Question:
Let \( \{ X_n \}, n \geq 1 \), be a sequence of random variables with the probability mass functions
\[
p_{X_n}(x) = \begin{cases}
\frac{n}{n+1}, & x = 0 \\
\frac{1}{n+1}, & x = n \\
0, & \text{elsewhere}
\end{cases}
\]
Let \( X \) be a random variable with \( P(X = 0) = 1 \). Then which of the following statements is/are true?
Let \( \{ X_n \}, n \geq 1 \), be a sequence of random variables with the probability mass functions
\[
p_{X_n}(x) = \begin{cases}
\frac{n}{n+1}, & x = 0 \\
\frac{1}{n+1}, & x = n \\
0, & \text{elsewhere}
\end{cases}
\]
Let \( X \) be a random variable with \( P(X = 0) = 1 \). Then which of the following statements is/are true?
Show Hint
Convergence in distribution and convergence in probability are related, but convergence in probability is a stronger condition. If a sequence of random variables converges almost surely, it also converges in probability.
Updated On: Dec 15, 2025
- \( X_n \) converges to \( X \) in distribution
- \( X_n \) converges to \( X \) in probability
- \( E(X_n) \to E(X) \)
- There exists a subsequence \( \{ X_{n_k} \} \) of \( \{ X_n \} \) such that \( X_{n_k} \) converges to \( X \) almost surely
Show Solution
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The Correct Option is A, B, D
Solution and Explanation
Step 1: Understand the sequence of random variables \( X_n \).
The probability mass function of \( X_n \) is given by: \[ P(X_n = 0) = \frac{n}{n+1}, \quad P(X_n = n) = \frac{1}{n+1}. \] As \( n \to \infty \), we observe the following: - \( P(X_n = 0) \to 1 \), since \( \frac{n}{n+1} \to 1 \) as \( n \) increases. - \( P(X_n = n) \to 0 \), since \( \frac{1}{n+1} \to 0 \) as \( n \to \infty \). Thus, the sequence \( X_n \) tends to 0 with probability 1 as \( n \to \infty \), which is the value of the random variable \( X \) (since \( P(X = 0) = 1 \)).
Step 2: Analyzing convergence in distribution.
For convergence in distribution, we check if the cumulative distribution function (CDF) of \( X_n \) converges to the CDF of \( X \). We know that \( X_n \to 0 \) almost surely as \( n \to \infty \), and since \( P(X = 0) = 1 \), the distribution of \( X_n \) converges to the distribution of \( X \) in distribution. Therefore, statement (A) is true: \( X_n \) converges to \( X \) in distribution.
Step 3: Analyzing convergence in probability.
Convergence in probability occurs when for any \( \epsilon>0 \), the probability \( P(|X_n - X| \geq \epsilon) \to 0 \) as \( n \to \infty \). Since \( X_n \) converges to 0 almost surely and \( X = 0 \), we can conclude that \( X_n \) converges to \( X \) in probability. Thus, statement (B) is also true.
Step 4: Analyzing expected values.
Since \( X_n \to 0 \) almost surely, we would expect that \( E(X_n) \to E(X) \), where \( E(X) = 0 \). However, the expected values \( E(X_n) \) do not converge to \( E(X) \) as the variance of \( X_n \) does not go to zero. Therefore, statement (C) is false.
Step 5: Analyzing almost sure convergence for a subsequence.
By the Borel-Cantelli Lemma, there exists a subsequence \( \{ X_{n_k} \} \) of \( \{ X_n \} \) such that \( X_{n_k} \to 0 \) almost surely. This is because the probability of \( X_n \) being 0 tends to 1, and there are infinitely many terms where \( X_n = 0 \). Hence, statement (D) is true.
Step 6: Conclusion.
Thus, the correct answer is (A), (B), and (D).
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