Some Connections Between Cycles and Permutations that Fix a Set and Touchard Polynomials and Covers of Multisets

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by  Ross G. Pinsky

SOME CONNECTIONS BETWEEN CYCLES AND
PERMUTATIONS THAT FIX A SET AND TOUCHARD
POLYNOMIALS AND COVERS OF MULTISETS
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0
ROSSG. PINSKY
2

n
a Abstract. Wepresentanewproofofafundamentalresultconcerning
J
cycles of random permutations which gives some intuition for the con-
0
2 nectionbetweenTouchardpolynomialsandthePoissondistribution. We

also introduce a rather novelpermutation statistic and study its distri-
]
R bution. This quantity, indexed by m, is the number of sets of size m
P
fixedbythepermutation. Thisleadstoanewandsimplerderivationof
.
h
the exponential generating function for the numberof covers of certain
t
a
multisets.
m
[

2
v
5 1. Introduction and Statement of Results
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8
In this paper, we present a new and simpler proof of a fundamentalresult
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0
concerningcycles ofrandompermutationswhichgives someintuition forthe
.
1
connection between TouchardpolynomialsandthePoisson distribution. We
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alsointroducearathernovelpermutationstatisticandstudyitsdistribution.
1
v: This quantity, indexed by m, is the number of sets of size m fixed by the
i
X permutation. This leads to a new and simpler derivation of the exponential
r
a generating function for the number of covers of certain multisets.
We begin by recalling some basic facts concerning Bell numbers and
Touchard polynomials, and their connection to Poisson distributions. The
facts noted below without proof can be found in many books on combina-
torics; for example, in [11], [12]. The Bell number B denotes the number
n
2000 Mathematics Subject Classification. 60C05, 05A05, 05A15.
Key words and phrases. permutations that fix a set, covers of multisets, Touchard
Polynomials, Dob´ınski’s formula, Bell numbers, cycles in random permutations, Ewens
sampling distribution .
1
2 ROSSG.PINSKY
of partitions of a set of n distinct elements. Elementary combinatorial rea-
soning yields the recursive formula
n
n
(1.1) B = B , n ≥ 0,
n+1 k
(cid:18)k(cid:19)
Xk=0
where B = 1. Let
0
∞
B
(1.2) E (x) = nxn,
B
n!
nX=0
denotetheexponentialmomentgeneratingfunctionof{B }∞ . Using(1.1)
n n=0
it is easy to show that E′ (x) = exE (x), from which it follows that
B B
(1.3) E (x) = eex−1.
B
A random variable X has the Poisson distribution Pois(λ), λ > 0, if
P(X = k) = e−λλk, k = 0,1,.... Let M (t) = EetX denote the moment
k! λ
generating function of X, and let µ = EXn denote the nth moment of
n;λ
(n)
X. Since M (0) = µ , we have
λ n;λ
∞
µ
(1.4) M (t)= n;λtn.
λ
n!
nX=0
However a direct calculation gives
∞ λk
(1.5) M(t) = etke−λ = eλ(et−1).
k!
Xk=0
From(1.2)-(1.5),itfollowsthatthenthmomentµ ofaPois(1)-distributed
n;1
random variable satisfies
(1.6) µ = B .
n;1 n
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
The Stirling number of the second kind n denotes the number of par-
k
(cid:8) (cid:9)
titions of a set of n distinct elements into k nonempty sets. Elementary
combinatorial reasoning yields the recursive formula
n+1 n n
= k + .
(cid:26) k (cid:27) (cid:26)k(cid:27) (cid:26)k−1(cid:27)
From this, it is not hard to derive the formula
n
n
(1.8) xn = (x) ,
j
(cid:26)j(cid:27)
Xj=0
where (x) = x(x−1)···(x−j +1) is the falling factorial, and one defines
j
(x) = 0 = 1. Now using (1.8) and the fact that (k) = 0 for j > k, we
0 0 j
(cid:8) (cid:9)
can write the nth moment µ of a Pois(λ)-distributed random variable as
n;λ
∞ λk ∞ λk n n
µ = (e−λ )kn = (e−λ ) (k) =
n;λ j
k! k! (cid:26)j(cid:27)
Xk=0 Xk=0 (cid:16)Xj=0 (cid:17)
∞ λk k∧n n n n ∞ (k)
(1.9) (e−λ ) (k) = e−λ λk j =
j
k! (cid:26)j(cid:27) (cid:26)j(cid:27) k!
Xk=0 (cid:16)Xj=0 (cid:17) Xj=0 Xk=j
n
n
λj.
