Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling...

Touchard Polynomials, Stirling Numbers andRandom Permutations

Ross Pinsky

Department of Mathematics, Technion32000 Haifa, ISRAEL

September 3, 2018

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Notation:

[n] = {1, 2, · · · , n}

Sn = permutations of [n]

|Sn| = n!

S6 3 σ1 =

(1 2 3 4 5 62 4 6 1 5 3

)= (124)(36)(5),

S6 3 σ2 =

(1 2 3 4 5 63 4 6 5 1 2

)= (136245) = (624513)

σ2 is a 6-cycle in S6. There are 5! different 6-cycles in S6.

There are (n − 1)! different n-cycles in Sn.

Notation:

[n] = {1, 2, · · · , n}

|Sn| = n!

S6 3 σ1 =

(1 2 3 4 5 62 4 6 1 5 3

)= (124)(36)(5),

S6 3 σ2 =

(1 2 3 4 5 63 4 6 5 1 2

)= (136245) = (624513)

Notation:

[n] = {1, 2, · · · , n}

|Sn| = n!

S6 3 σ1 =

(1 2 3 4 5 62 4 6 1 5 3

)= (124)(36)(5),

S6 3 σ2 =

(1 2 3 4 5 63 4 6 5 1 2

)= (136245) = (624513)

Notation:

[n] = {1, 2, · · · , n}

|Sn| = n!

S6 3 σ1 =

(1 2 3 4 5 62 4 6 1 5 3

)= (124)(36)(5),

S6 3 σ2 =

(1 2 3 4 5 63 4 6 5 1 2

)= (136245) = (624513)

Notation:

[n] = {1, 2, · · · , n}

|Sn| = n!

S6 3 σ1 =

(1 2 3 4 5 62 4 6 1 5 3

)= (124)(36)(5),

S6 3 σ2 =

(1 2 3 4 5 63 4 6 5 1 2

)= (136245) = (624513)

Notation:

[n] = {1, 2, · · · , n}

|Sn| = n!

S6 3 σ1 =

(1 2 3 4 5 62 4 6 1 5 3

)= (124)(36)(5),

S6 3 σ2 =

(1 2 3 4 5 63 4 6 5 1 2

)= (136245) = (624513)

Rising Factorials

x (n) := x(x +1) · · · (x +n−1)

n∑k=1

ankxk), x ∈ R, n ≥ 1. (1)

Unsigned Stirling Numbers of the First Kind:|s(n, k)| :=the number of permutations in Sn with exactly k cycles, 1 ≤ k ≤ nRecurrence relation

|s(n, n)| = 1, |s(n, 1)| = (n − 1)!, n ≥ 1;

|s(n + 1, k)| = n|s(n, k)|+ |s(n, k − 1)|, 2 ≤ k ≤ n.(2)

It is easy to check that the coefficients {ank} in (1) also satisfy(2); thus, ank = |s(n, k)|. Therefore

x (n) =n∑

|s(n, k)|xk . (3)

Rising Factorials

x (n) := x(x +1) · · · (x +n−1)(≡

n∑k=1

ankxk), x ∈ R, n ≥ 1. (1)

|s(n, n)| = 1, |s(n, 1)| = (n − 1)!, n ≥ 1;

|s(n + 1, k)| = n|s(n, k)|+ |s(n, k − 1)|, 2 ≤ k ≤ n.(2)

x (n) =n∑

|s(n, k)|xk . (3)

Rising Factorials

x (n) := x(x +1) · · · (x +n−1)(≡

n∑k=1

ankxk), x ∈ R, n ≥ 1. (1)

Unsigned Stirling Numbers of the First Kind:|s(n, k)| :=the number of permutations in Sn with exactly k cycles, 1 ≤ k ≤ n

Recurrence relation

|s(n, n)| = 1, |s(n, 1)| = (n − 1)!, n ≥ 1;

|s(n + 1, k)| = n|s(n, k)|+ |s(n, k − 1)|, 2 ≤ k ≤ n.(2)

x (n) =n∑

|s(n, k)|xk . (3)

Rising Factorials

x (n) := x(x +1) · · · (x +n−1)(≡

n∑k=1

ankxk), x ∈ R, n ≥ 1. (1)

|s(n, n)| = 1, |s(n, 1)| = (n − 1)!, n ≥ 1;

|s(n + 1, k)| = n|s(n, k)|+ |s(n, k − 1)|, 2 ≤ k ≤ n.(2)

x (n) =n∑

|s(n, k)|xk . (3)

Rising Factorials

x (n) := x(x +1) · · · (x +n−1)(≡

n∑k=1

ankxk), x ∈ R, n ≥ 1. (1)

|s(n, n)| = 1, |s(n, 1)| = (n − 1)!, n ≥ 1;

|s(n + 1, k)| = n|s(n, k)|+ |s(n, k − 1)|, 2 ≤ k ≤ n.(2)

It is easy to check that the coefficients {ank} in (1) also satisfy(2); thus, ank = |s(n, k)|.

