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What is the uncertainty (68% confidence level) of $N/M$, where $N$ is the number of entries that pass a cut and $M$ is the total number of entries? ($N$ and $M$ are both integers, and I'm interested in the extreme where $N$ or $M - N$ is a small integer, maybe zero.)

In the past, I've always assumed a Binomial model, where $N$ is the number of coin tosses that come up heads and $M$ is the total number of coin tosses. Following this logic, I've used the variance $Mp(1-p)$ to conclude that the uncertainty is $\sqrt{\frac{p(1-p)}{M}}$ (with $p=N/M$). However, I'm beginning to think this is flawed: this "uncertainty" is exactly zero if $p=0$ or $p=1$. Getting five heads in a row shouldn't lead one to conclude that the coin will always yield heads with perfect certainty.

In general, I think the upper uncertainty will be different from the lower uncertainty; that it should come from some integration that has a hard cut-off at $p=0$ and $p=1$, introducing an asymmetry close to the border. Should this come from a Bayesian formalism because I'm making inferences about an unknown distribution from measurement?

(I'm also surprised that I haven't found an answer online: I would have thought it to be a very common problem. Putting uncertainties on trigger efficiencies in physics, for instance.)

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    $\begingroup$ You may be interested in this thread: Confidence interval around binomial estimate of 0 or 1. $\endgroup$ May 30 '16 at 0:10
  • $\begingroup$ Thanks! That led me to R's binom package and en.wikipedia.org/wiki/Binomial_proportion_confidence_interval where I found (too many) answers to my question. Do you mind if I write it up as an answer to my own question? $\endgroup$ May 30 '16 at 0:44
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    $\begingroup$ Jim -- it would be great if you answer your own question (it's certainly okay to do so under the network rules); just make sure to credit your sources and give the source of any words you directly quote. However, if your answer is pretty much covered by the answers to an existing question you should instead flag this as a duplicate, so people can use searches that find your question to locate the answer (which will be linked-to from here if it's closed that way) $\endgroup$
    – Glen_b
    May 30 '16 at 0:48
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    $\begingroup$ Sure, you can answer your own question. If you think the linked thread has the answers to your question, then your Q can be considered a duplicate, so we should close it as such. (But you can always answer first.) $\endgroup$ May 30 '16 at 0:49
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Apparently, there are many answers to this question: it has its own Wikipedia page and R package. The uncertainty range I described above is the "normal approximation interval":

$\displaystyle p + z \sqrt{\frac{p(1 - p)}{M}}$

where $z$ is signed, $z=0$ is the central value ($p = N/M$), $z=1$ is "one sigma" (68% confidence level) above the central value and $z=-1$ is "one sigma" below the central value. It has the failures I discussed, and they stem from the fact that the distribution of true values around an observation are not themselves binomial.

There are many ways to correct the problem; this blog shows an overlay plot of the different estimators (drawn from the R package). Each has different properties and a different justification.

The simplest correction (if you're not running R) is the Wilson score interval:

$\displaystyle \frac{1}{1+z^2/M} \left(p + \frac{z^2}{2M} + z \sqrt{\frac{p(1-p)}{M} + \frac{z^2}{4M^2}} \right)$

Note that $z \to 0$ in the above yields $p$, so the central value is unchanged, but the positive and negative errors ($z = 1$ and $z = -1$) are now asymmetric and not equal to each other as $p \to 0$ or $1$.

I would take this as a simple formula to use for putting error bars on plots.

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I'd like to point out that the "exact" solution to this problem in the frequentist sense is the Clopper-Pearson interval. As the author noted, there are a number of nice approximate intervals with different characteristics, and the Wilson score interval is often a good choice. However I thought I would also point out that Bayesian inference with a uniform prior on $p$ yields a very simple formula for the posterior distribution for $p$

$$P(p|N,M) = \frac{(M+1)!}{N!(M-N)!} p^N (1-p)^{M-N}$$

with the following features:

$${\rm mode}(p|N,M) = \frac{N}{M}$$ $${\rm mean}(p|N,M) = \frac{N+1}{M+2}$$ $${\rm variance}(p|N,M)= \frac{(N+1) (N+2) }{(M+2)(M+3) } -\frac{(N+1)^2}{(M+2)^2} $$

Using [mean $\pm \sqrt{\rm variance}$] gives easy-to-compute 1$\sigma$ confidence intervals with very nice properties even when $N=0$ or $N=M$. One can even show that these intervals have descent frequentist coverage properties, especially at larger $M$, and only undercover when the "true" value of $p$ is unmeasurably close to 0 or 1.

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