Estimation of population size based on capture recapture designs and evaluation of the estimation reliability

In this paper, we start with the statistical formulation of the estimation problem. In section 2, we define the framework of our estimators under linear and non-linear constraints. Specifically, we develop the estimators under each of the following identification assumptions: no K-way interaction in a linear model, independence among samples, conditional independence among samples, and no K-way interaction in a log-linear model. In section 3, we derive the efficient estimator of the target parameter, and the statistical inference. In section 4 we provide the targeted learning updates for the smoothed estimators under the no K-way interaction log-linear model. In section 5, we illustrate the performance of existing and proposed estimators in various situations, including high-dimensional finite sample settings. In section 6, we show the performance of the estimators under violations of various identification assumptions. In section 7, we apply the estimators to surveillance data on a parasitic infectious disease in a region in southwestern China [17]. In section 8, we summarise the characteristics of the proposed estimators and state the main findings of the paper.

We define the capture-recapture experiments in the following manner. One takes a first random sample from the population of size $n_{1}$, records identifying characteristics of the individuals captured and an indicator of their capture, then repeats the process a total of $K$ times, resulting in $K$ samples of size $n_{1},\ldots,n_{K}$.

Each individual $i$ in the population, whether captured or not, defines a capture history as a vector $B^{*}_{i}=(B^{*}_{i}(1),\ldots,B^{*}_{i}(K))$, where $B^{*}_{i}(k)$ denotes the indicator that this individual $i$ is captured by sample $k$. We assume the capture history vectors $B^{*}_{i}$ of all $i=1,\dots,N$ individuals independently and identically follow a common probability distribution, denoted by $P_{B^{*}}$. Thus $P_{B^{*}}$ is defined on $2^{K}$ possible vectors of dimension $K$.

Note that, for the individuals contained in our observed sample of size $n=\sum_{k=1}^{K}n_{k}$, we actually observe the realization of $B^{*}$. For any individual $i$ that is never captured, we know that $B^{*}_{i}=0$, where $0=(0,\ldots,0)$ is the $K$ dimensional vector in which each component equals $0$. However, for these $N-n$ individuals we do not know the identity of these individuals and we also do not know how many there are (i.e., we do not know $N-n$).

Therefore, we can conclude that our observed data set $B_{1},\ldots,B_{n}$ are $n$ independent and identically distributed draws from the conditional distribution of $B^{*}$, given $B^{*}\not=0$. Let’s denote this true probability distribution with $P_{0}$ and its corresponding random variable with $B$.

So we can conclude that under our assumptions we have that $B_{1},\ldots,B_{n}\sim_{iid}P_{0}$, where $P_{0}(b)=P_{B^{*},0}(b\mid B\not=0)$ for all $b\in\{0,1\}^{K}$. Since the probability distribution $P$ of $B$ is implied by the distribution $P^{*}$ of $B^{*}$ we also use the notation $P=P_{P^{*}}$ to stress this parameterization.

Let ${\cal M}^{F}$ be a model for the underlying distribution $P_{B^{*},0}$. The full-data target parameter $\Psi^{F}:{\cal M}^{F}\rightarrow\hbox{${\rm I\kern-1.99997ptR}$}$ is defined as

$$\Psi^{F}(P_{B^{*}})=P_{B^{*}}(B\not=0).$$

In other words, we want to know the proportion of $N$ that on average will not be caught by the combined sample. An estimator $\psi^{F}_{n}$ of this $\psi^{F}_{0}=\Psi^{F}(P_{B^{*},0})$ immediately translates into an estimator of the desired size $N$ of the population of interest: $N_{n}=\frac{n}{\psi^{F}_{n}}$. We will focus on full-data models defined by a single constraint, where we distinguish between linear constraints defined by a function $f$ and a non-linear constraint defined by a function $\Phi$. Specifically, for a given function $f$, we define the full-data model

$${\cal M}^{F}_{f}=\left\{P_{B^{*}}:\sum_{b}f(b)P_{B^{*}}(b)=0,0<P^{*}(0)<1\right\}.$$

