Probabilistic degenerate Stirling polynomials of the second kind and their applications

ABSTRACT

The aim of this paper is to further study probabilistic versions of the degenerate Stirling polynomials of the second kind, namely the probabilistic degenerate Stirling polynomials of the second kind associated with $Y$, which are also degenerate versions of the probabilistic Stirling polynomials of the second kind investigated earlier. Here $Y$ is a random variable whose moment generating function exists in some neighbourhood of the origin. We derive some properties, explicit expressions and certain identities for those polynomials. In addition, we apply our results to the Bernoulli random variable with probability of success $p$, the normal random variable with parameters $\mu =0,{\sigma }^2=1$, and to the uniform random variable on $(0,1)$.

KEYWORDS:

• Probabilistic degenerate Stirling polynomials of the second kind
• normal random variable
• uniform random variable

1. Introduction

Carlitz initiated a study of degenerate versions of special polynomials and numbers with his work on the degenerate Bernoulli and degenerate Euler polynomials and numbers in [1]. Recently, some mathematicians have regained their interests in these explorations for various degenerate versions of many special numbers and polynomials not only with their number-theoretic or combinatorial interests but also with their applications to other areas, including probability, quantum mechanics and differential equations. It is amusing that many different tools, like generating functions, combinatorial methods, $p$-adic analysis, umbral calculus, operator theory, differential equations, special functions, probability theory and analytic number theory, are employed in the course of this quest (see [2–10] and the references therein).

Assume that $Y$ is a random variable such that the moment generating function of $Y$

(1) $E[e^{tY}]=\sum _{n=0}^{\mathrm{\infty }}E[Y^n]\frac{t^n}{n!},(|t|<r)\mathrm{e}\mathrm{x}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{s}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{s}\mathrm{o}\mathrm{m}\mathrm{e}r>0.$(1)

The aim of this paper is to further study probabilistic versions of the degenerate Stirling polynomials of the second kind, namely the probabilistic degenerate Stirling polynomials of the second kind associated with $Y$ (see [2]), which are also degenerate versions of the probabilistic Stirling polynomials of the second investigated in [11]. Then we derive some properties, explicit expressions and certain identities for those polynomials. In addition, we apply our results to the Bernoulli random variable with probability of success $p$, the normal random variable with parameters $\mu =0,{\sigma }^2=1$, and to the uniform random variable on $(0,1)$.

The outline of this paper is as follows. In Section 1, we recall the degenerate exponentials, the degenerate Stirling numbers of the second kind, the $\lambda$-analogues of the Stirling numbers of the first kind and the Cauchy numbers (of the first kind) of order $\alpha$. Then we remind the reader of the degenerate Stirling polynomials of the second kind and the Hermite polynomials. Assume that $Y$ is a random variable satisfying the moment condition in (1). Let $(Y_j{)}_{j\ge 1}$ be a sequence of mutually independent copies of the random variable $Y$, and let $S_k=Y_1+\cdots +Y_k,(k\ge 1)$, with $S_0=0$. Then we recall the probablistic Stirling polynomials of the second kind associated with $Y$, $S_Y(n,k|x)$, which are defined in terms of the $n$th moments of $S_j+x,(j=0,1,\dots ,k)$ (see [2,11])., Section 2 is the main result of this paper. Let $(Y_j{)}_{j\ge 1},S_k,(k=0,1,\dots )$ be as in the above. Then we recall from [2] the probabilistic degenerate Stirling polynomials of the second kind associated with $Y$, $S_{2,\lambda }^Y(n,k|x)$, which are defined as a degenerate version of $S_Y(n,k|x)$. We state an explicit expression of those polynomials in Theorem 1. We derive a finite sum identity involving the difference operators and the degenerate Stirling polynomials of the second kind in Theorem 2. Assume that $(U_i{)}_{i\ge 1}$ are mutually independent uniform random variables on $(0,1)$, and that $(U_i{)}_{i\ge 1}$ and $(Y_i{)}_{i\ge 1}$ are mutually independent. In Theorem 3, we express $\frac{1}{m!}{\mathrm{\Delta }}_{y_1,y_2,\dots ,y_m}(x{)}_{n,\lambda }$ (see (18)) as a finite sum involving $E[(x+y_1{U}_1+\cdots +y_m{U}_m{)}_{n-k,\lambda }]$. In Theorem 4, we deduce four different expressions for $S_{2,\lambda }^Y(n,m|x)$. In Theorem 6, we obtain a finite sum involving $E[(x+S_k{)}_{n,\lambda }]$ and $S_{2,\lambda }^Y(n,m|x)$. We show that $S_{2,\lambda }^Y(n,m|x)={\alpha }^m{W}_{\alpha ,\lambda }(n,m|x)$, for $Y=\alpha$ (see (35)). In Section 3, we apply the results in the previous section to the Bernoulli random variable with probability of success $p$, the normal random variable with parameters $\mu =0,{\sigma }^2=1$, and to the uniform random variable on $(0,1)$. In Theorem 8, we show $S_{2,\lambda }^Y(n,m|x)=p^m{S}_{2,\lambda }(n,m|x)$, if $Y$ is the Bernoulli random variable with probability of success $p$. In Theorem 9, we deduce $E[(x+iY{)}^n]=H_n(x)$, if $Y$ is the normal random variable with parameters $\mu =0,{\sigma }^2=1$. Finally, in Theorem 10 we find a finite sum expression of $S_{2,\lambda }^Y(n,m|x)$ involving the Cauchy numbers of order $k,(k=0,1,\dots ,m)$, when $Y$ is the uniform random variable on $(0,1)$.

