mean thetransposition of matrices and vectors. Calculating the Euclidean norms, we obtain

||XKn||2 = (XKn)(XKn)′ = X(KnK′n)X ′

=∑σ∈S

(∑j≤n

γ(j, σ)xj

)2

. (1.26)

Comparing this with inequality (1.25), we obtain a matrix reformulation of Theorem1.8.1.

1.8.2 Theorem. For each vector X ∈ Rn,

||XKn||2 = X(KnK′n)X ′ ≤ 2n!||X||2 .


In addition, this shows that the maximal eigenvalue of the matrix KnK

′n, which

has the dimension n× n, grows not faster than 2n! as n→∞. It is proved (see [?])that 2 can not be substituted by a constant less than 3/2. In 2018 the author andhis student J.Klimavicius established the exact inequality substituting 2 by 3/2.

Continuing the investigation of the quadratic forms, we will derive a useful com-binatorial result. In the sequel, we will need a complex version of the above inequal-ities.

1.8.1 Lemma. Let Q = ((qij)), where 1 ≤ i ≤ m and 1 ≤ j ≤ n, be a real matrixand λ > 0. The inequality

n∑j=1

∣∣∣ m∑i=1

qijxi

∣∣∣2 ≤ λ

m∑i=1

|xi|2

holds for all vectors X = (x1, . . . , xm) ∈ Cm if and only if, for all Y = (y1, . . . , yn) ∈Cn, the following inequality

m∑i=1

∣∣∣ n∑j=1

qijyj

∣∣∣2 ≤ λn∑j=1

|yj|2

holds.

Proof. Using the argument applied above, we can rewrite the inequalities of thelemma in the matrix form:

||XQ||2 ≤ λ||X||2 (1.27)

and||Y Q′||2 ≤ λ||Y ||2 . (1.28)

Let us denote the scalar product by < ·, · >. It follows from Cauchy’s inequalityand (1.27) that

| < Y,XQ > |2 ≤ ||Y ||2||XQ||2 ≤ λ||Y ||2||X||2 .

The square of the scalar product on the left-hand side equals

|Y (XQ)′|2 = |Y Q′X ′|2 ,

The bar now stands for the complex conjugate vector. Hence

|Y Q′X ′|2 ≤ λ||Y ||2||X||2 (1.29)

for arbitrary X ∈ Cm and Y ∈ Cn. Choosing X = Y Q′, we obtain from inequality(1.29) that

|Y Q′(Y Q′)′|2 = ||Y Q′||4 ≤ λ||Y ||2||Y Q′||2 .If ||Y Q′|| = 0, inequality (1.28) is trivial. Otherwise, dividing by ||Y Q′||2, we

have inequality (1.28).By the same argument we could derive (1.27) from inequality (1.28).The lemma is proved.


1.8.3 Theorem. Let y(σ) ∈ C, σ ∈ Sn, be arbitrary. Then∑j≤n

j∣∣∣ ∑σ∈Sn

y(σ)(kj(σ)− 1

j

)∣∣∣2 ≤ 2n!∑σ∈Sn

| y(σ) |2 . (1.30)

Proof. Apply the lemma with the matrix Q = Kn. Considering the real andimaginary parts separately, from (1.25), we have

∑σ∈Sn

∣∣∣∑j≤n

γ(j, σ)xj

∣∣∣2 ≤ 2n!n∑j=1

|xj|2 .

Hence its inversion, by Lemma 1.8.1, becomes

∑σ∈Sn

∣∣∣ n∑j=1

γ(j, σ)y(σ)∣∣∣2 ≤ 2n!

∑σ∈Sn

| y(σ) |2 .

It remains to use the definition of γ(j, σ).The theorem is proved. Recalling Enkj(σ) = 1/j, we see that inequality (1.30) is an estimate of the

variance. Again, the constant 2 can not be substituted by 1; therefore, this showsthe stochastic dependence of kj(σ), 1 ≤ j ≤ n.

The just described direct method becomes rather involved estimating the mo-ments of higher order. There is an undirect approach based on the tail probabilityestimate given in the Corollary of Theorem 1.7.2. This method reduces the problemto that for sums of independent r.vs. For brevity we will use as an analog of O(·)with a constant depending on s only.

Rozenthal’s Inequality. Let X1, . . . , Xn be independent r.vs, EXj = 0, E|Xj|s <∞, 1 ≤ j ≤ n, and s ≥ 2. Then

E∣∣∣∑j≤n

Xj

∣∣∣s (∑j≤n

EX2j

)s/2+∑j≤n

E|Xj|s .

Proof can be found in V.V. Petrov’s book [?].

1.8.4 Theorem. Let h(σ) be an additive function, s ≥ 2, and

En(h): =∑kj≤n

hj(k)

jkk!, Dn(h; s): =

∑kj≤n

|hj(k)|s

jkk!.

ThenEn|h(σ)− En(h)|s Dn(h; s) + (Dn(h; 2))s/2 .

Proof. We apply Rozenthal’s inequality and the following formula

E|Y |s = s

∫ ∞0

ys−1P (|Y | ≥ u)du


centralizing by n, we have the zero central moments. In general, an additive functioncan have an expression h(σ) = h(σ) + λ`(σ) with some λ ∈ R. In such cases, it isbetter to get rid from the second summand by changing the centralizing sequencesand apply the just derived inequalities for the first summand.

Exercise. Find a sequence α(n) and estimate the moment En|h(σ)− α(n)|s, if0 < s < 2. The necessary results on sums of independent r.vs can be found in [?] orin [?].