We will require that $f$ satisfies the following condition:

$$f(b=0)\not=0.$$

(1)

In other words, ${\cal M}^{F}_{f}$ contains all probability distributions of $B^{*}$ for which $E_{P_{B^{*}}}f(B^{*})=0$. An example of interest is:

$$f_{I}(b)=(-1)^{K+\sum_{k=1}^{K}b_{k}}.$$

In this case, $E_{0}f_{I}(B^{*})=0$ is equivalent with

$\displaystyle\sum_{b}(-1)^{K+\sum_{k=1}^{K}b_{k}}P_{B^{*},0}(b)=0.$

(2)

We note the left-hand side represents a $K$-th way interaction term $\alpha_{1}$ in the saturated model

$$P_{B^{*}}(b)=\alpha_{0}+\sum_{b^{\prime}\not=0}\alpha_{b^{\prime}}\prod_{j:b_{j}^{\prime}=1}b_{j}.$$

For example, if $K=2$, then the latter model states:

$$P_{B^{*}}(b_{1},b_{2})=\alpha_{00}+\alpha_{10}b_{1}+\alpha_{01}b_{2}+\alpha_{11}b_{1}b_{2},$$

and the constraint $Ef_{I}(B^{*})=0$ states that $\alpha_{11}=0$.

One might also use a log-link in this saturated model:

$$\log P_{B^{*}}(b)=a_{0}+\sum_{b^{\prime}\not=0}a_{b^{\prime}}\prod_{j:b_{j}^{\prime}=1}b_{j}.$$

In this case, $a_{1}$ is the $K$-way interaction term in this log-linear model, and we now have

$$a_{1}\equiv\sum_{b}(-1)^{K+\sum_{k=1}^{K}b_{k}}\log P_{B^{*},0}(b)=0.$$

So, assuming $a_{1}=0$ corresponds with assuming

$$0=\sum_{b}(-1)^{K+\sum_{k=1}^{K}b_{k}}\log P_{B^{*},0}(b).$$

This is an example of a non-linear constraint $\Phi_{I}(P^{*})=0$, where

$\displaystyle\Phi_{I}(P^{*})\equiv\sum_{b}(-1)^{1+\sum_{k=1}^{K}b_{k}}\log P_{B^{*}}(b).$

(3)

We will also consider general non-linear constraints defined by such a function $\Phi$, so that for that purpose one can keep this example $\Phi_{I}$ in mind. As we will see the choice of this constraint has quite dramatic implications on the resulting statistical target parameter/estimand and thereby on the resulting estimator. As one can already tell from the definition of $\Phi_{I}$, $\Phi_{I}$ is not even defined if there are some $b$ for which $P_{B^{*}}(b)=0$, so that also an NPMLE of $\Psi^{F}_{\Phi_{I}}(P^{*}_{0})$ will be ill defined in the case that the empirical distribution $P_{n}(B=b)=0$ for some $b\not=0$. As we will see the parameter $\Psi^{F}_{f_{I}}(P^{*}_{0})$ is very well estimated by the MLE, even when $K$ is very large relative to $n$, but for $\Psi^{F}_{\Phi_{I}}(P^{*}_{0})$ we would need a so called TMLE, incorporating machine learning [14].

Since it is hard to believe that the $\Phi_{I}$ constraint is more realistic than the $f$-constraint in real applications, it appears that, without a good reason to prefer $\Phi_{I}$, the $f$-constraint is far superior. However, it also raises alarm bells that the choice of constraint, if wrong, can result in dramatically different statistical output, so that one should really try to make sure that the constraint that is chosen is known to hold by design.

