# Unequal probability inverse sampling Section 5. Unequal probability sampling with replacementUnequal probability inverse sampling Section 5. Unequal probability sampling with replacement

Unequal probability sampling is not really more difficult to process when the draw is with replacement. Now let ${p}_{ik}$ denote the probability of an occupation being drawn in each draw with

$\sum _{k\in L}\text{\hspace{0.17em}}{p}_{ik}=1.$

Let ${P}_{i}$ be the sum of ${p}_{ik}$ limited to the occupations in enterprise $i:$

${P}_{i}=\sum _{k\in {F}_{i}}\text{\hspace{0.17em}}{p}_{ik}.$

In this case, ${X}_{i}$ has a negative binomial distribution with parameters $r$ and ${P}_{i}.$ Therefore,

$\text{E}\left({X}_{i}\right)=\frac{r\left(1-{P}_{i}\right)}{{P}_{i}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{var}\left({X}_{i}\right)=\frac{r\left(1-{P}_{i}\right)}{{P}_{i}}.$

Let ${A}_{ik},k\in L$ be the number of times that unit $k$ is selected in the sample. In an unequal probability design with replacement of size $n,$ the values of ${A}_{ik}$ have a multinomial distribution. Therefore,

$\mathrm{Pr}\left({A}_{ik}={a}_{ik},k\in L\right)=n!\prod _{k\in L}\frac{{p}_{ik}^{{a}_{ik}}}{{a}_{ik}!},$

where ${A}_{ik}=0,\dots ,n,$ and

$\sum _{k\in L}\text{\hspace{0.17em}}{a}_{ik}=n.$

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

$Pr( A ik = a ik ,k∈ F i | ∑ k∈ F i A ik =r ) = Pr( A ik = a ik ,k∈ F i and ∑ k∈ F i A ik =r ) Pr( ∑ k∈ F i A ik =r ) = n! ( 1− P i ) ( n−r ) ( n−r )! ∏ k∈ F i p ik a ik a ik ! n! P i r ( 1− P i ) n−r r!( n−r )! =r! ∏ k∈ F i ( p ik P i ) a ik 1 a ik ! , MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiFu0Je9sqqrpepC0xbbL8F4rqqrpipC0xd9Wqpe0dd9 qqaqFeFr0xbbG8FaYPYRWFb9fi0lXxbvc9Ff0dfrpm0dXdHqps0=vr 0=vr0=fdbaqaaeGaciGaaiaabeqaamaabaabaaGcbaqbaeaabmGaaa qaaiGaccfacaGGYbWaaeWabeaacaWGbbWaaSbaaSqaaiaadMgacaWG Rbaabeaakiaai2dacaWGHbWaaSbaaSqaaiaadMgacaWGRbaabeaaki aaiYcacaWGRbGaeyicI4SaamOramaaBaaaleaacaWGPbaabeaakmaa eeqabaGaaGPaVpaaqafabeWcbaGaam4AaiabgIGiolaadAeadaWgaa adbaGaamyAaaqabaaaleqaniabggHiLdGccaWGbbWaaSbaaSqaaiaa dMgacaWGRbaabeaakiaai2dacaWGYbaacaGLhWoaaiaawIcacaGLPa aaaeaacaaI9aWaaSaaaeaaciGGqbGaaiOCamaabmqabaGaamyqamaa BaaaleaacaWGPbGaam4AaaqabaGccaaI9aGaamyyamaaBaaaleaaca WGPbGaam4AaaqabaGccaaISaGaam4AaiabgIGiolaadAeadaWgaaWc baGaamyAaaqabaGccaaMe8UaaGjbVlaabggacaqGUbGaaeizaiaays W7caaMe8+aaabuaeqaleaacaWGRbGaeyicI4SaamOramaaBaaameaa caWGPbaabeaaaSqab0GaeyyeIuoakiaaykW7caWGbbWaaSbaaSqaai aadMgacaWGRbaabeaakiaai2dacaWGYbaacaGLOaGaayzkaaaabaGa ciiuaiaackhadaqadeqaamaaqafabeWcbaGaam4AaiabgIGiolaadA eadaWgaaadbaGaamyAaaqabaaaleqaniabggHiLdGccaaMc8Uaamyq amaaBaaaleaacaWGPbGaam4AaaqabaGccaaI9aGaamOCaaGaayjkai aawMcaaaaaaeaaaeaacaaI9aWaaSaaaeaadaWcaaqaaiaad6gacaaI HaWaaeWaaeaacaaIXaGaeyOeI0IaamiuamaaBaaaleaacaWGPbaabe aaaOGaayjkaiaawMcaamaaCaaaleqabaWaaeWaaeaacaWGUbGaeyOe I0IaamOCaaGaayjkaiaawMcaaaaaaOqaamaabmaabaGaamOBaiabgk HiTiaadkhaaiaawIcacaGLPaaacaaIHaaaamaarafabeWcbaGaam4A aiabgIGiolaadAeadaWgaaadbaGaamyAaaqabaaaleqaniabg+Givd GcdaWcaaqaaiaadchadaqhaaWcbaGaamyAaiaadUgaaeaacaWGHbWa aSbaaeaacaWGPbGaam4AaaqabaaaaaGcbaGaamyyamaaBaaaleaaca WGPbGaam4AaaqabaGccaaIHaaaaaqaamaalaaabaGaamOBaiaaigca caWGqbWaa0baaSqaaiaadMgaaeaacaWGYbaaaOWaaeWaaeaacaaIXa GaeyOeI0IaamiuamaaBaaaleaacaWGPbaabeaaaOGaayjkaiaawMca amaaCaaaleqabaGaamOBaiabgkHiTiaadkhaaaaakeaacaWGYbGaaG yiamaabmaabaGaamOBaiabgkHiTiaadkhaaiaawIcacaGLPaaacaaI HaaaaaaaaeaaaeaacaaI9aGaamOCaiaaigcadaqeqbqabSqaaiaadU gacqGHiiIZcaWGgbWaaSbaaWqaaiaadMgaaeqaaaWcbeqdcqGHpis1 aOWaaeWaaeaadaWcaaqaaiaadchadaWgaaWcbaGaamyAaiaadUgaae qaaaGcbaGaamiuamaaBaaaleaacaWGPbaabeaaaaaakiaawIcacaGL PaaadaahaaWcbeqaaiaadggadaWgaaadbaGaamyAaiaadUgaaeqaaa aakmaalaaabaGaaGymaaqaaiaadggadaWgaaWcbaGaamyAaiaadUga aeqaaOGaaGyiaaaacaaISaaaaaaa@D3EF@$