1.9 The law of large numbers

As in the probabilistic laws of large numbers, we can seek conditions under which,for a real additive function h(σ), a real sequence α(n) and β(n) > 0,

νn(|h(σ)− α(n)| ≥ δβ(n))→ 0

for each δ > 0 as n → ∞. More generally, we can examine sequences of additivefunctions hn : Sn → R defined via

hn(σ) =n∑j=1

hnj(kj(σ)) . (1.33)

Now hnj(k) is even a three-dimension array, hnj(0) ≡ 0 if j ≤ n and n ∈ N. In thissection we will find sufficient conditions assuring the weak law of large numbers:

νn(|hn(σ)− α(n)| ≥ δ) = o(1) (1.34)

as n → ∞. Here α(n) is a real centralizing sequence and δ > 0 is arbitrary fixednumber. Observe that the values hnj(k) = 0, jk > n, can are not involved in h(σ)if σ ∈ Sn; therefore, they can be changed in an arbitrary way.

Let us compare our problem with the classical weak law of large numbers for thesum

Yn := Xn1 + · · · ,+Xnn

of independent r.vs Xnj, 1 ≤ j ≤ n. Recall that they are infinitesimal if

pn := maxj≤n

P (|Xnj| ≥ ε) = o(1) ,

for every ε > 0 as n → ∞. In what follows, we take Xnj = hnj(ξj), where aspreviously, ξj, j = 1, 2, ..., are independent Poisson r.vs and Eξj = 1/j for 1 ≤ j ≤ n.

1.9.1 Lemma. The r.vs Xnj, 1 ≤ j ≤ n, are infinitesimal if and only if

hnj(k) = o(1) (1.35)

for each fixed j ≥ 1 and k ≥ 1 as n→∞.

1.9. THE LAW OF LARGE NUMBERS 33

Proof. We have

pn = maxj≤n

∑k≥1

|hnj(k)|≥ε

e−1/j 1

jkk!.

If K ≥ 1 is an arbitrary fixed number, then, for arbitrary ε > 0, we can choosen ≥ n0 sufficiently large such that |hnj(k)| < ε for all j ≤ K and k ≤ K. Hence, ifn ≥ n0, we can examine the summands with j > K or k > K only. Consequently,

pn ≤ maxK<j≤n

∑k≥1

1

jkk!+ max

j≥1

∑k>K

1

jkk!= O

( 1

K

).

Hence pn → 0 as n→∞.The converse claim follows by obtaining a contradiction.The lemma is proved. Condition (1.35) will be frequently used in the sequel. It will fairly simplify our

calculations. Set for brevity u∗ = min|u|, 1sgnu and anj = hnj(1) where 1 ≤ j ≤ n.We first recall a special case of the classical probabilistic result (see, for instance,[?]).

1.9.1 Theorem. Let n→∞ and condition (1.35) hold. If

∑j≤n

a∗2

nj

j→ 0 , (1.36)

thenP (|Yn − An| ≥ δ) = o(1) ,

where

An :=∑j≤n

a∗njj,

It is worth to observe that, under condition (1.35), the values hnj(k) for k ≥ 2are negligible. The next assertion concerns the additive functions.

1.9.2 Theorem. Let the conditions of the previous theorem be satisfied. Then

νn(δ) := νn(|hn(σ)− An| ≥ δ) = o(1) ,

as n→∞.

Proof. It suffices to apply Theorem 1.7.2 and to obtain

νn(δ) ≤ CP (|Yn − An| ≥ δ/3) .

The claim now follows from Theorem 1.9.1. We now include a more transparent version of the law of large numbers, this

time formulated for one fixed additive function.


1.9.3 Theorem. Let h(σ) be an additive function defined via hj(k). As previously,set

En(h) =∑jk≤n

hj(k)

jkk!, Dn(h) := (Dn(h; 2))1/2 =

(∑jk≤n

h2j(k)

jkk!

)1/2

.

For arbitrary positive ψn →∞, we have

νn(|h(σ)− En(h)| ≥ ψnDn(h)) = o(1) ,

as n→∞.

Proof. It suffices to apply the previous theorem with

hn(σ) = h(σ)/(ψnDn(n)), anj = hj(1)/(ψnDn(h)) .

Condition (1.36) is evidently satisfied. Another choice of the centralizing sequencecan be substantiated by the inequalities∑

jk≤nk≥2

|hj(k)|jkk!

Dn(h) ,∑j≤n|anj |≥1

|anj|j≤ ψ−2

n = o(1) .

The theorem is proved. Necessary and sufficient conditions for the weak law of large numbers of additive

functions have been found in author’s paper [?]. Actually, we have exposed allideas leading to the sufficiency part of this result. The only needed refinement ismentioned in the exercise below.

Exercise. Extend Theorems 1.9.2 and 1.9.3 taking into account the possiblepart λ`(σ) = λn which hn(σ) could have.

1.10 The central limit theorem

We again examine a sequence of additive functions

hn(σ) =n∑j=1

hnj(kj(σ)) .

The problem is to find conditions when the relation

νn(hn(σ)− α(n) < x) =1√2π

∫ x

−∞e−u

2/2dx+ o(1) =: Φ(x) + o(1), x ∈ R ,

holds as n → ∞. Here α(n) ∈ R is a centralizing sequence. For simplicity, weassume condition (1.35). Under it, the problem reduces to the case of the completelyadditive functions defined by anj := hnj(1), that is, to

hn(σ) =∑j≤n

anjkj(σ) . (1.37)

1.10. THE CENTRAL LIMIT THEOREM 35

To substantiate this, we check that the difference of additive functions is additiveand, by Theorem 1.9.2, satisfies the weak law of large numbers

νn(|hn(σ)− hn(σ)| ≥ δ) = o(1)

for every δ > 0 as n→∞. This shows that the asymptotic distributions of hn(σ)−α(n) and hn(σ)− α(n) can exist only at the same time and coincide.

