Adaptive rectangular sampling: An easy, incomplete, neighbourhood-free adaptive cluster sampling design Section 2. Adaptive rectangular samplingAdaptive rectangular sampling: An easy, incomplete, neighbourhood-free adaptive cluster sampling design Section 2. Adaptive rectangular sampling

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Suppose a total population of $N$ units partitioned into $M$ primary sampling units (PSUs), each containing $N_h$ secondary sampling units (SSUs). Let $\left\{\left(h\mathrm{,}j\right)\mathrm{,}h\mathrm{<mo>=</mo>\; 1,2,}\dots \mathrm{,}M\mathrm{<mo>;</mo>}j\mathrm{<mo>=</mo>\; 1,2,}\dots \mathrm{,}{N}_h\right\}$ denote the $j^{\text{th}}$ unit in the $h^{\text{th}}$ primary unit, with the associated measurement or count $y_{hj}.$ Then, ${\tau }_h\mathrm{<mo>=</mo>}{\displaystyle {\sum }_{j=1}^{N_h}{y}_{hj}}$ is the total of the $y$ values for the $h^{\text{th}}$ PSU, and $\mu \mathrm{<mo>=</mo>}1/N{\displaystyle {\sum }_{h\mathrm{<mo>=</mo>\; 1}}^M{\tau }_h}$ is the population mean.

Adaptive rectangular sampling (ARS) can be performed in a two-stage procedure. The first stage of the ARS design consists of selecting a conventional random sample, $s_0,$ of size $m,$ with $M$ PSUs.

The first phase of the second stage consists of selecting an initial conventional sample, $s_{1h},$ of size $n_{1h}$ in the $h^{\text{th}}$ PSU, where $h\in s_0.$

In the second phase, all the SSUs around those in $s_{1h}$ that satisfy condition $C$ with the radius $R$ are adaptively added, where $R\in \left\{\mathrm{1,2,}\dots \mathrm{,}{M}_d\right\}$ and $M_d$ is the maximum diameter of each PSU. Here, the definition of radius is different from the conventional definition. “Radius,” for a cell, is defined based on all cells around it. For example, $R\mathrm{<mo>=</mo>\; 1}$ refers to the first-level nearest neighbourhood, which consists of the eight SSUs around the cell, and $R\mathrm{<mo>=</mo>\; 2}$ refers to the nearest and the second-nearest neighbourhoods, which consist of all 24 SSUs around the cell (8 SSUs for $R\mathrm{<mo>=</mo>\; 1},$ plus 16 SSUs added for $R\mathrm{<mo>=</mo>\; 2}).$ If the cell is in a corner or close to a border of the PSU, these numbers are reduced (see Figure 2.1). Therefore, in the $h^{\text{th}}$ PSU, there is an additional sample, $s_{2h},$ of random size $n_{2h}.$ Now, the final sample is $s_h\mathrm{<mo>=</mo>}{s}_{1h}\cup s_{2h},$ of random size $n_h\mathrm{<mo>=</mo>}{n}_{1h}+n_{2h},$ inside the $h^{\text{th}}$ PSU. This procedure can be performed under either an overlapping scheme and a non-overlapping scheme.

An estimator for the population mean

The $\pi ‑$estimator (the Horvitz $–$ Thompson estimator) is an estimator for the population mean that requires inclusion probabilities for all sampled units. If all the inclusion probabilities are available, the following are used to estimate the population mean:

$$\widehat{\mu }\mathrm{<mo>=</mo>}\frac{1}{N}{\displaystyle \sum _{h\in s_0}\frac{{\widehat{\tau }}_h}{{\pi }_h}},$$

where ${\pi }_h$ is the inclusion probability of the $h^{\text{th}}$ PSU, and

$${\widehat{\tau }}_h\mathrm{<mo>=</mo>}{\displaystyle \sum _{j\in s_h}\frac{y_{hj}}{{\pi }_{hj}}},$$

where ${\pi }_{hj}$ is the inclusion probability of the SSU $\left(h\mathrm{,}j\right).$ To use the $\pi ‑$estimator, ${\pi }_{hj}$ must be calculated for all $\left\{\left(h\mathrm{,}j\right)\mathrm{<mo>;</mo>}h\in s_0\mathrm{,}j\in s_h\right\}.$

