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!



We can write $h(x)=\frac 1{\sqrt{2\pi}}\int_{-\infty}^x \exp\left(\frac{x^2-y^2}2\right)dy$. Now put $t=x-y$. We get \begin{align} h(x)&=\frac 1{\sqrt{2\pi}}\int_0^{+\infty}\exp\left(\frac{x^2-(x-t)^2}2\right)dt\\\ &=\frac 1{\sqrt{2\pi}}\int_0^{+\infty}\exp\left(xt-\frac{t^2}2\right)dt. \end{align} We can differentiate under the integral thanks to the dominated convergence theorem. We get $$h'(x)=\frac 1{\sqrt{2\pi}}\int_0^{+\infty}t\exp\left(xt-\frac{t^2}2\right)dt\geq 0.$$ Added later: we don't need to diffentiate. If $x_1\leq x_2$ then for $t\geq 0$ we have $e^{tx_1}\leq e^{tx_2}$ therefore $ h(x_1)\leq h(x_2) $.

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    $\begingroup$ Nice argument... $\endgroup$ – Igor Rivin Oct 3 '11 at 10:09

This is just an alternative argument to Davide's nice one.

First, note that $h' = e^{x^2/2}(x \Phi + \Phi')$.

Since $\Phi'' = -x \Phi'$, monotonicity of the integral yields $$ x \Phi(x) \geq \int_{-\infty}^x u \Phi'(u) \mathrm{d}u = -\Phi'(x). $$ So, $h' \geq 0$, and we are done.


I haven't actually done the computation, but it seems to me that integrating the $\Phi(x)$ term by parts ad nauseam, you get a nice power series for $h(x).$

EDIT @Davide's argument is obviously the complete answer to the question as asked, but just as a coda, the series for $h(x)$ is quite cute:

In the odd part, the coefficients of $x^{2k+1}$ is $1/p(k)$ where $p(k)$ is the product of the first $k$ odd integers, while in the even part, the coefficient of $x^{2k}$ is $\sqrt{\pi/2}/q(k),$ where $q(k)$ is the product of the first $k$ even integers.

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    $\begingroup$ Thanks Igor Rivin, I am afraid of series, especially if you multiply two series, one is of $e^{-x^2}$ and the other is of $\Phi(x)$. Since Davide has given a nice solution, I am very content now. Thanks again.:-) $\endgroup$ – Anand Oct 3 '11 at 10:18
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    $\begingroup$ I am glad you are content, but you might want to get over that series phobia (although the series is not useful for answering your question, it seems...) $\endgroup$ – Igor Rivin Oct 3 '11 at 10:22

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.


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