Binomial Distribution: C(M,N) p^L (1-p)^(M-N)
Poisson's Binomial Distribution (PBD) : p_i are different
Hyper geometric : Prob ( n success in m trials, given N success in M)
= C(m,n) C(M-m,N-n)/ C(M,N)
= distribution of partial sum (up to m)
Q. What is the distribution of partial sum for PBD ?
@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.
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