Question

The joint probability density function of a two-dimensional random variable \((X,Y)\) is \[ f(x,y)= \begin{cases} 2, & 0\\ 0, & \text{otherwise}. \end{cases} \] Which of the following is correct?

• A. \(X\) and \(Y\) are independent
• B. Marginal density function of \(X\) is \(x,\;0<x<1\)
• C. Marginal density function of \(Y\) is \((1-y),\;0<y<1\)
• D. \(X\) and \(Y\) are not independent

Hint:

A quick way to detect dependence is to observe the support region. If the support is triangular, circular, or bounded by one variable in terms of another, the variables are generally not independent.

Answer 1

The Correct Option is D

Concept: For a two-dimensional random variable, the marginal density functions are obtained by integrating the joint density function over the appropriate variable. Two random variables \(X\) and \(Y\) are independent if and only if \[ f(x,y)=f_X(x)f_Y(y) \] for all points in the support. Therefore, we first determine the marginal densities and then test the independence condition.

Step 1: Determine the marginal density function of \(X\).
Given \[ f(x,y)=2, \qquad 0<y<x<1. \] For a fixed value of \(x\), \[ 0<y<x. \] Hence, \[ f_X(x) = \int_0^x 2\,dy. \] \[ = 2[y]_0^x. \] \[ = 2x, \qquad 0<x<1. \] Thus, \[ f_X(x)=2x. \] Therefore option (B) is incorrect.

Step 2: Determine the marginal density function of \(Y\).
For a fixed value of \(y\), \[ y<x<1. \] Hence, \[ f_Y(y) = \int_y^1 2\,dx. \] \[ = 2[x]_y^1. \] \[ = 2(1-y), \qquad 0<y<1. \] Thus, \[ f_Y(y)=2(1-y). \] Hence option (C) is also incorrect.

Step 3: Check whether \(X\) and \(Y\) are independent.
If \(X\) and \(Y\) were independent, then \[ f(x,y)=f_X(x)f_Y(y). \] But \[ f_X(x)f_Y(y) = (2x)\bigl[2(1-y)\bigr] = 4x(1-y). \] This is not equal to \[ f(x,y)=2. \] Moreover, the support region \[ 0<y<x<1 \] is triangular and not rectangular, which itself indicates dependence. Hence, \[ X \text{ and } Y \] are not independent. \[ \boxed{\text{\(X\) and \(Y\) are not independent}} \]