Continuing with our example population \(U\) of size \(N=5\), suppose that in the first phase a sample of \(n_a=2\) elements is selected according to a simple random sampling design. In the second phase, a subsample of \(n=1\) is selected according to a simple random sampling design.
For the first phase, using Example 2.1.1, the \(\binom{N}{n_a}\) possible samples, together with their respective selection probabilities, are
X1 X2 p_a
1 Yves Ken 0.1
2 Yves Erik 0.1
3 Yves Sharon 0.1
4 Yves Leslie 0.1
5 Ken Erik 0.1
6 Ken Sharon 0.1
7 Ken Leslie 0.1
8 Erik Sharon 0.1
9 Erik Leslie 0.1
10 Sharon Leslie 0.1
The inclusion probability in the first-phase sample, for each of the 5 elements of \(U\), is \[
\pi_{ak}=\frac{n_a}{N}=\frac{2}{5}
\]
For the second phase, there are \(\binom{n}{n_a}\) possible subsamples for each first-phase sample. The second-phase sampling design and the overall sampling design are defined as follows:
X1 X2 p_a S p( |s_a) p(s)
1 Yves Ken 0.1 Yves 0.5 0.05
Ken 0.5 0.05
2 Yves Erik 0.1 Yves 0.5 0.05
Erik 0.5 0.05
3 Yves Sharon 0.1 Yves 0.5 0.05
Sharon 0.5 0.05
4 Yves Leslie 0.1 Yves 0.5 0.05
Leslie 0.5 0.05
5 Ken Erik 0.1 Ken 0.5 0.05
Erik 0.5 0.05
6 Ken Sharon 0.1 Ken 0.5 0.05
Sharon 0.5 0.05
7 Ken Leslie 0.1 Ken 0.5 0.05
Leslie 0.5 0.05
8 Erik Sharon 0.1 Erik 0.5 0.05
Sharon 0.5 0.05
9 Erik Leslie 0.1 Erik 0.5 0.05
Leslie 0.5 0.05
10 Sharon Leslie 0.1 Sharon 0.5 0.05
Leslie 0.5 0.05
Note that, using the theorem of total probability, the final sampling design, which accounts for the probability dynamics of the first and second phases, is defined as follows: \[
p(s)=
\begin{cases}
0.2, &\text{if $s=\{\text{Yves}\}$},\\
0.2, &\text{if $s=\{\text{Ken}\}$},\\
0.2, &\text{if $s=\{\text{Erik}\}$},\\
0.2, &\text{if $s=\{\text{Sharon}\}$},\\
0.2, &\text{if $s=\{\text{Leslie}\}$}.
\end{cases}
\]
The inclusion probability of an element of \(S_a\) in the subsample of the last phase, conditional on the realization of a particular sample, is given by \[
\pi_{k|S_a}=\frac{n_a}{n}=\frac{1}{2}
\]
Then the conditional inclusion probability of an element of \(U\) given by \(\pi_k^*\) is \[
\pi_k^*=\pi_{ak}\pi_{k|S_a}=\frac{n_a}{N}\frac{n_a}{n}=\frac{n}{N}=\frac{1}{5}
\]
which, for this particular case, coincides with the element’s inclusion probability, properly speaking, given in (12.1.6). However, almost always \(\pi_k^* \neq \pi_k\), as shown by the following configuration induced by a sampling design with unequal selection probabilities.