For any nonzero $\lambda \in \mathbb{R}$, the degenerate exponentials are defined by

(2) $e_{\lambda }^x\left(t\right)=(1+\lambda t{)}^{\frac{x}{\lambda }}=\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(x\right)}_{n,\lambda }}{n!}{t}^n,e_{\lambda }\left(t\right)=e_{\lambda }^1\left(t\right),$(2)

(See [2–10,12–14]),

where

$(x{)}_{0,\lambda }=1,(x{)}_{n,\lambda }=x(x-\lambda )(x-2\lambda )\cdots (x-(n-1)\lambda ),(n\ge 1).$

Note that $\underset{\lambda \to 0}{lim}{e}_{\lambda }^x(t)=e^{xt}$.

In [3], the degenerate Stirling numbers of the second kind are defined by

(3) ${\left(x\right)}_{n,\lambda }=\sum _{k=0}^n{\left\{\begin{array}{c}n\\ k\end{array}\right\}}_{\lambda }{\left(x\right)}_k,\left(n\ge 0\right),$(3)

(See [2,4,6,14]),

where

$(x{)}_0=1,(x{)}_n=x(x-1)\cdots (x-n+1),(n\ge 1).$

Note that $\underset{\lambda \to 0}{lim}{\left\{\begin{array}{c}n\\ k\end{array}\right\}}_{\lambda }=\left\{\begin{array}{c}n\\ k\end{array}\right\}$. Here $\left\{\begin{array}{c}n\\ k\end{array}\right\}$ are the ordinary Stirling numbers of the second kind defined by

$x^n=\sum _{k=0}^n\left\{\begin{array}{c}n\\ k\end{array}\right\}{\left(x\right)}_k,\left(n\ge 0\right),$

(see [15–17]).

The $\lambda$-analogues of the Stirling numbers of the first kind are given by

(4) ${\left(x\right)}_{n,\lambda }=\sum _{k=0}^n{S}_{1,\lambda }\left(n,k\right)x^k,\left(n\ge 0\right),$(4)

(see [4,5,8]).

Note that ${lim}_{\lambda \to 1}{S}_{1,\lambda }(n,k)=S_1(n,k)$. Note that $S_1(n,k)$ are the ordinary Stirling numbers of the first kind given by

${\left(x\right)}_n=\sum _{k=0}^n{S}_1\left(n,k\right)x^k$

(See [1–30]).

We recall that the degenerate Stirling polynomials of the second kind are defined by

(5) ${\left(x+y\right)}_{n,\lambda }=\sum _{k=0}^n{S}_{2,\lambda }(n,k|x)(y{)}_k,\left(n\ge 0\right),$(5)

(see [2,3]).

When $x=0$, $S_{\lambda }(n,k|0)={\left\{\begin{array}{c}n\\ k\end{array}\right\}}_{\lambda },(n,k\ge 0)$. From (5), we note that

(6) $\frac{1}{k!}{e}_{\lambda }^x\left(t\right)(e_{\lambda }\left(t\right)-1{)}^k=\sum _{n=k}^{\mathrm{\infty }}{S}_{2,\lambda }(n,k|x)\frac{t^n}{n!},$(6)

(see [2,3]).

It is known that the Hermite polynomials are given by

(7) $e^{xt-\frac{1}{2}{t}^2}=\sum _{n=0}^{\mathrm{\infty }}{H}_n\left(x\right)\frac{t^n}{n!},$(7)

(see [15,17,20]).

Thus, by (7), we get

(8) $H_n\left(x\right)=n!\sum _{m=0}^{\left[\frac{n}{2}\right]}\frac{{\left(-1\right)}^m{x}^{n-2m}}{m!(n-2m)!2^m},$(8)

(see [17]).

The Cauchy numbers (of the first kind) of order $\alpha$ are defined by

(9) ${\left(\frac{t}{log\left(1+t\right)}\right)}^{\alpha }=\sum _{n=0}^{\mathrm{\infty }}{C}_n^{\left(\alpha \right)}\frac{t^n}{n!},$(9)

(see [16,17]).

Assume that $Y$ is a random variable such that the moment generating function of $Y,E[e^{tY}]={\sum }_{n=0}^{\mathrm{\infty }}E[Y^n]\frac{t^n}{n!}$, $(|t|<r)$, exists for some $r>0$. Let $(Y_j{)}_{j\ge 1}$ be a sequence of mutually independent copies of the random variable $Y$, and let $S_k=Y_1+\cdots +Y_k,(k\ge 1)$, with $S_0=0$.

