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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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It depends on whether this is supposed to be true for all $y \gt 0$, or whether your definition of unimodality is a density strictly increasing below the mode and strictly decreasing above $0$. If both can be answered "no" then there are counter-examples. – Henry Apr 24 '12 at 21:09
Strictly increasing below 0 and strictly decreasing above 0 (with maximum at 0) are assumptions I'm OK with. Thanks for taking a look..! – Johan Apr 24 '12 at 21:11
Are you wanting the equality to hold for all $y$ or is the condition that there exists some $y$ for which the two integrals are equal? – Dilip Sarwate Apr 24 '12 at 21:12
for all $y \ge 0$. Updating question to reflect these two corrections. – Johan Apr 24 '12 at 21:12
up vote 1 down vote accepted

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

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


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

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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The first proof, where $0 < y \le x$, is an approach I've taken, but I can't get it to handle the case of $y > x$. The second proof intrigues me, but I don't see how the limit y->$\infty$ addresses all y. Can you elaborate? Many thanks for the help. – Johan Apr 25 '12 at 1:06
The first proof is easily extended to $0 \lt y \le 2x$. For $y \gt 2x$ it is a little more complicated so I have added to my answer. In the second proof, if the difference were zero for all $y$ then it would also be zero in the limit, since you have convergence and a probability is bounded above by $1$. – Henry Apr 25 '12 at 1:22
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! – Johan Apr 25 '12 at 1:35

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