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This question originated from a discussion where a 1st random sample was unmanageablely large-- so we wanted to create a computationally manageable dataset by sampling the 1st random sample, then using that to estimate descriptive statistics about the population.

If the answer is too complicated to easily explain or is "it depends" pointing me in the direction of a textbook/literature that deals with this issue would be very much appreciated.

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  • $\begingroup$ Could you tell us what exactly the sampling procedures are? For simple random samples (with or without replacement), a sample of a sample is equivalent to a sample. In other cases, it's more complicated. $\endgroup$
    – whuber
    Commented Jul 27, 2021 at 18:45
  • $\begingroup$ The question was for a simple random sample! I'm interested in learning about more complex cases if you can point me to a resource. :) $\endgroup$ Commented Jul 27, 2021 at 21:01
  • $\begingroup$ A special case is answered at stats.stackexchange.com/questions/78446. $\endgroup$
    – whuber
    Commented Jul 27, 2021 at 21:34
  • $\begingroup$ thank you this was very helpful $\endgroup$ Commented Jul 27, 2021 at 22:04

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The operational definition of simple random sampling of a population is that every subset of individuals of the specified sample size $k$ has the same chance of being the sample (Wikipedia).

A simple random subsample of a simple random sample is a simple random sample.

I will give a formal mathematical proof below, but first here's some intuition. Create a marble for each individual in the population on which that individual's identifier is engraved. Put all the marbles into a jar. To obtain the small sample of size $l,$

  1. Mix the jar thoroughly and blindly withdraw $l$ marbles. The IDs on those marbles give a simple random sample of the population.

  2. Mix the jar thoroughly and blindly withdraw $k\ge l$ marbles. Now mix those $k$ marbles and blindly take $l$ of them. This determines a subsample of a simple random sample.

I ask you only to believe that the second mixing procedure in step (2) is superfluous: it does not change the chances that any particular set of $l$ individuals will be the sample. It should at least be obvious that neither of these two procedures favors any particular collection of individuals: everyone, and every possible group, is treated the same way.


One advantage of the mathematical demonstration to follow is that it shows how one can approach similar questions in more complex sampling situations where intuition might fail us. In general, a sampling procedure is completely determined by the probability distribution of the possible samples that could be obtained. To show that two sampling procedures are equivalent, you only have to show they yield the same probability distributions. In the present case, this is the uniform distribution on the collection of all possible samples of a given size. In other cases the distribution will not be uniform. For many examples of other sampling procedures, consult any edition of Steven Thompson's book Sampling (Wiley).


The question concerns obtaining a sample by first taking a simple random of size $k$ and then taking a simple random subsample of size $l\le k$ from that sample. To address this, consider an arbitrary subset of $l$ individuals in a population of size $n.$

  1. There are $\binom{n-l}{k-l}$ sets of $k$ individuals containing these particular $l$ individuals, because there are this many ways to choose an additional $k-l$ individuals out of the remaining $n-l$ in order to enlarge an $l$-subset into a $k$-subset.

  2. Each of those sets of $k$ individuals has a chance of $1/\binom{n}{k}$ of being the original sample. (That's what simple random sampling means.)

  3. Conditional on a particular one of the $k$-sets in $(1)$ being the original sample, this particular set of $l$ individuals has a chance of $1/\binom{k}{l}$ of being selected out of that $k$-set. (That's what simple random subsampling means.)

The laws of probability state the chance these $l$ individuals are the ones finally selected is the product of the chances, given by

$$\binom{n-l}{k-l} \times \frac{1}{\binom{n}{k}} \times \frac{1}{\binom{k}{l}} = \frac{(n-l)!\, k!(n-k)!\, l!(k-l)!}{(k-l)!((n-l)-(k-l))!\, n!\,k!} = \frac{(n-l)!l!}{n!}= \frac{1}{\binom{n}{l}}.$$

Because that is the correct chance for simple random sampling of size $l$ in a population of size $n,$ we are done.

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  • $\begingroup$ fantastic and well explained proof, I appreciate your help $\endgroup$ Commented Jul 28, 2021 at 4:19

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