Solved – Is a Markov chain with a limiting distribution a stationary process

Here's a simple example illustrating why the answer is no.

Let $$P = begin{pmatrix} 0.5 & 0.5 \ 0.5 & 0.5 end{pmatrix}$$ be the transition matrix for a first-order Markov process $X_t$ with state space $left{0, 1right}$. The limiting distribution is $pi = left(0.5, 0.5right)$. However, suppose you start the process at time zero with initial distribution $mu = left(1, 0right)$, i.e. $X_0 = 0$ with probability one.

We then have $mathbf{E}[X_0] = 0 neq mathbf{E}[X_1] = frac{1}{2}$, meaning the moments of $X_t$ depend on $t$, which violates the definition of stationarity.

Here's some R code illustrating a similar example with

$$P = begin{pmatrix} 0.98 & 0.02 \ 0.02 & 0.98 end{pmatrix}$$

p_stay <- 0.98 P <- matrix(1 - p_stay, 2, 2) diag(P) <- p_stay stopifnot(all(rowSums(P) == rep(1, nrow(P)))) mu <- c(1, 0) pi <- matrix(0, 100, 2) pi[1, ] <- mu for(time in seq(2, nrow(pi))) {     pi[time, ] <- pi[time - 1, ] %*% P } plot(seq(1, nrow(pi)), pi[, 1], type="l", xlab="time", ylab="Pr[X_t = 0]") abline(h=0.5, lty=2) 

The fact that $X_t$ is converging in distribution to some limit does not mean the process is stationary.