We again find an appropriate result in probability theory. As previously, let ξjbe the Poisson r.vs with Eξj = 1/j, 1 ≤ j ≤ n. Set Xnj := anjξj and

Yu :=∑j≤u

Xnj , Au :=∑j≤u

a∗njj, 1 ≤ u ≤ n.

By Va(x) we denote the distribution function of the degenerated at x = a law.

1.10.1 Theorem. Let n→∞. The relations

P (Yn − α(n) < x) = Φ(x) + o(1) (1.38)

uniformly in x ∈ R andanj = o(1) (1.39)

for each fixed j ≥ 1 hold if and only if the following conditions are satisfied:∑j≤n

a∗nj2

j1anj < x = V0(x) + o(1) , x ∈ R, (1.40)

andα(n) = An + o(1) . (1.41)

Proof. This is a partial case of the Lindeberg–Feller’s theorem (see [?]).

1.10.1 Remark. The Lindeberg type condition (1.40) is equivalent to∑j≤n

1

j1|anj| ≥ ε = o(1) (1.42)

for every ε > 0. This remains to hold for some ε = εn → 0. Hence, for each fixed0 < δ < 1,∑

δn<j≤n

a∗nj2

j= o(1) + ε2

n

∑δn<j≤n

1

j1|anj| < εn o(1) + ε2

n log(1/δ) = o(1) .

Now, by Theorem 1.9.1 with anj = 0 for j ≤ δn,

P (ε) := P (|Yn − Yδn − (An − Aδn)| ≥ ε) = o(1) (1.43)

for every ε > 0. This also remains to hold for some δ = δn → 0. In other words, theLindeberg condition assures also the convergence

P (Yr − Ar < x) = Φ(x) + o(1) (1.44)

uniformly in x ∈ R for some r = rn = o(n) as n→∞.


We will need the so-called lemma on the convergence of types of distributions.

1.10.1 Lemma. Let Fn(x), G(x) 6= Va(x) and F (x) 6= Va(x), where a ∈ R, bedistribution functions, α(n) and β(n) > 0 be sequences of constants, and let ⇒denote weak convergence. If

Fn(x)⇒ F (x) and Fn(β(n)x+ α(n))⇒ G(x),

then G(x) = F (βx+ α), β(n)→ β, and α(n)→ α, where α and β are constants.

Proof can be found in [?]. We now return to combinatorics.

1.10.2 Theorem. Let hn(σ) be a sequence of additive functions defined in (1.33)via hnj(k) which satisfies condition (1.35) and anj := hnj(1).

If conditions (1.40) and (1.41) are satisfied, then

νn(x): = νn(hn(σ)− α(n) < x) = Φ(x) + o(1) (1.45)

uniformly in x ∈ R as n→∞.Conversely, if for every 0 < δ < 1

∑δn<j≤n

a∗2

nj

j= o(1) , (1.46)

and (1.45) holds with some α(n) ∈ R, then conditions (1.40) and (1.41) are satisfied.

Proof. Condition (1.35) allows us to deal with the completely additive functiondefined via hnj(k) = anjk, 1 ≤ j ≤ n and k ≥ 0. This concerns either the sufficiencyor the necessity part of the proof.

Sufficiency. We first see that, by the central limit theorem above, relations(1.38) and (1.44) hold. As we have remarked, we also have (1.46), which followsfrom conditions (1.40). By Theorem 1.7.2, this leads to

νn(ε) := νn

(|(hn(σ)− hr(σ))− (An − Ar)| ≥ ε

) P (ε/3) = o(1) , (1.47)

for every ε > 0 for some r = o(n) as n→∞. Here and in the sequel

hr(σ) :=∑j≤r

anjkj(σ)

is a truncated additive function. Now, by Fundamental Lemma and (1.44),

νn(hr(σ)− Ar < x) = P (Yr − Ar < x) + o(1) = Φ(x) + o(1)

uniformly in x ∈ R as n→∞. Recalling the just proved estimate (1.47), we obtain

(1.45) with α(n) = An. The sufficiency is proved.

1.10. THE CENTRAL LIMIT THEOREM 37

Necessity. We now have (1.45). The a fortiori condition (1.46) implies (1.43).Hence there exists a sequence r = δn = o(n) such that, for n→∞,

Φ(x) + o(1) = νn(hn(σ)− α(n) < x)

= νn

(hr(σ)− (α(n)− An + Ar) + (hn(σ)− hr(σ)− An + Ar) < x

)= νn

(hr(σ)− αr < x) + o(1)

= P (Yr − αr < x) + o(1) ,

where αr := α(n) − An + Ar. In the last step, we used the Fundamental Lemma.

The central limit theorem for Yr holds with the centralizing sequence Ar; therefore,Lemma 1.10.1 implies

αr = α(n)− An + Ar = Ar + o(1) ,

that is, α(n) = An + o(1) as n→∞. Thus, we have proved relation (1.41).By Remark, condition (1.46) shows that the last but one relation can be extended

to (1.38). Now, Theorem 1.10.1 implies the necessity of condition (1.40).The theorem is proved. The just proved Theorem 1.10.2 becomes simpler if one normalized function h(σ)

is considered. In such cases, we examine the convergence of

νn(h(σ)− α(n) < xβ(n))

as n → ∞. By virtue of (1.40), one can define the normalizing sequence β(n) > 0,β(n)→∞, just solving asymptotically the relation

n∑j=1

( ajβ(n)

)∗2 1

j= 1 + o(1) .

