Markov chain with state space $[0, 3]$: Random walk with repulsive boundary Let $X_t$, $t=0,1,2\dots$ be a random walk on $[0, 3]$ defined like this:
$$X_{t+1} = X_t + S_t$$
where the step $S_t$ is a uniform random variable but the direction is away from the boundary if we are close to it. That is:


*

*If $X_t \in [0, 1)$,   $ S_t \sim U(0,1)$

*If $X_t \in [1, 2]$,   $ S_t \sim U(-1,1)$

*If $X_t \in (2, 3]$,   $ S_t \sim U(-1,0)$


I think this is a continuous-state Markov chain, but I'm unfamiliar with those. What I'm trying to find is the limitig distribution as the number of steps goes to infinity. How can this be done?
I have come up with the kernel
$$p(x, y) =
\begin{cases}
1, y\in[x, x+1], x\in[0, 1) \\
\frac{1}{2}, y\in[x-1, x+1], x\in [1, 2] \\
1, y\in[x-1, x], x\in (2, 3] \\
0, \text{otherwise}
\end{cases}$$

I have simulated the walk ($n=5000000$) and stored all the visited states and I got this bell shaped distribution (of course supported on $[0, 3]$):

The red curve is pdf of (scaled) Beta(2,2) distribution. I thought it looked rather similar but as you can see it has a little different shape.
 A: Let $\pi$ denote the PDF of the stationary distribution. The dynamics of the Markov chain suggests that the stationarity yields a different equation in each interval $(0,1)$, $(1,2)$ and $(2,3)$. 
Indeed, for every $x$ in $(2,3)$, $$\pi(x)=\int_x^3\pi+\frac12\int_{x-1}^2\pi$$ Likewise, for every $x$ in $(1,2)$, $$\pi(x)=\int_{x-1}^1\pi+\frac12\int_1^2\pi+\int_2^{x+1}\pi$$ These determine $\pi$ on $(0,1)$ as well since, by symmetry, for every $x$ in $(0,3)$, $$\pi(x)=\pi(3-x)$$
Solving these integro-differential equations (computations omitted), one gets $$\pi(x)=\left\{\begin{array}{ccc}\frac14x^2+\frac14x&\text{for}&x\in[0,1]\\-\frac12x^2+\frac32x-\frac12&\text{for}&x\in[1,2]\\\frac14x^2-\frac74x+3&\text{for}&x\in[2,3]\end{array}\right.$$
Thus, $\pi$ is not Beta$(2,2)$, $\pi$ is continuous on $[0,3]$, $\pi(0)=\pi(3)=0$, $\pi(1)=\pi(2)=\frac12$, the maximum of $\pi$ is at $\pi(\frac32)=\frac58$, and $\pi$ is differentiable everywhere except at $1$ and $2$.
You might want to check the expressions of $\pi$ above, assuming, to save you a part of the computations, that these are indeed polynomial of degree $2$ on each interval.
