Combining link-tracing sampling and cluster sampling to estimate the size of a hidden population in presence of heterogeneous link-probabilities 5. Sample size determinationCombining link-tracing sampling and cluster sampling to estimate the size of a hidden population in presence of heterogeneous link-probabilities 5. Sample size determination

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We will present a procedure to determine the initial sample size $n.$ This procedure is based on stringent assumptions, but, as was indicated by one of the reviewers of the paper, it could nevertheless be very useful for researchers who want to apply this sampling design.

The first step is to compute the asymptotic variances of the proposed estimators. Although the variances depend on several unknown parameters, we can simplify them by assuming that the effects ${\alpha }_i^{\left(k\right)}$ of the sampled sites are homogeneous, that is, ${\alpha }_i^{\left(k\right)}={\alpha }^{\left(k\right)},i=1,\dots \mathrm{,}n;k=1,2.$ Under this premise, the probabilities ${\pi }_x^{\left(k\right)}$ and ${\pi }_x^{\left(A_i\right)}$ that the vectors of link-indicator variables associated with randomly selected persons from $U_k-S_0$ and $S_0,$ respectively, equal $x$ depend only on the number of $1’\text{s}$ that appear in the vector $x.$ Thus, their Gaussian quadrature approximations, given by (3.3) and (3.4), are simplified to

$${\tilde{\pi }}_x^{\left(k\right)}\left({\theta }^{\left(k\right)}\right)={\tilde{\pi }}_x^{\left(k\right)}\left({\alpha }^{\left(k\right)}\mathrm{,}{\sigma }_k\right)={\displaystyle \sum _{t=1}^q}\frac{\exp \left[x\left({\alpha }^{\left(k\right)}+{\sigma }_k{z}_t\right)\right]}{{\left[1+\exp \left({\alpha }^{\left(k\right)}+{\sigma }_k{z}_t\right)\right]}^n}{\nu }_t\mathrm{,}x=0,1,\dots \mathrm{,}n\mathrm{,}$$

and

$${\tilde{\pi }}_x^{\left(A\right)}\left({\theta }^{\left(k\right)}\right)={\tilde{\pi }}_x^{\left(A\right)}\left({\alpha }^{\left(1\right)}\mathrm{,}{\sigma }_1\right)={\displaystyle \sum _{t=1}^q}\frac{\exp \left[x\left({\alpha }^{\left(1\right)}+{\sigma }_1{z}_t\right)\right]}{{\left[1+\exp \left({\alpha }^{\left(1\right)}+{\sigma }_1{z}_t\right)\right]}^n}{\nu }_t\mathrm{,}x=0,1,\dots \mathrm{,}n-\mathrm{1,}$$

where ${\theta }^{\left(k\right)}=\left({\theta }_1^{\left(k\right)}\mathrm{,}{\theta }_2^{\left(k\right)}\right)=\left({\alpha }^{\left(k\right)}\mathrm{,}{\sigma }_k\right).$

Following Sanathanan’s (1972) procedure we get that the asymptotic variances of the proposed estimators are given by

$$V\left({\widehat{\tau }}_k\right)={\tau }_k/\left(D_k-{B^{\prime }}_k{A}_k^{-1}{B}_k\right)\mathrm{,}k=1,2,\text{ and }V\left(\widehat{\tau }\right)=V\left({\widehat{\tau }}_1\right)+V\left({\widehat{\tau }}_2\right)\mathrm{,}$$

where $A_k=\left[a_{ij}^{\left(k\right)}\right]$ is a $2\times 2$ matrix whose $a_{ij}^{\left(k\right)}$ element is

$$a_{ij}^{(k)}=\{\begin{array}{ll}\left(1-\frac{n}{N}\right){\displaystyle \sum _{x=0}^n\left(\begin{array}{c}n\\ x\end{array}\right)\frac{1}{{\tilde{\pi }}_x^{\left(1\right)}\left({\theta }^{\left(1\right)}\right)}\left[\frac{\partial {\tilde{\pi }}_x^{\left(1\right)}\left({\theta }^{\left(1\right)}\right)}{\partial {\theta }_i^{(1)}}\right]\left[\frac{\partial {\tilde{\pi }}_x^{\left(1\right)}\left({\theta }^{\left(1\right)}\right)}{\partial {\theta }_j^{(1)}}\right]}\hfill & \hfill \\ +\frac{n}{N}{\displaystyle \sum _{x=0}^{n-1}\left(\begin{array}{c}n-1\\ x\end{array}\right)\frac{1}{{\tilde{\pi }}_x^{\left(A\right)}\left({\theta }^{\left(1\right)}\right)}\left[\frac{\partial {\tilde{\pi }}_x^{\left(A\right)}\left({\theta }^{\left(1\right)}\right)}{\partial {\theta }_i^{\left(1\right)}}\right]\left[\frac{\partial {\tilde{\pi }}_x^{\left(A\right)}\left({\theta }^{\left(1\right)}\right)}{\partial {\theta }_j^{\left(1\right)}}\right]}\hfill & \text{if}k=1\hfill \\ {\displaystyle \sum _{x=0}^n\left(\begin{array}{c}n\\ x\end{array}\right)\frac{1}{{\tilde{\pi }}_x^{\left(2\right)}\left({\theta }^{\left(2\right)}\right)}\left[\frac{\partial {\tilde{\pi }}_x^{\left(2\right)}\left({\theta }^{\left(2\right)}\right)}{\partial {\theta }_i^{\left(2\right)}}\right]\left[\frac{\partial {\tilde{\pi }}_x^{\left(2\right)}\left({\theta }^{\left(2\right)}\right)}{\partial {\theta }_j^{\left(2\right)}}\right]}\hfill & \text{if}k=2,\hfill \end{array}$$

