Part CCSIR NET December 2024independence-permits-distributional-but-not-probability-convergence

Independence permits distributional but not probability convergence

Let $X_{1}, X_{2} \dots$ be a sequence of independent random variables with $X_{i}$ following N(0, 1 + 1/i) distribution for all $i \in N$ . Let X be N(0, 1)-random variable independent of { $X_{n} : n \geq 1$ }. Then, which of the following statements are true?

A. $X_{n}$ converges to X in probability as $n \to \infty$
B. $X_{n}$ converges to X in distribution as $n \to \infty$ ✓
C. $X_{n} - X$ converges to Z in distribution as $n \to \infty$ , where Z follows the distribution N(0, 2).✓
D.X/| $X_{n}$ | converges to M in distribution as $n \to \infty$ , where M follows standard Cauchy distribution.✓

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward.

See pricing

50 are analysed free — try those first.

Independence permits distributional but not probability convergence

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