We recall that the probabilistic Stirling polynomials of the second kind associated with $Y$, $S_Y(n,k|x)$, are defined by (see [2,11])

(10) $S_Y(n,k|x)=\frac{1}{k!}\sum _{j=0}^k\left(\begin{array}{c}k\\ j\end{array}\right)(-1{)}^{k-j}E[(S_j+x{)}^n].$(10)

Equivalently, they are given by

(11) $\frac{1}{k!}{e}^{xt}(E[e^{tY}]-1{)}^k=\sum _{n=k}^{\mathrm{\infty }}{S}_Y(n,k|x)\frac{t^n}{n!}.$(11)

2. Probabilistic degenerate Stirling polynomials of the second kind and their applications

Throughout this section, we assume that $Y$ is a random variable such that the moment generating function of $Y,E[e^{tY}]=\sum _{n=0}^{\mathrm{\infty }}E[Y^n]\frac{t^n}{n!}$, $(|t|<r)$, exists for some $r>0$. We let $(Y_j{)}_{j\ge 1}$ be a sequence of mutually independent copies of the random variable $Y$, and let

(12) $S_0=0,S_k=Y_1+Y_2+\cdots +Y_k,(k\in \mathbb{N}).$(12)

We recall from [2] that the probabilistic degenerate Stirling polynomials of the second kind associated with the random variable $Y$ are defined by

(13) $\frac{1}{k!}{e}_{\lambda }^x(t)(E[e_{\lambda }^Y(t)]-1{)}^k=\sum _{n=k}^{\mathrm{\infty }}{S}_{2,\lambda }^Y(n,k|x)\frac{t^n}{n!},(n\ge 0).$(13)

The following theorem is shown in [2].

Theorem 1.

([2]) For $n,k\ge 0$, we have

$\frac{1}{k!}\sum _{j=0}^k\left(\begin{array}{c}k\\ j\end{array}\right)(-1{)}^{k-j}E[(S_j+x{)}_{n,\lambda }]=\left\{\begin{array}{cc}{S}_{2,\lambda }^Y(n,k|x),& ifn\ge k,\\ 0,& otherwise.\end{array}\right.$

Note that ${lim}_{\lambda \to 0}{S}_{2,\lambda }^Y(n,k|x)=S_Y(n,k|x)=\frac{1}{k!}\sum _{j=0}^k\left(\begin{array}{c}k\\ j\end{array}\right)(-1{)}^{k-j}E[(S_j+x{)}^n],(n\ge k).$

The forward difference operator $\mathrm{\Delta }$ is defined by $\mathrm{\Delta }f(x)=f(x+1)-f(x)$. From this, we see that

(14) ${\mathrm{\Delta }}^{m}f\left(x\right)=\sum _{k=0}^m\left(\begin{array}{c}m\\ k\end{array}\right){\left(-1\right)}^{m-k}f\left(x+k\right),\left(m\ge 0\right),$(14)

(see [2,4,5,11]).

It is easy to show that

(15) $(1+\mathrm{\Delta }{)}^{m}f(x)=\sum _{k=0}^m\left(\begin{array}{c}m\\ k\end{array}\right){\mathrm{\Delta }}^{k}f(x)=f(x+m),(m\ge 0).$(15)

Thus, by (15), we get

(16) $\sum _{k=0}^N(x+k{)}_{n,\lambda }=\sum _{k=0}^N(1+\mathrm{\Delta }{)}^k(x{)}_{n,\lambda }=\sum _{k=0}^N\sum _{m=0}^k\left(\begin{array}{c}k\\ m\end{array}\right){\mathrm{\Delta }}^m(x{)}_{n,\lambda }$(16)

$=\sum _{m=0}^N\left(\sum _{k=m}^N\left(\begin{array}{c}k\\ m\end{array}\right){\mathrm{\Delta }}^m(x{)}_{n,\lambda }\right)=\sum _{m=0}^N{\mathrm{\Delta }}^m(x{)}_{n,\lambda }\sum _{k=m}^N\left(\left(\begin{array}{c}k+1\\ m+1\end{array}\right)-\left(\begin{array}{c}k\\ m+1\end{array}\right)\right)$

$=\sum _{m=0}^N\left(\begin{array}{c}N+1\\ m+1\end{array}\right){\mathrm{\Delta }}^m(x{)}_{n,\lambda },$

where $N$ is a nonnegative integer.

From (5) and (16), we note that

(17) $\sum _{k=0}^N(x+k{)}_{n,\lambda }=\sum _{k=0}^N\sum _{m=0}^n{S}_{2,\lambda }(n,m|x)m!\left(\begin{array}{c}k\\ m\end{array}\right)$(17)

$=\sum _{m=0}^n{S}_{2,\lambda }(n,m|x)m!\sum _{k=0}^N\left(\begin{array}{c}k\\ m\end{array}\right)=\sum _{m=0}^n\left(\begin{array}{c}N+1\\ m+1\end{array}\right)m!S_{2,\lambda }(n,m|x).$

Thus, by (16) and (17), we obtain the following theorem.

Theorem 2.

For $n,m\in \mathbb{Z}$ with $n\ge m\ge 0$, we have

$\sum _{m=0}^N\left(\begin{array}{c}N+1\\ m+1\end{array}\right){\mathrm{\Delta }}^m(x{)}_{n,\lambda }=\sum _{m=0}^n\left(\begin{array}{c}N+1\\ m+1\end{array}\right)m!S_{2,\lambda }(n,m|x).$

Now, we define the operator ${\mathrm{\Delta }}_y$ as

(18) ${\mathrm{\Delta }}_{y}f(x)=f(x+y)-f(x),{\mathrm{\Delta }}_{y_1,y_2,\dots ,y_m}f(x)={\mathrm{\Delta }}_{y_1}\circ {\mathrm{\Delta }}_{y_2}\circ {\mathrm{\Delta }}_{y_3}\circ \cdots \circ {\mathrm{\Delta }}_{y_m}f(x),$(18)

where $y\in \mathbb{R},and\left(y_1,y_2,\dots ,y_m\right)\in {\mathbb{R}}^m,$ see [11] Let

$I_m(k)=\{(i_1,i_2,\dots ,i_k)\in \{1,2,\dots ,m\}|i_r\ne i_s,r\ne s\}.$

Then, we can show that

(19) ${\mathrm{\Delta }}_{y_1,y_2,\dots ,y_m}f\left(x\right)=(-1{)}^{m}f\left(x\right)+\sum _{k=1}^m{\left(-1\right)}^{m-k}\sum _{I_m\left(k\right)}f\left(x+y_{i_1}+\cdots +y_{i_k}\right),$(19)

(see [11]).

