Question

Let \( X \) and \( Y \) be continuous random variables with probability density functions \( P_X(x) \) and \( P_Y(y) \), respectively. Further, let \( Y = X^2 \) and \[ P_X(x) = \begin{cases} 1, & x \in (0,1] \\ 0, & \text{otherwise} \end{cases} \] Which one of the following options is correct?

• A. \( P_Y(y) = \begin{cases} \frac{1}{2\sqrt{y}}, & y \in (0,1] \\ 0, & \text{otherwise} \end{cases} \)
• B. \( P_Y(y) = \begin{cases} 1, & y \in (0,1] \\ 0, & \text{otherwise} \end{cases} \)
• C. \( P_Y(y) = \begin{cases} 1.5 \sqrt{y}, & y \in (0,1] \\ 0, & \text{otherwise} \end{cases} \)
• D. \( P_Y(y) = \begin{cases} 2y, & y \in (0,1] \\ 0, & \text{otherwise} \end{cases} \)

Hint:

When performing a variable transformation in probability, use the formula \( P_Y(y) = P_X(x) \cdot \left| \frac{dx}{dy} \right| \) where \( x = g^{-1}(y) \). Make sure to adjust the limits of support accordingly.

Answer 1

The Correct Option is A

We are given: \[ X \sim \text{Uniform}(0,1), \quad \text{and } Y = X^2 \] To find the probability density function \( P_Y(y) \), we use the transformation of variables method. Since the transformation is \( Y = g(X) = X^2 \), and \( X \in (0,1] \), this implies \( Y \in (0,1] \). We invert the transformation: \[ X = \sqrt{Y}, \quad \text{(only positive root since \( X > 0 \))} \] Then the transformed PDF is: \[ P_Y(y) = P_X(x) \cdot \left| \frac{dx}{dy} \right| = 1 \cdot \left| \frac{d}{dy} \sqrt{y} \right| = \frac{1}{2\sqrt{y}}, \quad y \in (0,1] \]