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.


2 Answers 2


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 $[n-1,n]$.

  • $\begingroup$ Thanks for your quick reply. What about unimodal distributions? $\endgroup$
    – Peter
    Aug 13, 2019 at 3:19
  • 2
    $\begingroup$ It looks like there is indeed a tendency for $F$ to be increasing. Your conjecture may indeed by true in the unimodal case. Can you share how you came up with this conjecture? $\endgroup$ Aug 13, 2019 at 3:28
  • $\begingroup$ I come across it when I study a problem of informed trading. The basic intuition is when more informed traders trade on the same information, their trading is detected more easily by others so it is less profitable for a single informed trader. g(x) here refers to the distribution of noisy trading that hides informed trading. a refers to the volume of informed trading. $\endgroup$
    – Peter
    Aug 13, 2019 at 4:12

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 $F_\beta(\alpha) = F(\alpha)$.

  • $\begingroup$ If $\lim_{x\to \pm\infty}g(x)$...what? Not clear what this limit should be. Otherwise this looks very convincing. $\endgroup$ Aug 13, 2019 at 13:58
  • $\begingroup$ @David Roberts: Thank you for your hint. Of course this must be $0$. $\endgroup$ Aug 13, 2019 at 14:05
  • $\begingroup$ Very nice answer! $\endgroup$ Aug 13, 2019 at 14:06
  • $\begingroup$ Thanks for your reply. What about a g(x) like this? It's zero over the negative domain and it's decreasing over the positive domain. $F(\alpha)$ is increasing in this case but it also satisfies that its limits are zero when x goes to positive and negative infinity. $\endgroup$
    – Peter
    Aug 13, 2019 at 16:42
  • $\begingroup$ @Peter: The formulation of my answer may be misleading. Actually I show that for a density $g$ with the property given at least of the densities $x \to g(x)$, $x \to g(-x)$, $x \to g(x+\beta)$ or $x \to g(\beta-x)$ has the property that $F$ is not increasing everywhere. An example of this is $g(x) = e^{-x}, x > 0$. Here $F(\alpha) = \frac{1}{e^{-\alpha}+k}$ is increasing, but for $g(x) = e^x, x < 0$ we have $F(\alpha) = \frac{1}{e^\alpha+k}$, which is decreasing. You have to add suitable further assumptions..My answer doesn't say that $F$ is never increaing. $\endgroup$ Aug 13, 2019 at 18:12

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