Considering the $n=1$ version of this MO question leads to the following random process $(x_n)_{n=0}^\infty$:
Let $x_0 = \frac14$. Given $x_n \in [0,\frac14]$,
with probability $\frac14-x_n$, $\ x_{n+1} = x_n$;
with probability $\frac34-x_n$, $\ x_{n+1} = 0$; and
with probability $2x_n$, $\ x_{n+1}$ is uniformly distributed in $[0,x_{n}].$
Mathematica data apparently show that, for all $n \geq 0$, $x_n$ is distributed according to $\mu_n$: $$\mu_n(x) = n(n+1) \left(\frac14 - x\right)^{n-1} dx + \left(1 - \frac{n+1}{4^n}\right)\delta_0.$$
(I.e., the first term is a measure whose Radon-Nikodym derivative with respect to Lebesgue measure is $n(n+1) \left(\frac14 - x\right)^{n-1}$, and $\delta_0$ is the Dirac delta measure at $0$.)
To show this, it seems natural to express $\mu_{n+1} = \mathcal L(\mu_n)$ for a suitable (linear) operator $\mathcal L$ acting on measures (i.e., that maps the distribution of $x_n$ onto that of $x_{n+1}$), interpreting $\mu_0 = \delta_{1/4}$. (No, this isn't my area!)
However, my guess at this (a convolution operator) gave a wildly different sequence of measures (that doesn't agree with the data).
The question: does there exist a suitable operator $\mathcal L$, and is my assertion for the distribution for $x_n$ correct?