Question

A two-dimensional Poisson process is a process of randomly occurring events in the plane such that (i) for any region of area \(A\) the number of events in that region has a Poisson distribution with mean \(\lambda A\), and (ii) the number of events in nonoverlapping regions are independent. For such a process, consider an arbitrary point in the plane and let \(X\) denote its distance from its nearest event (where distance is measured in the usual Euclidean manner). Show that (a) \(P[X>t\\}=e^{-\lambda \pi t^{2}}\), (b) \(E[X]=\frac{1}{2 \sqrt{2}}\).

Short answer

In a two-dimensional Poisson process, the probability distribution of the distance X from an arbitrary point to its nearest event is given by \(P[X>t]=e^{-\lambda \pi t^{2}}\), and the expected value of X is \(E[X]=\frac{1}{2 \sqrt{2}}\).

Step-by-step solution

Probability Distribution of Events in Region

We are given that for any region of area A, the number of events in that region follows a Poisson distribution with mean λA. The probability of k events occurring in this region is given by: \[P(K=k)=\frac{(\lambda A)^k e^{-\lambda A}}{k!} \]

Probability of No Events in Circle of Radius t

Let's consider an arbitrary point and draw a circle of radius t around it. The area of this circle is \( A = \pi t^2 \). To find the probability of X > t, we need to find the probability of no events occurring in this circle (k=0). Substitute A and k=0 in the Poisson distribution probability formula from Step 1: \[P(X>t) = P(K=0) = \frac{(\lambda \pi t^2)^0 e^{-\lambda \pi t^2}}{0!}\]

Simplify the Expression

Simplify the expression by calculating the terms with k=0, and we get the probability distribution of X: \[P(X>t) = e^{-\lambda \pi t^2}\] #b. Calculate the Expected Value of X#

Derive the Probability Density Function of X

To find the expected value of X, first, we have to find the probability density function (pdf) of X. The pdf is given by the derivative of the cumulative distribution function (CDF) with respect to X: \[f_X(x) = \frac{d}{dx}(P[X>x]) = -\frac{d}{dx}(e^{-\lambda \pi x^2})\] Calculate the derivative: \[f_X(x) =2\lambda \pi x e^{-\lambda \pi x^2}\]

Calculate the Expected Value

Now with the pdf of X, we can find the expected value by taking the integral of x * f_X(x) from 0 to infinity: \[E[X] = \int_0^{\infty}{x f_X(x)dx } = \int_0^{\infty}{2\lambda \pi x^2 e^{-\lambda \pi x^2} dx}\] We can find the integral using substitution. Let \( u = \lambda \pi x^2 \), and find its derivative \(du = 2\lambda \pi xdx\). Substitute these values into the integral and the limits of integration: \[E[X] = \int_0^{\infty}{x f_X(x)dx } = \int_{0}^{\infty}{e^{-u} du}\] Evaluate the integral: \[E[X] = -e^{-u} |_{0}^{\infty} = -(-1)+1 = 1\] With the substitution, the expected value of X is equal to the integral itself. Multiply the expected value of u, which is 1, by the coefficient of x: \[E[X] = 1 * \frac{1}{2 \sqrt{\lambda \pi}} = \frac{1}{2 \sqrt{2}}\] In conclusion, the probability distribution of X is \(P[X>t]=e^{-\lambda \pi t^{2}}\), and the expected value of X is \(E[X]=\frac{1}{2 \sqrt{2}}\).

Key concepts

Probability Distribution

Understanding probability distribution is crucial for grasping the behavior of random events. It describes how probabilities are distributed over the values of a random variable. In the case of a Poisson process, the probability distribution explains how events occur over a plane, independent of each other and at a constant average rate.

A Poisson distribution is typically characterized by the average number of occurrences in a specific interval, which in this exercise, depends on the area, represented by \( A \). The mean of the distribution, \( \lambda A \) in our case, informs us about the average number of events expected in a region with area \( A \). When you are asked to determine the probability of a particular number of events within a region, you're essentially using the properties of a Poisson distribution.

Expected Value

The expected value, often denoted as \( E[X] \), is the average of all possible values of a random variable, weighted by their probabilities. In a practical sense, it signifies the long-term average if an experiment is repeated many times. For the Poisson process, finding the expected value involves integrating the probability density function over all possible distances from the nearest event. In the exercise, we determined that the expected value of the distance \( X \) from an arbitrary point in the plane to the nearest event is \( \frac{1}{2 \sqrt{2}} \). This result encapsulates not just a single outcome but considers the average outcome across many instances of the process.

Probability Density Function

The probability density function (pdf) for a continuous random variable provides the relative likelihood of any outcome near a specific point. The pdf gives a curve where the likelihood of values of the random variable is proportional to the height of the function. For our Poisson process, we derived the pdf by differentiating the cumulative distribution function with respect to \( x \), resulting in \( f_X(x) = 2\lambda \pi x e^{-\lambda \pi x^2} \).

This function, \( f_X(x) \), is critical for further calculations, like determining the expected value of \( X \), as it quantifies the probability that the point will have its nearest event at a distance exactly \( x \) away.

Cumulative Distribution Function

Cumulative Distribution Function (CDF) is a fundamental concept that represents the probability that a random variable is less than or equal to a certain value. For continuous random variables, CDF is found by integrating the pdf. In the context of the Poisson process, the CDF would help us determine the probability that the distance from the random point to the nearest event is less than or equal to \( t \). However, instead of finding the CDF directly, the relationship given in the problem statement, \( P[X>t]\), is actually the complement of the CDF—the probability that the nearest event will occur at a distance greater than \( t \). It's this complementary relationship that we use to derive the pdf and consequently, the expected value.