So, I was doing my Sunday tinkering and I came across this continuous-space, discrete-time Markov Chain:
$X_0 \in (0,1)$, $k \in (0,1)$
Update rule:
With probability $p$, $X_{t+1} = k+(1-k)X_t$
With probability $1-p$, $X_{t+1} = X_t(1-k)$
I'm interested in learning what this Markov chain is and what its steady-state distribution is, from simulation - I'm fairly convinced its a Beta distribution, as always, proof is preferable!
I mostly understand what's going on when $k \ge \frac12$.
For a nice special case, let $k = \frac23$ and $p = \frac12$. Then our update rule is that $X_{t+1}$ first multiplies $X_t$ by $\frac13$, and then flips a coin to decide whether to add $\frac23$ or not. If we write $X_t$ as $0.x_1x_2x_3x_4\dots$ in ternary, then $X_{t+1}$ is equally likely to be either $0.0x_1x_2x_3x_4\dots$ or $0.2x_1x_2x_3x_4\dots$ in ternary.
As a result, the limiting distribution is to choose a random number in $[0,1]$ via its ternary representation, putting a $0$ or $2$ in each ternary digit with equal probability. This has a name: it's the Cantor distribution. The Cantor distribution is, informally, a uniform distribution on the Cantor set. It is a singular distribution, not even a little discrete (every point has probability $0$) nor even a little continuous (its support has measure $0$).
If we keep $p$ the same, but vary $p$, then as long as $p$ is not either $0$ or $1$, we get a different distribution on the Cantor set, with similar properties.
For other values of $k$ such that $k>\frac12$, we get something similar happening on a Cantor set with different proportions. The update rule of the Markov chain guarantees that $X_1 \in [0,1-k] \cup [k,1]$. Therefore $X_2 \in [0,(1-k)^2] \cup [k(1-k),1-k] \cup [k,k+(1-k)^2] \cup [k+k(1-k),1]$. For $X_t$, we get a similar support of $2^t$ intervals of total length $(2(1-k))^t$.
Looking at $k =\frac12$ is trickier, but the answer turns out to be similar. In this case, $X_{t+1}$ is $\frac12 + \frac12X_t$ with probability $p$ and $\frac12X_t$ otherwise. Written in binary, we shift $X_t$ over and put $1$ (with probability $p$) or $0$ (with probability $1-p$) in the first position. This leads to a random real number whose binary digits are chosen from a Bernoulli distribution, and people have thought about this a lot; see, for example, this question on MSE.
If $p = \frac12$, then this is just the uniform distribution. If $p \ne \frac12$, it is once again purely singular. Suppose, for example, $p = \frac23$. Then we expect that around $\frac23$ of the first $t$ binary digits of $X_t$ are $1$, and around $\frac13$ are $0$, and in the limiting distribution, we expect that this is almost true for almost all $t$. Let $X$ have that limiting distribution. Then with probability $1$, there is a number $N$ such that for all $t \ge N$, at least $0.66t$ of the first $t$ binary digits are $1$. But the set of real numbers in $[0,1]$ with this property has measure $0$.
When $k<\frac12$, the intervals $[0,1-k]$ and $[k,1]$ overlap, and the description becomes messy. I do not know what's going on here and I don't know if it's still purely singular. I'm pretty sure that it's not a Beta distribution, because when I looked at a special case, there were no parameters of the Beta distribution that led to the PDF being a fixed point of the update rule.
This is a variant of an $\text{AR}(1)$ process.
To be precise, let $(\epsilon_n)_{n\in\mathbb N}$ be a sequence of i.i.d. random variables with distribution $$\mathbb P[\epsilon_n=k]=p, \hspace{1cm}\mathbb P[\epsilon_n=0]=1-p.$$ Then our update rule can be rewritten as $X_{t+1}=(1-k)X_t+\epsilon_{t+1}$. Due to this recursion, we can express the solution as $$X_t=(1-k)^t\cdot X_0+\sum_{i=0}^{t-1} (1-k)^{t-i}\epsilon_{i}.$$
Note that regardless of the value/distribution of $X_0$, the first term vanishes as $t\to\infty$. However, as already pointed out by the other answer, it can become very tricky to compute the limiting distribution of the second term. I will only show that it exists, i.e. that the series converges almost-surely:
To do this, we may simply show that both the expected values and the variances of the sum converge. Convergence of the series then follows by Kolmogorov's Three Series Theorem.
It is easily seen that $$\lim_t\mathbb E\bigg[\sum_{i=0}^{t-1}(1-k)^{t-i}\epsilon_i\bigg]=\lim_t\sum_{i=0}^{t-1}(1-k)^{t-i}pk<\infty$$ since $k\in (0,1)$ and similarly $$\lim_t\text{Var}\bigg(\sum_{i=0}^{t-1}(1-k)^{t-i}\epsilon_i\bigg)=\lim_t\sum_{i=0}^{t-1}(1-k)^{2(t-i)}\text{Var}(\epsilon_1)<\infty$$ again, since $k\in(0,1)$. In both computations we have simply used the convergence of the geometric series.
