Question

Let \( X \) be a continuous random variable with CDF \[ F(x) = \begin{cases} 0, & x < 0 \\ a x^2, & 0 \le x < 2 \\ 1, & x \ge 2 \end{cases} \] for some real constant \( a \). Then \( E(X) \) is equal to:

• A. \(\frac{4}{3}\)
• B. \(\frac{1}{4}\)
• C. 1
• D. 0

Hint:

Always check CDF continuity at boundary points to determine unknown constants before differentiating to get the pdf.

Answer 1

The Correct Option is A

Step 1: Find \( a \) using CDF condition.
Continuity at \( x = 2 \): \( F(2^-) = F(2^+) = 1 \). Thus, \( a(2)^2 = 1 \Rightarrow a = \frac{1}{4} \).
Step 2: Find PDF.
Differentiate \( F(x) \): \[ f(x) = \frac{d}{dx}F(x) = \begin{cases} \dfrac{x}{2}, & 0 \le x < 2 \\ 0, & \text{otherwise} \end{cases} \]
Step 3: Compute \( E(X) \).
\[ E(X) = \int_0^2 x f(x)\,dx = \int_0^2 x \cdot \frac{x}{2} \,dx = \frac{1}{2}\int_0^2 x^2 \,dx = \frac{1}{2}\cdot\frac{8}{3} = \frac{4}{3}. \] Final Answer: \[ \boxed{\frac{4}{3}} \]