Unequal probability inverse sampling Section 4. Simple random sampling without replacementUnequal probability inverse sampling Section 4. Simple random sampling without replacement

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For the case without replacement, the notation used is the same as for the draw with replacement. The number of failures $X_i$ therefore has a negative hypergeometric distribution. This probability distribution is little known, to the point that it has been presented as a “forgotten” distribution by Miller and Fridell (2007). This distribution is the counterpart to the negative binomial for the draw without replacement. The general framework is as follows: We consider a population of size $M$ in which there are $M{p}_i$ favourable units, namely the occupations in the list that exist in the enterprise. If the draws are equal probability without replacement until $r$ favorable units appear, then the negative hypergeometric variable, $X_i\sim NH\left(M\mathrm{,}r\mathrm{,}M{p}_i\right)\mathrm{,}$ counts the number of failures before $r$ favourable events occur.

The probability distribution is

$$\mathrm{Pr}\left(X_i\mathrm{<mo>=</mo>}x\right)\mathrm{<mo>=</mo>}p\left(x\mathrm{<mo>;</mo>}M\mathrm{,}r\mathrm{,}M{p}_i\right)\mathrm{<mo>=</mo>}\frac{\left(\begin{array}{c}x+r-1\\ x\end{array}\right)\left(\begin{array}{c}M-x-r\\ M{p}_i-r\end{array}\right)}{\left(\begin{array}{c}M\\ M{p}_i\end{array}\right)}\mathrm{,}$$

where $x\in \left\{\mathrm{0,}\dots \mathrm{,}M\left(1-p_i\right)\right\},$ $M\in \left\{\mathrm{1,2,}\dots \right\},$ $M{p}_i\in \left\{\mathrm{1,2,}\dots \mathrm{,}M\right\},$ and $r\in \left\{\mathrm{1,2,}\dots \mathrm{,}M{p}_i\right\}.$

$$\text{E}\left(X_i\right)\mathrm{<mo>=</mo>}\frac{Mr\left(1-p_i\right)}{M{p}_i+1}\mathrm{,}\text{var}\left(X_i\right)\mathrm{<mo>=</mo>}\frac{rM\left(1-p_i\right)\left(M+1\right)\left(M{p}_i-r+1\right)}{{\left(M{p}_i+1\right)}^2\left(M{p}_i+2\right)}\mathrm{.}$$

Again, $A_{ik}$ denotes the number of times that unit $k$ is selected in the sample. Now, the value of $A_{ik}$ can be only 0 or 1. If $n$ units are selected using a simple design without replacement in $L,$ the sample design is defined as

$$\mathrm{Pr}\left(A_{ik}\mathrm{<mo>=</mo>}{a}_{ik}\mathrm{,}k\in L\right)\mathrm{<mo>=</mo>}{\left(\begin{array}{c}M\\ n\end{array}\right)}^{-1}\mathrm{,}$$

where $a_{ik}\in \left\{\mathrm{0,1}\right\}\mathrm{,}$ and

$${\displaystyle \sum _{k\in L}}{a}_{ik}\mathrm{<mo>=</mo>}n\mathrm{.}$$

If the vector of $A_{ik}$ is conditioned on a fixed size in one part of the population, we have

$$\begin{array}{ll}\mathrm{Pr}\left(A_{ik}\mathrm{<mo>=</mo>}{a}_{ik}\mathrm{,}k\in F_i|{\displaystyle \sum _{k\in F_i}}{A}_{ik}\mathrm{<mo>=</mo>}r\right)\hfill & \mathrm{<mo>=</mo>}\frac{\mathrm{Pr}\left(A_{ik}\mathrm{<mo>=</mo>}{a}_{ik}\mathrm{,}k\in F_i\text{and}{\displaystyle \sum _{k\in F_i}}{A}_{ik}\mathrm{<mo>=</mo>}r\right)}{\mathrm{Pr}\left({\displaystyle \sum _{k\in F_i}}{A}_{ik}\mathrm{<mo>=</mo>}r\right)}\hfill \\ \hfill & \mathrm{<mo>=</mo>}{\left[\frac{\left(\begin{array}{c}M{p}_i\\ r\end{array}\right)\left(\begin{array}{c}M-M{p}_i\\ n-r\end{array}\right)}{\left(\begin{array}{c}M\\ n\end{array}\right)}\right]}^{-1}{\displaystyle \sum _{\begin{array}{c}k\in D_i\\ {\displaystyle {\sum }_{k\in F_i}{A}_{ik}=n-r}\\ A_{ik}\in \left\{0,1\right\}\end{array}}\frac{1}{\left(\begin{array}{l}M\hfill \\ n\hfill \end{array}\right)}}\hfill \\ \hfill & \mathrm{<mo>=</mo>}{\left[\frac{\left(\begin{array}{c}M{p}_i\\ r\end{array}\right)\left(\begin{array}{c}M-M{p}_i\\ n-r\end{array}\right)}{\left(\begin{array}{c}M\\ n\end{array}\right)}\right]}^{-1}\frac{\left(\begin{array}{c}M-M{p}_i\\ n-r\end{array}\right)}{\left(\begin{array}{c}M\\ n\end{array}\right)}\hfill \\ \hfill & \mathrm{<mo>=</mo>}{\left(\begin{array}{c}M{p}_i\\ r\end{array}\right)}^{-1}\mathrm{,}\hfill \end{array}$$

