# Probabilistic derangement numbers and polynomials

Taekyun Kim  Department of Mathematics, Kwangwoon University, Seoul 139-701,
Republic of Korea<EMAIL_ADDRESS>and  Dae San Kim  Department of Mathematics,
Sogang University, Seoul 121-742, Republic of Korea<EMAIL_ADDRESS>
###### Abstract.

Let $Y$ be a random variable such that the moment generating function of $Y$
exists in a neighborhood of the origin. The aim of this paper is to study
probabilistic versions of the derangement polynomials, the derangement
polynomials of type 2 and the $r$-derangement numbers, namely the
probabilistic derangement polynomials associated with $Y$, the probabilistic
derangement polynomials of type 2 associated with $Y$ and the probabilistic
$r$-derangement numbers associated $Y$, respectively. We derive some
properties, explicit expressions, certain identities and recurrence relations
for those polynomials and numbers. In addition, we consider the special case
that $Y$ is the gamma random variable with parameters $\alpha,\beta>0$.

###### Key words and phrases:

probabilistic derangement polynomials; probabilistic Euler numbers;
probabilistic $r$-derangement numbers; probabilistic derangement polynomials
of type 2

###### 2010 Mathematics Subject Classification:

11B73; 11B83

##  1\. Introduction

In combinatorics, a derangement is a permutation that has no fixed points. The
number of derangements of an $n$ element set is called the $n$th derangement
number. For $0\leq r\leq n$, the $n$th $r$-derangement number is the number of
derangements of an $n+r$ element set under the restriction that the first $r$
elements are in disjoint cycles.

Let $Y$ be a random variable satisfying the moment condition (see (11)). The
aim of this paper is to study the probabilistic derangement polynomials
associated with $Y$, the probabilistic derangement polynomials of type 2
associated with $Y$ and the probabilistic $r$-derangement numbers associated
$Y$, as probabilistic versions of the derangement polynomials, the derangement
polynomials of type 2 and the $r$-derangement numbers, respectively. We derive
some properties, explicit expressions, certain identities and recurrence
relations for those polynomials and numbers. In addition, we consider the
special case that $Y$ is the gamma random variable with parameters
$\alpha,\beta>0$.

The outline of this paper is as follows. In Section 1, we recall the
derangement numbers, the $r$-derangement numbers, the derangement polynomials
$D_{n}(x)$ and the derangement polynomials of type 2. We remind the reader of
the Fubini polynomials $F_{n}(x)$, the Stirling numbers of the first kind, the
unsigned Stirling numbers of the first kind and the Stirling numbers of the
second kind ${n\brace k}$. Assume that $Y$ is a random variable such that the
moment generating function of $Y$,
$E[e^{tY}]=\sum_{n=0}^{\infty}\frac{t^{n}}{n!}E[Y^{n}],\quad(|t|<r)$, exists
for some $r>0$. Let $(Y_{j})_{j\geq 1}$ be a sequence of mutually independent
copies of the random variable $Y$, and let
$S_{k}=Y_{1}+Y_{2}+\cdots+Y_{k},\,\,(k\geq 1)$, with $S_{0}=0$. Then we recall
the probabilistic Stirling numbers of the second kind associated with $Y$, the
probabilistic Bell polynomials associated with $Y$, $\phi_{n}^{Y}(x)$, and the
gamma random variable with parameters $\alpha,\,\beta>0$. Section 2 is the
main results of this paper. Let $(Y_{j})_{j\geq
1},\,\,S_{k},\,\,(k=0,1,\dots)$ be as in the above. Then we first define the
probabilistic derangement polynomials associated $Y$, $D_{n}^{Y}(x)$. In
Theorem 2.1, we find two explicit expressions for $D_{n}^{Y}(x)$. We derive a
recurrence relation for $D_{n}^{Y}(x)$ in Theorem 2.2. In Theorem 2.3, we
express $\phi_{n}^{Y}(1-x)$ as a finite sum involving $D_{k}(x)$. We define
the probabilistic Euler numbers $E_{n}^{Y}$ in a natural manner. In Theorem
2.4, we express the convolution
$\sum_{m=0}^{n}\binom{n}{m}\phi_{m}^{Y}(1-x)E_{n-m}^{Y}$, as an infinite sum
involving $D_{m}(x)$. We define the probabilistic $r$-derangement numbers
$D_{n}^{(r,Y)}$. We derive finite sum expressions for $D_{n}^{(r,Y)}$ in
Theorems 2.5 and 2.6. In Theorem 2.7, the $n$th moment of $Y$ is expressed as
a finite sum involving $D_{k+r}^{(r,Y)}$. We express $D_{n+r}^{(r,Y)}$ as a
finite sum involving $D_{n-l}^{Y},\,\,(0\leq l\leq n)$ in Theorem 2.8. We
define the probabilistic derangement polynomials of type 2, $d_{n}^{Y}(x)$. In
Theorem 2.9, we find an explicit expression and a recurrence relation for
$d_{n}^{Y}(x)$. In Theorem 2.10, we derive a finite sum identity which is
given by $\sum_{m=0}^{n}d_{m}^{Y}(x){n\brace
m}=\sum_{m=0}^{n}\sum_{l=0}^{m}(-1)^{l}{m\brace
l}\binom{n}{m}E[Y^{l}]F_{n-m}(x)$. We deduce an explicit expression for
$d_{n}^{Y}(x)$ in Theorem 2.11. Finally, when $Y\sim\Gamma(1,1)$, we derive an
expression of $D_{n}^{Y}(x)$ as a finite sum involving $D_{l}(1-x),\,\,(0\leq
l\leq n)$, in Theorem 2.12. For the rest of this section, we recall the facts
that are needed throughout this paper.

