I would like to ask a follow up on a [question I asked some days ago][1].

Consider the function 

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

My goal then was to analyze  

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

and Iosif Pinelis showed that this expression is constant.

Now instead of convolving with $f_0(x)=e^{-x^2}$ one could convolve with a function that decays slower than $e^{-x^2}$ for example $g_0(x)=e^{-\vert x \vert}.$ That is, we then get 

$$ G(y):=\frac{(f_2*g_0)(y)}{(f_0*g_0)(y)}- \left(\frac{(f_1*g_0)(y) }{(f_0*g_0)(y)}\right)^2.$$

Mathematica shows that $G$ is now not at all constant but has a unique maximum at $0$ and decreases from there. 

On the other hand one can consider the faster decaying function $h_0(x)=e^{-x^4}$ and consider

$$ H(y):=\frac{(f_2*h_0)(y)}{(f_0*h_0)(y)}- \left(\frac{(f_1*h_0)(y) }{(f_0*h_0)(y)}\right)^2.$$

In this case, the function has a unique minimum at zero and increases from there. 

So to summarize for 

 - slow decay - $F$ has a unique maximum (Fig. 1).
 - medium decay-$G$ is a constant function.
 - fast decay - $H$ has a unique minimum (Fig. 3).

Can one explain this phase transition? I find it very non-obvious from the expressions.

[![$e^{-\vert x\vert}$][2]][2]

[![$e^{-x^4}$][3]][3]


  [1]: https://mathoverflow.net/questions/350415/log-concavity-of-function
  [2]: https://i.sstatic.net/jAVuh.jpg
  [3]: https://i.sstatic.net/qM06f.jpg