2
$\begingroup$

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).

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

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

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

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
2
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.