I'm trying to show that the Hessian of the negative of the log likelihood with two parameters is positive definite, but I'm not sure how to go about it once I compute the Hessian.
The function is: $-log L(\alpha) =-\sum\limits_{i=1}^N ((1-y_i)(-\alpha_0 - \alpha_1x_i) - log(1 + e^{-\alpha_0 - \alpha_1x_i}))$
and I believe that the Hessian is:
\begin{bmatrix} \frac{e^{-\alpha_{0}-\alpha{1}x_i}}{(e^{-\alpha_{0}-\alpha{1}x_i}+1)^2}&\frac{e^{-\alpha_{0}-\alpha{1}x_i}x_i}{(e^{-\alpha_{0}-\alpha{1}x_i}+1)^2}\\\frac{e^{-\alpha_{0}-\alpha{1}x_i}x_i}{(e^{-\alpha_{0}-\alpha{1}x_i}+1)^2}&\frac{e^{-\alpha_{0}-\alpha{1}x_i}x_i^2}{(1+e^{-\alpha_{0}-\alpha{1}x_i})^2}\end{bmatrix}
How can I show that this matrix is positive definite for all $\alpha$ in order to prove convexity?
The Hessian simplifies to: $$\frac{e^{-\alpha_{0}-\alpha{1}x_i}}{(1+e^{-\alpha_{0}-\alpha{1}x_i})^2}\begin{bmatrix} 1 & x_i\\x_i & x_i^2\end{bmatrix}.$$ The factor is positive and does not affect positive (semi)definiteness. The matrix has trace $1+x_i^2$ and determinant $0$. Therefore, the eigenvalues are $0$ and $1+x_i^2$. This matrix is therefore positive semidefinite.