TITLE: Is the function $e^{x^2/2} \Phi(x)$ monotone increasing?
QUESTION [11 upvotes]: Hello,
Here is an interesting problem. It looks elementary, but it has taken me some efforts without solving it. Let
$$
h(x) = e^{x^2/2} \Phi(x),\qquad \text{with}\quad \Phi(x):=\int_{-\infty}^x \frac{e^{-y^2/2}}{\sqrt{2\pi}} dy.
$$
The question is whether the function $h(x)$ is monotone increasing over $R$? Are there some work dealing with such function? 
It seems a quite easy problem. By taking the first derivative, we need to prove that
$$
h(x)' = h(x) x + \frac{1}{\sqrt{2\pi}}  \ge 0.
$$
which again, not obvious (for $x<0$). Some facts, that might be useful, are:
$$
\lim_{x\rightarrow -\infty} h(x) =0, \quad \lim_{x\rightarrow -\infty} h(x)' =0.
$$
Thank you very much for any hints!
Anand

REPLY [2 votes]: A useful inequality is
\begin{equation}
\tag{1}
\frac{1}{\sqrt{2 \pi}} \frac{x}{x^2+1} \mathrm{e}^{-x^2/2} \le \frac{1}{\sqrt{2\pi}} \int_x^\infty \mathrm{e}^{-y^2/2} \, \mathrm{d} y \le \frac{1}{\sqrt{2 \pi}} \frac{1}{x} \mathrm{e}^{-x^2/2}
\end{equation}
for $x >0$, which can be shown by elementary arguments (see below). For example, one can prove the law of the iterated logarithm (for the Browian motion) by means of this inequality. By symmetry, we have for $x <0$ that
$$|x \mathrm{e}^{x^2/2} \Phi(x)| = \frac{|x|\mathrm{e}^{x^2/2} }{\sqrt{2\pi}} \int_{|x|}^\infty \mathrm{e}^{-y^2/2} \, \mathrm{d} y \le \frac{1}{\sqrt{2\pi}}$$
and this proves $h'(x) \ge 0$.
The second inequality in $(1)$ can be proven by using $\exp(-y^2) \le y/x \exp(-y^2)$ for $y \ge x$ and the first one in $(1)$ follows by applying partial integration in the form
\begin{equation*}
\frac{1}{x^2} \int_x^\infty \mathrm{e}^{-y^2/2} \, \mathrm{d} y \ge \int_x^\infty \frac{1}{y^2} e^{-y^2/2} = \frac{1}{x} \mathrm{e}^{-x^2/2} - \int_x^\infty \mathrm{e}^{-y^2/2}
\end{equation*}
and properly rearranging.