# Sum of Bernoulli variables with different success probabilities [duplicate]

Let $x_i$ be independent Bernoulli random variables with success probabilities $p_i$. That is, $x_i=1$ with probability $p_i$ and $x_i=0$ with probability $1-p_i$.

Is there a closed expression or an approximate formula for the distribution of the sum $\sum_i x_i$?

## marked as duplicate by whuber♦Apr 15 '14 at 16:57

• If the $p_i$ are very small, you can use Poisson approximation. Let $X_i\sim \mbox{Be}(p_i)$ be independent and let $Y\sim\mbox{Po}(\lambda)$ with $\lambda=\sum_{i=1}^np_i$. In a classic paper by Hodges and Le Cam it is shown that $|\mbox{P}(\sum_{i=1}^n X_i\leq x)-\mbox{P}(Y\leq x)|=3\cdot (\max_{1\leq i\leq n}p_i)^{1/3}.$ If the $p_i$ are all close to 0, this difference is small. – MånsT Apr 15 '14 at 13:15
• In addition to the duplicate, solutions appear at stats.stackexchange.com/questions/41247 (computational methods) and stats.stackexchange.com/questions/5347 (approximations for large numbers of variables). – whuber Apr 15 '14 at 16:59
• @MånsT Hodges and Le Cam's result you state is incorrect. The equality is less than or equal to!!. – Chamberlain Foncha Oct 19 '17 at 20:38
• @Chamberlain: you are absolutely right! I can't edit it now though, as my comment is too old. – MånsT Oct 20 '17 at 6:23

\begin{align} E\left[\sum_i x_i\right] &= \sum_i E[x_i] = \sum_i p_i\\ V\left[\sum_i x_i\right] &= \sum_i V[x_i] = \sum_i p_i(1-p_i). \end{align}
• The CLT often fails (that is, does not even apply) in this circumstance unless the $p_i$ remain away from $0$ and $1$. – whuber Apr 15 '14 at 17:01