Another example of a non-linear constraint is the independence between two samples, i.e., that

$$P^{*}(B^{*}(1:2)=(0,0))=P^{*}(B^{*}(1)=0)P^{*}(B^{*}(2)=0),$$

i.e., that the binary indicators $B^{*}(1)$ and $B^{*}(2)$ are independent, but the remaining components can depend on each other and depend on $B^{*}(1),B^{*}(2)$. This corresponds with assuming $\Phi_{II}(P^{*})=0$, where

$$\Phi_{II}(P^{*})\equiv\sum_{b}I(b(1:2)=(0,0))P^{*}(b)-\sum_{b_{1},b_{2}}I(b_{1}(1)=b_{2}(2)=0)P^{*}(b_{1})P^{*}(b_{2}).$$

(4)

A third non-linear constraint example is conditional independence between two samples, given the others. Suppose we have $K$ samples in total, and the distribution of the $j^{th}$ sample $B_{j}$ is independent of the $m^{th}$ sample $B_{m}$ given all other samples (there’s no time ordering of $j,m$). Then we can derive the target parameter and efficient influence curve for this conditional independence constraint. The conditional independence constraint $\Phi_{CI}=0$ is defined as

	$\displaystyle\Phi_{CI,(j,m)}=P^{*}(B_{j}=1|B_{1}=b_{1},\cdots,B_{m}=0,\cdots,B_{K}=b_{k})$
	$\displaystyle-P^{*}(B_{j}=1|B_{1}=b_{1},\cdots,B_{m}=1,\cdots,B_{K}=b_{k}),b_{t}=0,1,t=1,\cdots,K.$

Because we must have the term $P^{*}(0,0,\cdots,0)$ in the constraint to successfully identify the parameter of interest, the constraint $\Phi_{CI}=0$ is sufficient. Thus the equation above can be presented as

	$\displaystyle\Phi_{CI,(j,m)}=P^{*}(B_{j}=1|B_{1}=0,\cdots,B_{m}=0,\cdots,B_{K}=0)$
	$\displaystyle-P^{*}(B_{j}=1|B_{1}=0,\cdots,B_{m}=1,\cdots,B_{K}=0).$		(5)

This is because for all the combinations of $\{B_{1}=b_{1},\cdots,B_{m-1}=b_{m-1},B_{m+1}=b_{m+1},\cdots,B_{K}=b_{K}\},\forall b_{t}\in\{0,1\},t=1,\cdots,K$, only the situation of $\{B_{1}=0,\cdots,B_{m-1}=0,B_{m+1}=0,\cdots,B_{K}\ =0\}$ can generate the required term $P^{*}(0,0,\cdots,0)$ in the constraint.

Given such a full-data model with one constraint defined by $f$ or $\Phi$, one will need to establish that for all $P^{*}\in{\cal M}^{F}_{f}$, we have (say we use $f$)

$$\Psi^{F}(P^{*})=\Psi_{f}(P).$$

for a known $\Psi_{f}:{\cal M}\rightarrow(0,1)$ and $P=P_{P^{*}}$, where the statistical model for $P_{0}$ is defined as

$${\cal M}_{f}=\{P_{P_{B^{*}}}:P_{B^{*}}\in{\cal M}^{F}_{f}\}.$$

Let’s first study this identifiability problem in the special case of our full data models defined by the linear constraint $P^{*}_{0}f=0$ with $f(0)\not=0$ (equation 2). It will show that we have identifiability of the whole $P_{B^{*}}$ from $P=P_{P_{B^{*}}}$. Firstly, we note that $P_{P^{*}}(b)=P^{*}(b)/\psi^{F}$ for all $b\not=0$. Thus, $P^{*}(b)=\psi^{F}P(b)$ for all $b\not=0$. We also have $P^{*}(0)=1-\psi^{F}$, so that $\sum_{b}f(b)P^{*}(b)=0$ yields:

	$\displaystyle 0$	$\displaystyle=$	$\displaystyle\sum_{b}f(b)P^{*}(b)=\sum_{b\not=0}f(b)\psi^{F}P(b)+f(0)P^{*}(0)$
		$\displaystyle=$	$\displaystyle\psi^{F}\sum_{b\not=0}f(b)P(b)+f(0)(1-\psi^{F}).$

