Consider a symmetric, unimodal distribution $f(x)$ such that $\int_0^\infty f(x)\,dx > 1/2$. I want to prove or disprove the following:

$$ \int_0^\infty \int_{-x}^0 f(x)f(y)\,dy\,dx > \int_0^\infty \int_{-\infty}^{-x} f(x)f(y)\,dy\,dx. $$

The problem is depicted for a jointly Gaussian distribution in the following figure: Inequality for jointly Gaussian distribution

The grey area in the figure corresponds to LHS, whereas the yellow area corresponds to RHS. My view is that since the distribution has higher values in the grey area, the inequality should be true. My first attempt at proving this was to use Gauss's Inequality, but have no idea how to apply it in this scenario. Any help or hint is appreciated.

A side question is do we really need the unimodality condition on the distribution or all symmetrical distributions can satisfy the inequality?


1 Answer 1


If $X$ and $Y$ are two independent samples of the distribution, you're asking whether $P(X > -Y \mid X > 0, Y < 0) > 1/2$. This is not always true.

Note that the question only depends on the conditional distribution of $X$ given $X > 0$ and of $Y$ given $Y < 0$. That is, it only matters what the two halves of the distribution look like, not their relative weight. To find a counterexample, we want to arrange things so that the distribution has a long tail on the left but is concentrated near 0 on the right. We can achieve this by giving the distribution a long tail on the left, but making $P(Y < 0)$ small, so the corresponding long tail on the right is not significant.

For example, we can take an unnormalized density function which interpolates linearly through the points

$$(-1/\varepsilon, 0), \; (0, \varepsilon^2), \; (1,1), \; (2,\varepsilon^2), \; \text{and }(2+1/\varepsilon,0).$$

Then as $\varepsilon \rightarrow 0$, the probability $P( X < 2 \mid X > 0)$ approaches 1, and the probability $P(-Y > 2 \mid Y < 0)$ also approaches 1. So the probability $P(X > -Y \mid X>0, Y<0)$ approaches 0; in particular it is less than $1/2$ for sufficiently small $\varepsilon$.

  • $\begingroup$ I don't understand your second statement, i.e., $P(-Y>2 | Y<0) > 2$ is almost 1 ? Assuming you are saying $P(-Y>2 | Y<0)$ is 1, the conditional distribution $f(y|Y<0)$ will still have some mass in the region $-2<Y<0$ and because of the unimodality, it will be significant. So, how can we say that $P(-Y>2|Y<0)$ is 1? Can you please elaborate the steps a little further? Also, can you comment on under what conditions you think the inequality will be true? $\endgroup$
    – ubaabd
    Jan 3, 2014 at 16:57
  • $\begingroup$ @ubaabd: I mean $P(−Y>2 \mid Y<0)\rightarrow 1$ as $\varepsilon \rightarrow 0$. I edited the answer to make it a bit clearer. Unimodality doesn't matter: we have a lot of room to the left of $Y=-2$ to work with. In fact, unimodality is not a natural notion to consider for this problem, because it's not preserved by reparametrizing the positive and negative reals. You can take a unimodal distribution and make a new mode by squeezing a wide region into a narrow one on both sides of the origin, but this doesn't affect whether the distribution satisfies the inequality in the problem. $\endgroup$ Jan 3, 2014 at 18:06

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