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The law of large numbers says that if I sample $n$ points independently from a probability density function $f$, then the number of points lying in a neighborhood of a point $x$ with area $\epsilon$ will be proportional to $\epsilon n f(x)$. My question is: what if I sample $mn$ points from $f$, and these points are grouped into sets of size $m$? (i.e. you have $S_1 = \{x_1,\dots,x_m\}, S_2 = \{x_{n+1},\dots,x_{2m}\},\dots,S_n=\{x_{(n-1)m+1},\dots,x_{mn}\}$ There are now $m^n$ ways to select one element from each of the $n$ sets, and many of those selections may look nothing like the original distribution $f$. Is there anything we can still say probabilistically about these selections?

As a simple example, suppose that I sample $mn$ points uniformly at random on the unit interval. Suppose that for each one of the $n$ sets (consisting of $m$ points), I select the largest element. Then, the distribution of these largest points has a density function $f(x)= mx^{m-1}$. There are other ways I could select these points that would give other densities. I'd like to know if there's anything I can say about these "other density" functions in terms of $m$.

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  • $\begingroup$ I don't think there's very much mileage in this question. If $m$ is large, you should be able to get very close to your favourite distribution by picking suitable points from each set. I think it's really destroying the probabilistic spirit of the LLN by allowing "the opponent" to make all of these choices. $\endgroup$ Apr 14, 2016 at 6:36
  • $\begingroup$ I agree, and your point is well-taken -- as $m$ becomes large, we are able to "generate" a more diverse set of distributions. The kinds of phenomena I'm interested in are, for example, the fact that I can use a uniform distribution with $m=2$ to "generate" samples with a distribution $f(x) = 2x$, but presumably I wouldn't be able to "generate" samples with a distribution $f(x) = 3x^2$. $\endgroup$ Apr 14, 2016 at 6:41
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    $\begingroup$ OK. So you have in mind: $m$ fixed; $n$ large. So here's my best guess so far: a distribution, $\nu$, is attainable with an $m$-step sample if $\nu(A)\in [\lambda(A)^m,1-\lambda(A^c)^m]$ for each subset $A$ of the interval. This is certainly a necessary condition. I'll think about sufficiency... By the way, applying this to a set $A$ of small measure immediately gives that $\nu\ll\lambda$ with density bounded above by $m$. $\endgroup$ Apr 14, 2016 at 16:07

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