(cid:26)j(cid:27)
Xj=0
The Touchard polynomials T (x), n ≥ 0, are defined by
n
n
n
T (x)= xj.
n
(cid:26)j(cid:27)
Xj=0
Thus, (1.9) gives the formula
(1.10) µ = T (λ).
n;λ n
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
Since T (1) = n n = B , the Dob´ınski formula (1.7) is contained in
n j=0 j n
P (cid:8) (cid:9)
(1.11).
4 ROSSG.PINSKY
Let S denote the set of permutations of [n] =: {1,...,n}. For σ ∈
n
(n)
S , let C (σ) denote the number of cycles of length m in σ. Let P
n m n
denote the uniform probability measure on S . We can now think of σ ∈ S
n n
(n)
as random, and of C as a random variable. Using generating function
m
techniques and/or inclusion-exclusion formulas, one can show that under
(n)
P , the distribution of the random variable C converges weakly to Z ,
n m 1
m
where Z has the Pois(1)-distribution; equivalently;
1 m
m
1
(1.12) nl→im∞Pn(Cm(n) = j) = e−m1 mjj!, j = 0,1....
Moregenerally,weconsidertheEwenssamplingdistributions,P ,θ > 0,
n;θ
on S as follows. Let N(n)(σ) denote the number of cycles in the per-
n
mutation σ ∈ S , and let s(n,k) = |{σ ∈ Sn : N(n)(σ) = k}| denote
n
the number of permutations in S with k cycles. It is known that the
n
polynomial n s(n,k)θk is equal to the rising factorial θ(n), defined by
k=1
θ(n) = θ(θ+P1)···(θ+n−1). For θ > 0, define the probability measure P
n;θ
on S by
n
θN(n)(σ)
P ({σ}) = .
n;θ θ(n)
Of course, P reduces to the uniform measure P . The following theorem
n;1 n
can be proven; see for example, [1], [10].
(n) (n) (n)
Theorem C. Under P , the random vector (C ,C ,··· ,C ,...) con-
n,θ 1 2 m
vergesweaklyto(Z ,Z ,··· ,Z ,···), wheretherandom variables{Z }∞
θ θ θ θ m=1
2 m m
are independent, and Z has the Pois(θ )-distribution:
θ m
m
(1.13) (C(n),C(n),···C(n),...) ⇒w (Z ,Z ,··· ,Z ···);
1 2 m θ θ θ
2 m
equivalently,
(1.14) nl→im∞Pn(C1(n) =j1,C2(n) = j2,...,Cm(n) = jm)= kYm=1e−kθ (kθjk)!jk,
for all m ≥ 1 and j ,...j ∈Z .
1 m +
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
thatis for theconnection between themoments 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
weredirected to any paperson this subject.) For σ ∈ S and A⊂ [n], define
n
(n)
σ(A) = {σ : j ∈ A}. If σ(A) = A, we will say that σ fixes A. Let E (σ)
j m
denote the number of sets of cardinality m that are fixed by σ. (Note that
(n) (n)
E (σ) = C (σ), the number of fixed points of σ.) A little thought reveals
1 1
that
m (n)
C (ω)
(1.15) E(n)(ω) = j .
m (cid:18) l (cid:19)
(l1,...,lm)X:Pmj=1jlj=m jY=1 j
lj≤Cj(n),j∈[m]
Forexample,ifσ ∈S iswrittenincyclenotationasσ = (379)(24)(16)(5)(8),
9
then E(9)(ω) = 5, with the sets A ⊂ [9] for which |A| = 4 and σ(A) = A
4
being {3,5,7,9},{3,7,8,9},{1,2,4,6},{2,4,5,8},{1,5,6,8}.
We consider the uniformmeasure P = P on S . From Theorem C and
n n;1 n
(n)
(1.15) it follows that the random variable E under P converges weakly
m n
as n → ∞ to the random variable
m Z
1
(1.16) Em =: j ,
(cid:18) l (cid:19)
(l1,...,lm)X:Pmj=1jlj=m jY=1 j
lj≤Z1,j∈[m]
j
where {Z }m are independent and Z has the Pois(1)-distribution.
1 j=1 1 j
j j
Remark. Note that E = Z , E = Z1 +Z , E = Z +Z Z + Z1 .
1 1 2 2 1 3 1 1 1 3
2 3 2
(cid:0) (cid:1) (cid:0) (cid:1)
For k,m ∈ N, consider the multiset consisting of m copies of the set [k].
A collection {Γ }r such that each Γ is a nonempty subset of [k], and such
l l=1 l
that each j ∈ [k] appears in exactly m from among the r sets {Γ }r , is
l l=1
called an m-cover of [k] of order r. Denote the total number of m covers of
[k], regardless of order, by v . Note that when m = 1, we have v = B ,
k;m k;1 k
thekthBellnumber,denotingthenumberofpartitionsofasetofkelements.
Also, it’s very easy to see that v =1 and v = m+1.
1;m 2,m
6 ROSSG.PINSKY
(n)
By calculating directly the moments of E , we will prove the following
m
theorem.
