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If I have cellular base-stations distributed as a PPP $\Phi_C$ with $\lambda_c$ density. Then the pdf of distribution is well known i.e. $$P[N = n] = \frac{(\lambda_c\pi r^2)^n}{n!}e^{-\lambda_c\pi r^2}$$

My Take (I will be glad for any alternate answer)

The probability that n=1 is then $$P[N=1] = (\lambda_c\pi r^2)e^{-\lambda_c\pi r^2} = 1$$ $$ln(r)r^2 = -\frac{1}{2\lambda_c^2\pi^2}$$ How can I proceed further to find $r$ (minimum distance between two points)?. Or should I take the Expectation with $N=1$?

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  • $\begingroup$ Your probability mass function (not density) seems to be for the number of base-stations within $r$ of a given location; if that is correct then you might state it. I do not understand how you got $P[N=1] = 1$. $\endgroup$
    – Henry
    Mar 1, 2018 at 10:27
  • $\begingroup$ I take $P[N=1]=1$ as the probability "for sure" that one base-station near the referenced base-station is present with distance $r$. This is the way I thought of solving. If you have any other method, I will be glad to learn. $\endgroup$
    – SJa
    Mar 1, 2018 at 10:32
  • $\begingroup$ You would actually want $P(N \ge 1) =1$. But this is never the case as $P(N=0) = e^{-\lambda_c\pi r^2} \gt 0$ for all $r$. There is no "for sure" radius $\endgroup$
    – Henry
    Mar 1, 2018 at 10:35
  • $\begingroup$ If you wanted to simulate this, I would first find the total area $a$ (which could be a circle or square or something else - this will make a difference), then simulate the total number of base stations $N$ with $P[N=n] = \frac{(\lambda_c a)^n}{n!}e^{-\lambda_c a}$, then simulate the $n$ locations uniformly across the area, then find the minimum distance between them. Then rinse and repeat several times to get a view of the distribution $\endgroup$
    – Henry
    Mar 1, 2018 at 10:36
  • $\begingroup$ Oh I understand your point about $P(N\geq1) \neq 1$... I also understand your point about simulation as well. But here actually I don't want a simulation (I can make do simulation in no time), I want some analytical result. It may be based on some expected value (as I understand $r$ is a RV) $\endgroup$
    – SJa
    Mar 1, 2018 at 10:40

1 Answer 1

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I doubt there is an easy answer to "minimum distance between two points" in a finite area (there is a positive probability there are not two base-stations), and it will have a limit of $0$ in an infinite area

There is an answer to the question about the distribution of the distance to nearest base-station from a given location (in an infinite area):

  • The probability there is no base-station within radius $r$ is $P_r(N=0)=e^{-\lambda_c\pi r^2}$

  • so the probability the nearest base-station is within radius $r$ is $P(R \le r) =P_r(N\ge 1)=1-e^{-\lambda_c\pi r^2}$

  • so the probability density for the distance to the nearest base station is $f(r)= 2 \lambda_c\pi r e^{-\lambda_c\pi r^2}$

  • so the expectation of the distance to the nearest base station is $E[R]=\dfrac{1}{2 \sqrt{\lambda_c}}$

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  • $\begingroup$ Actually, this was the thing I was looking after. Many thanks. I dont know how hard I try I could not crack math on my own. Simulation comes quite easy to me $\endgroup$
    – SJa
    Mar 1, 2018 at 12:45

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