A Markov chain is shown in the figure above.
I am writing the transition matrix as:
$$P = \begin{bmatrix}
0.6 & 0.4 & 0 & 0\\
0 & 0 & 0.5 & 0.5\\
0 & 0 & 0.25 & 0.75\\
0.7 & 0.3 & 0 & 0
\end{bmatrix}$$
I using given formula to find stationary distribution:
$$AP=A$$
where $A=[w_1 \; w_2\; w_3\; w_4]$ is the stationary distribution. The answer for stationary distribution is given as:
$$w_1=\frac{4}{12} \quad w_2=\frac{3}{12} \quad w_3=\frac{3}{12} \quad w_4=\frac{2}{12}$$
which I can easily check the formula $$AP=A$$ is not satisfying. Where am I going wrong?
Best Answer
I'm not convinced that the answer claimed is correct.
Your matrix indicates that the chain is regular, so there should be no issue approximating the stationary distribution by taking $P^n$ for large $n$. For instance, WolframAlpha gives
$$P^{500} \approx \begin{bmatrix} 0.396226 & 0.226415 & 0.150943 & 0.226415 \\ 0.396226 & 0.226415 & 0.150943 & 0.226415 \\ 0.396226 & 0.226415 & 0.150943 & 0.226415 \\ 0.396226 & 0.226415 & 0.150943 & 0.226415 \end{bmatrix}$$
suggesting the stationary distribution is
$$A \approx \begin{bmatrix} 0.396226 & 0.226415 & 0.150943 & 0.226415 \end{bmatrix}$$
and indeed, $AP \approx A$, suggesting this would be the correct stationary distribution.