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Consider $$x \mapsto \frac{\int_{x-b}^{x+b} e^{-\frac{z^2}{2} }\text d z}{\int_{x-c}^{x+c} e^{-\frac{z^2}{2} }\text d z}$$ decreasing in $x\in [0,\infty)$, if $c > b > 0$ ?

What I tried:

taking the derivative in $x$ yields:

$$\frac{\left( e^{-\frac{(x+b)^2}{2}} - e^{-\frac{(x-b)^2}{2}} \right)\int_{x-c}^{x+c} e^{-\frac{z^2}{2}}\text d z - \left( e^{-\frac{(x+c)^2}{2}} - e^{-\frac{(x-c)^2}{2}} \right)\int_{x-b}^{x+b} e^{-\frac{z^2}{2}}\text d z}{\left( \int_{x-c}^{x+c} e^{-\frac{z^2}{2}}\text d z \right)^2} $$ where the sign is determined by the numerator, so the question is if

$$\frac{e^{-\frac{(x+b)^2}{2}} - e^{-\frac{(x-b)^2}{2}}}{e^{-\frac{(x+c)^2}{2}} - e^{-\frac{(x-c)^2}{2}}} - \frac{\int_{x-b}^{x+b} e^{-\frac{z^2}{2} }\text d z}{\int_{x-c}^{x+c} e^{-\frac{z^2}{2} }\text d z} \geq 0 \ ?$$

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  • $\begingroup$ Why not applying Leibnitz rule? $\endgroup$
    – Bertrand
    Jul 7, 2019 at 13:17
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    $\begingroup$ I tried to consider the derivative. But even in "explicit" form I think its sign is hardly to determine (I have not managed it). So maybe someone else has another idea. $\endgroup$
    – Falrach
    Jul 7, 2019 at 13:42
  • $\begingroup$ It makes sense, intuitively, that it is decreasing. As $x$ increases, the slices of $e^{-\frac{x^2}{2}}$ get smaller. $\endgroup$ Jul 7, 2019 at 13:53
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    $\begingroup$ It seems that the last inequality whoud have the opposite sign, because $ e^{-\frac{(x+c)^2}{2}} - e^{-\frac{(x-c)^2}{2}}<0$. $\endgroup$ Jul 11, 2019 at 19:21
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    $\begingroup$ The equality for the derivative is equivalent to that for a fixed $x$ a function $f(b)=\frac{e^{-\frac{(x+b)^2}{2}} - e^{-\frac{(x-b)^2}{2}}}{\int_{x-b}^{x+b} e^{-\frac{z^2}{2} }dz}$ is non-decreasing for $b>0$ and the computer experiments suggest that this is true. $\endgroup$ Jul 11, 2019 at 19:21

1 Answer 1

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Let me just show you a few tricks so the next time you'll be able to figure such things out without pen or paper, not to mention MSE.

First, drop the common part. We need to show that $$ x\mapsto \frac{\int_{[x-b,x+b]}e^{-z^2/2}dz}{\int_{[x-c,x+c]\setminus[x-b,x+b]}e^{-z^2/2}dz} $$ is decreasing for $x\ge 0$.

Next, use symmetry and the fact that the integral of a family of decreasing functions over the parameter is still a decreasing function. Thus, it suffices to show that for every $0<s<b$, $$ x\mapsto \frac{e^{-(x-s)^2/2}+e^{-(x+s)^2/2}}{\int_{[x-c,x+c]\setminus[x-b,x+b]}e^{-z^2/2}\,dz} $$ is decreasing for $x\ge 0$.

Turn the fraction upside down and use the same logic to conclude that it suffices to show that for all $0<s<b<t$, $$ x\mapsto \frac{e^{-(x-t)^2/2}+e^{-(x+t)^2/2}}{e^{-(x-s)^2/2}+e^{-(x+s)^2/2}} $$ is increasing for $x\ge 0$.

Now open parentheses and cancel inessential stuff. We just need to show that for $0<s<t$, $$ x\mapsto \frac{e^{xt}+e^{-xt}}{e^{xs}+e^{-xs}} $$ is increasing for $x\ge 0$.

This can be done by hand, but here is a general principle that makes it a no-brainer. Let $f(x)=\sum_{k\ge 0}a_k x^k$ and $g(x)=\sum_{k\ge 0}b_k x^k$. If $a_k,b_k\ge 0$ and $a_k/b_k$ increases with $k$, then $f(x)/g(x)$ increases for $x\ge 0$. Indeed, let $\mu=f(x)/g(x)$ for some $x$. Then $0=f(x)-\mu g(x)=\sum_{k\ge 0}(a_k-\mu b_k)x^k$. Note that the coefficients $a_k-\mu b_k$ must change sign just once from $-$ to $+$ as $k$ increases. Let them change sign between $K$ and $K+1$. Then, if we replace $x$ by $x'>x$, all negative terms will be multiplied by at most $(x'/x)^K$ while all positive terms will be multiplied by at least $(x'/x)^{K+1}$, so the difference $f(x')-\mu g(x')$ will be positive.

The rest should be clear.

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  • $\begingroup$ Took me a while to verify all your steps. The last part I have done by hand. Somehow I do not understand how your general principle can be apllied here. By the way, it seems to be crucial to make use of the symmetry as you did. I tried it first without this symmetry argument, cause the resulting quotient is simpler. But then it does not work in the end. $\endgroup$
    – Falrach
    Jul 13, 2019 at 15:17
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    $\begingroup$ @Falrach "I do not understand how your general principle can be apllied here" We have $f(x)=e^{xt}+e^{-xt}=2\sum_{k=0}^\infty \frac{t^{2k}}{(2k)!}(x^2)^k$ and similarly for the denominator $g(x)$ with $s$ instead of $t$. Thus $a_k/b_k=(t/s)^{2k}$ (if you view $x^2$ as a new variable), which is clearly increasing. The symmetry is crucial, indeed :-) $\endgroup$
    – fedja
    Jul 13, 2019 at 19:34
  • $\begingroup$ Got it. Thanks for your help $\endgroup$
    – Falrach
    Jul 13, 2019 at 20:01

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