arXiv:1701.04855v2 [math.PR] 20 Jan 2017 apigdsrbto . in distribution sampling cycles numbers, Bell Dob´ınski’s formula, Polynomials, ois o xml,i 1] 1] h elnumber c Bell on The books [12]. many [11], in in found example, be for can torics; proof without dist below Poisson noted to facts connection their and polynomials, Touchard mult certain of covers of th number of the derivation for simpler and function new generating a to leads This permutation. by indexed quantity, stu This and statistic permutation novel rather a introduce di also Poisson the and polynomials Touchard i between some connection gives which permutations random of cycles concerning 2000 e od n phrases. and words Key ebgnb ealn oebscfcscnenn elnumbe Bell concerning facts basic some recalling by begin We fundamen a of proof simpler and new a present we paper, this In EMTTOSTA I E N TOUCHARD AND SET A FIX THAT PERMUTATIONS OECNETOSBTENCCE AND CYCLES BETWEEN CONNECTIONS SOME ahmtc ujc Classification. Subject Mathematics uin hsqatt,idxdby indexed quantity, stu This and statistic bution. permutation novel rather a introduce also distr Poisson fo the intuition and polynomials some Touchard between gives nection which permutations random of cycles Abstract. h xoeta eeaigfnto o h ubro cover of number deriv the multisets. simpler for and function new generating a exponential to the leads This permutation. the by fixed OYOIL N OESO MULTISETS OF COVERS AND POLYNOMIALS 1. nrdcinadSaeeto Results of Statement and Introduction epeetanwpofo udmna eutconcerning result fundamental a of proof new a present We emttosta xast oeso utst,Touchard multisets, of covers set, a fix that permutations m stenme fst fsize of sets of number the is , OSG PINSKY G. ROSS 1 m 00,0A5 05A15. 05A05, 60C05, stenme fst fsize of sets of number the is , admpruain,Ewens permutations, random B n eoe h number the denotes yisdistribution. its dy yisdistri- its dy bto.We ibution. fcertain of s h con- the r m the for ntuition isets. iuin.The ributions. tiuin We stribution. to of ation exponential e xdb the by fixed m a result tal ombina- sand rs 2 ROSS G. PINSKY of partitions of a set of n distinct elements. Elementary combinatorial rea- soning yields the recursive formula

n n (1.1) Bn = B , n ≥ 0, +1 k k Xk=0 where B0 = 1. Let ∞ B (1.2) E (x)= n xn, B n! nX=0 ∞ denote the exponential moment of {Bn}n=0. Using (1.1) ′ x it is easy to show that EB(x)= e EB(x), from which it follows that

ex−1 (1.3) EB(x)= e .

A random variable X has the Pois(λ), λ > 0, if −λ λk tX P (X = k) = e k! , k = 0, 1,... . Let Mλ(t) = Ee denote the moment n generating function of X, and let µn;λ = EX denote the nth moment of (n) X. Since Mλ (0) = µn;λ, we have ∞ µ (1.4) M (t)= n;λ tn. λ n! nX=0 However a direct calculation gives

∞ k λ t (1.5) M(t)= etke−λ = eλ(e −1). k! Xk=0

From (1.2)-(1.5), it follows that the nth moment µn;1 of a Pois(1)-distributed random variable satisfies

(1.6) µn;1 = Bn.

Since ∞ 1 µ = EXn = (e−1 )kn, n;1 k! Xk=0 we conclude that ∞ 1 kn (1.7) B = , n e k! Xk=0 which is known as Dob´ınski’s formula. PERMUTATION FIXING SET, TOUCHARD POLYNOMIAL, COVER OF MULTISET 3

n The of the second kind k denotes the number of par- titions of a set of n distinct elements intok nonempty sets. Elementary combinatorial reasoning yields the recursive formula n + 1 n n = k + .  k  k k − 1 From this, it is not hard to derive the formula n n n (1.8) x = (x)j, j Xj=0 where (x)j = x(x − 1) · · · (x − j + 1) is the falling factorial, and one defines 0 (x)0 = 0 = 1. Now using (1.8) and the fact that (k)j = 0 for j > k, we  can write the nth moment µn;λ of a Pois(λ)-distributed random variable as