We can now solve for $\psi^{F}$:

$$\psi^{F}=\frac{f(0)}{f(0)-\sum_{b\not=0}f(b)P(b)}.$$

At first sight, one might wonder if this solution is in the range $(0,1)$. Working backwards from the right-hand side, i.e., using $f(0)P^{*}(0)+\sum_{b\not=0}f(b)P^{*}(b)=0$ and $P^{*}(b)=P(b)/\psi^{F}$, it indeed follows that the denominator equals $f(0)+f(0)(1-\psi^{F})/\psi^{F}$, so that the right-hand side is indeed in $(0,1)$. Thus, we can conclude that:

$\displaystyle\Psi^{F}(P^{*})=\Psi_{f}(P)\equiv\frac{f(0)}{f(0)-Pf},$

(6)

where we use the notation $Pf=\int f(b)dP(b)$ for the expectation operator.

Suppose now that our the one-dimensional constraint in the full-data model is defined in general by $\Phi(P^{*})=0$ for some function $\Phi:{\cal M}^{F}_{\Phi}\rightarrow\hbox{${\rm I\kern-1.99997ptR}$}$. The full data model is now defined by ${\cal M}^{F}_{\Phi}=\{P^{*}:\Phi(P^{*})=0,0<P^{*}(0)<1\}$, and the corresponding observed data model can be denoted with ${\cal M}_{\Phi}$. To establish the identifiability, we still use $P^{*}(b)=\psi^{F}P(b)$ for all $b\not=0$. We also still have $P^{*}(0)=1-\psi^{F}$. Define $P^{*}_{P,\psi}$ as $P^{*}_{P,\psi}(0)=1-\psi$ and $P^{*}_{P,\psi}(b)=\psi P(b)$ for $b\not=0$. The constraint $\Phi(P^{*})=0$ now yields the equation

$$\Phi(P^{*}_{P,\psi^{F}})=0\mbox{ in $\psi^{F}$}$$

for a given $P=P_{P^{*}}$.

Consider now our particular example $\Phi_{I}$. Note that

$$\Phi_{I}(P^{*}_{P,\psi})=\sum_{b\not=0}(-1)^{K+\sum_{k}b_{k}}\log\{\psi P(b)\}+(-1)^{K}\log(1-\psi).$$

We need to solve this equation in $\psi$ for the given $P$. For $K$ is odd, we obtain

$$\Psi_{I}(P)=\frac{1}{1+\exp\left(\sum_{b\not=0}f_{I}(b)\log P(b)\right)},$$

and for $K$ even we obtain $\log(1-\psi)/\psi)=\sum_{b\not=0}\log P(b)$ and thus

$$\Psi_{I}(P)=\frac{1}{1+\exp\left(-\sum_{b\not=0}f_{I}(b)\log P(b)\right)}.$$

So, in general, this solution can be represented as:

$$\Psi_{I}(P)=\frac{1}{1+\exp\left((-1)^{K+1}\sum_{b\not=0}f_{I}(b)\log P(b)\right)}.$$

(7)

This solution $\Psi(P)$ only exists if $P(b)>0$ for all $b\not=0$. In particular, a plug-in estimator $\Psi(P_{n})$ based on the empirical distribution function $P_{n}$ would not be defined when $P_{n}(b)=0$ for some cells $b\in\{0,1\}^{K}$.

