Let $R(x)$ be the residual function associated to the normal probability density, i.e.

$$R(x)~=~\int_x^{+\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}dy, \mbox{ for all } x\in R.$$


$$\phi(s)~=~\int_0^1\log\Big(\frac{R(sx)}{R(s)}\Big)dx \mbox{ for all } s\ge 0.$$

Could we show the map $s\mapsto\phi(s)$ is increasing on $R_+$? Thanks for the reply!

Ps: Indeed, I have shown using Mathematica that the function $s\mapsto\phi(s)$ is increasing and moreover, for each $0\le x<1$, the maps $s\frac{R(sx)}{R(s)}$ is increasing, ex: $x=0, 0.1, 0.2, ..., 0.9$. Thus I strongly believe that it suffices to show the map $s\mapsto\frac{R(sx)}{R(x)}$ is increasing for $0\le x<1$. However, when I compute its derivative, it is still hard to determine its positivity. If someone has an idea to show that, many thanks!


Let's prove that $R(sx)/R(s)$ increases in $s$, i.e. that $R(s)/R(sx)$ decreases. Denote $s=e^t$, $x=e^{-\tau}$, we have to prove that $f(t)-f(t-\tau)$ decreases in $t$, where $f(t)=\log R(e^t)$. It follows (and is really equivalent) that $f$ is concave, i.e. that $f'(t)$ decreases. We have $f'(t)=e^tR'(e^t)/R(e^t)$. Again denote $e^t=s$, we have to prove that $sR'(s)/R(s)$ decreases in $s$. Derivative in $s$ equals $\frac{R(R'(s)+sR''(s))-sR'^2}{R^2}$, thus we have to prove that numerator is non-positive. It is equivalent to the following inequality: $(s^2-1)\int_{s}^{\infty} e^{-t^2/2}dt\leq se^{-t^2/2}$. This is true even with $s^2$ instead of $s^2-1$, i.e. $\int_{s}^{\infty} e^{-t^2/2}dt\leq s^{-1}e^{-s^2/2}$. In order to prove this, denote $h(s)=s^{-1}e^{-s^2/2}-\int_{s}^{\infty} e^{-t^2/2}dt$ and see that $h(+\infty)=0$ and $h'(s)=-s^{-2}e^{-s^2/2}<0$, so $h$ decreases.

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