From Wikipedia:

a stationary process (or strict(ly) stationary process or strong(ly) stationary process) is a stochastic process whose joint probability distribution does not change when shifted in time. Consequently, parameters such as the mean and variance, if they are present, also do not change over time and do not follow any trends.

If a given Markov chain admits a limiting distribution, does it mean this Markov chain is stationary?

Edit: to be more precise, can we say the unconditional moments of a Markov chain are those of the limiting (stationary) distribution, and then, since these moments are time-invariant, the process is stationary?

  • $\begingroup$ When did the process start? What was the initial distribution? $\endgroup$
    – Adrian
    Feb 24, 2016 at 12:27
  • $\begingroup$ I am referring to the case where the initial distribution (at a given period $t=0$) is not the limiting distribution. But I assume a markov chain that admits a limiting (stationary) distribution after a given number of periods. Can we say it's stationary ? $\endgroup$
    – James
    Feb 24, 2016 at 13:06
  • $\begingroup$ what did you have in mind when you wrote "after a given number of periods"? $\endgroup$
    – Adrian
    Feb 24, 2016 at 17:53
  • $\begingroup$ With only two outcomes possible as in your example, if the distribution at $t$ is $\pi_t=[\pi_{t1},\pi_{t2}]$, I was referring to the number of iterations over the equation $\pi_{t+1}=\pi_t P$ until convergence, i.e. until $\pi=\pi P$. $\endgroup$
    – James
    Feb 25, 2016 at 10:14
  • $\begingroup$ that number could easily be infinite $\endgroup$
    – Adrian
    Feb 25, 2016 at 11:01

1 Answer 1


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 $\left\{0, 1\right\}$. The limiting distribution is $\pi = \left(0.5, 0.5\right)$. However, suppose you start the process at time zero with initial distribution $\mu = \left(1, 0\right)$, 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.

  • $\begingroup$ Thanks a lot for your answer. The thing I have difficulty to understand is that, for instance in the case of a continuous state space, an AR(1) process defined by $X_t = c + \rho X_{t-1}+\epsilon_{t}$ where $\epsilon_{t}$ are i.i.d. is stationary if $|\rho|<1$. And an AR(1) process is a Markov process (stats.stackexchange.com/questions/23789/is-ar1-a-markov-process). So in this case, isn't it a stationary process? $\endgroup$
    – James
    Feb 24, 2016 at 14:39
  • $\begingroup$ No, it isn't necessarily stationary. It depends on whether the initial distribution is identical to the stationary distribution. It's very similar to the simple example I gave, just slightly more complex. $\endgroup$
    – Adrian
    Feb 24, 2016 at 15:29
  • $\begingroup$ Maybe we are not talking about the same kind of stationarity. For the AR(1) process I mentioned in the previous comment, the expected value is given by $E(X_t)=\frac{c}{1-\rho}$, which is not time-dependent. $\endgroup$
    – James
    Feb 25, 2016 at 10:30
  • $\begingroup$ @James you are assuming that the process started infinitely long ago, or finitely long ago with an initial distribution equal to the stationary distribution. If that assumption does not hold, the expected value of X_t will be time-dependent $\endgroup$
    – Adrian
    Feb 25, 2016 at 11:03

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