In addition, the variance of the estimator is

$$\text{Var}\left(\widehat{\mu }\right)\mathrm{<mo>=</mo>}\frac{1}{N^2}\left[{\displaystyle \sum _{h\mathrm{<mo>=</mo>\; 1}}^M}{\displaystyle \sum _{h^{\prime }=1}^M\left(\frac{{\pi }_{h{h}^{\prime }}-{\pi }_h{\pi }_{h^{\prime }}}{{\pi }_h{\pi }_{h^{\prime }}}\right){\tau }_h{\tau }_h^{{}^{\prime }}}+{\displaystyle \sum _{h\mathrm{<mo>=</mo>\; 1}}^M\frac{\text{Var}\left({\widehat{\tau }}_h\right)}{{\pi }_h}}\right],$$

where ${\pi }_{h{h}^{\prime }}$ is the joint inclusion probability of the $h^{\text{th}}$ and $h^{{}^{\prime }\text{th}}$ PSUs. An unbiased estimator for the above is

$$\widehat{\text{Var}}\left(\widehat{\mu }\right)\mathrm{<mo>=</mo>}\frac{1}{N^2}\left[{\displaystyle \sum _{h\in s_0}}{\displaystyle \sum _{h^{\prime }\in s_0}\left(\frac{{\pi }_{h{h}^{\prime }}-{\pi }_h{\pi }_h^{{}^{\prime }}}{{\pi }_h{\pi }_h^{{}^{\prime }}}\right)\frac{{\widehat{\tau }}_h{\widehat{\tau }}_h^{{}^{\prime }}}{{\pi }_{h{h}^{\prime }}}}+{\displaystyle \sum _{h\in s_0}\frac{\widehat{\text{Var}}\left({\widehat{\tau }}_h\right)}{{\pi }_h}}\right],$$

where $\widehat{\text{Var}}\left({\widehat{\tau }}_h\right)$ is

$$\widehat{\text{Var}}\left({\widehat{\tau }}_h\right)\mathrm{<mo>=</mo>}{\displaystyle \sum _{j\in s_{h1}}}{\displaystyle \sum _{j^{\prime }\in s_{h1}}\left(\frac{{\pi }_{hj{j}^{\prime }}-{\pi }_{hj}{\pi }_{h{j}^{\prime }}}{{\pi }_{hj}{\pi }_{h{j}^{\prime }}}\right)}\frac{y_{hj}{y}_{h{j}^{\prime }}}{{\pi }_{hj{j}^{\prime }}},$$

where ${\pi }_{hj{j}^{\prime }}$ is the joint inclusion probability of $\left(h\mathrm{,}j\right)$ and $\left(h\mathrm{,}{j}^{\prime }\right)$ in the $h^{\text{th}}$ PSU.

It is easy to calculate the inclusion probabilities for the first stage, especially when simple random sampling without replacement (SRSWOR) is used. In this situation, it is easy to see that

$${\pi }_h\mathrm{<mo>=</mo>}\frac{m}{M}\mathrm{,}{\pi }_{h{h}^{\prime }}\mathrm{<mo>=</mo>}\frac{m\left(m-1\right)}{M\left(M-1\right)}\mathrm{<mo>;</mo>}h\ne h^{\prime }\mathrm{,}{\pi }_{hh}\mathrm{<mo>=</mo>}{\pi }_h\mathrm{.}$$

To calculate ${\pi }_{hj},$ it is necessary to know how many of the cells around cell $\left(h\mathrm{,}j\right)$ within radius $R$ satisfy condition $C,$ because selecting them as the initial sample leads to selecting the cells around them as the final sample. It is necessary to introduce some new notations. In ARS, with the radius $R,$ $B_{hj}$ represents the event of unit $\left(h\mathrm{,}j\right)$ being selected as the final sample, and $A_{hj}$ represents the event of unit $\left(h\mathrm{,}j\right)$ satisfying condition $C$ and being selected as the initial sample. In addition, $s_{hj}$ represents all the units that satisfy condition $C$ and that would be adaptively added to the sample if $A_{hj}$ occurs, including unit $\left(h\mathrm{,}j\right),$ with the size $f_{hj}.$ The size $f_{hj}$ is partitioned as $f_{hj}\mathrm{<mo>=</mo>}{f}_{hj1}+f_{hj2},$ where the former indicates the number of cells in $s_{hj}$ that are available in the final sample $\left(s_h\right)$ and the latter is defined as $f_{hj2}\mathrm{<mo>=</mo>}{f}_{hj}-f_{hj1}.$ However, no information about its units is available. $F_{.}$ is defined like $f_{.},$ but for all units (those satisfying condition $C$ and those not satisfying condition $C).$

In addition, let $f_{hj{j}^{\prime }}$ be the size of $s_{hj{j}^{\prime }}\mathrm{<mo>=</mo>}{s}_{hj}\cup s_{h{j}^{\prime }},$ and $f_{hj{j}^{\prime }1},$ $f_{hj{j}^{\prime }2},$ and $F_{.}$ be defined the same.

