Let $p$ be a Lebesgue density function with infinite support (i.e. $p(x)>0 \forall x\in \mathbb{R}$ and $\int p(x) dx = 1$). Moreover, assume that $p$ is even (i.e. $p(x) = p(-x)$) and that $p(x) \geq p(x')$ for $0 \leq x \leq x'$ so $p$ is monotonically non-increasing after zero and monotonically non-decreasing before zero. Examples of density that satisfy these properties are the Gaussian, Cauchy and Laplace densities. Now define $$
H(\mu) := \int^{\infty}_{-\infty} \frac{p(x-\mu)}{0.5p(x-c) + 0.5p(x+c)} p(x) d x,
$$
for some $c > 0$. One can show that $H(-\mu) = H(\mu)$ (use substitution with $y=-x$). Using this, one can easily show that $2H(c) = H(c) + H(-c) = 2$ by writing the sum as one integral which simplifies a lot. This implies that $H(c) =1.$

I am now trying to show that $H(\mu) \leq 1$ for $|\mu| \geq c$. As $H$ is an even function, it is sufficient to show that $H(\mu) \leq 1$ for $\mu \geq c$.

This can be shown for Gaussians using an additional monotonicity assumption (see [this post][1]). Such assumption does not hold for the Cauchy densities. However, computing the integral using Monte Carlo approximations still seems to show that $H(\mu) \leq 1$ for $|\mu| \geq c$ for Cauchy.

I am not a 100% sure whether my current assumptions are sufficient. I at least haven't found a counterexample yet. If anyone knows one, this would be also highly valuable. 

Any other insights would also be much appreciated!


  [1]: https://math.stackexchange.com/questions/4997194/expectation-of-likelihood-ratio-involving-gaussian-pdfs?noredirect=1#comment10713216_4997194