X1 X2 p_a S p( |S_a) p(s)
1 Yves Ken 0.25 Yves 0.9 0.225
Ken 0.1 0.025
2 Yves Erik 0.15 Yves 0.8 0.120
Erik 0.2 0.030
3 Yves Sharon 0.15 Yves 0.7 0.105
Sharon 0.3 0.045
4 Yves Leslie 0.10 Yves 0.6 0.060
Leslie 0.4 0.040
5 Ken Erik 0.10 Ken 0.5 0.050
Erik 0.5 0.050
6 Ken Sharon 0.05 Ken 0.4 0.020
Sharon 0.6 0.030
7 Ken Leslie 0.05 Ken 0.3 0.015
Leslie 0.7 0.035
8 Erik Sharon 0.05 Erik 0.2 0.010
Sharon 0.8 0.040
9 Erik Leslie 0.05 Erik 0.1 0.005
Leslie 0.9 0.045
10 Sharon Leslie 0.05 Sharon 0.5 0.025
Leslie 0.5 0.025
Note that, for this configuration, and once again using the theorem of total probability, the final sampling design is defined as follows: \[
p(s)=
\begin{cases}
0.510, &\text{if $s=\{\text{Yves}\}$},\\
0.110, &\text{if $s=\{\text{Ken}\}$},\\
0.140, &\text{if $s=\{\text{Sharon}\}$},\\
0.095, &\text{if $s=\{\text{Erik}\}$},\\
0.145, &\text{if $s=\{\text{Leslie}\}$}.
\end{cases}
\]
In this case, for the first phase, the inclusion probability in the first-phase sample for each of the 5 elements of \(U\) is
\[
\pi_{ak}=
\begin{cases}
0.65, &\text{if $k=\text{Yves}$},\\
0.45, &\text{if $k=\text{Ken}$},\\
0.35, &\text{if $k=\text{Erik}$},\\
0.30, &\text{if $k=\text{Sharon}$},\\
0.25, &\text{if $k=\text{Leslie}$}.
\end{cases}
\]
The inclusion probability of an element of \(S_a\) in the second-phase subsample, conditional on the realization of a particular sample, is given by the following 10 cases, as many cases as first-phase samples:
- If \(S_a=S_1\), then \[
\pi_{k|S_a}=
\begin{cases}
0.90, &\text{if $k=\text{Yves}$},\\
0.10, &\text{if $k=\text{Ken}$}.
\end{cases}
\]
- If \(S_a=S_2\), then \[
\pi_{k|S_a}=
\begin{cases}
0.80, &\text{if $k=\text{Yves}$},\\
0.20, &\text{if $k=\text{Erik}$}.
\end{cases}
\]
- And so on, until
- If \(S_a=S_{10}\), then \[
\pi_{k|S_a}=
\begin{cases}
0.50, &\text{if $k=\text{Sharon}$},\\
0.50, &\text{if $k=\text{Leslie}$}.
\end{cases}
\]
Therefore, there will also be 10 cases for calculating the quantity \(\pi_k^*\), as follows:
- If \(S_a=S_1\), then \[
\pi_{k}^*=
\begin{cases}
0.65 \times 0.90 = 0.585, &\text{if $k=\text{Yves}$},\\
0.45 \times 0.10 = 0.045, &\text{if $k=\text{Ken}$}.
\end{cases}
\]
- If \(S_a=S_2\), then \[
\pi_{k}^*=
\begin{cases}
0.65 \times 0.80 = 0.520, &\text{if $k=\text{Yves}$},\\
0.35 \times 0.20 = 0.007, &\text{if $k=\text{Erik}$}.
\end{cases}
\]
- And so on, until
- If \(S_a=S_{10}\), then \[
\pi_{k}^*=
\begin{cases}
0.30 \times 0.50 = 0.150, &\text{if $k=\text{Sharon}$},\\
0.25 \times 0.50 = 0.125, &\text{if $k=\text{Leslie}$}.
\end{cases}
\]
The above shows that \(\pi_k^* \neq \pi_k\), because the inclusion probability is given by \[
\pi_{k}=
\begin{cases}
0.510, &\text{if $k=\text{Yves}$},\\
0.110, &\text{if $k=\text{Ken}$},\\
0.140, &\text{if $k=\text{Erik}$},\\
0.095, &\text{if $k=\text{Sharon}$},\\
0.145, &\text{if $k=\text{Leslie}$}.
\end{cases}
\]
Note that in practice, with fairly large populations, it is not possible to calculate \(\pi_k\). As an exercise, using the data from Example 2.1.3, the unbiasedness of the \(\pi_k^*\) estimator should be verified both in the first configuration and in this last one.