The sample $s_{hj},$ with the size $f_{hj},$ contains all the cells that lead to the selection of unit $\left(h\mathrm{,}j\right),$ and $s_{hj{j}^{\prime }},$ with the size $f_{hj{j}^{\prime }},$ contains all the cells that lead to the selection of at least one of $\left(h\mathrm{,}j\right)$ and $\left(h\mathrm{,}{j}^{\prime }\right)$ as the final sample.

Theorem 1: In ARS, with the radius $R,$  for the $h^{th}$  PSU and using SRSWOR to select the initial sample of size $n_{1h},$

$${\pi }_{hj}\mathrm{<mo>=</mo>\; 1}-\frac{\left(\begin{array}{c}{N}_h-f_{hj}\\ n_{1h}\end{array}\right)}{\left(\begin{array}{c}{N}_h\\ n_{1h}\end{array}\right)}(2.1)$$

$${\pi }_{hj{j}^{\prime }}\mathrm{<mo>=</mo>\; 1}-\frac{\left(\begin{array}{c}{N}_h-f_{hj}\\ n_{1h}\end{array}\right)}{\left(\begin{array}{c}{N}_h\\ n_{1h}\end{array}\right)}-\frac{\left(\begin{array}{c}{N}_h-f_{h{j}^{\prime }}\\ n_{1h}\end{array}\right)}{\left(\begin{array}{c}{N}_h\\ n_{1h}\end{array}\right)}+\frac{\left(\begin{array}{c}{N}_h-f_{hj{j}^{\prime }}\\ n_{1h}\end{array}\right)}{\left(\begin{array}{c}{N}_h\\ n_{1h}\end{array}\right)}.(2.2)$$

For the proof of the theorem, see the Appendix.

Here, only one problem arises: $f_{hj}$ is known only in the initial sample that satisfies the condition. However, other samples (those that are adaptively added) have partial information about $f_{hj}$ (i.e., $f_{hj1}).$ Let $\text{Bin}\left(n\mathrm{,}p\right)$ stand for a binomial distribution based on the independent Bernoulli variable $n$ with parameter $p.$ To estimate $f_{hj},$ $f_{hj2}$ represents the number of successes (the number of units that satisfy condition $C)$ in $F_{hj2}$ trials (by searching $F_{hj2}$ units); the independency and identicality (iid) of the trials are assumed. The latter assumption (iid) is for simplifying the calculations. This, of course, leads to bias in estimating some of the inclusion probabilities and, so, to bias for the respective $\pi ‑$estimator. With all the above assumptions, $F_{hj2}$ would be considered a random variable with a binomial distribution, as follows:

$$f_{hj2}\sim \text{Bin}\left(F_{hj2}\mathrm{,}{p}_{hj}\right),$$

where $p_{hj}$ is the probability of satisfying condition $C$ for all cells in the $h^{\text{th}}$ PSU around the $j^{\text{th}}$ cell with the radius $R.$ Then,

$$E\left(f_{hj}\right)\mathrm{<mo>=</mo>}{f}_{hj1}+E\left(f_{hj2}\right)\mathrm{<mo>=</mo>}{f}_{hj1}+p_{hj}{F}_{hj2}.$$

Calculating ${\pi }_{hj}$ leads to two situations:

• If the cell satisfies condition $C$ and belongs to the initial sample, or is adaptively added and is located in a place in the final sample that contains complete information about the area around it, then it is possible to calculate the inclusion probability precisely from the information in the final sample.
• If the final sample does not contain enough information to calculate the inclusion probability, two strategies are proposed:
  • There is partial information about $f_{hj},$ this means that everything is known about $F_{hj1},$ and only $F_{hj2}$ needs to be investigated. For $F_{hj2},$ only knowledge about how many of the cells satisfy the condition is required; there is no need for exact knowledge about $y.$ For example, if condition $C$ is defined as $y\mathrm{<mo>></mo>\; 0},$ it is necessary to know only how many cells of $F_{hj2}$ are nonempty. If this is easy to investigate, the exact inclusion probabilities for all of the units in the sample can be calculated.
  • It is not possible to calculate the inclusion probabilities, ${\pi }_{hj}$ can be estimated as (see equation 2.1)

$${\widehat{\pi }}_{hj}\mathrm{<mo>=</mo>\; 1}-\frac{\left(\begin{array}{c}{N}_h-E\left(f_{hj}\right)\\ n_{1h}\end{array}\right)}{\left(\begin{array}{c}{N}_h\\ n_{1h}\end{array}\right)}\mathrm{<mo>=</mo>\; 1}-\frac{\left(\begin{array}{c}{N}_h-\left(f_{hj1}+p_{hj}{F}_{hj2}\right)\\ n_{1h}\end{array}\right)}{\left(\begin{array}{c}{N}_h\\ n_{1h}\end{array}\right)}\mathrm{.}(2.3)$$