For $n\geq 0$, the derangement numbers $D_{n},\ (n\geq 0)$ are given by

(1)
$D_{n}=n!\sum_{k=0}^{n}\frac{(-1)^{k}}{k!}=\sum_{k=0}^{n}\binom{n}{k}(n-k)!(-1)^{k},\quad(\mathrm{see}\
[5,6,8,10,18,19,21,24]).$

From (1), we easily derive the following equation

(2)
$\frac{1}{1-t}e^{-t}=\sum_{n=0}^{\infty}D_{n}\frac{t^{n}}{n!},\quad(\mathrm{see}\
[18,19,21,24,29]).$

For $0\leq r\leq n$, the $r$-derangement numbers $D_{n}^{(r)}$ are given by

(3)
$\sum_{n=0}^{\infty}D_{n}^{(r)}\frac{t^{n}}{n!}=\frac{t^{r}}{(1-t)^{r+1}}e^{-t},\quad(\mathrm{see}\
[18,19,21,24]).$

The derangement polynomials are defined by

(4)
$\frac{e^{xt}}{1-t}e^{-t}=\sum_{n=0}^{\infty}D_{n}(x)\frac{t^{n}}{n!},\quad(\mathrm{see}\
[11,18,19,21,24,30]),$

and the derangement polynomials of type 2 are given by

(5)
$\frac{1}{1-xt}e^{-t}=\sum_{n=0}^{\infty}d_{n}(x)\frac{t^{n}}{n!},\quad(\mathrm{see}\
[18]).$

Note that

$D_{n}(0)=D_{n}\quad\mathrm{and}\quad d_{n}(1)=D_{n},\ (n\geq 0).$

It is well known that the Fubini polynomials are defined by

(6) $F_{n}(x)=\sum_{k=0}^{n}{n\brace k}k!x^{k},\quad(n\geq
0),\quad(\mathrm{see}\ [6,16,17,20,23,30]),$

where ${n\brace k}$ are the Stirling numbers of the second kind given by

(7) $\frac{1}{k!}\big{(}e^{t}-1\big{)}^{t}=\sum_{n=k}^{\infty}{n\brace
k}\frac{t^{n}}{n!},\quad(k\geq 0),\quad(\mathrm{see}\ [1-32]).$

From (6), we have

(8)
$\frac{1}{1-x(e^{t}-1)}=\sum_{n=0}^{\infty}F_{n}(x)\frac{t^{n}}{n!},\quad(\mathrm{see}\
[6,17,23]).$

The Stirling numbers of the first kind are given by

(9)
$\frac{1}{k!}\big{(}\log(1+t)\big{)}^{k}=\sum_{n=k}^{\infty}S_{1}(n,k)\frac{t^{n}}{n!},\quad(\mathrm{see}\
[6-30]).$

For $n\geq k\geq 0$, the unsigned Stirling numbers of the first kind are
defined by

