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 $$ f(x, y)=\left\\{\begin{array}{cc} \frac{1}{\pi R^{2}} & x^{2}+y^{2} \leq R^{2} \\ 0 & \text { otherwise } \end{array}\right. $$ 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

Understand the Problem

An ecologist is trying to select a random point within a circle of radius \(R\) centered at the origin \((0,0)\). The uniform distribution implies that every point has an equal probability of being chosen within the circle. Given the joint pdf \(f(x, y)\), which equals \(\frac{1}{\pi R^2}\) inside the circle and \(0\) outside, we want to find certain probabilities based on different constraints.

02

Part a: Calculate Probability within $R/2$

Draw a picture of the circle with a smaller circle of radius \(R/2\) inside it also centered at the origin. The probability is the area of this smaller circle divided by the area of the larger circle. The area of the circle with radius \(R/2\) is \(\pi (R/2)^2 = \frac{\pi R^2}{4}\). The area of the larger circle is \(\pi R^2\). Thus, the probability is \(\frac{\text{Area of smaller circle}}{\text{Area of larger circle}} = \frac{\frac{\pi R^2}{4}}{\pi R^2} = \frac{1}{4}\). Thus, the probability that the point is within \(R/2\) of the center is \(0.25\).

03

Part b: Probability within Square Region $R/2$

Consider the square region where \(X\) and \(Y\) both range from \(-R/2\) to \(R/2\). This forms a square with side length \(R\) inside the circle. Since the square is completely inside the circle of radius \(R\), the probability is simply the area of the square over the area of the circle. The area of the square is \((R/2)^2 \times 4 = R^2\). The probability is then the area of the square divided by the area of the circle: \(\frac{R^2}{\pi R^2} = \frac{1}{\pi}\).

04

Part c: Adjusted Region for $R/\sqrt{2}$

If \(X\) and \(Y\) differ from 0 by at most \(R/\sqrt{2}\), the square formed will have side lengths of \(R\sqrt{2}\). However, this square's area is \(2R^2\), which exceeds the circle's area, so the maximum square that fits within the circle is when the square's diagonal equals \(R\), i.e., the corners touching the circle - a square of side \(R/\sqrt{2}\). Using the part a logic since it covers the whole area within the circle, we integrate the squared region by adjusting limits. The region now will form the maximum possible area inside circle, hence considering full intersection, probability covers entire inside circle and is the circle's area of intersection, still calculated directly as for the maximum value as \(1\) in ideal uniform.

05

Part d: Find Marginal PDFs and Independence

To find the marginal pdf of \(X\), integrate the joint density function over \(y\): \(f_X(x) = \int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} \frac{1}{\pi R^2} \, dy = \frac{2\sqrt{R^2-x^2}}{\pi R^2}\).Similarly, for \(Y\): \(f_Y(y) = \int_{-\sqrt{R^2-y^2}}^{\sqrt{R^2-y^2}} \frac{1}{\pi R^2} \, dx = \frac{2\sqrt{R^2-y^2}}{\pi R^2}\).Since the marginal distributions depend on each other, \(X\) and \(Y\) are not independent.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Uniform Distribution

In probability theory, a uniform distribution means that every outcome in a particular range is equally likely. For this exercise, we're dealing with a circular region where every point within the circle is equally likely to be chosen.

This concept is represented by the joint probability density function (pdf) for the coordinates of a point. If the circle is centered at \( (0,0) \) and has a given radius \( R \), then the joint pdf for selecting a point \( (X, Y) \) within the circle is uniform. It's denoted as \( f(x, y) = \frac{1}{\pi R^{2}} \) when \( x^{2}+y^{2} \leq R^{2} \) and zero otherwise.

• Uniformity ensures each section of the circle is as likely to contain the selected point as any other section of the same size.
• This simplifies calculations involving probable regions within the circle.
• The uniform distribution ultimately depends on the shape of the distribution region (here, circular).

Marginal Probability Density Function

To understand the distribution of just one variable, like \( X \) or \( Y \), we use marginal probability density functions. These involve summing or integrating out the other variable.

For a circle of radius \( R \), the marginal pdf of \( X \) is found by integrating the joint pdf over all values of \( Y \). For example:

• The marginal pdf of \( X \) is given by: \( f_X(x) = \int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} \frac{1}{\pi R^2} \, dy \)
• Similarly for \( Y \): \( f_Y(y) = \int_{-\sqrt{R^2-y^2}}^{\sqrt{R^2-y^2}} \frac{1}{\pi R^2} \, dx \)
• These help in understanding the behavior of each variable separately, still under the influence of being inside the circular region.

Independence of Random Variables

In probability, two random variables are independent if the occurrence of one does not affect the probability of the other.

For a circular region sampled uniformly, \( X \) and \( Y \) are considered:

• The marginal pdfs of \( X \) and \( Y \) depend on circular boundary constraints, meaning they are linked by the circle's geometry.
• Given \( f_X(x) \) and \( f_Y(y) \, \) are influenced by their cross-dependencies through the circle equation \( x^2 + y^2 \leq R^2 \), they are not independent.
• If \( X \) were independent of \( Y \), the joint pdf could be expressed as the product of the marginal pdfs, which is not the case here.

Circular Regions in Probability

In this context, a circular region describes the set of all possible points within a circle of a given radius \( R \). This concept is crucial because it defines the boundary of uniform probability spread.

Here are key aspects:

• The main characteristic is the boundary: \( x^2 + y^2 \leq R^2 \, \) delineating the region where probabilities are uniform.
• Any smaller circular region (like a circle with radius \( R/2 \) inside the full circle) provides a way to calculate probabilities by comparing areas.
• Circles are a natural way to represent regions of constant probability in two dimensions due to their symmetry and simplicity.

Considering circular regions allows one to derive probabilities by leveraging area ratios, keeping computations straightforward when dealing with uniform distributions.