Using $p_{hj},$ or, in other words, assuming different probabilities for different cells, leads to tedious calculations. Estimating $p_{hj}$ can be done based on the spatial pattern of the population. For example, in the case of clustered populations, it may be reasonable to assume two kinds of $p_{hj},$ one for units in the sample satisfying condition $C$ and one for units not satisfying condition $C,$ so that greater probability is provided for the units satisfying condition $C.$ A wide class of spatial patterns can be assumed in estimating $p_{hj},$ but, here, to have a simple and understandable strategy, $p_{hj}\mathrm{<mo>=</mo>}{p}_h$ is assumed for all units in the $h^{\text{th}}$ PSU. Therefore, $p_h$ is the probability of satisfying condition $C$ for units in the $h^{\text{th}}$ PSU. It may be possible to guess $p_h$ from previous information or to estimate it without bias from the initial sample as the portion of the units in the initial sample in the $h^{\text{th}}$ PSU that satisfy condition $C.$

Estimating $p_h$ based on the initial sample is a common procedure in adaptive designs (for example, see Brown et al. [2008]). For rare populations, such estimations might be imprecise. Practically, however, this is not a serious problem in ARS, because, for initial-sample units that satisfy the condition, it is possible to calculate inclusion probabilities without error $(p_h$ is not required). Furthermore, $p_h$ is not required in adaptively added units with $y\mathrm{<mo>=</mo>\; 0.}$ For some adaptively added units with $y\mathrm{<mo>></mo>\; 0},$ $p_h$ has an insignificant role in calculating inclusion probabilities. The example in the next subsection and the simulation results in Section 3 confirm such assertions.

For ${\pi }_{hj{j}^{\prime }}$ (as for ${\pi }_{hj}),$ if, depending on the final sample, there is enough information to calculate $f_{hj},$ $f_{h{j}^{\prime }}$ and $f_{hj{j}^{\prime }},$ then it is enough to use equation 2.2. If there is partial information about $f_{hj{j}^{\prime }},$ then

$$f_{hj{j}^{\prime }2}\sim \text{Bin}\left(F_{hj{j}^{\prime }2}\mathrm{,}{p}_h\right),$$

and then

$$E\left(f_{hj{j}^{\prime }}\right)\mathrm{<mo>=</mo>}{f}_{hj{j}^{\prime }1}+E\left(f_{hj{j}^{\prime }2}\right)\mathrm{<mo>=</mo>}{f}_{hj{j}^{\prime }1}+p_h{F}_{hj{j}^{\prime }2}.$$

And, it is enough to replace the respective $“f_{·}”\mathrm{s}$ with $“E\left(f_{·}\right)”\mathrm{s}$ in equation 2.2. Without the assumption of $p_{hj}\mathrm{<mo>=</mo>}{p}_h,$ $p_{hj{j}^{\prime }}$ should perhaps be used instead of $p_h$ for estimating $f_{hj{j}^{\prime }}.$ This would make the calculations more difficult. A reduction in precision (assuming constant $p_h$ for all the units in the $h^{\text{th}}$ PSU) allows such a simple sampling strategy to be presented.

Discussing an example can help clarify all of the above formulas and calculations.

Discussion of an example

For this example, see the top-left PSU in the right plot in Figure 2.1, where $N\mathrm{<mo>=</mo>\; 112},$ $N_h\mathrm{<mo>=</mo>\; 28},$ $h\mathrm{<mo>=</mo>\; 1}$ and $n_{11}\mathrm{<mo>=</mo>\; 2.}$ Assume that it is necessary to calculate ${\pi }_{hj}$ for two units in Figure 2.1, with $R\mathrm{<mo>=</mo>\; 1.}$ First, for the initial sample with $y\mathrm{<mo>=</mo>\; 6},$ it is easy to see that $F_{hj}\mathrm{<mo>=</mo>\; 6},$ that it has five cells around it plus itself, and that five of them satisfy the condition $\left(f_{hj}\mathrm{<mo>=</mo>\; 5}\right).$ This information is available at the end of the sampling in the final sample. Therefore,

$${\pi }_{y\mathrm{<mo>=</mo>\; 6}}\mathrm{<mo>=</mo>\; 1}-\frac{\left(\begin{array}{c}28-5\\ 2\end{array}\right)}{\left(\begin{array}{c}28\\ 2\end{array}\right)}\simeq \mathrm{0.33.}$$