∞ k ∞ k n −λ λ n −λ λ n µ = (e )k = (e ) (k)j = n;λ k! k! j  Xk=0 Xk=0  Xj=0 

∞ k k∧n n ∞ λ n n (k)j (1.9) (e−λ ) (k) = e−λ λk = k! j  j j  k! Xk=0  Xj=0  Xj=0 Xk=j n n λj. j  Xj=0

The Touchard polynomials Tn(x), n ≥ 0, are defined by n n j Tn(x)= x . j  Xj=0 Thus, (1.9) gives the formula

(1.10) µn;λ = Tn(λ).

Since ∞ λk µ = (e−λ )kn, n;λ k! Xk=0 we conclude from (1.10) that ∞ kn (1.11) T (x)= e−x xk. n k! Xk=0 n n Since Tn(1) = j=0 j = Bn, the Dob´ınski formula (1.7) is contained in (1.11). P  4 ROSS G. PINSKY

Let Sn denote the set of permutations of [n] =: {1,...,n}. For σ ∈ (n) Sn, let Cm (σ) denote the number of cycles of length m in σ. Let Pn denote the uniform probability measure on Sn. We can now think of σ ∈ Sn (n) as random, and of Cm as a random variable. Using generating function techniques and/or inclusion-exclusion formulas, one can show that under (n) Pn, the distribution of the random variable Cm converges weakly to Z 1 , m 1 where Z 1 has the Pois( )-distribution; equivalently; m m

(n) − 1 1 (1.12) lim Pn(Cm = j)= e m , j = 0, 1 .... n→∞ mjj!

More generally, we consider the Ewens sampling distributions, Pn;θ, θ > 0, (n) on Sn as follows. Let N (σ) denote the number of cycles in the per- n (n) mutation σ ∈ Sn, and let s(n, k) = |{σ ∈ S : N (σ) = k}| denote the number of permutations in Sn with k cycles. It is known that the n k (n) polynomial k=1 s(n, k)θ is equal to the rising factorial θ , defined by (n) P θ = θ(θ + 1) · · · (θ + n − 1). For θ > 0, define the probability measure Pn;θ on Sn by n θN ( )(σ) P ({σ})= . n;θ θ(n)

Of course, Pn;1 reduces to the uniform measure Pn. The following theorem can be proven; see for example, [1], [10].

(n) (n) (n) Theorem C. Under Pn,θ, the random vector (C1 ,C2 , · · · ,Cm ,...) con- ∞ verges weakly to (Zθ,Z θ , · · · ,Z θ , · · · ), where the random variables {Z θ }m=1 2 m m θ are independent, and Z θ has the Pois( )-distribution: m m

(n) (n) (n) w (1.13) (C1 ,C2 , · · · Cm ,...) ⇒ (Zθ,Z θ , · · · ,Z θ · · · ); 2 m equivalently,

m θ jk (n) (n) θ ( ) (n) − k k lim Pn(C1 = j1,C2 = j2,...,Cm = jm)= e , n→∞ jk! (1.14) kY=1

for all m ≥ 1 and j1,...jm ∈ Z+.

