Chapter 7, Survey Sampling
Solution 27
To show that the procedure given will generate a simple random sample of size $\, n \,$, we need to prove that the probability of selecting an element from the total population to form a random sample is same for every element and is equal to $\, \frac n N \,$ which is same as in the case of simple random sampling.
Let $\, u_i = 1 \,$ denotes that $\, i^{th} \,$ element of the population is in the sample after the procedure completes and $\, u_i = 0 \,$ otherwise.
Thus in the given procedure, we need to check that in the end when procedure completes, the probability $\, P(u_i = 1) \,$ should be $\, \frac n N \,$ irrespective of $\, i \,$.
Let $\, u_{ik} = 1 \,$ if $\, i^{th} \,$ element of the population stays in the sample after the $\, k^{th} \,$ step of the procedure, assuming that it stayed in the sample in $\, (k-1)^{th} \,$ state. Thus $\, P(u_i = 1) \,$ is equal to the probability that $\, P(u_{ik} = 1) \,$ for all steps k.
Let $\, s_k = 1 \,$ if $\, (n+k)^{th} \,$ element is selected from the list of population. Thus by definition, $\, P(s_k = 1) = \frac n {n+k} \,$ and $\, P(s_k = 0) = (1-\frac n {n+k}) \,$.
Thus we have two cases to compute the probability, $\, P(u_i = 1) \,$:
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Case $\, i \le n \,$(note that here $\, i < n+k \,$), Using LOTP, we have: $\, \begin{align*} P(u_{ik} = 1) \\ &= P(u_{ik} = 1 \,\vert\, s_k = 1) \times P(s_k = 1) + P(u_{i k} = 1 \,\vert\, s_k = 0) \times P(s_k = 0) \\ &= \frac {n-1} n \times \frac n {n+k} + 1 \times (1 - \frac n {n+k}) = \frac {n+k-1} {n+k} \\ \text{Thus we can compute} \\ P(u_i = 1) \\ &= \prod_{k=1}^{k=N-n} P(u_{ik} = 1) \\ &= \frac n {n+1} \times \frac {n+1} {n+2} \times ... \times \frac {n+N-n-1} {n+N-n} \\ &= \frac n N \end{align*} \,$
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Case $\, i > n \,$, In this case unlike in the last case $\, P(u_i = 1) = P(s_i = 1) \times \prod_{k=i+1}^{k=N-n} P(u_{ik} = 1) \,$. Note that $\, k \,$ starts from $\, i+1 \,$. Thus here also like in previous case $\, i < n+k \,$. Thus we can use the result $\, P(u_{ik} = 1) = \frac {n+k-1} {n+k} \,$ from the previous case. Thus we get $\, P(u_i = 1) = \frac n {n+i} \times \frac {n+i} {n+i+1} \times ... \times \frac {n+N-n-1} {n+N-n} = \frac n N \,$.
Thus in both cases, we have $\, P(u_i =1) = \frac n N \,$. Thus the procedure is equivalent to simple random sampling.
$$\tag*{$\blacksquare$} $$