How to use Maple derive the 4th central moment for this distribution? Let $X_i$ be a random variable, and the probability density function of X_i is
$$P(X_i=k)=\frac{{k-1 \choose i-1}{m-k \choose n-i}}{m \choose n}$$
For more about the distribution, please refer to http://www.randomservices.org/random/urn/OrderStatistics.pdf
The mean of $X_i$ is $$E(X_i)=i\frac{m+1}{n+1}$$
I want to calculate the fourth central moment (kurtosis) of the $X_i$.
Is it possible to use Maple to derive it ?
It is actually a 3-variable hyper geometric distribution.
 A: It seems the best way to describe the moments of $X_i$ is to use ascending factorials. Using Pochhammer's notation 
$$
(x)_p=x(x+1)\cdots(x+p-1)
$$ 
for every nonnegative integer $p$, one gets
$$
E((X_i)_p)=(i)_p\frac{(m+1)_p}{(n+1)_p}.
$$
Since $(x)_1=x$, this yields your formula for $E(X_i)$. Using the shorthand
$$
a_p=(i+p-1)\frac{m+p}{n+p},
$$
one gets $E((X_i)_p)=a_1a_2\cdots a_{p}$. This yields (I think) the value of the kurtosis of $X_i$ defined by
$$
\beta_2(X_i)=\frac{E((X_i-E(X_i))^4)}{E((X_i-E(X_i))^2)^2}
$$ 
as the ratio
$$
\beta_2(X_i)=\frac{a_2a_3a_4-6a_2a_3+7a_2-1-4a_1(a_2a_3-3a_2+1)+6a_1^2(a_2-1)-3a_1^3}{a_1(a_2-a_1-1)^2}.
$$
And I am well aware that this solution does not use Maple...
Edit To get the formula for $E((X_i)_p)$ written above, one can denote by $X_i^{m,n}$ the random variable considered by the OP and sum over every possible value of $k$ the algebraic identity
$$
P(X_i^{m,n}=k)\cdot (k)_p=P(X_{i+p}^{m+p,n+p}=k+p)\cdot (i)_p\frac{(m+1)_p}{(n+1)_p}.
$$
The LHS of the resulting equality is $E((X_i^{m,n})_p)$ and the RHS is the total mass of the distribution of $X_{i+p}^{m+p,n+p}$, which equals $1$, times the factor $(i)_p(m+1)_p/(n+1)_p$. This proves the formula.
