Estimation in Complex Sampling Designs Based on Resampling Methods

Abstract

Generally, to select a representative sample of the population, we use a combination of several probabilistic sampling methods which is called a complex sampling design. A complex sampling design usually needs very sophisticated mathematical calculations to provide unbiased estimators of the population parameters. Therefore, only a limited number of sampling designs are commonly used in practice. In the present study, to overcome this complexity, we propose a general method of estimation based on resampling that is suitable for all standard designs, either conventional or adaptive. In this method, we calculate Murthy estimator as an unbiased estimator for the population mean and its variance estimator without intensive mathematical calculations. Using this method, researchers can perform any probability design with the guarantee that the estimator is unbiased. To show this ability and as an application of the method, we introduce Adaptive Random Walk Sampling as a complex and efficient sampling design, proper for the quadrat-based environmental population. Despite the complexity of this design, the method proposed in this paper provides unbiased estimator for the population mean based on the design and then makes it a practical design. Simulations confirm the expected performance of the method.

Appendix

Proof 1

For ${\hat{p}}_s$ we have

$$\begin{aligned} {\hat{p}}_s=\frac{n(s)+1}{K+1};\;\;\; n(s)=\sum \limits _{k= 1}^KI_s(k);\;\;\; I_s(1),I_s(2),\ldots ,I_s(K)\sim ^{iid}\hbox {Bernoulli}(p_s) \end{aligned}$$

where $I_s(k)=1$ if $s_k=s$ and iid indicates the sample units are independent and identically distributed.

Then Based on Strong Law of Large Numbers and $E(|I_s(k)|)<\infty$ (Karr 1993, pp. 188), as $K\xrightarrow { }\infty$ we have

$$\begin{aligned} {\bar{I}}_s=\frac{n(s)}{K}\xrightarrow {a.s.}p_s \end{aligned}$$

and then

$$\begin{aligned} {\hat{p}}_s=\frac{n(s)+1}{K+1}=\frac{{\bar{I}}_s+1/K}{1+1/K}\xrightarrow {a.s.}p_s. \end{aligned}$$

proofs for ${\hat{p}}_{s,i},\; {\hat{p}}_i,\; {\hat{p}}_{s,i,j}$ and ${\hat{p}}_{i,j}$ are the same as the proof of ${\hat{p}}_s$.

Also

$$\begin{aligned} {\hat{\mu }}_{\text{ K }}\xrightarrow {a.s.}{\hat{\mu }}; \;\;\; {\hat{V}}_{1\text{ K }}\xrightarrow {a.s.}{\hat{V}}_1;\;\;\; {\hat{V}}_{2\text{ K }}\xrightarrow {a.s.}{\hat{V}}_2 \end{aligned}$$

are satisfied because of the continuity of their functions and a.s. convergence of their elements (for more details see Karr 1993 pp. 150). $\square$

Proof 2

First please $E_d$ and $E_r$ denote expectations according to design and resampling, respectively. In MBR it is easy to show that:

$$\begin{aligned} n(s,i)|n(s)\sim B(n(s),\frac{p_{s,i}}{p_s}). \end{aligned}$$

Also, if $X\sim B(K,p)$,

$$\begin{aligned} E_r\left( \frac{X}{X+1}\right) =1-\frac{1-(1-p)^{(K+1)}}{(K+1)p}, \end{aligned}$$

then

$$\begin{aligned} E_r\left( \frac{n(s,i)}{n(s)+1}\right) = \frac{p_{s,i}}{p_s}-\left[ \frac{1-(1-p_s)^{K+1}}{(K+1)p_s}\frac{p_{s,i}}{p_s}\right] , \end{aligned}$$

and then

$$\begin{aligned} \frac{|E({\hat{\mu }}_{\text{ K }})-\mu |}{\mu }= & {} \left| \frac{1}{\mu }E_d\left( \frac{1-(1-p_s)^{K+1}}{(K+1)p_s}\frac{1}{N}\sum \limits _{i\in s}\frac{p_{s,i}}{p_sp_i}y_i\right) \right| \\\le & {} \max \limits _{s}{\frac{1-(1-p_s)^{K+1}}{(K+1)p_s}} \le {\frac{1}{(K+1)p_{s^*}}}. \end{aligned}$$

