Unequal probability inverse sampling Section 3. Simple random sampling with replacementUnequal probability inverse sampling Section 3. Simple random sampling with replacement

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Assume that enterprise $i$ has proportion $p_i$ of the occupations in the list in the enterprise. If the sample of occupations is drawn with replacement in enterprise $i$ until $r$ occupations in the enterprise have been identified, then $X_i$ has a negative binomial distribution denoted by $X_i\sim NB\left(r\mathrm{,}{p}_i\right).$ In that case,

$$\mathrm{Pr}\left(X_i\mathrm{<mo>=</mo>}{x}_i\right)\mathrm{<mo>=</mo>}\left(\begin{array}{c}r+x_i-1\\ x_i\end{array}\right)p_i^r{\left(1-p_i\right)}^{x_i}\mathrm{,}$$

with $x_i\in \mathbb{N}\mathrm{<mo>=</mo>}\left\{\mathrm{0,1,2,3,}\dots \right\}\mathrm{,}{p}_i\in \left[\mathrm{0,1}\right]\mathrm{,}r\in {\mathbb{N}}^{\mathrm{<mo>*</mo>}}\mathrm{<mo>=</mo>}\left\{\mathrm{1,2,3,}\dots \right\}.$ Furthermore,

$$\text{E}\left(X_i\right)\mathrm{<mo>=</mo>}\frac{r\left(1-p_i\right)}{p_i}\text{and}\text{var}\left(X_i\right)\mathrm{<mo>=</mo>}\frac{r\left(1-p_i\right)}{p_i^2}\mathrm{.}$$

Let $A_{ik}\mathrm{,}k\in L\mathrm{,}$ be the number of times that unit $k$ is selected in the sample taken from enterprise $i.$ In a simple design with replacement of size $n,$ the values of $A_{ik}$ have a multinomial distribution. Therefore,

$$\mathrm{Pr}\left(A_{ik}\mathrm{<mo>=</mo>}{a}_{ik}\mathrm{,}k\in L\right)\mathrm{<mo>=</mo>}\frac{n\mathrm{<mo>!</mo>}}{M^n}{\displaystyle \prod _{k\in L}}\frac{1}{a_{ik}\mathrm{<mo>!</mo>}}\mathrm{,}$$

where $A_{ik}\mathrm{<mo>=</mo>\; 0,}\dots \mathrm{,}n\mathrm{,}$ and

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

If this multinomial vector is conditioned on a fixed size in a given part of the population, then

$$\begin{array}{ll}\mathrm{Pr}\left(A_{ik}\mathrm{<mo>=</mo>}{a}_{ik}\mathrm{,}k\in F_i|{\displaystyle \sum _{k\in F_i}}{A}_{ik}\mathrm{<mo>=</mo>}r\right)\hfill & \mathrm{<mo>=</mo>}\frac{\mathrm{Pr}\left(A_{ik}\mathrm{<mo>=</mo>}{a}_{ik}\mathrm{,}k\in F_i\text{et}{\displaystyle \sum _{k\in F_i}}{A}_{ik}\mathrm{<mo>=</mo>}r\right)}{\mathrm{Pr}\left({\displaystyle \sum _{k\in F_i}}{A}_{ik}\mathrm{<mo>=</mo>}r\right)}\hfill \\ \hfill & \mathrm{<mo>=</mo>}\frac{\frac{n\mathrm{<mo>!</mo>}{\left(1-p_i\right)}^{\left(n-r\right)}}{\left(n-r\right)\mathrm{<mo>!</mo>}{M}^r}{\displaystyle \prod _{k\in F_i}}\frac{1}{a_{ik}\mathrm{<mo>!</mo>}}}{\frac{n\mathrm{<mo>!</mo>}{p}_i^r{\left(1-p_i\right)}^{n-r}}{r\mathrm{<mo>!</mo>}\left(n-r\right)\mathrm{<mo>!</mo>}}}\hfill \\ \hfill & \mathrm{<mo>=</mo>}r\mathrm{<mo>!</mo>}{\left(\frac{1}{M{p}_i}\right)}^r{\displaystyle \prod _{k\in F_i}}\frac{1}{a_{ik}\mathrm{<mo>!</mo>}}\mathrm{,}\hfill \end{array}$$

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, the distribution remains multinomial and conditionally there is still a simple design with replacement.

With the procedure in which we draw with replacement until we obtain $r$ occupations in enterprise $i,$ we have

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

In fact, conditionally on $X_i,$ in $F_i$ of size $M{p}_i\mathrm{,}$ $r$ occupations are selected and, in $D_i$ of size $M\left(1-p_i\right),$ $X_i$ occupations are selected.

In the case with replacement, what is calculated is not really an inclusion probability, but rather the expected value of $A_{ik}$ which is denoted as ${\pi }_{k|i},$

$${\pi }_{k|i}\mathrm{<mo>=</mo>}\text{EE}\left(A_{ik}|X_i\right)\mathrm{<mo>=</mo>}\frac{r}{M{p}_i}\mathrm{,}$$

$k\in L\mathrm{.}$ The problem is that we know $M\mathrm{,}r$ and $X_i,$ but not $p_i.$ We can estimate $p_i$ using the method of moments by solving $\text{E}\left(X_i\right)\mathrm{<mo>=</mo>}{X}_i,$ which yields

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

and therefore

$${\widehat{p}}_{i1}\mathrm{<mo>=</mo>}\frac{r}{X_i+r}\mathrm{.}$$

The maximum likelihood method provides the same estimator as the method of moments, but this estimator is biased (Mikulski and Smith 1976; Johnson, Kemp and Kotz 2005, page 222). If $r\ge \mathrm{2,}$ the unbiased minimum variance estimator of $p_i$ is

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

However, $1/{\widehat{p}}_{i1}$ is unbiased for $1/p_i.$

Since we are using weights that are inverses of ${\pi }_{k|i},$ the inverses of ${\pi }_{k|i}$ are thus estimated as follows:

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

However, the case with replacement is not very satisfactory, because selecting $r$ occupations with replacement does not necessarily result in $r$ distinct occupations, since the same occupation may be selected more than once. Furthermore, sampling may be especially long if $M{p}_i$ is small. Therefore, sampling without replacement is preferred.

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