My initial attempt to solve this type of problem can be found here, however, to not overcomplicate the conversation and prior results, I am starting a clean post with my newest results to this problem.
I am interested in the limiting distribution where a successive transformation is performed in a distribution, $Y$. The transformation is of the form
\begin{equation} X_{j+1} = \frac{n_j}{n_j+X_j} \end{equation}
where $n_j \sim U(0,2)$ such that
\begin{equation} Y \sim \lim_{j\rightarrow \infty} X_j \end{equation}
This is similar to a continued fraction that acts as a transformation on a random variate of a distribution. From prior work and from numerical simulations, we observe that the distribution settles down into a stationary form after successive iterations. Reasonable approximations to the stationary distribution are reached after about 4 to 5 iterations already. Below is a visualization of the distribution:
Partial Solution
We can determine that the range of the transformation is $[0,1]$. The next step is to determine which values of $n$ and $X$ can map to which regions within the n-by-X or 2-by-1 region. The graph below is partitioned into "Regions". These Regions are marked as $R_{st}$, where $s$ marks the region where $X_j$ belongs and $t$ marks the region where $X_{j+1}$ belongs.
By inspecting the 2-by-1 grid, we can divide the region into 7 distinct regions. Those regions in the blue map to the $\left[0,\frac{2}{3}\right]$ region, the region in green maps to the $\left[\frac{3}{4},1\right]$ region, and finally, those regions in the purple map to the $\left[\frac{2}{3},\frac{3}{4}\right]$ region on the $[0,1]$ number line.
To determine how much of each region distribution contributes to the total probability, we need to find the limiting distributions of the transition probabilities. We can understand this as a Markov process where the transition probabilities are the individual areas. Calculating these we get the transition matrix, $\mathbf{T}$ as follows:
\begin{equation} \mathbf{T} = \begin{bmatrix}\frac{4}{9}&\frac{2}{9}&\frac{1}{3}\\\frac{17}{24}&\frac{7}{24}&0\\\frac{7}{8}&\frac{1}{8}&0\end{bmatrix} \end{equation}
The limiting distribution of $R_1$, $R_2$, and $R_3$, can be found easily then as $\frac{17}{29}$, $\frac{19}{87}$, and $\frac{17}{87}$, respectively.
From the prior work at the aforementioned link, we can determine the distribution between $\left[0,\frac{2}{3}\right]$ as follows:
\begin{equation} p_1(x) = \frac{17}{58} \frac{1}{(x-1)^2} I_{\left[0,\frac{2}{3}\right]}(x) \end{equation}
Now to find the distribution for region $\left[\frac{3}{4},1\right]$, we can leverage $p_1(x)$ with the transformation if and only if $n_j \sim U(3X,2)$ $\forall j$. Doing this provides us with $p_3(x)$ as follows:
\begin{equation} p_3(x) = \frac{68}{87 (3 \ln 3 - 2)} \frac{2-2x+(3x-2)\ln\left(3-\frac{2}{x}\right)}{4(x-1)^2(3x-2)}I_{\left[\frac{3}{4},1\right]}(x) \end{equation}
Current Challenge: Region 2
The region $\left[\frac{2}{3},\frac{3}{4}\right]$ presents an additional difficulty as it seems that its distribution is the aggregation of regions 1 and 3. It is not clear to me how to determine $p_2(x)$ and it is here where I would appreciate some insight and perhaps assistance.
I would expect that we should be able to complete a similar transformation using $p_1(x)$ and $p_3(x)$. I would expect that $p_2(x)$ should be the aggregate distribution of the following:
Using $p_1(x)$ and the transformation where $n \sim U(2X,3X)$
Using $p_3(x)$ and the transformation where $n \sim U(2X,2)$
It seems reasonable but it does not seem to work out.
