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There can be many ways and ad hoc variants to perform sampling without replacement from a limited population.

Consider we have $k$ categories (types of objects) and the k-length vector of their corresponding probabilities, kprob. These weights are seen as background "demand" probabilities. (In the simplest case of k categories being k individual objects, the vector is the constant $\text {kprob}_i=1/k$.) And we have limited $F$-size population, or "stock", of the k categories with frequency (proportion) distribution characterized by corresponding k-length vector stock: $\text {stock}_i=f_i/F$. These are "availability" proportions.

We are drawing randomly obiects one by one without replacement in order to collect sample of size $n$. At each step $c$ (drawing $c$th object) probability $\text{prob}_{(c)i}$ to select object of category $i$ is estimated as directly proportional to the product

$\text{kprob}_{i} \cdot \text{stock}_{(c-1)i}$,

where $\text{stock}_{(c-1)}$ is the proportion distribution of the population observed prior step $c$. And an object is then selected by generating one value from k-categorical distribution with probability parameter vector $\text{prob}_{(c)}$. Upon the object (of some fallen category $i$) is selected, hence removed from the population, vector stock is updated. (Above saying "proportional to the product" implies that the exact probabilities are obtained in normalizing these k computed products to sum=1.)

My question: is there any name for the modification of the described process whereby $\text{prob}_{(c)i}$ is estimated as likewise proportional to

$[1-\text{prob}_{(c-1)i}] \cdot \text{stock}_{(c-1)i}$,

that is, instead of the fixed "demand" kprob, 1 - "floating demand" is used? (At $c=1$ step, the [] multiplier is taken equal to kprob.)

What do you know about the usage and properties of the method?

The method I'm asking about may be useful as a "smoothing" alternative to the former one. It somewhat facilitated drawing from small-populated categories and complicates drawing from large-populated categories. Consider an example with input kprob = 5 equiprobable categories and input population stock frequencies is {1000, 2000, 3000, 4000, 5000}, that is, percents {6.667, 13.333, 20.000, 26.667, 33.333} and we are drawing w/o replacement a sample of size, say, 3000. An example of such sample for the latter method is:

enter image description here

which clearly tempers the utmost percentages, whereas the former, classic method:

enter image description here

follows the population highly unbalanced percentages (it follows it because drawing 3000 out from 15000 did not empty any of the 5 categories in the stock prematurely).

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  • $\begingroup$ 1) Is the first display equation: $\text{kprob}_i \cdot \text{stock}_{(c-1)i}$ supposed to sum to 1 (summing over $i$)? Is this an implicit constraint that you are placing on $\text{stock}_{(c-1)i}$, or is it somehow explicit? 2) Is your first sampling scheme something besides a multivariate hypergeometric scheme? $\endgroup$ – Andrew M Jan 31 '17 at 16:06
  • $\begingroup$ @AndrewM, 1. Since I said "proportional to" (the product), it doesn't matter what is the exact sum: the prob(c) vector is obtained as the normalization of these products to sum to 1 across all k i's. 2. I didn't quite understand this question of yours. The schemes (both) are what I described them. Are they something "besides" something - ? $\endgroup$ – ttnphns Jan 31 '17 at 16:13
  • $\begingroup$ Without going into details, my gut reaction is, "this is probably wrong", in the sense that it does not give the desirable probabilities. Unequal probability sampling without replacement is counterintuitive, and requires weird adjustments. I mean, this also depends on what your target of inference is; if you need to make inference with respect to your finite population, you need to form a Horvitz-Thompson estimator... for which you need to know the exact probabilities of selection. And they are difficult to compute a lot of the times. $\endgroup$ – StasK Jan 31 '17 at 21:46
  • $\begingroup$ As an education officer of the Survey Research Methods Section of the American Statistical Association, I just organized a webinar by David Haziza less than a week ago, Jan 26 2017. He talked about the mathematical reasons why "simple" approaches tend to fail. $\endgroup$ – StasK Jan 31 '17 at 21:47
  • $\begingroup$ @ttnphs: 1) thanks for the clarification. 2) I don't entirely understand your notation. (eg what is $f_i$ in the second paragraph?) So I was trying to see if there is some other analogy or related process that could help readers understand the stochastic process you describe. $\endgroup$ – Andrew M Feb 1 '17 at 2:49

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