Here aj := hj(1). If the latter values are comparatively small, one can use sequences

α(n) = An: =n∑j=1

ajj, β(n) = Bn: =

( n∑j=1

a2j

j

)1/2

.

In some sense, this idea has been aproved in the moment analysis and by the weaklaw of large numbers. Check that, if Bn →∞ as n→∞, then hnj(k) = hj(k)/Bn =o(1) for each fixed j, k ≥ 1. As in the proof of Theorem 1.10.2, we can confineourselves to the case of completely additive functions.

1.10.3 Theorem. Assume that Bn →∞ and n→∞. If

1

B2n

∑j≤n

a2j

j1|aj| ≥ εBn = o(1) (1.48)

for every ε > 0, then

νn(h(σ)− An < xBn) = Φ(x) + o(1)

uniformly in x ∈ R as n→∞. Conversely, the last relation and the extra conditionBr ∼ Bn for some r = o(n) as n→∞, imply (1.48).


Proof. As an exercise, we leave it for the Reader.

1.10.1 Example. Examine asymptotic distribution of

w(σ): =n∑j=1

1kj(σ) ≥ 1

expressing the number of different lengths of cycles in a random σ ∈ Sn. Thisfunction is additive and defined via the array

wj(k) = 1k ≥ 1, j ≥ 1, k ≥ 0 .

Now aj ≡ 1 and An = B2n ∼ log n as n→∞. Theorem 1.10.3 implies

νn(w(σ)− log n < x√

log n) = Φ(x) + o(1)

as n→∞.

In contrast to the central limit theorem for independent r.vs, our Theorem 1.10.2is not a final solution of the posed problem. The necessity part is proved only underthe extra condition (1.46). The author jointly with G.J. Babu [?] have constructedan instance of additive function which does not satisfy (1.40), nevertheless its asymp-totic distribution, after an appropriate normalization, is normal. The problem stillwaits for new ideas.

Apart from the normal law, one can seek other limit distributions. Recent au-thor’s results [?] give necessary and sufficient conditions if the normalizing sequenceβ(n) satisfies some additional regularity conditions. There is some progress in find-ing the necessary conditions for the convergence of νn(hn(σ) < x) to the Poisson orother discrete laws if hnj(k) ∈ Z (see [?] and [?]).

Exercise. Find the normalizing sequences α(n) and β(n) > 0 such that

νn

( n∑j=1

kj(σ) log j − α(n) < xβ(n))

= Φ(x) + o(1), n→∞.

1.11 Mean values of multiplicative functions

A multiplicative function f : Sn → C is defined via a two-dimensional array ofcomplex numbers fj(k), where j ≥ 1 and k ≥ 0, setting

f(σ) =n∏j=1

fj(kj(σ)) .

Here also the condition fj(0) ≡ 1, j ≥ 1, is assumed. If fj(k) = bkj with somebj ∈ C for all j ≥ 1 and k ≥ 0, keeping the agreement 00 := 1, we have a completelymultiplicative function.

1.11. MEAN VALUES OF MULTIPLICATIVE FUNCTIONS 39

The asymptotic behavior of the mean values

Mn := Mn(f) :=1

n!

∑σ∈Sn

f(σ)

as n → ∞ is now the main target. The previously touched problems on the valuedistribution of sequences of additive functions, using Fourier transforms, can bereduced to these mean values. Then the function f(σ) depends on the sequenceparameter n and t ∈ R, and we need asymptotic formulae with the uniform withrespect to t ∈ ∆ estimates of the remainders for various ∆ ⊂ R. This possibleobstacle is taken into account in this section.

1.11.1 Lemma. Let f(σ) be a multiplicative function defined via fj(s), j ≥ 1and s ≥ 0, where fj(0) ≡ 1, then

Mn = Mn(f) =∑`(s)=n

n∏j=1

fj(sj)

jsjsj!. (1.49)

Here the summation is over s = (s1, . . . , sn) ∈ Zn+ such that `(s) = n.

Proof. Group the summands in Mn over the classes Sn(s) which are indexedby s. All values of f(σ) for σ ∈ Sn(s) are the same. Applying Theorem 1.1.1 wecomplete the proof.

1.11.2 Lemma. In the notation of Lemma 1.11.1, we have the following formalpower series relation:

M(z) =∞∑m=0

Mmzm =

∞∏j=1

(1 +

∞∑s=1

fj(s)

jss!zjs).

For completely multiplicative functions defined via fj(s) = bsj, this reduces to

M(z) = m(z) := exp ∞∑

j=1

bjjzj

=: exp B(z) . (1.50)

Proof. Multiply the given formal power series term by term and use Lemma1.11.1 to compare the coefficients.

1.11.3 Lemma. In the case of completely multiplicative function f(σ), we have thefollowing recurrence relation:

Mn =1

n

n−1∑k=0

Mkbn−k , M0: = 1 . (1.51)

Proof. Exploit the formal equality

M ′(z) = M(z)B′(z) .


The main traditional ways to investigate Mn are based on the complex analysis,

Cauchy’s formula, and contour integration involving the function M(z) providedthat it is well defined in a nontrivial vicinity of zero. We now present a comparativelynew approach elaborated by V. Zacharovas in his thesis [?]. The first ingredient isthe following inequality raising an interest in itself.

1.11.4 Lemma. Let the series

g(z) =∞∑k=0

gkzk, gk ∈ C ,

converge in |z| < 1. If

S(m) =m∑k=0

kgk, m ≥ 0

then, for every n ≥ 1,∣∣∣ n∑k=0

gk − g(e−1/n)− S(n)

n

∣∣∣ ≤ 2

n

n∑k=1

|S(k)|k

+2

n

∑k>n

|S(k)|k

e−k/n.