For general $\Phi$, one needs to assume that this one-dimensional equation $H(\psi^{F},P)\equiv\Phi(P^{*}_{P,\psi})=0$ in $\psi$, for a given $P\in{\cal M}$, always has a unique solution, which then proves the desired identifiability of $\psi^{F}(P^{*})$ from $P_{P^{*}}$ for any $P^{*}\in{\cal M}^{F}_{\Phi}$. This solution is now denoted with $\Psi_{\Phi}(P)$. So, in this case, $\Psi_{\Phi}:{\cal M}_{\Phi}\rightarrow\hbox{${\rm I\kern-1.99997ptR}$}$ is defined implicitly by $H(\Psi_{\Phi}(P),P)=0$. If $\Phi$ is one-dimensional, one will still have that ${\cal M}_{\Phi}$ is nonparametric.

Let’s now consider the $\Phi_{II}$ constraint which assumes that $B^{*}(1)$ is independent of $B^{*}(2)$. Again, using $P^{*}(b)=P(b)\psi$ for $b\not=0$ and $P^{*}(0)=(1-\psi)$, the equation $\Phi_{II}(P^{*})=0$ yields the following quadratic equation in $\psi$:

$$a_{II}(P)\psi^{2}+b_{II}(P)\psi=0,$$

where

	$\displaystyle a_{II}(P)$	$\displaystyle=$	$\displaystyle-\sum_{b_{1}\not=0,b_{2}\not=0}I(b_{1}(1)=b_{2}(2)=0)P(b_{1})P(b_{2})$
			$\displaystyle+\sum_{b_{2}\not=0}I(b_{2}(2)=0)P(b_{2})+\sum_{b_{1}\not=0}I(b_{1}(1)=0)P(b_{1})-1$
	$\displaystyle b_{II}(P)$	$\displaystyle=$	$\displaystyle\sum_{b\not=0}I(b(1:2)=(0,0))p(b)-\sum_{b_{2}\not=0}I(b_{2}(2)=0)P(b_{2})$
			$\displaystyle-\sum_{b_{1}\not=0}I(b_{1}(1)=0)P(b_{1})+1.$

Since $\psi\not=0$, this yields the equation $a_{II}(P)\psi+b_{II}(P)=0$ and thus

$$\Psi_{II}(P)=\frac{-b_{II}(P)}{a_{II}(P)}.$$

A more helpful way this parameter can be represented is given by:

$$\Psi_{II}(P)=\frac{1-P(B(1)=0)-P(B(2)=0)+P(B(1:2)=(0,0))}{1-P(B(1)=0)-P(B(2)=0)+P(B(1)=0)P(B(2)=0)}.$$

(8)

This identifiability result relies on $\Psi_{II}(P)\in(0,1)$, i..e, that $P(B(1)=B(2)=0)<P(B(1)=0)P(B(2)=0)$. In particular, we need $0<P(B(1)=0)<1$ and $0<P(B(2)=0)<1$ and thereby that each of the three cells $(1,0),(0,1),(1,1)$ has positive probability under the bivariate marginal distribution of $B(1),B(2)$ under $P$. It follows trivially that the inequality holds for $K=2$ since in that case $P(B(1)=B(2)=0)=0$. It will have to be verified if this inequality constraint always holds for $K>2$ as well, or that this is an actual assumption in the statistical model. Since it would be an inequality constraint in the statistical model, it would not affect the tangent space and thereby the efficient influence function for $\Psi_{II}:{\cal M}\rightarrow\hbox{${\rm I\kern-1.99997ptR}$}$ presented below, but the MLE would now involve maximizing the likelihood under this constraint so that the resulting MLE of $P_{0}$ satisfies this constraint.

Similarly, for the conditional independence assumption, the parameter $\Psi_{CI}$ can be derived as

$$\Psi_{CI}=\frac{P(B_{m}=1,B_{j}=1,0,...,0)}{C_{0}(P)},$$

(9)

where

	$\displaystyle C_{0}(P)$	$\displaystyle=$	$\displaystyle P(B_{m}=1,B_{j}=1,0,...,0)$		(10)
			$\displaystyle+P(B_{m}=0,B_{j}=1,0,...,0)P(B_{m}=1,B_{j}=0,0,...,0).$

Details on this derivation are presented in the appendix 9.4.