We will use the method of moments to give a new and simpler proof of Theorem C, which will give intuition for (1.10), or equivalently, for (1.11); PERMUTATION FIXING SET, TOUCHARD POLYNOMIAL, COVER OF MULTISET 5 that is for the connection between the moments of Poisson random variables and Touchard polynomials. We now consider a permutation statistic that hasn’t been studied much. (Indeed, it was only after completing the first version of this paper that we were directed to any papers on this subject.) For σ ∈ Sn and A ⊂ [n], define (n) σ(A) = {σj : j ∈ A}. If σ(A) = A, we will say that σ fixes A. Let Em (σ) denote the number of sets of cardinality m that are fixed by σ. (Note that (n) (n) E1 (σ)= C1 (σ), the number of fixed points of σ.) A little thought reveals that

m (n) (n) Cj (ω) (1.15) Em (ω)= .  lj  (l ,...,l ):Pm jl =m j 1 m Xj=1 j Y=1 (n) lj ≤Cj ,j∈[m]

For example, if σ ∈ S9 is written in cycle notation as σ = (379)(24)(16)(5)(8), E(9) then 4 (ω) = 5, with the sets A ⊂ [9] for which |A| = 4 and σ(A) = A being {3, 5, 7, 9}, {3, 7, 8, 9}, {1, 2, 4, 6}, {2, 4, 5, 8}, {1, 5, 6, 8}.

We consider the uniform measure Pn = Pn;1 on Sn. From Theorem C and (n) (1.15) it follows that the random variable Em under Pn converges weakly as n →∞ to the random variable m Z 1 (1.16) Em =: j ,  l  (l ,...,l ):Pm jl =m j j 1 m Xj=1 j Y=1 lj ≤Z 1 ,j∈[m] j

m 1 where {Z 1 } are independent and Z 1 has the Pois( )-distribution. j j=1 j j

Z1 Z1 Remark. Note that E1 = Z1, E2 = + Z 1 , E3 = Z 1 + Z 1 Z1 + . 2 2 3 2 3   For k,m ∈ N, consider the multiset consisting of m copies of the set [k]. r A collection {Γl}l=1 such that each Γl is a nonempty subset of [k], and such r that each j ∈ [k] appears in exactly m from among the r sets {Γl}l=1, is called an m-cover of [k] of order r. Denote the total number of m covers of

[k], regardless of order, by vk;m. Note that when m = 1, we have vk;1 = Bk, the kth , denoting the number of partitions of a set of k elements.

Also, it’s very easy to see that v1;m =1 and v2,m = m + 1. 6 ROSS G. PINSKY

(n) By calculating directly the moments of Em , we will prove the following theorem.

Theorem 1. For m, k ∈ N,

k EEm = vk;m.

2 In particular, EEm = 1 and EEm = m + 1; thus, Var(Em)= m.

Remark. It is natural to suspect that Em converges weakly to 0 as m →∞; that is, limn→∞ P (Em ≥ 1) = 0. This is in fact a hard problem. In [8] it was (n) 1 − 100 n shown that Pn(Em ) ≤ Am , for 1 ≤ m ≤ 2 and n ≥ 2. Thus indeed,

Em converges weakly to 0 as m → ∞. A lower bound on P (Em ≥ 1) of the log m form A m was obtained in [5]. These results were dramatically improved −δ+o(1) in [9] where it was shown that P (Em ≥ 1) = m as m → ∞, where 1+log log 2 δ = 1 − log 2 ≈ 0.08607. And very recently, in [7], this latter bound has −δ − 3 −δ − 3 been refined to A1m (1 + log m) 2 ≤ P (Em ≥ 1) ≤ A2m (1 + log m) 2 .

Let ∞ v V (x)= k;m xk m k! Xk=1 ∞ denote the exponential generating function of the sequence {vk;m}k=1. Of course, by Theorem 1 Vm is also the moment generating function of the

xEm random variable Em: Vm(x)= Ee . Using (1.16) and Theorem 1, we will give an almost immediate proof of the following representation theorem for m ∞ m Vm(x). We use the notation [z ]P (z)= am, where P (z)= m=0 amz . P Theorem 2.

m −u m 1 j j xEm − Pj xγm(u) (1.17) Vm(x)= Ee = e =1 j e , uj! u1,...,uXm≥0 jY=1  where m m j uj γm(u) = [z ] (1 + z ) . jY=1 PERMUTATION FIXING SET, TOUCHARD POLYNOMIAL, COVER OF MULTISET 7