For variance, as

$$\begin{aligned} n(s,i,j)|n(s)\sim B(n(s),\frac{p_{s,i,j}}{p_s}), \end{aligned}$$

and since $(h+1)x(1-x)^{h}\le 1$ for any positive integer h and $x\in [0,1]$, we have

$$\begin{aligned} E_r\left( \frac{n(s,i,j)}{n(s)+1}\right) -\frac{p_{s,i,j}}{p_s}= \left[ \frac{(1-p_s)^{K+1}-1}{(K+1)p_s}\frac{p_{s,i,j}}{p_s}\right] \le \frac{1}{(K+1)(K+2)p^2_s}\frac{p_{s,i,j}}{p_s}, \end{aligned}$$

and

$$\begin{aligned} \frac{p_{s,i,j}}{p_s}-E_r\left( \frac{n(s,i,j)}{n^*(s)}\right) = \left[ \frac{1-(1-p_s)^{K+1}}{(K+1)p_s}\frac{p_{s,i,j}}{p_s}\right] \le \frac{1}{(K+1)p_s}\frac{p_{s,i,j}}{p_s}, \end{aligned}$$

then

$$\begin{aligned} |E_r\left( \frac{n(s,i,j)}{n^*(s)}\right) -\frac{p_{s,i,j}}{p_s}|\le \frac{1}{(K+1)p^2_s}\frac{p_{s,i,j}}{p_s}. \end{aligned}$$

Now we have

$$\begin{aligned} \frac{|E({\hat{V}}_{1\text{ K }})-V_1|}{V_1}= & {} \frac{|E_d(E_r({\hat{V}}_{1\text{ K }})-{\hat{V}}_1)|}{V_1}\\\le & {} \frac{E_d\left( \frac{1}{N^2}\sum \limits _{i\in s}\sum \limits _{j<i\in s}\frac{1}{{p}_{i,j}}\left| E_r\left( \frac{{\hat{p}}_{s,i,j}}{{\hat{p}}_s}\right) -\frac{p_{s,i,j}}{p_s}\right| \left( \frac{y_i}{{p}_i}-\frac{y_j}{{p}_{j}}\right) ^2{p}_i{p}_{j}\right) }{V_1}\\\le & {} \frac{\frac{1}{(K+1)p^2_{s^*}}E_d\left( \frac{1}{N^2}\sum \limits _{i\in s}\sum \limits _{j<i\in s}\frac{p_{s,i,j}}{p_{i,j}p_s}\left( \frac{y_i}{{p}_i} -\frac{y_j}{{p}_{j}}\right) ^2{p}_i{p}_{j}\right) }{V_1}\\= & {} \frac{1}{(K+1)p^2_{s^*}}\frac{E_d({\hat{V}}_1)}{V_1}=\frac{1}{(K+1)p^2_{s^*}}, \end{aligned}$$

and for $V_2$ we have

$$\begin{aligned} n(s,i)n(s,j)|n(s)\sim MB\left( n(s),\frac{p_{s,i}}{p_s},\frac{p_{s,j}}{p_s},1 -\frac{p_{s,i}}{p_s}-\frac{p_{s,j}}{p_s}\right) , \end{aligned}$$

where MB denotes the Multinomial distribution. Then

$$\begin{aligned} E_r\left( \frac{n(s,i)n(s,j)}{(n(s)+1)^2}\right)= & {} E_r\left[ E_r\left( \frac{n(s,i)}{n(s)+1}\frac{n(s,j)}{n(s)+1}|n(s)\right) \right] \\= & {} E_r\left[ E_r\left( \frac{n(s,i)}{n(s)+1}|n(s)\right) E_r\left( \frac{n(s,j)}{n(s)+1}|n(s)\right) \right. \nonumber \\&\left. + Cov_r\left( \frac{n(s,i)}{n(s)+1}\frac{n(s,j)}{n(s)+1}|n(s)\right) \right] \\&\quad \frac{p_{s,i}p_{s,j}}{p^2_s}E_r\left[ \frac{n(s)^2}{(n(s)+1)^2} -\frac{n(s)}{(n(s)+1)^2}\right] \\\le & {} \frac{p_{s,i}p_{s,j}}{p_s^2} E_r\left[ \frac{n(s)^2}{(n(s)+1)^2}\right] \le \frac{p_{s,i}p_{s,j}}{p_s^2}E_r\left[ \frac{n(s)}{n(s)+1}\right] \\= & {} \frac{p_{s,i}p_{s,j}}{p_s^2}\left( 1+\frac{(1-p_s)^{K+1}-1}{(K+1)p_s}\right) , \end{aligned}$$