One could also define a model by a multivariate $\Phi:{\cal M}^{F}\rightarrow\hbox{${\rm I\kern-1.99997ptR}$}^{d}$. In this case, the full data model is restricted by $d$ constraints $\Phi(P^{*})=0$ and the observed data model will not be saturated anymore. Let’s consider the example in which we assume that $(B^{*}(1),\ldots,B^{*}(K))$ are $K$ independent Bernoulli random variables. In this case, we assume that $P^{*}(B^{*}=b)=\prod_{k=1}^{K}P^{*}(B^{*}(k)=b(k))$ for all possible $b\in\{0,1\}^{K}$. This can also be defined as stating that for each 2 components $B^{*}(j_{1}),B^{*}(j_{2})$, we have that these two Bernoulli’s are independent. Let $\Phi_{,II,j_{1},j_{2}}(P)$ be the constraint defined as in (4) but with $B(1)$ and $B(2)$ replaced by $B(j_{1})$ and $B(j_{2})$, respectively. Then, we can define $\Phi_{III}(P)=(\Phi_{II,j_{1},j_{2}}(P):(j_{1},j_{2})\in\{1,\ldots,K\}^{2},j_{1}\not=j_{2})$, a vector of dimension $K(K-1)/2$. This defines now the model ${\cal M}^{F}_{III}=\{P^{*}:\Phi_{III}(P^{*})=0\}$, which assumes that all components of $B^{*}$ are independent. We will also work out the MLE and efficient influence curve of $\psi_{0}^{F}$ for this restricted statistical model.

We have now defined the statistical estimation problem for linear and non-linear constraints. For linear constraints defined by a function $f$, we observe $B_{1},\ldots,B_{n}\sim P_{0}\in{\cal M}_{f}=\{P_{P^{*}}:P^{*}\in{\cal M}^{F}\}$, ${\cal M}^{F}=\{P^{*}:P^{*}f=0,0<P^{*}(0)<1\}$, and our statistical target parameter is given by $\Psi_{f}:{\cal M}\rightarrow\hbox{${\rm I\kern-1.99997ptR}$}$, where

$\displaystyle\Psi_{f}(P)=\frac{f(0)}{f(0)-Pf}.$

(11)

The statistical target parameter satisfies that $\Psi^{F}(P^{*})=P^{*}(B^{*}\not=0)=\Psi_{f}(P_{P^{*}})$ for all $P^{*}\in{\cal M}^{F}$.

Since the full-data model only includes a single constraint, it follows that ${\cal M}_{f}$ consists of all possible probability distributions of $B$ on $\{b:b\not=0\}$, so that it is a nonparametric/saturated model.

Similarly, we can state the statistical model and target parameter for our examples using non-linear one-dimensional constraints $\Phi$.

In general, we have a statistical model ${\cal M}=\{P_{P^{*}}:P^{*}\in{\cal M}^{F}\}$ for some full data model ${\cal M}^{F}$ for the distribution of $B^{*}$, and, we would be given a particular mapping $\Psi:{\cal M}\rightarrow\hbox{${\rm I\kern-1.99997ptR}$}$, satisfying $\Psi^{F}(P^{*})=\Psi(P_{P^{*}})$ for all $P^{*}\in{\cal M}^{F}$. In this case, our goal is to estimate $\Psi(P_{0})$ based on knowing that $P_{0}\in{\cal M}$.

An estimator of $\Psi$ is efficient if and only if it is asymptotically linear with influence curve equal to canonical gradient of pathwise derivative of $\Psi$. Therefore it is important to determine this canonical gradient. It teaches us how to construct an efficient estimator of $\Psi(P)$ in model ${\cal M}^{F}$. In addition, it provides us with Wald type confidence intervals based on an efficient estimator. As we will see for most of our estimation problems $\Psi$ with a nonparametric model, $P$ and $\Psi$ can be estimated with the empirical measure $P_{n}$ and $\Psi(P_{n})$. However for constraint $\Phi_{I}$, an NPMLE is typically not defined due to empty cells, so that smoothing and bias reduction is needed.