Remark. When m = 2, 3, the above formula reduces to

∞ (r)x − 3 1 ex e 2 V (x)= e 2 e 2 ; 2 r! Xr=0 r 1 rx x ∞ ( )x+ e − 11 e e 3 2 V (x)= e 6 e 3 . 3 r! Xr=0

The formula for m = 2 was proved by Comtets [3] and the formula for m = 3 was proved by Bender [2]. The case of general m was proved by Devitt and Jackson [4]. They also prove that there exists a number c such that the extraction of the coefficient vk;m from the exponential generating function m Vm(x) can be done in no more than ck log k arithmetic operations. In section 2 we will give our new proof of Theorem C via the method of moments. In section 3 we prove Theorems 1 and 2.

2. A proof of Theorem C via the method of moments

∞ If a sequence of nonnegative random variables {Xn}n=1 satisfies supn≥1 EXn < ∞, then the sequence is tight, that is, pre-compact with respect to weak con- vergence. Let X be distributed as one of the accumulation points. If for N k k+1 some k ∈ , limn→∞ EXn exists and equals µk, and supn≥1 EXn < ∞, k ∞ k then the {Xn}n=1 are uniformly integrable, and thus EX = µk. Thus, if

k N (2.1) µk =: lim EXn exists for all k ∈ , n→∞

k then EX = µk, for all k. The Stieltjes moment theorem states that if

1 µ k (2.2) sup k < ∞, k≥1 k

∞ then the sequence {µk}k=1 uniquely characterizes the distribution [6]. We ∞ conclude then that if a sequence of nonnegative random variables {Xn}n=1 satisfies (2.1) and (2.2), then the sequence is weakly convergent to a random k variable X satisfying EX = µk. 8 ROSS G. PINSKY

An extremely crude argument shows that the Bell numbers satisfy Bk ≤ kk; thus

1 B k (2.3) sup k < ∞. k≥1 k θ By (1.10), the kth moment µ θ of the Pois( )-distributed random variable k; m m θ θ Z θ is equal to Tk( ). Now Tk( ) is bounded from above by Tk(θ), for all m m m k m ≥ 1, and Tk(θ) ≤ Bk max(1,θ ). Thus, in light of (2.3) and the previous paragraph, if we prove that

(n) k θ N (2.4) lim En;θ(Cm ) = Tk( ), k,m ∈ , n→∞ m where En;θ denotes the expectation with respect to Pn;θ, then we will have (n) proved that Cm under Pn;θ converges weakly to Z θ , for all m ∈ N. And if m we then prove that m m (n) kj θ (2.5) lim En;θ (C ) = Tk ( ), m ≥ 2, kj ∈ N, j = 1,...,m, n→∞ j j j jY=1 jY=1 then we will have completed the proof of Theorem C. We first prove (2.4). In fact, we will first prove (2.4) in the case of the uniform measure, Pn = Pn;1. Once we have this, the case of general θ will follow after a short explanation. Assume that n ≥ mk. For D ⊂ [n] with

|D| = m, let 1D(σ) be equal to 1 or 0 according to whether or not σ ∈ Sn possesses an m-cycle consisting of the elements of D. Then we have

(n) (2.6) Cm (σ)= 1D(σ), DX⊂[n] |D|=m and k (n) k (2.7) En(Cm ) = En 1Dj .

Dj ⊂[nX],|Dj|=m jY=1 j∈[k] k Now En j=1 1Dj 6= 0 if and only if for some l ∈ [k], there exist disjoint l Q k l sets {Ai}i=1 such that {Dj }j=1 = {Ai}i=1. If this is the case, then k (n − lm)!((m − 1)!)l (2.8) E 1 = . n Dj n! jY=1 PERMUTATION FIXING SET, TOUCHARD POLYNOMIAL, COVER OF MULTISET 9