Assume that $(U_i{)}_{i\ge 1}$ are mutually independent uniform random variables on $(0,1)$, and that $(U_i{)}_{i\ge 1}$ and $(Y_i{)}_{i\ge 1}$ are mutually independent. From (19), viewing $e_{\lambda }^x(t)$ as a function of $x$ we note that

(20) ${\mathrm{\Delta }}_{y_1,y_2,\dots ,y_m}{e}_{\lambda }^x(t)={\mathrm{\Delta }}_{y_1,y_2,\dots ,y_{m-1}}{e}_{\lambda }^x(t)(e_{\lambda }^{y_m}(t)-1)={\mathrm{\Delta }}_{y_1,\dots ,y_{m-2}}{e}_{\lambda }^x(t)(e_{\lambda }^{y_m}(t)-1)(e_{\lambda }^{y_{m-1}}(t)-1)$(20)

$=\cdots =(e_{\lambda }^{y_m}(t)-1)(e_{\lambda }^{y_{m-1}}-1)\cdots (e_{\lambda }^{y_1}(t)-1)e_{\lambda }^x(t).$

Now, we observe that

(21) $E[e_{\lambda }^{x+y_1{U}_1+y_2{U}_2}(t)]=e_{\lambda }^x(t){\int }_0^1{\int }_0^1{e}_{\lambda }^{x+y_1{t}_1+y_2{t}_2}(t)d{t}_{1}d{t}_2$(21)

$={\int }_0^1{e}_{\lambda }^{y_2{t}_2}(t)\frac{\lambda }{y_{1}log(1+\lambda t)}(e_{\lambda }^{y_1+x}(t)-e_{\lambda }^x(t))d{t}_2$

$=\frac{\lambda }{y_{1}log(1+\lambda t)}{\mathrm{\Delta }}_{y_1}{e}_{\lambda }^x(t){\int }_0^1{e}_{\lambda }^{y_2{t}_2}(t)d{t}_2$

$=\frac{\lambda }{y_{1}log(1+\lambda t)}\frac{\lambda }{y_{2}log(1+\lambda t)}{\mathrm{\Delta }}_{y_1}{e}_{\lambda }^x(t)(e_{\lambda }^{y_2}(t)-1)$

$=\frac{1}{y_1{y}_2}{\left(\frac{\lambda }{log(1+\lambda t)}\right)}^2{\mathrm{\Delta }}_{y_1}(e_{\lambda }^{x+y_2}(t)-e_{\lambda }^x(t))$

$=\frac{1}{y_1{y}_2}{\left(\frac{\lambda }{log(1+\lambda t)}\right)}^2{\mathrm{\Delta }}_{y_1,y_2}{e}_{\lambda }^x(t).$

Thus, by (21), we get

${\mathrm{\Delta }}_{y_1,y_2}{e}_{\lambda }^x(t)=y_1{y}_2{\left(\frac{log(1+\lambda t)}{\lambda }\right)}^{2}E[e_{\lambda }^{x+y_1{U}_1+y_2{U}_2}(t)].$

Continuing this process, we have

(22) ${\mathrm{\Delta }}_{y_1,y_2,\dots ,y_m}{e}_{\lambda }^x(t)=y_1{y}_2\cdots y_m{\left(\frac{log(1+\lambda t)}{\lambda }\right)}^{m}E[e_{\lambda }^{x+y_1{U}_1+\cdots +y_m{U}_m}(t)].$(22)

By (22), we get

(23) $\sum _{n=0}^{\mathrm{\infty }}{\mathrm{\Delta }}_{y_1,y_2,\dots ,y_m}(x{)}_{n,\lambda }\frac{t^n}{n!}$(23)

$=m!y_1{y}_2\cdots y_m\sum _{k=m}^{\mathrm{\infty }}{S}_1(k,m)\frac{{\lambda }^{k-m}}{k!}{t}^k\sum _{l=0}^{\mathrm{\infty }}E[(x+y_1{U}_1+\cdots +y_m{U}_m{)}_{l,\lambda }]\frac{t^l}{l!}$

$=\sum _{n=m}^{\mathrm{\infty }}(m!y_1{y}_2\cdots y_m\sum _{k=m}^n\left(\begin{array}{c}n\\ k\end{array}\right)S_1(k,m){\lambda }^{k-m}E[(x+y_1{U}_1+\cdots +y_m{U}_m{)}_{n-k,\lambda }])\frac{t^n}{n!}.$

Therefore, by comparing the coefficients on both sides of (23), we obtain the following theorem.

Theorem 3.

