Suppose $g(x)$ is a pdf function and k is a positive real number. Let $F(\alpha)=\int_{-\infty}^{\infty}\frac{1}{\frac{g(x+\alpha)}{g(x)}+k}g(x)dx$, where $\alpha$ is positive.
I feel $F(\alpha)$ is increasing in $\alpha$. But I don't know how to prove it for general $g(x)$. Or maybe it is only right for some $g(x)$.
Could anyone provide some ideas on this property? Thanks!
Thank Iosif Pinelis for showing it is not true for general distributions. Maybe we should focus on unimodal distributions.
This conjecture is not true in general. Indeed, let $a:=\alpha$, $k:=1$, $n:=100$, and $g(x) :=\frac12\,1_{0 < x < 1} + \frac12\,1_{n < x < n+1}$. Then $F(a)=\frac{n+3-a}{4}$ for $a\in[n-1,n]$, so that $F$ is decreasing on $[n-1,n]$.
If $\lim_{x \to \pm \infty} g(x) = 0$ then the assertion is false. Define $h(x) := g(-x)$, $H(\alpha) := \int_{-\infty}^\infty \frac{1}{\frac{h(x+\alpha)}{h(x)}+k} ~ h(x) ~ dx = \int_{-\infty}^\infty \frac{1}{\frac{g(-x-\alpha)}{g(-x)}+k} ~ g(-x) ~ dx = \int_{-\infty}^\infty \frac{1}{\frac{g(x-\alpha)}{g(x)}+k} ~ g(x) ~ dx$. (We only need this since $\alpha \geq 0$.) Then again $h$ is a density and by Lebesgue's theorem on dominated convergence $\lim_{\alpha \to \infty} F(\alpha) = \frac{1}{k} = \lim_{\alpha \to \infty} H(\alpha)$. Define $F(\alpha) := H(-\alpha)$ for $\alpha <= 0$. If the assertion is always true then $\alpha \to F(\alpha)$ is always decreasing for $\alpha \leq 0$ and increasing for $\alpha \geq 0$.
But this then also holds true for $F_\beta(\alpha) := \int_{-\infty}^\infty \frac{1}{\frac{g(x+\beta-\alpha)}{g(x+\beta)}+k} ~ g(x+\beta) ~ dx = F(\alpha-\beta)$, $\beta \in \mathbb{R}$. As a consequence any $\beta \in \mathbb{R}$ is a minimum point of $F$, i.e. $F \equiv \frac{1}{k}$. This is only possible if $g \equiv 0$, a contradiction.
This sort of reasoning of course holds for unimodal densities too.
Edit: This answer is not correct. Actually $F_\beta(\alpha) = F(\alpha)$.
