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\begin{center}
	\vskip 1cm{\LARGE\bf Probabilistic Proofs of Some Beta-Function \\
	\vskip .2in
		Identities }
	\vskip 1cm
	\Large
	Palaniappan Vellaisamy\\
	Department of Mathematics \\
	Indian Institute of Technology Bombay\\
	Powai, Mumbai-400076\\
	India \\
	\href{mailto:pv@math.iitb.ac.in}{\tt pv@math.iitb.ac.in} \\
	\ \\
	Aklilu Zeleke\\
	 Lyman Briggs College \&amp; Department of Statistics and Probability \\
	 Michigan State University\\
  East Lansing, MI 48825	\\
	USA \\
	\href{mailto:zeleke@msu.edu}{\tt zeleke@msu.edu}\\
\end{center}

\vskip .2 in

\begin{abstract}
  Using a probabilistic approach, we derive some interesting identities involving beta functions. These results generalize certain well-known combinatorial identities involving binomial coefficients and gamma functions.
\end{abstract}

\section{Introduction}
There are several interesting combinatorial identities involving binomial coefficients, gamma functions, and hypergeometric functions  (see, for example, Riordan \cite{Riordan},  Bagdasaryan \cite{Bagdasaryan},  Vellaisamy \cite{Vellaisamy}, and the references therein). One of these is the following famous identity that involves the convolution of the central binomial coefficients: \begin{equation} \label{eqn1.1}
 \sum_{k=0}^{n} \binom{2k}{k}\binom{2n-2k}{n-k}=4^n.
\end{equation}   
In recent years, researchers have provided several proofs of \eqref{eqn1.1}. 
A proof that uses generating functions can be found in Stanley \cite{Stanley}. Combinatorial proofs can also be found, for example, in Sved \cite{Sved}, De Angelis \cite{De Angelis}, and Miki\'{c} \cite{Mickic}. 
A related and an interesting identity for the alternating convolution of central binomial coefficients is 
\begin{align}\label{eqn1.2}
	\sum_{k=0}^{n} (-1)^k \binom{2k}{k}  \binom{2n-2k}{n-k} =
	\begin{cases}
		2^n \binom{n}{\frac{n}{2}}, &amp; \text{if $n$ is even;}  \\
		0, &amp; \text{if  $n$ is odd.}
	\end{cases} 
\end{align}
 Nagy \cite{Nagy}, Spivey \cite{Spivey}, and  Miki\'{c} \cite{Mickic} discussed the combinatorial proofs of the above identity.
Recently, there has been considerable interest in finding simple probabilistic proofs for combinatorial identities (see Vellaisamy and Zeleke \cite{VellaisamyZeleke} and the references therein). Pathak \cite{Pathak} gave a probabilistic proof of the  identity in \eqref{eqn1.2}. Chang and  Xu \cite{ChungXu} extended the result in \eqref{eqn1.1} and presented a probabilistic proof of the identity
\begin{equation}\label{eqn1.3}
 \sum_{\substack{k_j \geq 0,\; 1 \leq j \leq m;\\ \sum_{j=1}^m  k_j=n}} \binom{2k_1}{k_1}\binom{2k_2}{k_2}\cdots\binom{2k_m}{k_m}=\frac{4^n}{n!} \frac{\Gamma(n+\frac{m}{2})}{\Gamma(\frac{m}{2})},
\end{equation}
 where $k_1, \ldots, k_m$ are nonnegative integers,
 and  $m$ and $n$ are positive integers. Miki\'{c} \cite{Mickic} discussed a combinatorial proof of \eqref{eqn1.3} based on the method of recurrence relations and telescoping. Duarte and Guedes de Oliveira \cite{Duarte} discussed
 a generalization of the result in \eqref{eqn1.3} and proved the following identity (see their Theorem 2),  using combinatorial
 arguments,
\begin{equation}\label{eqn1.4}
\sum_{\substack{\sum_{j=1}^m  k_j=n}} \binom{2k_1+l_1}{k_1}\binom{2k_2+l_2}{k_2}\cdots\binom{2k_m+l_m}{k_m}=
4^n \binom{n+ \frac{m}{2}-1}{n},
\end{equation}
where $l_1, \ldots, l_m$ are reals such that $ l_1+ \cdots+l_m=0$.