For $n,m\in \mathbb{Z}$ with $n\ge m\ge 0$, we have

$\frac{1}{m!}{\mathrm{\Delta }}_{y_1,y_2,\dots ,y_m}(x{)}_{n,\lambda }=y_1{y}_2\cdots y_m\sum _{k=m}^n\left(\begin{array}{c}n\\ k\end{array}\right)S_1(k,m){\lambda }^{k-m}E[(x+y_1{U}_1+\cdots +y_m{U}_m{)}_{n-k,\lambda }].$

From (19), we note that

(24) $E\left[{\mathrm{\Delta }}_{Y_1,Y_2,\dots ,Y_m}f\left(x\right)\right]=\sum _{k=0}^m\left(\begin{array}{c}m\\ k\end{array}\right){\left(-1\right)}^{m-k}E\left[f\left(x+S_k\right)\right],$(24)

(see [11]).

Let $f(x)=(x{)}_{n,\lambda },(n\ge 0)$. Then, from (24), we have

(25) $E[{\mathrm{\Delta }}_{Y_1,Y_2,\dots ,Y_m}(x{)}_{n,\lambda }]=\sum _{k=0}^m\left(\begin{array}{c}m\\ k\end{array}\right)(-1{)}^{m-k}E[(x+S_k{)}_{n,\lambda }].$(25)

From (13), we have

(26) $\sum _{n=m}^{\mathrm{\infty }}{S}_{2,\lambda }^Y(n,m|x)\frac{t^n}{n!}=\frac{1}{m!}{e}_{\lambda }^x(t)(E[e_{\lambda }^Y(t)]-1{)}^m$(26)

$=\frac{1}{m!}{e}_{\lambda }^x(t)E[e_{\lambda }^{Y_1}(t)-1]\cdots E[e_{\lambda }^{Y_m}(t)-1]$

$=\frac{1}{m!}E[(e_{\lambda }^{Y_1}(t)-1)\cdots (e_{\lambda }^{Y_m}(t)-1)e_{\lambda }^x(t)]$

$=\frac{1}{m!}E[{\mathrm{\Delta }}_{Y_1,\dots ,Y_m}{e}_{\lambda }^x(t)]=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{m!}E[{\mathrm{\Delta }}_{Y_1,\dots ,Y_m}(x{)}_{n,\lambda }]\frac{t^n}{n!}.$

Thus, by comparing the coefficients on both sides of (26), we get

(27) $\frac{1}{m!}E[{\mathrm{\Delta }}_{Y_1,Y_2,\dots ,Y_m}(x{)}_{n,\lambda }]=S_{2,\lambda }^Y(n,m|x),(n\ge m\ge 0).$(27)

Now, we observe that

(28) $e_{\lambda }^x(t)E[Y_1{Y}_2\cdots Y_m{e}_{\lambda }^{Y_1{U}_1+\cdots +Y_m{U}_m}(t)]$(28)

$=e_{\lambda }^x(t)(E[e_{\lambda }^{Y_1}(t)]-1)(E[e_{\lambda }^{Y_2}(t)]-1)\cdots (E[e_{\lambda }^{Y_m}(t)]-1)(\frac{\lambda }{log(1+\lambda t)}{)}^m$

$=e_{\lambda }^x(t)(E[e_{\lambda }^Y(t)]-1{)}^m(\frac{\lambda }{log(1+\lambda t)}{)}^m.$

Thus, by (13) and (28), we get

(29) $\sum _{n=m}^{\mathrm{\infty }}{S}_{2,\lambda }^Y(n,m|x)\frac{t^n}{n!}=\frac{1}{m!}{e}_{\lambda }^x(t)(E[e_{\lambda }^Y(t)]-1{)}^m$(29)

$=\frac{1}{{\lambda }^m}\frac{1}{m!}(log(1+\lambda t){)}^m{e}_{\lambda }^x(t)E[Y_1{Y}_2\cdots Y_m{e}_{\lambda }^{Y_1{U}_1+\cdots +Y_m{U}_m}(t)]$

$=\sum _{k=m}^{\mathrm{\infty }}{S}_1(k,m){\lambda }^{k-m}\frac{t^k}{k!}\sum _{l=0}^{\mathrm{\infty }}E[Y_1{Y}_2\cdots Y_m(x+Y_1{U}_1+\cdots +Y_m{U}_m{)}_{l,\lambda }]\frac{t^l}{l!}$

$=\sum _{n=m}^{\mathrm{\infty }}\sum _{k=m}^n{S}_1(k,m){\lambda }^{k-m}\left(\begin{array}{c}n\\ k\end{array}\right)E[Y_1{Y}_2\cdots Y_m(x+Y_1{U}_1+\cdots +Y_m{U}_m{)}_{n-k,\lambda }]\frac{t^n}{n!}.$

From (3), we note that

(30) $(x+S_k{)}_{n,\lambda }=\sum _{j=0}^n{\left\{\begin{array}{c}n\\ j\end{array}\right\}}_{\lambda }(x+S_k{)}_j,(n\ge 0).$(30)

From (25), (27) and (30), we get

(31) $S_{2,\lambda }^Y(n,m|x)=\frac{1}{m!}\sum _{j=0}^n{\left\{\begin{array}{c}n\\ j\end{array}\right\}}_{\lambda }\sum _{k=0}^m\left(\begin{array}{c}m\\ k\end{array}\right)(-1{)}^{m-k}E[(x+S_k{)}_j].$(31)

Therefore, by (25), (27), (29) and (31), we obtain the following theorem.

Theorem 4.

