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

The median survives where the mean does not

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

A. $X_{1}$ and $(X_{1} + X_{2}) / 2$ have the same distribution.
B. $(1/ n) \sum (i = 1.. n) X_{i}$ converges to 0 in probability, as $n \to \infty$ .
C.Median of { $X_{1}, X_{2}, \dots, X (2 n + 1)$ } converges to 0 in probability, as $n \to \infty$ .✓
D.E(| $X_{1}$ | $^{(} 3/4)) = \infty$

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.

See pricing

50 are analysed free — try those first.

The median survives where the mean does not

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