Integral of product of distribution functions

Question

We know that definite integrals over $[0,L]$ of non-uniform distribution functions $f(x)$ and $g(x)$ equal $1$ for both. The functions can overlap or not, but the distribution isn't otherwise specified, it could be lognormal, gaussian etc. Does it hold for some case that integral over $[0,L]$ for product function $f(x)g(x)$ is greater than one?

Answer 1

In a nutshell, it can be any number.

Example 1. Consider the case where $a > 0$ and $f = g$ is the density of the uniform distribution over $[-\frac{a}{2}, \frac{a}{2}]$. Then

$$ \int_{-\infty}^{\infty} f(x)g(x) \, dx = \int_{-a/2}^{a/2} \frac{1}{a^2} \, dx = \frac{1}{a}. $$

Example 2. For each $a > 0$ we can check that

$$ f(x) = g(x) = \dfrac{(-\log x)^{a}}{\Gamma(a+1)} \mathbf{1}_{(0,1)}(x) $$

is a density of some distribution, i.e., it is non-negative and $\int_{0}^{1}f(x) \, dx = \int_{0}^{1} g(x) \, dx = 1$. Then it follows that

$$ \int_{-\infty}^{\infty} f(x)g(x) \, dx = \int_{0}^{1} \frac{(-\log x)^{2a}}{\Gamma(a+1)^2} \, dx = \frac{\Gamma(2a+1)}{\Gamma(a+1)^2}. $$

Again this can be made arbitrarily large by choosing large $a$.

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Remark. Both examples are created by considering distributions that contrates near $0$. Although mathematically nonsensical, the corresponding mental image is that

$$ \int_{-\infty}^{\infty} \delta(x)^2 \, dx \ {``}=\text{''} \ \infty $$

for the "Dirac delta function" $\delta(x)$ which is the "density" of the point mass concentrated at $0$. Both examples can be recognized as a perturbed version of this heuristics.