To prove the lemma, it suffices to show that the Hessian ∇ $^{2}$G is positive semidefinite. We need to show that all the principal minors of ∇ $^{2}$G in (76) are nonnegative. However, because we already know that all the principal minors of the matrix in (75) are nonnega- tive, it only remains to show that ∂ $^{2}$G ∂µ 2 ≥ 0 and det ∇ $^{2}$G ≥ 0 . First, we show that ∂ $^{2}$G ∂µ 2 ≥ 0 Find the first derivative of G. Find the second derivative of G . The denominator of (79) is positive. Hence, it only remains to show that the numerator is nonneg- 45 ∂ $^{2}$G ∂µ 2 = 1 µ ∑ N i $_{=1}$θ$_{i}$ e$^{ω}$ i µ ∑ N i $_{=1}$θ$_{i}$ ω 2 i µ 2 e$^{ω}$ i µ − ∑ N i $_{=1}$θ$_{i}$ ω$_{i}$ µ 2 e$^{ω}$ i µ ∑ N i $_{=1}$θ$_{i}$ ω$_{i}$e$^{ω}$ i µ [ ∑ N i $_{=1}$θ$_{i}$ e$^{ω}$ i µ ] 2 = ∑ N i $_{=1}$θ$_{i}$ e$^{ω}$ i µ ∑ N i $_{=1}$θ$_{i}$ ω 2 i e$^{ω}$ i µ − [ ∑ N i $_{=1}$θ$_{i}$ ω$_{i}$e$^{ω}$ i µ ] 2 µ 3 [ ∑ N i $_{=1}$θ$_{i}$ e$^{ω}$ i µ ] 2 (79) ∂ $^{2}$G ∂µ 2 = ∑ N i $_{=1}$θ$_{i}$ − ω$_{i}$ µ 2 e$^{ω}$ i µ N i $_{=1}$θ$_{i}$ e$^{ω}$ i µ + 1 µ 2 ∑ N i $_{=1}$θ$_{i}$ ω$_{i}$e$^{ω}$ i µ N i $_{=1}$θ$_{i}$ e$^{ω}$ i µ − 1 µ ∑ N i $_{=1}$θ$_{i}$ e$^{ω}$ i µ ∑ N i $_{=1}$θ$_{i}$ − ω 2 i µ 2 e$^{ω}$ i µ − ∑ N i $_{=1}$θ$_{i}$ − ω$_{i}$ µ 2 e$^{ω}$ i µ ∑ N i $_{=1}$θ$_{i}$ ω$_{i}$e$^{ω}$ i µ [ ∑ N i $_{=1}$θ$_{i}$ e$^{ω}$ i µ ] 2 (78) ∂G ∂µ = ln ( N ∑ i =1 θ$_{i}$e$^{ω}$ i µ ) + µ ∑ N i $_{=1}$θ$_{i}$ − ω$_{i}$ µ 2 e$^{ω}$ i µ N i $_{=1}$θ$_{i}$ e$^{ω}$ i µ = ln ( N ∑ i =1 θ$_{i}$e$^{ω}$ i µ ) − 1 µ ∑ N i $_{=1}$θ$_{i}$ ω$_{i}$e$^{ω}$ i µ N i $_{=1}$θ$_{i}$ e$^{ω}$ i µ (77) ∇ $^{2}$G = ω$_{1}$ ω$_{N}$ µ ω$_{1}$ ... ω$_{N}$ µ     ∂ $^{2}$G ∂ω 2 1 ... ∂ $^{2}$G ∂ω$_{1}$∂ω$_{N}$ ∂ $^{2}$G ∂ω$_{1}$∂µ ... ... ∂ $^{2}$G ∂ω$_{N}$ ∂ω$_{1}$ ... ∂ $^{2}$G ∂ω 2 N ∂ $^{2}$G ∂ω$_{N}$ ∂µ ∂ $^{2}$G ∂µ∂ω$_{1}$ ... ∂ $^{2}$G ∂µ∂ω$_{N}$ ∂ $^{2}$G ∂µ 2     (76)