I am looking for the distribution of a random variable $Z$ defined as

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

Here the $X_k$'s are i.i.d. and have same distribution as $X$.

1. Update

I am looking to find a simple distribution for $X_k$, that results in a simple distribution for the nested square root $Z$. Thus my idea to investigate distributions stable under some particular transformations. But this may not be the easiest way.

I tried a Bernoulli (with parameter $\frac{1}{2}$) for $X_k$, but this leads to some very difficult, nasty stuff, and a distribution on $[1, \frac{1+\sqrt{5}}{2}]$ full of gaps - some really big - for $Z$. So far the most promising result is the following.

Use a discrete distribution for $X_k$, taking on three possible values $0, 1, 2$ with the probabilities

  • $P(X_k = 0) = p_1$
  • $P(X_k = 1) = p_2$
  • $P(X_k = 2) = p_3 = 1-p_1-p_2$.

Now the resulting domain for $Z$'s distribution is $[1, 2]$, and the gaps are eliminated. The resulting distribution is still very wild, unless $p_1, p_2, p_3$ are carefully chosen. Consider

  • $p_1=\sqrt{5\sqrt{2}-1}-2$,
  • $p_2=\sqrt{5\sqrt{3}-1}-\sqrt{5\sqrt{2}-1}$,
  • $p_3=3-\sqrt{5\sqrt{3}-1}$.

I was naively thinking that this would lead to $Z$ being uniform on $[1, 2]$, based on the table featured in my article Number Representation Systems Explained in One Picture (published here, see column labeled "nested square root", with row labeled "digits distribution".) But $Z$ does not appear to be uniform, though it does appear to be well behaved: it looks like $F_Z(z)$ is a polynomial of degree 2 if $z\in [1, 2]$. Then I modified a bit the values of $p_1, p_2, p_3$, removing 0.02 to $p_1$ and adding 0.02 to $p_3$. The result for $Z$ looks much closer to uniform on $[1, 2]$ this time.

Anyway, that's where I am now. My re-formulated question is: with appropriate values for $p_1, p_2, p_3$ (and what would these values be?) can we have a simple distribution for $Z$? (uniform or polynomial on $[1,2]$)

Note: With the particular discrete distribution in question, the support domain for $Z$ is $[1, 2]$. Sure, if all $X_k$ are zero, then $Z=0$ but that happens with probability zero. If all but one of the $X_k$ is zero, then $Z\geq 1$.

2. Second update

Regarding my statement I was naively thinking that this would lead to $Z$ being uniform on $[1, 2]$. I think the reason that it doesn't is because for this to happen, the $X_k$'s would need to have the right auto-correlation structure required to form a normal number in the numeration system based on infinite nested radicals. In my experiment, I used i.i.d. $X_k$'s. But for normal numbers (in that system) lag-1 auto-correlation between successive digits (the $X_k$'s being the digits) is close to zero, but not exactly zero. By contrast, in the binary numeration system, the digits $X_k$'s of normal numbers are not correlated, and thus if $X_k$ is Bernouilli of parameter $p=\frac{1}{2}$, then $Z = \sum_{k=0}^\infty X_k \cdot 2^{-k}$ is uniform on $[0, 1]$. But if $p\neq \frac{1}{2}$, then the distribution of $Z$ is pretty wild, see here.

3. Third update

Assume the $X_k$'s are i.i.d. with the discrete distribution mentioned earlier. Then the density $f$ associated with $Z$, if it exists, must satisfy:

  • $z \in ]1,\sqrt{2}[\Rightarrow f(z) = 2p_1 z f(z^2)$

  • $z \in ]\sqrt{2},\sqrt{3}[\Rightarrow f(z) = 2p_2 z f(z^2-1)$

  • $z \in ]\sqrt{3},2[\Rightarrow f(z) = 2p_3 z f(z^2-2)$

This excludes the possibility that $Z$'s distribution is as simple as a finite polynomial, regardless of $p_1, p_2, p_3$. Also, at $z=1, \sqrt{2}, \sqrt{3}$ or $2$, $f(z)$ may be zero, infinite, not exist or be discontinuous.

