Here is a proposal for a model of a random, mountainous, skyline: let $Z = (Z_1, \ldots, Z_n)$ be multivariate Gaussian with mean zero and covariance matrix as follows. We imagine these as arranged around a circle, and let $d(i,j) = \min(|i-j|, n-|i-j|)$ be the circular distance between $i$ and $j$. Then for each $1 \le i \le n$ let $\text{var}[Z_i] = 1.01$ and for $i \neq j$ let $$ \text{cov}[Z_i, Z_j] = \frac{1}{1 + \left(\frac{10 \, d(i,j)}{n}\right)^2} . $$
Simulate a few random landscapes from this process with different values of $n$ between 50 and 500. To do this: (a) compute the covariance matrix, $\Sigma$; (b) let $K$ be the Cholesky decomposition (a square root) of $\Sigma$, and (c) let $Z = K W$, where $W$ is a vector of iid Normals.
For $0 \le s < t \le 1$ let $U(s, t) = \frac{1}{n(t-s)} \sum_{i=ns}^{nt} Z_i$: in other words, $U(s,t)$ is the average of the values between positions $s$ and $t$ around the circle, if the circle has diameter 1. Find an approximation for large $n$ for $\text{cov}[U(0), U(s,t)]$ and check your answer using simulation with a few values of $s$ and $t$.
Let $Y(t)$ be the diffusion that solves the SDE $$ dY(t) = - 4 Y(t)\left( Y(t)^2 - 1 \right) dt + dB(t) . $$
Write down the generator for $Y(t)$, and show that it satisfies one of the definitions of the generator.
Show that if $h(y) = y^2 (y^2 - 2)$, then the stationary distribution of $Y(t)$ is proportional to $e^{-2h(y)}$.
Simulate from the process and check at least qualitatively your answer.
Let $B(t)$ be standard Brownian motion and $$ X(t) = \frac{1}{1 + \exp(-B(t))} .$$