Our goal in this paper is to generalize the combinatorial identities in \eqref{eqn1.2} and \eqref{eqn1.3}, using a simple probabilistic approach.  Indeed, we derive certain identities involving beta functions. Our method uses 
the Dirichlet-multinomial distribution and also the moments of  the difference of two gamma random variables.



\section{Identities involving beta functions}

Let the  random variable $T$ follow the beta distribution with parameters  $a, b&gt;0$ and with  probability density 
\begin{equation*}
f(t)= \frac{1}{B(a,b)}  t^{a-1}(1-t)^{b-1}, ~ t&gt;0,  
\end{equation*}
where $B(a, b)$ is the beta function. Note that the beta function  $B(x, y)$ is related to the gamma function by
\begin{equation*}
B(x, y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)},
\end{equation*}
where $\Gamma(x)=\int_0^\infty t^{x-1}e^{-t} dt, x&gt;0.$ The beta function is symmetric (that is, $B(x,y)=B(y,x)$) and satisfies the basic identity 
\begin{equation} \label{eqn2.3}
 B(x,y)=B(x,y+1)+B(x+1,y),~ \text{for}~x, y&gt;0. 
 \end{equation}
It is easy to see that the derivative of the beta function is  
\begin{equation} \label{neqn2.4}
\frac{\partial}{\partial y} B(x,y)=B(x,y) \left( \psi(y)-\psi(x+y)\right),
\end{equation}
where $\psi(x)={\Gamma'(x)}/{\Gamma(x)}$ is the digamma function.


 We start with a result  that relates binomial coefficients and beta functions on one side to a simple rational expression on the other side. 

\begin{theorem}  For $s &gt; 0$ and an integer $n \geq 0$, 
\begin{align*}
\sum_{j=0}^{n}\sum_{i=0}^j (-1)^{j}\binom{n}{j} \frac{B(j+1,s)}{s+i}= \frac{1}{(s+n)^2} \cdot
\end{align*}
\end{theorem}

\begin{proof} Let the random variable $Y$ follow the
beta distribution with parameters $1$ and $n+s$. Using the density of $Y$, we get
\begin{alignat}{2}
&amp; \int_{0}^{\infty} (1-y)^{n+s-1} dy = B(1, n+s) &amp;&amp;= \frac{1}{n+s}; \nonumber \\
&amp;\Longrightarrow  \int_{0}^{\infty} \left(\sum_{j=0}^{n}(-1)^{j}\binom{n}{j}  y^j\right) (1-y)^{s-1} dy &amp;&amp;= \frac{1}{n+s}; \nonumber \\
&amp;\Longrightarrow \sum_{j=0}^{n}(-1)^{j}\binom{n}{j} \int_{0}^{\infty} y^j (1-y)^{s-1} dy &amp;&amp;= \frac{1}{n+s}; \nonumber \\
&amp;\Longrightarrow \sum_{j=0}^{n}(-1)^{j}\binom{n}{j}B(j+1,s) &amp;&amp;= \frac{1}{n+s} \cdot \label{eqn2.6}
\end{alignat} 
Differentiate both sides of \eqref{eqn2.6} with respect to $s&gt;0$ to get
\begin{align}
\frac{-1}{(s+n)^2} =&amp; \sum_{j=0}^{n}(-1)^{j}\binom{n}{j}\frac{\partial}{\partial s} B(j+1,s)\nonumber\\=&amp; \sum_{j=0}^{n}(-1)^{j}\binom{n}{j}  B(j+1,s) \left( \psi(s)-\psi(j+1+s)\right), \label{neqn2.6}
\end{align}
using \eqref{neqn2.4}. Further, it is known that the digamma function $ \psi(x)$ satisfies 
\begin{equation} \label{neqn2.7}
\psi(x+1)-\psi(x)=\frac{1}{x}. 
\end{equation}
Using \eqref{neqn2.7} iteratively leads to 
\begin{equation} \label{neqn2.8}
\psi(x+j+1)-\psi(x)=\displaystyle \sum_{i=0}^j \frac{1}{x+i},
\end{equation}
for a nonnegative integer $j$.  The result follows by putting \eqref{neqn2.8} into \eqref{neqn2.6}.
\end{proof}

