Let the Markov-process $(X_n)_{n\in\mathbb N}$ be defined by $$ X_{n+1} \sim \mathcal U[0, 2X_{n}], $$ where $\mathcal U$ is the uniform distribution and $X_0$ is positive. I'm interested in finding the stationary distribution $\pi$.
My attempt: At first, I found that $\mathbb E[X_n] =\mathbb E[X_0]$ and that $\mathbb E [X_n^k] = \left(\frac{2^k}{k+1}\right)^{n-1} \mathbb E [X_0^k]$, thus the moments greater than $1$ of the stationary distribution do not exist.
I also found out that if $X_0$ follows a semi-Cauchy distribution, i.e., $f_{X_0}(x) = \frac{2}{\pi(1+x^2)}, x>0$, then the pdf of $X_n$ is $f_{X_n}(x) = \frac{-\mathrm{Li}_n\left(-\frac{2^{n+3}}{x^2}\right)}{2^{n+2} \pi }, x>0$, where $\mathrm{Li}$ is the polylogarithm function. Thus, I expect that the stationary distribution must be equal to the following limit. \begin{align} \pi(x) = \lim_{n\to\infty} \frac{-\mathrm{Li}_n\left(-\frac{2^{n+3}}{x^2}\right)}{2^{n+2} \pi } , \quad x>0 \tag 1. \end{align} However, I could not solve it.
Another attempt was that, from the definition of the process, \begin{align} \pi(y) = \int_{y/2}^{\infty} \frac{\pi(x)}{2x}\,\mathrm d x, \quad y>0. \end{align} Thus, the stationary distribution must satisfy $2x\pi'(2x) = -\pi(x)$.
However, I could not solve this either.
Additional comment: I integrate the differential equation by parts on $[0,\infty]$ and got that \begin{align} \int_{0}^{\infty} 2x\pi'(2x) \,\mathrm d x &= (x\pi(2x))_0^{\infty} - \int_{0}^{\infty} \pi(2x) \,\mathrm d x = 0 - 1/2. \end{align} However, $-\int_{0}^{\infty} \pi(x) \,\mathrm d x = -1$. Thus, $-1/2=-1$ ???
Probably, $x\pi(2x)$ does not converge to $0$ as $x$ tends to $0$ or $\infty$.
We can realize $X_n$ as $X_n=X_0 U_1 U_2 \cdots U_n$, where $U_i$'s are i.i.d. $\sim\mathcal{U}([0,2])$. Then $$\log X_n=\log X_0+\sum_{i=1}^{n}\log U_i$$ is a random walk with drift $\mathbf{E}[\log U_i]=\frac{\log 2 - 1}{2}<0$, hence $\log X_n\to -\infty$ a.s. by SLLN. This means that $X_n \to 0$ a.s., so $\pi=\delta_0$ is a stationary distribution.