sums of series

[lavoisier]

Hi,

I'm unsuccessfully looking for the proof of some sums.

I'm sure they are very well known, so I wonder if some mathematician could redirect me to useful resources on the web, or just give me a hint about where to start for solving them.

 

The first is the famous binomial. If:

 

PP(i) = p^i * (1-p)^(s-i) * s! / ( i! * (s-i)! )

 

with 0<p<1, i and s positive integers, then I know the proof for:

 

Sum(PP(i), i, 0, s) = 1

 

But I can't work out the proof for:

 

Sum(i * PP(i), i, 0, s) = s*p

 

and for:

 

Sum((i-s*p)^2 * PP(i), i, 0 , s) = p*(1-p)*s

 

A related problem leads to the definition of:

 

PR(i) = Product((n*p-j)*(s-j)/((n-j)*(j+1)), j, 0, i-1) * Product((n-n*p-j)/(n-i-j), j, 0, s-i-1)

 

where 0<p<1, 1<=s<=n, s and n positive integers.

 

Here I don't have a clue for any of the three:

 

Sum(PR(i), i, 0, s) = 1

 

Sum(i * PR(i), i, 0, s) = s*p

 

Sum((i-s*p)^2 * PR(i), i, 0 , s) = p*(1-p)*s*(n-s)/(n-1)

 

Any help will be greatly appreciated.

[the tree]

Sum((i-s*p)^2 * PR(i), i, 0 , s) = p*(1-p)*s*(n-s)/(n-1)

Your format is really confusing, and I can't tell what you mean.

On these forums we use the LaTeX system to create properly typset equations like this:

[math]\sum_{i=0}^{s} (i-sp)^{2} \times P_R (i) = sp(1-p)\times \frac{n-s}{n-1}[/math]

(I completely guessed that that'd be what you meant, so don't be surprised if it's wrong)

 

You can click on a LaTeX image to see what code was used and I'd recomend reading this tutorial.

 

Thanks.

[lavoisier]

Sorry, I'm new in this forum. I didn't mean to confuse anybody. If and when I post new threads I will use LaTex. However, your guess was correct.

 

I've found a partial answer to my question, i.e. the function PR(i) is the hypergeometric distribution, just expressed in a different format.

 

Mathworld explains how the mean and the variance are calculated, starting from the Bernoulli distribution. It looks very easy in that way.

But it also gives for granted that some awful sums are easy to prove, which was my initial doubt.

[lavoisier]

OK, I have the proof. It's in Planetmath, both for the mean and the variance. Thank you anyway for your help.

[the tree]

To be honest, my knowlege of anything to do with stats is simply appauling. But you've reminded me that there may well be some interesting elements to it and I really ought to learn it by the time I next get to take exams, so thankyou.