3.8.2 Time-Space Poisson Process

Problem 3.26 applies this concept.

Example 15: Distribution of Travel Distance ("Nearest Neighbor")

Suppose that emergency response units are distributed throughout a large region as a two-dimensional Poisson process with intensity parameter y units per square mile. We wish to know the pdf of the travel distance D between an incident, whose position is selected independently of response unit positions, and the nearest response unit. Assume Euclidean travel distance. (This is sometimes known as a "nearest-neighbor" problem; in three-dimensional space this problem has been used to determine the distribution of distance between stars in a galaxy.)

Solution

We use the never-fail cumulative distribution method in conjunction with our new knowledge of spatial Poisson processes.

    This is a Rayleigh pdf with parameter Thus, the mean and variance are

    Question: How could you extend these ideas to obtain other interesting properties of the system?

Example 16: Nearest Neighbor with Right-Angle Travel Distance

If travel distance is right-angle, rather than Euclidean, the analysis in Example 15 follows straight through, except instead of a circle of radius r we have a square rotated at 45°, centered at (x, y), with area equal to 2r² . Following the same steps in the solution,

This is a Rayleigh pdf with parameter The mean and variance are

Question: In Example 4 in this chapter we derived that in an isotropic environment a responseunit traveling according to the right-angle distance metric travels 4/ = 1.273 times farther (on the average) than a unit traveling "as the crow flies." Thus, one might be tempted to think that the ratio of the mean right-angle to Euclidean distances computed in Examples 15 and 16 would be 1.273. In fact, the ratio is . Why?

Hint: See Problems 3.9 and 3.10.

Further work: Problems 3.25 and 3.26.