Question: Using the var(GLS) expression below (12.30) and var(Within) = 2 W 1 XX, show that (var(GLS))1 (var(Within))1 = 2BXX/2 which is positive semi-definite.

Using the var(βGLS) expression below (12.30) and var(βWithin) = σ2

νW

−1 XX, show that

(var(βGLS))−1 − (var(βWithin))−1 = φ2BXX/σ2

ν

which is positive semi-definite. Conclude that var(βWithin)− var(βGLS) is positive semi-definite.

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