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Chapter 11: Problem 26

Let \(R\) denote a region in the two-dimensional plane. Show that for a twodimensional Poisson process, given that there are \(n\) points located in \(R\), the points are independently and uniformly distributed in \(R\) - that is, their density is \(f(x, y)=c,(x, y) \in R\) where \(c\) is the inverse of the area of \(R\).

Short Answer

Expert verified
Given a two-dimensional Poisson process with \(n\) points in region \(R\), we found the joint probability density function as \(f_\mathbf{X}(x) = \left(\frac{1}{A(R)}\right)^n \) for \(x \in R^n\). This confirms that the points are independently and uniformly distributed in \(R\) with density function \(f(x, y) = \frac{1}{A(R)}\) for \((x, y) \in R\).

Step by step solution

01

Understanding Poisson process assumptions

A key assumption of a Poisson process is that the events (points in this case) occur independently of each other. Another assumption is that the probability of an event occuring in an infinitesimally small region is proportional to the size of that region. We will use these assumptions as the basis for our proof.
02

Define probability variable

Let \(X_i=(X_{i1}, X_{i2})\) be the random variable representing the Cartesian coordinates of the \(i\)-th point, with \(i \in \{1, 2, ... , n\}\).
03

Determine the joint probability density function

For the Poisson process, we want to find the joint probability density function of the points within the region \(R\). Since the points are independently distributed, the joint probability density function \(f_\mathbf{X}(x)\) is given as the product of the marginal probability density functions of the individual points: \[f_\mathbf{X}(x)=\prod_{i=1}^{n} f_{X_i}(x_i),\] where \(x=(x_1, x_2, ... , x_n)\) and \(f_{X_i}(x_i)\) is the probability density function of the coordinates \((x_{i1}, x_{i2})\) of the \(i\)-th point.
04

Individual point probability density functions

Since the Poisson process assumption states that the probability of an event occurring in an infinitesimally small region is proportional to the size of that region, we can write the individual point probability density functions as: \[f_{X_i}(x_i)=c \quad (x_i \in R),\] where \(c\) is a constant.
05

Substitute individual point probability density functions in joint probability density function

Now we can substitute the individual point probability density functions back into the joint probability density function: \[f_\mathbf{X}(x)=c^n \quad (x \in R^n).\]
06

Calculating the value of c

Since \(R^n\) denotes product of region \(R\) repeated \(n\) times, the area of \(R^n\) is equal to the area of the original region \(R\) raised to the power of \(n\). Therefore, the area of \(R^n\) is given by: \[A(R^n) = [A(R)]^n.\] We can normalize the joint probability density function by dividing by the area of \(R^n\): \[\int_{R^n} f_\mathbf{X}(x) dx = \int_{R^n} c^n dx = c^n A(R^n).\] Since the joint probability density function should integrate to 1: \[c^n A(R^n)=1 \quad \Rightarrow c= \frac{1}{A(R)}.\]
07

Final result

Finally, we can plug the value of \(c\) back into the joint probability density function: \[f_\mathbf{X}(x)=\left(\frac{1}{A(R)}\right)^n \quad (x \in R^n),\] which confirms that the points are independently and uniformly distributed in \(R\) with density function \(f(x, y)=\frac{1}{A(R)}\) for \((x, y) \in R\).

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