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\title{\bf A simple proof of an identity of Lacasse}

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% the address should include the country, and does not have to include
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\author{Yidong Sun \\
\small Department of Mathematics, \\[-0.8ex]
\small Dalian Maritime University, 116026 Dalian, P.R. China\\[-0.8ex]
%\small North Kentucky, U.S.A.\\
\small\tt sydmath@yahoo.com.cn \\
%\and
%Forgotten Second Author \qquad  Forgotten Third Author\\
%\small School of Hard Knocks\\[-0.8ex]
%\small University of Western Nowhere\\[-0.8ex]
%\small Nowhere, Australasiaopia\\
%\small\tt \{fsa,fta\}@uwn.edu.ao
}

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% \small Mathematics Subject Classifications: comma separated list of
% MSC codes available from http://www.ams.org/mathscinet/freeTools.html}

\date{\dateline{Jan 5, 2013}{Apr 9, 2013}{Apr 17, 2013}\\
\small Mathematics Subject Classifications: 05A15, 05A19.}

\begin{document}

\maketitle

\begin{abstract}
In this note, using the derangement polynomials and their umbral
representation, we give a simple proof of an identity conjectured by
Lacasse in the study of the PAC-Bayesian machine learning theory.

  % keywords are optional
  \bigskip\noindent \textbf{Keywords:} Derangement polynomial; Umbral operator.
\end{abstract}

\section{Introduction}


In his thesis \cite{Lacasse}, Lacasse introduced the functions
$\xi(n)$ and $\xi_2(n)$ in the study of the PAC-Bayesian machine
learning theory, where
\begin{eqnarray*}
\xi(n)  \hskip-.22cm &=&\hskip-.22cm \sum_{k=0}^{n}\binom{n}{k}\left(\frac{k}{n}\right)^k\left(1-\frac{k}{n}\right)^{n-k}, \\
\xi_2(n)\hskip-.22cm &=&\hskip-.22cm
\sum_{k=0}^{n}\sum_{j=0}^{n-k}\binom{n}{k}\binom{n-k}{j}\left(\frac{k}{n}\right)^k\left(\frac{j}{n}\right)^j\left(1-\frac{k}{n}-\frac{j}{n}\right)^{n-k-j}.
\end{eqnarray*}
Based on numerical verification, Lacasse presented the following
conjecture.
\begin{conjecture}
For any integer $n\geq 1$, there holds
\begin{eqnarray}\label{eqn 1.1}
\xi_2(n)\hskip-.22cm &=&\hskip-.22cm \xi(n)+n.
\end{eqnarray}
\end{conjecture}
Recently, by applying a multivariate Abel identity due to Hurwitz,
Younsi \cite{Younsi} gave an algebraic proof of this conjecture.
Later, using a decomposition of triply rooted trees into three
doubly rooted trees, Chen, Peng and Yang \cite{Chen} gave it a nice
combinatorial interpretation. A very short proof was also obtained
by Prodinger \cite{Prodinger}, based on the study of the tree
function, with links to Lambert's W-function and Ramanujan's
Q-function.

In this note, using the derangement polynomials and their umbral
representation, we provide another simple proof of (\ref{eqn 1.1}).


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{The derangement polynomials and the proof of (\ref{eqn 1.1})}

Recall that the derangement polynomials
$\{\mathcal{D}_{n}(\lambda)\}_{n\geq 0}$ are defined by
\begin{eqnarray}\label{eqn 2.1}
\mathcal{D}_{n}(\lambda) \hskip-.22cm &=&\hskip-.22cm
\sum_{k=0}^{n}\binom{n}{k}D_k\lambda^{n-k}.
\end{eqnarray}
where $\mathcal{D}_{n}(1)=n!$ and $\mathcal{D}_n(0)=D_n$ is the
$n$-th derangement number, counting permutations on
$[n]=\{1,2,\dots, n\}$ with no fixed points. The derangement
polynomials $\mathcal{D}_{n}(\lambda)$, also called
$\lambda$-factorials of $n$, have been considerably investigated by
Eriksen, Freij and W$\ddot{a}$stlund \cite{Eriksen}, Sun and Zhuang
\cite{SunZhuang}. They have a basic recursive relation
\cite{Eriksen} and an Abel-type formula \cite{SunZhuang},
\begin{eqnarray}
\mathcal{D}_{n}(\lambda+\mu) \hskip-.22cm &=&\hskip-.22cm \sum_{k=0}^{n}\binom{n}{k}\mathcal{D}_{k}(\lambda)\mu^{n-k}, \label{eqn 2.2}\\
\mathcal{D}_{n}(\lambda+\mu) \hskip-.22cm &=&\hskip-.22cm
\sum_{k=0}^{n}\binom{n}{k}(\lambda+k)^{k}(\mu-k-1)^{n-k}, \label{eqn
2.3}
\end{eqnarray}
and obey the following property \cite{SunZhuang},
\begin{eqnarray}\label{eqn 2.4}
\sum_{k=0}^{n}\binom{n}{k}\mathcal{D}_{k}(\lambda)\mathcal{D}_{n-k}(\mu+1)
\hskip-.22cm &=&\hskip-.22cm
(\lambda+\mu-1)^{n+1}+(n-\lambda-\mu+2)\mathcal{D}_n(\lambda+\mu).
\end{eqnarray}
Denote by $\mathbf{D}$ the umbral operator defined by
$\mathbf{D}^n=D_n$ for $n\geq 0$ (See \cite{Gesselb, Roman, RomRota}
for more information on the umbral calculus), then by (\ref{eqn
2.1}) $\mathcal{D}_{n}(\lambda)$ can be represented as
\begin{eqnarray*}
\mathcal{D}_{n}(\lambda)  \hskip-.22cm &=&\hskip-.22cm
(\mathbf{D}+\lambda)^n.
\end{eqnarray*}