Proof. For the finite sum and the infinite series we apply Abel’s summation. Theneeded reordering in the infinite series is justified by the absolute convergence onthe circumference |z| = e−1/n. Since gk = S(k)− S(k − 1), S(0) = 0, we obtain

R: =n∑k=0

ak − g(e−1/n)− S(n)

n=

n−1∑k=1

S(k)(1

k− 1

k + 1

)−∞∑k=1

S(k)(e−k/n

k− e−(k+1)/n

k + 1

). (1.52)

The modulo of the remainder sum over k ≥ n of the last series does not exceed∑k≥n

|S(k)|k

e−k/n∣∣∣1− e−1/n +

e−1/n

k + 1

∣∣∣ ≤ 2

n

∑k≥n

|S(k)|k

e−k/n.

The remaining summands on the right-hand side of (1.52) can be added. They give

R1: =n−1∑k=1

S(k)[1

k

(1− e−k/n

)− 1

k + 1

(1− e−(k+1)/n

)].

Including the terms ±e−(k+1)/n/k under the brackets and grouping appropriately,we see the absolute value of the quantity within them equals∣∣∣1

ke−k/n(e−1/n − 1) +

1

k(k + 1)(1− e−(k+1)/n)

∣∣∣ ≤ 2

kn.


Consequently,

|R1| ≤2

n

n−1∑k=1

|S(k)|k

.

Inserting the obtained estimates into formula (1.52) and observing that one of thesummands is attributed to another sum, we complete the proof.

The next theorem gives an approximation of the mean values of completelymultiplicative functions. Apart from the combinatorial meaning the result providesan asymptotic formula for the Taylor coefficients of a power series if some informationon the Taylor coefficients of its logarithmic derivative is available.

1.11.1 Theorem. Assume that bj ∈ C, |bj| ≤ 1, j ≥ 1, n ≥ 1, and

m(z): =∞∑n=0

mnzn = exp

∞∑j=1

bjjzj.

Then ∣∣∣mn − exp∑j≤n

bj − 1

j

∣∣∣ ≤ 5

n

∑j≤n

|bj − 1|(

1 + logn

j

).

Moreover, for every p > 1,∣∣∣mn − exp∑j≤n

bj − 1

j

∣∣∣ ≤ (2 +3p

p− 1

)( 1

n

∑j≤n

|bj − 1|p)1/p

. (1.53)

Proof. We will apply Lemma 1.11.4 for the function

g(z): =∞∑n=0

gnzn: = (1− z)m(z) = exp

∞∑j=1

bj − 1

jzj,

which is defined in the disk |z| < 1. The values bj for j > n have no influence tomn; therefore, we put 1 instead of them and obtain g(z) in the whole complex plane.Now S(k) = 1g1 + · · ·+ kgk and

∞∑k=1

S(k)zk =zg′(z)

1− z=

1

1− zexp

∞∑j=1

bj − 1

jzj ∞∑

j=1

(bj − 1)zj

= m(z)∞∑j=1

(bj − 1)zj .

By (1.49), the condition |bj| ≤ 1, j ≥ 1, implies |mn| ≤ 1. Consequently,

|S(k)| =∣∣∣ k∑j=1

(bj − 1)mk−j

∣∣∣ ≤ k∑j=1

|bj − 1| ≤ min

2k,n∑j=1

|bj − 1|


for every k ≥ 1. The conditions of Lemma 1.11.4 are satisfied. It yields

|mn − g(1)| =∣∣∣mn − exp

n∑j=1

gj − 1

j

∣∣∣≤ |g(e−1/n)− g(1)|+ |S(n)|

n

+2

n

n∑k=1

|S(k)|k

+2

n

∑k>n

|S(k)|k

e−k/n . (1.54)

Applying the above estimate of |S(k)|, we obtain

|g(1)− g(e−1/n)| ≤∫ 1

e−1/n

|g′(x)| dx ≤∫ 1

e−1/n

1− xx

∞∑k=1

|S(k)|xk dx

≤∑j≤n

|bj − 1|∫ 1

e−1/n

1− xx

∞∑k=1

xk dx

= (1− e−1/n)∑j≤n

|bj − 1| ≤ 1

n

∑j≤n

|bj − 1| =: ρn .

As we have observed, |S(n)|/n ≤ ρn. The last but one error in (1.54) does notexceed

2

n

n∑k=1

1

k

k∑j=1

|bj − 1| = 2

n

n∑j=1

|bj − 1|∑j≤k≤n

1

k≤ 2

n

n∑j=1

|bj − 1|(

1 + logn

j

).

Finally, the last term of (1.54) does not exceed

2

n

n∑j=1

|bj − 1|∞∫

1

e−v

vdv ≤ (2/e)ρn .

Adding the just obtained estimates we complete the proof of the first claim of thetheorem.

To prove the second claim, it suffices to apply (1.54) and the inequality

|S(k)| ≤ k1/q(∑j≤k

|bj − 1|p)1/p

≤ k1/q(∑j≤n

|bj − 1|p)1/p

,

where q = p/(p − 1) and k ≥ 1. The appearing extra sums of monotonicallydecreasing summands can be approximated by integrals as, for instance, this one:

∑k≤n

k−1/p ≤ 1 +

n∫1

u−1/p du ≤ 1 + qn1/q − q ≤ qn1/q.