Let $\Psi_{1f}(P)=\sum_{b\not=0}f(b)P(b)$ so that $\Psi_{f}(P)=\frac{f(0)}{f(0)-\Psi_{1f}(P)}$. Note that $\Psi_{1f}(P)=Pf$ is simply the expectation of $f(B)$ w.r.t its distribution. $\Psi_{1f}:{\cal M}_{f}\rightarrow\hbox{${\rm I\kern-1.99997ptR}$}$ is pathwise differentiable parameter at any $P\in{\cal M}$ with canonical gradient/efficient influence curve given by:

$$D^{*}_{1f}(P)(B)=f(B)-\Psi_{1f}(P).$$

By the delta-method the efficient influence curve of $\Psi_{f}$ at $P$ is given by:

$\displaystyle D^{*}_{f}(P)(B)=\frac{f(0)}{\{f(0)-\Psi_{1f}\}^{2}}D^{*}_{1f}(P)(B).$

(12)

In general, if $\Psi:{\cal M}\rightarrow\hbox{${\rm I\kern-1.99997ptR}$}$ is the target parameter and the model ${\cal M}$ is nonparametric, then the efficient influence curve $D^{*}(P)$ is given by $D^{*}(P)=d\Psi(P)(P_{n=1}-P)$, where

$$d\Psi(P)(h)=\left.\frac{d}{d\epsilon}\Psi(P+\epsilon h)\right|_{\epsilon=0}$$

is the Gateaux derivative in the direction $h$, and $P_{n=1}$ is the empirical distribution for a sample of size one $\{B\}$, putting all its mass on $B$ [18].

Consider now the model ${\cal M}_{\Phi}$ for a general univariate constraint function $\Phi:{\cal M}^{F}_{NP}\rightarrow\hbox{${\rm I\kern-1.99997ptR}$}$ that maps any possible $P^{*}$ into a real number. Recall that $\Psi_{\Phi}:{\cal M}_{\Phi}\rightarrow\hbox{${\rm I\kern-1.99997ptR}$}$ is now defined implicitly by $\Phi(P^{*}_{P,\psi})=0$. For notational convenience, in this particular paragraph we suppress the dependence of $\Psi$ on $\Phi$. The implicit function theorem implies that the general form of efficient influence curve is given by:

$$D^{*}_{\Phi}(P)(B)=-\left\{\frac{d}{d\psi}\Phi(P^{*}_{P,\psi})\right\}^{-1}\frac{d}{dP}\Phi(P^{*}_{P,\psi})(P_{n=1}-P).$$

where the latter derivative is the directional derivative of $P\rightarrow\Phi(P^{*}_{P,\psi})$ in the direction $P_{n=1}-P$.

Note that

$$\frac{d}{d\psi}\Phi(P^{*}_{P,\psi})=d\Phi(P^{*}_{P,\psi})\frac{d}{d\psi}P^{*}_{P,\psi}.$$

We now use that $P^{*}_{P,\psi}(b)=P(b)I(b\not=0)+(1-\psi)I(b=0)$. So we obtain:

$$\frac{d}{d\psi}\Phi(P^{*}_{P,\psi})=d\Phi(P^{*}_{P,\psi})(-1I_{0}),$$

where $I_{0}(b)$ is the function in $b$ that equals 1 if $b=0$ and zero otherwise, and

$$d\Phi(P^{*0})(h)\equiv\left.\frac{d}{d\epsilon}\Phi(P^{*0}+\epsilon h)\right|_{\epsilon=0}$$

is the directional/Gateaux derivative of $\Phi$ in the direction $h$. We also have:

$$\frac{d}{dP}\Phi(P^{*}_{P,\psi})(P_{n=1}-P)=d\Phi(P^{*}_{P,\psi})\frac{d}{dP}P^{*}_{P,\psi}(P_{n=1}-P).$$

