# Solving $F(x) = F(e^x) - F(-e^x)$

What are non-trivial functions such that for all $x \in \mathbb{R}$, $$F(x) = F(e^x) - F(-e^x)?$$

I obtained this equations after seeking for random variable $X$ with CDF $F$ such that $\log |X|$ is also following $F$.

Edit: Here is the CDF $F$ I'm looking for:

and the related density:

ECDF and empirical density have been obtained as follows in R (note that the results appear to not depend on the initial sampled distribution)

N = 100000
x = runif(N, -10, 10)
for(i in 1:100) {
x = log(abs(x))
}
hist(x, breaks = 100, xlim = c(-10,10), probability = TRUE)
plot(ecdf(x), ylab = "F(x)")

• So you want $f$ be positive and Lebesgue-integrable as well? – user228113 Mar 21 '18 at 10:33
• Are you sure about your functional equation? I don't see how it follows from your requirement. – Raskolnikov Mar 21 '18 at 10:37
• User @Raskolnikov has a good point. Truth to be told, it seems like you that your $f$ is the CDF, rather than the density itself. – user228113 Mar 21 '18 at 10:40
• Thank you, yes it is CDF, so I just edited the message. I would like a general solution if possible. If few CDF verify this condition, maybe they also have some name – ahstat Mar 21 '18 at 12:14
• I could only find one article about this dynamic system. Maybe one of the references contains what you're looking for. – Raskolnikov Mar 23 '18 at 19:39

So, I found the CDF it is

$$F(x)=\exp(-\exp(-\exp(x)))-\exp(-\exp(\exp(x)) \; .$$

Just kidding. It's not quite that, but I think the chance of having a closed form for this distribution is small. This is however a very good approximation of the distribution and I'm going to explain how you can get even better approximations and how I came up with the procedure.

I started by studying the iterates of function $\ln|x|$ in Desmos. This is two iterations:

This is three iterations:

This is after ten iterations:

I realized that if I could somehow capture the density of the image of those iterates, I would find the distribution. I also realized that since pretty much any starting value, except those on the orbit leading to $+\infty$ and some repelling orbits, would lead to the full distribution, I probably only needed to look at a few iterations and look at the density of one "branch" of the iterate. I tried for three iterations and the result was not so good. I tried with 4 iterations and the result was worse. So I went back to 2 iterations and inverted the function on the interval $[0,1]$ getting $\exp(-\exp(x))$. (Actually, I went one step further interpreting everything in terms of distributions but it would take me too much time to explain the whole thought process.) So I got as a distribution:

$$F(x)=1-\exp(-\exp(x))$$

which gives the following approximation:

Which is really good, but obviously doesn't satisfy the requirement

$$F(x)=F(\exp(x))-F(-\exp(x))$$

But that's where the second step of my thought process kicked in. Since we want the function to satisfy this property, let's reinterpret this equation as a recursion

$$F_{n+1}(x)=F_n(\exp(x))-F_n(-\exp(x))$$

Applying one step of this recursion to the CDF $F(x)=1-\exp(-\exp(x))$ gives the approximation I have in the beginning. And obviously you can repeat the process and this seems like it results in a stable recursion. You might run into problems though with having towers of exponentials from a numerical point of view. But it is clear that the limiting function has to be somehow a combination of infinite exponential towers.

Surely, there's more to say about this, but being no expert on dynamical systems and the stability of recursions on functions, I don't know how to make the whole more rigourous. Hope you liked the answer though.

• I liked this answer :) I wish we could express the CDF as a special function, for example combining some Lambert W-like functions. – ahstat Mar 25 '18 at 13:50

My only idea is, suppose that $F$ is analytic. Then $F = \sum a_n x^n$.

$$F = \sum a_n x^n = \sum a_n e^{nx}-\sum a_n (-1)^ne^{nx} = \sum_{n \text{ even}} 2a_n e^{nx}.$$

Now recall that $e^x=\sum \frac{x^m}{m!}$ and substitute.

$$F = \sum a_n x^n = \sum_{n \text{ even}} 2a_n e^{nx} = \sum_{n \text{ even}} 2a_n \sum_m \frac{(nx)^m}{m!} = \sum_m \sum_{n \text{ even}} \frac{2a_nn^m}{m!}x^m.$$

Thus, it looks to me that one gets $$a_n =\frac{1}{n!} \sum_{k \text{ even}} 2a_kk^n.$$

In the case of $a_0$, just to focus on something precise, one gets:

$$a_0 = - \sum_{k >0 \text{ & even }} 2a_k.$$

This specifies an infinite matrix: $$A = (a_{mn})=\frac{1}{m!}2n^m$$ if $m$ is even, otherwise is $0$ and the requirement is equivalent to ask for an $1$-eigenvector of $A$.

Does this help?!