Question:

Let \( \{ X_n \}_{n \geq 1} \) be a sequence of independent random variables and \( X_n \xrightarrow{a.s.} 0 \) as \( n \to \infty \). Then which of the following options is/are necessarily correct?

Show Hint

Almost sure convergence implies that the sequence converges pointwise for almost all outcomes. Additionally, properties of sums and series involving probabilities can be used to analyze convergence in probability and almost surely.
Updated On: Apr 9, 2025
  • \( E(X_n^3) \to 0 \) as \( n \to \infty \)
  • \( X_n^7 \xrightarrow{P} 0 \) as \( n \to \infty \)
  • For any \( \epsilon>0 \), \( \sum_{n=1}^{\infty} \Pr(|X_n| \geq \epsilon)<\infty \)
  • \( X_n^2 + X_n + 5 \xrightarrow{a.s.} 5 \) as \( n \to \infty \)
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is B, C, D

Solution and Explanation

Step 1: Convergence of \( E(X_n^3) \)
Since \( X_n \to 0 \) almost surely, \( E(X_n^3) \) will also tend to 0. This is because the third moment of a sequence that converges to 0 almost surely will also tend to 0.
Thus, Option (A) is incorrect because we need to show that the third moment tends to zero, but it is not guaranteed under the given conditions.
Step 2: Convergence in Probability of \( X_n^7 \)
By the convergence in probability of \( X_n \) to 0, we can raise \( X_n \) to any positive power, including 7, and it will still converge to 0 in probability.
Thus, Option (B) is correct.
Step 3: Series Convergence for \( |X_n| \geq \epsilon \)
The Borel-Cantelli Lemma tells us that if \( \sum_{n=1}^{\infty} \Pr(|X_n| \geq \epsilon) < \infty \), then \( X_n \to 0 \) almost surely. Since \( X_n \to 0 \) almost surely, this condition holds.
Thus, Option (C) is correct.
Step 4: Almost Sure Convergence of \( X_n^2 + X_n + 5 \)
As \( X_n \to 0 \) almost surely, we have \( X_n^2 + X_n + 5 \to 5 \) almost surely.
Thus, Option (D) is correct.
Final Answer:
The correct answers are \( \boxed{B, C, D} \).
Was this answer helpful?
0
0

Questions Asked in GATE ST exam

View More Questions