Therefore

x (n) =n∑

|s(n, k)|xk . (3)

Rising Factorials

x (n) := x(x +1) · · · (x +n−1)(≡

n∑k=1

ankxk), x ∈ R, n ≥ 1. (1)

|s(n, n)| = 1, |s(n, 1)| = (n − 1)!, n ≥ 1;

|s(n + 1, k)| = n|s(n, k)|+ |s(n, k − 1)|, 2 ≤ k ≤ n.(2)

x (n) =n∑

|s(n, k)|xk . (3)

Falling Factorials

(x)n := x(x − 1) · · · (x − n + 1), x ∈ R, n ≥ 1. (4)

Substituting −x for x in (4), we have

(−x)n = −x(−x−1) · · · (−x−n+1) = (−1)nx(x+1) · · · (x+n−1)= (−1)nx (n).Substituting above −x for x , we have

(x)n = (−1)n(−x)(n)from (3)

= (−1)nn∑

|s(n, k)|(−x)k =

n∑k=1

(−1)n−k |s(n, k)|xk .(5)

Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called theStirling Numbers of the First Kind. From (5) we have

(x)n =n∑

s(n, k)xk . (6)

Falling Factorials

(x)n := x(x − 1) · · · (x − n + 1), x ∈ R, n ≥ 1. (4)

(−x)n = −x(−x−1) · · · (−x−n+1) =

(−1)nx(x+1) · · · (x+n−1)= (−1)nx (n).Substituting above −x for x , we have

(x)n = (−1)n(−x)(n)from (3)

= (−1)nn∑

|s(n, k)|(−x)k =

n∑k=1

(−1)n−k |s(n, k)|xk .(5)

(x)n =n∑

s(n, k)xk . (6)

Falling Factorials

(x)n := x(x − 1) · · · (x − n + 1), x ∈ R, n ≥ 1. (4)

(−x)n = −x(−x−1) · · · (−x−n+1) = (−1)nx(x+1) · · · (x+n−1)

= (−1)nx (n).Substituting above −x for x , we have

(x)n = (−1)n(−x)(n)from (3)

= (−1)nn∑

|s(n, k)|(−x)k =

n∑k=1

(−1)n−k |s(n, k)|xk .(5)

(x)n =n∑

s(n, k)xk . (6)

Falling Factorials

(x)n := x(x − 1) · · · (x − n + 1), x ∈ R, n ≥ 1. (4)

(−x)n = −x(−x−1) · · · (−x−n+1) = (−1)nx(x+1) · · · (x+n−1)= (−1)nx (n).

Substituting above −x for x , we have

(x)n = (−1)n(−x)(n)from (3)

= (−1)nn∑

|s(n, k)|(−x)k =

n∑k=1

(−1)n−k |s(n, k)|xk .(5)

(x)n =n∑

s(n, k)xk . (6)

Falling Factorials

(x)n := x(x − 1) · · · (x − n + 1), x ∈ R, n ≥ 1. (4)

(x)n = (−1)n(−x)(n)from (3)

= (−1)nn∑

|s(n, k)|(−x)k

n∑k=1

(−1)n−k |s(n, k)|xk .(5)

(x)n =n∑

s(n, k)xk . (6)

Falling Factorials

(x)n := x(x − 1) · · · (x − n + 1), x ∈ R, n ≥ 1. (4)

(x)n = (−1)n(−x)(n)from (3)

= (−1)nn∑

|s(n, k)|(−x)k =

n∑k=1

(−1)n−k |s(n, k)|xk .(5)

(x)n =n∑

s(n, k)xk . (6)

Falling Factorials

(x)n := x(x − 1) · · · (x − n + 1), x ∈ R, n ≥ 1. (4)

(x)n = (−1)n(−x)(n)from (3)

= (−1)nn∑

|s(n, k)|(−x)k =

n∑k=1

(−1)n−k |s(n, k)|xk .(5)

Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called theStirling Numbers of the First Kind.

From (5) we have

(x)n =n∑

s(n, k)xk . (6)

Falling Factorials

(x)n := x(x − 1) · · · (x − n + 1), x ∈ R, n ≥ 1. (4)

(x)n = (−1)n(−x)(n)from (3)

= (−1)nn∑

|s(n, k)|(−x)k =

n∑k=1

(−1)n−k |s(n, k)|xk .(5)

(x)n =n∑

s(n, k)xk . (6)

Stirling Numbers of the Second Kind

S(n, k) := number of ways to partition the set [n] into k nonemptysets.

Example: S(4, 3) = 5:{1, 2}, {3}, {4}{1, 3}, {2}, {4}{1, 4}, {2}, {3}{2, 3}, {1}, {4}{2, 4}, {1}, {3}

S(n, k) := number of ways to partition the set [n] into k nonemptysets.Example: S(4, 3) = 5:

{1, 2}, {3}, {4}{1, 3}, {2}, {4}{1, 4}, {2}, {3}{2, 3}, {1}, {4}{2, 4}, {1}, {3}

S(n, k) := number of ways to partition the set [n] into k nonemptysets.Example: S(4, 3) = 5:{1, 2}, {3}, {4}{1, 3}, {2}, {4}{1, 4}, {2}, {3}{2, 3}, {1}, {4}{2, 4}, {1}, {3}

S(n, k) := number of ways to partition the set [n] into k nonemptysets.

Theorem.

xn =n∑

S(n, k)(x)k , (7)

Proof. Consider functions f : [n]→ X , where |X | = x ∈ N.How many such functions are there?The answer by direct count: xn.

An alternative indirect count:For k = 1, 2, · · · , n, let ck denote the number of such functionswith |Im(f )| = k . (If x < n, then ck = 0 for x < k ≤ n.)So the answer by this indirect count is

∑nk=1 ck .

Thus, xn =∑n

k=1 ck . To complete the proof we now show thatck = S(n, k)(x)k .

S(n, k) := number of ways to partition the set [n] into k nonemptysets.Theorem.

xn =n∑

S(n, k)(x)k , (7)

∑nk=1 ck .

Thus, xn =∑n

xn =n∑

S(n, k)(x)k , (7)

Proof. Consider functions f : [n]→ X , where |X | = x ∈ N.

