Question

Which of the following functions is not a p.d.f. of a continuous random variable \( X \)? \[ F_1 \text{ given by} \quad f(x) = e^{-x} \quad \text{if} \quad x \geq 0, \quad f(x) = 0 \quad \text{otherwise}, \] \[ F_2 \text{ given by} \quad f(x) = \frac{1}{4} x^{-1/2} \quad \text{if} \quad 0 \leq x \leq 4, \quad f(x) = 0 \quad \text{otherwise}, \] \[ F_3 \text{ given by} \quad f(x) = 6x(1 - x) \quad \text{if} \quad 0 \leq x \leq 1, \quad f(x) = 0 \quad \text{otherwise}, \] \[ F_4 \text{ given by} \quad f(x) = \frac{x}{2} \quad \text{if} \quad -2 \leq x \leq 2, \quad f(x) = 0 \quad \text{otherwise}, \]

• A. \( F_3 \)
• B. \( F_4 \)
• C. \( F_1 \)
• D. \( F_2 \)

Hint:

To check if a function is a valid p.d.f., ensure that the integral over its entire range equals 1, and that the function is non-negative for all \( x \).

Answer 1

The Correct Option is B

Step 1: Check for valid p.d.f.
For a function to be a valid probability density function (p.d.f.), the function must satisfy two conditions: 1. The function must be non-negative for all values of \( x \). 2. The total integral of the function must equal 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1. \]
Step 2: Evaluate the functions.
- \( F_1 \) is a valid p.d.f. because it satisfies both conditions. - \( F_2 \) is a valid p.d.f. because it satisfies both conditions. - \( F_3 \) is a valid p.d.f. because it satisfies both conditions. - \( F_4 \) is **not** a valid p.d.f. because the integral of \( f(x) = \frac{x}{2} \) over the interval \( -2 \leq x \leq 2 \) does not equal 1. It results in a total probability greater than 1, violating the normalization condition.
Step 3: Conclusion.
Thus, the correct answer is option (B), \( F_4 \).