To determine the nature of the set \( S \) where the joint probability density function (PDF) of independent variables \( X \) and \( Y \) is uniform, we will consider the properties of independent random variables and their joint distributions.
Independence between \( X \) and \( Y \) implies that the joint PDF of \( X \) and \( Y \), denoted \( f_{X,Y}(x,y) \), can be expressed as the product of their marginal PDFs:
\[ f_{X,Y}(x,y) = f_X(x) f_Y(y) \]
If \( f_{X,Y}(x,y) \) is uniform over the set \( S \), it must integrate to 1 over that set. Since each marginal density must also integrate to 1, this implies that \( S \) must inherently allow for the formulation of independent marginals.
Now, let's analyze the given options:
A. Must be a square.
- This is not necessarily true. The region \( S \) does not have to be a square; it could have other shapes (e.g., a rectangle).
B. Must be a set of the form {(x, y) | x ∈ A, y ∈ B} (known as the Cartesian product of two sets A and B).
- This is true. Since \( X \) and \( Y \) are independent, the joint distribution can be represented as a Cartesian product of their support intervals. Thus, \( S \) can be expressed as \( A \times B \), where \( A \) is the support of \( X \) and \( B \) is the support of \( Y \).
C. Can be any set.
- This is false if we consider the fact that the joint PDF must be defined in a way that reflects the independence of \( X \) and \( Y \). While \( S \) can technically be any set, for \( X \) and \( Y \) to be independent and uniform in the context described, \( S \) must have a specific structure that conforms to the properties outlined in option B.
Given this analysis, the correct choice is:
B. Must be a set of the form {(x, y) | x ∈ A, y ∈ B} (known as the Cartesian product of two sets A and B).