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 $p_i$?
The mean probability in a binomial event.
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probability
probability-theory
1 Answers
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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 $$ N^2\cdot\text{Var}(\hat p)=\sum\limits_{i=1}^Np_i(1-p_i). $$
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0i don't understand :( can you post here a proof or a reference? – 2012-04-10
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0What part? $ $ $ $ – 2012-04-10
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0last two equations. – 2012-04-10
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0Do 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? – 2012-04-10
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0yes. ...and so? – 2012-04-11
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0And so, by linearity of the expectation, E(W) is the sum of the expectations of these zero-or-one random variables. The expectation of the $i$th is $p_i$, this gives the formula for E(W). – 2012-04-11
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0ok, you right. one comment. you postulated that the $p_i$ are a sequence of 1 and 0's. That is true if you consider the $p_i$ as the actual frequency over one event. Now i can say that the $p_i$ are the theoretical probabilities. Now your statements are still valid?. – 2012-04-11
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0*you postulated that the $p_i$ are a sequence of 1 and 0's*... Absolutely not. – 2012-04-11