\begin{remark}
The identity in \eqref{eqn2.6} itself is an interesting identity. When $n=2$, it reduces to 
\begin{align*}
B(1,s)-2 B(2,s)+B(3,s)=\frac{1}{s+2},
\end{align*}
 which can also be verified using the basic identity in \eqref{eqn2.3}. 
\end{remark}

Next we extend the identity in \eqref{eqn1.2}.
 Let $ X$ be a gamma  random variable with parameter $p &gt;0$, denoted by $ X \sim G(p)$. The density of $X$ is given by $$f(x|p)=\frac{1}{\Gamma(p)}e^{- x}x^{p-1},\;x&gt;0,\; p &gt;0.$$  Then, it follows  (see Rohatgi and Saleh \cite{RohatgiSaleh}) that
\begin{align*}
E(X^n)=\frac{1}{\Gamma(p)} \int_0^\infty e^{- x}x^{p+n-1}dx=\frac{\Gamma(p+n)}{\Gamma(p)}. 
\end{align*}

\begin{theorem} 
Let $p&gt;0$ and $n$ be a positive integer. Then	
\begin{displaymath}
 \sum_{k=0}^{n} (-1)^k \binom{n}{k} B(p+k, p+n-k)=
\begin{cases}
\displaystyle{\frac {n! \Gamma(p) \Gamma(p+\frac{n}{2})} {\Gamma(\frac{n}{2}+1) \Gamma(2p+n)}}, &amp; \text{if $n$ is even;}\\
  0,  &amp; \text{if $n$ is odd.}
\end{cases}	
\end{displaymath}	
\end{theorem}

\begin{proof}Consider the random variable $X= X_1-X_2$, where $X_1$ and $X_2$ are independent gamma random variables with the same parameter $p&gt;0$, that is, $X_i \sim G(p),$ for $ i=1, 2$.
Since $X_1$ and $X_2$ are independent and identically distributed, we have   $X\stackrel{d}{=}-X$ (that is, $X$ and $-X$ have the same distributions on $\mathbb{R}$). This implies the 
density of $X$ is symmetric about zero. Hence, $E(X^n)=0$ if $n$ is an odd integer.

 Next we compute the even moments of $X$. Finding the moments of $X$ using the probability density function is tedious. This is because the  density of $X$ is very complicated and it involves Whittaker's W-function (see Mathai \cite{Mathal}). Therefore, we use the moment generating function (MGF) approach to find the moments of $X$.

