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:

F(x) CDF I'm looking for

and the related density:

Density related to CDF F

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)")
  • $\begingroup$ So you want $f$ be positive and Lebesgue-integrable as well? $\endgroup$ – user228113 Mar 21 '18 at 10:33
  • $\begingroup$ Are you sure about your functional equation? I don't see how it follows from your requirement. $\endgroup$ – Raskolnikov Mar 21 '18 at 10:37
  • 2
    $\begingroup$ 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. $\endgroup$ – user228113 Mar 21 '18 at 10:40
  • $\begingroup$ 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 $\endgroup$ – ahstat Mar 21 '18 at 12:14
  • 1
    $\begingroup$ I could only find one article about this dynamic system. Maybe one of the references contains what you're looking for. $\endgroup$ – Raskolnikov Mar 23 '18 at 19:39

So, I found the CDF it is

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

Plot of CDF and PDF

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:

Two iterations

This is three iterations:

Three iterations

This is after ten iterations:

enter image description here

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:


which gives the following approximation:

enter image description here

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


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


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.

  • $\begingroup$ I liked this answer :) I wish we could express the CDF as a special function, for example combining some Lambert W-like functions. $\endgroup$ – 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?!


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