For an adaptively added sample, like $y\mathrm{<mo>=</mo>\; 248},$ as discussed earlier, there is partial information (see Figure 2.2). Here, $F_{hj}\mathrm{<mo>=</mo>\; 9}$ (the blue cells in part B) and $F_{hj1}\mathrm{<mo>=</mo>\; 6}$ (the orange cells in part C). From the final sample, $f_{hj1}\mathrm{<mo>=</mo>\; 5}$ is also known (the positive response in the orange cells in part C). In addition, $F_{hj2}\mathrm{<mo>=</mo>\; 3}$ (the blue cells in part C), but $f_{hj2}$ is not known. To estimate it, $p_1$ would be estimated from the initial sample as $p_1\mathrm{<mo>=</mo>}1/2\mathrm{<mo>=</mo>\; 0.5}$ (see the green cells in the first PSU in Figure 2.1), then

$${\widehat{\pi }}_{y\mathrm{<mo>=</mo>\; 248}}\mathrm{<mo>=</mo>\; 1}-\frac{\left(\begin{array}{c}28-\mathrm{<mo\; stretchy="false">(</mo>\; 5}+0.5\times \mathrm{3\; <mo\; stretchy="false">)</mo>}\\ 2\end{array}\right)}{\left(\begin{array}{c}28\\ 2\end{array}\right)}\simeq \mathrm{0.42.}$$

But, according to the population, the following can be calculated:

$${\pi }_{y\mathrm{<mo>=</mo>\; 248}}\mathrm{<mo>=</mo>\; 1}-\frac{\left(\begin{array}{c}28-7\\ 2\end{array}\right)}{\left(\begin{array}{c}28\\ 2\end{array}\right)}\simeq 0.44,$$

which is very close to the estimation.

Now, assume that the goal is to calculate a joint probability, ${\pi }_{y\mathrm{<mo>=</mo>\; 6,}{y}^{\prime }\mathrm{<mo>=</mo>\; 5}}$ (see Figure 2.3). Here, according to equation 2.2, $f_{hj}\mathrm{<mo>=</mo>\; 5}$ (part B), $f_{h{j}^{\prime }}\mathrm{<mo>=</mo>\; 3}+f_{h{j}^{\prime }2},$ $F_{h{j}^{\prime }2}\mathrm{<mo>=</mo>\; 5}$ (the blue cells in part C), $f_{hj{j}^{\prime }}\mathrm{<mo>=</mo>\; 5}+f_{hj{j}^{\prime }2}$ and $F_{hj{j}^{\prime }2}\mathrm{<mo>=</mo>\; 5}$ (part D), and $E\left(f_{hj{j}^{\prime }2}\right)\mathrm{<mo>=</mo>\; 5}\times \mathrm{0.5\; <mo>=</mo>\; 2.5.}$ From the information in the sample,

$${\widehat{\pi }}_{y\mathrm{<mo>=</mo>\; 6,}{y}^{\prime }\mathrm{<mo>=</mo>\; 5}}\mathrm{<mo>=</mo>\; 1}-\frac{\left(\begin{array}{c}28-5\\ 2\end{array}\right)}{\left(\begin{array}{c}28\\ 2\end{array}\right)}-\frac{\left(\begin{array}{c}28-\left(3+0.5\times 5\right)\\ 2\end{array}\right)}{\left(\begin{array}{c}28\\ 2\end{array}\right)}+\frac{\left(\begin{array}{c}28-\left(5+0.5\times 5\right)\\ 2\end{array}\right)}{\left(\begin{array}{c}28\\ 2\end{array}\right)}\simeq \mathrm{0.22.}$$

With complete information about the population,

$${\pi }_{y\mathrm{<mo>=</mo>\; 6,}{y}^{\prime }\mathrm{<mo>=</mo>\; 5}}\mathrm{<mo>=</mo>\; 1}-\frac{\left(\begin{array}{c}28-5\\ 2\end{array}\right)}{\left(\begin{array}{c}28\\ 2\end{array}\right)}-\frac{\left(\begin{array}{c}28-6\\ 2\end{array}\right)}{\left(\begin{array}{c}28\\ 2\end{array}\right)}+\frac{\left(\begin{array}{c}28-8\\ 2\end{array}\right)}{\left(\begin{array}{c}28\\ 2\end{array}\right)}\simeq 0.22,$$

which shows no error to two decimal places. By writing a code for $“f_{.}”\mathrm{s}$ and $“F_{.}”\mathrm{s},$ the $\pi ‑$estimator can be calculated easily.

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