(10) ${n\brack k}=(-1)^{n-k}S_{1}(n,k),\quad(\mathrm{see}\ [6-30]).$

Let $Y$ be a random variable such that the moment generating function of $Y$

(11)
$E\big{[}e^{tY}\big{]}=\sum_{n=0}^{\infty}E[Y^{n}]\frac{t^{n}}{n!},\quad(|t|<r)\quad\textrm{exists
for some $r>0$.}$

Assume that $(Y_{j})_{j\geq 1}$ is a sequence of mutually independent copies
of $Y$, and that $S_{k}=Y_{1}+Y_{2}+\cdots+Y_{k},\ (k\geq 1),\ S_{0}=0$, (see
[2,3,14,31]). The probabilistic Stirling numbers of the second kind associated
with random variable $Y$ are defined by

${n\brace
k}_{Y}=\frac{1}{k!}\sum_{j=0}^{k}\binom{k}{j}(-1)^{k-j}E\big{[}S_{j}^{n}\big{]},\quad(n\geq
0),\quad(\mathrm{see}\ [2,3,14,22,32]).$

Equivalently, they are given by

(12) $\frac{1}{k!}(E[e^{tY}]-1)^{k}=\sum_{n=k}^{\infty}{n\brace
k}_{Y}\frac{t^{n}}{n!}.$

When $Y=1,\ {n\brace k}_{Y}={n\brace k},\ (n\geq k\geq 0)$. The probabilistic
Bell polynomials associated with random variable $Y$ are defined by

(13) $\phi_{n}^{Y}(x)=\sum_{k=0}^{n}{n\brace k}_{Y}x^{k},\quad(n\geq
0),\quad(\mathrm{see}\ [3,14,32]).$

From (13), we have

(14) $e^{x(E[e^{Yt}]-1)}=\sum_{n=0}^{\infty}\phi_{n}^{Y}(x)\frac{t^{n}}{n!}.$

A continuous random variable $X$ whose density function is given by

$f(x)=\left\\{\begin{array}[]{ccc}\frac{\beta}{\Gamma(\alpha)}e^{-\beta
x}(\beta x)^{\alpha-1},&\textrm{if $x\geq 0$,}\\\ 0,&\textrm{if
$x<0$,}\end{array}\right.$

for some $\alpha,\beta>0$ is said to be the gamma random variable with
parameters $\alpha,\beta$, which is denoted by $X\sim\Gamma(\alpha,\beta),$
(see [28,31]).

In particular, if $X\sim\Gamma(1,1)$, then we have

$\displaystyle E\Big{[}e^{(X-1+p)t}\Big{]}$
$\displaystyle=\int_{0}^{\infty}e^{(x-1+p)t}f(x)dx=\int_{0}^{\infty}e^{(x-1+p)t}e^{-x}dx$
$\displaystyle=\frac{e^{-t}}{1-t}e^{pt}=\sum_{n=0}^{\infty}D_{n}(p)\frac{t^{n}}{n!},\quad(t<1).$

Thus, we get

$E\big{[}(X-1+p)^{n}\big{]}=D_{n}(p),\quad(n\geq 0).$

##  2\. Probabilistic derangement numbers and polynomials associated with
random variables

Let $(Y_{k})_{k\geq 1}$ be a sequence of mutually independent copies of the
random variable $Y$, and let

$S_{0}=0,\,\,S_{k}=Y_{1}+Y_{2}+\cdots+Y_{k},\quad(k\geq 1).$

Now, we define the probabilistic derangement polynomials associated with $Y$
by

(15)
$\frac{e^{xt}}{1-t}E\big{[}e^{-tY}\big{]}=\sum_{n=0}^{\infty}D_{n}^{Y}(x)\frac{t^{n}}{n!}.$

In particular, for $x=0,\ D_{n}^{Y}=D_{n}^{Y}(0)$ are called the probabilistic
derangement numbers associated with $Y$.

When $Y=1$, we have $D_{n}^{Y}(x)=D_{n}(x)$, $(n\geq 0)$. From (15), we have

(16) $\displaystyle\sum_{n=0}^{\infty}D_{n}^{Y}(x)\frac{t^{n}}{n!}$
$\displaystyle=\frac{E\big{[}e^{-tY}\big{]}}{1-t}e^{xt}=\sum_{l=0}^{\infty}D_{l}^{Y}\frac{t^{l}}{l!}\sum_{m=0}^{\infty}x^{m}\frac{t^{m}}{m!}$
$\displaystyle=\sum_{n=0}^{\infty}\sum_{l=0}^{n}\binom{n}{l}D_{l}^{Y}x^{n-l}\frac{t^{n}}{n!}.$

