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Part CCSIR NET June 2024divide-by-the-right-power-of-n

Divide by the right power of n

Let {XnX_{n}}n≥1_{n}\ge_{1} be a sequence of independent and identically distributed random variables with E(X1)=0E(X_{1}) = 0 and Var(X1)=1Var(X_{1}) = 1. Which of the following statements are true?

  1. A.lim⁡(n→∞)P((n⋅∑i\lim(n\to\infty) P((\sqrt{n}\cdot\sum_{i}₌1nXi)/(∑i_{1}^{n}X_{i})/(\sum_{i}₌1nXi2)≤0)=1/2_{1}^{n}X_{i}^{2}) \le 0) = 1/2✓ Correct
  2. B.(∑i(\sum_{i}₌1nXi)/(∑i_{1}^{n}X_{i})/(\sum_{i}₌1nXi2)_{1}^{n}X_{i}^{2}) converges in probability to 0 as n→∞n \to \infty✓ Correct
  3. C.(1/n)∑i(1/n)\sum_{i}₌1nXi2_{1}^{n}X_{i}^{2} converges in probability to 1 as n→∞n \to \infty✓ Correct
  4. D.lim⁡(n→∞)P((∑i\lim(n\to\infty) P((\sum_{i}₌1nXi)/n≤0)=1/2_{1}^{n}X_{i})/\sqrt{n} \le 0) = 1/2✓ 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 limit assumed to exist.

See pricing

50 are analysed free — try those first.

The trap it tests

Limit assumed to exist

Reasoning with a limit before establishing there is one. limsup is not lim.

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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