and therefore

$$\begin{aligned} E_r\left( \frac{n(s,i)n(s,j)}{(n(s)+1)^2}\right) -\frac{p_{s,i}p_{s,j}}{p_s^2}\le & {} \frac{p_{s,i}p_{s,j}}{p_s^2}\left( \frac{(1-p_s)^{K+1}-1}{(K+1)p_s}\right) \\\le & {} \frac{p_{s,i}p_{s,j}}{p_s^2}\left( \frac{1}{(K+1)(K+2)p_s^2}-\frac{1}{(K+1)p_s}\right) \\\le & {} \frac{p_{s,i}p_{s,j}}{p_s^2}\left( \frac{1}{(K+1)(K+2)p_s^2}\right) \\\le & {} \frac{p_{s,i}p_{s,j}}{p_s^2}\left( \frac{1}{(K+1)p_s^2}\right) . \end{aligned}$$

Now as

$$\begin{aligned}&E_r\left( \frac{n(s,i)n(s,j)}{(n(s)+1)^2}\right) \\&\quad = \frac{p_{s,i}p_{s,j}}{p_s^2} E_r\left[ \frac{n(s)(n(s)-1)}{(n(s)+1)^2}\right] \ge \frac{p_{s,i}p_{s,j}}{p_s^2} E_r\left[ \frac{(n(s)-1)^2}{(n(s)+1)^2}\right] \\&\quad \ge \frac{p_{s,i}p_{s,j}}{p_s^2} E_r\left[ \frac{(n(s)-1)^2}{(n(s)+1)^2}\right] \ge \frac{p_{s,i}p_{s,j}}{p_s^2} E_r\left[ \frac{(n(s)-1)^2}{(n(s)+1)(n(s)+2)}\right] \\&\quad =\frac{p_{s,i}p_{s,j}}{p_s^2}\\&\qquad \times \frac{-9(1-p_s)^{K+2}-4(K+2)p_s(1-p_s)^{K+1} +(K+2)p_s(1-p_s)+((K+2)p_s-3)^2}{(K+1)(K+2)p_s^2}, \end{aligned}$$

therefore

$$\begin{aligned}&\frac{p_{s,i}p_{s,j}}{p_s^2}-E_r\left( \frac{n(s,i)n(s,j)}{(n(s)+1)^2}\right) \le \frac{p_{s,i}p_{s,j}}{p_s^2}\\&\qquad \times \left( 1-\frac{-9(1-p_s)^{K+2}-4(K+2)p_s(1-p_s)^{K+1}+(K+2)p_s(1-p_s) +((K+2)p_s-3)^2}{(K+1)(K+2)p_s^2}\right) \\&\quad \le \frac{p_{s,i}p_{s,j}}{p_s^2}\left( 1+\frac{9}{(K+1)(K+2)p_s^2} +\frac{4}{(K+2)(K+1)p^2_s}-\frac{K+2}{(K+1)}\right. \\&\qquad \left. -\frac{9}{(K+1)(K+2)p_s^2}+\frac{6}{(K+1)p_s}-\frac{1-p_s}{(K+1)p_s}\right) \\&\quad =\frac{p_{s,i}p_{s,j}}{p_s^2}\left( \frac{4}{(K+2)(K+1)p^2_s} -\frac{1}{(K+1)}+\frac{6}{(K+1)p_s}\right) \le \frac{p_{s,i}p_{s,j}}{p_s^2}\left( \frac{10}{(K+1)p^2_s}\right) , \end{aligned}$$

and then

$$\begin{aligned} \left| E_r\left( \frac{n(s,i)n(s,j)}{(n(s)+1)^2}\right) -\frac{p_{s,i}p_{s,j}}{p_s^2}\right| \le \frac{10}{(K+1)p_s^2}, \end{aligned}$$

and similar to $V_1$ we have

$$\begin{aligned} \frac{|E({\hat{V}}_{2\text{ K }})-V_2|}{V_2}= & {} \frac{|E_d(E_r({\hat{V}}_{2\text{ K }})-{\hat{V}}_2)|}{V_2}\\\le & {} \frac{E_d\left( \frac{1}{N^2}\sum \limits _{i\in s}\sum \limits _{j<i\in s}\left| E_r\left( \frac{{\hat{p}}_{s,i}{\hat{p}}_{s,j}}{{\hat{p}}^2_s}\right) -\frac{p_{s,i}p_{s,j}}{p^2_s}\right| \left( \frac{y_i}{{p}_i}-\frac{y_j}{{p}_{j}}\right) ^2\right) }{V_2}\\\le & {} \frac{\frac{10}{(K+1)p^2_{s^*}}E_d\left( \frac{1}{N^2}\sum \limits _{i\in s}\sum \limits _{j<i\in s}\frac{p_{s,i}p_{s,j}}{p^2_s}\left( \frac{y_i}{{p}_i}-\frac{y_j}{{p}_{j}}\right) ^2\right) }{V_2}\\= & {} \frac{10}{(K+1)p^2_{s^*}}\frac{E_d({\hat{V}}_2)}{V_2}=\frac{10}{(K+1)p^2_{s^*}}. \end{aligned}$$