Finally, if $f(z)$ is properly defined (not zero or infinite) at $z=(1+\sqrt{5})/2$, then we have $p_2 = 1/(1+\sqrt{5})$: this is a direct result of the second equation in the above mathematical formula. Using the same equations with $z=\sqrt{2}$ and $z=\sqrt{3}$ yields $p_2/p_1=p_3/p_2$, if $f(1)$ and $f(2)$ are well defined. Combined with the value for $p_2$ and the fact that $p_1+p_2+p_3 =1$, we easily obtain interesting values: $p_1 = 1/2, p_2 = 1/(1+\sqrt{5}), p_3= (3-\sqrt{5})/4$. The following of this section is split into three cases.

Case 1:

If $p_1 = 1/2$ and $f(1)$ is well defined, one would assume that if $z \in ]1,\sqrt{2}[$ and the density is continuous, then $f(z) = f(1) / z$, because of the first formula resulting in

$$f(z) = f(\sqrt{z})/\sqrt{z} = f(z^{1/2^n})\cdot\Big(z^{\frac{1}{2}+\frac{1}{2^2}+\cdots +\frac{1}{2^n}}\Big)^{-1} \rightarrow \frac{f(1)}{z}.$$

Case 2:

The case $z\in ]\sqrt{2},\sqrt{3}[$ is quite interesting. Let's use $p_2 = 1/(1+\sqrt{5})$ and let $\phi = 2p_2$. Also, let us define $$R_1(z) =\sqrt{1+z}, R_2(z) =\sqrt{1+\sqrt{1+z}},R_3(z) =\sqrt{1+\sqrt{1+\sqrt{1+z}}}$$ and so on. Using the formula $f(z) = \phi\cdot\sqrt{1+z}\cdot f(\sqrt{1+z})$ iteratively, one gets $$f(z)=f(R_n(z))\cdot\phi^n\cdot\prod_{k=1}^n R_k(z).$$ The expression on the right-hand size converges as $n\rightarrow\infty$. Note that $R_n(z) \rightarrow \phi^{-1}$.

Note that if $z\in ]2^{1/4}, 3^{1/4}[$ then $f(z)$ can be computed either using case 1, or as follows: $f(z) = 2p_1 z f(z^2)$ and since $z^2 \in ]\sqrt{2}, \sqrt{3}[$ you can compute $f(z^2)$ using case 2. If the two different methods produce different results, the likely explanation is that $f(1)$ does not exist: $f$ oscillates infinity many times around $z=1$, making case 1 useless. This is something I have yet to explore.

Case 3:

Here $z\in ]\sqrt{3},2[$. I haven't checked it yet.

  • 1
    $\begingroup$ The lognormal distribution satisfies your first requirement (but not the second one). $\endgroup$ Oct 18, 2019 at 20:36
  • 1
    $\begingroup$ So as Bernoulli $\endgroup$
    – gunes
    Oct 18, 2019 at 20:37
  • 4
    $\begingroup$ Generalized gamma also works for the first requirement and in some special cases, the second. $\endgroup$
    – soakley
    Oct 18, 2019 at 20:44
  • 4
    $\begingroup$ It depends what you call a family of distributions. If this family contains all probability distributions, this is always true. $\endgroup$
    – Xi'an
    Oct 19, 2019 at 12:42
  • 2
    $\begingroup$ You can create families of such distributions by closing any set of distributions under this operation. That reduces the question to characterizing those families; especially, to showing there are some "interesting" ones. As an example of where this can lead, one such family consists of all (necessarily discrete) distributions supported on the non-negative algebraic numbers whose degrees are powers of two. This, however, does not satisfy the more restrictive limiting condition imposed on $Z$ in the question. It does raise a question, though: are you looking for a finitely parameterized family? $\endgroup$
    – whuber
    Oct 19, 2019 at 13:30

1 Answer 1


My answer has three parts. Part 1 is related to using the discrete distribution investigated earlier for $X_k$. Part 2 is related to finding a family of distributions to meet the requirements of the original question. Part 3 is a generalization to nested cubic roots and continued fractions.

Part 1: using the discrete distribution for $X_k$

Using the discrete distribution discussed earlier for $X_k$ (that is, with $p_1=1/2$, $p_2 = 1/(1+\sqrt{5})$ and $ p_3 = (3-\sqrt{5})/4$) then $Z$'s distribution is much smoother than with various other combinations of $p_1, p_2, p_3$. Yet it is deeply chaotic in the sense that it might be differentiable nowhere. In short, $f(z)$ seems to be defined nowhere, and formulas based on limits as in case 1 and case 2, make no real sense.

