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

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.

$e^{-\vert x\vert}$" />

$e^{-x^4}$" />