Here is a calculus problem which bored me for sometime. Let $a>0$ and $b<0$ be fixed.Define the following function (EDIT: Following the comment by Barry Cipra, you may only consider the case where $a=1$)

$$ W_{a,b} (x)= e^{-2 b x}\left( \Phi^2\left(\frac{a b -x}{\sqrt{a}}\right)-\Phi\left(\frac{2 a b -x}{\sqrt{a}}\right)\right), $$


$$ \Phi(x): = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^x e^{-y^2/2} d y = \frac{1}{2} \left(Erf\left(\frac{x}{\sqrt{2}}\right)+1\right). $$

The question is whether the following equation has only three zeors: $x=0$ and $x=\pm\infty$:

$$ W_{a,b}(x) = W_{a,b}(-x). $$

Plotting the function $W_{a,b}(x)$ suggests that the answer is right. But to find a proof seems quite hard.

Here are some graphs of the functions: $W_{1,-1}(x)$ (the blue one), $W_{1,-1}(-x)$ (the red one), and $W_{1,-1}(x)-W_{1,-1}(-x)$ (the one crossing the origin).

alt text

Here is the original problem. Define

$$ E_{a,b}(x) = e^{-b x}\Phi\left(\frac{ab-x}{\sqrt{a}}\right)+e^{b x}\Phi\left(\frac{ab+x}{\sqrt{a}}\right). $$

We wish to prove that for $a>0$, $b<0$,

$$ E_{a,b}^2(x)\ge E_{a,2b}(x),\quad\text{for all $x\in R$.} $$

If one define

$$ F_{a,b}(x) =E_{a,b}^2(x)- E_{a,2b}(x), $$


$$ \frac{d F_{a,b}(x)}{d x} = -b \left( W_{a,b}(x) - W_{a,b}(-x)\right). $$

Hence, this problem reduces to the above question.

Thank you very much for any suggestions!


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    $\begingroup$ It looks like the derivative of $W_{a, b}(x)-W_{a, b}(-x)$ isn't too nasty, can show that it only changes sign twice? $\endgroup$ – Noah Schweber Mar 26 '13 at 15:08
  • $\begingroup$ Dear Noah S. The derivatives of $W_{a,b}(x)-W_{a,b}(-x)$ is a bit nasty. To prove it changes sign twice is not that easy, there are some recursions. $\endgroup$ – Anand Mar 26 '13 at 15:18
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    $\begingroup$ This may be of no help in solving the problem, but writing $b=c/\sqrt{a}$ and $x=\sqrt{a}u$ certainly simplifies the look of the expression. $\endgroup$ – Barry Cipra Mar 26 '13 at 16:22
  • $\begingroup$ Dear Barry Cipra, thanks for your comments. Please simply choose $a=1$ and a negative $b$. $\endgroup$ – Anand Mar 26 '13 at 16:24

I'll try to address the original problem directly.

It is a bit obscure with all that $\Phi$ notation, but, if I deciphered the meaning of it all correctly (please, let me know if I'm wrong), you will be completely satisfied with showing that the ratio $$ \frac{(Ee^{-b|x-\xi|})^2}{Ee^{-2b|x-\xi|}}\,, $$ where $b>0$ and $\xi$ is the standard Gaussian random variable on the line, is decreasing in $x$ when $x>0$, so its infimum is attained at infinity, where it equals $e^{-b^2}$.

Taking the (minus) logarithmic derivative with respect to $x$ and shifting the variable by $x$, we can rewrite it as $$ \frac{\int_0^\infty{e^{-bt}e^{-(x-t)^2/2}}dt-\int_0^\infty{e^{-bt}e^{-(x+t)^2/2}}dt} {\int_0^\infty{e^{-bt}e^{-(x-t)^2/2}}dt+\int_0^\infty{e^{-bt}e^{-(x+t)^2/2}}dt} \ge \frac{\int_0^\infty{e^{-2bt}e^{-(x-t)^2/2}}dt-\int_0^\infty{e^{-2bt}e^{-(x+t)^2/2}}dt} {\int_0^\infty{e^{-2bt}e^{-(x-t)^2/2}}dt+\int_0^\infty{e^{-2bt}e^{-(x+t)^2/2}}dt}\,, $$ which can be restated as $$ \frac{\int_0^\infty{e^{-bt}e^{-(x+t)^2/2}}dt} {\int_0^\infty{e^{-bt}e^{-(x-t)^2/2}}dt} \le \frac{\int_0^\infty{e^{-2bt}e^{-(x+t)^2/2}}dt} {\int_0^\infty{e^{-2bt}e^{-(x-t)^2/2}}dt}\,, $$ or, equivalently, $$ \frac{\int_0^\infty{e^{-bt}e^{-(x+t)^2/2}}dt} {\int_0^\infty{e^{-2bt}e^{-(x+t)^2/2}}dt} \le \frac{\int_0^\infty{e^{-bt}e^{-(x-t)^2/2}}dt} {\int_0^\infty{e^{-2bt}e^{-(x-t)^2/2}}dt}\,. $$ Now, for any measure $\mu$ on $(0,+\infty)$, we have $$ \log\left(\int e^{-bt}d\mu(t)\right)-\log\left(\int e^{-2bt}d\mu(t)\right)= \int_b^{2b}\frac{\int te^{-ct}\,d\mu(t)}{\int e^{-ct}\,d\mu(t)}\,dc\,. $$ Thus it suffices to show that the ratio $\frac{\int t\,d\nu(t)}{\int 1\,d\nu(t)}$ is less for $\nu_+$ than for $\nu_-$ where $d\nu_{\pm}(t)=e^{-ct}e^{-(x\pm t)^2/2}$. To this end, it suffices to show that for every $y>0$, we have $$ \frac{\nu_+([0,y])}{\nu_+([0,+\infty))}\ge \frac{\nu_-([0,y])}{\nu_-([0,+\infty))} $$ or, equivalently, $$ \frac{\nu_+([0,y])}{\nu_+([y,+\infty))}\ge \frac{\nu_-([0,y])}{\nu_-([y,+\infty))} $$ But this is obvious because the density of $\nu_-$ is just $e^{2xt}$ times that of $\nu_+$, so $\nu_-([0,y])\le e^{2xy}\nu_+([0,y])$ while $\nu_-([y,\infty])\ge e^{2xy}\nu_+([y,\infty])$.

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