with

$\sum _{k\in {F}_{i}}\text{\hspace{0.17em}}{a}_{ik}=r.$

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 an unequal probability design with replacement.

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

$\text{E}\left({A}_{ik}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}{X}_{i}\right)=\left\{\begin{array}{ll}\frac{r{p}_{ik}}{{P}_{i}}\hfill & \text{if}\text{\hspace{0.17em}}\text{\hspace{0.17em}}k\in {F}_{i}\hfill \\ \frac{{X}_{i}{p}_{ik}}{1-{P}_{i}}\hfill & \text{if}\text{\hspace{0.17em}}\text{\hspace{0.17em}}k\in {D}_{i}.\hfill \end{array}$

The expected value of ${A}_{ik}$ is

${\pi }_{k\text{\hspace{0.17em}}|\text{\hspace{0.17em}}i}=\text{EE}\left({A}_{ik}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}{X}_{i}\right)=\frac{r{p}_{ik}}{{P}_{i}},$

$k\in L.$ The problem is that we know ${p}_{ik},r$ and ${X}_{i},$ but not ${P}_{i}.$ We can estimate ${P}_{i}$ using the method of moments by solving $\text{E}\left({X}_{i}\right)={X}_{i},$ which gives

${X}_{i}=\frac{r\left(1-{\stackrel{^}{P}}_{i}\right)}{{\stackrel{^}{P}}_{i}}$

and therefore

${\stackrel{^}{P}}_{i1}=\frac{r}{{X}_{i}+r}.$

The maximum likelihood method provides the same estimator as the method of moments, but this estimator is biased (Mikulski and Smith 1976; Johnson et al. 2005, page 222). In fact, the unbiased minimum variance estimator is

${\stackrel{^}{P}}_{i2}=\frac{r-1}{{X}_{i}+r-1}.$

However, $1/{\stackrel{^}{P}}_{i1}$ is unbiased for ${P}_{i}.$

Again, since we are using weights that are inverses of ${\pi }_{k\text{\hspace{0.17em}}|\text{\hspace{0.17em}}i}.$ The inverses of ${\pi }_{k\text{\hspace{0.17em}}|\text{\hspace{0.17em}}i}$ are thus estimated as follows:

$\stackrel{^}{1/{\pi }_{k\text{\hspace{0.17em}}|\text{\hspace{0.17em}}i}}=\left\{\begin{array}{lll}\frac{{\stackrel{^}{P}}_{i2}}{r{p}_{ik}}\hfill & =\frac{r-1}{\left({X}_{i}+r-1\right)r{p}_{ik}}\hfill & \text{if}\text{\hspace{0.17em}}\text{\hspace{0.17em}}k\in {F}_{i}\hfill \\ \frac{1-{\stackrel{^}{P}}_{i2}}{{X}_{i}{p}_{ik}}\hfill & =\frac{1}{\left({X}_{i}+r-1\right){p}_{ik}}\hfill & \text{if}\text{\hspace{0.17em}}\text{\hspace{0.17em}}k\in {D}_{i}.\hfill \end{array}\text{ }\text{ }\text{ }\text{ }\text{ }\left(5.1\right)$

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