The density $f_Z$ may not exist, but the distribution $F_Z$ does. It clearly has three legs: $z \in [1, \sqrt{2}]$, $z \in [\sqrt{2}, \sqrt{3}]$ and $z \in [\sqrt{3}, 2]$. Based on case 1 that suggests $f_Z(z) \propto 1/z$ if $z\leq \sqrt{2}$, I decided to compute the integral to get a "best bet" for $F_Z(z) = P(Z<z)$, resulting in $F_Z(z) \propto \log z$.

Even though that step does not really make sense (since $f_Z$ does not exist), it yields a very good approximation for $F_Z$. Indeed, $F_Z(z)$ is very well approximated by $\log_2 z$, especially if $z \in [1,\sqrt{2}]$. The picture below shows $F_Z(z)$ in blue, and its approximation by $\log_2 z$ in red. The X-axis represents $z$, the Y-axis $F_Z(z)$.

enter image description here

The chart below shows the approximation error $E(z) = F_z(z) - \log_2 z$. Note that the error is maximum at $z = (1+\sqrt{5})/2$. Notable local minima for $E(z)$ include (among infinitely many others) $z=1, 2^{1/4}, \sqrt{2}, \sqrt{3}$ and $z=2$. Also, the curve below seems to be differentiable nowhere, indeed it has some of the patterns of a Brownian motion. In particular, one can see a fractal behavior, with the successive double-bumps (followed and preceded by a big dip all the way down to $E(z)=0$) repeating themselves over time but being amplified as $z$ increases. The maximum attained at each double-bump seems to be exactly 2 times the maximum reached at the previous double-bump.

enter image description here

Furthermore, it seems that the median is $\sqrt{2}$, though I haven't checked. Now if you switch the values of $p_1$ and $p_3$, then it looks like the median becomes $\sqrt{3}$. And if $p_1=p_2=p_3 = 1/3$ (a very chaotic case), it looks like the median becomes $(1+\sqrt{5})/2$.

Part 2: finding $X$ and $Z$ using characteristic functions

This is still a work in progress, but the idea is as follows. If $\phi_2$ is the characteristic function (CF) of $Z^2$, $\phi_1$ is the CF of $Z$, and $\phi$ is the CF of $X_k$, and if $\phi = \frac{\phi_2}{\phi_1}$, then the distribution of the nested square root of the $X_k$'s is also the distribution of $Z$.

The idea is to first find some $Z$ (that is, $\phi_2$ and $\phi_1$), compute the ratio of the two CF's. If this ratio is the CF of some distribution $X$, then we solved the problem (in a backward way, by specifying the limit $Z$ first, and then finding $X_k$.)

Note that $Z$ can not have a log-normal distribution unfortunately, because $Z$ can not be lower than 1 (prove it, this is an easy exercise.) A potential candidate for $Z$'s distribution is uniform on $[1, 2]$, or log-log-normal, that is $\log\log Z$ is normal.

Below is a chart based on $X$ being log-normal (see here for more.) It looks like $\log \log Z$ is almost normal, but it is not exactly normal.

enter image description here

Perhaps the easiest solution is considering $f_z(z) = \frac{2}{3} z$ with $z \in [1,2]$. Then $\mbox{CF}(Z) =\frac{2}{3}\int_1^2 z \exp(i t z)dz$ and $\mbox{CF}(Z^2) =\frac{2}{3}\int_1^2 z \exp(i t z^2)dz$. These two CF's are easy to compute and result in $$\mbox{CF}(X) = \frac{it}{2}\cdot\frac{e^{3it}-1}{e^{it}(1-2it)+it -1}.$$ But is the latter really a CF? It does not appear to be bounded. And is the support domain for $X$ equal to $[0, 2]$ as expected?

Part 3: Generalization to nested cubic roots and continued fractions

This can be generalized to nested cubic roots or continued fractions as follows. Consider $Z_{k+1}=(X_k + Z_k)^{\alpha}$ with $Z=\lim_{k\rightarrow\infty} Z_k, Z_0=0$ and the $X_k$'s are i.i.d.Then we have $\phi = \frac{\phi_\alpha}{\phi_1}$ where $\phi$ is the CF of $X_k$, $\phi_1$ is the CF of $Z$, and $\phi_\alpha$ is the CF of $Z^{1/\alpha}$. The most popular cases are:

  • $\alpha = 1/2$: Nested square roots,
  • $\alpha = 1/3$: Nested cubic roots,
  • $\alpha = -1$: continued fractions.

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