# Mathematical Statistics and Data Analysis - Solutions

### 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, $\, P(u_i = 1) \,$:

• Case $\, i \le n \,$(note that here $% $), Using LOTP, we have: %

• 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 $% $. 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.

$$\tag*{\blacksquare}$$