On the other hand, by (15), we also have

(17) $\displaystyle\sum_{n=0}^{\infty}D_{n}^{Y}(x)\frac{t^{n}}{n!}$
$\displaystyle=\frac{E\big{[}e^{-tY}\big{]}}{1-t}e^{xt}=\frac{1}{1-t}E\big{[}e^{t(x-Y)}]$
$\displaystyle=\sum_{l=0}^{\infty}t^{l}\sum_{m=0}^{\infty}E[(x-Y)^{m}]\frac{t^{m}}{m!}$
$\displaystyle=\sum_{n=0}^{\infty}n!\sum_{m=0}^{n}\frac{E\big{[}(x-Y)^{m}\big{]}}{m!}\frac{t^{n}}{n!}.$

Thus, by comparing the coefficients on both sides of (16) and (17), we obtain
the following theorem.

######  Theorem 2.1.

For $n\geq 0$, we have

$D_{n}^{Y}(x)=n!\sum_{m=0}^{n}\frac{E\big{[}(x-Y)^{m}\big{]}}{m!}=\sum_{l=0}^{n}\binom{n}{l}D_{l}^{Y}x^{n-l}.$

From (15), we note that

(18) $\displaystyle E\big{[}e^{t(x-Y)}\big{]}$
$\displaystyle=(1-t)\sum_{n=0}^{\infty}D_{n}^{Y}(x)\frac{t^{n}}{n!}$
$\displaystyle=\sum_{n=1}^{\infty}\big{(}D_{n}^{Y}(x)-nD_{n-1}^{Y}(x)\big{)}\frac{t^{n}}{n!}+1.$

On the other hand, by Taylor expansion, we also get

(19) $\displaystyle E\big{[}e^{t(x-Y)}\big{]}$
$\displaystyle=\sum_{n=0}^{\infty}E\big{[}(x-Y)^{n}\big{]}\frac{t^{n}}{n!}$
$\displaystyle=1+\sum_{n=1}^{\infty}E\big{[}(x-Y)^{n}\big{]}\frac{t^{n}}{n!}.$

Therefore, by (18) and (19), we obtain the following theorem.

######  Theorem 2.2.

For $n\geq 1$, we have

$D_{n}^{Y}(x)-nD_{n-1}^{Y}(x)=E\big{[}(x-Y)^{n}\big{]}.$

In particular, for $x=0$, we get

$D_{n}^{Y}-nD_{n-1}^{Y}=(-1)^{n}E[Y^{n}].$

Replacing $t$ by $1-E[e^{Yt}]$ in (4), we get

(20) $\displaystyle e^{(1-x)(E[e^{Yt}]-1)}$
$\displaystyle=\sum_{k=0}^{\infty}D_{k}(x)\frac{1}{k!}\big{(}1-E[e^{Yt}]\big{)}^{k}E[e^{Yt}]$
$\displaystyle=\sum_{k=0}^{\infty}D_{k}(x)(-1)^{k}\sum_{l=k}^{\infty}{l\brace
k}_{Y}\frac{t^{l}}{l!}\sum_{m=0}^{\infty}E[Y^{m}]\frac{t^{m}}{m!}$
$\displaystyle=\sum_{l=0}^{\infty}\sum_{k=0}^{l}D_{k}(x)(-1)^{l}{l\brace
k}_{Y}\frac{t^{l}}{l!}\sum_{m=0}^{\infty}E[Y^{m}]\frac{t^{m}}{m!}$
$\displaystyle=\sum_{n=0}^{\infty}\sum_{l=0}^{n}\sum_{k=0}^{l}\binom{n}{l}{l\brace
k}_{Y}D_{k}(x)(-1)^{l}E[Y^{n-l}]\frac{t^{n}}{n!}.$

Therefore, by (14) and (20), we obtain the following theorem.