 It is known (see Rohatgi and Saleh \cite{RohatgiSaleh}) that the  MGF of $X_1$ is
  \begin{equation*}
  M_{X_1}(t)= E(e^{tX_1})= (1-t)^{-p}.
  \end{equation*}
 Hence, the MGF of $X$ is
 \begin{equation*}
  M_{X}(t) = M_{X_1}(t)M_{X_2}(-t) =(1-t)^{-p} (1+t)^{-p} 
 = (1-t^2)^{-p},      
 \end{equation*}
which exists for $|t|&lt;1$. Using the result  
\begin{equation*}
(1-q)^{-p}= \sum_{n=0}^{\infty} \frac{\Gamma(n+p) q^n}{\Gamma(n+1) \Gamma(p)},\;\mbox{for}\;p&gt;0\; \mbox{and}\;  |q|&lt;1,
\end{equation*}
we have
\begin{equation} \label{eqn2.19}
M_{X}(t)= (1-t^2)^{-p}= \sum_{n=0}^{\infty} \frac{\Gamma(n+p) t^{2n}}{\Gamma(n+1) \Gamma(p)}.
\end{equation}
Hence, for  $n \geq 1$, we have from \eqref{eqn2.19}
\begin{equation*} 
E(X^{2n})=M_{X}^{(2n)}(t)|_{t=0}=  \frac{\Gamma(n+p) {(2n)!}}{\Gamma(n+1) \Gamma(p)},
\end{equation*}	
where $f^{(k)}$ denotes the $k$-th derivative of $f$. Thus,
we have shown that	
\begin{align} \label{eqn2.21}
E(X^{n})=
\begin{cases}
\displaystyle  \frac{n! \Gamma(\frac{n}{2}+p) }{\Gamma(\frac{n}{2}+1) \Gamma(p)}, &amp; \text{if $n$ is even;}\\
\phantom{0}0,  &amp; \text{if $n$ is odd.}
\end{cases}
\end{align}
Next, we compute the moments of $X$, using the binomial theorem. Note that
\begin{align}\label{eqn2.22}
E(X^n) =&amp; E(X_1-X_2)^n= \sum_{k=0}^{n} (-1)^k \binom{n}{k} E(X_1^k) E(X_2^{n-k}) \nonumber \\
=&amp; \sum_{k=0}^{n}  (-1)^k \binom{n}{k} \left(\frac{\Gamma(p+k)}{\Gamma(p)}\right) \left(\frac{\Gamma(p+n-k)}{\Gamma(p)}\right).
\end{align}
Equating \eqref{eqn2.21} and \eqref{eqn2.22}, we get
\begin{align}\label{eqn2.23}
\sum_{k=0}^{n} (-1)^k \binom{n}{k} {\Gamma(p+k)}{\Gamma(p+n-k)}=
\begin{cases}
 \displaystyle \frac{n!\Gamma(\frac{n}{2}+p) \Gamma(p)}{\Gamma(\frac{n}{2}+1)}, &amp; \text{if $n$ is even;}\\ 
 0,  &amp; \text{if $n$ is odd,} 
\end{cases}
\end{align}
which is an interesting identity involving gamma functions and 
binomial coefficients.
Dividing both sides of \eqref{eqn2.23}  by $\Gamma(2p+n)$, the result follows.
\end{proof}

 We will show now that the identity in \eqref{eqn1.2} follows as a special case.
\begin{corollary}  Let $n$ be a positive integer. Then 
\begin{displaymath}
\sum_{k=0}^{n} (-1)^k \binom{2k}{k} \binom{2n-2k}{n-k} =
\begin{cases}
2^n \binom{n}{\frac{n}{2}}, &amp;\text{if $n$ is even;}\\
0, &amp; \text{if  $n$ is odd.}
\end{cases} 
\end{displaymath}
\end{corollary}
 
\begin{proof} Let $p=\frac{1}{2}$ in \eqref{eqn2.23} and it suffices to consider the case when $n$ is even. Then
\begin{align*}
\sum_{k=0}^{n}  (-1)^k \binom{n}{k} {\Gamma\left(k+\frac{1}{2}\right)}{\Gamma \left(n-k+\frac{1}{2}\right)}=
\displaystyle \frac{{n!}\Gamma(\frac{n}{2}+\frac{1}{2}) \Gamma(\frac{1}{2})}{\Gamma(\frac{n}{2}+1)}.
\end{align*}
That is,
\begin{align} \label{eqn2.25}
\sum_{k=0}^{n}  (-1)^k \binom{n}{k} \left(\frac{\Gamma(k+\frac{1}{2})}{\Gamma(\frac{1}{2})}\right)
\left(\frac{\Gamma(n-k+\frac{1}{2})}{\Gamma(\frac{1}{2})}\right)  =
\displaystyle \frac{{n!}\Gamma(\frac{n}{2}+\frac{1}{2}) }{{\Gamma(\frac{1}{2})}\Gamma(\frac{n}{2}+1)}.
\end{align}
Note  that, 
\begin{align}
\frac{\Gamma(n+\frac{1}{2})}{\Gamma(\frac{1}{2})}&amp;=\frac{\left(n-\frac{1}{2}\right) \left(n-\frac{3}{2}\right) \cdots \left(\frac{3}{2}\right) \left(\frac{1}{2}\right) \Gamma \left(\frac{1}{2} \right)}{\Gamma \left(\frac{1}{2} \right)}\nonumber\\&amp;=
\frac{(2n-1) \cdot (2n-3) \cdots 3 \cdot 1}{2^n} \nonumber \\ &amp;=\frac{(2n)!}{ n! 4^n}. \label{eqn2.26}
\end{align}
 Using \eqref{eqn2.26} in \eqref{eqn2.25}, we get
\begin{align*} 
\displaystyle \sum_{k=0}^{n}  (-1)^k  \frac{n!}{k! (n-k)!}\frac{ (2k)!}{4^k k!} \frac{(2n-2k)!}{4^{(n-k)} (n-k)! } =&amp; \displaystyle
 \frac{n! n!}{4^{\frac{n}{2}} (\frac{n}{2})!(\frac{n}{2})!}. 
\end{align*}
That is,
\begin{equation}\label{eqn2.27}
\sum_{k=0}^{n} (-1)^k \binom{2k}{k}  \binom{2n-2k}{n-k} =
\displaystyle \frac{n! 4^n }{(\frac{n}{2})!(\frac{n}{2})! 4^{\frac{n}{2}}} \nonumber \\
  = \displaystyle 2^n \binom{n}{\frac{n}{2}},
\end{equation}
 which proves the result.
\end{proof}


