I have a series of N binomial (win/loss) events $e_i$ that can happens with different probabilities $p_i$. If i count the times $W$ that an event happens (the number of wins), the mean probability will be $\hat p = \frac{W}{N}$. What is the relation between $\hat p$ and $p_i$?


In distribution, $\hat p$ can follow a variety of distributions on $\{0,1,\ldots,N\}$ since, for every $0\leqslant k\leqslant N$, $$ \mathrm P(N\cdot\hat p=k)=\sum\limits_{|I|=k}\prod_{i\in I}p_i, $$ where the sum runs over every subset $I$ of $\{0,1,\ldots,N\}$ of size $k$. In the mean, $$N\cdot\mathrm E(\hat p)=\mathrm E(W)=\sum\limits_{i=1}^N\mathrm P(\text{event}\ e_i\ \text{is a win})=\sum\limits_{i=1}^Np_i. $$ Finally, by independence, the variance is such that $$ N^2\cdot\text{Var}(\hat p)=\sum\limits_{i=1}^Np_i(1-p_i). $$

  • $\begingroup$ i don't understand :( can you post here a proof or a reference? $\endgroup$ – emanuele Apr 10 '12 at 18:04
  • $\begingroup$ What part? $ $ $ $ $\endgroup$ – Did Apr 10 '12 at 19:14
  • $\begingroup$ last two equations. $\endgroup$ – emanuele Apr 10 '12 at 20:26
  • $\begingroup$ Do you agree that $W$ is a sum of $N$ random zeroes and ones, the $i$th term being 1 if and only if event $e_i$ happens? $\endgroup$ – Did Apr 10 '12 at 20:41
  • $\begingroup$ yes. ...and so? $\endgroup$ – emanuele Apr 11 '12 at 6:42

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