######  Theorem 2.3.

For $n\geq 0$, we have

$\phi_{n}^{Y}(1-x)=\sum_{l=0}^{n}\sum_{k=0}^{l}\binom{n}{l}{l\brace
k}_{Y}D_{k}(x)(-1)^{l}E[Y^{n-l}].$

The probabilistic Euler numbers are given by

(21) $\frac{2}{E[e^{tY}]+1}=\sum_{n=0}^{\infty}E_{n}^{Y}\frac{t^{n}}{n!}.$

When $Y=1$, $E_{n}^{Y}=E_{n},\ (n\geq 0)$, where $E_{n}$ are the ordinary
Euler numbers given by

$\frac{2}{e^{t}+1}=\sum_{n=0}^{\infty}E_{n}\frac{t^{n}}{n!},\quad(\mathrm{see}\
[30]).$

Replacing $t$ by $-E[e^{tY}]$ in (4), we have

(22) $\displaystyle\frac{1}{1+E[e^{Yt}]}e^{(1-x)E[e^{Yt}]}$
$\displaystyle=\sum_{m=0}^{\infty}D_{m}(x)\frac{(-1)^{m}}{m!}\Big{(}E[e^{Yt}]\Big{)}^{m}$
$\displaystyle=\sum_{m=0}^{\infty}D_{m}(x)\frac{(-1)^{m}}{m!}E\Big{[}e^{(Y_{1}+\cdots+Y_{m})t}\Big{]}$
$\displaystyle=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}D_{m}(x)\frac{(-1)^{m}}{m!}E[S_{m}^{n}]\frac{t^{n}}{n!}.$

On the other hand, by (21), we get

(23) $\displaystyle\frac{1}{1+E[e^{Yt}]}e^{(1-x)E[e^{Yt}]}$
$\displaystyle=\frac{e^{1-x}}{2}\frac{2}{1+E[e^{Yt}]}e^{(1-x)(E[e^{Yt}]-1)}$
$\displaystyle=\frac{e^{1-x}}{2}\sum_{l=0}^{\infty}E_{l}^{Y}\frac{t^{l}}{l!}\sum_{m=0}^{\infty}\phi_{m}^{Y}(1-x)\frac{t^{m}}{m!}$
$\displaystyle=\frac{e^{1-x}}{2}\sum_{n=0}^{\infty}\sum_{m=0}^{n}\binom{n}{m}\phi_{m}^{Y}(1-x)E_{n-m}^{Y}\frac{t^{n}}{n!}.$

Therefore, by (22) and (23), we obtain the following theorem.

######  Theorem 2.4.

For $n\geq 0$, we have

$\sum_{m=0}^{n}\binom{n}{m}\phi_{m}^{Y}(1-x)E_{n-m}^{Y}=2e^{x-1}\sum_{m=0}^{\infty}\frac{D_{m}(x)}{m!}(-1)^{m}E[S_{m}^{n}].$

For $0\leq r\leq n$, we consider the probabilistic $r$-derangement numbers
associated with $Y$, which are defined by

(24)
$\frac{t^{r}}{(1-t)^{r+1}}E\big{[}e^{-Yt}\big{]}=\sum_{n=r}^{\infty}D_{n}^{(r,Y)}\frac{t^{n}}{n!}.$

When $Y=1$, $D_{n}^{(r,Y)}=D_{n}^{(r)},\ (n\geq 0)$. From (24), we have

(25) $\displaystyle\frac{t^{r}}{(1-t)^{r+1}}E\big{[}e^{-Yt}\big{]}$
$\displaystyle=\sum_{k=0}^{\infty}\binom{k+r}{r}t^{k+r}\sum_{m=0}^{\infty}(-1)^{m}E[Y^{m}]\frac{t^{m}}{m!}$
$\displaystyle=\sum_{k=r}^{\infty}\binom{k}{r}t^{k}\sum_{m=0}^{\infty}(-1)^{m}E[Y^{m}]\frac{t^{m}}{m!}$
$\displaystyle=\sum_{n=r}^{\infty}n!\sum_{k=r}^{n}\binom{k}{r}(-1)^{n-k}\frac{E[Y^{n-k}]}{(n-k)!}\frac{t^{n}}{n!}.$

Therefore, by (24) and (25), we obtain the following theorem.

######  Theorem 2.5.

For $n\geq r\geq 0$, we have

$D_{n}^{(r,Y)}=n!\sum_{k=r}^{n}\binom{k}{r}(-1)^{n-k}\frac{E[Y^{n-k}]}{(n-k)!}.$

From (24), we note that

(26) $\displaystyle\sum_{n=r}^{\infty}D_{n}^{(r,Y)}\frac{t^{n}}{n!}$
$\displaystyle=\bigg{(}\frac{t}{1-t}\bigg{)}^{r}\frac{1}{1-t}E[e^{-Yt}]$
$\displaystyle=\sum_{l=r}^{\infty}\binom{l-1}{r-1}t^{l}\sum_{m=0}^{\infty}D_{m}^{Y}\frac{t^{m}}{m!}$
$\displaystyle=\sum_{n=r}^{\infty}n!\sum_{l=r}^{n}\binom{l-1}{r-1}\frac{D_{n-l}^{Y}}{(n-l)!}\frac{t^{n}}{n!}.$

By comparing the coefficients on both sides of (26), we obtain the following
theorem.