Theorem 1. For m,k ∈ N,
EEk = v .
m k;m
In particular, EE = 1 and EE2 = m+1; thus, Var(E ) =m.
m m m
Remark. It is naturalto suspectthat E converges weakly to 0 as m → ∞;
m
that is, lim P(E ≥ 1) = 0. This is in fact a hard problem. In [8] it was
n→∞ m
shown that Pn(Em(n)) ≤ Am−1010, for 1 ≤ m ≤ n2 and n ≥ 2. Thus indeed,
E converges weakly to 0 as m → ∞. A lower bound on P(E ≥ 1) of the
m m
form Alogm was obtained in [5]. These results were dramatically improved
m
in [9] where it was shown that P(E ≥ 1) = m−δ+o(1) as m → ∞, where
m
δ = 1− 1+loglog2 ≈ 0.08607. And very recently, in [7], this latter bound has
log2
been refined to A1m−δ(1+logm)−23 ≤ P(Em ≥ 1) ≤ A2m−δ(1+logm)−32.
Let
∞
v
V (x) = k;mxk
m
k!
Xk=1
denote the exponential generating function of the sequence {v }∞ . Of
k;m k=1
course, by Theorem 1 V is also the moment generating function of the
m
random variable E : V (x) = EexEm. Using (1.16) and Theorem 1, we will
m m
give an almost immediate proof of the following representation theorem for
V (x). We use the notation [zm]P(z) = a , where P(z) = ∞ a zm.
m m m=0 m
P
Theorem 2.
(1.17) Vm(x) = EexEm = e−Pmj=1 1j m j−uj exγm(u),
u !
u1,..X.,um≥0(cid:0)jY=1 j (cid:1)
where
m
γ (u) = [zm] (1+zj)uj.
m
jY=1
PERMUTATION FIXING SET, TOUCHARD POLYNOMIAL, COVER OF MULTISET 7
Remark. When m = 2,3, the above formula reduces to
V2(x) = e−23e12ex ∞ e(r2)x;
r!
Xr=0
V3(x) = e−161ee3x ∞ e(r3)x+21erx.
r!
Xr=0
Theformulaform = 2wasprovedbyComtets [3]andtheformulaform = 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 v from the exponential generating function
k;m
V (x) can be done in no more than ckmlogk arithmetic operations.
m
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
Ifasequenceofnonnegativerandomvariables{X }∞ satisfiessup EX <
n n=1 n≥1 n
∞,thenthesequenceistight, thatis,pre-compactwithrespecttoweakcon-
vergence. Let X be distributed as one of the accumulation points. If for
some k ∈ N, lim EXk exists and equals µ , and sup EXk+1 < ∞,
n→∞ n k n≥1 n
then the {Xk}∞ are uniformly integrable, and thus EXk = µ . Thus, if
n n=1 k
(2.1) µ =: lim EXk exists for all k ∈ N,
k n
n→∞
then EXk = µ , for all k. The Stieltjes moment theorem states that if
k
1
µk
(2.2) sup k <∞,
k
k≥1
then the sequence {µ }∞ uniquely characterizes the distribution [6]. We
k k=1
conclude then that if a sequence of nonnegative random variables {X }∞
n n=1
satisfies (2.1) and(2.2), then thesequence is weakly convergent to a random
variable X satisfying EXk = µ .
k
8 ROSSG.PINSKY
An extremely crude argument shows that the Bell numbers satisfy B ≤
k
kk; thus
1
Bk
(2.3) sup k < ∞.
k
k≥1
By (1.10), thekth momentµ ofthePois(θ )-distributedrandomvariable
k;θ m
m
Z is equal to T (θ ). Now T (θ ) is bounded from above by T (θ), for all
θ k m k m k
m
m ≥ 1, and T (θ) ≤ B max(1,θk). Thus, in light of (2.3) and the previous
k k
paragraph, if we prove that
θ
(2.4) lim E (C(n))k = T ( ), k,m ∈ N,
n→∞ n;θ m k m
where E denotes the expectation with respect to P , then we will have
n;θ n;θ
proved that C(n) under P converges weakly to Z , for all m ∈ N. And if
m n;θ θ
m
we then prove that
m m
θ
(2.5) lim E (C(n))kj = T ( ), m ≥ 2,k ∈ N,j = 1,...,m,
n→∞ n;θ j kj 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, P = P . Once we have this, the case of general θ will
n n;1
follow after a short explanation. Assume that n ≥ mk. For D ⊂ [n] with
|D| = m, let 1 (σ) be equal to 1 or 0 according to whether or not σ ∈ S
D n
possesses an m-cycle consisting of the elements of D. Then we have
(2.6) C(n)(σ) = 1 (σ),
m D
X
D⊂[n]
|D|=m
and
k
(2.7) E (C(n))k = E 1 .
