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I am trying to find solutions to the following functional equation: $$g(kx)^2=g(x)\text.$$ 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.

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    $\begingroup$ 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. $\endgroup$ Commented Dec 4, 2018 at 19:37
  • $\begingroup$ 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! $\endgroup$
    – Phil
    Commented Dec 4, 2018 at 21:19
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    $\begingroup$ 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 $0$. $\endgroup$
    – user593746
    Commented Dec 6, 2018 at 14:35

1 Answer 1

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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 $0<k<1$, 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.

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    $\begingroup$ 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 $$\hat{f}(x) = \int_{-\infty}^\infty e^{ixt}f(t)dt.$$ $\endgroup$
    – Phil
    Commented Dec 4, 2018 at 4:40
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    $\begingroup$ @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. $\endgroup$ Commented Dec 4, 2018 at 4:43

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