Question

Let \(X\) be a random variable with \[ f(x) = \frac{1}{2} e^{-|x|}, \quad -\infty<x<\infty \] Then, which of the following statements is FALSE?

• A. \(E(X|X|) = 0\)
• B. \(E(X|X|^2) = 0\)
• C. \(E(|X|\sin(\frac{X}{|X|})) = 0\)
• D. \(E(|X|\sin^2(\frac{X}{|X|})) = 0\)

Hint:

For even pdfs, expectations of odd functions vanish, but expectations involving even transformations remain positive.

Answer 1

The Correct Option is D

Step 1: Note the symmetry of \(f(x)\).
The given pdf is even, \(f(x) = f(-x)\). Therefore, any odd function of \(X\) will have zero expectation.
Step 2: Check each expectation.
(A) \(E(X|X|)\): Function \(X|X|\) is odd ⇒ expectation = 0.
(B) \(E(X|X|^2)\): Function \(X^3\) is odd ⇒ expectation = 0.
(C) \(E(|X|\sin(\frac{X}{|X|}))\): Here \(\sin(\frac{X}{|X|})\) = \(\sin(1)\) for \(x>0\) and \(\sin(-1) = -\sin(1)\) for \(x<0\). Thus, overall function is odd ⇒ expectation = 0.
(D) \(E(|X|\sin^2(\frac{X}{|X|}))\): Since \(\sin^2(\frac{X}{|X|}) = \sin^2(1)\), which is constant and positive, \[ E(|X|\sin^2(1)) = \sin^2(1)E(|X|) = \sin^2(1) \neq 0 \] Hence, (D) is false. Final Answer: \[ \boxed{(D) \; E(|X|\sin^2(\frac{X}{|X|})) = 0 \text{ is false.}} \]