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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/iii/paper-46/1/i/solution

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Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 46 / 1 / i / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

Put

D_{1} = X_{1} - Y_{1}

,

D_{2} = X_{2} - Y_{2}

and

B = (1001 - 1 0 0 - 1) .

The difference vector is

D = BZ

. Its mean is zero because the two sons have the same mean vector. Its covariance matrix is

B V B^{T} = 2 (a - c b - c b - c a - c) .

For example,

Cov (D_{1}, D_{2}) = b - c - c + b = 2 (b - c)

. Thus

(D_{1} D_{2}) \sim N_{2} ((00), 2 (a - c b - c b - c a - c)) .

The positive definiteness assumption supplies

a - b > 0

and

a + b - 2 c > 0

: these are eigenvalues of the original matrix on within-son contrast and between-son difference directions. In particular the resulting difference covariance is positive definite.

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