How many such functions are there?The answer by direct count: xn.

∑nk=1 ck .

Thus, xn =∑n

xn =n∑

S(n, k)(x)k , (7)

Proof. Consider functions f : [n]→ X , where |X | = x ∈ N.How many such functions are there?

The answer by direct count: xn.

∑nk=1 ck .

Thus, xn =∑n

xn =n∑

S(n, k)(x)k , (7)

∑nk=1 ck .

Thus, xn =∑n

xn =n∑

S(n, k)(x)k , (7)

An alternative indirect count:

For k = 1, 2, · · · , n, let ck denote the number of such functionswith |Im(f )| = k . (If x < n, then ck = 0 for x < k ≤ n.)So the answer by this indirect count is

∑nk=1 ck .

Thus, xn =∑n

xn =n∑

S(n, k)(x)k , (7)

An alternative indirect count:For k = 1, 2, · · · , n, let ck denote the number of such functionswith |Im(f )| = k . (If x < n, then ck = 0 for x < k ≤ n.)

So the answer by this indirect count is∑n

k=1 ck .Thus, xn =

∑nk=1 ck . To complete the proof we now show that

ck = S(n, k)(x)k .

xn =n∑

S(n, k)(x)k , (7)

∑nk=1 ck .

Thus, xn =∑n

xn =n∑

S(n, k)(x)k , (7)

∑nk=1 ck .

Thus, xn =∑n

k=1 ck .

To complete the proof we now show thatck = S(n, k)(x)k .

xn =n∑

S(n, k)(x)k , (7)

∑nk=1 ck .

Thus, xn =∑n

Consider functions f : [n]→ X , where |X | = x ∈ N. Let ckdenote the number of such functions with |Im(f )| = k.Proof that ck = S(n, k)(x)k :

For f with |Im(f )| = k, let Im(f ) = {x1, · · · , xk}. Then

f −1({x1}), f −1({x2}), · · · , f −1({xk}) is a partition of [n] into knon-empty sets, which we will denote by {B1, · · · ,Bk}.

Now:1. There are S(n, k) ways to choose the sets {B1, · · · ,Bk};

2. There are(xk

)was to choose the {x1, · · · , xk};

3. There are k! ways to make the correspondence betweenf −1({x1}), f −1({x2}), · · · , f −1({xk}) and {B1, · · · ,Bk}.

Thus ck = S(n, k)×(xk

)× k! = S(n, k) x!

(x−k)! = S(n, k)(x)k . �

For f with |Im(f )| = k, let Im(f ) = {x1, · · · , xk}.

2. There are(xk

)× k! = S(n, k) x!

(x−k)! = S(n, k)(x)k . �

2. There are(xk

)× k! = S(n, k) x!

(x−k)! = S(n, k)(x)k . �

2. There are(xk

)× k! = S(n, k) x!

(x−k)! = S(n, k)(x)k . �

2. There are(xk

)× k! = S(n, k) x!

(x−k)! = S(n, k)(x)k . �

2. There are(xk

)× k! = S(n, k) x!

(x−k)! = S(n, k)(x)k . �

2. There are(xk

)× k!

= S(n, k) x!(x−k)! = S(n, k)(x)k . �

2. There are(xk

)× k! = S(n, k) x!

(x−k)!

= S(n, k)(x)k . �

2. There are(xk

)× k! = S(n, k) x!

(x−k)! = S(n, k)(x)k . �

xn =∑n

k=1 S(n, k)(x)k , n ≥ 1

(x)n =∑n

k=1 s(n, k)xk , n ≥ 1

S(n, k) (Stirling number of the second kind) is the number ofways to partition [n] into k nonempty subsets

|s(n, k)| (unsigned Stirling number of the first kind) is thenumber of permutations of [n] which have k cycles

s(n, k) = (−1)n−k |s(n, k)| (Stirling number of the first kind)

Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.

Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id.

That is,∑∞

j=1 S(n, j)s(j ,m) =

{1, if n = m;

0, if n 6= m.

xn =∑n

k=1 S(n, k)(x)k , n ≥ 1

(x)n =∑n

k=1 s(n, k)xk , n ≥ 1

That is,∑∞

j=1 S(n, j)s(j ,m) =

{1, if n = m;

0, if n 6= m.

xn =∑n

k=1 S(n, k)(x)k , n ≥ 1

(x)n =∑n

k=1 s(n, k)xk , n ≥ 1

That is,∑∞

j=1 S(n, j)s(j ,m) =

{1, if n = m;

0, if n 6= m.

xn =∑n

k=1 S(n, k)(x)k , n ≥ 1

(x)n =∑n

k=1 s(n, k)xk , n ≥ 1

Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).

Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id.

That is,∑∞

j=1 S(n, j)s(j ,m) =

{1, if n = m;

0, if n 6= m.

xn =∑n

k=1 S(n, k)(x)k , n ≥ 1

(x)n =∑n

k=1 s(n, k)xk , n ≥ 1

Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).

Corollary. Ss = sS = Id.

That is,∑∞

j=1 S(n, j)s(j ,m) =

{1, if n = m;

0, if n 6= m.

xn =∑n

k=1 S(n, k)(x)k , n ≥ 1

(x)n =∑n

k=1 s(n, k)xk , n ≥ 1

That is,∑∞

j=1 S(n, j)s(j ,m) =

{1, if n = m;

0, if n 6= m.

xn =∑n

k=1 S(n, k)(x)k , n ≥ 1

(x)n =∑n

k=1 s(n, k)xk , n ≥ 1

That is,∑∞

j=1 S(n, j)s(j ,m) =

{1, if n = m;

0, if n 6= m.

xn =∑n

k=1 S(n, k)(x)k , n ≥ 1 (*)(x)n =

∑nk=1 s(n, k)xk , n ≥ 1 (**)

Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id.Proof.

