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Part CCSIR NET December 2024the-transform-is-uniform-not-cauchy

The transform is uniform not cauchy

Let X be a random variable with probability density function f(x)=1/(π(1+x2)),x∈Rf(x) = 1/(\pi(1 + x^{2})), x \in \mathbb{R}. If Z=1/2+(1/π)tan⁡Z = 1/2 + (1/\pi)\tan⁻1(X)^{1}(X), then which of the following statements are true?

  1. A.E(Zᵐ) = 1/(m + 1), for all m∈Nm \in \mathbb{N}✓ Correct
  2. B.E(ΦE(\Phi⁻1(Z))=0^{1}(Z)) = 0, where Φ(⋅)\Phi(\cdot) is the cumulative distribution function of standard normal random variable.✓ Correct
  3. C.Z is degenerate at 0.
  4. D.If Z1Z_{1} and Z2Z_{2} are independent and identically distributed (i.i.d.) random variables having distribution same as the distribution of Z, then Z has the same distribution as (Z1+Z2)/2(Z_{1} + Z_{2})/2.

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 standard counterexample.

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The trap it tests

Standard counterexample

There is a canonical object that settles this. Recognising it is the whole question.

Drill statements like this

Related counterexample: Uncorrelated ⇒ independent

More on this topic

The chapter behind this: Random variables, moments and generating functions — free to read

From Probability › Random variables, distributions, moments, MGF

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