# Random variables $X, Z$ such that $Z$ and $\sqrt{X + Z}$ have the same distribution?

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.

• The lognormal distribution satisfies your first requirement (but not the second one). Oct 18, 2019 at 20:36
• So as Bernoulli Oct 18, 2019 at 20:37
• Generalized gamma also works for the first requirement and in some special cases, the second. Oct 18, 2019 at 20:44
• It depends what you call a family of distributions. If this family contains all probability distributions, this is always true. Oct 19, 2019 at 12:42
• 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?
– whuber
Oct 19, 2019 at 13:30

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 $$F_Z(z) = P(Z, resulting in $$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 $$F_Z(z)$$. 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. 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. 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.