Question

Let \( X \) be a random variable having uniform distribution on \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \). Then which one of the following statements is NOT true?

• A. \( Y = \cot X \) follows standard Cauchy distribution
• B. \( Y = \tan X \) follows standard Cauchy distribution
• C. \( Y = -\log_e \left( \frac{1}{2} + \frac{X}{\pi} \right) \) has moment generating function \( M(t) = \frac{1}{1 - t}, \, t<1 \)
• D. \( Y = -2 \log_e \left( \frac{1}{2} + \frac{X}{\pi} \right) \) follows central chi-square distribution with one degree of freedom

Hint:

When working with transformations of uniform distributions, use the moment generating function or known distribution properties to identify the resulting distribution.

Answer 1

The Correct Option is D

We are given a uniform distribution for \( X \) on the interval \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \). Step 1: Check each option.
- Option (A) is correct because \( \cot X \) for a uniformly distributed variable on \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \) follows a standard Cauchy distribution.
- Option (B) is correct because \( \tan X \) also follows a standard Cauchy distribution.
- Option (C) is correct because the transformation \( Y = -\log_e \left( \frac{1}{2} + \frac{X}{\pi} \right) \) gives a moment generating function as described.
- Option (D) is incorrect because \( Y = -2 \log_e \left( \frac{1}{2} + \frac{X}{\pi} \right) \) does not follow a central chi-square distribution with one degree of freedom. Final Answer: \[ \boxed{Y = -2 \log_e \left( \frac{1}{2} + \frac{X}{\pi} \right) \, \text{does not follow central chi-square distribution}}. \]