Unequal probability inverse sampling Section 6. Unequal probability sampling without replacementUnequal probability inverse sampling Section 6. Unequal probability sampling without replacement

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6.1 Sequential sampling without replacement

For the draw without replacement, the first problem is determining the design. One option is to use the method by Ohlsson (1995) called sequential Poisson sampling. This method involves generating $M$ uniform random variables in the interval $\left[\mathrm{0,1}\right],$ denoted $u_{ik}.$ Next, we select the $n$ units corresponding to the smallest values of $u_{ik}/{\pi }_{k\mathrm{<mo>|</mo>}i}\mathrm{.}$ This method has the advantage of being usable for any sample size and providing a sequence of samples that are included in each other. Unfortunately, it only satisfies approximately the fixed inclusion probabilities. However, the approximations are very accurate according to the simulations given in Ohlsson (1995).

Methods have also been proposed by Sampford (1962) and Pathak (1964). We propose an exact solution to the problem in the sense that the inclusion probabilities are exactly satisfied. We begin by calculating the inclusion probabilities for a design of fixed size $n$ with inclusion probabilities proportional to a strictly positive auxiliary variable $b_k\mathrm{,}k\in L\mathrm{.}$ The probabilities are determined by

$${\pi }_{k|i}\left(n\right)\mathrm{<mo>=</mo>}\min \left(\mathrm{1,}{C}_n\frac{b_k}{{\displaystyle \sum _{\mathcal{l}\in L}}{b}_{\mathcal{l}}}\right)\mathrm{,}$$

where $C_n$ is determined such that

$${\displaystyle \sum _{k\in L}}{\pi }_{k|i}\left(n\right)\mathrm{<mo>=</mo>}{\displaystyle \sum _{k\in L}}\min \left(\mathrm{1,}{C}_n\frac{b_k}{{\displaystyle \sum _{\mathcal{l}\in L}}{b}_{\mathcal{l}}}\right)\mathrm{<mo>=</mo>}n\mathrm{.}$$

A simple algorithm for calculating these probabilities is described in Tillé (2006, page 19), among others. The probabilities can be calculated simply using the function inclusionprobabilities in the R sampling package.

A sequential selection method must therefore select a sample of size $n$ with inclusion probabilities ${\pi }_{k|i}\left(n\right)\mathrm{.}$ It must then make it possible to go from size $n$ to size $n+1$ by simply selecting an additional unit such that the completed sample has an inclusion probability of ${\pi }_{k|i}\left(n+1\right)\mathrm{.}$ It appears that the only method that allows that to be achieved is the elimination method (Tillé 1996). This method starts with the entire population (the list of occupations) and eliminates one unit in each step. In step $j\mathrm{<mo>=</mo>\; 1,}\dots \mathrm{,}N\mathrm{,}$ the unit is eliminated from among the remaining units with the probability

$$1-\frac{{\pi }_{k|i}\left(N-j\right)}{{\pi }_{k|i}\left(N-j+1\right)}\mathrm{.}$$

This method can thus be used to create a sequence of samples included in each other that verify the inclusion probabilities in relation to their size.

Therefore, we can simply apply the elimination method for sample size $n\mathrm{<mo>=</mo>\; 1}$ so that the algorithm successively eliminates all the units. Taking them in the reverse order of elimination, we obtain a sequence of units. The first $n$ units of the sequence are selected with inclusion probability ${\pi }_{k|i}\left(n\right)\mathrm{.}$ The appendix contains a function written in R that can be used to generate this sequence. The code is executed in a simulation that shows that the probabilities obtained through simulations by applying this function are equal to the fixed inclusion probabilities for all sample sizes.

6.2 Inverse or negative design with unequal probabilities

Now that the design is defined, the inverse design can be defined. The units in the list of occupations are taken using the elimination method until $r$ occupations in the enterprise are selected. In this case, the probability distribution of the number of failures $X_i$ seems impossible to calculate. Calculating the conditional inclusion probability $\text{E}\left(A_{ik}|X_i\right)$ is also problematic.

However, we can proceed by analogy and estimate the inclusion probabilities on the basis of expression (5.1) developed for the case with replacement, where $p_{ik}$ can simply be replaced by

$$\frac{{\pi }_{k|i}\left(r+X_i\right)}{r+X_i}\mathrm{.}$$

Therefore, we obtain

$$\widehat{1/{\pi }_{k|i}}\mathrm{<mo>=</mo>}\{\begin{array}{ll}\frac{\left(r-1\right)\left(r+X_i\right)}{r\left(X_i+r-1\right){\pi }_{k|i}\left(r+X_i\right)}\hfill & \text{if}k\in F_i\hfill \\ \frac{r+X_i}{\left(X_i+r-1\right){\pi }_{k|i}\left(r+X_i\right)}\hfill & \text{if}k\in D_i\mathrm{.}\hfill \end{array}$$

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