For $n,m\in \mathbb{Z}$ with $n\ge m\ge 0$, we have

$S_{2,\lambda }^Y(n,m|x)=\frac{1}{m!}E[{\mathrm{\Delta }}_{Y_1,Y_2,\dots ,Y_m}(x{)}_{n,\lambda }]$

$=\sum _{k=m}^n\left(\begin{array}{c}n\\ k\end{array}\right)S_1(k,m){\lambda }^{k-m}E[Y_1\cdots Y_m(x+Y_1{U}_1+\cdots +Y_m{U}_m{)}_{n-k,\lambda }]$

$=\frac{1}{m!}\sum _{k=0}^m\left(\begin{array}{c}m\\ k\end{array}\right)(-1{)}^{m-k}E[(x+S_k{)}_{n,\lambda }]$

$=\frac{1}{m!}\sum _{j=0}^n{\left\{\begin{array}{c}n\\ j\end{array}\right\}}_{\lambda }\sum _{k=0}^m\left(\begin{array}{c}m\\ k\end{array}\right)(-1{)}^{m-k}E[(x+S_k{)}_j].$

Taking $Y=1$, we obtain the following corollary.

Corollary 5.

For $n,m\in \mathbb{Z}$ with $n\ge m\ge 0$, we have

$S_{2,\lambda }(n,m|x)=\frac{1}{m!}{\mathrm{\Delta }}^m(x{)}_{n,\lambda }$

$=\sum _{k=m}^n\left(\begin{array}{c}n\\ k\end{array}\right)S_1(k,m){\lambda }^{k-m}E[(x+U_1+\cdots +U_m{)}_{n-k,\lambda }]$

$=\frac{1}{m!}\sum _{k=0}^m\left(\begin{array}{c}m\\ k\end{array}\right)(-1{)}^{m-k}(x+k{)}_{n,\lambda }$

$=\frac{1}{m!}\sum _{j=0}^n{\left\{\begin{array}{c}n\\ j\end{array}\right\}}_{\lambda }\sum _{k=0}^m\left(\begin{array}{c}m\\ k\end{array}\right)(-1{)}^{m-k}(x+k{)}_j.$

As is well known, the binomial inversion is given by

(32) $a_k=\sum _{m=0}^k\left(\begin{array}{c}k\\ m\end{array}\right)b_m\iff b_k=\sum _{m=0}^k(-1{)}^{k-m}\left(\begin{array}{c}k\\ m\end{array}\right)a_m.$(32)

Then, from (32) and (25), we have

(33) $E[(x+S_k{)}_{n,\lambda }]=\sum _{m=0}^k\left(\begin{array}{c}k\\ m\end{array}\right)E[{\mathrm{\Delta }}_{Y_1,Y_2,\dots ,Y_m}(x{)}_{n,\lambda }].$(33)

Thus, by (33) and Theorem 4, we get

(34) $\sum _{k=0}^{N}E[(x+S_k{)}_{n,\lambda }]=\sum _{k=0}^N\sum _{m=0}^k\left(\begin{array}{c}k\\ m\end{array}\right)E[{\mathrm{\Delta }}_{Y_1,\dots ,Y_m}(x{)}_{n,\lambda }]$(34)

$=\sum _{m=0}^{N}E[{\mathrm{\Delta }}_{Y_1,\dots ,Y_m}(x{)}_{n,\lambda }]\sum _{k=m}^N\left(\begin{array}{c}k\\ m\end{array}\right)=\sum _{m=0}^N\left(\begin{array}{c}N+1\\ m+1\end{array}\right)E[{\mathrm{\Delta }}_{Y_1,\dots ,Y_m}(x{)}_{n,\lambda }]$

$=\sum _{m=0}^{N}m!S_{2,\lambda }^Y(n,m|x)\left(\begin{array}{c}N+1\\ m+1\end{array}\right).$

Therefore, by (34) and noting ${\mathrm{\Delta }}_{Y_1,\dots ,Y_m}(x{)}_{n,\lambda }=0$, for $n<m$ (see (23)), we obtain the following theorem.

Theorem 6.

For $N\ge 0$, we have

$\sum _{k=0}^{N}E[(x+S_k{)}_{n,\lambda }]=\sum _{m=0}^{min\{N,n\}}m!S_{2,\lambda }^Y(n,m|x)\left(\begin{array}{c}N+1\\ m+1\end{array}\right).$

For $\alpha \in \mathbb{R}$, we recall that the degenerate $\alpha$-Whitney polynomials of the second kind are given by

(35) $\sum _{n=m}^{\mathrm{\infty }}{W}_{\alpha ,\lambda }(n,m|x)\frac{t^n}{n!}=\frac{e_{\lambda }^x\left(t\right)}{m!}{\left(\frac{e_{\lambda }^{\alpha }\left(t\right)-1}{\alpha }\right)}^m,$(35)

(see [4,5]).

Thus, by (35), we get

(36) $\sum _{n=m}^{\mathrm{\infty }}{W}_{\alpha ,\lambda }(n,m|x)\frac{t^n}{n!}=\frac{e_{\lambda }^x}{m!}\frac{1}{{\alpha }^m}\sum _{k=0}^m\left(\begin{array}{c}m\\ k\end{array}\right)(-1{)}^{m-k}{e}_{\lambda }^{\alpha k}(t)$(36)

$=\sum _{n=0}^{\mathrm{\infty }}(\frac{1}{{\alpha }^m}\frac{1}{m!}\sum _{k=0}^m\left(\begin{array}{c}m\\ k\end{array}\right)(-1{)}^{m-k}(\alpha k+x{)}_{n,\lambda })\frac{t^n}{n!}.$

Comparing the coefficients on both sides of (36), we have

(37) ${\alpha }^m{W}_{\alpha ,\lambda }(n,m|x)=\frac{1}{m!}\sum _{k=0}^m\left(\begin{array}{c}m\\ k\end{array}\right)(-1{)}^{m-k}(\alpha k+x{)}_{n,\lambda },(n\ge m\ge 0).$(37)

Let $Y=\alpha$ in Theorem 1. Then, from (37), we have

(38) $S_{2,\lambda }^Y(n,m|x)=\frac{1}{m!}\sum _{k=0}^m\left(\begin{array}{c}m\\ k\end{array}\right)(-1{)}^{m-k}(\alpha k+x{)}_{n,\lambda }$(38)

$={\alpha }^m{W}_{\alpha ,\lambda }(n,m|x),(n\ge m\ge 0).$

Theorem 7.

