Skip to content
Part BCSIR NET December 2024the-median-survives-where-the-mean-does-not

The median survives where the mean does not

Suppose that {XnX_{n}}(n∈N)(n\in\mathbb{N}) is a sequence of independent and identically distributed (i.i.d.) random variables with the common probability density function f(x)=1/(π(1+x2)),x∈Rf(x) = 1/(\pi(1 + x^{2})), x \in \mathbb{R}. Then which of the following statements is true?

  1. A.X1X_{1} and (X1+X2)/2(X_{1} + X_{2})/\sqrt{2} have the same distribution.
  2. B.(1/n)∑(i=1..n)Xi(1/n)\sum(i=1..n)X_{i} converges to 0 in probability, as n→∞n \to \infty.
  3. C.Median of {X1,X2,…,X(2n+1)X_{1}, X_{2}, \dots, X(2n+1)} converges to 0 in probability, as n→∞n \to \infty.✓ Correct
  4. D.E(|X1X_{1}|(3/4))=∞^(3/4)) = \infty

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 moments and tails.

See pricing

50 are analysed free — try those first.

The trap it tests

Moments and tails

A moment assumed to exist, or tail behaviour assumed to be tame.

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.

ShareWhatsAppTelegram