Consider the real vector space V of polynomials of theform

∑nk=1 ckx

k , n ≥ 1, ck ∈ R. .

Of course,B1 := {x , x2, x3, · · · } is a basis for V .

By (*), B2 := {(x)1, (x)2, (x)3, · · · } is also a basis for V .

By (*) and (**), the matrices S and s transform between thetwo basis, and thus Ss = sS = I . �

xn =∑n

k=1 S(n, k)(x)k , n ≥ 1 (*)(x)n =

∑nk=1 s(n, k)xk , n ≥ 1 (**)

Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id.Proof. Consider the real vector space V of polynomials of theform

∑nk=1 ckx

k , n ≥ 1, ck ∈ R.

xn =∑n

k=1 S(n, k)(x)k , n ≥ 1 (*)(x)n =

∑nk=1 s(n, k)xk , n ≥ 1 (**)

∑nk=1 ckx

k , n ≥ 1, ck ∈ R. .

xn =∑n

k=1 S(n, k)(x)k , n ≥ 1 (*)(x)n =

∑nk=1 s(n, k)xk , n ≥ 1 (**)

∑nk=1 ckx

k , n ≥ 1, ck ∈ R. .

xn =∑n

k=1 S(n, k)(x)k , n ≥ 1 (*)(x)n =

∑nk=1 s(n, k)xk , n ≥ 1 (**)

∑nk=1 ckx

k , n ≥ 1, ck ∈ R. .

Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id

That is,∑∞

j=1 S(n, j)s(j ,m) =

{1, if n = m;

0, if n 6= m.

Example: n = 8,m = 4:0 =

∑∞j=1 S(8, j)s(j , 4) =

S(8, 4)−S(8, 5)|s(5, 4)|+S(8, 6)|s(6, 4)|−S(8, 7)|s(7, 4)|+|s(8, 4)|.

Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id

That is,∑∞

j=1 S(n, j)s(j ,m) =

{1, if n = m;

0, if n 6= m.Example: n = 8,m = 4:0 =

∑∞j=1 S(8, j)s(j , 4) =

S(8, 4)−S(8, 5)|s(5, 4)|+S(8, 6)|s(6, 4)|−S(8, 7)|s(7, 4)|+|s(8, 4)|.Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Recalling the Poisson distribution

Let X be a random variable with the distribution Poiss(λ), λ > 0:

P(X = j) = e−λ λj

j! , j = 0, 1, 2, · · · .

nth moment: µn;λ := EX n

Recalling the Poisson distribution

j! , j = 0, 1, 2, · · · .

nth moment: µn;λ := EX n

j! , j = 1, 2, · · · .nth moment: µn;λ := EX n

µn;λ =∞∑j=0

(e−λ

(7)=∞∑j=0

(e−λ

) n∑k=1

S(n, k)(j)k

=∞∑j=0

(e−λ

)min(j ,n)∑k=1

S(n, k)(j)k(because (j)k = 0, k > j)

= e−λn∑

S(n, k)∞∑j=k

j!(j)k .

But∑∞

j=kλj

j! (j)k =∑∞

j=kλj

(j−k)! = λk∑∞

j=kλj−k

(j−k)! = λk∑∞

l=0λl

λkeλ. Thus,

µn;λ =n∑

S(n, k)λk .

µn;λ =∞∑j=0

(e−λ

(7)=∞∑j=0

(e−λ

) n∑k=1

S(n, k)(j)k =

=∞∑j=0

(e−λ

)min(j ,n)∑k=1

S(n, k)(j)k

(because (j)k = 0, k > j)

= e−λn∑

S(n, k)∞∑j=k

j!(j)k .

But∑∞

j=kλj

j! (j)k =∑∞

j=kλj

(j−k)! = λk∑∞

j=kλj−k

(j−k)! = λk∑∞

l=0λl

λkeλ. Thus,

µn;λ =n∑

S(n, k)λk .

µn;λ =∞∑j=0

(e−λ

(7)=∞∑j=0

(e−λ

) n∑k=1

S(n, k)(j)k =

=∞∑j=0

(e−λ

)min(j ,n)∑k=1

= e−λn∑

S(n, k)∞∑j=k

j!(j)k .

But∑∞

j=kλj

j! (j)k =∑∞

j=kλj

(j−k)! = λk∑∞

j=kλj−k

(j−k)! = λk∑∞

l=0λl

λkeλ. Thus,

µn;λ =n∑

S(n, k)λk .

µn;λ =∞∑j=0

(e−λ

(7)=∞∑j=0

(e−λ

) n∑k=1

S(n, k)(j)k =

=∞∑j=0

(e−λ

)min(j ,n)∑k=1

= e−λn∑

S(n, k)∞∑j=k

j!(j)k .

But∑∞

j=kλj

j! (j)k =∑∞

j=kλj

(j−k)! = λk∑∞

j=kλj−k

(j−k)! = λk∑∞

l=0λl

λkeλ. Thus,

µn;λ =n∑

S(n, k)λk .

µn;λ =∞∑j=0

(e−λ

(7)=∞∑j=0

(e−λ

) n∑k=1

S(n, k)(j)k =

=∞∑j=0

(e−λ

)min(j ,n)∑k=1

= e−λn∑

S(n, k)∞∑j=k

j!(j)k .