Let $Y=\alpha$. For $n\ge m\ge 0$, we have

$S_{2,\lambda }^Y(n,m|x)={\alpha }^m{W}_{\alpha ,\lambda }(n,m|x).$

3. Applications to several random variables

In this section, we apply the results in the previous section to the Bernoulli random variable with probability of success $p$, the normal random variable with parameters $\mu =0,{\sigma }^2=1$, and to the uniform random variable on $(0,1)$.

Let $Y$ be the Bernoulli random variable with probability of success $p$. Then we have

(39) $E[e_{\lambda }^Y(t)]-1=1-p+p{e}_{\lambda }(t)-1=p(e_{\lambda }(t)-1).$(39)

Thus, by (39), we have

(40) $\sum _{n=m}^{\mathrm{\infty }}{S}_{2,\lambda }^Y(n,m|x)\frac{t^n}{n!}=\frac{e_{\lambda }^x(t)}{m!}(E[e_{\lambda }^Y(t)]-1{)}^m$(40)

$=p^m\frac{1}{m!}{e}_{\lambda }^x(t)(e_{\lambda }(t)-1{)}^m$

$=\sum _{n=m}^{\mathrm{\infty }}{p}^m{S}_{2,\lambda }(n,m|x)\frac{t^n}{n!}.$

Therefore, by (40), we obtain the following theorem.

Theorem 8.

Let $Y$ be the Bernoulli random variable with probability of success $p$. Then we have

$S_{2,\lambda }^Y(n,m|x)=p^m{S}_{2,\lambda }(n,m|x),(n\ge m\ge 0).$

Let $Y$ be the normal random variable with parameters $\mu \in \mathbb{R},{\sigma }^2\in {\mathbb{R}}_{>0}$, which is denoted by $Y\sim N(\mu ,{\sigma }^2)$. Then the probability density function of $Y$ is given by

(41) $\rho \left(x\right)=\frac{1}{\sqrt{2\pi }\sigma }{e}^{-{\left(x-\mu \right)}^2/2{\sigma }^2},x\in \left(-\mathrm{\infty },\mathrm{\infty }\right),$(41)

(see [27,28]).

By simple calculation, we easily get

(42) ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{y}^{2n+1}{e}^{-\frac{y^2}{2}}dy=0,\mathrm{a}\mathrm{n}\mathrm{d}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{y}^{2n}{e}^{-\frac{y^2}{2}}dy=\sqrt{2\pi }1\cdot 3\cdot 5\cdots (2n-1),$(42)

where $n$ is a nonnegative integer.

Let $i=\sqrt{-1}$. For $Y\sim N(0,1)$, by (8), we get

(43) $E[(x+iY{)}^n]=\frac{1}{\sqrt{2\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}(x+iy{)}^n{e}^{-\frac{y^2}{2}}dy=\frac{1}{\sqrt{2\pi }}\sum _{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)x^{n-k}{i}^k{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{y}^k{e}^{-\frac{y^2}{2}}dy$(43)

$=\sum _{k=0}^{[\frac{n}{2}]}\left(\begin{array}{c}n\\ 2k\end{array}\right)x^{n-2k}\frac{{(-1)}^k}{\sqrt{2\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{y}^{2k}{e}^{-\frac{y^2}{2}}dy+i\sum _{k=0}^{[\frac{n-1}{2}]}\left(\begin{array}{c}n\\ 2k+1\end{array}\right)x^{n-2k-1}\frac{{(-1)}^k}{\sqrt{2\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{y}^{2k+1}{e}^{-\frac{y^2}{2}}dy$

$=\sum _{k=0}^{[\frac{n}{2}]}\left(\begin{array}{c}n\\ 2k\end{array}\right)x^{n-2k}\frac{{(-1)}^k}{\sqrt{2\pi }}\sqrt{2\pi }(2k-1)(2k-3)\cdots 3\cdot 1$

$=\sum _{k=0}^{[\frac{n}{2}]}\frac{{(-1)}^{k}n!}{(2k)!(n-2k)!}\frac{(2k)!}{(2k)2(k-1)\cdots 2}{x}^{n-2k}=n!\sum _{k=0}^{[\frac{n}{2}]}\frac{{(-1)}^k{x}^{n-2k}}{(n-2k)!k{!2}^k}=H_n(x),$

where $n$ is a positive integer.

From (43), we note that

(44) $E[(x+iY{)}_{n,\lambda }]=\sum _{k=0}^n{S}_{1,\lambda }(n,k)E\left[(x+iY{)}^k\right]$(44)

$=\sum _{k=0}^n{S}_{1,\lambda }(n,k)H_k(x),(n\ge 0).$

Therefore, by (44), we obtain the following theorem.

Theorem 9.