As we see, we were rather pedantically calculating the constants in Theorem1.11.1. This was done to stress that the result is valid even uniformly in the functionsm(z) satisfying the conditions in Theorem 1.11.1. The instance below demonstrateshow we can extend the classical problem on counting of permutations missing somecycles. In it, we can take the forbidden set of the cycle lengths depending on n.

1.11.1 Example. We seek an asymptotic formula for the number of permutationsσ ∈ Sn which cycle lengths belong to J := Jn ⊂ 1, . . . , n = Nn. If, as previously,kj(σ) ≥ 0 denotes the number of cycles of lengths j, then the problem is to examinethe cardinality

Sn(J): = σ ∈ Sn: kj(σ) = 0,∀j 6∈ J .

The indicator of such set of permutations is the completely multiplicative functiondefined via bj = 1 if j ∈ J and bj = 0 if j ∈ J := Nn \ J . Hence

1 +∞∑m=1

|Sn(J)|m!

zm = exp∑

j∈J

zj

j

.

Theorem 1.11.1 yields the inequality∣∣∣ |Sn(J)|n!

− exp−∑j∈J

1

j

∣∣∣ ≤ 5

n

∑j∈J

(1 + log

n

j

).

The result is nontrivial if |J | is not large.

The function m(z) examined in Theorem 1.11.1 has a very special shape. Formany applications, we need asymptotic formulae for the Taylor coefficients of theproduct

M(z) :=∞∑n=o

Mnzn = m(z)H(z) := exp

∞∑j=1

bjzj

j

∞∑k=0

Hkzk ,

where H(z) is a function defined in the closed disk |z| ≤ 1 and, in some sense,”better” than B(z). We start with a very simple case.

1.11.2 Theorem. Assume that bj ∈ C, |bj| ≤ 1 if j ≥ 1, and

rn :=1

n

∑j≤n

|bj − 1|2 .

If|Hk| ≤ C3/(k + 1)2 (1.55)

for some constant C3 > 0 and all k ≥ 0, then∣∣∣Mn −H(1) exp∑j≤n

bj − 1

j

∣∣∣ ≤ 4C3

n+

7π2C3

3

√rn .


Proof. It suffices to apply estimate (1.53) in the first part of the sum

Mn =( ∑k≤n/2

+∑

n/2<k≤n

)Hkmn−k

and to estimate the second part. Since |mk| ≤ 1 for all k ≥ 0, the contribution ofthe second sum is not greater than C3

∑k>n/2 1/(1+k)2 ≤ 2C3/n. Inserting formula

(1.53) with p = 2, we have∣∣∣ ∑k≤n/2

Hkmn−k −∑k≤n/2

Hk exp ∑j≤n−k

bj − 1

j

∣∣∣ ≤ 8√rn∑k≤n/2

|Hk| ≤ (4/3)π2C3

√rn .

Here we have inserted the value ζ(2) = π2/6 of the Riemann zeta-function ζ(s) =∑n≥1 n

−s. If k ≤ n/2 and n ≥ 1, then∑n−k<j≤n

|bj − 1|j

≤ 2 log(

1− k

n

)−1

≤ log 4 .

Hence, by the inequality |ez − 1| ≤ |z|e|z|, where z ∈ C,∣∣∣∣ exp ∑n−k<j≤n

1− bjj

− 1

∣∣∣∣ ≤ 4(∑j≤n

|bj − 1|2)1/2( ∑

j>n−k

1

j2

)1/2

≤ 6√rn

and ∑k≤n/2

Hk exp ∑j≤n−k

bj − 1

j

=∑k≤n/2

Hk exp∑j≤n

bj − 1

j

(1 + 6θ

√rn

)=

(H(1) +

2θC3

n

)exp

∑j≤n

bj − 1

j

+ C3θπ

2√rn

with some |θ| ≤ 1, not the same at different places. Collecting all estimates weobtain the desired result.

The theorem is proved. Theorem 1.11.2 enables to find necessary and sufficient conditions under which

a multiplicative function f(σ) such that |f(σ)| ≤ 1 posses the limit mean valuelimn→∞Mn(f). We have just to analyze condition (1.55) more carefully. Afterwardswe will exploit some technical lemmata which are essentially due to H.-K. Hwang[?].

Let [zm]q(z) denote the m-th Taylor coefficient of a function q(z) defined in avicinity of zero.

1.11.5 Lemma. Let n ≥ 0. If |[zn]q(z)| ≤ C4/n! and |[zn]g(z)| ≤ C5/n!, then

[zn](q(z)g(z)) ≤ C4C52n/n! .

If |[zn]q(z)| ≤ C4(n+ 1)−2 and |[zn]g(z)| ≤ C5(n+ 1)−2, then

[zn](q(z)g(z)) ≤ 8ζ(2)C4C5(n+ 1)−2, [zn] exp g(z) ≤ C6(n+ 1)−2 .

Here C6 is a positive constant depending on C4 and C5 only.


Proof. The first claim is easy. To prove the second one, we exploit the givenconditions as follows:

[zn](q(z)g(z)) ≤ C4C5

( ∑k≤n/2

+∑

n/2<k≤n

) 1

(k + 1)2(n− k + 1)2≤ 8ζ(2)C4C5(n+1)−2 .

Further, by induction, we obtain

[zn](g(z)r) ≤ (8ζ(2))r−1Cr5(n+ 1)−2

for each r ≥ 1 and n ≥ 0. Now, if g(0) = 0, then

|[zn] exp g(z)| =∣∣∣ n∑r=0

[zn]gr(z)1

r!

∣∣∣≤ (n+ 1)−2

∞∑r=0

(8ζ(2))r−1Cr5

r!

