## Is the function $e^{x^2/2} \Phi(x)$ monotone increasing?

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

-
Dawson's integral seems closely related mathworld.wolfram.com/DawsonsIntegral.html – Aaron Hoffman Oct 3 2011 at 12:58
@Aaron Hoffman, thank you very much. I also noticed latter that it is closely related to this special function. Actually, in Chapter 41 of the book: books.google.com/… – Anand Oct 4 2011 at 8:45

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 $h(x_1)\leq h(x_2)$.

-
Nice argument... – Igor Rivin Oct 3 2011 at 10:09
@Davide Giraudo, Thanks a lot! :-) – Anand Oct 3 2011 at 10:11

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.

-
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.:-) – Anand Oct 3 2011 at 10:18
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...) – Igor Rivin Oct 3 2011 at 10:22
Thanks Igor Rivin. You are right. :-) – Anand Oct 3 2011 at 10:24

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.

-
 Thanks Cardinal. It is an equally nice solution. :-) – Anand Oct 17 2011 at 20:29