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I am trying to prove the following proposition:

Proposition. Given a unimodal probability distribution $f(u)$, symmetric around $u=0$, strictly increasing for u<0 and strictly decreasing for $u>0$, then for all $y \ge 0$, $ \int_{x-y}^x f(u)du - \int_x^{x+y} f(u)du = 0 \hspace{0.1in} \Leftrightarrow \hspace{0.1in} x=0.$

The proof of $\Leftarrow$ is simple (plug in $x=0$), but the proof of $\Rightarrow$ is escaping me. Even though it feels very intuitive, I can't seem to nail it down rigorously. Is it true?

(In terms of strategy, I'm trying is to break it up in three cases: $x>0$, $x=0$, x<0, and try and find a contradiction for the two $x \ne 0$ cases, but I'm not getting it.)

(Assumptions about continuous differentiability of $f$ are fine if necessary.)

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    for all $y \ge 0$. Updating question to reflect these two corrections.2012-04-24

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This first proof uses strictly decreasing and shows you cannot have the equality for any $x$ and $y$ with $0 \lt y \le x$

If, for $x \gt 0$, $f(u)$ is strictly decreasing for positive $u$ then for $0 \lt z \le y \le 2x$ you have $f(x-z) \gt f(x) \gt f(x+z)$ so

$\int_{x-y}^x f(u)du \gt \int_{x-y}^x f(x)du = y f(x) = \int_x^{x+y} f(x)du \gt \int_x^{x+y} f(u)du$

so

$\int_{x-y}^x f(u)du - \int_x^{x+y} f(u)du \gt 0.$

For $y\gt 2x \gt 0$ it gets a little longer as:

$\int_{x-y}^x f(u)du -\int_x^{x+y} f(u)du $ $= \int_{-x}^x f(u)du + \int_{x-y}^{-x} f(u)du -\int_x^{x+y} f(u)du $ $= \int_{-x}^x f(u)du + \int_{x}^{y-x} f(u)du - \int_x^{x+y} f(u)du $ $= \int_{-x}^x f(u)du - \int_{y-x}^{x+y} f(u)du$ $\gt 2x f(x) -2xf(x) = 0$

If $x \lt 0$, do something similar, reversing the inequalities where necessary.


Alternatively, not using strictly decreasing, but using all $y \gt 0$,

$\lim_{y \to +\infty} \left( \int_{x-y}^x f(u)du - \int_x^{x+y} f(u)du \right) $ $ = \Pr(X \lt x)-\Pr(X \gt x) = \Pr(X \gt -x) - \Pr(X \gt x) = \Pr(-x \lt X \le x) $

which is positive if $x \gt 0$ and $X$ has a positive probability for any open interval including $0$, which it must do if $0$ is the mode.

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    Thanks for the second proof. It seems to me that the statement holding in the limit does not imply that it must hold for all y>0 (your statement is the reverse implication), but I went through a more careful bookkeeping of the y-dependent tails without a limit, and the proof still holds. Your trick of switching to an integral on (-x,x) was key. Many thanks!2012-04-25