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Suppose I have a dataset $D$ and let it be split into 2 smaller datasets $D_1$ and $D_2$ such that $D = D_1 + D_2 $. Thus we can also say that $D_1=D - D_2$.

Will we get a simple random sample of $D_1$ if we took a simple random sample of $D$ and removed from it those data points obtained by taking a random sample of $D_2$?

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    $\begingroup$ Are you sampling randomly and separately from $D$ and $D_2$? I don't see how you'd get a simple sample of $D_1$ by subtracting points in $D_2$ from different points in $D$...E.g., if all points in $D_1$ happen to equal 1, you won't get $D_1$ by randomly sampling a 6 from $D$ and subtracting a randomly sampled 3 from $D_2$...you'd have to sample the corresponding 5 from $D_2$ to be sure of getting a data point sampled from $D_1$. I.e., I think you'd only get to sample randomly from $D$ or $D_2$, not both. $\endgroup$ – Nick Stauner Mar 16 '14 at 23:38
  • $\begingroup$ Thank you. I think that gave me a better intuition now. You can't get a random sample of $D_1$ by taking a random sample of $D$ and subtracting the points that are the random sample of $D_2$. However, if you take a stratified sample of $D$ and $D_2$ and take their difference, you would get a stratified sample of $D_1$. I still need to work out the math and make sure this works. $\endgroup$ – Budhapest Mar 16 '14 at 23:45
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    $\begingroup$ You didn't answer @Nick Stauner's question clearly. You refer to "random sample of $D_2$. If you mean a different random sample of $D_2$ taken independently of the random sample of $D$, the answer is "No", as Nick stated. If you mean that you remove from the sample of $D$ thos elements which come from $D_2$, the answer is "Yes". $\endgroup$ – Steve Samuels Mar 17 '14 at 22:36
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I don't see how you'd be guaranteed a simple sample of $D_1$ by subtracting points in $D_2$ from different points in $D$. For example, consider these distributions:

$$D=\{1,2,...100\}\\D_1=\{1,1,...1\}\\D_2=\{0,1,...99\}$$

If you happen to randomly sample the first five values in a row (it could happen!) from $D$ and subtract five randomly sampled values from $D_2$, you're pretty unlikely to end up with a sample of $D_1$; you'll probably get a bunch of negative values, none of which belong to $D_1$ of course.

I.e., I think you'd only get to sample randomly from $D$ or $D_2$, not both. Same goes for stratified sampling: it'll only work if you sample systematically within strata, not randomly – at least, not randomly with both samples. E.g., you could sample randomly from $D$, with or without stratification, but if you happen to sample the 8th, 48th, and 88th values from $D$, the only way I can see to guarantee that you'll get values from $D_1$ by subtracting values sampled from $D_2$ is to systematically sample the same 8th, 48th, and 88th values from $D_2$.

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    $\begingroup$ I don't understand this example, which contradicts the conditions in @Niru's first paragraph, namely that $D$ is "split" into $D_1$ and $D_2$ and that $D_1 = D - D_2$, so that $$ D_1 \cup D_2 = D $$ $$ D_1 \cap D_2 = \emptyset $$ $\endgroup$ – Steve Samuels Mar 17 '14 at 23:19
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    $\begingroup$ I don't think that @Niku has answered your question about what the "sample of $D_2$" is. Another interpretation is this: that there is a SRS from $D$ of size $n$, and it contains $n_1$ elements from $D_1$ and $n_2$ from $D_2$. If this is the case, then conditional on $n_1>0$, the elements from $D_1$ do constitute a simple random sample of size $n_1$ from $D_1$. The proof is given on page 35 of Cochran (1977). Reference: Cochran, W. G. (1977). Sampling techniques (3rd ed.). New York: Wiley. $\endgroup$ – Steve Samuels Mar 17 '14 at 23:27

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