How to show this function is convex, increasing function:
Define $f(z)=\frac{T(z)}{g(z)}$, where $T(z)=\int_{-\infty}^z \Phi(x)\, dx-\alpha\int_{-\infty}^{\frac{z}{\alpha}} \Phi(x)\, dx$ and $g(z)=Pr(Zi<z,Zt>\frac{z}{\alpha})$ and $\Phi(x)$ is the CDF of standard normal distribution and $(Zi,Zt)$ are bivariate normal distribution with mean (0,0) and variances of (1,1) and correlation of $\alpha$ where $0<\alpha<1$. (Pr is "probability of").
How can we show $f$ is increasing and convex (numerical graph of the function is increasing and convex)?
Note that $f(z)$ seems related to hazard rate of normal distribution. Also $g'(z)=\frac{1}{\sqrt{2\pi}}\Phi'(\frac{z}{\alpha})\left(\frac{1-\Phi(\sqrt{\frac{1-\alpha^2}{\alpha^2}}\,z)}{\Phi'(\sqrt{\frac{1-\alpha^2}{\alpha^2}}\,z)}-\sqrt\frac{\pi}{2}\frac{1}{\alpha}\right)$
Also $T(z)>0$ and $T(z)$ can be simplified $T(z)=\Phi'(z)+z\, \Phi(z)-\alpha\, \Phi'(\frac{z}{\alpha})-z \Phi(\frac{z}{\alpha})$.
I think the steps of the proof should be as follows:
1- show $\frac{\int_{-\infty}^{z} \Phi(x)\, dx}{\Phi(z)}$ is positive, increasing, convex, with positive higher derivatives.
2- show $\frac{\Phi(\frac{z}{\alpha})}{g(z)}=\frac{1}{Pr(Zi<z|Zt>\frac{z}{\alpha})}$ is positive increasing, convex, with positive higher derivatives.
3- $\frac{\alpha\int_{-\infty}^{\frac{z}{\alpha}} \Phi(x)\, dx}{\Phi(\frac{z}{\alpha})}$ and $\frac{\Phi(\frac{z}{\alpha})}{g(z)}=\frac{1}{Pr(Zi<z|Zt>\frac{z}{\alpha})}$ as functions of $\alpha$ are incrasing, convex, with positive higher derivatives.