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Say $k \sim Bin(N, p)$. What is the distribution of $N$, given fixed $p$ and $k$? Looks like Poisson but starting at k instead of zero (???)

Thanks.

EDIT: Application: I have some real number of birds ($N$), each is seen with probability $p$. I see $k$ birds and want to estimate the real number of birds ($N$).

mpiktas
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Tomas
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    $N$ has no distribution unless you assume one or unless you have some kind of stopping rule for a sequential experiment. What's your situation? – whuber Aug 23 '11 at 19:27
  • My situation is as I described - I have `k` which I know is a draw from some `Bin(N, p)`, I also know p and I want to know distribution of N... – Tomas Aug 23 '11 at 21:28
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    That's not sufficient information to answer the question unambiguously. If that's *all* you know, there are two classes of answer. One is that $N$ does not have a distribution (but you could at least compute a confidence interval for $N$). The other is that if you assume some prior distribution for $N$, you can use Bayes' Theorem to update it based on $k$ and $p$. – whuber Aug 23 '11 at 21:31
  • Well, I'm afraid that's all I know now, I also don't have any prior, moreover [I don't much believe in priors yet](http://stats.stackexchange.com/questions/14721/how-is-the-bayesian-framework-better-in-interpretation-when-we-usually-use-uninfo) :-) – Tomas Aug 23 '11 at 21:40
  • Say that I would use as much uninformative prior as possible... like uniform distrubution from 0 to some very high number or something like that... how the distribution of N would look like then? – Tomas Aug 23 '11 at 21:44
  • @whuber I don't think he's looking to perform inference and estimate N here; I think you may be confusing him by introducing priors at all (but of course this is an exercise in mind-reading). Tomas: we need more information about where this data comes from if you want a precise answer. My guess is that you are sampling from a Bernoulli distribution with parameter p until you reach k successes, and want to know the distribution of the total number of trials until this occurs. Is that correct? – bnaul Aug 23 '11 at 22:06
  • @bnaul, no. I have some real number of birds (`N`), each is seen with probability `p`. I see `k` birds and want to know the real number of birds (`N`). Thank you both for reading my mind and trying to understand my question :-) – Tomas Aug 23 '11 at 22:22
  • You are asking to *estimate* $N$. You might also want to assess the error of your estimate by computing a *standard error of estimate* and/or a *confidence interval* for $N$. However, I doubt that a binomial sampling model is accurate. How do you know each of the $N$ birds has an equal chance of being observed and that the observations are mutually independent? Do you even know these birds are distinct? – whuber Aug 23 '11 at 22:31
  • Guess I was way off then. In that case, I agree that a binomial model seems inappropriate here. If you're trying to estimate a population size, then there are many methods for doing so, but I am not an expert in any of them. Perhaps you should be asking about that more general problem instead. – bnaul Aug 24 '11 at 04:44
  • Possible duplicate of [Estimator for a binomial distribution](https://stats.stackexchange.com/questions/16686/estimator-for-a-binomial-distribution) – kjetil b halvorsen Mar 01 '19 at 16:43
  • I don't think this is a duplicate - here $p$ is fixed and the question is asking for a (posterior?) distribution while the possible duplicate is about point estimators for $n,p$. – Juho Kokkala Mar 02 '19 at 11:51

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Your question is somewhat ambiguous, but I think that you might be talking about the negative binomial distribution:

...the negative binomial distribution is a discrete probability distribution of the number of successes in a sequence of Bernoulli trials before a specified (non-random) number r of failures occurs.

It seems that you're asking about the number of trials needed before k successes, which is the same as N-k failures. It actually is quite close to the Poisson distribution, so your intuition was not far off.

bnaul
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    @Tomas It's impossible to say whether this answer is correct until you clarify the question. – whuber Aug 23 '11 at 21:27
  • Thanks bnaul! Negative binomial seems great! So the distribution for `N` would then be `N ~ k + NBin(k, 1 - p)`, right? – Tomas Jan 18 '21 at 20:58