Question

Let \( X \) and \( Y \) be two independent standard normal random variables. Then the probability density function of \( Z = \frac{|X|}{|Y|} \) is

• A. \(f(z) = \begin{cases} \dfrac{\sqrt{1/2}}{\sqrt{\pi}}\, e^{-z/2}\, z^{-1/2}, & z > 0, \\ 0, & \text{otherwise}. \end{cases} \)
• B. \( f(z) = \begin{cases} \dfrac{2}{\sqrt{2\pi}} e^{-z^2/2}, & z > 0, \\ 0, & \text{otherwise}. \end{cases} \)
• C. \( f(z) = \begin{cases} e^{-z}, z > 0, \\ 0, & \text{otherwise}. \end{cases} \)
• D. \( f(z) = \begin{cases} \dfrac{2}{\pi}\dfrac{1}{(1 + z^2)}, z > 0, \\ 0, & \text{otherwise}. \end{cases} \)

Hint:

The ratio of two independent standard normal variables follows the Cauchy distribution.

Answer 1

The Correct Option is D

Step 1: Recognize the form of the distribution. 
The variable \( Z = \frac{|X|}{|Y|} \) is a ratio of two independent standard normal random variables. The distribution of the ratio of two independent normal variables follows the Cauchy distribution.

Step 2: Cauchy distribution density. 
The probability density function of a standard Cauchy distribution is: \( f(z) = \begin{cases} \frac{1}{\pi(1 + z^2)}, z > 0, \\ 0, & \text{otherwise}. \end{cases} \)

Step 3: Conclusion. 
The correct answer is (D) \( f(z) = \begin{cases} \frac{2}{\pi(1 + z^2)}, z > 0, \\ 0, & \text{otherwise}. \end{cases} \)