# Whether does the following equation have only one finite zero?

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 $$a=1$$)

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

where

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

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).

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

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

then

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

Hence, this problem reduces to the above question.

Thank you very much for any suggestions!

Anand

• 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? – Noah Schweber Mar 26 '13 at 15:08
• 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. – Anand Mar 26 '13 at 15:18
• 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. – Barry Cipra Mar 26 '13 at 16:22
• Dear Barry Cipra, thanks for your comments. Please simply choose $a=1$ and a negative $b$. – Anand Mar 26 '13 at 16:24

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 $$\nu_-([y,\infty])\ge e^{2xy}\nu_+([y,\infty])$$.