Setting $\lambda=0, \mu=n+1$ in (\ref{eqn 2.4}), we have
\begin{eqnarray*}
n^{n+1}+\mathcal{D}_{n}(n+1)
\hskip-.22cm &=&\hskip-.22cm  \sum_{k=0}^{n}\binom{n}{k}\mathcal{D}_{k}(0)\mathcal{D}_{n-k}(n+2) \\
\hskip-.22cm &=&\hskip-.22cm  \sum_{k=0}^{n}\binom{n}{k}\mathcal{D}_{k}(0)\mu^{n-k}|_{\mu=(\mathbf{D}+n+2)} \\
\hskip-.22cm &=&\hskip-.22cm  \mathcal{D}_{n}(\mu)|_{\mu=(\mathbf{D}+n+2)}  \hskip2cm \mbox{by (\ref{eqn 2.2})}  \\
\hskip-.22cm &=&\hskip-.22cm  \sum_{k=0}^{n}\binom{n}{k}k^{k}(\mu-k-1)^{n-k}|_{\mu=(\mathbf{D}+n+2)}  \hskip1cm \mbox{by (\ref{eqn 2.3})}  \\
\hskip-.22cm &=&\hskip-.22cm
\sum_{k=0}^{n}\binom{n}{k}k^k\mathcal{D}_{n-k}(n-k+1),
\end{eqnarray*}
which proves (\ref{eqn 1.1}), if one notices that $\xi(n)$ and
$\xi_2(n)$, by (\ref{eqn 2.3}), can be rewritten as
\begin{eqnarray*}
\xi(n)=\frac{1}{n^n}\mathcal{D}_{n}(n+1) \hskip.2cm
\mbox{and}\hskip.2cm \xi_2(n)=
\frac{1}{n^n}\sum_{k=0}^{n}\binom{n}{k}k^k\mathcal{D}_{n-k}(n-k+1).
\end{eqnarray*}


\begin{remark}
By the nontrivial property of $\mathbf{D}$ \cite{SunZhuang},
\begin{eqnarray*}
(\mathbf{D}+\lambda)(\mathbf{D}+\lambda+n+1)^{n}\hskip-.22cm
&=&\hskip-.22cm (n+\lambda)^{n+1},
\end{eqnarray*}
one can get another expression for $\xi_2(n)$,
\begin{eqnarray*}
\xi_2(n) \hskip-.22cm &=&\hskip-.22cm  \xi(n)+n= \frac{1}{n^n}\big(\mathcal{D}_{n}(n+1)+n^{n+1}\big) \\
         \hskip-.22cm &=&\hskip-.22cm  \frac{1}{n^n}\big((\mathbf{D}+n+1)^n+\mathbf{D}(\mathbf{D}+n+1)^n \big) \\
         \hskip-.22cm &=&\hskip-.22cm  \frac{1}{n^n}\big((\mathbf{D}+1)(\mathbf{D}+n+1)^n \big) =\frac{1}{n^n}\sum_{k=0}^{n}\binom{n}{k}(\mathbf{D}+1)^{k+1}n^{n-k}  \\
         \hskip-.22cm &=&\hskip-.22cm  \frac{1}{n^n}\sum_{k=0}^{n}\binom{n}{k}\mathcal{D}_{k+1}(1)n^{n-k}=\frac{1}{n^n}\sum_{k=0}^{n}\binom{n}{k}(k+1)!n^{n-k}.
\end{eqnarray*}
This expression has been obtained by Younsi by using the Hurwitz
identity on multivariate Abel polynomials and plays a critical role
in his proof.
\end{remark}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{Acknowledgements}
The author is grateful to the referee for the helpful suggestions
and comments. The work was partially supported by The National
Science Foundation of China and by the Fundamental Research Funds
for the Central Universities (2012QN061 and 2012TD032).

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\begin{thebibliography}{10}


\bibitem{Chen} W.~Y.~C. Chen, J.~F.~F. Peng, and H.~R.~L. Yang.
\newblock Decomposition of triply rooted trees. 
\newblock {\em Electronic J. Combinatorics 20(2), $\#$P10}, 2013. 

\bibitem{Eriksen} N. Eriksen, R. Freij, and J. W$\ddot{a}$stlund. \newblock Enumeration of
derangements with descents in prescribed positions. \newblock {\em
Electronic J. Combinatorics 16(1), $\#$R32}, 2009.



\bibitem{Gesselb} I.~M. Gessel. \newblock Applications of the classical umbral calculus.
\newblock {\em Algebra Universalis, 49:397--434}, 2003.


\bibitem{Lacasse} A. Lacasse. \newblock  Bornes PAC-Bayes et algorithmes d'apprentissage. Ph.D.
Thesis, Universite Laval, Quebec, 2010.


\bibitem{Prodinger} H. Prodinger. \newblock  An identity conjectured by Lacasse via the tree
function. \newblock {\em arXiv:1301.3669}, 2013.

\bibitem{Roman} S. Roman. \newblock The Umbral Calculus. \newblock {\em Academic Press, Orlando, FL}, 1984.

\bibitem{RomRota} S. Roman and G.-C. Rota. \newblock The umbral calculus.
\newblock {\em Adv. Math. 27:95--188}, 1978.

\bibitem{SunZhuang} Y. Sun and J. Zhuang. \newblock  
$\lambda$-factorials of $n$.
\newblock {\em Electronic J. Combinatorics 17, $\#$R169}, 2010.

\bibitem{Younsi} M. Younsi. \newblock  Proof of a combinatorial conjecture coming from the PAC-Bayesian
machine learning theory.  \newblock {\em arXiv:1209.0824}, 2012.


\end{thebibliography}

\end{document}
