by partial sum do you mean sum of 'm' identical  PBD variables?

@subrata chakraborty The whole point of asking the question about PBD is that they are independent but not identical. I kind of feel the question is about whether the CLT holds or not. 

Ok may be the following link is of some help

http://www.cs.berkeley.edu/~jordan/courses/210B-spring08/lectures/stat210b_lecture_2.pdf

this might help you

good luck

http://www.google.com.sa/url?sa=t&rct=j&q=&esrc=s&frm=1&source=web&cd=6&cad=rja&uact=8&ved=0CEoQFjAF&url=http%3A%2F%2Fgalton.uchicago.edu%2F~lalley%2FCourses%2F312%2FPoissonProcesses.pdf&ei=TPBMVcvhAYurswHPv4GIAQ&usg=AFQjCNFUi_hqqsOcJhxJiN-wKg0mShnHbQ

I'm assuming that you mean the following:  the random variable X =X_1 + X_2 + .. + X_n, where each X_i = 1 with probability p_i, 0  with probability 1-p_i.  Then  X is PBD.

Let Y_j = X_1 + X+2 + .. + X_j: you are asking for the distribution of Y_j given the value of X.

Is this a correct restatement of your question? 

If so, it is fairly easy to write down a formula involving p_i's: but I don't expect at first glance for the formula to have a nice closed form except in special cases such as small values of X or Y_j.

 The exact values of the point probabilities can be calculated by the following formula. Let $ S=\sum_{i=1}^{n}X_{i} $ denote a sum of independent Bernoulli random variables, I.e. S has a Poisson binomial distribution. Then

\[

P\left(S=k\right)=s_{k}^{n}\left(\frac{p_{1}}{1-p_{1}},\frac{p_{2}}{1-p_{2}},\dots,\frac{p_{n}}{1-p_{n}}\right)\cdot\prod_{i=1}^{n}\left(1-p_{i}\right)

\]

A Poisson binomial distributions can be approximated by binomial distribution by choosing the success probability and the number parameter of the binomial distribution in such way that it gets the same mean and variance as the Poisson binomial distribution. A more accurate approximation can be obtained by a sum of two binomial distribution.

It is a non-trivial fact that hypergeometric distributions are Poisson binomial distributions.

Dear  Neil, 

Thank  you for restating the question.

Yes,     you are right; it is simple to write  it; but is there a    GOOD looking formula  or function ?   When p_i=1/2,   we have      hyper geometric distribution 

C(n,y) C(n-j, x-y)/C(n,x)

For  generic {p_i} :     =  j  x/n,   what is   s_j = -^2 ?  It has to be 0  for j=0, and  j=n .

Thanks.  PK. 

Dear Mohanty,

In my previous comment I forgot to mention that s_{k}^{n} denotes one of the elementary symmetric functions. If there existed a simpler way to write the point probabilities of a Poisson binomial distribution there would also exist a simpler way to write the elementary symmetric functions.So in a sense the formula that I gave is the simplest possible formula. I do not understand your last question.

Dear  Harremoës,

Thank you very much for  drawing my attention to symmetric functions.

The question was  rather,

If   sum   $\sum_{i=1}^{N }X_{i}=S$   what is   the distribution  $s =  \sum_{i=1}^{n } X_{i}$ ?

where,  X_i are nonidentical Bernoulli random variables   and   N, S and   n  are

constants.   Of course   0

I guess that such a partial sum is also a Poisson binomial distribution. I have never seen a theorem stating that, so one has to check which may in fact be as complicated as checking that a hypergeometric distribution is a Poisson binomial distribution. In order to estimate the distribution one should calculate mean and variance and use a Gaussian approximation.

Mohanty --- the symmetric functions are going to be crucial, as is the order in which you pick your variables,  I'd be extraordinarily surprised if anything that had been written in this forum was simplified further in the general setting.  For special cases of the probabilities (monotone especially, but probably linear) there might be nice things to say.