# Solving a functional equation arising from a probability problem.

I am trying to find solutions to the following functional equation: $$(g(kx))^2=g(x).$$ Here, $$x$$ is in $$\mathbb{R}$$ and $$k$$ is a constant. In particular, I'm looking for solutions for $$k=2^{-1/4}$$. Furthermore, I need that the Fourier inverse of $$g$$ is a density (i.e., nonnegative and integral over $$\mathbb{R}$$ is 1). The function $$g(x)=e^{-x^4}$$ satisfies the equation, but does not have a nonnegative Fourier inverse. I have deduced that $$g(0)$$ must be one, but have little experience with functional equations and am at a loss at how to proceed. (It may very well be the case that there are no other solutions.)

The context is the following. I am tasked with finding (or showing that there exist none) i.i.d. random variables $$X$$ and $$Y$$ such that $$\frac{X+Y}{2^{1/4}}\sim X$$. If you assume that $$X$$ and $$Y$$ have density $$f$$, then standard Fourier arguments show that $$\hat{f}=g$$ must satisfy the functional equation above. The requirement that the Fourier inverse of $$g$$ be nonnegative comes from the fact that $$f$$ is a density.

Any hint, either with the functional equation or the original problem, would be greatly appreciated. In particular, with regards to the original problem, I have been able to deduce that $$E[X]$$ is either 0 or infinite, and in either case, $$E[X^2]$$ is infinite, but I am not sure how to proceed to show either existence or non-existence.

• I can tell you that the Dirac delta-distribution works in your case (very trivial, I know, but it answers the question). If $X$ and $Y$ are random variables with this distribution, then $aX+bY$ has this distribution for any $a,b\in\mathbb{C}$. Statistic independence is not needed here. However, if you assume that the random variables have absolutely continuous distribution (i.e., with probability density functions), then I am still not sure whether they exist, and if they do, which distribution they have. – Batominovski Dec 4 '18 at 19:37
• I will post more details later, but I believe (without proof yet, but hopefully soon) that no such (nontrivial) random variables exist. It seems, then that the functional equation approach is no good. I appreciate the work you put in, however! – Steve L Dec 4 '18 at 21:19
• The only non-zero solution $g$ that is analytic in an open neighborhood of $0$ to $$\Biggl(g\left(\frac{x}{2^{1/4}}\right)\Biggr)^{2}=g(x)$$ is $$g(x)=e^{\alpha x^4}.$$ I don't know what happens if $g$ is not analytic near $0$. – user593746 Dec 6 '18 at 14:35

Unfinished Attempt: I shall return!

For a fixed $$k\in\mathbb{R}$$, I shall find the general solution $$g:\mathbb{R}\to\mathbb{R}$$ to the functional equation$$\big(g(kx)\big)^2=g(x)\text{ for all }x\in\mathbb{R}\,.\tag{*}$$ It can be easily seen that $$g(0)\in\{0,1\}$$. If $$k=0$$, then it follows immediately that $$g\equiv 0$$ and $$g\equiv 1$$ are the only solutions.

If $$k=\pm1$$, then we get $$g(x)\in\{0,1\}$$ for all $$x\in\mathbb{R}$$ (with the additional requirement that $$g$$ be an even function in the case $$k=-1$$). From now on, assume that $$k\notin\{-1,0,+1\}$$. Clearly, we have $$g(x)\geq 0$$ for every $$x\in\mathbb{R}$$.

If $$k>1$$, then $$g$$ is completely determined by its behavior on $$[+1,+k)$$ and $$(-k,-1]$$ using (*), as well as the exceptional value $$\epsilon:=g(0)\in\{0,1\}$$. Let $$g_+:[+1,+k)\to\mathbb{R}_{\geq 0}$$ and $$g_-:(-k,-1]\to\mathbb{R}_{\geq 0}$$ be arbitrary. We have, for each $$x\in\mathbb{R}$$, $$g(x)=\begin{cases}\sqrt[2^{n(k,x)}]{g_+\left(\frac{x}{k^{n(k,x)}}\right)}&\text{if }x>0\,, \\\epsilon&\text{if }x=0\,,\\ \sqrt[2^{n(k,x)}]{g_-\left(\frac{x}{k^{n(k,x)}}\right)}&\text{if }x<0\,. \end{cases}\tag{#}$$ Here, $$n(k,x):=\left\lfloor\frac{\ln|x|}{\ln(k)}\right\rfloor\text{ for each }x\in\mathbb{R}\,.$$

If $$0, then $$g$$ is completely determined by its behavior on $$(+k,+1]$$ and $$[-1,-k)$$, as well as the value $$\epsilon:=g(0)\in\{0,1\}$$. Let $$g_+:(+k,+1]\to\mathbb{R}_{\geq 0}$$ and $$g_-:[-1,-k)\to\mathbb{R}_{\geq 0}$$ be arbitrary. Then, $$g$$ is also given by (#).

This leaves the case $$k<0$$ (with $$k\neq -1$$). I still need to look into the Fourier transform condition.

• For what it's worth, we must assume that $g(0)=\hat{f}(0)=1$, because $\hat{f}=0$ implies that the integral of $f$ over $\mathbb{R}$ is 1, while $g(0) = 0$ implies that the integral over $\mathbb{R}$ is 0. Also, to clarify, my definition of the Fourier transform is $$\hat{f}(x) = \int_{-\infty}^\infty e^{ixt}f(t)dt.$$ – Steve L Dec 4 '18 at 4:40
• @SteveL Thanks for the elaboration. I was first trying to solve the functional equation in the most general form (i.e., without assuming that $g$ is the Fourier transform of some probability density). Then, I will see what I can do after this. The Fourier transform condition is the most challenging part of the problem. – Batominovski Dec 4 '18 at 4:43