Finally, we discuss an extension of the identity given in \eqref{eqn1.3}.
Let $ p_i&gt;0$ for $1\leq i \leq m$. Let
\begin{equation*}
B(p_1,\ldots,p_m)= \frac{\Gamma(p_1)\Gamma(p_2)\cdots\Gamma(p_m)}{\Gamma(p_1+\cdots+p_m)}
\end{equation*}
denote the beta function of $m$ variables, and $\binom{n}{k_1,\ldots,k_m}= \frac{n!}{k_1! \cdots k_m!}$ denote the multinomial coefficient.

Let $X=(X_1, \ldots, X_m)$ be a discrete nonnegative random vector and $ Y=(Y_1, \ldots, Y_m)$ be a continuous positive random vector such that $\sum_{1}^{m} X_i=n $ and $ \sum_{1}^{m} Y_i =1.$
Let $(X|Y) \sim {\rm MN}(n; Y_1, \ldots, Y_m)$,  the multinomial distribution, with
\begin{align*}
P\left(X_1=k_1, \ldots, X_m=k_m| Y_1=y_1, \ldots,  Y_m=y_m \right) =\binom {n} {k_1,\ldots, k_m} y_1^{k_1}\ldots y_m^{k_m}
\end{align*}
and $Y \sim {\rm Dir} (p_1, \ldots, p_m),$ the Dirichlet distribution, with density
 \begin{equation*}
  f(y_1, \ldots, y_m)=  \frac{1}{B(p_1,\ldots,p_m)} y_1^{p_1-1} \cdots y_m^{p_m-1}.
 \end{equation*}
 Then the
marginal distribution of $X$  follows the Dirichlet-multinomial distribution with 
\begin{align*}
P\left(X_1=k_1, \ldots, X_m=k_m \right)\\
=&amp;~\frac{1}{B(p_1,\ldots,p_m)} 
\binom {n} {k_1,\ldots, k_m} B(k_1+p_1,\ldots,k_m+p_m),
\end{align*}
where $k_1 + \cdots+k_m=n$.

 The next result follows trivially as the sum of the
above probabilities is unity.

\begin{theorem}  Let $p_1, \ldots, p_m &gt; 0.$ Then for any nonnegative integer $n$, 
\begin{equation}\label{eqn3.3}
\sum_{\substack{k_j\geq 0,~ 1 \leq j \leq m; \\ \sum_{j=1}^m  k_j=n}} \binom {n} {k_1,\ldots, k_m} B(k_1+p_1,\ldots,k_m+p_m)=B(p_1,\ldots,p_m).
\end{equation}
 \end{theorem} 

 It is interesting to note  that the identity  in  \eqref{eqn1.3} follows as a special case.

