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Let's uniformly sample $n$ times from the unit square. Define a polygon contained within the unit square with area $A$. Surely as $n \to \infty$, the probability that the polygon contains at least $\lfloor{nA}\rfloor$ samples and at most $\lceil{nA}\rceil$ samples tends to 1. What I'm interested in is if there are $m$ polygons, $m = kn$, where the polygons may overlap arbitrarily, does the probability that every polygon has at least $\lfloor{nA_i}\rfloor$ and at most $\lceil{nA_i}\rceil$ samples tend to 1 as well?

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  • $\begingroup$ Assuming $n\ge 0$ and $A \ge 0$, you will have $\lfloor{nA}\rfloor\ge 0$ so you do not need $\max(0,...)$ $\endgroup$
    – Henry
    Dec 3, 2023 at 12:53
  • $\begingroup$ It is more usual to write $m |n$ rather than $n|m$ to indicate $n$ is a multiple of $m$. $\endgroup$
    – Henry
    Dec 3, 2023 at 13:00
  • $\begingroup$ Actually I meant $m$ is a multiple of $n$ $\endgroup$
    – user1261526
    Dec 3, 2023 at 13:11
  • $\begingroup$ Oh yeah thanks. I removed the $max(0, ...)$ $\endgroup$
    – user1261526
    Dec 3, 2023 at 13:12
  • $\begingroup$ If $m$ is a multiple of $n$, what is increasing? Is $n$ fixed and $m$ increasing? Or $m=kn$ so they are increasing together with a fixed multiple? Or something else? Do the new polygons overlap or are they getting smaller? $\endgroup$
    – Henry
    Dec 3, 2023 at 13:16

1 Answer 1

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Your statement

Surely as $n \to \infty$, the probability that the polygon contains at least $\max(0, \lfloor{nA}\rfloor)$ samples and at most $\lceil{nA}\rceil$ samples tends to $1$

is incorrect. It would be correct to say that the number contained in the polygon divided by $n$ converges towards $A$ in a law-of-large-numbers sense, but that is not saying the same thing. Flip a fair coin $1$ million times, and the proportion of heads will probably be close to $\frac12$ but the probability you see exactly $500\,000$ heads is about $0.0008$.

The number sampled in the polygon has a binomial distribution with parameters $n$ and $A$, so expectation $nA$ and variance $nA(1-A)$ which is increasing with $n$. The probability that the numbered sampled in the polygon is somewhere from $\lfloor{nA}\rfloor$ through to $\lceil{nA}\rceil$ converges to $0$ as $n$ increases.

With $m$ polygons, the probability is of course lower that they are all in this interval than that one specific one is.

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  • $\begingroup$ Ah yes. I have a follow up question which I'll ask as part of a separate Question. Thank you! $\endgroup$
    – user1261526
    Dec 3, 2023 at 13:18
  • $\begingroup$ Hi Henry, my follow up question is: math.stackexchange.com/questions/4819343/… $\endgroup$
    – user1261526
    Dec 3, 2023 at 14:24
  • $\begingroup$ Your next question (with a wider band, but still of fixed size) is going to face the same issue as this one: for any $A$ with positive area and any $d>0$, the probability the number sampled in $A$ being in $[nA-d,nA+d]$ will converge to $0$ in the same sense. You need something like $[n(A-d),n(A+d)]$ for convergence to $1$ $\endgroup$
    – Henry
    Dec 3, 2023 at 16:23
  • $\begingroup$ Hi Henry, actually $d$ was expressed as a function of $n$ and $A$. Would you be so kind as to take a look at math.stackexchange.com/questions/4820616/… please? This is the last question in this chain of questions I have, with it being the least abstract $\endgroup$
    – user1261526
    Dec 5, 2023 at 10:53

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