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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.

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/Ï€ inside the circle) and zero outside. This makes calculating probabilities and finding marginal densities more straightforward.

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.

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.