with

$${\displaystyle \sum _{k\in F_i}}{a}_{ik}\mathrm{<mo>=</mo>}r\mathrm{.}$$

This shows that, if the sum of $A_{ik}$ is conditioned on one part of the population, we still have a simple design without replacement. In the procedure in which we draw without replacement until we obtain $r$ occupations in enterprise $i,$ we therefore have

$$\text{E}\left(A_{ik}|X_i\right)\mathrm{<mo>=</mo>}\{\begin{array}{ll}\frac{r}{M{p}_i}\hfill & \text{if}k\in F_i\hfill \\ \frac{X_i}{M-M{p}_i}\hfill & \text{if}k\in D_i\mathrm{.}\hfill \end{array}$$

The inclusion probability is therefore

$${\pi }_{k|i}\mathrm{<mo>=</mo>}\text{EE}\left(A_{ik}|X_i\right)\mathrm{<mo>=</mo>}\{\begin{array}{ll}\frac{r}{M{p}_i}\hfill & \text{if}k\in F_i\hfill \\ \frac{\text{E}\left(X_i\right)}{M-M{p}_i}\mathrm{<mo>=</mo>}\frac{r}{M{p}_i+1}\hfill & \text{if}k\in D_i\mathrm{,}\hfill \end{array}$$

for all $k\in L\mathrm{.}$ Again, the problem is that we know $M\mathrm{,}r$ and $X_i,$ but not $p_i.$ We can estimate $p_i$ using the maximum likelihood method, through a numerical method.

Using the method of moments, an estimate can be obtained by solving for $p_i$ in the equation $X_i\mathrm{<mo>=</mo>}\text{E}\left(X_i\right)$ , that is,

$$X_i\mathrm{<mo>=</mo>}\frac{Mr\left(1-{\widehat{p}}_i\right)}{M{\widehat{p}}_i+1}\mathrm{.}$$

Hence

$${\widehat{p}}_{i1}\mathrm{<mo>=</mo>}\frac{Mr-X_i}{M\left(r+X_i\right)}\mathrm{.}$$

However, in a few lines it is verified that, if $r\ge \mathrm{2,}$

$${\widehat{p}}_{i2}\mathrm{<mo>=</mo>}\frac{r-1}{r+X_i-1}$$

is unbiased for $p_i\mathrm{.}$

Again, since we are using weights that are inverses of ${\pi }_{k|i}.$ The inverses of the inclusion probabilities are thus estimated as follows:

$$\widehat{1/{\pi }_{k|i}}\mathrm{<mo>=</mo>}\{\begin{array}{lll}\frac{M{\widehat{p}}_{i2}}{r}\hfill & \mathrm{<mo>=</mo>}\frac{M\left(r-1\right)}{r\left(X_i+r-1\right)}\hfill & \text{if}k\in F_i\hfill \\ \frac{M\left(1-{\widehat{p}}_{i2}\right)}{X_i}\hfill & \mathrm{<mo>=</mo>}\frac{M}{X_i+r-1}\hfill & \text{if}k\in D_i\mathrm{.}\hfill \end{array}$$

These weights are also used in the estimator by Murthy (1957), which is unbiased (see also Salehi and Seber 2001). If $M{p}_i\mathrm{<mo><</mo>}r\mathrm{,}$ all occupations will be selected in enterprise $i$ and the estimated inclusion probabilities are then equal to 1.

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