Capture–recapture estimation based upon the geometric distribution allowing for heterogeneity

Abstract

Capture–Recapture methods aim to estimate the size of an elusive target population. Each member of the target population carries a count of identifications by some identifying mechanism—the number of times it has been identified during the observational period. Only positive counts are observed and inference needs to be based on the observed count distribution. A widely used assumption for the count distribution is a Poisson mixture. If the mixing distribution can be described by an exponential density, the geometric distribution arises as the marginal. This note discusses population size estimation on the basis of the zero-truncated geometric (a geometric again itself). In addition, population heterogeneity is considered for the geometric. Chao’s estimator is developed for the mixture of geometric distributions and provides a lower bound estimator which is valid under arbitrary mixing on the parameter of the geometric. However, Chao’s estimator is also known for its relatively large variance (if compared to the maximum likelihood estimator). Another estimator based on a censored geometric likelihood is suggested which uses the entire sample information but is less affected by model misspecifications. Simulation studies illustrate that the proposed censored estimator comprises a good compromise between the maximum likelihood estimator and Chao’s estimator, e.g. between efficiency and bias.

Appendices

Appendix 1: Proof of theorems

Theorem 1

Let $k_y(p)=(1-p)^yp$ for $y=0,1,\cdots$ and $p \in (0,1)$.

• (a) Let $\log L(p) = f_1\log (\pi _1)+f_2\log (\pi _2)$ with $\pi _1 = 1/(2-p)$ and $\pi _2 =(1-p)/(2-p)$ being the geometric probabilities truncated to counts of ones and twos. Then $\log L(p)$ is maximized for $\hat{p} = (f_1-f_2)/f_1.$

• (b) $E(f_0|f_1,f_2;\hat{p})=f_1^2/f_2,$ for $\hat{p}= (f_1-f_2)/f_1$.

Proof

For the first part, it is clear that $f_1\log (\pi _1)+f_2\log (\pi _2)$ is maximal for $\hat{\pi }_1 =f_1/(f_1+f_2)=1/(2-\hat{p})$, which is attained for $\hat{p} = (f_1-f_2)/f_1$. For the second part, we see that with $e_y=E(f_y|f_1,f_2;p)= k_y(p) N$ we have the following:

$$\begin{aligned} e_y = k_y(p) N=k_y(p)\left(e_0+f_1+f_2+\sum _{j=3}^\infty e_j\right) \end{aligned}$$

so that

$$\begin{aligned} e_0+e_3^+=[1-k_1(p)-k_2(p)] (e_0+e_3^+) + [1-k_1(p)-k_2(p)] (f_1+f_2) \end{aligned}$$

with $e_3^+= \sum _{j=3}^{\infty } e_j$. Hence

$$\begin{aligned} e_0+e_3^+=\frac{1-k_1(p)-k_2(p)}{k_1(p)+k_2(p)}(f_1+f_2) \end{aligned}$$

and

$$\begin{aligned} e_0=k_0(p)(f_1+f_2+ e_0+e_3^+)&= k_0(p)(f_1+f_2)\left[1+\frac{1-k_1(p)-k_2(p)}{k_1(p)+k_2(p)}\right]\\&= \frac{k_0(p)}{k_1(p)+k_2(p)}(f_1+f_2)=\frac{f_1+f_2}{(1-p)(2-p)}. \end{aligned}$$

Plugging in the maximum likelihood estimate $\hat{p}=(f_1-f_2)/f_1$ for $p$ yields

$$\begin{aligned} \frac{f_1+f_2}{(1-\hat{p})(2-\hat{p})}=\frac{f_1+f_2}{\frac{f_2}{f_1}\frac{f_1+f_2}{f_1}}=f_1^2/f_2, \end{aligned}$$

the desired result. $\square$

Theorem 2

Let $k_y(p)=(1-p)^yp$ for $y=0,1,\cdots$ and $p \in (0,1)$. Then,

$$\begin{aligned} \lim _{N\rightarrow \infty }\frac{E(\hat{N})}{N} =1 \end{aligned}$$

for $\hat{N}=\hat{N}_{ML}, \hat{N}_{C}$, or $\hat{N}_{Cen}$.