But∑∞

j=kλj

j! (j)k =∑∞

j=kλj

(j−k)!

= λk∑∞

j=kλj−k

(j−k)! = λk∑∞

l=0λl

λkeλ. Thus,

µn;λ =n∑

S(n, k)λk .

µn;λ =∞∑j=0

(e−λ

(7)=∞∑j=0

(e−λ

) n∑k=1

S(n, k)(j)k =

=∞∑j=0

(e−λ

)min(j ,n)∑k=1

= e−λn∑

S(n, k)∞∑j=k

j!(j)k .

But∑∞

j=kλj

j! (j)k =∑∞

j=kλj

(j−k)! = λk∑∞

j=kλj−k

(j−k)!

= λk∑∞

l=0λl

λkeλ. Thus,

µn;λ =n∑

S(n, k)λk .

µn;λ =∞∑j=0

(e−λ

(7)=∞∑j=0

(e−λ

) n∑k=1

S(n, k)(j)k =

=∞∑j=0

(e−λ

)min(j ,n)∑k=1

= e−λn∑

S(n, k)∞∑j=k

j!(j)k .

But∑∞

j=kλj

j! (j)k =∑∞

j=kλj

(j−k)! = λk∑∞

j=kλj−k

(j−k)! = λk∑∞

l=0λl

λkeλ. Thus,

µn;λ =n∑

S(n, k)λk .

µn;λ =∞∑j=0

(e−λ

(7)=∞∑j=0

(e−λ

) n∑k=1

S(n, k)(j)k =

=∞∑j=0

(e−λ

)min(j ,n)∑k=1

= e−λn∑

S(n, k)∞∑j=k

j!(j)k .

But∑∞

j=kλj

j! (j)k =∑∞

j=kλj

(j−k)! = λk∑∞

j=kλj−k

(j−k)! = λk∑∞

l=0λl

λkeλ.

µn;λ =n∑

S(n, k)λk .

µn;λ =∞∑j=0

(e−λ

(7)=∞∑j=0

(e−λ

) n∑k=1

S(n, k)(j)k =

=∞∑j=0

(e−λ

)min(j ,n)∑k=1

= e−λn∑

S(n, k)∞∑j=k

j!(j)k .

But∑∞

j=kλj

j! (j)k =∑∞

j=kλj

(j−k)! = λk∑∞

j=kλj−k

(j−k)! = λk∑∞

l=0λl

λkeλ. Thus,

µn;λ =n∑

S(n, k)λk .

j! , j = 1, 2, · · · .nth moment: µn;λ := EX n =

∑∞j=0

(e−λ λj

µn;λ =∑n

k=1 S(n, k)λk

Touchard Polynomial: Tn(x) :=∑n

k=1 S(n, k)xk

µn;λ = Tn(λ)(Compare Tn with (6).)

∞∑j=0

(e−λ

)jn = EX n = µn;λ = Tn(λ) :=

n∑k=1

S(n, k)λk

When λ = 1:

∞∑j=0

j!= EX n = µn;1 = Tn(1) :=

n∑k=1

S(n, k) := Bn,

where Bn is the nth Bell number:Bn = the total number of partitions of [n].

∑∞j=0

(e−λ λj

µn;λ =∑n

k=1 S(n, k)λk

k=1 S(n, k)xk

∞∑j=0

(e−λ

)jn = EX n = µn;λ = Tn(λ) :=

n∑k=1

S(n, k)λk

When λ = 1:

∞∑j=0

j!= EX n = µn;1 = Tn(1) :=

n∑k=1

S(n, k) := Bn,

∑∞j=0

(e−λ λj

µn;λ =∑n

k=1 S(n, k)λk

k=1 S(n, k)xk

µn;λ = Tn(λ)

(Compare Tn with (6).)

∞∑j=0

(e−λ

)jn = EX n = µn;λ = Tn(λ) :=

n∑k=1

S(n, k)λk

When λ = 1:

∞∑j=0

j!= EX n = µn;1 = Tn(1) :=

n∑k=1

S(n, k) := Bn,

∑∞j=0

(e−λ λj

µn;λ =∑n

k=1 S(n, k)λk

k=1 S(n, k)xk

∞∑j=0

(e−λ

)jn = EX n = µn;λ = Tn(λ) :=

n∑k=1

S(n, k)λk

When λ = 1:

∞∑j=0

j!= EX n = µn;1 = Tn(1) :=

n∑k=1

S(n, k) := Bn,

∑∞j=0

(e−λ λj

µn;λ =∑n

k=1 S(n, k)λk

k=1 S(n, k)xk

∞∑j=0

(e−λ

)jn = EX n = µn;λ = Tn(λ) :=

n∑k=1

S(n, k)λk

When λ = 1:

∞∑j=0

j!= EX n = µn;1 = Tn(1) :=

n∑k=1

S(n, k) := Bn,

∑∞j=0

(e−λ λj

µn;λ =∑n

k=1 S(n, k)λk

k=1 S(n, k)xk

∞∑j=0

(e−λ

)jn = EX n = µn;λ = Tn(λ) :=

n∑k=1

S(n, k)λk

When λ = 1:

∞∑j=0

j!= EX n = µn;1 = Tn(1) :=

n∑k=1

S(n, k)

:= Bn,

∑∞j=0

(e−λ λj

µn;λ =∑n

k=1 S(n, k)λk

k=1 S(n, k)xk

∞∑j=0

(e−λ

)jn = EX n = µn;λ = Tn(λ) :=

n∑k=1

S(n, k)λk

When λ = 1:

∞∑j=0

j!= EX n = µn;1 = Tn(1) :=

n∑k=1

S(n, k) := Bn,

µn;λ =∑n

k=1 S(n, k)λk

k=1 S(n, k)xk

µn;λ = Tn(λ)

∞∑j=0

(e−λ

)jn = EX n = µn;λ = Tn(λ) :=

n∑k=1

S(n, k)λk

When λ = 1:

∞∑j=0

j!= EX n = µn;λ = Tn(1) :=

n∑k=1

S(n, k) := Bn, (8)

where Bn is the nth Bell number:Bn = the total number of partitions of [n].(8) is known as Dobinski’s formula.

