Let me start with the following remark, which only serves to motivate what follows:
- If $(V_n)$ is a sequence of independent random variables uniformly distributed in $\lbrace 0,1\rbrace$ (i.e., with Bernoulli distribution with $p=\frac{1}{2}$) and $U := \sum_{n=1}^{+\infty} V_n/2^n$, then $U$ is uniformly distributed in $[0,1]$. (This follows from the fact that the binary digits of $U$ are being chosen uniformly at random.)
Now we define the Fabius function and distribution:
- If $(U_n)$ is a sequence of independent random variables uniformly distributed in $[0,1]$ and $Z := \sum_{n=1}^{+\infty} U_n/2^n$, then $Z$ has a distribution which we call the Fabius distribution in $[0,1]$, and its cumulative distribution function is the Fabius function $f:[0,1] \to [0,1]$, a curious $C^\infty$-but-nowhere-analytic function satisfying the functional equation $f'(t)=2f(2t)$ (see here, here and here for basic facts about the Fabius function, including graphs and special values).
This suggests the following definition (which we could then take to further levels, but let's start just one):
- If $(Z_n)$ is a sequence of independent random variables each following the Fabius distribution (defined above) in $[0,1]$ and $Y := \sum_{n=1}^{+\infty} Z_n/2^n$, then $Y$ has a distribution which we call the hyperfabius distribution in $[0,1]$, and its cumulative distribution function is the hyperfabius function $g:[0,1] \to [0,1]$.
Question(s): Does the hyperfabius function $g$ satisfy an interesting functional equation (analogous to the $f'(t) = 2f(2t)$ equation satisfied by the Fabius function)? How can we compute it numerically? Does it appear somewhere in the literature?
Note that we can define the Fabius distribution by taking an array $(V_{i,j})$ (with $i,j\geq 1$) of independent random variables uniformly distributed in $\lbrace 0,1\rbrace$ and letting $Z := \sum_{n=1}^{+\infty} \frac{1}{2^n} \sum_{i+j=n} V_{i,j}$, where the $\sum_{i+j=n} V_{i,j}$ are independent and now follow a binomial distribution with $n-1$ trials and success probability $\frac{1}{2}$; similarly, we can define the hyperfabius distribution by taking an array $(U_{i,j})$ of independent random variables uniformly distributed in $[0,1]$ and letting $Y := \sum_{n=1}^{+\infty} \frac{1}{2^n} \sum_{i+j=n} U_{i,j}$, where $\sum_{i+j=n} U_{i,j}$ satisfy an Irwin-Hall distribution with mean $(n-1)/2$; or equivalently, by taking a three-dimensional array $(V_{i,j,k})$ of independent random variables uniformly distributed in $\lbrace 0,1\rbrace$ and letting $Y := \sum_{n=1}^{+\infty} \frac{1}{2^n} \sum_{i+j+k=n} V_{i,j,k}$, where $\sum_{i+j+k=n} V_{i,j,k}$ now follows a binomial distribution with $(n-1)(n-2)/2$ trials. But I don't think these various reformulations are of much use in answering my question.