Proof

Let $\hat{N}=\hat{N}_{ML}=n/(1-n/S)$. Note that $E(n)=Np$ and $E(S/N)=(1-p)/p$ so that

$$\begin{aligned} \frac{E(n/(1-n/S))}{N} \mathop {\rightarrow }\limits _{N\rightarrow \infty } \frac{p}{1-\frac{p}{p/(1-p)}}=1. \end{aligned}$$

Let $\hat{N}=\hat{N}_{C}=n + f_1^2/f_2$. Note that $E(f_1)=Np(1-p)$ and $E(f_2)=Np(1-p)^2$ so that

$$\begin{aligned} \frac{E(n+f_1^2/f_2)}{N} \mathop {\rightarrow }\limits _{N\rightarrow \infty } (1-p) + \frac{p^2(1-p)^2}{p(1-p)^2}=1. \end{aligned}$$

Finally, let $\hat{N}=\hat{N}_{Cen}=\frac{n}{1-f_1/n}$. Using the above we have

$$\begin{aligned} \frac{E\left(\frac{n}{1-f_1/n}\right)}{N} \mathop {\rightarrow }\limits _{N\rightarrow \infty } \frac{1-p}{1-\frac{(1-p)p}{(1-p)}}=1, \end{aligned}$$

which ends the proof. $\square$

Appendix 2: Standard errors

Let $\hat{N}$ be the estimator of the population size $N$ of interest, the latter being a fixed but unknown quantity. Also, let the random quantity $n$ be the observed number of units. We will make use of the result

$$\begin{aligned} \text{ Var}(\hat{N}) = E_n \{\text{ Var}(\hat{N}|n)\} + \text{ Var}_n\{E(\hat{N}|n)\}, \end{aligned}$$

(9)

where $\hat{N}|n$ refers to the distribution of $\hat{N}$ conditional upon $n$ and $E_n(.)$ and $\text{ Var}_n(.)$ refer to the first and second (central) moment w.r.t. the distribution of $n$. For more details see Böhning (2008).

1.1 Maximum likelihood estimator

We consider the maximum likelihood estimator $\hat{p}_{ML}=n/S$ and the associated population size estimator $\hat{N}=\hat{N}_{ML}=n/(1-n/S)$. We start with the second term in (9) and have that $E(\hat{N}|n)\approx n/(1-p)$, approximately, so that

$$\begin{aligned} \text{ Var}_n [n/(1-p)] = \frac{1}{(1-p)^2}Np(1-p). \end{aligned}$$

Note that $N(1-p)$ can be estimated by $n$ and $p$ by the maximum likelihood estimator $n/S$, so that the variance estimator $\frac{Sn^2}{(S-n)^2}$ arises.

For the first term in (9), we use the $\delta$-method to determine $\text{ Var}(\hat{N}|n)$ as

$$\begin{aligned} \frac{n^2}{(1-p)^4} \text{ Var}_n(\hat{p}_{ML}) \end{aligned}$$

and, using the Fisher information for $p$, we can determine $\text{ Var}_n(\hat{p}_{ML})$ as

$$\begin{aligned} \text{ Var}_n(\hat{p}_{ML}) \approx \frac{n(S-n)}{S^3}. \end{aligned}$$

The expected value $E_n \{\text{ Var}(\hat{N}|n)\}$ is then replaced by its moment estimate $\text{ Var}(\hat{N}|n)$ to achieve the total variance

$$\begin{aligned} \frac{Sn^2}{(S-n)^2}+\frac{n^2}{(1-n/s)^4}\frac{n(S-n)}{S^3} = \frac{S^2n^2}{(S-n)^3} \end{aligned}$$

(10)

1.2 Censored estimator

We consider the censored estimator $\hat{p}_{Cen}=f_1/n$ and the associated population size estimator $\hat{N}=\hat{N}_{Cen}=n/(1-f_1/n)$. We have $E(\hat{N}|n)\approx n/(1-p)$, approximately, so that, as before, $\text{ Var}_n n/(1-p) = \frac{1}{(1-p)^2}Np(1-p)$, which can be estimated as $\frac{f_1}{(1-f_1/n)^2}$ by replacing $N(1-p)$ by $n$ and $p$ by $f_1/n$.

