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Let $X_i$ be a $\{0,1\}$-valued r.v. with $P(X_i=1)=p_i$, $i=1,\dots, n$. $\{X_i, i=1, \dots, n\}$ are independent.

  1. Is there a name for the distribution of $\sum_{i=1}^n X_i$?
  2. How is $P(\sum_{i=1}^n X_i = m)$ determined?

    Some said $P(\sum_{i=1}^n X_i = m)$ is the coefficient of $x^m$ in $\prod_{i=1}^n (1-p_i + p_ix)$. I wonder why?

Thanks!

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Your "Some said" question may depend on an understanding of generating functions and how they relate to convolutions –  Henry Oct 24 '12 at 18:37
    
@Henry: Thanks! How does it depend? –  Tim Oct 24 '12 at 18:37

2 Answers 2

up vote 1 down vote accepted

Note that $(1-p_i + p_ix)$ can also be written as $(1-p_i)x^0 + p_ix^1$.

Now imagine multiplying $n$ of those together and looking at the terms with coefficient $x^m=x^{(n-m)\times 0 +m\times 1 }$. So each of the terms in the sum will have $n-m$ multiplicands of the form $(1-p_i)$ and $m$ terms of the form $p_i$ with all the ${n \choose m}$ combinations of possible $0$s and $1$s for the different $i$ appearing such that exactly $m$ of them are $1$.

But this is precisely how you would work out the probability in answer to question 2.

For example, if $n=3$ and the $p_i$ are $0.1,0.2,0.3$, and you wanted the probability of exactly two $1$s then you would calculate $0.9 \times 0.2 \times 0.3 + 0.1 \times 0.8 \times 0.3 + 0.1 \times 0.2 \times 0.7$ but this is the same as the coefficient of $x^2$ in $(0.9+0.1 \times x)(0.8+0.2 \times x)(0.7+0.3 \times x)$

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There is a name. It's called a Poisson binomial distribution. Formulas can be found on Wikipedia.

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