< (n+ 1)−2 exp 8ζ(2)C5 =: (n+ 1)−2C7 .

In the remaining case, we have

[zn]eg(z) = eg(0) [zn]eg(z)−g(0) ≤ eC5C7 .

The lemma is proved.

1.11.6 Lemma. Assume that

χj(z) =∑n≥2

ajnzn, j ≥ 1,

are entire functions satisfying |ajn| ≤ Cn/n! for all j ≥ 1 and n ≥ 0, where C > 0is a constant. Then

An := [zn]∏j≥1

(1 + χj(zj/j)) ≤ C1

n2, n ≥ 1,

where C1 is a positive constant depending on C only.

Proof. We have

An =∑`(k)=nkj 6=1

n∏j=1

ajkjjkj≤∑`(k)=nkj 6=1

n∏j=1

(Cj

)kj 1

kj!

= [zn]∏j≥1

(1 +

(eCz

j/j − 1− Czj

j

))≤ [zn] exp

∑j≥1

(eCz

j/j − 1− Czj

j

).


Check that

[zn]∑j≥1

(eCz

j/j − 1− Czj

j

)=∑

d|n,d≥2

Cd

(n/d)dd!

≤ 2C2

n2

∑d|n,d≥2

Cd−2

(d− 2)!

(dn

)d−2

≤ 2C2eC

n2.

The desired estimate now follows from the previous lemma.The lemma is proved.

1.11.3 Theorem. Let f(σ) be a complex-valued multiplicative function defined via|fj(s)| ≤ 1 for j, s ≥ 1. Denote

rn =1

n

∑j≤n

|fj(1)− 1|2 .

Then

Mn(f)−∏j≤n

e−1/j

(1 +

∞∑s=1

fj(s)

jss!

)√rn +

1

n. (1.56)

The constant implied in the symbol is absolute.

Proof. In the previous notation, the generating series M(z) defined in Lemma1.11.2 can be written as M(z) = m(z)H(z) with

H(z) =∏j≥1

e−bjzj/j

(1 +

∞∑s=1

fj(s)

jss!zsj)

=:∏j≥1

(1 + χj(zj/j))

and bj := fj(1). Here

χj(z) =∑n≥2

zn

n!

(−bj)n − n(−bj)n + n!∑r+s=nr≥0,s≥2

(−bj)rfj(s)r!s!

=:∑n≥2

ajnzn

are entire functions. Moreover,

|ajn| ≤1

n!

(1 + n+

n∑s=0

(ns

))≤ 4n

n!.

By Lemma 1.11.6, this implies condition (1.55). Theorem 1.11.2 now yields theapproximation

Mn(f)−H(1) exp∑j≤n

bj − 1

j

1

n+√rn.

Since the values fj(s) for j > n and s ≥ 1 do not have any influence to Mn(f) wemay assume that they are equal to 1. The last formula then coincides with (1.56).


Exercise. Find a proof of the classical Tauber theorem based on Lemma 1.11.4.

1.12. THE THREE SERIES THEOREM 47

1.12 The three series theorem

Given a real-valued additive function h(σ), its distribution can be treated by theuse of the Fourier transform∫

Reitxdνn(h(σ) < x) =

1

n!

∑σ∈Sn

eith(σ) =: ϕn(t), t ∈ R .

This leads to a problem on asymptotic mean values for multiplicative functions asn→∞. We now can apply Theorem 1.11.1.

Recall that u∗ = min|u|, 1sgnu. For simplicity, examine the completely addi-tive function

h(σ) =n∑j=1

ajkj(σ) .

1.12.1 Theorem. The distribution functions νn(h(σ) < x) weakly converge to alimit distribution if and only if the series

∞∑j=1

a∗2

j

j(1.57)

and∞∑j=1

a∗jj

(1.58)

converge. Under these conditions the characteristic function of the limit law is

ϕ(t) = exp ∞∑

j=1

eitaj − 1

j

.

Proof. Define a completely multiplicative function f(σ) via bj = eitaj , wherej ≥ 1 and t ∈ R.

Sufficiency. We have to prove that, under conditions (1.57) and (1.58), thecharacteristic functions

ϕn(t) :=1

n!

∑σ∈Sn

eith(σ)

converge uniformly in |t| ≤ T for every T > 0 as n → ∞. Nevertheless, we willprove even more. Namely, assuming only (1.57), we will derive that

ϕn(t) := exp − itAnϕn(t) = ϕ(t) + o(1) (1.59)

with the same uniformity, where

An =n∑j=1

a∗jj.


Afterwards we will use the inequalities |1− z|2 ≤ 2(1−<z) if |z| ≤ 1 and

1− cosαt ≤ 21|α| ≥ 1+ |t|21|α| < 1, α, t ∈ R .

If (1.57) holds, then∑j≤n

1− cos tajj

≤ 2∑|aj |≥1

1

j+ T 2

∑|aj |<1

a2j

j≤ C(T ) <∞

uniformly in |t| ≤ T . Moreover,

1

n

∑j≤n

|1− eitaj |2 ≤ 4 log n

n+ 2

∑logn<j≤n

1− cos tajj

= o(1)

as n→∞ in the same region. Further, from Theorem 1.11.1, we obtain∣∣∣ϕn(t)−exp−itA(n)+

∑j≤n

eitaj − 1

j

∣∣∣ ≤ 12( 1

n

∑j≤n

(1−cos taj))1/2

= o(1) . (1.60)

The real part of the sum in the exponential function on the left-hand side convergesuniformly in |t| ≤ T .

The remaining sums can be rewritten as:

i∑j≤n|aj |<1

sin taj − tajj

+ i∑j≤n|aj |≥1

sin taj − tsgn ajj

.