(Here we have used the assumption that n ≥ mk, since otherwise n − ml will be negative for certain l ∈ [k].) The number of ways to construct l l disjoint, ordered sets {Ai}i=1, each of which consists of m elements from n! l [n], is (m!)l(n−lm)! . Given the {Ai}i=1, the number of ways to choose the k k l sets {Dj }j=1 so that {Dj}j=1 = {Ai}i=1 is equal to the Stirling number k l , the number of ways to partition a set of size k into l nonempty parts. From these facts along with (2.7) and (2.8), we conclude that for n ≥ mk,

k l (n) k (n − lm)!((m − 1)!) n! k En(C ) = = m n! (m!)l(n − lm)! l  Xl=1   (2.9) k 1 k 1 = T ( ), ml l  k m Xl=1 proving (2.4) in the case θ = 1. For the case of general θ, we note that the only change that must be made in the above proof is in (2.8). Recalling that s(n, k) denotes the number of permutations in Sn with k cycles, we have (2.10) k l n−ml k+l (n−ml) l ((m − 1)!) k=1 s(n − lm,k)θ θ ((m − 1)!) l E 1D = = θ ∼ n;θ j P θ(n) θ(n) jY=1 k (n − ml)!((m − 1)!)l θl = θlE 1 , as n →∞. n! n Dj jY=1

Thus, instead of (2.9), we have

k (n) k θ l k θ En(C ) ∼ ( ) = T ( ), as n →∞. m m l  k m Xl=1

We now turn to (2.5). The method of proof is simply the natural extension of the one used to prove (2.4); thus, since the notation is cumbersome we will suffice with illustrating the method by proving that

(n) k1 (n) k2 1 1 (2.11) lim En(Cm1 ) (Cm2 ) = Tk1 ( )Tk2 ( ). n→∞ m1 m2 10 ROSS G. PINSKY

Let n ≥ m1k1 + m2k2. By (2.6), we have

k1 k2 (n) k1 (n) k2 (2.12) EnC ) (C ) = En 1 1 1 2 . m1 m2 Dj Dj D1 nX, D1 m D2 nX, D2 m jY=1 jY=1 j ⊂[ ] | j |= 1 j ⊂[ ] | j |= 2 j∈[k1] j∈[k2]

k1 k2 Now En 1 1 1 2 6= 0 if and only if for some l1 ∈ [k1] and some j=1 Dj j=1 Dj Q Q 1 l1 2 l2 r kr l2 ∈ [k2], there exist disjoint sets {Ai }i=1, {Ai }i=1 such that {Dj }j=1 = r lr {Ai }i=1, r = 1, 2. If this is the case, then

k k 1 2 l1 l2 (n − l1m1 − l2m2)!((m1 − 1)!) (m2 − 1)!) (2.13) En 1D1 1D2 = . j j n! jY=1 jY=1

1 l 2 l The number of ways to construct disjoint, ordered sets {Ai }i=1, {Ai }i=1, 1 2 with the Ai each consisting of m1 elements from [n] and the Ai each n! consisting of m2 elements from [n], is l l . Given (m1!) 1 (m2!) 2 (n−l1m1−l2m2)! 1 l 2 l 1 k1 2 k2 {Ai }i=1, {Ai }i=1, the number of ways to choose the ordered sets {Dj }j=1, {Dj }j=1 so that {D1}k1 = {A1}l1 and {D2}k2 = {A2}l2 is equal to k1 k2 . j j=1 i i=1 j j=1 i i=1 l1 l2 We have  

(n − l m − l m )!((m − 1)!)l1 (m − 1)!)l2 n! 1 1 1 2 2 1 2 = . n! (m !)l1 (m !)l2 (n − l m − l m )! l1 l2 1 2 1 1 2 2 m1 m2

From these fact along with (2.12) and (2.13), we conclude that for n ≥ m1k1 + m2k2,

k k 2 1 1 k k 1 1 E C(n) k1 C(n) k2 1 2 T T , n m1 ) ( m2 ) = l l = k1 ( ) k2 ( ) m 1 m 2 l1 l2  m1 m2 lX2=1 lX1=1 1 2 proving (2.11). 