We have $\frac{d}{dP}P^{*}_{P,\psi}(h)=I_{0^{c}}h$, where $I_{0^{c}}(b)$ is the function in $b$ that equals zero if $b=0$ and equals $1$ otherwise. So we obtain:

$$\frac{d}{dP}\Phi(P^{*}_{P,\psi})(P_{n=1}-P)=d\Phi(P^{*}_{P,\psi})(I_{0^{c}}(P_{n=1}-P)).$$

We conclude that

$$D^{*}_{\Phi}(P)=-\{d\Phi(P^{*}_{P,\psi})(-1I_{0})\}^{-1}d\Phi(P^{*}_{P,\psi})(I_{0^{c}}(P_{n=1}-P)).$$

Let’s now consider our special example $\Phi_{I}$. In this case, the statistical target parameter $\Psi_{I}:{\cal M}\rightarrow\hbox{${\rm I\kern-1.99997ptR}$}$ is given by (7). It is straightforward to show that the directional derivative of $\Psi$ at $P=(P(b):b)$ in direction $h=(h(b):b)$ is given by:

$$d\Psi_{I}(P)(h)=(-1)^{K}\Psi_{I}(P)(1-\Psi_{I}(P))\sum_{b\not=0}\frac{f_{I}(b)}{P(b)}h(b).$$

As one would have predicted from the definition of $\Psi_{I}$, this directional derivative is only bounded if $P(b)>0$ for all $b\not=0$. The efficient influence curve is thus given by $d\Psi_{I}(P)(P_{n=1}-P)$:

	$\displaystyle D^{*}_{\Phi_{I}}(P)$	$\displaystyle=$	$\displaystyle(-1)^{K}\Psi_{I}(P)(1-\Psi_{I}(P))\sum_{b\not=0}\frac{f_{I}(b)}{P(b)}\{I_{B}(b)-P(b)\}$		(13)
		$\displaystyle=$	$\displaystyle(-1)^{K}\Psi_{I}(P)(1-\Psi_{I}(P))\left\{\frac{f_{I}(B)}{P(B)}+f_{I}(0)\right\},$

where we use that $\sum_{b\not=0}f_{I}(b)=-f_{I}(0)$ so that indeed the expectation of $D^{*}_{\Phi_{I}}(P)$ equals zero (under $P$).

The efficient influence function for $\Psi_{II}:{\cal M}\rightarrow\hbox{${\rm I\kern-1.99997ptR}$}$ (8), corresponding with the constraint $\Phi_{II}(P^{*})=0$, is the influence curve of the empirical plug-in estimator $\Psi_{II}(P_{n})$, and can thus be derived from the delta-method:

	$\displaystyle D^{*}_{\Phi_{II}}(P)$	$\displaystyle=$	$\displaystyle C_{2}(P)\{I(B(1)=0)-P(B(1)=0)\}$		(14)
			$\displaystyle+C_{3}(P)\{I(B(2)=0)-P(B(2)=0)\}$
			$\displaystyle+C_{4}(P)\{I(B(1)=B(2)=0)-P(B(1:2)=0)\},$

where

$$\begin{array}[]{l}C_{2}(P)=\frac{1-P(B(1)=0)-P(B(2)=0)-P(B(1)=B(2)=0)}{(1-P(B(1)=0)-P(B(2)=0)-P(B(1)=0)P(B(2)=0))^{2}}P(B(2)=0)\\ C_{3}(P)=\frac{1-P(B(1)=0)-P(B(2)=0)-P(B(1)=B(2)=0)}{(1-P(B(1)=0)-P(B(2)=0)-P(B(1)=0)P(B(2)=0))^{2}}P(B(1)=0)\\ C_{4}(P)=-\frac{1}{1-P(B(1)=0)-P(B(2)=0)-P(B(1)=0)P(B(2)=0)}.\end{array}$$