I have a finite set $X$ of unknown size. From that I independently draw a multiset of samples $S\subset X$. For each sample $s\in S$ I can give the exact probability $p(s)$ with which it has been drawn from $X$ such that $\sum_{x\in X}p(x)=1$ and how often it has been drawn. I also have a lower bound $l\le p(s)$ for the probability of each sample.

Given this information, how can I estimate the size of $X$?

A mark and recapture approach is not going to work because $|S| \ll |X|$ by several orders of magnitude. I have drawn about a billion of samples so far without encountering any duplicates.

  • 1
    $\begingroup$ Statement not clear (to me anyhow). Perhaps look at mark-recapture $\endgroup$ – BruceET Oct 8 '19 at 16:13
  • $\begingroup$ @BruceET The population $X$ is much larger than the sample size I can reasonably achieve. I have so far drawn a few billion samples without any duplicates, so I don't think mark-recapture is feasible. What part of the statement is unclear? $\endgroup$ – fuz Oct 8 '19 at 16:20
  • $\begingroup$ "For each sample $S∈s$ I can give the probability $p(S)$ it has to be drawn from $X$ such that ..." Maybe give examples from your work. How do you get the probabilities? Exact or estimated? $\endgroup$ – BruceET Oct 8 '19 at 16:28
  • $\begingroup$ @BruceET I can compute these probabilities exactly. I am working on the same problem this question is about (sampling vertices of a graph that have a given distance from a fixed vertex). Once I found a vertex with the right distance, I can exactly compute the probability with which it was found. $\endgroup$ – fuz Oct 8 '19 at 16:33
  • $\begingroup$ @BruceET Also, I'm sorry for the wrong spelling and grammar. I have fixed the post. $\endgroup$ – fuz Oct 8 '19 at 16:35

DISCLAIMER: Simulation has shown that this is not a good estimator but it might be helpful for further ideas

I propose the following estimator: With sample space $S$ and sample $s_1, ..., s_n$, we could use $\widehat{\#S}= n \times\frac{1}{\sum_{i=1}^np(s_i)}$ as an estimator for $\#S$

The theoretical idea is the following:

We have a random Variable $X$ that has the probability values $p_s:=p(s)$ as the sample space and a distribution, such that $P(X=p_s)=p(s)$, i.e. the numeric value of the sample and its probability coincide.

We then estimate the mean of this Random Variable $X$, i.e. $E[X]=\sum_{s\in S}p(s)^2$ by taking the sample mean $\sum_{i=1}^np(s_i)$ In addition to that we use the fact that $nE[X]=1 \iff n = \frac{1}{E[X]}$

So we use an unbiased estimator for $E[X]$ and plug it into the formula $n=\frac{1}{E[X]}$ (which means that the estimator for $\#S$ is not necessarily unbiased anymore.

Maybe we could also estimate the distribution of $X$ in order to get a correction-factor of our estimate that takes into consideration how much $1/E[X]$ deviates from $E[1/X]=n$.

| cite | improve this answer | |
  • $\begingroup$ I do draw individual elements. $S\subset X$ is the set of samples I draw of which the probability $p(s)$ for each $s\in S$ is known. Here, $p(s)$ is the probability of drawing $s$ from $X$. $\endgroup$ – fuz Oct 8 '19 at 20:02
  • $\begingroup$ Ok then the last paragraph is unnecessary but the rest might still help you. I will simulate this procedure tomorrow to check how well it works and update my post. $\endgroup$ – Sebastian Oct 8 '19 at 20:07
  • $\begingroup$ I actually thought about something similar. We do know about the distribution of probabilities because we sampled them, so I guess there should be a way to make that exact. $\endgroup$ – fuz Oct 8 '19 at 20:32
  • $\begingroup$ Re your formula: although this statistic is an estimator, what properties does it have? Why should it be any good at all? I don't follow your remarks about "uniformly distributed," because if that assumption is true, then all you have to do is compute a single probability $p_i$ and recover the population size as $1/p_i.$ $\endgroup$ – whuber Oct 8 '19 at 21:09
  • $\begingroup$ As in the first line: "This is not a known method (at least to me) but just an idea that makes sense to me and might be worth exploring." $\endgroup$ – Sebastian Oct 8 '19 at 21:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.