3. Proofs of Theorems 1 and 2

(n) Proof of Theorem 1. Since Em converges weakly to Em, it follows from the discussion in the first paragraph of section 2 that it suffices to show that

(n) k (3.1) lim En(Em ) = vk;m. n→∞ PERMUTATION FIXING SET, TOUCHARD POLYNOMIAL, COVER OF MULTISET 11

Let n ≥ km. For D ⊂ [n], let 1D(σ) equal 1 or 0 according to whether or not σ ∈ Sn induces an embedded permutation on D. Then we have

(n) (3.2) Em (ω)= 1D(ω), DX⊂[n] |D|=m and

k (n) k (3.3) En(Em ) = En 1Dj .

Dj ⊂[nX],|Dj|=m jY=1 j∈[k]

k There is a one-to-one correspondence between collections {Dj}j=1, satisfying

Dj ⊂ [n] and |Dj| = m, and collections {AI }I⊂[k] of disjoint sets satisfying

AI ⊂ [n] and satisfying

(3.4) lI = m, for all i ∈ [k], IX:i∈I where

(3.5) lI = |AI |, I ⊂ [k].

The correspondence is through the formula

(3.6) AI = ∩i∈I Di ∩ ∩i∈[k]−I ([n] − Di) , I ⊂ [k].   k Now j=1 1Dj (ω) = 1 if and only if for all I ⊂ [k], σ induces an embedded Q permutation on AI . Thus, we have

k (n − I⊂[k] lI )! I⊂[k] lI ! (3.7) E 1 = . n Dj P n! Q jY=1

Given the values lI = |AI |, I ⊂ [k], the number of ways to construct the n! disjoint sets {AI }I⊂[k] is . Using this with (3.3)-(3.7), (n−PI⊂[k] lI )! QI⊂[k] lI ! (n) k it follows that En(Em ) equals the number of solutions {lI }I⊂[k] to (3.4). To complete the proof, we will show that the number of solutions to (3.4) k is vk;m. Consider the set ∪j=1Dj ⊂ [n]. Label the elements of this set r r by {xi}i=1. Of course, m ≤ r ≤ km. Now construct the sets {Γi}i=1 by r Γi = {j : xi ∈ Dj}. By construction, the sets {Γi}i=1 form an m-cover of 12 ROSS G. PINSKY

[k] (of order r). There is a one-to-one correspondence between solutions to

(3.4) and m covers of [k]; indeed, lI = |{i ∈ [k]:Γi = I}|. 

xEm Proof of Theorem 2. We use (1.16) to calculate Vm(x)= Ee . We have m Z 1 Vm(x)= E exp(xEm)= E exp x j =  l  (l ,...,l ):Pm jl =m j j 1 m Xj=1 j Y=1  lj ≤Z 1 ,j∈[m] (3.8) j m m 1 1 − 1 uj ( )uj e j exp x . j uj!  lj  u , ,u j (l ,...,l ):Pm jl =m j 1≥0X··· m≥0 Y=1 1 m Xj=1 j Y=1  lj ≤uj ,j∈[m] Thus, to complete the proof, we only need to show that m uj (3.9) = γm(u),  lj  (l ,...,l ):Pm jl =m j 1 m Xj=1 j Y=1 lj ≤uj ,j∈[m] where m m j uj (3.10) γm(u) = [z ] (1 + z ) . jY=1 Expanding with the binomial formula, we have u m m j u (3.11) (1 + zj)uj = j zjlj .  lj  jY=1 jY=1 lXj =0  From (3.11) and (3.10), it follows that (3.9) holds. 

Acknowlegement. The author thanks Ron Holzman and Roy Meshulam for a discussion and references with regard to multisets, and Ron Holzman for the reference [7].

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Department of Mathematics, Technion—Israel Institute of Technology, Haifa, 32000, Israel E-mail address: [email protected] URL: http://www.math.technion.ac.il/~pinsky/