# 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

$$P(X_i=k)=\frac{{k-1 \choose i-1}{m-k \choose n-i}}{m \choose n}$$

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.

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(Wiki)[en.wikipedia.org/wiki/Hypergeometric_distribution] has the kurtosis. Is that enough or you want to use maple to reproduce it? – Fabian Mar 1 '11 at 17:30
@Fabian It is somewhat different with my formula. Mine is a 3-variable hyper geometric distribution. Do you think moment generating function could solve the problem? – Fan Zhang Mar 1 '11 at 17:41
The generating function definitely solves the problem. Do you have it? By the way what is the range of $k$? – Fabian Mar 1 '11 at 17:53
@Fabian I do not have the generating function. I give a wrong link in my question, I have updated it. Please refer to it. And the reference lists the E(X_i) and Var(X_i). The range of k should be i<=k<=m-n+i according to the reference. – Fan Zhang Mar 1 '11 at 18:00
@Bryon: please do not use answers to make comments. – Qiaochu Yuan Mar 2 '11 at 17:21

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.
I can not understand why $E((X_i)_p)=(i)_p*(m+1)_p/(n+1)_p$, actually I have tested the case when p=2, it's right. But I don't know how do you derive the equality. I can understand all but this. – Fan Zhang Mar 3 '11 at 17:20