n m n Dj
Dj⊂[nX],|Dj|=m jY=1
j∈[k]
Now E k 1 6= 0 if and only if for some l ∈ [k], there exist disjoint
n j=1 Dj
sets {A }l Qsuch that {D }k = {A }l . If this is the case, then
i i=1 j j=1 i i=1
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
disjoint, ordered sets {A }l , each of which consists of m elements from
i i=1
[n], is n! . Given the {A }l , the number of ways to choose the
(m!)l(n−lm)! i i=1
sets {D }k so that {D }k = {A }l is equal to the Stirling number
j j=1 j j=1 i i=1
k , the number of ways to partition a set of size k into l nonempty parts.
l
(cid:8) (cid:9)
From these facts along with (2.7) and (2.8), we conclude that for n ≥ mk,
k (n−lm)!((m−1)!)l n! k
E (C(n))k = =
n m n! (m!)l(n−lm)! (cid:26)l(cid:27)
Xl=1(cid:0) (cid:1)(cid:0) (cid:1)
(2.9)
k
1 k 1
= T ( ),
ml(cid:26)l(cid:27) k m
Xl=1
proving (2.4) in the case θ = 1.
For thecaseof general θ,wenotethattheonly changethatmustbemade
in the above proof is in (2.8). Recalling that s(n,k) denotes the number of
permutations in S with k cycles, we have
n
(2.10)
k ((m−1)!)l n−mls(n−lm,k)θk+l θ(n−ml)((m−1)!)l
E 1 = k=1 = θl ∼
n;θ Dj P θ(n) θ(n)
jY=1
(n−ml)!((m−1)!)l k
θl = θlE 1 , as n → ∞.
n! n Dj
jY=1
Thus, instead of (2.9), we have
k
θ k θ
E (C(n))k ∼ ( )l = T ( ), as n → ∞.
n m m (cid:26)l(cid:27) k m
Xl=1
Wenowturnto(2.5). Themethodofproofissimplythenaturalextension
of the one used to prove (2.4); thus, since the notation is cumbersome we
will suffice with illustrating the method by proving that
1 1
(2.11) lim E (C(n))k1(C(n))k2 = T ( )T ( ).
n→∞ n m1 m2 k1 m1 k2 m2
10 ROSSG.PINSKY
Let n ≥ m k +m k . By (2.6), we have
1 1 2 2
k1 k2
(2.12) E C(n))k1(C(n))k2 = E 1 1 .
n m1 m2 n Dj1 Dj2
Dj1⊂[nX],|Dj1|=m1Dj2⊂[nX],|Dj2|=m2 jY=1 jY=1
j∈[k1] j∈[k2]
Now E k1 1 k2 1 6= 0 if and only if for some l ∈ [k ] and some
n j=1 D1 j=1 D2 1 1
j j
l ∈ [k ], tQhere existQdisjoint sets {A1}l1 , {A2}l2 such that {Dr}kr =
2 2 i i=1 i i=1 j j=1
{Ar}lr , r = 1,2. If this is the case, then
i i=1
k1 k2 (n−l m −l m )!((m −1)!)l1(m −1)!)l2
1 1 2 2 1 2
(2.13) E 1 1 = .
n Dj1 Dj2 n!
jY=1 jY=1
The number of ways to construct disjoint, ordered sets {A1}l ,{A2}l ,
i i=1 i i=1
with the A1 each consisting of m elements from [n] and the A2 each
i 1 i
consisting of m elements from [n], is n! . Given
2 (m1!)l1(m2!)l2(n−l1m1−l2m2)!
{A1}l ,{A2}l ,thenumberofwaystochoosetheorderedsets{D1}k1 ,{D2}k2
i i=1 i i=1 j j=1 j 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
(cid:8) (cid:9)(cid:8) (cid:9)
We have
(n−l m −l m )!((m −1)!)l1(m −1)!)l2 n! 1
1 1 2 2 1 2
= .
n! (m1!)l1(m2!)l2(n−l1m1−l2m2)! ml1ml2
1 2
From these fact along with (2.12) and (2.13), we conclude that for n ≥
m k +m k ,
1 1 2 2
k2 k1 1 k k 1 1
E C(n))k1(C(n))k2 = 1 2 = T ( )T ( ),
n m1 m2 lX2=1lX1=1ml11ml22(cid:26)l1(cid:27)(cid:26)l2(cid:27) k1 m1 k2 m2
proving (2.11). (cid:3)
3. Proofs of Theorems 1 and 2
(n)
Proof of Theorem 1. Since E converges weakly to E , it follows from the
m m
discussion in the first paragraph of section 2 that it suffices to show that
(3.1) lim E (E(n))k = v .
n m k;m
n→∞