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](https://en.wikipedia.org/wiki/Logarithmically_concave_function#Operations_preserving_log-concavity)


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.



Log-concavity of function
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. 


