On the Turing estimator in capture–recapture count data under the geometric distribution

Abstract

We introduce an estimator for an unknown population size in a capture–recapture framework where the count of identifications follows a geometric distribution. This can be thought of as a Poisson count adjusted for exponentially distributed heterogeneity. As a result, a new Turing-type estimator under the geometric distribution is obtained. This estimator can be used in many real life situations of capture–recapture, in which the geometric distribution is more appropriate than the Poisson. The proposed estimator shows a behavior comparable to the maximum likelihood one, on both simulated and real data. Its asymptotic variance is obtained by applying a conditional technique and its empirical behavior is investigated through a large-scale simulation study. Comparisons with other well-established estimators are provided. Empirical applications, in which the population size is known, are also included to further corroborate the simulation results.

Appendix: Proof of Proposition 1

According to the conditional technique, we have

$$\begin{aligned} \textit{Var}(\hat{N}_{\textit{TG}}) = \textit{Var}_{n} \left\{ E(\widehat{N}_{\textit{TG}}|n) \right\} + E_{n} \left\{ \textit{Var}(\widehat{N}_{\textit{TG}}|n) \right\} . \end{aligned}$$

(8)

Starting from the first term on the right hand side of (8), the delta method we have $E(\widehat{N}_{\textit{TG}}|n)\approx \frac{n}{1-\kappa _0}$ and, accordingly,

$$\begin{aligned} \textit{Var}_{n} \left\{ E(\widehat{N}_{\textit{TG}}|n) \right\}\approx & {} \textit{Var}_{n} \left\{ \frac{n}{1-{\kappa }_0} \right\} = \frac{1}{(1-{\kappa }_0)^{2}} \textit{Var}(n) = \frac{N(1-{\kappa }_0){\kappa }_0}{(1-{\kappa }_0)^{2}}. \end{aligned}$$

(9)

Since $E(n) = N(1-{\kappa }_0)$ and $\widehat{\kappa }_{0(TG)} = \sqrt{\frac{f_{1}}{S}}$, the variance in (9) can be estimated as:

$$\begin{aligned} \widehat{\textit{Var}}_{n}\left\{ E(\widehat{N}_{\textit{TG}}|n) \right\} = \frac{n \sqrt{\frac{f_{1}}{S}}}{\left( 1- \sqrt{\frac{f_{1}}{S}}\right) ^{2} }. \end{aligned}$$

Additionally,

$$\begin{aligned} \textit{Var}(\widehat{N}_{\textit{TG}}|n)= & {} \textit{Var}\left( \frac{n}{1-\sqrt{\frac{f_{1}}{S}}}|n \right) = n^{2} \textit{Var}\left( \frac{1}{1-\sqrt{\frac{f_{1}}{S}}}\right) . \end{aligned}$$

We know that $\textit{Var}\Big ( \frac{1}{1-\sqrt{\frac{f_{1}}{S}}}\Big )$ can be approximated by the delta-method. Hence, let $y= \frac{f_{1}}{S}$ and we take $h(y)=\frac{1}{1-\sqrt{y}}$. Then,

$$\begin{aligned} h'(y)=-(1-y^{1/2})^{-2} \left( -\frac{1}{2}y^{-1/2} \right) =\frac{1}{2\sqrt{y}(1-\sqrt{y})^{2}}. \end{aligned}$$

Furthermore,

$$\begin{aligned} \textit{Var}\left( \frac{1}{1-\sqrt{\frac{f_{1}}{S}}}|n\right)\approx & {} \left( \frac{1}{2\sqrt{y}(1-\sqrt{y})^{2}} \right) ^{2}{} \textit{Var}\left( \frac{f_{1}}{S} \right) \\= & {} \left( \frac{1}{4\frac{f_{1}}{S}\left( 1-\sqrt{\frac{f_{1}}{S}}\right) ^{4}} \right) \textit{Var}\left( \frac{f_{1}}{S} \right) . \end{aligned}$$

As next step, using the conditional variance technique to estimate $\textit{Var}\left( \frac{f_{1}}{S} \right)$, we have that

$$\begin{aligned} \textit{Var}\left( \frac{f_{1}}{S} \right)= & {} \textit{Var}_{f_{1}}\left\{ E\left( \frac{f_{1}}{S}\right) |f_{1} \right\} +E_{f_{1}}\left\{ \textit{Var} \left( \frac{f_{1}}{S}|f_{1} \right) \right\} . \end{aligned}$$

(10)

With the approximation $E\left( \frac{f_{1}}{S}|f_{1} \right) = f_{1}E(\frac{1}{S}) \approx \frac{f_{1}}{S}$, we have that

$$\begin{aligned} \textit{Var}_{f_{1}}\left\{ E\left( \frac{f_{1}}{S} |f_{1}\right) \right\}\approx & {} \textit{Var}_{f_{1}}\left( \frac{f_{1}}{S} \right) = \frac{1}{S^{2}} \textit{Var}(f_{1}) = \frac{1}{S^{2}} Np_{1}(1-p_{1}) \nonumber \\= & {} \frac{1}{S^{2}}\left( N\frac{f_{1}}{N} \left( 1-\frac{f_{1}}{N}\right) \right) = \frac{f_{1}}{S^{2}}\left( 1-\frac{f_{1}}{N}\right) . \end{aligned}$$