Random Permutations

Uniform probability measure on Sn: Pn(σ) := 1n! , σ ∈ Sn

We now define a family of non-uniform probability measures on Sn:

Let kn(σ) = the number of cycles in σ.

For θ > 0, define the following probability measure on Sn:

P(θ)n (σ) = θkn(σ)

Nn,θ, where Nn,θ =

∑σ∈Sn θ

kn(σ) is the normalizingconstant.

We have Nn,θ =∑

σ∈Sn θkn(σ) =

∑nk=1 θ

k |s(n, k)| (3)= θ(n).

Thus, P(θ)n (σ) = θkn(σ)

θ(n), σ ∈ Sn.

θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measure

Random Permutations

∑σ∈Sn θ

We have Nn,θ =∑

σ∈Sn θkn(σ) =

∑nk=1 θ

k |s(n, k)| (3)= θ(n).

θ(n), σ ∈ Sn.

Random Permutations

∑σ∈Sn θ

We have Nn,θ =∑

σ∈Sn θkn(σ) =

∑nk=1 θ

k |s(n, k)| (3)= θ(n).

θ(n), σ ∈ Sn.

Random Permutations

∑σ∈Sn θ

We have Nn,θ =∑

σ∈Sn θkn(σ) =

∑nk=1 θ

k |s(n, k)| (3)= θ(n).

θ(n), σ ∈ Sn.

Random Permutations

∑σ∈Sn θ

We have Nn,θ =∑

σ∈Sn θkn(σ) =

∑nk=1 θ

k |s(n, k)| (3)= θ(n).

θ(n), σ ∈ Sn.

Random Permutations

∑σ∈Sn θ

We have Nn,θ =∑

σ∈Sn θkn(σ) =

∑nk=1 θ

k |s(n, k)|

(3)= θ(n).

θ(n), σ ∈ Sn.

Random Permutations

∑σ∈Sn θ

We have Nn,θ =∑

σ∈Sn θkn(σ) =

∑nk=1 θ

k |s(n, k)| (3)= θ(n).

θ(n), σ ∈ Sn.

Random Permutations

∑σ∈Sn θ

We have Nn,θ =∑

σ∈Sn θkn(σ) =

∑nk=1 θ

k |s(n, k)| (3)= θ(n).

θ(n), σ ∈ Sn.

Random Permutations

∑σ∈Sn θ

We have Nn,θ =∑

σ∈Sn θkn(σ) =

∑nk=1 θ

k |s(n, k)| (3)= θ(n).

θ(n), σ ∈ Sn.

Let C(n)j (σ) = the number of j-cycles in σ ∈ Sn, j = 1, · · · , n.

Then for each θ > 0, we can think of {C (n)j }nj=1 as random

variables on the probability space (Sn,P(θ)n ).

Theorem. For any j, the distribution of C(n)j under P

(θ)n converges

weakly to the distribution Poiss( θj ) as n→∞. That is,

limn→∞ P(θ)n (C

(n)j = m) = e−

θj( θj)m

m! , for m = 0, 1, · · · .

Remark. C(n)1 (σ) is the number of fixed points of σ:

C(n)1 = |{k ∈ [n] : σk = k}|. So under P

(θ)n , the distribution of

the number of fixed points in a permutation in Sn converges asn→∞ to the Poisson distribution with parameter θ. Inparticular, under the uniform measure (θ = 1), it converges tothe Poisson distribution with parameter 1.

θ(n), σ ∈ Sn.

θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measureLet C

(n)j (σ) = the number of j-cycles in σ ∈ Sn, j = 1, · · · , n.

(θ)n converges

(n)j = m) = e−

θj( θj)m

m! , for m = 0, 1, · · · .

θ(n), σ ∈ Sn.

(θ)n converges

(n)j = m) = e−

θj( θj)m

m! , for m = 0, 1, · · · .

θ(n), σ ∈ Sn.

(θ)n converges

weakly to the distribution Poiss( θj ) as n→∞.

That is,

(n)j = m) = e−

θj( θj)m

m! , for m = 0, 1, · · · .

θ(n), σ ∈ Sn.

(θ)n converges

(n)j = m) = e−

θj( θj)m

m! , for m = 0, 1, · · · .

θ(n), σ ∈ Sn.

(θ)n converges

(n)j = m) = e−

θj( θj)m

m! , for m = 0, 1, · · · .

(θ)n converges

(n)j = m) = e−

( θj)m

m! , for m = 0, 1, · · · .

Proof for θ = 1 (the same method works with virtually no extrawork for all θ).

Method of moments: It is enough to show that for all m, the

mth moment of C(n)j converges to the mth moment of the

distribution Poiss(1j ).

