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.

All functions $F,G$ and $H$ are positive by the Cauchy-Schwarz inequality.

So to summarize for

• slow decay - $G$ has a unique minimum (Fig. 1)

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

• medium decay-$F$ is a constant function.
• fast decay - $H$ has a unique maximum (Fig. 2).

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

I also checked the function $e^{-\vert x \vert^{2+10^{-5}}}$, i.e. with an exponent slightly above $2$

$e^{-\vert x \vert^{2+10^{-5}}}$" />

and it seems the function has the expected behaviour, i.e. there exists a unique maximum at zero and the function decays from there.

One sees the opposite effect by plotting

Addendum:

I also checked for log-convexity/log-concavity. Apparently, the functions $G,H$ are neither log-convex nor log-concave.

See here a plot for $\log(G)$

And finally, I computed the first derivative for convolving with $e^{-\vert x \vert^p}$ at $p=2$ by numerically differentiating (so be careful about the result please