How do you find the stationary distribution of a Markov chain?

The equations derived from $\pi P = \pi$ are linearly dependent because the rows of the transition matrix $P$ sum to 1. This implies that the sum of the equations is identically zero, leaving only $N-1$ independent constraints for $N$ states.

Conditions: $P$ is a stochastic matrix (rows sum to 1).; $\pi$ is a row vector of probabilities.

The stationary distribution is the row vector $\pi = [4/11, 7/11]$.

Conditions: The transition matrix is $P = \begin{bmatrix} 0.3 & 0.7 \\ 0.4 & 0.6 \end{bmatrix}$.; $\pi$ is a row vector satisfying $\pi P = \pi$ and $\sum \pi_i = 1$.

A stationary distribution (also called an invariant distribution) is a probability distribution row vector $\pi$ that remains unchanged when multiplied by the transition matrix $P$ of a Markov chain. This means that if the system starts in this distribution, it will remain in this distribution at all future time steps.

Conditions: $\pi$ is a row vector.; $P$ is the transition matrix of the Markov chain.; The elements of $\pi$ are non-negative and sum to 1.

We add the equation $\pi(a)+\pi(b)=1$ because $\pi$ is a valid probability distribution, meaning its elements must sum to 1. This provides the second independent equation needed to solve the underdetermined system created by the redundant balance equations.

Conditions: $\pi$ is a probability distribution over the states.; The state space has exactly two states, a and b.

By substituting $\pi(b)=7/4 \pi(a)$ into $\pi(a)+\pi(b)=1$, we get $\pi(a) + 7/4 \pi(a) = 1$. Combining the terms yields $11/4 \pi(a) = 1$, and solving for $\pi(a)$ gives $\pi(a) = 4/11$.

Conditions: The balance equation gives $4\pi(b) = 7\pi(a)$, which rearranges to $\pi(b) = 7/4 \pi(a)$.; The normalization equation is $\pi(a) + \pi(b) = 1$.