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

Independence permits distributional but not probability convergence

Let X1,X2…X_{1}, X_{2} \dots be a sequence of independent random variables with XiX_{i} following N(0, 1 + 1/i) distribution for all i∈Ni \in \mathbb{N}. Let X be N(0, 1)-random variable independent of {Xn:n≥1X_{n} : n \ge 1}. Then, which of the following statements are true?

  1. A.XnX_{n} converges to X in probability as n→∞n \to \infty
  2. B.XnX_{n} converges to X in distribution as n→∞n \to \infty✓ Correct
  3. C.Xn−XX_{n} - X converges to Z in distribution as n→∞n \to \infty, where Z follows the distribution N(0, 2).✓ Correct
  4. D.X/|XnX_{n}| converges to M in distribution as n→∞n \to \infty, where M follows standard Cauchy distribution.✓ Correct

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, and why this one tests dependence misread.

See pricing

50 are analysed free — try those first.

The trap it tests

Dependence misread

Independence, pairwise vs mutual, or correlation vs dependence.

Drill statements like this

Related counterexample: Convergence in probability implies almost sure convergence

More on this topic

The chapter behind this: Modes of convergence and the limit theorems — free to read

From Limit Theorems and Markov Chains › Modes of convergence, WLLN, SLLN, CLT

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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