Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.math.uah.edu/stat/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.

share|improve this question
    
(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
1  
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
1  
@Bryon: please do not use answers to make comments. –  Qiaochu Yuan Mar 2 '11 at 17:21
show 1 more comment

1 Answer 1

up vote 5 down vote accepted

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.

share|improve this answer
    
I could not fully understand your approach. Is there any reference about your approach? –  Fan Zhang Mar 3 '11 at 9:46
    
@Fan OK. Could you specify what point(s) in my post you do not fully understand? –  Did Mar 3 '11 at 16:21
    
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
    
@Fan See the edit. –  Did Mar 3 '11 at 21:07
    
I understand your proof, it's a beautiful proof. Do you have any idea about how to calculate the any order of the central moment for this distribution? –  Fan Zhang Mar 4 '11 at 6:36
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.