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 (???)


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 ($N$).

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    $\begingroup$ $N$ has no distribution unless you assume one or unless you have some kind of stopping rule for a sequential experiment. What's your situation? $\endgroup$ – whuber Aug 23 '11 at 19:27
  • $\begingroup$ 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... $\endgroup$ – Curious Aug 23 '11 at 21:28
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    $\begingroup$ 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 $p$. $\endgroup$ – whuber Aug 23 '11 at 21:31
  • $\begingroup$ 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 :-) $\endgroup$ – Curious Aug 23 '11 at 21:40
  • $\begingroup$ 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? $\endgroup$ – Curious Aug 23 '11 at 21:44

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.

  • $\begingroup$ OK, so the answer would then be N ~ k + NBin(k, 1 - p), is that correct? $\endgroup$ – Curious Aug 23 '11 at 21:18
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    $\begingroup$ @Tomas It's impossible to say whether this answer is correct until you clarify the question. $\endgroup$ – whuber Aug 23 '11 at 21:27

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