Discussion
Suppose as \(n \to \infty\), \(X_{n} \to 3\) in probability and \(Y_{n} \to Y\) in distribution. Which of the following is true as \(n \to \infty\) ?
Answer: B
The true statement is \(\frac{n(X_{n} - 4) + Y_{n}}{n} \to -1\) in probability.
Split the expression:
\[ \begin{aligned} \frac{n(X_{n} - 4) + Y_{n}}{n} &= (X_{n} - 4) \\ &\quad + \frac{Y_{n}}{n} \end{aligned} \]
\(X_{n} - 4 \to -1\) in probability. \(Y_{n}\) converges in distribution, so it is bounded in probability (tight), and dividing by n sends \(\frac{Y_{n}}{n} \to 0\) in probability. The sum tends to \(-1\) in probability.
Options A and C use Slutsky's theorem correctly for the limit but claim convergence in probability, when \(Y_{n}\) converges only in distribution, so only convergence in distribution is guaranteed. Option D gives \((Y_{n} - 4) + \frac{X_{n}}{n} \to Y - 4\) in distribution, not the constant \(-4\).
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