That is, it is enough to show that

limn→∞

)m= µm; 1

(θ)n converges

(n)j = m) = e−

( θj)m

m! , for m = 0, 1, · · · .

limn→∞

)m= µm; 1

(θ)n converges

(n)j = m) = e−

( θj)m

m! , for m = 0, 1, · · · .

limn→∞

)m= µm; 1

Proof that limn→∞ E(1)n

)m= µm; 1

We will show that

)m= µm; 1

), for n ≥ mj .

For D ⊂ [n] with |D| = j , define

1D(σ) =

{1, if σ has a cycle consisting of the elements of D;

0, otherwise.

Then C(n)j (σ) =

∑D⊂[n]:|D|=j 1D(σ) and(

C(n)j (σ)

)m=( ∑

D1⊂[n]:|D1|=j

1D1(σ))· · ·( ∑

Dm⊂[n]:|Dm|=j

1Dm(σ))

∑(D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

m∏l=1

1Dl(σ).

SoE(1)n

)m=∑(

D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

(∏ml=1 1Dl

)m= µm; 1

We will show that

)m= µm; 1

), for n ≥ mj .

1D(σ) =

0, otherwise.

Then C(n)j (σ) =

∑D⊂[n]:|D|=j 1D(σ) and(

C(n)j (σ)

)m=( ∑

D1⊂[n]:|D1|=j

1D1(σ))· · ·( ∑

Dm⊂[n]:|Dm|=j

1Dm(σ))

∑(D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

m∏l=1

1Dl(σ).

SoE(1)n

)m=∑(

D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

(∏ml=1 1Dl

)m= µm; 1

We will show that

)m= µm; 1

), for n ≥ mj .

1D(σ) =

0, otherwise.

Then C(n)j (σ) =

∑D⊂[n]:|D|=j 1D(σ) and(

C(n)j (σ)

)m=( ∑

D1⊂[n]:|D1|=j

1D1(σ))· · ·( ∑

Dm⊂[n]:|Dm|=j

1Dm(σ))

∑(D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

m∏l=1

1Dl(σ).

SoE(1)n

)m=∑(

D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

(∏ml=1 1Dl

)m= µm; 1

We will show that

)m= µm; 1

), for n ≥ mj .

1D(σ) =

0, otherwise.

Then C(n)j (σ) =

∑D⊂[n]:|D|=j 1D(σ)

(n)j (σ)

)m=( ∑

D1⊂[n]:|D1|=j

1D1(σ))· · ·( ∑

Dm⊂[n]:|Dm|=j

1Dm(σ))

∑(D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

m∏l=1

1Dl(σ).

SoE(1)n

)m=∑(

D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

(∏ml=1 1Dl

)m= µm; 1

We will show that

)m= µm; 1

), for n ≥ mj .

1D(σ) =

0, otherwise.

Then C(n)j (σ) =

∑D⊂[n]:|D|=j 1D(σ) and(

C(n)j (σ)

)m=( ∑

D1⊂[n]:|D1|=j

1D1(σ))· · ·( ∑

Dm⊂[n]:|Dm|=j

1Dm(σ))

∑(D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

m∏l=1

1Dl(σ).

SoE(1)n

)m=∑(

D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

(∏ml=1 1Dl

)m= µm; 1

We will show that

)m= µm; 1

), for n ≥ mj .

1D(σ) =

0, otherwise.

Then C(n)j (σ) =

∑D⊂[n]:|D|=j 1D(σ) and(

C(n)j (σ)

)m=( ∑

D1⊂[n]:|D1|=j

1D1(σ))· · ·( ∑

Dm⊂[n]:|Dm|=j

1Dm(σ))

∑(D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

m∏l=1

1Dl(σ).

SoE(1)n

)m=∑(

D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

(∏ml=1 1Dl

∑(D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

( m∏l=1

1Dl= 1).

Now∏m

l=1 1Dl6≡ 0, (that is, there exists some σ ∈ Dn such that∏m

l=1 1Dl(σ) 6= 0),

if and only if for each pair i1, i2 ∈ [n], either Di1 = Di2 orDi1 ∩ Di2 = ∅.

That is, if and only if there exist a k ∈ [m] and disjoint sets{Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1.

In this case,∏m

l=1 1Dl=∏k

i=1 1Ai, and

(∏ml=1 1Dl

= P(1)n

(∏ki=1 1Ai

((j−1)!

)k(n−kj)!

(Here we use the assumption that n ≥ mj , which insures thatn ≥ kj .)

∑(D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

( m∏l=1

1Dl= 1).

Now∏m

l=1 1Dl6≡ 0,

(that is, there exists some σ ∈ Dn such that∏ml=1 1Dl

(σ) 6= 0),if and only if for each pair i1, i2 ∈ [n], either Di1 = Di2 orDi1 ∩ Di2 = ∅.

In this case,∏m

l=1 1Dl=∏k

i=1 1Ai, and

(∏ml=1 1Dl

= P(1)n

(∏ki=1 1Ai

((j−1)!

)k(n−kj)!

∑(D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

( m∏l=1

1Dl= 1).

Now∏m

l=1 1Dl(σ) 6= 0),

In this case,∏m

l=1 1Dl=∏k

i=1 1Ai, and

(∏ml=1 1Dl

= P(1)n

(∏ki=1 1Ai

((j−1)!

)k(n−kj)!

∑(D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

( m∏l=1

1Dl= 1).

Now∏m

l=1 1Dl(σ) 6= 0),

In this case,∏m

l=1 1Dl=∏k

i=1 1Ai, and

(∏ml=1 1Dl

= P(1)n

(∏ki=1 1Ai

((j−1)!

)k(n−kj)!

∑(D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

( m∏l=1

1Dl= 1).

