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Part CCSIR NET December 2023spearman-for-bivariate-normal

Spearman for bivariate normal

Let (X1,Y1),(X2,Y2),(X3,Y3)(X_{1}, Y_{1}), (X_{2}, Y_{2}), (X_{3}, Y_{3}) be i.i.d. bivariate normal with mean (0, 0) and correlation matrix [[1,ρ],[ρ,1]][[1, \rho], [\rho, 1]], |ρ\rho| < 1. Let Sρ=3E(S_\rho = 3E(sgn(X1−X2)(Y1−Y3))(X_{1} - X_{2})(Y_{1} - Y_{3})). Which of the following statements are true?

  1. A.If X1X_{1} and Y1Y_{1} are independent, then Sρ=0S_\rho = 0✓ Correct
  2. B.Sρ=(6/π)sin⁡S_\rho = (6/\pi) \sin⁻1(ρ/2)^{1}(\rho/2)✓ Correct
  3. C.If Sρ=0S_\rho = 0, then X1X_{1} and Y1Y_{1} are independent✓ Correct
  4. D.If X1X_{1} and Y1Y_{1} are independent, then Sρ=1/2S_\rho = 1/2

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 hypothesis dropped.

See pricing

50 are analysed free — try those first.

The trap it tests

Hypothesis dropped

A theorem applied without checking one of its preconditions.

Drill statements like this

Related counterexample: If every marginal is normal then the vector is multivariate normal

More on this topic

The chapter behind this: Multivariate normal and Wishart — free to read

From Linear Models and Multivariate › Multivariate normal distribution

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

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