How to show this function is increasing convex function:
Define $f(z)=\frac{T(z)}{g(z)}$, where
$T(z)=\phi(z)-\alpha \phi(\frac{z}{\alpha})+z(\Phi(z)-\Phi(\frac{z}{\alpha}))\,,$
$g(z)=\Phi(z)-\frac{1}{2}\Phi(\frac{z}{\alpha})-\int_{-\infty}^{z}\phi(x)\Phi(\sqrt{\frac{1-\alpha^2}{\alpha^2}}\,x)) dx$
and $\Phi(z)=\int_{-\infty}^z\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$ is the CDF of standard normal distribution and $\phi(z)=\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}$ is the PDF of standard normal distribution and $0<\alpha<1$.
How can we show $f(z)$ is increasing and convex (numerical graph of the function is increasing and convex)? (maybe it suffices to show for $z<0$, $\frac{T'(z)}{T(z)}-\frac{g'(z)}{g(z)}$ is increasing and positive. Note that even $\log f(z)$ looks convex increasing).