# Is the stationary distribution in a Markov chain just an average or will this probability distribution actually be reached?

So I know that a connected Markov chain has a stationary distribution $$\pi$$ that satisfies

$$\lim_{t \rightarrow \infty} a_t = \pi,$$

where $$a_t$$ is the average probability distribution at time $$t$$. So this average converges to our stationary distribution. However, does that mean that $$\lim_{t \rightarrow \infty} p_t = \pi$$ (where $$p_t$$ is the probability vector of where the Markov chain is after $$t$$ steps)?

Just going off of the average definition, couldn't it be that the Markov chain bounces around two or more different probability distributions, so that the average is still $$\pi$$, but $$\pi$$ itself is never a probability distribution for the states of the Markov chain?

For example, say we have 3 states and $$\pi = (0.2, 0.3, 0.5)$$, then $$p_{(1)} = (0.1, 0.2, 0.7)$$ and $$p_{(2)} = (0.3, 0.4, 0.3)$$. Say for even $$i$$, $$p_i = p_{(1)}$$, and for uneven $$i$$, $$p_i = p_{(1)}$$, which means that the distribution of the process oscillates between $$p_{(1)}$$ and $$p_{(2)}$$. So the average $$\lim_{t \rightarrow \infty} a_t$$ is indeed $$\pi$$, but not $$\lim_{t \rightarrow \infty} p_t = \pi$$, since $$p_t$$ always switches between $$p_{(1)}$$ and $$p_{(2)}$$.

This is obviously a simple example, but I hope you understand my general question.

• I think you need to add aperiodic to guarantee stationarity. Now if there exists a stationarity distribution, it is necessarily unique. Jul 31, 2023 at 16:05

where$$S_n(f)=\sum_{i=1}^n f(\it\Phi_i)$$