######  Theorem 2.6.

For $n\geq r\geq 0$, we have

$D_{n}^{(r,Y)}=n!\sum_{l=r}^{n}\binom{l-1}{r-1}\frac{D_{n-l}^{Y}}{(n-l)!}.$

By (24), we get

(27) $\displaystyle E\big{[}e^{-Yt}\big{]}$
$\displaystyle=(1-t)^{r+1}\sum_{k=r}^{\infty}D_{k}^{(r,Y)}\frac{t^{k-r}}{k!}$
$\displaystyle=(1-t)^{r+1}\sum_{k=0}^{\infty}D_{k+r}^{(r,Y)}\frac{t^{k}}{(k+r)!}$
$\displaystyle=\sum_{l=0}^{\infty}(-1)^{l}\binom{r+1}{l}t^{l}\sum_{k=0}^{\infty}D_{k+r}^{(r,Y)}\frac{t^{k}}{(k+r)!}$
$\displaystyle=\sum_{n=0}^{\infty}n!\sum_{k=0}^{n}\frac{D_{k+r}^{(r,Y)}}{(k+r)!}(-1)^{n-k}\binom{r+1}{n-k}\frac{t^{n}}{n!}.$

Therefore, by (27), we obtain the following theorem.

######  Theorem 2.7.

For $n\geq 0$, we have

$E\big{[}Y^{n}\big{]}=n!\sum_{k=0}^{n}\frac{D_{k+r}^{(r,Y)}}{(k+r)!}(-1)^{k}\binom{r+1}{n-k}.$

Now, we observe that

(28) $\displaystyle\frac{1}{(1-t)^{r+1}}E\big{[}e^{-Yt}\big{]}$
$\displaystyle=\frac{1}{t^{r}}\frac{t^{r}}{(1-t)^{r+1}}E\big{[}e^{-tY}\big{]}$
$\displaystyle=\sum_{n=0}^{\infty}\frac{D_{n+r}^{(r,Y)}}{(n+r)!}t^{n}=\sum_{n=0}^{\infty}\frac{D_{n+r}^{(r,Y)}}{\binom{n+r}{r}r!}\frac{t^{n}}{n!}.$

On the other hand, by binomial expansion, we get

(29) $\displaystyle\frac{1}{(1-t)^{r+1}}E\big{[}e^{-Yt}\big{]}$
$\displaystyle=\bigg{(}\frac{1}{1-t}\bigg{)}^{r}\frac{1}{1-t}E\big{[}e^{-Yt}\big{]}$
$\displaystyle=\sum_{l=0}^{\infty}\binom{r+l-1}{l}t^{l}\sum_{m=0}^{\infty}D_{m}^{Y}\frac{t^{m}}{m!}$
$\displaystyle=\sum_{n=0}^{\infty}n!\sum_{l=0}^{n}\binom{r+l-1}{l}\frac{D_{n-l}^{Y}}{(n-l)!}\frac{t^{n}}{n!}.$

Thus, by (28) and (29), we get the following theorem.

######  Theorem 2.8.

For $n\geq 0$, we have

$D_{n+r}^{(r,Y)}=(n+r)!\sum_{l=0}^{n}\binom{r+l-1}{l}\frac{D_{n-l}^{Y}}{(n-l)!}.$

Now, we define the probabilistic derangement polynomials of type 2 associated
with $Y$ as

(30)
$\frac{1}{1-xt}E\big{[}e^{-Yt}\big{]}=\sum_{n=0}^{\infty}d_{n}^{Y}(x)\frac{t^{n}}{n!}.$

When $Y=1$, we have $d_{n}^{Y}(x)=d_{n}(x),\ (n\geq 0)$. In particular, for
$x=1$, $d_{n}^{Y}(1)=D_{n}^{Y},\ (n\geq 0)$.