$\square$

Proof 3

According to the condition (3) of Theorem 3

$$\begin{aligned} E_r\left( \frac{n(s,i)}{n(s)}\right) =\frac{p_{s,i}}{p_s} \end{aligned}$$

and then

$$\begin{aligned} E({\hat{\mu }}_{\text{ K }})=E_d\left( \frac{1}{N}\sum \limits _{i\in s}\frac{p_{s,i}}{p_sp_i}y_i\right) =\mu . \end{aligned}$$

For variance, as

$$\begin{aligned} E_r\left( \frac{n(s,i,j)}{n(s)}\right) =\frac{p_{s,i,j}}{p_s} \end{aligned}$$

then

$$\begin{aligned} E({\hat{V}}_{1\text{ K }})=V_1 \end{aligned}$$

and for $V_2$ we have

$$\begin{aligned} E_r\left( \frac{n(s,i)n(s,j)}{n(s)^2}\right) =\frac{p_{s,i}p_{s,j}}{p_s^2} \left( 1-\frac{1}{K}\right) \end{aligned}$$

then we have

$$\begin{aligned} \frac{|E({\hat{V}}_{2\text{ K }})-V_2|}{V_2}= & {} \frac{|E_d\left( E_r({\hat{V}}_{2\text{ K }})-{\hat{V}}_2\right) |}{V_2}\\= & {} \frac{E_d\left( \frac{1}{N^2}\sum \limits _{i\in s} \sum \limits _{j<i\in s}\left| E_r\left( \frac{{\hat{p}}_{s,i}{\hat{p}}_{s,j}}{{\hat{p}}^2_s}\right) -\frac{p_{s,i}p_{s,j}}{p^2_s}\right| \left( \frac{y_i}{{p}_i}-\frac{y_j}{{p}_{j}}\right) ^2\right) }{V_2} \\= & {} \frac{\frac{1}{K}E_d\left( \frac{1}{N^2}\sum \limits _{i\in s} \sum \limits _{j<i\in s}\frac{p_{s,i}p_{s,j}}{p^2_s}\left( \frac{y_i}{{p}_i} -\frac{y_j}{{p}_{j}}\right) ^2\right) }{V_2}=\frac{1}{K}. \end{aligned}$$

$\square$

Proof 4

Consider s as the result of a standard sampling design with equally likely sample space $L=\{s(1),s(2),\ldots ,s(M)\}$. Then

$$\begin{aligned} p_s=\frac{N_L(s)}{M};\;\;\;p_{s,i}=\frac{N_L(s,i)}{M}, \end{aligned}$$

where $N_L(s)$ and $N_L(s,i)$ are the number of outcomes in L that lead to s and s with i as the first unit, respectively. Therefore for $p_{i|s}$ we have

$$\begin{aligned} p_{i|s}=\frac{p_{s,i}}{p_s}=\frac{N_L(s,i)}{N_L(s)}. \end{aligned}$$

Now executing the design on s will lead to

$$\begin{aligned} L^*=\{s^*(1),s^*(2),\ldots ,s^*(M^*)\}, \end{aligned}$$

and as the design is a standard ($P(s|{\mathbf {y}})$ is not dependent on y values of $U-s$) with equally likely sample space, then $N_L(s)=N_{L^*}(s)$ and $N_L(s,i)=N_{L^*}(s,i)$. Therefore

$$\begin{aligned} p^*_s=\frac{N_L^*(s)}{M^*}=\frac{N_L(s)}{M^*};\;\;\;p^*_{s,i}=\frac{N_L^*(s,i)}{M^*}=\frac{N_L(s,i)}{M^*}, \end{aligned}$$

and then

$$\begin{aligned} p^*_{i|s}=\frac{p^*_{s,i}}{p^*_s}=\frac{N_L(s,i)}{N_L(s)}=p_{i|s}. \end{aligned}$$

The same holds for $p_{i,j|s}$. $\square$