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Consider the function

$$f_{n}(x)=e^{-x^2}x^n.$$

My goal is to show that

$$ G(y):=\frac{(f_2*f_0)(y)}{(f_0*f_0)(y)}- \left(\frac{(f_1*f_0)(y) }{(f_0*f_0)(y)}\right)^2$$

is log-concave.

Let us first observe that indeed $G(y) \ge 0.$

This just follows from a Cauchy-Schwarz

$$(f_1*f_0)(y) \le \sqrt{(f_2*f_0)(y)(f_0*f_0)(y)}$$

so everything is well-defined.

Usually one can say a lot when convolutions are involved about log-concavity due to standard theorems see wikipedia

but this combination looks a bit tricky.

Addendum I should add that I am in particular very interested in theoretical insights why this particular expression has to be log-concave.

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Direct calculations show that $$(f_2*f_0)(y)=\frac{1}{4} \sqrt{\frac{\pi }{2}} e^{-\frac{y^2}{2}} \left(y^2+1\right), $$ $$(f_1*f_0)(y)=\frac{1}{2} \sqrt{\frac{\pi }{2}} e^{-\frac{y^2}{2}} y, $$ $$(f_0*f_0)(y)=\sqrt{\frac{\pi }{2}} e^{-\frac{y^2}{2}} $$ for all real $y$, so that $G$ is the constant $1/4$ and hence log concave.

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  • $\begingroup$ thank you. although i must admit i was curious whether there is something behind my expression that explains why it has to be log-concave. $\endgroup$
    – Landauer
    Commented Jan 14, 2020 at 17:32
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    $\begingroup$ @Martinique : I doubt that a "behind-the-expression" reasoning exists, in particular given that $G$ is a constant and thus just "barely" log concave. $\endgroup$
    – Iosif Pinelis
    Commented Jan 14, 2020 at 17:36
  • $\begingroup$ I should say that- although not related to log-concavity-there is some interesting structure that I noticed which I ask about in this post mathoverflow.net/questions/350524/… $\endgroup$
    – Landauer
    Commented Jan 16, 2020 at 15:07

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