\begin{corollary} \label{cor3.1} 
When $p_1=p_2=\cdots=p_m=\frac{1}{2}$, the identity in \eqref{eqn3.3} reduces to
\begin{equation}\label{eqn3.6}
 \sum_{\substack{k_j \geq 0,\; 1 \leq j \leq m;\\ \sum_{j=1}^m  k_j=n}} \binom{2k_1}{k_1}\binom{2k_2}{k_2}\cdots\binom{2k_m}{k_m}=\frac{4^n}{n!} \frac{\Gamma(n+\frac{m}{2})}{\Gamma(\frac{m}{2})},
 \end{equation}
  for all integers $m,n \geq 1$. 
\end{corollary}

\begin{proof}
Putting $p_1=p_2=\cdots =p_m=\frac{1}{2}$ in \eqref{eqn3.3}, we obtain 
$$\sum_{\substack{k_j \geq 0,\; 1 \leq j \leq m;\\ \sum_{j=1}^{m} k_j=n}} \binom {n} {k_1,\ldots, k_m} B\left(\frac{1}{2}+k_1,\ldots, \frac{1}{2}+k_m\right)=B\left(\frac{1}{2},\ldots, \frac{1}{2}\right).$$
This implies, 
\begin{align*}
 \sum_{\substack{k_j \geq 0,\; 1 \leq j \leq m;\\ \sum_{j=1}^{m}  k_j=n}} \binom {n} {k_1,\ldots, k_m} \frac{\Gamma\left(\frac{1}{2}+k_1\right)\cdots \Gamma \left(\frac{1}{2}+k_m\right)}{\Gamma \left(n+\frac{m}{2}\right)}&amp;=\frac{\Gamma \left(\frac{1}{2}\right) \cdots \Gamma \left(\frac{1}{2}\right)}{\Gamma \left(\frac{m}{2}\right)},
\end{align*}
or, equivalently, 
\begin{align*}
 \sum_{\substack{k_j \geq 0,\; 1 \leq j \leq m;\\ \sum_{j=1}^{m} k_j=n}} \binom {n} {k_1,\ldots, k_m} \frac{\Gamma\left(\frac{1}{2}+k_1\right)\cdots \Gamma \left(\frac{1}{2}+k_m\right)}{\Gamma \left(\frac{1}{2}\right) \cdots \Gamma \left(\frac{1}{2}\right)}=\frac{\Gamma \left(n+\frac{m}{2}\right)}{\Gamma \left(\frac{m}{2}\right)}.
\end{align*}
Using \eqref{eqn2.26}, we get 
\begin{align*} 
 \sum_{\substack{k_j \geq 0,\; 1 \leq j \leq m;\\ \sum_{j=1}^{m} k_j=n}} \binom {n} {k_1,\ldots, k_m} \frac{(2k_1)! \cdots (2k_m)!}{4^{k_1}(k_1)! \cdots 4^{k_m}(k_m)!} 
=\frac{\Gamma \left(n+\frac{m}{2}\right)}{\Gamma \left(\frac{m}{2}\right)},
\end{align*}
which is equivalent to the identity in \eqref{eqn3.6}.
\end{proof}

\begin{remark} \label{rem3.2} 
	Obviously, when $m=2$, the identity in \eqref{eqn3.3} reduces to 
\begin{align*}
\displaystyle \sum_{\substack{k_j\geq 0,~ 1 \leq j \leq 2; \\ k_1+k_2=n}} \binom {n} {k_1, k_2}&amp; B(p_1+k_1, p_2+k_2)\\
=&amp;\sum_{k=0}^n \binom{n}{k}B(p_1+k,p_2+n-k)=B(p_1,p_2).
\end{align*}	
 Also, in view of  Corollary \ref{cor3.1},  when 
$p_1=p_2=\frac{1}{2}$, the above equation 
reduces to \eqref{eqn1.1}.	
\end{remark}