(11)

Again, estimating $E_{f_{1}} \left\{ \textit{Var}\left( \frac{f_{1}}{S}|f_{1} \right) \right\}$ by $\textit{Var}\left( \frac{f_{1}}{S}|f_{1} \right)$ we have that

$$\begin{aligned} E_{f_{1}} \left\{ \textit{Var}\left( \frac{f_{1}}{S}|f_{1} \right) \right\}\approx & {} \textit{Var}\left( \frac{f_{1}}{S}|f_{1} \right) = f_{1}^{2} \textit{Var}\left( \frac{1}{S}\right) \end{aligned}$$

Using the delta method, we achieve that

$$\begin{aligned} \textit{Var}\left( \frac{1}{S}\right)\approx & {} \frac{1}{S^{4}} \textit{Var}(N\bar{X}) = \frac{1}{S^{4}}N^{2} \textit{Var}(\bar{X}) = \frac{1}{S^{4}}N^{2} \frac{\textit{Var}(X)}{N}. \end{aligned}$$

Since $X\sim \textit{Geo}(p)$ we have that $E(X)=\frac{1-p}{p}$ and $\textit{Var}(X)=\frac{1-p}{p^{2}}$.

$$\begin{aligned} \textit{Var}\left( \frac{1}{S}\right)\approx & {} \frac{1}{S^{4}}N^{2} \frac{ \left( \frac{1-p}{p^{2}}\right) }{N} = \frac{1}{S^{4}}N^{2} \frac{ \left( \frac{E(X)}{p}\right) }{N} =\frac{1}{S^{4}}N^{2} \frac{ \left( \frac{E(S/N)}{p}\right) }{N} \approx \frac{1}{pS^{3}}. \end{aligned}$$

Let us note that

$$\begin{aligned} E\left( \frac{S}{N}\right) = \frac{1-p}{p}; \quad \frac{S}{N} \approx \frac{1-p}{p} \quad \mathrm{or} \quad p(S+N)\approx & {} N \quad \mathrm{or} \quad \frac{1}{p} \approx \frac{S+N}{N}.\qquad \end{aligned}$$

(12)

Hence,

$$\begin{aligned} \widehat{\textit{Var}}\left( \frac{f_{1}}{S}|f_{1} \right) =\frac{f_{1}^{2}}{S^{3}}\left( \frac{S+N}{N} \right) . \end{aligned}$$

Substituting (11) and (12) into (10), this leads to

$$\begin{aligned} \widehat{\textit{Var}} \left( \frac{f_{1}}{S} \right)= & {} \frac{1}{S^{2}}\left\{ f_{1}\left( 1-\frac{f_{1}}{N} \right) \right\} +\frac{f_{1}^{2}}{S^{3}}\left( \frac{S+N}{N} \right) \\= & {} \frac{f_{1}}{S^{2}} \left\{ \frac{N+f_{1}}{N}+\frac{f_{1}}{S}\left( \frac{S+N}{N} \right) \right\} \nonumber \\= & {} \frac{f_{1}}{S^{2}}\left\{ \frac{NS-Sf_{1}+f_{1}S+f_{1}N}{NS} \right\} =\frac{f_{1}}{S^{2}} \left\{ \frac{N(S+f_{1})}{NS} \right\} =\frac{f_{1}S+f_{1}^{2}}{S^{3}}. \end{aligned}$$

We have that

$$\begin{aligned} \widehat{\textit{Var}} \left( \frac{1}{1-\sqrt{\frac{f_{1}}{S}}} \right)= & {} \left\{ \frac{1}{\frac{4f_{1}}{S} \left( 1-\sqrt{\frac{f_{1}}{S}} \right) ^{4}} \right\} \left\{ \frac{f_{1}S+f_{1}^{2}}{S^{3}} \right\} = \widehat{\textit{Var}} \left( \frac{1}{1-\sqrt{\frac{f_{1}}{S}}} \right) \nonumber \\= & {} \left\{ \frac{S}{4f_{1}\left( 1-\sqrt{\frac{f_{1}}{S}} \right) ^{4}} \right\} \left\{ \frac{f_{1}S+f_{1}^{2}}{S^{3}} \right\} = \frac{Sf_{1}+f_{1}^{2}}{4f_{1}S^{2}\left( 1-\sqrt{\frac{f_{1}}{S}} \right) ^{4}} \nonumber \\= & {} \frac{S+f_{1}}{4S^{2}\left( 1-\sqrt{\frac{f_{1}}{S}} \right) ^{4}} . \end{aligned}$$