From (30), we have

(31) $\displaystyle\sum_{n=0}^{\infty}d_{n}^{Y}(x)\frac{t^{n}}{n!}$
$\displaystyle=\frac{1}{1-xt}E[e^{-Yt}]$
$\displaystyle=\sum_{m=0}^{\infty}x^{m}t^{m}\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!}E[Y^{k}]t^{k}$
$\displaystyle=\sum_{n=0}^{\infty}n!\sum_{k=0}^{n}\frac{(-1)^{k}}{k!}E\big{[}Y^{k}\big{]}x^{n-k}\frac{t^{n}}{n!},$

and

(32) $\displaystyle\sum_{n=0}^{\infty}(-1)^{n}E[Y^{n}]\frac{t^{n}}{n!}$
$\displaystyle=E\big{[}e^{-Yt}\big{]}=(1-xt)\sum_{n=0}^{\infty}d_{n}^{Y}(x)\frac{t^{n}}{n!}$
$\displaystyle=d_{0}^{Y}(x)+\sum_{n=1}^{\infty}\bigg{(}d_{n}^{Y}(x)-nxd_{n-1}^{Y}(x)\bigg{)}\frac{t^{n}}{n!}.$

Thus, by (32), we get

(33) $d_{0}^{Y}(x)=1,\quad
d_{n}^{Y}(x)=nxd_{n-1}^{Y}(x)+(-1)^{n}E[Y^{n}],\quad(n\geq 1).$

Therefore, by (31) and (33), we obtain the following theorem.

######  Theorem 2.9.

For $n\geq 0$, we have

$d_{n}^{Y}(x)=n!\sum_{k=0}^{n}\frac{(-1)^{k}}{k!}E[Y^{k}]x^{n-k}.$

Moreover, we have

$d_{0}^{Y}(x)=1,\quad
d_{n}^{Y}(x)=nxd_{n-1}^{Y}(x)+(-1)^{n}E[Y^{n}],\quad(n\geq 1).$

Replacing $t$ by $e^{t}-1$ in (30), we get

(34) $\displaystyle\frac{1}{1-x(e^{t}-1)}E\big{[}e^{-Y(e^{t}-1)}\big{]}$
$\displaystyle=\sum_{m=0}^{\infty}d_{m}^{Y}(x)\frac{1}{m!}(e^{t}-1)^{m}$
$\displaystyle=\sum_{n=0}^{\infty}\sum_{m=0}^{n}d_{m}^{Y}(x){n\brace
m}\frac{t^{n}}{n!}.$

On the other hand, by (8), we obtain

(35) $\displaystyle\frac{1}{1-x(e^{t}-1)}E\big{[}e^{-Y(e^{t}-1)}\big{]}$
$\displaystyle=\sum_{k=0}^{\infty}F_{k}(x)\frac{t^{k}}{k!}\sum_{l=0}^{\infty}(-1)^{l}E[Y^{l}]\frac{(e^{t}-1)^{l}}{l!}$
$\displaystyle=\sum_{k=0}^{\infty}F_{k}(x)\frac{t^{k}}{k!}\sum_{l=0}^{\infty}(-1)^{l}E[Y^{l}]\sum_{m=l}^{\infty}{m\brace
l}\frac{t^{m}}{m!}$
$\displaystyle=\sum_{k=0}^{\infty}F_{k}(x)\frac{t^{k}}{k!}\sum_{m=0}^{\infty}\sum_{l=0}^{m}(-1)^{l}E[Y^{l}]{m\brace
l}\frac{t^{m}}{m!}$
$\displaystyle=\sum_{n=0}^{\infty}\sum_{m=0}^{n}\sum_{l=0}^{m}(-1)^{l}\binom{n}{m}{m\brace
l}E[Y^{l}]F_{n-m}(x)\frac{t^{n}}{n!}.$

Therefore, by (34) and (35), we obtain the following theorem.

