$N$ points are placed randomly according to a uniform distribution in a $1 \times 1$ square. If $M$ is the number of points that have a distance more than $c/\sqrt{N}$ to others, then prove $\exists c,\alpha>0$ such that $$\lim\limits_{N\to \infty}\mathbb{P}(M>\alpha N)=1$$

Besides this question, please let me know if you can find the average of $M$.

This is follow up question for this problem


Let $X_i$ be the indicator variable of the event that the $i$-th point "is free", i.e., it has a distance more than $c/\sqrt{N}$ to all others.

Then, assuming $N$ is large and we ignore "border effects" (points near the border of the square),

$$P(X_i=1)=E(X_i)= 1- \pi \frac{c^2}{N}$$

Then the expected number of total "free" points is

$$E(Y)=\sum_i E(X_i)= N -\pi c^2$$

Update: This answer was writen before the OP edited the question: This is useless now

  • $\begingroup$ Thanks so much. I changed the problem a little bit. Sorry. $\endgroup$ – Susan_Math123 Apr 25 '17 at 3:31
  • $\begingroup$ I changed the problem again to cover your answer. I'd appreciate it if you could also answer my updated question. $\endgroup$ – Susan_Math123 Apr 25 '17 at 3:34
  • $\begingroup$ I can change it back to the original question if you want. $\endgroup$ – Susan_Math123 Apr 25 '17 at 3:48

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