Question

Two continuous random variables \(\mathrm{X}\) and \(\mathrm{Y}\) may also be jointly distributed. Suppose \((X, Y)\) has a distribution which is uniform over a unit circle centered at \((0,0)\). Find the joint density of \((\mathrm{X}, \mathrm{Y})\) and the marginal densities of \(\mathrm{X}\) and \(\mathrm{Y}\). Are \(\mathrm{X}\) and \(\mathrm{Y}\) independent?

Short answer

The joint density of (X,Y) is given by $f_{X,Y}(x,y) = \frac{1}{\pi}$ if $x^2 + y^2 \leq 1$ and $0$ otherwise. The marginal densities are $f_X(x) = \frac{2}{\pi}\sqrt{1-x^2}$ and $f_Y(y) = \frac{2}{\pi}\sqrt{1-y^2}$. Since $f_{X,Y}(x, y) \neq f_{X}(x) \cdot f_Y(y)$, the random variables X and Y are not independent.

Step-by-step solution

Find the joint density of (X, Y)

Since the distribution is uniform over a unit circle centered at (0,0), the joint probability density function (pdf) will have a constant value inside the circle and zero outside the circle. The area of the unit circle is given by A = π. To find the constant value, let c be the value of the joint density function inside the circle. Then, the total probability should be equal to 1. \[\iint_{circle} c \, dx \, dy = 1\] Since the area A = π, we can rewrite the integral equation in terms of the radius: \[c \cdot A = c \cdot \pi = 1\] Solving for c, we get: \[c = \frac{1}{\pi}\] So the joint density function is: \[f_{X,Y}(x,y) = \begin{cases} \frac{1}{\pi} & \text{if} \: x^2 + y^2 ≤ 1 \\ 0 & \text{otherwise} \end{cases}\]

Find the marginal density of X and Y

To find the marginal densities of X and Y, we need to integrate the joint density function over each of the random variables. For X: \[f_X(x) = \int_{-1}^{1} f_{X,Y}(x, y) dy\] Since the joint density is defined inside the circle, the limits of integration for y will be from −√(1−x^2) to √(1−x^2): \[f_X(x) = \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \frac{1}{\pi} dy = \frac{1}{\pi} \left[ y \right]^{\sqrt{1-x^2}}_{-\sqrt{1-x^2}} = \frac{2}{\pi}\sqrt{1-x^2}\] Similarly, for Y: \[f_Y(y) = \int_{-1}^{1} f_{X,Y}(x, y) dx = \frac{2}{\pi}\sqrt{1-y^2}\]

Check if X and Y are independent

To check if X and Y are independent, their joint density must be equal to the product of their marginal densities: \[f_{X,Y}(x, y) = f_{X}(x) \cdot f_Y(y)\] We already know the joint density function: \[f_{X,Y}(x,y) = \begin{cases} \frac{1}{\pi} & \text{if} \: x^2 + y^2 ≤ 1 \\ 0 & \text{otherwise} \end{cases}\] And the product of the marginal densities is: \[f_X(x) \cdot f_Y(y) = \frac{2}{\pi}\sqrt{1-x^2} \cdot \frac{2}{\pi}\sqrt{1-y^2} = \frac{4}{\pi^2}\sqrt{1-x^2}\sqrt{1-y^2}\] Comparing both functions, we see that they are not equal; the joint density function has an additional condition on x and y. Hence, X and Y are not independent.

Key concepts

Continuous Random Variables

Continuous random variables are variables that can take any value within a given range. Unlike discrete random variables, which have specific distinct values, continuous random variables have an infinite number of possible values within a specific interval.
In statistical terms, the probabilities for continuous random variables are hence described by a probability density function (pdf), rather than a probability mass function (pmf). The pdf describes the likelihood that the random variable takes on any given value or falls within a specific range of values. For our exercise, the continuous random variables in question are X and Y, which can represent any point within the unit circle centered at (0,0).
Continuous random variables like X and Y are commonly used to model real-world processes where measuring distinct values is impractical or impossible, such as temperature, time, or in this case, spatial location. Depicting them with pdf allows us to understand their behavior over a continuous domain.

Uniform Distribution

A uniform distribution is one where all outcomes are equally likely within a certain range. This type of distribution is straightforward as it maintains a constant probability over its defined interval, making it easy to comprehend.
In the exercise, X and Y are jointly distributed uniformly over a unit circle. This means that any point within the circle is just as likely to be chosen as any other point. Unlike the typical uniform distribution on a line interval where the same constant value is maintained across a straightforward range, here it applies to a circular region.
The concept of uniformity in this situation simplifies understanding the joint density function, as it results in a constant value within the circular boundary (the function is 1/π inside the circle) and zero outside. This makes calculating probabilities and finding marginal densities more straightforward.

Marginal Density Function

The marginal density function allows us to determine the probability distribution of one of the random variables from a joint distribution. It effectively summarizes the possible outcomes of one variable, disregarding the influence of the other.
To find the marginal distribution in this problem, we integrate out the other variable from the joint density function. For example, to find the marginal density of X, we integrate the joint pdf over all possible values of Y. Similarly, for Y, we integrate over all X values.
This process gives us functions that describe the probability density for X and Y independently:

• For X: \[ f_X(x) = \frac{2}{\pi}\sqrt{1-x^2} \]
• For Y: \[ f_Y(y) = \frac{2}{\pi}\sqrt{1-y^2} \]

The marginal density functions describe how the probability distribution looks from the perspective of each variable on its own, offering insights into their individual behavior apart from each other.

Independence of Random Variables

Two random variables are considered independent if the occurrence of one does not affect the probability distribution of the other. In statistical terms, X and Y are independent if their joint pdf is the product of their individual marginal pdfs.
In this exercise, the independence test requires comparing the joint density function to the product of the marginal density functions.
Upon comparison, we find:

• The joint density function inside the unit circle is\[ f_{X,Y}(x,y) = \frac{1}{\pi} \]
• The product of the marginal density functions is\[ \frac{4}{\pi^2}\sqrt{1-x^2}\sqrt{1-y^2} \]

Since these two expressions are not equal, X and Y do not satisfy the condition of independence. Instead, they show a dependency within the context of their distribution, as the joint distribution imposes an additional condition on the relationship between X and Y.