To prove the lemma, it suffices to show that function F is convex. The Hessian is: Because z$_{i}$ is negative, ∇ $^{2}$F is positive semidefinite. Therefore, F is convex. The next step is to show that the objective function µ ln ( ∑ N i $_{=1}$θ$_{i}$ e$^{ω}$ i µ ) is convex. Define a new function. Before proceeding with the proof, we present a very useful result from Boyd and Vandenberghe (2004). $^{11}$. Lemma 4 Let the function f ( y ) be defined as (73) where a$_{k}$ , y ∈ R $^{n}$, b$_{k}$ ∈ R . f ( y ) is a convex function. Therefore, the Hessian of f ( y ) must be positive semidefinite. Lemma 5 Function G , which is defined as (72), is convex. Proof. Note that function G can be expressed as (74). With µ fixed, the structure of G as a function of ω$_{1}$, ω$_{2}$, ..., ω$_{N}$ is exactly analogous to (73). With µ fixed, G is a convex function of ω$_{1}$, ω$_{2}$, ..., ω$_{N}$ . Hence all the principal minors of the matrix in (75) are nonnegative . $^{11}$See equation (4.44) on page 162 44     ∂ $^{2}$G ∂ω 2 1 ∂ $^{2}$G ∂ω$_{1}$∂ω$_{2}$ ... ∂ $^{2}$G ∂ω$_{1}$∂ω$_{N}$ ∂ $^{2}$G ∂ω$_{2}$∂ω$_{1}$ ∂ $^{2}$G ∂ω 2 2 ... ∂ $^{2}$G ∂ω$_{2}$∂ω$_{N}$ ... ∂ $^{2}$G ∂ω$_{N}$ ∂ω$_{1}$ ∂ $^{2}$G ∂ω$_{N}$ ∂ω$_{2}$ ... ∂ $^{2}$G ∂ω 2 N     (75) G ( ω$_{1}$, ω$_{2}$, ..., ω$_{N}$ , µ ) = µ ln ( N ∑ i =1 e 1 $_{µ}$ω$_{i}$ +ln θ$_{i}$ ) (74) f ( y ) = ln ( K ∑ k =1 e a T $_{k}$y + b$_{k}$ ) (73) G ( ω$_{1}$, ω$_{2}$, ..., ω$_{N}$ , µ ) = µ ln ( N ∑ i =1 θ$_{i}$e$^{ω}$ i µ ) (72) ∇ $^{2}$F =   ∂ $^{2}$F ∂ω 2 i ∂ $^{2}$F ∂ω$_{i}$∂ζ$_{i}$ ∂ $^{2}$F ∂ω$_{i}$∂ζ$_{i}$ ∂ $^{2}$F ∂ζ 2 i   =   0 0 0 − z$_{i}$e ζ$_{i}$   (71)