For the first term in (9), $\text{ Var}(\hat{N}|n)$, using the $\delta -$method once more we achieve the approximation

$$\begin{aligned} \text{ Var}\left(\frac{n}{1-f_1/n}|n\right)\approx \frac{n^2}{(1-f_1/n)^4} \text{ Var}\left(\frac{f_1}{n}|n\right), \end{aligned}$$

from where the variance estimator $\frac{f_1(1-f_1/n)}{(1-f_1/n)^4}=\frac{f_1}{(1-f_1/n)^3}$ arises. In total, taking both variance terms into account, we achieve the variance estimator

$$\begin{aligned} \frac{f_1}{(1-f_1/n)^2}+\frac{f_1}{(1-f_1/n)^3} = \frac{f_1}{(1-f_1/n)^2}\frac{2n-f_1}{n-f_1} \end{aligned}$$

(11)

1.3 Chao’s estimator

Finally, we consider the Chao-type estimator $\hat{N}= \hat{N}_{C}=n+f_1^2/f_2$. Note that it differs from the original Chao-estimator $n+f_1^2/(2f_2)$ for which a variance estimator is provided in Chao (1987). If we would be only interested in a variance estimator of $\hat{f}_0$ we could simply multiply the Chao-variance-estimator by a factor of 4. However, interest is usually in the population size estimator $\hat{N}$ for which this simple adjustment is not valid. Hence we provide a full analysis in the following, again using the conditioning technique (9).

We have $E(\hat{N}|n)= E(n+\frac{f_1^2}{f_2})\approx n + (g_1^+)^2 n/g_2^+=n(1+(g_1^+)^2/g_2^+)$, approximately. Recall that $g_y^+=g_y/(1-g_0)$ for $y=1,2,\ldots$. (Note that $E(\hat{N}|n)$ refers to the conditional count distribution $g_1^+, g_2^+, \ldots$ which is estimated by $f_1/n, f_2/n, \ldots$. Hence $\text{ Var}_n \{n(1+(g_1^+)^2/g_2^+)\}= (1+(g_1^+)^2/g_2^+)^2 Ng_0(1-g_0)$ which can be estimated as follows. $Ng_0$ can be estimated as $\hat{f}_0 =f_1^2/f_2$ and $(1-g_0)$ as $1-f_1^2/(\hat{N} f_2)=\frac{f_2n}{f_2n+f_1^2}$, so that in total the estimate $(1+\frac{f_1^2}{f_2 n})^2 \frac{f_1^2n}{f_2n+f_1^2}$ arises, which we can simplify as

$$\begin{aligned} \left(1+\frac{f_1^2}{f_2n}\right)^2 \frac{f_1^2n}{f_2n+f_1^2}= f_1^2/f_2+f_1^4/(f_2^2n). \end{aligned}$$

(12)

For the first term in (9), $\text{ Var}(\hat{N}|n)$, using the bivariate $\delta -$method, we achieve the approximation

$$\begin{aligned} \text{ Var}(\hat{N}|n) \approx \nabla \phi _0(f_1,f_2)^T cov(f_1,f_2) \nabla \phi _0(f_1,f_2) \end{aligned}$$

where $\phi _0(f_1,f_2)=f_1^2/f_2$ and $\nabla \phi _0(f_1,f_2)$ is the two-vector of partial derivatives with respect to $f_1$ and $f_2$:

$$\begin{aligned} \nabla \phi _0(f_1,f_2)^T=(2f_1/f_2,-f_1^2/f_2^2). \end{aligned}$$

The covariance matrix, conditional on $n$, is $cov(f_1,f_2)=n (\text{ dia}({{\mathbf g}^{+}}) -{{\mathbf g}^{+}}{{\mathbf g}^{+T}})$, where ${\mathbf g}^{+}$ is the two-vector of probabilities, conditional on $n$, for observing a one or a two, respectively. Also, $\text{ dia}({{\mathbf g}^+})$ is the diagonal $2\times 2$ matrix with $g_1^+$ and $g_2^+$ on the main diagonal. This matrix is estimated by

$$\begin{aligned} \left(\begin{array}{ll} f_1-f_1^2/n&-f_1f_2/n\\ -f_1f_2/n&f_2-f_2^2/n \end{array}\right). \end{aligned}$$

Hence we find for

$$\begin{aligned} \nabla \phi _0(f_1,f_2)^T \widehat{cov}(f_1,f_2) \nabla \phi _0(f_1,f_2)=\frac{4f_1^3}{f_2^2}+\frac{f_1^4}{f_2^3} -\frac{f_1^4}{f_2^2n}. \end{aligned}$$

(13)

Ultimately, taking (12) and (13) together, we achieve the variance estimator for $\hat{N} =\hat{N}_C=n+f_1^2/f_2$ as

$$\begin{aligned} \frac{f_1^4}{f_2^3}+\frac{4f_1^3}{f_2^2}+\frac{f_1^2}{f_2}, \end{aligned}$$

(14)

being of remarkably simple form.