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The main object of my question is this: if $X$ has a log-normal distribution, $Y = X + Z$ and $Y$ has the same distribution as that of $Z^2$ (in other words, $F_{Z^2} = F_{X+Z}$) and $X, Z$ are independent, then what is the distribution of $Z$? See below a potential example with $X$ being log-normal, and $\log \log Z$ being almost, maybe exactly normal (its CDF pictured below, based on empirical calculations):

enter image description here

Below are details explaining why I came up with this problem, and why I am interested in it. Let's say that $Y = X + Z$, with $X$ and $Z$ being independent, and $Y$ having the same distribution as $Z^2$. In addition, $X \geq 0$ and $Y, Z \geq 1$. The latter condition is needed for reasons that are due to the nature of my problem, and explained below.

If the distribution of $Z$ is known, how do I find the distribution of $X$? Or conversely? Of course one can use the characteristic functions and the convolution theorem. I am interested here in finding the most simplest distributions for $X$ and $Y$, satisfying all the requirements.

The problem is as follows. We are dealing with

$$Z=\sqrt{X_1+\sqrt{X_2+\sqrt{X_3+\cdots}}},$$

where the $X_k$'s are i.i.d. with same distribution as $X$. I am trying to find $X$ such that $Z$ has a simple distribution. The problem seems easier to solve backward: finding $Z$ such that $X$ has a simple distribution.

I tried a log-normal distribution for $X$, more precisely $X=\exp(T)$ with $T \sim$ Normal$(0,1)$. It is pretty obvious that $Z \geq 1$ with probability one. Based on empirical evidence, $\log \log \beta Z$ is pretty well approximated if $\beta = 1$. Is it possible to compute the actual distribution of $Z$? Is my approximation exact? (Update, not, it's just an approximation, not an exact match) If not I am interested in any combination of $X$ and $Z$ where both distributions are simple ones. This is a follow up to my previous question on CV, see here. (Update: a summary about all this research can be found here.)

Vincent Granville
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  • How can $Y=_d Z^2$ and yet $Y=X+Z$? – AdamO Oct 22 '19 at 22:07
  • @AdamO: What makes you think it is impossible? Maybe it's impossible, but would like to see an explanation. – Vincent Granville Oct 23 '19 at 00:30
  • It may be worth noting in your first paragraph that Y and Z are dependent (as your later explanation reveals); you don't say they aren't (so it's natural to consider that they might be), but the particular form of dependence revealed later matters. – Glen_b Oct 23 '19 at 03:22
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    Cool question. The convergence for a number of X distributions I tried looks pretty rapid. – Glen_b Oct 23 '19 at 03:35
  • @Glen_b: Could you elaborate more on this? My question is by no mean trivial, I've spent days on this and so far I was only able to get approximations to a solution. Very good approximations for sure, but not a solution -- if there is one to begin with. Of course, $Z^2= X+ Z$, but if it was that simple, we would have $Z=(1+\sqrt{4X})/2$. That's not the case. – Vincent Granville Oct 23 '19 at 04:17
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    What is it you're asking me to elaborate more on? I can't tell whether it's the first comment or the second. In the first case I was hoping to avoid the case where someone inadvertently tried to proceed from Y=X+Z by starting with Y and Z (and perhaps implicitly treating them as independent without recognizing that it matters). If you mean the second comment, I am not sure there's much more I can say -- if you try a few common distributions for $X_i$ that fit the conditions, the sequence $Z_i=\sqrt{X_i+Z_{i-1}}$ progresses quite rapidly to what appears to be a stationary distribution ... ctd – Glen_b Oct 23 '19 at 04:21
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    ctd... some with means that appear tantalizingly close to simple expressions. Be careful writing $Z^2=X+Z$. You don't mean that! You mean instead that $F_{Z^2} = F_{X+Z}$ – Glen_b Oct 23 '19 at 04:22
  • Of course, yes Glen. – Vincent Granville Oct 23 '19 at 04:27
  • Yes Glen, as you mentioned, it's about $Z_i =\sqrt{X_i + Z_{i-1}}$ with $Z= \lim_{i\rightarrow \infty} Z_i$, assuming the limit (which is about convergence in distribution) exists. – Vincent Granville Oct 23 '19 at 04:32
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    I believe the answer to your question is that such variables $(X,Y)$ do not exist. That is simply because sums of lognormals are not lognormal, even though squares and square roots of lognormals are lognormal. – whuber Oct 24 '19 at 21:18
  • @Whuber: You might be right. See however my chart at the bottom of this post https://math.stackexchange.com/questions/3405042/question-about-the-log-normal-distribution: $X$ is log-normal, and $\log \log Z$ is almost normal, but maybe not exactly. – Vincent Granville Oct 24 '19 at 23:22
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    The key is "not exactly." Indeed, in the geostatistical literature there is a concept of "conservation of lognormality" that recognizes both (a) sums of lognormals are not lognormal but (b) for many analytical purposes they can be modeled with lognormal distributions. – whuber Oct 25 '19 at 12:36
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    @Whuber: I believe you. There was actually another instance involving nested square roots where $F_Z(z) = \log_2 z (1\leq z \leq 2)$ almost exactly, but NOT exactly (using a discrete distribution for $X$). – Vincent Granville Oct 25 '19 at 15:52
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    I think you can prove this negative result in a straightforward manner: assuming the sum of independent lognormal variables $Z=X+Y$ is also lognormal, you get two formulas for the moments of $Z$ (one from the sum and the other from its lognormality). Comparing means and variances will show the mean and variance of $Y$ are determined by those of $X$ (already demonstrating the general result, but not ruling out some special cases); then by comparing the third moments you can show there are no solutions. – whuber Oct 25 '19 at 16:08
  • @Whuber: I plan on posting a solution that does not involve log-normal, rather $f_Z =\frac{2}{3} z$ with $z\in [1, 2]$. This could be the only case not involving a very complicated characteristic function for $X$, and even then, I'm not sure $X$'s CF is really is a characteristic function to begin with. – Vincent Granville Oct 25 '19 at 16:36
  • @Whuber: I posted my tentative solution at https://stats.stackexchange.com/questions/432136/random-variables-x-z-such-that-z-and-sqrtx-z-have-the-same-distribut/ – Vincent Granville Oct 25 '19 at 17:00

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