$B_k$ is a bi-dimensional vector whose elements are

$$b_i^{\left(k\right)}=-\left[\partial {\tilde{\pi }}_0^{\left(k\right)}\left({\theta }^{\left(k\right)}\right)/\partial {\theta }_i^{\left(k\right)}\right]/{\tilde{\pi }}_0^{\left(k\right)}({\theta }^{\left(k\right)}),i=1,2$$

and $D_k$ is a real number given by

$$D_k=\{\begin{array}{ll}\left[1-\left(1-n/N\right){\tilde{\pi }}_0^{\left(1\right)}\left({\theta }^{\left(1\right)}\right)\right]/\left[\left(1-n/N\right){\tilde{\pi }}_0^{\left(1\right)}\left({\theta }^{\left(1\right)}\right)\right]\hfill & \text{if }k=1\hfill \\ \left[1-{\tilde{\pi }}_0^{\left(2\right)}\left({\theta }^{\left(2\right)}\right)\right]/{\tilde{\pi }}_0^{\left(2\right)}\left({\theta }^{\left(2\right)}\right)\hfill & \text{if }k=2.\hfill \end{array}$$

It is worth noting that in the derivation of the asymptotic variances we have made the assumptions that ${\tau }_k\to \infty ,k=1,2,m_i\to \infty ,i=1,\dots \mathrm{,}N,$ and $N$ and $n$ are fixed numbers.

To obtain numerical values of the variances we need to specify values for ${\tau }_k,{\alpha }^{\left(k\right)},{\sigma }_k$ and $n.$ One way to do this is to assign values to ${\tau }_k,{\sigma }_k$ and to the proportion ${\tilde{\pi }}_{1+}^{\left(k\right)}$ of people in $U_k$ who are linked at least to a particular site $A_i,$ which is common to all the sites and it is easier to specify than ${\alpha }^{\left(k\right)}.$ Then, for a given $n,$ the value of ${\alpha }^{\left(k\right)}$ is the solution to the equation

$${\tilde{\pi }}_{1+}^{\left(k\right)}-{\displaystyle \sum _{x=1}^n}\left(\begin{array}{c}n-1\\ x-1\end{array}\right){\tilde{\pi }}_x^{\left(k\right)}\left({\alpha }^{\left(k\right)}\mathrm{,}{\sigma }_k\right)=0.$$

Once ${\alpha }^{\left(k\right)},k=1,2,$ is obtained for a given $n,$ we can compute the numerical values of the variances $V\left({\widehat{\tau }}_k\right)$ and $V\left(\widehat{\tau }\right);$ the square roots of the relative variances $\sqrt{V\left({\widehat{\tau }}_k\right)/{\tau }_k^2}$ and $\sqrt{V\left(\widehat{\tau }\right)/{\tau }^2},$ and the sampling fractions $f_1=1-\left(1-n/N\right){\tilde{\pi }}_0^{\left(1\right)},f_2=1-{\tilde{\pi }}_0^{\left(2\right)}$ and $f=\left(f_1\times {\tau }_1+f_2\times {\tau }_2\right)/\tau .$ If the values of the square roots of the relative variances were not satisfactory, we could try different values of $n$ until we get satisfactory values.

We have programmed this procedure in the R software programming language and it is available to the interested readers by requesting to the authors. To illustrate the procedure, let us suppose that we have a sampling frame of $N=150$ sites and we assign the values ${\tau }_1=\text{1,200},$ ${\tau }_2=400,{\sigma }_1={\sigma }_2=1,$ ${\tilde{\pi }}_{1+}^{\left(1\right)}=0.05$ and ${\tilde{\pi }}_{1+}^{\left(2\right)}=0.04,$ then for $n=15$ we get that $V\left({\widehat{\tau }}_1\right)=\text{4,780}\text{.8},$ $V\left({\widehat{\tau }}_2\right)=\text{11,525}\text{.3},$ $V\left(\widehat{\tau }\right)=\text{16,306}\text{.1},$ $\sqrt{V\left({\widehat{\tau }}_1\right)}/{\tau }_1=0.06,$ $\sqrt{V\left({\widehat{\tau }}_2\right)}/{\tau }_2=0.27,$ $\sqrt{V(\widehat{\tau })}/\tau =0.08,$ $f_1=0.50,$ $f_2=0.38$ and $f=\mathrm{0.47.}$

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