######  Theorem 2.10.

For $n\geq 0$, we have

$\sum_{m=0}^{n}d_{m}^{Y}(x){n\brace
m}=\sum_{m=0}^{n}\sum_{l=0}^{m}(-1)^{l}{m\brace
l}\binom{n}{m}E[Y^{l}]F_{n-m}(x).$

From (30), we note that

(36) $\displaystyle\sum_{n=0}^{\infty}d_{n}^{Y}(x)\frac{t^{n}}{n!}$
$\displaystyle=e^{-\log(1-xt)}E\big{[}e^{-Yt}\big{]}=\sum_{l=0}^{\infty}\frac{(-1)^{l}}{l!}\big{(}\log(1-xt)\big{)}^{l}E\big{[}e^{-Yt}\big{]}$
$\displaystyle=\sum_{j=0}^{\infty}\sum_{l=0}^{j}{j\brack
l}x^{j}\frac{t^{j}}{j!}\sum_{m=0}^{\infty}(-1)^{m}E[Y^{m}]\frac{t^{m}}{m!}$
$\displaystyle=\sum_{n=0}^{\infty}\sum_{j=0}^{n}\sum_{l=0}^{j}{j\brack
l}\binom{n}{j}x^{j}(-1)^{n-j}E[Y^{n-j}]\frac{t^{n}}{n!}.$

Therefore, by comparing the coefficients on both sides of (36), we obtain the
following theorem.

######  Theorem 2.11.

For $n\geq 0$, we have

$d_{n}^{Y}(x)=\sum_{j=0}^{n}\sum_{l=0}^{j}{j\brack
l}\binom{n}{j}x^{j}(-1)^{n-j}E[Y^{n-j}].$

Let $Y\sim\Gamma(1,1)$. Then, for $t>-1$, we have

(37) $\displaystyle E\big{[}e^{(x-Y)t}\big{]}$
$\displaystyle=e^{xt}\int_{0}^{\infty}e^{-yt}e^{-y}dy=\frac{1}{1+t}e^{t}e^{(x-1)t}$
$\displaystyle=\sum_{l=0}^{\infty}(-1)^{l}D_{l}(1-x)\frac{t^{l}}{l!}.$

From (37), we have

(38) $\displaystyle\sum_{n=0}^{\infty}D_{n}^{Y}(x)\frac{t^{n}}{n!}$
$\displaystyle=\frac{e^{xt}}{1-t}E\big{[}e^{-Yt}\big{]}=\frac{1}{1-t}\frac{1}{1+t}e^{t}e^{(x-1)t}$
$\displaystyle=\frac{1}{1-t}\sum_{l=0}^{\infty}(-1)^{l}D_{l}(1-x)\frac{t^{l}}{l!}=\sum_{n=0}^{\infty}n!\sum_{l=0}^{n}\frac{(-1)^{l}}{l!}D_{l}(1-x)\frac{t^{n}}{n!}.$

Therefore, by comparing the coefficients on both sides of (38), we obtain the
following theorem.

######  Theorem 2.12.

Let $Y\sim\Gamma(1,1)$. For $n\geq 0$, we have

$D_{n}^{Y}(x)=n!\sum_{l=0}^{n}\frac{(-1)^{l}}{l!}D_{l}(1-x).$

##  3\. Conclusion

By means of generating functions we studied probabilistic versions of the
derangement polynomials, the $r$-derangement numbers and the derangement
polynomials of type 2, namely the probabilistic derangement polynomials
$D_{n}^{Y}(x)$ associated with $Y$, the probabilistic $r$-derangement numbers
$D_{n}^{(r,Y)}$ associated $Y$ and the probabilistic derangement polynomials
of type 2 $d_{n}^{Y}(x)$ associated with $Y$. Here $Y$ is a random variable
such that the moment generating function of $Y$ exists in a neighborhood of
the origin. In more detail, we derived an explicit expression for
$D_{n}^{Y}(x)$ (see Theorem 2.1) and that in terms of derangement polynomials
$D_{l}(x)$ for the special case of $Y\sim\Gamma(1,1)$ (see Theorem 2.12). We
deduced a recurrence relation for $D_{n}^{Y}(x)$ (see Theorem 2.2). We
expressed $\phi_{n}^{Y}(1-x)$ and the convolution of that with the
probabilistic Euler numbers $E_{n}^{Y}$ in terms of $D_{k}(x)$ (see Theorems
2.3, 2.4). We found explicit expressions for $D_{n}^{(r,Y)}$ (see Theorems
2.5, 2.6) and two identities involving those numbers (see Theorems 2.7, 2.8).
We obtained explicit expressions for $d_{n}^{Y}(x)$ (see Theorems 2.9, 2.11),
a recurrence relation for $d_{n}^{Y}(x)$ (see Theorem 2.9) and an identity
involving those polynomials (see Theorem 2.10).

As one of our future projects, we would like to continue to study
probabilistic versions of many special polynomials and numbers and to find
their applications to physics, science and engineering as well as to
mathematics.

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