91Ó°ÊÓ

An ecologist wishes to select a point inside a circular sampling region according to a uniform distribution (in practice this could be done by first selecting a direction and then a distance from the center in that direction). Let \(X=\) the \(x\) coordinate of the point selected and \(Y=\) the \(y\) coordinate of the point selected. If the circle is centered at \((0,0)\) and has radius \(R\), then the joint pdf of \(X\) and \(Y\) is $$ f(x, y)=\left\\{\begin{array}{cl} \frac{1}{\pi R^{2}} & x^{2}+y^{2} \leq R^{2} \\ 0 & \text { otherwise } \end{array}\right. $$ a. What is the probability that the selected point is within \(R / 2\) of the center of the circular region? [Hint: Draw a picture of the region of positive density D. Because \(f(x, y)\) is constant on \(D\), computing a probability reduces to computing an area.] b. What is the probability that both \(X\) and \(Y\) differ from 0 by at most \(R / 2\) ? c. Answer part (b) for \(R / \sqrt{2}\) replacing \(R / 2\). d. What is the marginal pdf of \(X\) ? Of \(Y\) ? Are \(X\) and \(Y\) independent?

Step by step solution

01

Understanding the problem

We are given a joint probability density function (pdf) for a uniform distribution within a circle of radius \( R \), centered at the origin \((0,0)\). Our task is to find specific probabilities and verify the independence of the coordinates \(X\) and \(Y\).

02

Analyzing Part (a)

We need to find the probability that the selected point lies within a circle of radius \( R/2 \). The area of this circle is \( \pi (R/2)^2= \pi R^2/4 \). Since the probability is the ratio of the area of the smaller circle to the larger one, the probability is \( \frac{\pi R^2/4}{\pi R^2} = \frac{1}{4} \).

03

Analyzing Part (b)

For part (b), we consider the square with vertices \((-R/2, -R/2)\), \((R/2, -R/2)\), \((R/2, R/2)\), \((-R/2, R/2)\). The area of this square is \((R/2)\times(R/2) = R^2/4\). So, the probability is given by \( \frac{\text{Area of Square}}{\text{Area of Circle}} = \frac{R^2/4}{\pi R^2} = \frac{1}{4\pi} \).

04

Analyzing Part (c)

Now, replace \(R/2\) with \(R/\sqrt{2}\) in part (b). The side length of the square becomes \(R/\sqrt{2}\) and its area will be \[ \left(\frac{R}{\sqrt{2}}\right)^2 = \frac{R^2}{2} \]. The probability is \( \frac{R^2/2}{\pi R^2} = \frac{1}{2\pi} \).

05

Analyzing Part (d)

For the marginal pdfs, integrate \(f(x,y)\) over the other variable. The marginal pdf of \(X\) is \( f_X(x) = \int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} \frac{1}{\pi R^2} \, dy = \frac{2 \sqrt{R^2 - x^2}}{\pi R^2} \). Similarly, \( f_Y(y) = \frac{2 \sqrt{R^2 - y^2}}{\pi R^2} \). Since the marginals depend on each other, \(X\) and \(Y\) are not independent.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Uniform Distribution

Uniform distribution is a probability distribution where all outcomes are equally likely within a defined range. This means that every point within a region has the same chance of being selected.
In terms of selecting a point within a circular region, as described in the problem, a uniform distribution implies that no particular direction or distance from the center is preferred. This ensures fairness in the selection process.
The formula for joint probability density function (pdf) we are using, \[ f(x, y)=\begin{cases} \frac{1}{\pi R^{2}} & x^{2}+y^{2} \leq R^{2} \ 0 & \text{otherwise} \end{cases} \] ensures that every segment of the circle has equal probability, providing a uniform chance across the entire area. This characteristic is crucial when analyzing random processes over continuous sample spaces like our circular region.

Circular Region

A circular region, in mathematical terms, is the area enclosed by a circle. In our specific context, it is defined as a circle of radius \( R \) centered at the origin \((0,0)\).
This region represents the entire space where the points can be selected. When we say the distribution is uniform over this circular region, it means it covers every spot within the circle equally.
A critical aspect is determining the area of various sections within this circle, which is key to finding probabilities for subregions, as seen when calculating the probability that a point falls within a circle of smaller radius, such as \( R/2 \). This smaller circle's probability is found by comparing its area, \( \pi (R/2)^2 \), to the area of the entire circular region, \( \pi R^2 \). These formulae demonstrate how circular regions are approached in problems involving uniform distributions.

Marginal Probability Density Function

The marginal probability density function (pdf) describes the probability distribution of a subset of variables within a joint distribution. In our case, the joint pdf is given for \((X, Y)\) within a circle.
To find out how each coordinate alone behaves, we look at their marginal pdfs. This involves integrating out the other variable in the joint pdf. For the \(x\) coordinate, this is accomplished by integrating over \(y\), resulting in: \[ f_X(x) = \int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} \frac{1}{\pi R^2} \, dy = \frac{2 \sqrt{R^2 - x^2}}{\pi R^2}. \] Similarly for \(y\), the integral over \(x\) gives \[ f_Y(y) = \frac{2 \sqrt{R^2 - y^2}}{\pi R^2}. \] These marginal pdfs give us insights into the distribution of points along each axis, showing how values are spread independently within the circular region.

Coordinate Independence

Coordinate independence refers to whether the coordinates \(X\) and \(Y\) are statistically independent. Independence implies that knowing the value of one coordinate does not provide any information about the other.
In the problem's context, we analyze whether the marginal distributions of \(X\) and \(Y\) can exist independently. However, due to the uniform circular distribution, the marginal pdfs we derived are dependent on each other. This is evident because both functions, \( f_X(x) \) and \( f_Y(y) \), include the term \( \sqrt{R^2 - x^2} \) or \( \sqrt{R^2 - y^2} \), linking them to the \( R \) constraint.
Thus, knowing one coordinate constrains the possible values of the other, indicating that \(X\) and \(Y\) are not independent. This dependency is intrinsic to uniformly distributed points within a circular region and highlights an essential characteristic of two-dimensional random variables.