Discussion
Suppose that X and Y are two random variables with c.d.f. \(F_{1}\) and \(F_{2}\) respectively, where
\[ F_{1}(x) = \begin{cases} 0 & \text{if } x \lt 0 \\ 2/3 & \text{if } 0 \le x \lt 1 \\ 1 & \text{if } x \ge 1 \end{cases} \]
and
\[ F_{2}(x) = \begin{cases} 0 & \text{if } x \lt 0 \\ \sqrt{x} & \text{if } 0 \le x \lt 1 \\ 1 & \text{if } x \ge 1 \end{cases} \]
Then, which of the following statements is true ?
Answer: A
The true statement is E(X) = E(Y); both equal \(\frac{1}{3}\).
\(F_{1}\) has jumps of \(\frac{2}{3}\) at 0 and \(\frac{1}{3}\) at 1, so X is discrete with \(E(X) = 0 \cdot \frac{2}{3} + 1 \cdot \frac{1}{3} = \frac{1}{3}\).
\(F_{2}(x) = \sqrt{x}\) on [0, 1) is continuous with density \(\frac{1}{2\sqrt{x}}\), so
\[ \begin{aligned} E(Y) &= \int_{0}^{1} x \cdot \frac{1}{2\sqrt{x}}\,dx \\ &= \frac{1}{2}\int_{0}^{1} \sqrt{x}\,dx = \frac{1}{3} \end{aligned} \]
Option B takes the jump at 0 as the mean of X. Options C and D have the right E(X) but the wrong E(Y): \(\frac{1}{2}\) is the mean of a uniform variable, and \(\frac{1}{6}\) halves the correct value.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development