Question

Choose a possible probability density function from the given functions:

• A. $f(x) = \begin{cases} 1, & 0 \le x \le 2 \\ 0, & \text{otherwise} \end{cases}$
• B. $f(x) = \begin{cases} e^{-x}, & x \ge 0 \\ 0, & x < 0 \end{cases}$
• C. $f(x) = \begin{cases} \frac{6}{5}x(1+x), & x \ge 0 \\ 0, & x < 0 \end{cases}$
• D. $f(x) = \begin{cases} x(1-x), & 0 \le x \le 1 \\ 0, & \text{elsewhere} \end{cases}$

Hint:

An exponential function profile of the form $f(x) = \lambda e^{-\lambda x}$ for $x \ge 0$ is the standard model for the Exponential Distribution. Recognizing this pattern with $\lambda = 1$ instantly identifies it as a valid normalized probability distribution.

Answer 1

The Correct Option is B

Concept: For a continuous function $f(x)$ to qualify as a valid Probability Density Function (PDF), it must fulfill two strict mathematical criteria: 1) Non-negativity: The function value must be greater than or equal to zero for all real values of $x$: \[ f(x) \ge 0 \quad \forall x \in \mathbb{R} \] 2) Total Normalization: The total area under the curve across the entire domain $(-\infty, \infty)$ must equal exactly $1$: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] Let us systematically evaluate each option using the integration condition.

Step 1: Analyze Option (A).
The function is $f(x) = 1$ for $0 \le x \le 2$. Let us integrate over its non-zero domain: \[ \int_{0}^{2} 1 \, dx = [x]_{0}^{2} = 2 - 0 = 2 \neq 1 \] Since the area equals 2, this is not a valid PDF.

Step 2: Analyze Option (B).
The function is $f(x) = e^{-x}$ for $x \ge 0$. Integrating over this semi-infinite domain: \[ \int_{0}^{\infty} e^{-x} \, dx = \left[ -e^{-x} \right]_{0}^{\infty} = \left( -e^{-\infty} \right) - \left( -e^{0} \right) = 0 - (-1) = 1 \] Since $e^{-x} \ge 0$ for all $x \ge 0$ and the total integrated area is exactly 1, this fulfills both fundamental constraints.

Step 3: Analyze Option (C).
The function is $f(x) = \frac{6}{5}x(1+x)$ for $x \ge 0$. Integrating over this domain: \[ \int_{0}^{\infty} \frac{6}{5}(x + x^2) \, dx = \frac{6}{5} \left[ \frac{x^2}{2} + \frac{x^3}{3} \right]_{0}^{\infty} = \infty \neq 1 \] Since the integral diverges to infinity, it cannot be a valid PDF.

Step 4: Analyze Option (D).
The function is $f(x) = x(1-x)$ for $0 \le x \le 1$. Integrating over this interval: \[ \int_{0}^{1} (x - x^2) \, dx = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1} = \left( \frac{1}{2} - \frac{1}{3} \right) - 0 = \frac{1}{6} \neq 1 \] Since the area under the curve is $\frac{1}{6}$, it fails the normalization criterion. Hence, only Option (B) is a valid PDF.