Problem 142 - An Introduction to Mathematical Statistics and Its Applications

Question

A point is chosen at random from the interior of a circle whose equation is \(x^{2}+y^{2} \leq 4\). Let the random variables \(X\) and \(Y\) denote the \(x\) - and \(y\)-coordinates of the sampled point. Find \(f_{X, Y}(x, y)\).

Step-by-Step Solution

Answer

The joint probability density function is \(f_{X, Y}(x, y) = 1/(4\pi)\) for \(x^{2}+y^{2} \leq 4\), and 0 otherwise.

Step 1: Identify the Circle's Parameters

The equation \(x^{2}+y^{2} \leq 4\) represents a circle centered at the origin (0,0) with radius \(r^2 = 4\), thus the radius \(r = 2\).

Step 2: Calculate the Area of the Circle

Use the formula for the area of a circle, which is \(\pi r^{2}\). With \(r=2\), the area is \(\pi * 2^{2} = 4\pi\). This value represents the total possible outcomes when picking a random point within the circle.

Step 3: Determine the Joint Probability Density Function

The joint probability density function \(f_{X,Y}(x,y)\) represents the probability that the point chosen falls in an infinitesimally small region around (x, y). Since the point is equally likely to be in each infinitesimal part of the area, we can use the definition of a uniform distribution and divide 1 by the total area. So, \(f_{X, Y}(x, y) = 1/(4\pi)\) for \(x^{2}+y^{2} \leq 4\) and 0 otherwise.

Key Concepts

Uniform Distribution

When we describe the concept of uniform distribution, we're referring to a situation where every outcome in a range is equally likely to occur. Imagine throwing a dart at a perfectly flat and smooth dartboard; the chance of the dart landing on any point on the dartboard is the same. This is the essence of a uniform distribution in the context of a two-dimensional space such as the dartboard, or, more relevant to our exercise, the area of a circle.

In our problem, since the point can be anywhere within the circle, and every point is equally likely to be chosen, this is an example of a two-dimensional uniform distribution. The joint probability density function for a uniform distribution over a region is always a constant value, as the likelihood of landing on any infinitesimal area within the region is the same. This is why when designing an exercise or solution, it's crucial that students understand the property that defines uniform distribution: every outcome in the specified range is equally probable.

Random Variables

Random variables are a fundamental concept in probability and statistics. They are variables whose values are determined by the outcomes of a random phenomenon. In our textbook exercise, the random variables are denoted as \(X\) and \(Y\), and they represent the \(x\)- and \(y\)-coordinates of a randomly chosen point inside the circle.

Thinking of \(X\) and \(Y\) as random variables means that we do not have one fixed value for them. Instead, they can take on any value within specific bounds determined by the properties of the circle. When you're guided through an exercise, it helps to imagine taking a handful of points and scattering them across the circle - each point represents a different potential outcome for our random variables \(X\) and \(Y\).

Understanding the role of random variables in joint probability density functions is critical. It allows us to quantify the probability of simultaneous occurrences: for instance, 'what is the likelihood that \(X\) is within a certain range while \(Y\) falls in another?' It's this sort of inquiry that random variables allow us to explore and quantify.

Area of a Circle

The concept of the area of a circle is pivotal in many geometric and probabilistic applications. The formula to calculate the area is \(A = \(pi\)r^2\), where \(r\) is the radius of the circle.

In our exercise, the circle with equation \(x^{2}+y^{2} \leq 4\) has a radius of 2 units. Applying the area formula gives us \(A = \(pi\) \cdot (2)^2 = 4\(pi\)\), which is the area in which our random variables \(X\) and \(Y\) can operate. This area is critical because it serves as the denominator in the calculation of the joint probability density function for a uniform distribution over a circular region.

Why is the area important?

Because it represents the total 'space' of all possible outcomes. When teaching students how to approach problems involving areas and probabilities, highlighting the area as the space for potential 'events' can make the abstract concept of probability more concrete and understandable.