Probits

We define the normal density $$\phi(x)=\frac{1}{\sqrt{2\pi}}\exp(-\frac12 x^2),$$ and the normal distribution function $$\Phi(x)=\int_{-\infty}^x\phi(z)dz$$ in the usual way. In addition we define $$f(x)=-\log\Phi(x).$$

Clearly $$f'(x)=-\frac{\phi(x)}{\Phi(x)}$$ $$f''(x)=\frac{x\phi(x)}{\Phi(x)}+\left[\frac{\phi(x)}{\Phi(x)}\right]^2.$$ We can get more insight into these derivatives by rewriting them as conditional expectations. If $u=\phi(z)$ then $du=-z\phi(z)dz$ and thus $$\int_{-\infty}^x z\phi(z)dz=-\int_0^{\phi(x)}du=-\phi(x),$$ which implies $$f'(x)=\frac{\int_{-\infty}^xz\phi(z)dz}{\int_{-\infty}^x\phi(z)dz}=\mathbf{E}(z|z<x).$$ This shows that $f'(x)<0$ and thus $f$ is decreasing.

Now in the same way we can define $u=z\phi(z)$ and use $du=(1-z^2)\phi(z)$ to derive $$\int_{-\infty}^x (1-z^2)\phi(z)dz=\int_0^{x\phi(x)}du=x\phi(x),$$ which implies $$1-\mathbf{E}(z^2|z<x)=\frac{x\phi(x)}{\Phi(x)},$$ and thus $$f''(x)=1-[\mathbf{E}(z^2|z<x)+\mathbf{E}(z|z<x)]=1-\mathbf{V}(z|z<x).$$ This shows that $0<f''(x)<1$, and thus $f$ is convex and has a bounded second derivative. Moreover $f''(x)$ is decreasing, which implies that $f'$ is concave. Also $$\begin{align*} \lim_{x\rightarrow-\infty}f''(x)&=1,\\ \lim_{x\rightarrow+\infty}f''(x)&=0. \end{align*}$$

A function $g$ majorizes our function $f$ in a point $y$ if $g(x)\geq f(x)$ for all $x$ and $g(y)=f(y)$. A quadratic function $$g(x)=c+b(x-y)+\frac12 a (x-y)^2$$ majorizes $f$ in $y$ if and only if $c=f(y)$, $b=f'(y)$, and $$a\geq A(y)=\sup_{x\not= y}\delta(x|y),$$ where $$\delta(x|y)=\frac{f(x)-f(y)-f'(y)(x-y)}{\frac12 (x-y)^2}.$$ We find the \emph{best quadratic majorization} of $f$ in $y$ by choosing $a=A(y)$.

Since $\delta(x|y=f''(z)$ for some $z$ between $x$ and $y$, we see that $\delta(x|y)<1$ for all $x$. On the other hand $$\lim_{x\rightarrow-\infty}\delta(x,y)=1,$$ and consequently $A(y)=1$ for all $y$. Thus the best quadratic majorization is actually the uniform quadratic majorization $$g(x)=f(y)+f'(y)(x-y)+\frac12 (x-y)^2.$$