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I'm trying to estimate the number of apples in an apple tree by repeatedly kicking the tree and counting how many apples fall down. This process, I believe, is called removal sampling.

The only assumption I'm making is that there is a constant probability $p$ that an apple falls down when I kick the tree.

Given that, e.g., [100, 10, 1, 0] apples fall down, I'm prone to believe that $p \sim 0.9$ and that $N$, the total number of apples, is 111.

I'm trying to estimate $N$ using MCMC, but I seem to struggle to get it right.

First, I've picked $N$ to be a uniform discrete prior in $[0, 1000]$ and $p$ is chosen to be a flat uniform over $[0, 1]$.

But, suppose that I remove, e.g., [19, 17, 13, 1, 1], I used

$$ \begin{align} N &= \text{DiscreteUniform}(0, 1000)\\ p &= \text{Uniform}(0, 1)\\ q_1 &= \text{Binomial}(N, p, \text{observed}=19)\\ q_2 &= \text{Binomial}(N - q_1, p, \text{observed}=17)\\ q_3 &= \text{Binomial}(N - (q_1 + q_2), p, \text{observed}=13)\\ q_4 &= \text{Binomial}(N - (\sum_j^{3} q_j), p, \text{observed}=1)\\ q_5 &= \text{Binomial}(N - (\sum_j^{4} q_j), p, \text{observed}=1) \end{align} $$

With this model, I end up with an estimate on $N \sim 54.5$. A different model I tried consistently get $N \leq 53$.

My question is the following:

Are my $q_i$ correctly modeled, or are they possibly the reason my model is over-estimating?

Do you see any obvious mistakes?


Simulating the aforementioned removal example

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    $\begingroup$ I don't know how this affects the posterior but N should be discrete for the binomial and your prior for N is continuous? $\endgroup$ – einar Oct 9 '20 at 10:47
  • $\begingroup$ Yes, N should be discrete, and I just used TruncatedNormal because I didn't know what else to use. This is literally all I know of my apple trees: N have been estimated previously to [6, 38, 53, 112, 140, 145, 264]. $\endgroup$ – Pål GD Oct 9 '20 at 11:52
  • $\begingroup$ When I tried with a flat discrete prior, it over-estimated much more, especially if the upper bound was too high (e.g. 1000). $\endgroup$ – Pål GD Oct 9 '20 at 11:53
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    $\begingroup$ I believe it over-estimates because there are two industry standard ways of estimating, and they both give lower numbers than my MCMC. $\endgroup$ – Pål GD Oct 9 '20 at 12:27
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    $\begingroup$ Although you write you "end up with an estimate," I cannot find any description of how you make that estimate. What is your method?? Running a simulation cannot succeed because the logic is circular: in order to simulate you need to supply estimates in the first place. $\endgroup$ – whuber Oct 9 '20 at 13:59
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Your results look reasonable given your model and your other assumptions. I can't speak to whether the model (and the assumptions) are themselves reasonable.

I'm going to change the notation a bit because I like to use the "$p$" to denote a probability density (or mass) function. So I'll use $\theta$ as the probability of "success" instead.

The observations are given by \begin{equation} q_{1:K} = (q_1, \ldots, q_K) . \end{equation} The likelihood is \begin{equation} p(q_{1:K}|N,\theta) = p(q_1|N,k)\prod_{k=2}^K p(q_k|q_{1:k-1},N,\theta) , \end{equation} where \begin{align} p(q_1|N,\theta) &= \textsf{Binomial}(q_1|N,\theta) \\ p(q_k|q_{1:k-1},N,\theta) &= \textsf{Binomial}\left(q_k\Big|N - \sum_{j=1}^{k-1} q_j,\theta\right) . \end{align} The prior for the latent variables $N$ and $\theta$ is flat. Therefore, the posterior is proportional to the likelihood: \begin{equation} p(N,\theta|q_{1:K}) \propto p(q_{1:K}|N,\theta) . \end{equation}

We can apply the Metropolis-within-Gibbs sampler. Let $(\theta^{(r)},N^{(r)})$ denote the current state of the chain. The full-conditional posterior distribution for $\theta$ delivers the following: \begin{equation} \theta^{(r+1)} \sim p(\theta|q_{1:K},N^{(r)}) = \textsf{Beta}(\theta|a,b^{(r)}) \end{equation} where \begin{align} a &= 1 + \sum_{k=1}^K q_k \\ b^{(r)} &= 1 + K\,N^{(r)} - \sum_{k=1}^K (K-k+1)\,q_k . \end{align} To draw $N^{(r+1)}$ we can take a Metropolis step, using a symmetric uniform discrete proposal. Let $N' = N^{(r)} + \delta$ where $\delta \sim \textsf{Uniform}(-3,3)$ for example. Then \begin{equation} N^{(r+1)} = \begin{cases} N' & M \ge u \\ N^{(r)} & \text{otherwise} \end{cases} , \end{equation} where $u \in \textsf{Uniform}(0,1)$ and \begin{equation} M = \frac{p(q_{1:K}|N',\theta^{(r+1)})}{p(q_{1:K}|N^{(r)},\theta^{(r+1)})} . \end{equation} Given the draws $\{(\theta^{(r)},N^{(r)})\}_{r=1}^R$ one can produce figures that are similar to those in the question (except that the distribution for $N$ should be discrete). The posterior mode for $N$ is 53 and the posterior mean is about 55. There's about a 10% chance that $N\ge 60$.

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    $\begingroup$ Notational quibble: should it not be $p(q_k | N, q_1, \ldots, q_{k-1}, \theta) \sim Binomial(.)$? The full likelihood being p(q_{1:k}|N,\theta), with $q_i$s not independent but can be factorized $p(q_1 | N, \theta)p(q_2 | N, q_1, \theta),$ etc.? $\endgroup$ – einar Oct 14 '20 at 8:52
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    $\begingroup$ @einar Quite right. Thank you. $\endgroup$ – mef Oct 14 '20 at 9:09

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