\begin{remark} 
 Let $m$ be even so that $m=2l$ for some positive integer $l$. Then the right hand side of \eqref{eqn3.6} is 
$$ \frac{4^n}{n!}\frac{\Gamma (n+l)}{\Gamma (l)}=4^n \binom{n+l-1}{n}=4^n \binom{n+\frac{m}{2}-1}{n}.$$ 
Similarly, when $m$ is odd, say $m=2l+1$, 
\begin{align*}
\frac{4^n}{n!}\frac{\Gamma (n+\frac{m}{2})}{\Gamma (\frac{m}{2})}&amp;= \frac{4^n}{n!}\frac{\Gamma (n+l+\frac{1}{2})}{\Gamma (l+\frac{1}{2})}=\frac{4^n}{n!}\left(\frac{\Gamma (n+l+\frac{1}{2})}{\Gamma(\frac{1}{2})}\right)\left(\frac{\Gamma(\frac{1}{2})}{\Gamma (l+\frac{1}{2})}\right)\\
&amp;= \binom{2n+2l}{2n}\left(\frac{(2 n)!}{n!\; n!}\right) \left(\frac{l!\; n!}{(n+ l)!}\right)\; (\mbox{using}\; \eqref{eqn2.26})\\
&amp;= \displaystyle \frac{\binom{2n+2l}{2n} \binom{2n}{n}}{\binom{n+l}{n}}\\
&amp;=  \displaystyle \frac{\binom{2n+m-1}{2n} \binom{2n}{n}}{\binom{n+\frac{m-1}{2}}{n}}, 
\end{align*}
since $2l={m-1}$. Thus, we have, from \eqref{eqn3.6},
\begin{displaymath} 
   \sum_{\substack{k_j \geq 0,\; 1 \leq j \leq m;\\ \sum_{j=1}^{m} k_j=n}} \binom {2k_1} {k_1}  \binom {2k_2} {k_2} \cdots \binom {2k_m} {k_m}=
     \begin{cases} \displaystyle
       4^n \binom{n+\frac{m}{2}-1}{n}, &amp; \text{if $m$ is even;}\\
       &amp;\\
      \displaystyle \frac{\binom{2n+m-1}{2n} \binom{2n}{n}}{\binom{n+\frac{m-1}{2}}{n}}, &amp;  \text{if $m$ is odd,}
     \end{cases}
\end{displaymath}
which is equation $(3)$ of Miki\'{c} \cite{Mickic}. Indeed, Miki\'{c} \cite{Mickic}
provided a combinatorial proof of the above result based on recurrence relations.
\end{remark}

\begin{corollary} \label{cor32} 
Let $ k_1, \ldots, k_m $ be nonnegative integers and $ l_1, \ldots, l_m $ be integers such that  $ 0 \leq k_i+l_i \leq n$ and $ \sum_{i=1}^{m} l_i=0.$
Then
\begin{equation*}
\sum_{\substack{ \sum_{j=1}^m  k_j=n}} \binom{2k_1+2l_1}{k_1+l_1}\binom{2k_2+2l_2}{k_2+l_2}\cdots\binom{2k_m+2l_m}{k_m+l_m}=\frac{4^n}{n!} \frac{\Gamma(n+\frac{m}{2})}{\Gamma(\frac{m}{2})},
\end{equation*}
for all integers $m,n \geq 1$.	
\end{corollary}	

 The above corollary, which follows from Corollary \ref{cor3.1}, is similar to
\eqref{eqn1.4}. It is not clear if the identity in \eqref{eqn1.4} can be obtained through probabilistic considerations.

\section{\bf Acknowledgments} The authors are grateful to the editor-in-chief and the referee for several helpful 
comments and also for pointing out the paper of Duarte and Guedes de Oliveira \cite{Duarte}. This work was completed while the first author was visiting  the Department of Statistics and Probability at Michigan State University during the summer of 2019. We are grateful to Professor Frederi Viens for his support and encouragements. 


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\noindent 2010 {\it Mathematics Subject Classification}: Primary
05A19; Secondary 05A10, 33B15.

\noindent \emph{Keywords:}
combinatorial identity, beta distribution, gamma distribution, moment, probabilistic proof.

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\vspace*{+.1in}
\noindent
Received  April 12 2019;
revised versions received May 14 2019; August 23 2019; August 31 2019;
September 2 2019; September 10 2019.
Published in {\it Journal of Integer Sequences}, September 23 2019.

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