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This question already has an answer here:

EDITED

John is playing a game on $n$ days, each day being independent.

On each day $i$, his probability of success is $p_i$.

We have $\frac{1}{n} \sum_{i=1}^n p_i = p$, and typically, the standard deviation $\sigma$ of these $p_i$ is small. $\sigma$ is known.

So I have a succession of $Bernoulli(p_i)$. I want to model the probability of having k success after n trials, using $p$ and $\sigma$ for instance.

I can approximate this with a Binomial distribution, but even though it's close it still leads to biases. Any thoughts on how to improve this?

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marked as duplicate by Tim Jan 25 at 14:19

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  • $\begingroup$ Sorry but your edit is still unclear. If probability of success is fixed to p and you have n independent Bernoulli trials, then the distribution of n trials is Bernoulli(n, p). Why isn't this satisfactory? $\endgroup$ – Tim Jan 24 at 23:02
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    $\begingroup$ Could you explain in what sense the sum of 0/1 variables can be interpreted as a "probability of success"? What shall we do about the fact that your Normal model for $\epsilon$ places some chance of $p$ lying outside the interval $[0,1],$ which is impossible? Regardless, is $\sigma^2$ known or not--and if not known, what data do you have to estimate it? This appears to be an effort to simplify some real problem, but the simplification seems to be creating confusion: what is the actual problem? $\endgroup$ – whuber Jan 24 at 23:19
  • $\begingroup$ Thanks for your comments, I fixed the definition problem of $p$. Also, $\sigma$ is known. $\endgroup$ – MaximeKan Jan 25 at 3:55
  • $\begingroup$ Your current question is answered here stats.stackexchange.com/q/177199/35989 Please don't make such drastic changes in your questions. After edits, this is a third question you ask, all of them were different. Rather then asking completely different question, you should rather ask this as a new question. $\endgroup$ – Tim Jan 25 at 5:52
  • $\begingroup$ Thanks @Tim, this is exactly what I was looking for. Sorry for the edits, in my mind these were just reformulations of the same question, which is why I just reedited until it gets clearer. $\endgroup$ – MaximeKan Jan 25 at 14:14