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Part CCSIR NET December 2024the-independence-variance-is-one-over-n-minus-one

The independence variance is one over n minus one

Let (X1,Y1),(X2,Y2),…,(Xn,Yn)(X_{1}, Y_{1}), (X_{2}, Y_{2}), \dots, (X_{n}, Y_{n}) be a random sample of size n from a bivariate distribution F(X,Y) with absolutely continuous marginal distribution functions F_X and F_Y of X and Y, respectively. Let r_S be the Spearman's rank correlation coefficient defined with ranks of XiX_{i} and ranks of Yi,i=1,2,…,nY_{i}, i = 1, 2, \dots, n. Then, which of the following statements are true?

  1. A.If F(X,Y)(x, y) = F_X(x)F_Y(y), ∀(x, y), then E(r_S) = 0, ∀n ≥ 2.✓ Correct
  2. B.If n = 3, then P(r_S = 0) = 0.✓ Correct
  3. C.If F(X,Y)(x, y) = F_X(x)F_Y(y), ∀(x, y), then Var(r_S) = 1/n, ∀n ≥ 2.
  4. D.If n = 4 and F(X,Y)(x, y) = F_X(x)F_Y(y), ∀(x, y), then P(r_S = 0) = 1/24.

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 execution slip.

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50 are analysed free — try those first.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Related counterexample: Any interval of the form [X̄ − (S/√n)t, ∞) with a 90% quantile t is a 90% confidence interval

More on this topic

The chapter behind this: Standard tests and confidence intervals — free to read

From Hypothesis Testing › Likelihood ratio and standard tests

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