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I have come across this particular problem in my research. I have a background in mathematical statistics and have attempted to solve this problem for some time but alas I have not made much progress. The problem I wish to solve is of the following nature:

Let $X_0, X_1,..., X_j \overset{iid}{\sim} U(0,1)$ and let $Y_j = \frac{X_{j}}{X_j+Y_{j-1}}$, $Y_0 = X_0$, and $j\in \mathbb{N}$. Is there a way to write $Y_j$ analytically as $j \rightarrow \infty$?

I am fairly certain that this process will settle into a stationary distribution with three distinct parts.

Part 1

We can rewrite the equation for $Y_j$ as $Y_j(X_j + Y_{j-1}) = X_j$, which can give us $\frac{Y_j Y_{j-1}}{1-Y_j}$. As $j\rightarrow \infty$, $Y_j = Y_{j-1}$, so $\frac{Y^2_j}{1-Y_j} \sim U(0,1)$. Solving for $Y_j$ then provides, $f_Y(Y=y) = \frac{1}{(Y-1)^2}-1$, $Y\in (0,\varphi-1]$.

Part 2

We can now apply the transformation to the result from Part 1

\begin{align} \frac{3 \sqrt{5}-1-8 \text{csch}^{-1}(2)}{\left(\sqrt{5}-3\right)^2} && 2 X+1=\sqrt{5} \\ \frac{3 \sqrt{5}-1-8 \text{csch}^{-1}(2)}{4 (Y-1)^2} && Y>0\land 2 Y+1<\sqrt{5} \\ -\frac{\frac{1}{Y^2}-\frac{2}{Y}+\frac{1}{1-2 Y}+2 \log \left(\frac{Y}{2 Y-1}\right)+2}{2 (Y-1)^2} && \frac{1}{2} \left(\sqrt{5}-1\right)<Y<1 \\ 0 && Y\geq 1\lor Y<0 \\ \text{Indeterminate} && \text{True} \end{align}

We see that we obtain a new functional form and retain the functional form from Part 1 albeit with a different scaling factor.

Part 3

This is now where things get tricky - at least for me. It seems as if the solutions for each begin to overlap one another and begin to blend together.

From my empirical investigation, it seems as if this blending reduces the domain of the first functional form to $[0,\frac{1}{2}]$ and reduces the domain for the second functional form to $[\varphi-1,1]$. The third part, which overlaps and blends together covers the remaining part of the domain, from $(\frac{1}{2}, \varphi-1)$.

It is the functional form of this blended part and the proper factoring coefficients, which I am having difficulties to determine and for which I am seeking the community's insight and guidance.

I appreciate any thoughts you may have.

dsmalenb
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    I doubt the statment $Y_{i}=Y_{i+1}$ except possibly in the distribution. I think it is better to try and work with the pdf and get a recursion for that. – user619894 Oct 23 '21 at 06:00
  • It is true that for all $\omega\in\Omega$, $Y_j(\omega)$ is positive and strictly decreasing, hence converges to a limit. However, that is different from $Y_j=Y_{j+1}$ as $j\to\infty$ (i.e., $Y_j$ is eventually constant) that you are asserting in part 1. – user10354138 Oct 23 '21 at 06:28
  • $Y_j$ is a distribution. Saying that $Y_j = Y_{j-1}$ as $j\rightarrow \infty$ is simply stating that the distribution converges after an infinite number of same transformations applied to it. – dsmalenb Oct 23 '21 at 15:24
  • @dsmalenb My point was that $\frac{Y^2_j}{1-Y_j} \sim U(0,1)$ is incorrect, even though $Y_j, Y_{j-1}$ have the same distribution doesn't mean that they are equal (unless you can show that they are totally correlated). – user619894 Oct 24 '21 at 07:44

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