(a) Show that E(s2) = 2tr( PX)/(n K) = 2. Hint: Follow the same proof given below equation (7.6) of Chapter 7, but

(a) Show that E(s2) = σ2tr(Ω ¯ PX)/(n − K) = σ2. Hint: Follow the same proof given below equation (7.6) of Chapter 7, but substitute σ2Ω instead of σ2In.

(b) Use the fact that PX and Σ are non-negative definite matrices with tr(ΣPX) ≥ 0 to show that 0 ≤ E(s2) ≤ tr(Σ)/(n − K) where tr(Σ) =

n i=1 σ2i with σ2i

= var(ui) ≥ 0. This bound was derived by Dufour (1986). Under homoskedasticity, show that this bound becomes 0 ≤ E(s2) ≤ nσ2/(n − K). In general, 0 ≤ {mean of n − K smallest characteristic roots of

Σ} ≤ E(s2) ≤ {mean of n−K largest characteristic roots of Σ} ≤ tr(Σ)/(n−K), see Sathe and Vinod (1974) and Neudecker (1977, 1978).

(c) Show that a sufficient condition for s2 to be consistent for σ2 irrespective of X is that

λmax = the largest characteristic root of Ω is o(n), i.e., λmax/n → 0 as n → ∞ and plim

(uu/n) = σ2. Hint: s2 = u ¯ PXu/(n−K) = uu/(n−K)−uPXu/(n−K). By assumption, the first term tends in probability limits to σ2 as n→∞. The second term has expectation

σ2tr(PXΩ)/(n −K). Now PXΩ has rank K and therefore exactly K non-zero characteristic roots each of which cannot exceed λmax. This means that E[uPXu/(n−K)] ≤ σ2Kλmax/(n−

K). Using the condition that λmax/n → 0 proves the result. See Kr¨amer and Berghoff (1991).

(d) Using the same reasoning in part (a), show that s∗2 given in (9.6) is unbiased for σ2.