Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

January 2006

3 (34 points)

Assume that the locations of cellular telephone towers can be accurately modeled by a 2-dimensional homogeneous Poisson process for which the following two facts are know to be true:

1. The number of towers in a region of area A is a Poisson random variable with mean \lambda A , where \lambda>0 .

2. The number of towers in any two disjoint regions are statistically independent.

Assume you are located at a point we will call the origin within this 2-dimensional region, and let $R_{\left(1\right)}<R_{\left(2\right)}<R_{\left(3\right)}<\cdots$ be the ordered distances between the origin and the towers.

(a)

Show that $R_{\left(1\right)}^{2},R_{\left(2\right)}^{2},R_{\left(3\right)}^{2},\cdots$ are the points of a one-dimensional homogeneous Poisson process.

$P\left(R_{\left(k+1\right)}^{2}-R_{\left(k\right)}^{2}>r\right)=P\left(\text{there is no tower in area }\pi r\right)=\frac{\left(\lambda\pi r\right)^{0}}{0!}e^{-\lambda\pi r}=e^{-\lambda\pi r}.$

$P\left(R_{\left(k+1\right)}^{2}-R_{\left(k\right)}^{2}\leq r\right)=P\left(\left\{ \text{there is at least one tower in area }\pi r\right\} \right) $$=1-e^{-\lambda\pi r}\text{: CDF of exponential random variable}. ref. You can see the expressions about exponentail distribution [CS1ExponentialDistribution]. R_{\left(k+1\right)}^{2}-R_{\left(k\right)}^{2} is an exponential random variable with parameter \lambda\pi . \therefore R_{\left(1\right)}^{2},R_{\left(2\right)}^{2},R_{\left(3\right)}^{2},\cdots are the points of a one-dimensional homogeneous Poisson process. (b) What is the rate of the Poisson process in part (a)? \lambda\pi . cf. The mean value for the exponential random variable is \frac{1}{\lambda\pi} . (c) Determine the density function of R_{\left(k\right)} , the distance to the k -th nearest cell tower. F_{k}\left(x\right)\triangleq P\left(R_{\left(k\right)}\leq x\right)$$ =P\left(\text{There are at least }k\text{ towers within the distance between origin and }x\right) $$=P\left(N\left(0,x\right)\geq k\right)=1-P\left(N\left(0,x\right)\leq k-1\right)=1-\sum_{j=0}^{k-1}\frac{\left(\lambda\pi x^{2}\right)^{j}}{j!}e^{-\lambda\pi x^{2}}. f_{k}\left(x\right)=\frac{dF_{k}\left(x\right)}{dx}=-\sum_{j=0}^{k-1}\left\{ \frac{j\left(\lambda\pi x^{2}\right)^{j-1}\left(2\lambda\pi x\right)}{j!}e^{-\lambda\pi x^{2}}+\frac{\left(\lambda\pi x^{2}\right)^{j}}{j!}e^{-\lambda\pi x^{2}}\left(-2\lambda\pi x\right)\right\}$$ =\left(2\lambda\pi x\right)e^{-\lambda\pi x^{2}}\cdot\left\{ -\sum_{j=1}^{k-1}\frac{\left(\lambda\pi x^{2}\right)^{j-1}}{\left(j-1\right)!}e^{-\lambda\pi x^{2}}+\sum_{j=0}^{k-1}\frac{\left(\lambda\pi x^{2}\right)^{j}}{j!}\right\} $$=\left(2\lambda\pi x\right)e^{-\lambda\pi x^{2}}\cdot\left\{ -\sum_{j=0}^{k-2}\frac{\left(\lambda\pi x^{2}\right)^{j}}{j!}+\sum_{j=0}^{k-1}\frac{\left(\lambda\pi x^{2}\right)^{j}}{j!}\right\}$$ =\left(2\lambda\pi x\right)e^{-\lambda\pi x^{2}}\cdot\frac{\left(\lambda\pi x^{2}\right)^{k-1}}{\left(k-1\right)!}.$

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