I recently came across the following recursion relation for a function $n(x)$:
$$n(x) = \begin{cases} 1 + \sum_{i=1}^m n(r_i x) ~,~ x>x_c \\ 0 ~, ~ x\leq x_c\end{cases}$$
here the $r_i\in[0,1]$'s have joint density function $\eta_m(r_1,r_2,\dots r_m)$ with the property that the marginal distribution of any of the $r_i$ is independent of $i$ and denoted $\eta_1(r)$.
Is it possible to understand the distribution of $n(x)$ for fixed $x_c$ and given $\eta_m$ ? Say for the case $\eta_m(r_1,r_2,\dots,r_m) = 1$? What about the average or variance of $n(x)$? How would I tackle this problem? How would I formally treat this problem? I think I am having a problem in looking at the problem in the right way and understanding what the probability space is...
EDIT: An example to clarify the problem. Suppose we choose $x_c=0.5$ and $m=2$:
- $x=1, r_1=0.8, r_2 = 0.2$ hence $n(1) = 1+ n(0.8) + n(0.2)$ now $n(0.2)=0$ because $0.2<0.5$.
- Hence we have to evaluate $n(0.8)$. Let $r_1$ and $r_2$ be such that $0.8r_1=0.7$ and $0.8r_2=0.1$. In this case: $n(0.8) = 1+n(0.7) + n(0.1)$. Again $n(0.1) = 0$ and we are left with $n(0.7)$.
- Let the random numbers in this case be such that $0.7r_1=0.3$ and $0.7r_2=0.4$ so we get $n(0.7) = 1 + n(0.4) + n(0.3)$. Because both 0.3 and 0.4 are less than $x_c=0.5$ we eventually have
$n(1) = 1 + 1 + 1 = 3$.
The thing is next time $n(1)$ can evaluate to something else. The recursive expansion happens based on the random numbers.
Below Python code that defines the function for the case $m=2$ and $\eta_2(r_1,r_2)$ uniform on all $(r_1,r_2)$ with $r_1+r_2=1$:
from numpy.random import uniform
def func(x,xc=.5):
if x>xc:
r = uniform()
return 1+ func(r*x) + func((1-r)*x)
else:
return 0