For $Y\sim N(0,1)$, and $n\ge 0$, we have

$E[(x+iY{)}^n]=H_n(x),E[(x+iY{)}_{n,\lambda }]=\sum _{k=0}^n{S}_{1,\lambda }(n,k)H_k(x),$

where $i=\sqrt{-1}$, and $S_{1,\lambda }(n,k)$ are the $\lambda$-analogues of the Stirling numbers of the first kind.

Let $Y$ be the uniform random variable on $(0,1)$. Then we have

(45) $(E[e_{\lambda }^Y(t)]{)}^k=(\frac{\lambda t}{log(1+\lambda t)}{)}^k\frac{k!}{t^k}\frac{1}{k!}(e_{\lambda }(t)-1{)}^k$(45)

$=\sum _{l=0}^{\mathrm{\infty }}{C}_l^{(k)}{\lambda }^l\frac{t^l}{l!}\sum _{m=0}^{\mathrm{\infty }}{\left\{\begin{array}{c}m+k\\ k\end{array}\right\}}_{\lambda }\frac{k!m!}{(m+k)!}\frac{t^m}{m!}$

$=\sum _{l=0}^{\mathrm{\infty }}{C}_l^{(k)}{\lambda }^l\frac{t^l}{l!}\sum _{m=0}^{\mathrm{\infty }}\frac{{\left\{\begin{array}{c}m+k\\ k\end{array}\right\}}_{\lambda }}{\left(\begin{array}{c}m+k\\ k\end{array}\right)}\frac{t^m}{m!}$

$=\sum _{n=0}^{\mathrm{\infty }}\sum _{m=0}^n\left(\begin{array}{c}n\\ m\end{array}\right){\lambda }^{n-m}{C}_{n-m}^{(k)}\frac{{\left\{\begin{array}{c}m+k\\ k\end{array}\right\}}_{\lambda }}{\left(\begin{array}{c}m+k\\ m\end{array}\right)}\frac{t^n}{n!}.$

On the other hand, by (12), we get

(46) $(E[e_{\lambda }^Y(t)]{)}^k=E[e_{\lambda }^{Y_1}(t)]E[e_{\lambda }^{Y_2}(t)]\cdots E[e_{\lambda }^{Y_k}(t)]$(46)

$=E[e_{\lambda }^{Y_1+\cdots +Y_k}(t)]=E[e_{\lambda }^{S_k}(t)]=\sum _{n=0}^{\mathrm{\infty }}E[(S_k{)}_{n,\lambda }]\frac{t^n}{n!}.$

From Theorem 1, we note that

(47) $S_{2,\lambda }^Y(n,m)=\frac{1}{m!}\sum _{k=0}^m\left(\begin{array}{c}m\\ k\end{array}\right)(-1{)}^{m-k}E[(S_k{)}_{n,\lambda }],(n\ge m\ge 0),$(47)

where $S_{2,\lambda }^Y(n,m)=S_{2,\lambda }^Y(n,m|0)$.

By (45), (46) and (47), we get

(48) $S_{2,\lambda }^Y(n,m)=\frac{1}{m!}\sum _{k=0}^m\left(\begin{array}{c}m\\ k\end{array}\right)(-1{)}^{m-k}\sum _{j=0}^n\left(\begin{array}{c}n\\ j\end{array}\right){\lambda }^{n-j}{C}_{n-j}^{(k)}\frac{{\left\{\begin{array}{c}j+k\\ k\end{array}\right\}}_{\lambda }}{\left(\begin{array}{c}j+k\\ k\end{array}\right)}$(48)

$=\sum _{j=0}^n{\lambda }^{n-j}\frac{n!}{(m+j)!(n-j)!}\sum _{k=0}^m\left(\begin{array}{c}m+j\\ j+k\end{array}\right)(-1{)}^{m-k}{\left\{\begin{array}{c}j+k\\ k\end{array}\right\}}_{\lambda }{C}_{n-j}^{(k)}.$

Therefore, by (48), we obtain the following theorem.

Theorem 10.

Let $Y$ be the uniform random variable on $(0,1)$. For $n,m\in \mathbb{Z}$ with $n\ge m\ge 0$, we have

$S_{2,\lambda }^Y(n,m)=\sum _{j=0}^n{\lambda }^{n-j}\frac{n!}{(m+j)!(n-j)!}\sum _{k=0}^m\left(\begin{array}{c}m+j\\ j+k\end{array}\right)(-1{)}^{m-k}{\left\{\begin{array}{c}j+k\\ k\end{array}\right\}}_{\lambda }{C}_{n-j}^{(k)}.$

Remark. Recently, new applicable research results related to this paper were published. Researchers interested in such studies should refer to references [31,32].

4. Conclusion

In this paper, we further studied by using generating functions probabilistic degenerate Stirling polynomials of the second associated with $Y$ as degenerate versions of the probabilistic Stirling polynomials of the second associated with $Y$ (see [11]). Here $Y$ is a random variable satisfying the moment condition in (1). In more detail, we derived several explicit expressions for $S_{2,\lambda }^Y(n,k|x)$ (see Theorems 1, 4, 7) and some related finite sum identities (see Theorems 2, 3, 6). Furthermore, we applied our results to the Bernoulli random variable with probability of success $p$, the normal random variable with parameters $\mu =0,{\sigma }^2=1$, and to the uniform random variable on $(0,1)$.

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

Disclosure statement

No potential conflict of interest was reported by the author(s).