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Hi everyone

The question is the following:

A certain event may or may not take place. So we say that if we focus on it one time, it has a probability p of being satisfied (0 <= p < 1)

If we observe it multiple times, and we find out that it occurred zero times among our n observation, what can we say about p? What is the most likely value for p? Or better, find a function k(x, n) that returns the probability that p = x

How does it all change if among our n observation the event occurred m times? I know that if n tens to infinity then p tends to m/n but that is only part of the question

Thank you

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Probably this is off-topic for this site. In any case, no answer can be given without more information, such as the joint distribution of the observations. –  Gerald Edgar Sep 29 '12 at 13:24
    
Thanks for the comment, but I am not a mathematician yet, so I do not understand what's missing.. –  Ant Sep 29 '12 at 13:45
    
If the events are not independent, what happens? For example, if all observations will surely be the same. Then: observing 0 multiple times is no better than observing it once. –  Gerald Edgar Sep 29 '12 at 17:30
    
well I should have mentioned that but of course I meant that the events are independent from each other.. –  Ant Sep 29 '12 at 18:00
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closed as off topic by Steven Landsburg, Gerald Edgar, Bill Johnson, George Lowther, Anthony Quas Sep 29 '12 at 18:11

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1 Answer

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I'll interpret "most likely" as Maximum Likelihood Estimation, that is, given some observations, what is the value of $p$ that makes the probability of those observations the largest. For example, if we observe something occurring $0$ out of $n$ times, the maximum likelihood estimate for $p$ is $0$, because that gives our observations a probability of $1$. This is only unreasonable if you have some prior information about $p$ (such as "it's probably around $0.5$, but maybe a little higher or lower" or "it's very close to either $1$ or $0$, but I don't know which").

Suppose we run $n$ trials, and find that the event occurred $n$ times, and we assume that each trial is independent with the same probability $p$ of success. Then the probability of our observation is $$P(p) = {n \choose m} p^m (1-p)^{n-m},$$ since the probability of each of the $m$ successes is $p$, the probability of the $n-m$ failures is $1-p$, and there are ${n\choose m}$ ways for $m$ of the $n$ trials to be the successful trials. Assuming $m$ and $n-m$ are both nonzero, this probability vanishes when $p=0$ or $1$, so the maximizing value of $p$ will be somewhere in between. Then we can find this maximizing value of $p$ by taking the derivative of $P$ and setting the result equal to $0$. We can take the derivative: $$P'(p) = {n\choose m} \left[mp^{m-1}(1-p)^{n-m} - (n-m)p^m(1-p)^{n-m-1}\right]$$ $$ = {n\choose m} \left[m(1-p) - (n-m)p\right]p^{m-1}(1-p)^{n-m-1}$$ This vanishes when $m(1-p)-(n-m)p=0$, i.e. when $m = np$ or $p = m/n$.

So if you have no prior information about what $p$ should be, but you observe $m$ successes in $n$ independent trials, the value of $p$ that best matches your observation is $p=m/n$.

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That was easier than i thought, thank you! Though I do not understand why they closed the question as off-topic since it's not; one on mathoverflow can only ask something about Abelian rings to be answered? –  Ant Sep 30 '12 at 9:28
    
@Ant, I'd say that it was closed because it's quite far from research level... –  Yvan Velenik Sep 30 '12 at 13:05
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