Discussion
Suppose X is a real random variable on a probability space \((\Omega, A, P)\) and \(\{A_{n}, n \ge 1\}\) is a sequence of sets in A such that :
\[ A_{n} \to \phi \text{ as } n \to \infty \]
Answer: A
The conclusion is \(XI_{A_{n}} \to 0\) in probability.
\(A_{n} \to \varnothing\) means each \(\omega\) lies in only finitely many \(A_{n}\), so \(I_{A_{n}}(\omega) \to 0\) for every \(\omega\) and \(X I_{A_{n}} \to 0\) pointwise, hence almost surely. Almost sure convergence implies convergence in probability. Directly: for \(\varepsilon \gt 0\), \(P(|XI_{A_{n}}| \gt \varepsilon) \le P(A_{n}) \to 0\) by continuity of probability.
The limit is 0, not X or 1, so options B and C fail unless X itself is 0. Convergence in law to X (option D) would need X to have the same law as the constant 0.
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