A property about probability distribution 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.  
 A: 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 $[n-1,n]$. 
A: 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 $F_\beta(\alpha) = F(\alpha)$.
