Chapter 7, Survey Sampling

Solution 27

To show that the procedure given will generate a simple random sample of size , 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 which is same as in the case of simple random sampling.

Let denotes that element of the population is in the sample after the procedure completes and otherwise.

Thus in the given procedure, we need to check that in the end when procedure completes, the probability should be irrespective of .

Let if element of the population stays in the sample after the step of the procedure, assuming that it stayed in the sample in state. Thus is equal to the probability that for all steps k.

Let if element is selected from the list of population. Thus by definition, and .

Thus we have two cases to compute the probability, :

  • Case (note that here ), Using LOTP, we have:

  • Case , In this case unlike in the last case . Note that starts from . Thus here also like in previous case . Thus we can use the result from the previous case. Thus we get .

Thus in both cases, we have . Thus the procedure is equivalent to simple random sampling.

$$\tag*{$\blacksquare$} $$