91Ó°ÊÓ

. An ecologist selects a point inside a circular sampling region according to a uniform distribution. 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 Circular Region

The given circle has a radius \(R\) and is centered at \((0,0)\). The uniform distribution inside the circle means that each point has the same likelihood, which is reflected in the constant density function \(f(x, y)=\frac{1}{\pi R^{2}}\) for any point \((x, y)\) within the circle, i.e., \(x^2 + y^2 \leq R^2\).

02

Finding Probability for a Circle of Radius \(R/2\)

The region within \(R/2\) is a smaller circle centered at \(0,0\). The area of this smaller circle is \(\pi\left(\frac{R}{2}\right)^2 = \frac{\pi R^2}{4}\). Since the density is uniform, the probability is the ratio of this area to the area of the larger circle. Hence, the probability is \(\frac{\pi R^2/4}{\pi R^2} = \frac{1}{4}\).

03

Probability of \(X\) and \(Y\) within \(R/2\) from Zero

The range \(-R/2 \leq X \leq R/2\) and \(-R/2 \leq Y \leq R/2\) form a square with area \(R^2/4\). This square is entirely within the circle. The probability is the ratio of the square's area to the circle's area: \(\frac{R^2/4}{\pi R^2}\). Calculating this gives \(\frac{1}{4\pi}\).

04

Probability for \(X\) and \(Y\) within \(R/\sqrt{2}\)

Now consider \(-R/\sqrt{2} \leq X \leq R/\sqrt{2}\) and \(-R/\sqrt{2} \leq Y \leq R/\sqrt{2}\). This square has area \(R^2/2\). The probability is \(\frac{R^2/2}{\pi R^2} = \frac{1}{2\pi}\).

05

Finding Marginal Densities and Independence

To find the marginal pdf of \(X\), integrate the joint pdf over \(y\) from \(-\sqrt{R^2-x^2}\) to \(\sqrt{R^2-x^2}\). This results in \(f_X(x) = \frac{2\sqrt{R^2 - x^2}}{\pi R^2}\). Similarly, for \(Y\), the marginal pdf is the same: \(f_Y(y) = \frac{2\sqrt{R^2 - y^2}}{\pi R^2}\). Since the product of marginal densities doesn't equal the joint pdf, \(X\) 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

In this exercise, the ecologist is tasked with selecting a point inside a circular region that is distributed uniformly. What this means is that every point within the circle has an equal likelihood of being chosen.
This is reflected in the uniform joint probability density function (pdf) for random variables \(X\) and \(Y\), which gives us:

• \( f(x, y) = \frac{1}{\pi R^2} \) over the area where \( x^2 + y^2 \leq R^2 \)
• \( f(x, y) = 0 \) otherwise

This constant pdf illustrates the fundamental property of a uniform distribution: no point is preferred over another within the specified region.

Circular Sampling Region

The circular sampling region discussed in this exercise is a defined space where random sampling occurs. The circle is centered at \((0,0)\) with a radius \(R\). Understanding this region is crucial as it forms the basis of calculating probabilities.
Imagine the circle as a dartboard, where the dart represents a randomly selected point. Any point that fulfills the condition \( x^2 + y^2 \leq R^2 \) is considered inside the circle, contributing to the positive density.
The exercise further discusses probabilities within smaller circles. For example, the probability of a point lying within a smaller circle of radius \(R/2\) involves finding the area of this smaller circle and dividing it by the area of the larger circle. Here, the smaller circle has an area of \( \frac{\pi R^2}{4} \), leading to a simple probability calculation.

Marginal Probability Density Function

To understand specific characteristics of the random variables \(X\) and \(Y\), we calculate their marginal probability density functions (pdfs). The marginal pdf is found by integrating over all possible values of the other variable.
For \(X\), the marginal pdf \(f_X(x)\) is obtained by integrating the joint pdf over \(y\):

• \[ f_X(x) = \int_{\-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} \frac{1}{\pi R^2} \, dy = \frac{2\sqrt{R^2 - x^2}}{\pi R^2} \]

This expression signifies the distribution of \(X\) when all potential \(Y\) values are considered.
Similarly, the marginal pdf for \(Y\) is \(f_Y(y) = \frac{2\sqrt{R^2 - y^2}}{\pi R^2}\). Both variables have similar forms due to the symmetric nature of the circle.

Variable Independence

Variable independence is a critical concept that tells us whether two random variables influence each other. In this exercise, we determine if \(X\) and \(Y\) are independent.
For independence, the joint pdf should equal the product of individual marginal pdfs. In this case:

• The joint pdf is \( \frac{1}{\pi R^2} \)
• The product of the marginal pdfs does not equate to the joint pdf

Thus, we conclude that \(X\) and \(Y\) are not independent. Their dependence stems from the constraint \( x^2 + y^2 \leq R^2 \), suggesting that knowing one variable impacts the distribution of the other.