Part CCSIR NET June 2025for-negative-rho-the-leading-eigenvalue-is-1-minus-rho-not-1-plus-2-rho

For negative rho the leading eigenvalue is 1 minus rho not 1 plus 2 rho

Let the random vector $(X_{1}, X_{2}, X_{3})^{T}$ have the positive definite dispersion matrix with rows $(1, ρ, ρ), (ρ, 1, ρ)$ and $(ρ, ρ, 1)$ . Then, which of the following statements are true?

A. $ρ$ may be −0.47✓
B.The first principal component can only explain 32% of the total variation for some $ρ$
C.The second principal component can explain more than 32% of the total variation for any $ρ$
D.The variance of the first principal component is $1 + 2 ρ$ for any $ρ$

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 boundary and endpoint.

See pricing

50 are analysed free — try those first.

The trap it tests

Boundary and endpoint

The statement turns at the edge of the interval, the domain, or the parameter range.

Drill statements like this

For negative rho the leading eigenvalue is 1 minus rho not 1 plus 2 rho

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