Now∏m

l=1 1Dl(σ) 6= 0),

In this case,∏m

l=1 1Dl=∏k

i=1 1Ai, and

(∏ml=1 1Dl

= P(1)n

(∏ki=1 1Ai

((j−1)!

)k(n−kj)!

∑(D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

( m∏l=1

1Dl= 1).

Now∏m

l=1 1Dl(σ) 6= 0),

In this case,∏m

l=1 1Dl=∏k

i=1 1Ai,

(∏ml=1 1Dl

= P(1)n

(∏ki=1 1Ai

((j−1)!

)k(n−kj)!

∑(D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

( m∏l=1

1Dl= 1).

Now∏m

l=1 1Dl(σ) 6= 0),

In this case,∏m

l=1 1Dl=∏k

i=1 1Ai, and

(∏ml=1 1Dl

= P(1)n

(∏ki=1 1Ai

((j−1)!

)k(n−kj)!

∑(D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

( m∏l=1

1Dl= 1).

Now∏m

l=1 1Dl(σ) 6= 0),

In this case,∏m

l=1 1Dl=∏k

i=1 1Ai, and

(∏ml=1 1Dl

= P(1)n

(∏ki=1 1Ai

((j−1)!

)k(n−kj)!

)m=∑(

D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

(∏ml=1 1Dl

Now∏m

l=1 1Dl6≡ 0 if and only if there exist a k ∈ [m] and disjoint

sets {Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1. In this case,∏ml=1 1Dl

i=1 1Ai, and

(∏ml=1 1Dl

= P(1)n

(∏ki=1 1Ai

((j−1)!

)k(n−kj)!

The number of ways to construct k disjoint (ordered) sets(A1, · · · ,Ak), each with j elements from [n], is(nj

)(n−jj

)· · ·(n−(m−1)j

(j!)k (n−jk)! .

Given {Ai}ki=1, the number of ways to choose {Dl}ml=1 so that{Dl}ml=1 = {Ai}ki=1 is equal to S(m, k). So

m∑k=1

((j − 1)!

)k(n − kj)!

n!× n!

(j!)k(n − jk)!× S(m, k) =

m∑k=1

)kS(m, k) = Tm(

j) = µm; 1

)m=∑(

D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

(∏ml=1 1Dl

Now∏m

i=1 1Ai, and

(∏ml=1 1Dl

= P(1)n

(∏ki=1 1Ai

((j−1)!

)k(n−kj)!

n! .The number of ways to construct k disjoint (ordered) sets(A1, · · · ,Ak), each with j elements from [n], is(nj

)(n−jj

)· · ·(n−(m−1)j

(j!)k (n−jk)! .

m∑k=1

((j − 1)!

)k(n − kj)!

n!× n!

(j!)k(n − jk)!× S(m, k) =

m∑k=1

)kS(m, k) = Tm(

j) = µm; 1

)m=∑(

D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

(∏ml=1 1Dl

Now∏m

i=1 1Ai, and

(∏ml=1 1Dl

= P(1)n

(∏ki=1 1Ai

((j−1)!

)k(n−kj)!

)(n−jj

)· · ·(n−(m−1)j

(j!)k (n−jk)! .

Given {Ai}ki=1, the number of ways to choose {Dl}ml=1 so that{Dl}ml=1 = {Ai}ki=1 is equal to S(m, k).

m∑k=1

((j − 1)!

)k(n − kj)!

n!× n!

(j!)k(n − jk)!× S(m, k) =

m∑k=1

)kS(m, k) = Tm(

j) = µm; 1

)m=∑(

D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

(∏ml=1 1Dl

Now∏m

i=1 1Ai, and

(∏ml=1 1Dl

= P(1)n

(∏ki=1 1Ai

((j−1)!

)k(n−kj)!

)(n−jj

)· · ·(n−(m−1)j

(j!)k (n−jk)! .

m∑k=1

((j − 1)!

)k(n − kj)!

n!× n!

(j!)k(n − jk)!× S(m, k)

m∑k=1

)kS(m, k) = Tm(

j) = µm; 1

)m=∑(

D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

(∏ml=1 1Dl

Now∏m

i=1 1Ai, and

(∏ml=1 1Dl

= P(1)n

(∏ki=1 1Ai

((j−1)!

)k(n−kj)!

)(n−jj

)· · ·(n−(m−1)j

(j!)k (n−jk)! .

m∑k=1

((j − 1)!

)k(n − kj)!

n!× n!

(j!)k(n − jk)!× S(m, k) =

m∑k=1

)kS(m, k)

= Tm(1

j) = µm; 1

)m=∑(

D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

(∏ml=1 1Dl

Now∏m

i=1 1Ai, and

(∏ml=1 1Dl

= P(1)n

(∏ki=1 1Ai

((j−1)!

)k(n−kj)!

)(n−jj

)· · ·(n−(m−1)j

(j!)k (n−jk)! .

m∑k=1

((j − 1)!

)k(n − kj)!

n!× n!

(j!)k(n − jk)!× S(m, k) =

m∑k=1

)kS(m, k) = Tm(

= µm; 1j.

)m=∑(

D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

(∏ml=1 1Dl

Now∏m

i=1 1Ai, and

(∏ml=1 1Dl

= P(1)n

(∏ki=1 1Ai

((j−1)!

)k(n−kj)!

)(n−jj

)· · ·(n−(m−1)j

(j!)k (n−jk)! .

m∑k=1

((j − 1)!

)k(n − kj)!

n!× n!

(j!)k(n − jk)!× S(m, k) =

m∑k=1

)kS(m, k) = Tm(

j) = µm; 1

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