By inequality | sinx − x| |x|3, x ∈ R, and (1.57), either of the sums convergesas n→∞ with the desired uniformity. The proof of (1.59) is completed. Estimate(1.60) also shows that

ϕ(t) = exp ∞∑

j=1

eitaj − 1− ita∗jj

, t ∈ R .

If both series in Theorem 1.12.1 converge, the same argument applies to provethe sufficiency part of the theorem and it gives the claimed expression for ϕ(t) aswell.

Necessity. For each T > 0, if |t| ≤ T , we have

ϕm(t) = ϕ(t) + o(1), (1.61)

as m → ∞. Here ϕ(t) is a characteristic function. There exists a T > 0 such that|ϕ(t)| ≥ 1/2 for t ∈ [−T, T ]. We will use (1.61) for m = n, n− 1, . . . , n−K, whereK ≥ 1 is an arbitrary fixed natural number.

We now do a very unexpected step in applying Theorem 1.8.3. If y(σ) = f(σ) =eith(σ) and bj = eitaj , inequality (1.30) yields∑

j≤K

j

∣∣∣∣ ∑σ∈Sn

f(σ)(kj(σ)− 1

j

)∣∣∣∣2 ≤ 2(n!)2 (1.62)


for every 1 ≤ K ≤ n− 1. Calculating the sums, we first reckon the summands withkj(σ) = 1, where j ≤ K. We obtain

1

n!

∑σ∈Snkj(σ)=1

f(σ) =∑`(s)=nsj=1

n∏i=1

(bii

)si 1

isisi!

=bjj

∑`(s)=n−jsj=0

n∏i=1

(bii

)si 1

isisi!.

Observe that, if m < j ≤ n, then it follows from `(s) = m that sj = 0. Hence

1

n!

∑σ∈Snkj(σ)=1

f(σ) =bjj

∑1s1+···+(n−j)sn−j=n−j

sj=0

n−j∏i=1

(bii

)si 1

isisi!

=bj

j(n− j)!∑σ∈Sn−jkj(σ)=0

f(σ)

=bjjEn−jf(σ) +R1;

here

|R1| =

∣∣∣∣ −bjj(n− j)!

∑σ∈Sn−jkj(σ)≥1

f(σ)

∣∣∣∣ ≤ 1

j(n− j)!∑σ∈Sn−jkj(σ)≥1

1

=1

j

∑1≤m≤(n−j)/j

νn−j(kj(σ) = m) .

Theorem 1.2.1 now yields

|R1| ≤1

j

∑1≤m≤(n−j)/j

1

jmm!

∑0≤s≤((n−j)/j)−m

(−1)s

jss!

≤ e−1/j

j

∑m≥1

1

jmm!≤ 1

j(1− e−1/j) ≤ 1

j2.

Similarly,

|R2| :=1

n!

∣∣∣ ∑σ∈Snkj(σ)≥2

f(σ)kj(σ)∣∣∣

≤∑

2≤m≤n/j

mνn(kj(σ) = m)

=∑

2≤m≤n/j

1

jm(m− 1)!

∑0≤s≤(n/j)−m

(−1)s

jss!

≤ e−1/j

j

∑r≥1

1

jrr!≤ 1

j2.


In the just introduced notation, inequality (1.62) can be rewritten as∑j≤K

j

∣∣∣∣bjj En−jf(σ) +R1 +R2 −1

jEnf(σ)

∣∣∣∣2 ≤ 2 .

Hence using |a+ b|2 ≤ 2|a|2 + 2|b|2, where a, b ∈ C, we obtain∑j≤K

j∣∣∣bjjEn−jf(σ)− 1

jEnf(σ)

∣∣∣2≤ 4 + 2

∑j≤K

j(|R1|+ |R2|)2

≤ 4 + 8∞∑j=1

1

j3≤ C <∞ .

It follows from (1.61) that Emf(σ) = ϕ(t) + o(1) uniformly in |t| ≤ T and n−K ≤m ≤ n. Consequently, ∑

j≤K

j

∣∣∣∣bjj ϕ(t)− 1

jϕ(t) + o(1)

∣∣∣∣2 ≤ C .

Since |ϕ(t)| ≥ 1/2, as n→∞, we obtain∑j≤K

|eitaj − 1|2

j≤ 4C

for each K ≥ 1. In other words, the series

∞∑j=1

1− cos tajj

(1.63)

converges uniformly in |t| ≤ T .We now claim that this implies the convergence of series (1.57). Indeed, applying

1− cosx ≥ 2π−2x2, where |x| ≤ π/2, we obtain the convergence of∑|aj |≤π/(2T )

a2j

j. (1.64)

We nay assume here that π/(2T ) ≥ 2. Integrating the part of series (1.63) with|aj| ≥ π/(2T ) in the interval [0, T ] and dividing by T , we derive the inequality∑

|aj |≥π/(2T )

1

j

(1− sin taj

Taj

)≤ 4C

which implies ∑|aj |≥π/(2T )

1

j≤ 4C

(1− 2

π

)−1

=: C1 <∞ .


Hence and from (1.64) we obtain

∑|aj |≥1

1

j≤ C1 +

∑|aj |≤π/(2T )

a2j

j<∞ .

Now, using (1.64), we have convergence of (1.57). By the sufficiency part, this leadsto

νn(h(σ)− An < x)⇒ G(x) ,

here G(x) is a distribution function with the characteristic function ϕ(t). Now therelation

expitAn = ϕ(t)/ϕ(t) + o(1)

as n→∞ gives (1.58).The theorem is proved.

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